ELEMIENTS OF ALGEBRA: ON TIlE BASIS OF HI. BOURDON: EMBRACING STURM'S AND TIORNER'S THEOREMS, AND PRACTICAL EXAMPLES. BY CIHARLES DAVIES, LL.D. AUTIIOl OF ARITIIIMETIC1 ELEMENTARY ALGEBRA ELEMIENTARY GEOMETRhY PRACTICAL GEOMIETRY, ELEVENTS OF SURVEYIONG ELEMENTS OF DESCRIPTIVE LhD ANALYTICAL GEOMETRY, ELEMEINTS OF DIFFERENTIAL AND INTEGRAL CALCULUS, AND A TREATISE ON SHADES1 SHADOWS AND PERSPECTIVE. A. S. BjtAI:NES &1 COOMPANY, NEW YORK AN D CHICAIIGO. 1871. DAVIES' MATHEMATICS. I'I EiT a Q Q aI And Only Thorough and Complete Mathematical Series. I\T 71-= aD ID Zes2rW I. COM0S/ON 8 O1/0L COURSE. Davien' Primary Arithn otic. -The findamcntal principles displayedl in Object Lessons.:Davies'; tnteliectual JArithmetic, —Iteferriang all operations to thc unit 1 a3 the only tangible basis for logical development. Davies' Eilements of W%.itten Arithsanetic.-A practical introduction to the whole subject. Theory subordinated to Practice. Davies' Practical Arithnietic:*-The most successful combination of Theory and Practice, clear, exact, brief, and comprehensive. i/. ACADEMIC COURSE. Davies' University Arithmetic.-I-Treating the subject exhaustively as a science, in a logical series of connected propositions. Davies':Elementary Algebra.* —A connecting link, conducting the pupil easily from arithmetical processes to abstract analysis. BDavies' University Algebra. —For institutions desiring a more complete but not the fullest course in pure Algebra. Davies' Practizal athematics. — Th science practically applied to the tiscful arts, as Drawing, Architecture, Surveying, Mechanics, etc. Daxries' Elementary Goeometry.-The important principles in simple form, but with all the exactness of vigorous reasonin'g. BDavies' Elements of Surveying.-Rte-written in 1380. The simplest and mnost practical presentation for youths of 12 to 10. /11. COLLEGIATE COUIRSE. Davies' 13ourdond s Algebra.t":-IEmbracing Sturm's Theorem, and a most exhaustive and scholarly course. Davies''University Ag1.ebra.' —A shorter course than Dourdon, for Institutions have less time to give the subject. Da~vies' Zegendre's Geomeltry.-Acl nowledged tCe only satisfactory treatise of its grade. 310,000 copies have been sold. Davies' Analytical Geometry and Clcuaas. —The shorter treatises, combined in one volume, are more available for American courses of study. Davies' Analytical Geometry. tThe original compendiums, for those deDavies' Diff. & Ent. Calculus. iring to give full time to each branch. Davies' Descriptive Geometry.-With application to Spherical Trigonometry, Spherical Projections, and Warped Surfaces. Davie' B~hades, Shadows, anld Oerspec;tivo. —A asuccinct exposition of the mathematical principles involved. Davies' Science o f ~Mathematicn. —For teachers, embracing I. GRAMMARI Or ARITHMETIC, III. LOGIC AND UTILIT rY OF MATnEMxATICS, II. OUTLrNES 0sF MIATIIE IATICS, IV. MATHEMATIC&L DICTIONARY. * Keys may be obtained from the Publishers by Teachers only. Entered, according to Act of Congress, in the year i85I, by CHARLES DAVIES, in the Clerk's Office of the District Court of the United States for the Southern District of New York. T1. PREFACE THE Treatise on Algebra, by Mi. Bourdon, is a work of sillgular excellence and merit. In Franlce, it has long been one of the standard Text books. Shortly after its first publication, it passed through several editions, and has formed the basis of every subsequent work on the subject of Algebra, both in Europe and in this country. The oriiginal work is, however, a full and complete treatise oil the subject of Algebra, the later editions containing about eight hundred pages octavo. The time which is given to the study of Algebra, in this country, even in those seminaries where the course of mathematics is the fullest, is too short to accomplish so voluminous a work, and hence it has been found necessary either to modify it essentially, or to abandon it altogether. In the following work, the original Treatise of Bourdon has been regarded only as a model. The order of arrangement, in many parts, has been changed; new rules. and new methods have been introduced: the modifications indicated by its use, for twenty years, as a text book 4 PREFACE. in the Military Academy have been freely made, for the purpose of giving to the work a more practical character, and bringing it into closer harmony with the trains of thought and improved systems of instruction which prevail in that institution. But the work, in its present form, is greatly indebted to the labors of' \Tilliain G. Peck, A. Al.) U. S. Topographical Engineers, and Assistant Professor of Mlathematics in the Military Academy. Many of the new definitions, new rules and improved m;ethods of illustration, are his. Htis experience as a teacher of mathematics has enabled him to bestow upon the work *much valuable labor which will be found to bear the mnarks of profound study and the freshness of daily instruction. A Key to th:is volume has been prepared for thle use of cTac/e)t8 oI i72 g6. CONTENTS. CHAPTER I. DEFINITIONS AND PRELIMINARY REMARKS. AxELaa A —Definitions-Explanation of the Algebraic Signs....... 1-28 Similar Terms-Reduction of Similar Terms............2.......... 8 —30'Theorems-Problems-Definition of —Problem.................... 30-31 CHAPTER II. ADDITION, SUBTRACTIONS MULTIPLICATION, AND DIVISION. Addition-Rule............................................... 31- 3 Subtraction-Rule-Remark................................... 35 —41. Multiplication-Rule for IMonomials and Signs................... 41-45 Rule for Polynomials........................................ 45-46 Remarks-Theorems Proved.................................... 46-49 Division of Monomials-Rule................................... 49-53 Signification of the Symbol a................................... 53-55 Division of Polynomials-Rule.................................. 55-58 Remarks on the division of Polynomials......................... 58-59 Of Factoring Pylynomials..................................... 59-60 When m is entire, am-bn is divisible by a-b.................. 60-62 CHAPTER III. ALGEBRAIC FRACTIONS. Definition-Entire Quantity-Mixed Quantity.................... 62-68 Reduction of Fractions.............................. 68-69 To Reduce a Fraction to its Simplest Form.................. 68 —L To Reduce a Mixed Quantity to a Fraction...............-.... 6S-II. To Reduce a Fraction to an entire or Mixed Quantity.......... 68 —III To Reduce Fractions to a Common Denominator................ 68-IV To Add Fractions............................................ 68 —V To Subtract Fractions...................................... 68 —V 86 Z~ONTENTS. ARTICLNS To Multiply Fraetiors..............68-VII. To Divide Fractions.........................,............68-VIII Results from adding to both Terms of a Fraction............ 70-71 Symbols 0, Mc and a................................. 71-.. CHAPTER IV. EQUATIONS OF THE FIRST DEGREE. Definition of an Equation-Different Kinds-Properties of Equations 72-77 Solution of Equations...................................... 77 —78 Transformation of Equations-First and Second............... 18-80 Resolution of Equations of the First Degree-Rule.............. 81 Problems involving Equations of the First Degree............... 81 Equations with two or more Unknown Quantities................ 82-83 Elimination-By Addition-By Subtraction-By Comparison......... 83 —88 Problems giving rise to Simultaneous Equations...............Page 96 Indetermintte Eouations and Indeterminate Problems.............. 88-89 Interpretation of' Negative Results.............................. 89-91 Discussion of Problems................,................. 91-92 Inequslities................................................... 92-93 CHAPTER V. EXTRACTION OF TIIE SQUARE ROOT OF NUMBERS.-OF ALGEBRAIC QUANTITIES. —TRANSFORMATION OF RADICALS OF THE SECOND DEGREE. Extraction of the Square Root of Numbers..................... 93-96 Extraction of the Square Root of Fractions..................... 96-100 Extraction of the Square Root of Algebraic Quantities...........100-104 Of Monomials........................................100-101 Of Polynomials........................................101-104 Radicals of the Second Degree................................. 104-106 Addition and Subtraction-Of Radicals....................... 106-107 Ilultiplication, Division, and TransformatioLn...............107-11 1 CHAPTER VI. EQUATIONS OF TtIE SECOND DEGREE. Equations of the Second Degree................ 110 —112 Incomplete Equations-S-lution of...............................112 —114 Solution of Complete Eliations of the Second Degree...........114 —115 I)iscussion of Equations of the Second Degree...............1....15-117 Of the Four Forms........................................... 117-121 Iroblem of the Lights........................................1. 21-122 Of Trinonmial Equations....................................... 122-125 Extraction of the Square Root of the Binomial a -4-/.......125-126 Equations with two or more Unlmown Quantities............126-128 CONTENTS.' CIIAPTER VII. FORMATION OF POWERS, BINOMIAL TIIEOREM~ EXTR &CTION OF ROOTS OF ANY DEGREE WHATEVER.-OF Z[ADICALS. Formation of Powers,.......................................128 —13G Theory of Permutations and Combinations.1............130-136 Binomial Theorem............................136-1 41 Extraction of the Cube Roots of Numbers.141-14..............141-142 To Extract the nth Root of a Whole Number.,.........142-144 Extraction of Roots by Approximation......144-145 Extraction of the nth root of Fractions..............145-146 Cube Root of Decimal Fractions.............................. 146 —147 Extraction of Roots of Algebraic Quantities..... 147-148 Of Polynomials............................................. 148-150 Transformation of Radicals.................................. 150-15~ Addition and Subtraction of Radical;....................... 155-15b Multiplication of Radicals................................... 156-15~ Division of Radicals.........................................157-158 Formation of Powers of Radicals..................... 158-159 Extraction of Roots............................................159-160 Different Roots of the Same Power.............................160-162 Modifications of the Rules for Radicals.......................... 162-164 Theory of Fractional and Negative Exponents.................. 164-171 CHAPTER VIII. OF S1ERIES.-ARITHMETICAL PROGRESSION.-GEOMETRICAL PROFORTION AND PROGRESSION.-REC URRING SERIES.-BINOMIAL FORMULA.SUMMATION OF SERIES. —PILING OF SHOT AND SHELLS. Series Defined...............-..............................171-172 Arithmetical Progression-Defined..............................172-173 Expression for the General Term............................... 174-175 Sum of any two Terms. 175 —176 Sum of all the Term~s......................................176 —.1. Sum of all the Terms. 176-177 Formulas and Examplres.............................. 1.77-1 St Ratio and Geometrical ilPlportion..............1.............. 181 —186 Geometrical Progression —Defined..........................-17.... Expression for any Term...........................8....... 187 —1 8 Sum of n Termns-Formulas and Examples.....................188-193 Indeterminate Co-efficients..............19..............193-199 Recurring Series.............................................199-202 General Demonstration of Binomial Theorem...................202-204 Applications of the Binomial Formula.........................204-208 Summation of Series...........................,...... 208-209 Method of Diffeiences..................................... 20)9-210 Piling of Balls........................................ 21( —21U 8 CONTENTS. CHAPTER IX. CONGTINUED FRACTIONS.-EXPONENTIAL QUANTITIES,~ -LOGARITHMS.Continued IFractions................................. 15-224 Exponential Quantities..........................................2-1 227 Theory of Logarithms...................................... 227 229 General Properties of Logarithms.....................2......... 229-236 Logarithmic Series —Modulus................................... 236 —241 Transforination of Scries..........4................241-242 Of Ilterp'lation...... I................. 242-243 J9f Interest................................................... 243-244 CHAPTER X. GENERAL THEORY OF EQUATIONS. General Properties of Equations.......................... 244 —251 Composition of Equations..................................... 251-252 Of the Greatest common Divisor........................... 252-262 Transformation of Equations..................................262-264 Formation of Derived Polynomials..........................264-266 Properties of Derived Polynomials............................ 266-267 Equal Roots......................................267-270 Eli mination.................................,......... 270-275 CHAPTER XI. SOLUTION OF NUMEhCICAL EQUATIONS.-STURM S THEOREM. —CARDANS RULE. —HORNER) S METHOD. General Principles.............................................275-277 First Principle................................................. 277-279 Second Principle............................................... 279-280 Third Principle.................................... 280-281 Limits of Real Roots...................................... 281-284 Ordinary Limits of Positive Roots............................... 284-285 Smallest Limit in Entire Numbers....................285-286 Superior Limit of Negative Roots-Inferior Limit of Positive and Negative Roots................................... 286-281 Consequences........................................28l-293 Descartes' Rule...........................................293-295 Commensurable Roots of Numerical Equations................295-298 atmurum's Theorem..............................................298-308 CLaurldan9's R lule. *...................................................08 309 Preliminaries to Horner's Method.............................. 309-310 Mlultiplication by Detached Co-effieients.......................... 10- 311 Division by Detached Co-efficients.......................... 311-312 Synthetical Division....................................... 312-313 Method of Transformation.3.....................313-314 lorney's Methld,...................................e314 INTRODUCTIO N. QUvANTITY is a general term applicable to everything which can be increased or diminished, and measured. There are two kinds of quantity; 1st. Abstract quantity, or quantity, the conception of which does not involve the idea of matter; and, 2dly. Concrete quantity, which embraces every thing that is material. Mathematics is the science of quantity; that is, the science which treats of the measurement of quantities, and of their relations to each other. It is divided into two parts: 1st. The Pure Mathematics, embracing the principles of the science and all explanations of the processes by which these principles are derived from the abstract quantities, Number and Space: and, 2d. The Mixed Mathematics, embracing the applications of these principles to all investigations involving the laws of matter, to the discussion of all questions of a practical nature, and to the solution of all problems, whether they relate to abstract or concrete quantity.* *Davies' Logic and Utility of Mathematics. Book TI. 10 INTRODUCTION. There are three operations of the mind which are imrme diately concerned in the investigations of mathematical science: lst. Apprehension; 2d. Judgment; 3d. Reasoning, 1st. Apprehension is the notion, or conception of an idea in the mind, analogous to the perception by the senses. 2d. Judgment is the comparing together, in the mind, two of the ideas which are the objects of Apprehension, and pro uouncing that they agree or disagree with each other. Judg iment, therefore, is either affirmative or negative. 3d. Reasoning is the act of proceeding from one judgment to another, or of deducing unknown truths from principles al ready known. Language affords the signs by which these opela. lions of the mind are expressed and communicated. An appre hension, expressed in language, is called a term; a judgment, expressed in language, is called a proposition; and a pro. ers if reasoning, expressed in language, is called a dernonsira. tfion.* The reasoning processes, in Logic, are conducted usually by;means of words, and in all complicated cases, can take place in no other way. The words employed are signs of ideas,:and are also one of the principal instruments or helps of thought; and any imperfection in the instrument, or in the node of using it, will destroy all ground of confidence inll the result. So, in the science of mathematics, the meaning of the,terms employed are accurately defined, while the language arising from the use of the symbols, in each branch, has a iefinite and precise signification. * Whateley's Logic,-of tie operations of the mind and senses. INTRODUCTION. 11 In the science of unumbers, the ten characters, called figures, are the alphabet of the arithmetical language; the combinations of these characters constitute the pure language of arithmetic; and the principles of numbers which are unfolded by means of this, in connection with our common language, constitute the science. IJr Geometry, the signs which are employed to indicate thb boundaries and forms of portions of space, are simply the straight line and the curve; and these, in connection with our common language, make up the language of Geometry: a science which treats of space, by comparing portions of it with each other, for the purpose of pointing out their proper ties and mutual relations. Analysis is a general term embracing that entire portion of mathematical science in which the quantities considered are represented by letters of the alphabet, and the operations to be performed on them are indicated by signs. Algebra, which is a branch of Analysis, is also a species of universal arithmetic, in which letters and signs are employed to abridge and generalize all processes involving numbers. It is divided into two parts, corresponding to the science and art of Arithmetic: 1st. That which has for its object the investigation of the properties of numbers, embracing all the processes of reasoning, by which new properties are inferred from known ones; and, 2d. The solution of all problems or questions involving the determination of certain numbers which are unknown, from their connection with certain others which are known or given. 12 INTRODUCTION. Inl arithmetic, all Quantity is regarded as c:r;Sisting of parts, which can be numbered exactly or approximatively, and in this respect, possesses all the properties of numbers. Proposi. tions, therefore, concerning numbers, have this remarkable pecu. liarity, that they are propositions concerning all quantities whatever. Algebra extends the generalization still further. A number is a collection of things of the same kind, without reference to the nature of the thing, and is generally expressed by figures. Algebraic symbols may stand for alf nulnbers, or for all quantities which numbers represent, or even for quantities which cannot be exactly expressed numerically. In Geometry, each geometrical figure stands for a class; and when we have demonstrated a property of a figurle, that property is considered proved for every figure of the class. In Algebra, all numbers, all lines, all surfaces, all solids, may be denoted by a single symbol, a or x. Hence, the conclusions deduced by means of those symbols are true of all things whatever, and not like those of number and Geometry, true only for particular classes of things. The symbols of Algebra, therefore, should not excite in our minds ideas of particular things. The written characters, a, b, c, d, x, y, z, serve as the representatives of things in general, whether abstract or conorete, whether known or unknown, whether finite or infinite. In the various uses which we make of these symbols, anid the processes of reasoning carried on by means of them, the mind insensibly comes to regard them as things, and not as muere signs; and we constantly predicate of them the properties of things in general, without pausing to inquire what kind of INTRODUCTION. 13 thing is implied. All this we are at liberty to do, since the symbols being the representatives of quantity in genera!, there is no necessity of keeping the idea of quantity continually alive in the mrind; and the processes of thought may, without danger, be allowed to rest on the symbols themselves, and therefore, become to that extent, merely mechanical. But when we look back and see on what the reasoning is based, and how the processes have been conducted, we shall find that every step was taken on the supposition that we were actually dealing with things, and not with symbols; and that without this understanding of the language, the whole system is without signification, and fails.* The quantities which are the subjects of the algebraic analysis may be divided into two classes: those which are known or given, and those which are unknown or sought. The known are uniformly represented by the first letters of the alphabet, a, b, c, d, &c.; and the unknown by the final letters, x, y, a, v, C&. Five operations, only, can be performed upc n a quantity that will give results differing from the quantity itself: viz. 1st. To add a quantity:' to it; 2d. To subtract a quantity from it; 3d. To multiply it by a quantity; 4th. To divide it; 5th. Tc extract a root of it. Five signs only, are employed to denote these operatiors, They are too well known to be repeated here. These, with * Davies' Logic and Utility of Mathematics.,i 278. 14 INTRODUCTION. the signs of equality and inequality, together with the letters of the alphabet, are the elements of the algebraic language. The interpretation of the language of Algebra is the first dhing to which the attention of a pupil should be directed; and he should be drilled in the meaning and import of the aymnbols, until their significations and uses are as familiar as the sounds of the letters of the alphabet. All the apprehensions, or elementary ideas, are conveyed to the mind by means of definitions and arbitrary signs; and every judgment is the result of a comparison of such impressions. Hence, the connection between the symbols and the ideas which,hey stand for, should be so close and intimate, that the one shall always suggest the other; and thus, the processes of Algebra become chains of thought, in which each link lulfils the double office of a distinct and connecting propor Lion. ELEMENTS OF ALGEBRA.l CHAPTER I. DEFINITIONS AND PRELIMINARY REMARKS. 10 QUANTITY is anything which can be increased or diminished, anld measurled. 2a MIATIIHMATICS is the Science which treats of the measurement and relations of quantities. 3. ALGEBRA is a branch of mathematics, in which the quantities considered are represented by letters, and the operations to be performed upon them are indicated by signs. The letters and signs are called symbols. 4. In algebra two kinds of quantities are considered: 1st. Klowan quantities, or those whose values are known or given. These are represented by the leading letters of the alplabet, as, a, b, c, &c. 2d. Unknown quantities, or those whose values are not given. lThey are denoted by the final letters of the alphabet, as, x. y, z, &c. Letters employed to represent quantities are sometimes written with one or more dashes, as, a', b", c"', x', y", &c., and are read, a prime, b second, c third, x prime, y second, &c. 5. The sign ~-, is called plus, and when placed between two quantities, indicates that the one on the right is to be adde-d to the cle on the left. Thus, a + b is read a plus b, and ",dicates 16 ELEMENTS OF ALGEBRA. [CHAP. I. that the quantity represented by b is to be added to the quantity represented by a. 6. The sign.-, is called minus, and when placed between two quantities, indicates that the one on the right is to be subtracted frojm the one on the left. Thus, c —d is read c minus d, and indicates that the quantity represented by d is to be subtracted fiom thle quantity represented by c. The sign +-, is sometimes called the positive sign, and the quantity before which it is placed is said to be positive. The sign -, is called the negative sign, and quantities affected by it are said to be negative. 7. The sign X, is called the sign of multiplication, and when placed between two quantities, indicates that the one on the left is to be multiplied by the one on the light. Thus, a x b, indi eates that a is to be multiplied by b. The multiplication of quantities may also be indicated by placing a simple point between them, as a.b, which is read a multiplied by b. The multiplication of quantities, which are represented by letters, is generally indicated by simply writing the letters one after another,' without interposing any sign. Thus, ab is the same as a X b, or a.b; and abc, the same as a x b X c, or a.b.c. It is plain that the notation last explained cannot be employed when the quantities are represented by figures. For, if it were required to indicate that 5 was to be multiplied by 6, we could not write 5 6, without confounding the product with the number 56. The result of a multiplication is called the product, and each ~f thc quantities employed, is called a factor. In the product of several letters, each single letter is called a lileral factor. Thus, in the product ab there are two literal factors a and b; in the product bcd there are three, b, c and d. 8,. The sign —, is called the sign of division, and when placed between two quantities, indicates that the one on the left is to be divided by the one on the right. Thus, a -- b indicates that a is to CHAP. I.] DEFINITIONS AND REMARKS. 17 be divided by b. The same operation may be indicated by writing b under a, and drawing a line between them, as -; or by writing 6b n the right of a, and drawing a line between them, as alb. 9. The sign =, is called the sign of equality, and indicates that Ino two quantities between which it is placed are equal to each other. Thus, a-b = c + d, indicates that a diminished by b is equal to c increased by d. 10. The sign >, is called the sign of inequality, and is used to indicate that one quantity is greater or less than another. Thus, a > b is read, a greater than b; and a < b is read, a les3 than b; that is, the opening of the sign is turned toward the greater quantity. 11. The sign. is sometimes employed tc indicate the difference Xf two quantities when it is not known which is the greater. Thus, a - b, indicates the difference between a and b, without Thowing which is to be subtracted from the other. 12. The sign cc, is used to indicate that one quantity varies as 1.1 to another. Thus a oc T' indicates that a varies as 13. The signs: and:, are called the signs of proportion; the first is read, is to, and the second is read, as. Thus, a:: c: d, is read, a is to b, as c is to d. The sign.'., is read hence, or consequently. 140 If a quantity is taken several times, as a - a + a + a ~+ a, it is generally written but once, and a number is then placed, before it, to show how many times it is taken. Thus, a.- a + a + a -- a may be written 5a. The number 5 is called the co-efficient of a, and denotes that a is. taken 5 times. Hence, a co-efficient is a number prefixed to a quantity denoting: the number of times which the quantity is taken. 2 18 ELEMENTS OF ALGEBRA. [CHAP. I. When no co-eflEcient is written, the co-efficient 1 is always understood; thus, a is the same as la. 15, If a quantity is taken several times as a factor, the product mnay be expressed by writing the quantity once, and placing a nunmber to the right and above it, to show how many times it i' taken as a factor. Thus, a x a x a x a x a may be written a5. The number 5 is called an ex.ponent, and indicates that a is taken 5 times as a factor. ihence, an exponent is a number written to the right and above a quantity, to show how many times it is taken as a factor. If no exponent is wnritten, the exponent 1 is understood. Thus, a is the same as a'. 16. If a quantity be taken any number of times as a factor, the resulting product is called a pozwer of that quantity: the exponent denotes the degree of the power.'For example, al - a is the first power of a, a2 = a X a is the second power, or square of a, a3-= a x a x a is the third power, or cube of a, a4 —a a X a a X a is the fourth power of a, a5 = a X a X a X a X a is the fifth power of a, in which the exponents of the powers are, 1, 2, 3, 4 and 5; and the powers themselves, are the results of the multiplications. It should be observed that the exponent of a power is always greater'by one than the number of multiplications. The exponent ef a power of a qtuantity is somet;imes, for the sake of brevity, called tke ep2onent of the quantity. 17. "As an example of the use of the exponent in algebra, let:it be required to express that a number a is to be multiplied three times by itself; that this product is then to be multiplied three times by b, and this new product twice by c; we shaolld write a X a X a X a X b X b X b X c X c = a4b32. If it were fiurther required to take this result a certain inulbe,of times, say sevein, we should simply write 7a6b:-'~ CHAP. I.] DEFINITIONS AND REMAtRKS 19 i o A root of a quantity, is a quantity which being taken a certain numbzr of times, as a factor, will produce the given quantity. The sign Iis called the radical sign, and when placed over a qua ntity, indicates that its root is to be extracted. Thus, 2 a or simplly a denotes the square loot of a. 3V- denotes the cube root of a. 42a denotes the fourth root of a. hlle number placed over the radical sign is called tho index of the root. Thus, 2 is the index of the square root, 3 of the cube root, 4 of the fourth root, &c. 19. Thle recirocal of a quantity, is 1 divided by that quantity. Thus, - is the reciprocal of a; and is the reciprocal of a + b. a~b 20o Every quantity written in algebraic language, that is, by the aid of letters and signs, is called an algebraic q~uantity, or the algebraic expression of a quantity. Thus, 3 is the algebraic expression of three times the quantity denoted by a; 5n2 I is the algebraic expression of five times the square of a;'7032 is the algebraic expression of seven times the product of the cube of a and the square of b; -30 56$ is the algebraic expression of the difference between three times a and five times 6; is the algebraic expression of twice the square a2 2- 30 +q 4b21 of a, diminished by three times the product - of a and b, augmented by four times the square of b. 2,. A single algebraic expression, not connected with any other by the sign of addition or subtraction, is called a nonomnial, ox aimply, a term. 20 ELEMENTS OF ALGEBRA. LCHAP. L Thus, 3a, 5a2, 7a3b2, are monomials, or single terms. An algebraic expression composed of two or more terms con. nected by the sign + or -, is called a polynomial.:For example, 3a - 5b and 2a2 - 3cb +- 42, are polynomials. A polynomial of two terms, is called a binomial; and one of three terms, a trinomiaci. 22. The numerical value of an algebraic expression, is the num ber obtained by giving a particular value to each letter which enters it, and performing the operations indicated. This numerical value will depend on the particular values attributed to the le tters, and will generally vary with them. For example, the numerical value of 2a3, will be 54 if we make a -3; for, 33-3 X 3 X 3 =27, and 2 X 27-54. The numerical value of the same expression is 250 when we make a = 5; for, 53 = 5 x 5 x 5-125, and 2 x 125 = 250. We say that the numerical value of an algebraic expression generally varies with the values of the letters which enter it; it does not, however, always do so. Thus, in the expression a - b, so long as a and b are increased or diminished by the same number, the value of the expression will not be changed. For example, make a = 7 and b = 4: there results a — b = 3. Now, make a -=7 + 5 = 12, and b = 4 + 5 = 9, and there results, as before, a - b = 12 - 9 = 3. 23. Of the different terms which compose a polynomial, some are preceded' by the sign +, and others by the sign -. The former are called additive terms, the latter, si.biru-cive terms. When the first term of a polynomial is plus, tile sign is gene. rally omitted; and when no sign is written before a term: it is always understood to have the sign -. 24. The numerical value of a polynomial is not affected by clanging the order of its terms, provided the signs of all the terms remain unchanged. For example, the polynomial 4a' - 3a2b ~ 5ac2 = 5ac2 - 3a2b + 4a3 - - 3a2 - 5ac2 q- 4a. 25. Each literal factor which enters a term, is called a dimesion of the term;. and the degree of a term is indicated by the number of these factors or dimensions. Thus CHAP. I.] DEFINITIONS AND REMARKS. 2t. 3a is a term of one dimension, or of the first degree. 5cib is a term of two dimensions, or of the second degree. 7a3bc2 = 7ccaabcc is of six dimensions, or of the sixth degree. In general, the degree of a term is determined by taking the sum of the e.xonents of the letters which enter it. For example, the term Sa2bcd3 is of the seventh degree, since the sum of the expoenllts, 2+ + 1 +3, is equal to 7. 26. A polynomial is said to be homogeneous, when all of its terms are of the same degree. The polynomial 3a - 23 + c is homogeneous and of the first degree. - 4ab + 62 is homogeneous and of the second degree. 5a2c - 4C3 + 2c2d is homogeneous and of the third degree. 8a3 - 4ab - c is not homogeneous. 27. A vinculum -, parenthesis (), brackets [], { }, t bar 1, may be used to indicate that all the quantities which they connect are to be considered together. Thus, a b + c X x, (a + b L C) X x, [a+b+c] X x, or {a b -c}x, indicate that the trinomial a + -b - c is to be multiplied by x. When the parenthesis or brackets are used, the sign of Inultiplication may be omitted: as, (a + b + c). The bar is used in some cases, and differs from the vinculum in being placed vertically, as + a X. + b 28, Terms which contain the same letters affected with equal exponents are said to be similar. Thus, in the polynomial, 7ab + 3ab - 4ca32 + 5a3b2, the termns 7ab and c3a, are simnilar, and so also are the terms — 4a3b' and 5a3b2, the letters in each being the sam.e, and tha same letters being affected with equal exponents. But in the binomial 8a2)b + 7ab2, the terms are not similar; for, although, they contain the same letters, yet the. same letters are not affected with equal oexponents. 22 ILEIKEN'TS AF LERB. P. iC ktB. Ik 29. When a polynomial o)lntins similar terms, it may be reduced to a simpler form by forming a single term fiom each set of similar terms. It is said to be in its simplest formn, xhlen it contailns the fewest terms to which it can be reduced. If we take the polynomial 2abc2- 4a3bc2 + 6azc2 - 8a3bc2 + l-a'bc2, we know, from the definition of a co-efficiellt, that the literal part a3bc2 is to be taken additively, 2 +6 + 11, or 19 times; and subtraetively, 4 - 8, or 12 times. Hience, the given polynomial reduces to 19ac3c2 - 12aUbc2 - 7a3sbc2. It -may happen that the co-efficient of the subtractive term, obtained as above, will exceed thcat of the additive term. In that ease, sutbract the positive co-eficient friom the nqegative, pref/x the minus signz to the remainder, and then anlnex the literal part. In the polynomial 3a2b + 2a2b - - 502b - 3a2b, we have, + 3ca2b - 5a2b + 2a2b - 3126 + 5a2b -8a2b But, - 8a2b - -5u2 -- 3a2b: hence 5a26 - Sa2b -- 5a2b -- ab =- 3a2b. In like manner we may reduce the similar terms of any poly. Uomial. hIence, for the reducltion of a po(lytnomial contanliillg sets of similar terms, to its silpel,,st tlni, Mwe have the following R U L E. I. Add together the co-ef/cietxs o? all tie odditive t7erm7s f ealch selt and anl7ex to their su?m the literal 2part': jbr.n a single slbtrw(tive term in thle mie mtonner. 11. Then, subtract the less co-efficier t frj'o the greater, and to the remainder prefix the sigyn of the grealer co-efficienl, anld anc:e tihl literal part. CHAP. I.] REDUCTION 0F POLYNOMtALS. 2 EXAMPLES. 1. Reduce the polynomial 4a2b - 8a2b 9a2b + -1a'b to its simplest form. Ans. - 2a2b. 2. Reduce the polynomial 7abc2 - abc2 - 7abc- - 8abc2 -+ 6abc to its simplest form. Ans. — 3abc2 3. Reduce the polynomial 9cb3 - Sac2 + 15cb3 -t Sca + 9ac2 -24cb3 to its simplest form. Ans. ac2 + 8ca. 4. Reduce the polynomial 6ac2 - 5a63 + 7ac2 - 3ab3 — 13a62 - 18ab3 to its simplest form. Ans. 10ab3. 5. Reduce the polynomial abc2 - abe - 5cc2 - 9abc2 + Gabc -Sac2 to its simplest form. Ans. - Sabcz + 5abc — 3ac2. 6. Reduce the polynomial 3a262 - 7a3b +- 5ab - 9a.2b2 - 9a3b 4- 3ab to its simplest form. Ans. - 6a2b2 + 2a3b + Sab. 7. Reduce the,polynomial 3acb4 - 7a0c2b3 - 6a4b5 - 3acb4 q 6a3c23 - 6acb4 + 4a4b5 + 2a4b5 to its simplest form. Ans. - a3C2b3 - 6acb4. S. Reduce the polynomial - 7a2b2c2 + 9a5bC2 + 6a2b2C2 + a2b2ac - 5ab2-be- b5c5 to its simplest form. Ans. 4a5bc2 - b5. 9. Reduce the polynomial - 10a3b 4- 6a2b2 + 7a3b - 5a2b2 - 5a.36 3a2b2 to its simplest form. Ans. - 8a3, - 4a2b2. REMXAR. —It should be observed that the reduction affects only the co-efficients, and not the exponents. 30. A THEOP,;M is a general truth, which is made evident by a course of reasoning called a demonstration. A PROBLEM is a question proposed which requires a solution. 31. We shall now illustrate the utility and brevity of algebraic language by solving the following PROBLEM. The sium of tOo numbers is 67, and their dif'erence is 19; wMtal rae the numbers? Let us first indicate, by the aid of algebraic symbols, tbhe relation which exists betwee: the given and unknown numbers of the problem. 24 ELE:MENTS OF ALGEBRA. [CHAP. L If the less of the two numbers were klnown, the greater could be found by adding to it the difference 19; or in other words, thlle less number, plus 19, is equal to the greattr. 1f, then, we denote the less number by x, x -- 19 will denote the greater, and 2x + 19 will denote the sum. But fiom the enunciation, this suin is to be equal to 67.'Theie fiore, 2x + 19 = 67. Now, if 2x augmented by 19, is equal to 67, 2x alone is equal to 67 minus 19, or 2x = 67 - 19, -or performing the subtraction, 2x =48. Hence, x is equal to half of 48, that is 48 - = 24. The less number being 24, the greater is x + 19 — 24 + 19 = 43. And, indeed, we have 43 - 24 = 67, and 43-24 = 19. GENERAL SOLUTION. The sum of two numbers is a, and their difference is b IhaR: are the two numbers? Let x denote the less nlumber; Then will x + b denote the greater number. Now, from the conditions of the problem, x + x + b, or 2x +b Uill be equal to the sum of the two numbers: her te, 2x — = -a. Now, if 2x + b is equal to a, 2x alone must be equal to a{- b and a- b a b'2 2 2 CIHAP. I.] SOLUTION OF PROBLEMS. 25 If the value of x be increased by b, we shall have the greater number: that is, a b a b x -- b= 2 2 -b- +; a 6 heace, x + = + - = the greater number, and a b x= the less number. That is, the greater of two slumbers is equal to half their sutm increased by half their diference; ad thle less is equal to half their sum diminished by half their difference. As the formn of these results is independent of any particular values attributed to the letters a and b, the expressions are called formzulas, and may be regarded as comprehending the solution of all problems of the same kind, differing only in the numerical values of the given quantities. Hence, A formula is the algebraic expression of a general rule, or principle. To apply these formulas to the case in which the sum is 237 and difference 99, we have 237 99 237 1 99 3)36 the greater nunmber = 2 F - -- = 106; 237 99 237 99 1383 and the less = 92 - 69; and these are the true numbers; for, 168 -F 69 =. 237 which is the given sum, anfd 168 -o9 = 99 whic' is the given difierenec CIHAPTER IL aDDITION, SI.'BTt AC rON, MULTIPLICATION, ~ND DIVISION. ADDITION. 31, ADDITION, in algebra, is the operation of finding the simn plest equivalent expression for the aggregate of two or more algebraic quantities. Such equivalent expression is called their sum. 32. If the quantities to be added are dissimilar, no reductions can be made among the terms. We then write them one a-fter the other, each with its proper sign, and the resulting polynomial will be the simplest expression for the suim. For example, let it be required to add together the mono-,nials 3a, 5b and 2c; we connect them by the sign of addition, oa + 53 + 2c, a result which cannot be reduced to a simpler form. 33. If some of the quantities to be added have similar terms, we connect thle quantities by the silgn of addition as before,,red then reduce the resulting polynomial Ito its simlplest form,!)b the rule already given. This reduction will, in general, be atore readily accomrplished if we write down the qluantities to Ite added, so that similar termns shall fall in the same columnn. Thus; Let it be required to find the sum of i 2. -- 3ab q- b2 the quantities, 2a-Their su, fter reduc5ing ( 9), is Tl1eir sum, after reducing (Art. 29), is. 5a2 - 5ab.- 4b2 MCAP II.] ADDITION. 27 34. As operations similar to the above apply to all algebraic expre.;sions, we deduce, for the addition of algebraic quantities, the following general RULE. 1. fi5ite down the quantities to be added, with their respective signs, so that the similar terms shall fall in, the same colzmn. 11. Redulce the similar terms, and annex to the results tfhose tlerm, vwhich cannot be reduced, giving to each term its respective sign. EXAMPLES. 1. Add together the polynomials, 3a2 -- 22 - 4ab, 5a2 - 2 -t 2ab ald 3ab 3c2 - 2b2 The term 3a2 being similar to 5a2 3(2 - b - 2,h2 we write 8a2 for the result of the reduction of these two terms, at the same+ - time slightly crossing them as in the ~ 3u6 - 2 2 8a2 +- ab -- 5b2 - 3c2 terms of the example. Passing then to the term - 4ab, which is similar to the tvwo terms +- 2ab and + 3ab, the three ieduce to + ab, which is placed after- Sc2, and the terms crossed like the first ternm. Passing then to the terms involvinlg b2, we find their sum to be -5b2, cfter which we write -3c2. The marks are drawin across the terms, that none of them may be overlooked and omitted. (2). (3). 7x +- 3ab -- 2c 16a2b2 + 6c - 2abe - 3z - 3ab - 5c - 4a2b2 - 9bc + 6abc 5x- 9ab- 9c -.-9a2b + bc - abc Suml.. 9.c - 9ah - 12c 3a2b2 - 7bc +- 5abc (4). (5). a - ca - cd + f 6ab +- cd — d Oa +- 5fb - -cd - f 3b +- 5cd- y - 5a - 6ab - 6cd-7f - 4ab - 6cd +- x + ab- cd+ 4f - 5ab — 12cd + y;,,iSn — 2a -- ab -F- - 3f 6 0 + x - d 28 ELEMENTS OF ALG EBRA. LC.HAP. IL 6. Add together 3a + b, 3a + 3b, - 9a - 7b, 6a + 9b and Sa + 3b + Sc. Ains. Ila + 9b -fSc. 7. Add together 3ax - 3Cac + f, - 9ax -- 7a c- (1, +- 6 4ax - 3at - 3f, 8ax + 13ac + 9f anld - 14f +- 3a x. Ans. IlCx + 1 9ac —'+ 7a 4 d. 8. Add together the polynomials, 3a62c + 5ab, 7a2c- 3ab -4 3a 5,Pc - Gab t- 9ac and - 8a2c + ab - 12ac..lns. 7a2c - 3ob. 9. Add the polynomials, 19a2x3b - 12aC3b, 5a2.3b + 15ah3b - Oax, - 2a2x3b - 13a3cb and - 18a2X3b -- 12a3eb + 9 ax. Ans. 4a2x3b - 22c3cb - Cx. 10. Add together 3a + b + c, 5a + 2b + 3ac, a - c -+ ec and -3a- 9ac - Sb. Ans. 6a - 5b) + 2c - 5ac. 11. Add together 5a2b + 6cx + 9b5c, 7cx - Sa26 and -- 15c.x -- 9bc2 + 2a2b. Ans. - a2b — 2cx. 12. Add together 8ctx + 5ab + 3a2b2C2, - 18a. + 6a 2 4- 1Oab and 10x - 15ab - 6a2b2c2. Ans. - 3a2b2c2 + 6a2. 13. What is the sum of 41a3b2c - 27a6c -14a2y and O0a3Vb( 4- 9abc? Ans. 51 a3b2c - 18abc -14a2y. 14. WVhat is the sumn of Sabc - 9ab - G6c2 - 3c + 9ax and 9abuc- 3c - 9ax? Als. 27abc - 9ab ~ 6ch. 15. What is the sum of 8abec + 63a - 2cx - Gxy and 7ca - xy - 13b3a? Ans. 8abc - 12b3a + 5cx - 7ry. 16. ~What is the sumr of 9a2c —14aby- 15a+ h2 and — acC -8a2b2 A,s. 8a2c - 14aby + 7a2b2. 17. What is the sutm of 17a52 +9C3b- 3a, - 2, - 14a%52 +- 7 - 9a3, - 15a3b + 7a52 - a3 and 14a3b - 19a3b A is. 18. What is the sum of 3ax2 -,,x3 -- 17czxy, + 9az2 -- I8CZx' + 34caxy and 7ab +- 3,t3 - 7a.x2 - 4bcx? Ans. 19. Add together 38c 1- 5aC22c2 - 9a2x. 7a2 - 8a2b2C2 - 10a3s and O1ab 4- 16a%2+ c2 +- 1-)c3x. Al1s. +10a2 + 13a262c2 - lOab. 20. Add together 7ac2b - 3abc - Sb2c - 9c3 A- Cd2, Scc - 5a2b + 3C3 - 4b2C + cd2, and 4ab - S8c3 + 9b2c - 3d3. Anls. 6a2b + (-ce U- — 14c3 + 2cd -. 3d3. CMfAP, II.J SUBTRACTION. 29 21. Add together - 18a3b + 2ab4 +- 6a262, -- 8abt4 + 7a3b - 5a2b2 and -- 5a6 6- 6ab 1 1a2b2. Ans. - 1Gab + 12aC22. 22. What is ithe sum of 3bc c- 16a4 - 9ax3d, + Qa32c.- ctx3d +.7a4x and + 16ax3d - a4x - 8a3b2c? Azns. a362c + ctx3d. 23. What is the sum of the following terms: viz., 8a5 - 10a4b -. 1k-362 + 4a263 - 12a6b 4- 15a362 - 4a2b3 -- 6ab4 -- 16a36 4 1'4.(2b3 + 32ab4 - b5? kAns. 8a - 22(a4b - 17aSb2 +- 48a2b3 + 26ab4 - 8b5. SUBTRACTION. 35, SUBTRACTION, in algebra, is the operation for finding the simplest expression for the difference between two algebraic quantities. This difference is called the remainder. 36. Let it be required to subtract 4b from 5a. Here, as the quantities are not similar, their difference can only be indicated, and we write 5a - 4b. Again, let it be required to subtract 4a3b from 7a3b. These terms being similar, one of them may be taken from the other and their true difference is expressed by 7a3b - 4a3b - 3a3b. 37. Generally, if from one polynomial we wish to subtract another, the operation may be indicated by enclosing the second in a parenthesis, prefixing the minus sign, and then writing it after the first. To deduce a rule for performing the operation thus indicated, let us represent the sum of all the terms in the first polynomial by a. Let c represent the suml of all'hole tad, ditive terms in the other polynomial, and — d the suil of the snbtractive terms; then this polynomial will be represented by c — d. The ^peration may t;hen be indicated thus, aC-(c -d); where it is required t,) subtract from a the d'fference betweer c and d. 30 ELEMENTS OF ALGEBRA. rCOAP. Ii. If, now, we dimllWl!: the quantity a by the quantity c, the result a- will be too sma.l by the qulaltity d, since c should have been diminished by d before taking it from a. Ience, to obtain the true remainder, we must increase the first result by d, which gives the expression a - c + d, and this is the true remainder. By comparing this remainder with the given polynomials, e see that we have changed the signs of all the terms of the quantity to be subtracted, and added the result to the other quantity. To facilitate the operation, similar quantities are written in the same column.:Hence, for the subtraction of algebraic quantities, we have the Collowing RULE. I. Write th'e quantity to be subtracted.under that fromn which at is to be taken, placing the similar termns, if there are any, in the same colzrmn. II. Change the signs of all the ternms of the quantity to be subtracted, or conceive them to be changed, awnd then add the result to the other quantity. EnXAMPPLES. (1 ). (1). From - Gac - 5ab + c2 r 6ac - 5ab + c2 Take- gab - 3cc a -7 3-ac - 3ab - 7c Remainder oac — 8ab + c2 + 7c.'' 3ac - 8ab c2 + 7c. (2). (3). From 16a2 — 5bc ~ 7ac 19abe - Gax --- 5axy Take. 14a2 + be + 8a- Sc 7abc l - 7ax- l5adx Reaint nder 2(2 -- 1Obe - ac 2abe - 23axs -- lOazy (4). (5). Froln 5a3 - 4a2b + 3b2c ab - cd 4- 3at2 Takle - -- 3a2b - 862c 5ab - 4cd -+- a t- 5b' Remaindlr:ia:3 a- 7a. 3cd~ q 0 5- 5A CHAP. II.] S UBTIRACTION 81 6. From 3a2x - 13abc + 7wa2, take 9ac2- 1gabc. A is. -- Oci2r -- 72. 7. From 51ab2c - 18abc - 14a2y, take 41a'3b2c - 27abc 14a2y. Alns. 10a3b2c -- 9abe. S. From 27abc - 9ab + 6c2, take 9alc + 3c - 9ax. Ans1 18abc - 9ab + 6c2 - 3c + 9aXr 9. From Sabc - 12b3a + 5cx - 7.y, take 7cx - xy - 13b'a. Ans. Saitc -- b3a - 2cx -- 6xy. 10. From 8a2c - 14aby + 7a2b2, take 9a2c - 14aby + 15ab2. AIns. - a2c - SC2b2. I1. From 9a6x2 - 13 + 20ab3x - 4b6cx2, take 36bcx2 + 9a6X2 - -- 3ab3x. IAns. 17ab.r -- 7b6cx2 - 7. 12. From 5a4 -- 7a362 -3cd2 -: 7d, take 34 - 3a2 7c5d2 -- 15a:;b2. Alns. 2a4 ~- 8a3b' + 4 cSd2 ~- 7d + 3a0. 13. From 51a2b2 - 48a3b -t- l10a, take 10a4 - Sa3b - Oa2b2. Inus. 57aTh2 - 40aC3. 14. From 21xz3y 2 + 25x2y.3 + 68xy4 - 40y5, take 64x2y3 - 48.Ty4 - 40y5. Ans. 20xy4 - 39x2y + 21xSy2. 15. From 53y2-, -- _15.2y3 -- 1Sx4 - 5, take - 15x-ay+ I 8x3y2 + 24Xzy. Ans. 35x3y2 - 42x4y - 56x5. 38. From what has preceded, we see that polynomials may be riulbjeeted to certain transformations. For exampl - - 6a2 - 3ab + 2b2 - 2bc, -may be written - - 6a2 - (3ab - 2b2 + 2bc). Il like mnanner - - Satb - 4b2c + b3, rimy be written- - - 7a3 - (Sa2b - 4b2c - 6b); t'i'l. again,- - 7a3 - Sa2b - (4b2c - 6b3). \i) so, 8a -.6a.2 a + 5a2b,3,..... a2 - (6a2b - 5a2b3). A 5,_-...a2c3 4 8a4 + b2 _- C. miay be written - - - 9a2c3 - (8Cw -- b2 + C); (or, it may be written - 9at2c3 - b-2 (Saw + c). These truansformations consist in separating a polynomial intu tvo parts, and then connecting the parts by the minus sign. 312 ELEMENTS OF ALGEBRA. [CHAP. II. it will be observed that the sign of each term is changed when the term is placed within the parenthesis. Hence, if we have one or more terms included within a parenthesis having the mitrnus sign before it, the signs of all the terms must be changed when the ptarenthesis is omitted. Thus, 4a - (6ab — 3c - 26), is equal to 4a - 6ab + 3c +- 2b. Also, Gab - (- 4ac + 3d - 4ab), is equal to 6ab + 4ac - 3d +- 4ab. 39, LEMAsrK. —From what has been shown in addition and subtraction, we cleduce the following principles. 1st. In Algebra, the words add and sum do not always, as in arithmetic, convey the idea of augmentation. For, if to a we add - b, the sum is expressed by a - b, and this is, properly speaking, the arithmetical difference between the number of units expressed by a, and the number of units expressed by b. Consequently, this result is actually less than a. To distinguish this sum from an arithmetical sum, it is called the algebraic sum. Thus, the polynomial, 2a3 -3a2b + 3b2c, is an algebraic sum, so long as it is considered as the result of the union of the monomials 2a3, - 3a2b, - 3b2c, with their respective signs; but, in its proper acceptation, it is the arithmetical difference between the sum of the units contained in the additive terms, and the units contained in the subtractive term. It follows from this, that an algebraic sum may, in the numer ical applications, be reduced to a rneyative expression. 2d. T'he words subtraction and diference, do not always convey the idea of diminution. For, the difference between -+ a and -. being a - (- b) = a + b, is numerically greater than a. This result is an algebraic differ eane. CHI P. I1.1 MULTTPLICATION S3 40, It frequently occurs in Algebra, that the algebraic sighl + or -, which is written, is not the true sign of the term before which it is placed. Thus, if it were required to subtract -b from a, we should write a -(- b) a b. Here the truhe sign of the second term of the binomial is. plus, alr,hough its algebraic sign is -. This minus sign, operating upon the sign of b, which is also negative, produces a plus sign for b in the result. The sign which results, after combining the algebraic sign with the sign of the quantity, is called the esserstial sign of the ternm, and is often different from the algebraic sign. MULTIPLICATION. 41. MULTIPLICATION, in Algebria, is the operation of finding the product of two algebraic quantities. The quantity to be multiplied is called the nmultiplicand; the quantity by which it is multiplied is called the multiplier; and both are called factors. 42. Let us first consider the case in which both factors are monomials. Let it be required to multiply 7a3b2 by 4a2b; the operation may be indicated thus, 7a3b2 X 4a2b, or by resolving both multiplicand and multiplier into their simple factors, 7aaabb X 4aab. Now, it has been shown in arithmetic, that the value of a product is not changed by changing the 3rder of ts factors; hence, we may write the product as follows: 7 X 4aaaaabbb, which is equivalent to 28a5b3. Comparing this result with the given factors, we see that the o efficient in the product is equal to the product of the co-efflcients of the multiplicand and multiplier; and that the exponeiit of each letter is equal to the sum of the exponents of that letter in both multiplicand and multiplier. 84 ELEMENTS OF ALGEBRA. L[CHAP. IL And since the same course of reasoning may be appliecd to any two monomials, we have, for the multiplication of mono m:als, the following RULE. I. M.lltip~ the co-efcients toelether for a newv co-eficzient. II Trilte after this co-ef/icient all the letters which enlter into thit 9xultiplicand and 9n7tltiplier, giving to each an ex.%onent equai bA t.e sum of its exponents in both factors. EXAM]PLES. (1) - - 8a2bc2 X 7abd2 56a3b2c2d2, ~2) - 21a3b-dc X Sabc3 _ 168a4b3c4d. (3) (4) (5) 6) Multiply - 3a26 - 12a2X - - 6xyz - a2xy by - 2ba2 - - 2x2y ay2z - - 2xy2 Ga4b2 144lac'xa3y 6axy3z2 2a2x2 y3. 7. Multiply Sa5b2c by 7a%5ecd. Ans. 56al3b7cid. 8. Multiply 5abd3 by 12cd5..Ans. D60acd8. 9. Multiply 7a4bd2c3 by abdc.. -Ans. 7aSb2d1c4. 43. We will now proceed to the multiplication of polynomials. in order to explain the most general case, we will suppose thle multiplicand and multiplier each to contain additive and subtractive terms. Let a represent the sum of all the additive terms of the multi.:plicand, and — b the sum of the subtractive terms; c the sum of the additive terms of the multiplier, and -- d the sum of -the subtractive terms. The multiplicand will then be represented by a - b and the multiplier, by c - d. We will now show how the multiplication expressed by (a - b) X (c - d) can be effected. The required product is equal to a - b a -- talken as many times as there are units in c -- d. Let us first multiply by c; ac —bc that is, take a - b as many times as there are units in c. We begin by writ- - ac - bC -- ad +- d. u'g ac, which is too great by b taken CHAP. Ii.) MULTIPL, CATION. 35 c times; lbr it is only the difterence between a and b, that is first to be multiplied by c. Hence, ac - bc is the product of a —b by c. But the true product is a - b taken c - d times: hence, the last product is too great by a —b taken d times; that is, by ad- bd, which must, therefore, be subtracted. Suaitracting this from the first product (Art. 37), we have (a - ) X (c - d) -ac - oc -ad + bd: If we suppose a and c each equal to 0, the prcduct will xi duce to t bd. 44. By considering the product of a - b by c - d, we may deduce the following rule for signs, in multiplication. Wlhen two terms of th.e multiplicand and d ultiplier are a.fected with the same sign, their product will be afected wit/h the sign +, and when they are affected with contrary signs, their product will be 3ffzcted wvith, the sig n -. We say, in algebraic language, that + multiplied by + or - multiplied by -, gives +; - multiplied by +, or + mul tiplied by -, gives -. But since mere signs cannot be multiplied together, this last enunciation does not, in itself, express a distinct idea, and should only be considered as an abbreviation of the preceding. This is not the only case in which algebraists, for the sake of brevity, employ expressions in a technical sense in order to secure the advantage of fixing tlre rules in the memory. 45. We have, then, for the multiplication of polynomials, the following RULE. M~ltlp)ly all thle terms of the multiplicand by each ter?n of the ml;!tiplier in succession, affecting the product of any two termzs with the sign plhts, when, tlheir signs are alike, and wvith the sign MtiMsR when their siygns are unlike. Then redume the polynomiaz rcsull to its simlilest form. ,8'6 ELEMENTS OF ALGEBRA. ICHAP. IEL EXAMPLES. 1. Multipll.- -. 3a2 + 4ab + 62 by - - - 2a + 5b 6a3 + Sa2b + 2ab2 + 15a2b - 20ab2 + 5b5 Product - 6a3 + 23a2b + 22ab2 + 5b3 (2). (3). x2 4_. y2 x5 -+ xy6 - 7ax x -y ax +- 5ax x3 + x2 ax6 + ax2y6 + 7a2x2 _ x2y y3 + 5ax6 + 5ax2y6 + 35a2x2 x3 + xy2 - X2y _ y3 6ax6 + 6ax2y6 + 42a2x2. 4. ]Multiply x2 + 2ax + a2 by x + a. Ans. x3 + 3ax2 + 3a2x + a3. 5. MIultiply x2 + y2 by x -- y. Ans. x3 + xy2 + x2y + y3. 6. Multiply 3ab2 + 6a2c2 by 3ab2 + 3a2c2. Ans. 9a2b4 + 27a3b2c2 + 18a4c4. 7. Multiply 4x2 - 2y by 2y. Ans. 82y - 4y2. 8. Multiply 2x + 4y by 2x - 4y. Ans. 4x2 - 16y2. 9. Multiply x3 + x2y + xy2 -- y3 by x - y. Ans. 10. Multiply x2 + xy + y2 by x2 - y +- y2. Ans. X4 + x2y2 + y4. In order to bring together the similar terms, in the product o two polynomials, we arrange the terms of each polynomial witn reference to a particular letter; that is, we arrange them so tha the exponents of that letter shall go cn diminishing from left to right. 11 Mult'ply 4a3 - 5a2b - 8ab2 + 2b3 by 2a2 — 3ab - 46b2 8a5 - 10a4b - 6Ga3b2 + 4a2b3 - 12a4b + 15a.b2 + 24a2b3 - 6ab4 - 1laSb2 + 20a263 + 32ab4 - 8b5 Sa5 - 22a4b - 17a3b2 + 48a2b3 + 26ab4 - 8b5. CHAP. II.] MULTIPLICATION. 37 After having arranged the polynomials, with reference to the letter a, multiply each term of the first, by the term 2a2 of the second; this gives the polynomial 8a5 - 10a4b - 16a'b2 + 4a2b3, in which the signs of the terms are the same as in the multi. plicand. Passing then to the term - 3ab of the multiplier, mul tiply each term of the multiplicand by it, and as it is affected with the sign -, affect each product with a sign contrary to that of the corresponding term in the multiplicand; this gives -12a4b + 15a3b2 ~ — 24a2b3 - 6ab4. Multiplying the multiplicand by - -4b2, gives - 16a3b2 + 20a2B3 + 32ab4 - 8b5. The product is then reduced, and we finally obtain, for the most simple expression of the product, 8a5 - 22a4b - 17a3b2 + 48a23 + 2Gab4 -b5. 12. Multiply 2a2 - 3ax ~- 4x2 by 5a2 - 6ax - 2x2. Ans. 10a4- 27a3 + 34a2X2 -18ax3 - 8S4. 13. Multiply 3x2 - 2yx + 5 by x2 + 2xy - 3. Ans. 3x4 + 4x37y - 4X2 - 4x2y2 + 16xy - 15. t4. Multiply 3x3 + 2x2y2 + 3y2 by 2x3 - 3x2y2 + 5y3. Ans. x6- 55y2 -6x4y4 + 6x3y2 + 15x3yl Ans -- 9x2y4 + 102y5 + 15y5. 15. 1Multiply 8ax - 6ab - c by 2ax + ab +- c. An.s. 16a2x2 - 4a2bx - 6a2b2 + 6acx - 7abc - c2. 16. Multiply 3a2 - 562 + 3c2 by a2 - b3. Ans. 3a4 - 5a2b2 + 3a22 - 3a2b3 + 5b5 - 3b3c2. 17. Multiply 3a2 - 5bd +- cf by - 5a2 2+ 4bd- 8cf. Product - 15a4 + 37a26d - 29ca2qf - 2062d2 + 44bcdf — 8c: 18, {Multiply 4a3b2 - 5a262c + Sa2bc2 - 3a2c3 - 7abc3 by 2ab2 - 4abc - 2bc2 + c3. 8a4b4 - 10a3b64c 28a3b3C2 - 34alb2C3 Product -- 4a2b3c3- 16a4b3c + 12a3bc4 +- 7a 22c4 + 14a2bc5 + 14ab2c5 - 3a2c6 - 7abc6. 88 ELEMENTS OF ALGEBRA. [CHAP. IL 46. REMARKS ON THE MULT'IPLICATION OF POLYNOMIALS. 1st. If both mgultiplicand and mztltiplier are homongeneous, the prodchlt will be homogeneouts, and thle degree of cazy term oqf the prodtlct will be i)ndicated by the stem of the nzmzltmbers which, i'zdic ato the degrees of its two factors. Thu s, in example 18th, each term of the multiplicalld is of the 5th degree, and each term of the multiplier of the Od dei gree: hence, each term of the product is of the Sth degree. This remark serves to discover any errors in the addition of the exponents. 2d. If no two terms of the produlct are similar, there wvill be no reduction amongst them; angd the nutfmber of tetrms in the producft will then be equsal to the number of terms in the multiplicand, multi plied by thle number of terms iz, the multiplier. This is evident, since each term of the multiplier will produce is many terms as there are terms in the lultiplicand. Thus, in example 16th, there are three terms in the multiplicand and two in the multiplier: hence, the number of terms in the,''os1uct is equal to 3 2 = G6. 3d. Anzon.g the termns of the product there are always two which cannot be reduced with any others. For, let us consider the product with reference to any letter common to the multiplicand and multil'.'er: Then the irreducible terms are, 1st. The term produced by the multiplication of the two terms of the multiplicand and multiplier which contain the highest power of this letter; and 2d. The term produced by the multiplication of the two terms which contain the lowest power of this letter. F01or, these two partial products will contain this letter, to a highwerl and to a lower power than either of the other partial pro ducts, and consequently, they cannot be similar to any of them. This remarl, the truth of which is deduced from the law of thit exponents, will be very useful in division. vutAP. II,. MULTIPLICATION. 89 EXAMhI PLE. Muiltiply 5a4b2 + 3aO2 - ab4- 2ab3 by ab - ab2 roduct 5ab3 +- 3a4b2 - a3b5 - 2a34 5 — f55x4 - 3a3b3 + a266 +- 2a2b5. If we examine the multiplicand and multiplier, with reference to a, we see that the product of 5a4b2 by c2b, mlust be irre. ducible; also, the product of - 2ab3 by ab2. If we consider the letter b, we see that:the product of - cb4 by - ab2, must be irreducible, also that of 3a2b by a2b. 47. The following formulas dependirg upon the rule for multiplication, will be found useful inz the practical operations of algebra. Let a and b represent any two quantities; then a + b will represent their sum, and a- b their difference. I. W~e have (a, + b) = (a + b) x (a + b), or performing the multiplication indicated, (a + b)2 =a2 - ab6 -- b2; that is, Thle sautare of the sumn of two quantilies is equal to the squtare if the first, };lus twice the product of ihe flrSt by the second, plus the square of the second. To apply this formula to finding the square of the binomial 5a2 + Sc2b, we have (5a2 + 8a2b)2 25at +~ 80a4b + 64a4b2. Also, (Ga4b + 9a63)2 -- 36aWb2 + 108a5b4 + 81a266. II. We have, (a - b) = (a- b) (a -), or performing the multiplication indicated, (a — b)2 -= a2 - 2ab + b2; that is, hLe square of thZe difereice betweenl two quantities is equal to the square of the first, mzinus tluice the p2roduct of th~e.first by the ueco?2d, plus the sqluare of the second. To apply this to an example, we nave (7a2b - 12ab3)2 =- 49a4b4 -- 18a365 + 144a2b6. Also, (4s;o3b3 7c2d3)2 -l 16a66 - 56(t3b3c2d3 - 49c4d8. 4C ELEMENTS OF ALGEBRA. [CHAP. IL..Il. We have (a + b) x (a - b) -= a2 - b2 by performing the. multiplication; that is, The sum of two quantities multiplied by their diJerence is equal to the digference of their squares. To apply this formula to an example, we have (8a3 + 7ab2) x (8a3 - 7ab2) = 64a6 - 49a2b4. 48. By considering the last three results, it is perceived teat their composition, or the manner in which they are formed from the multiplicand and multiplier, is entirely independent of ally particular values that may be attributed to the letters a and b, which enter the two factors. The manner in which an algebraic product is formed from its two factors, is called the law of the product; and this law remains always the same, whatever values may be attributed to thl' otters which enter into the two factors. DIVISION. 49, DIVISION, in algebra, is the operation for finding from two given quantities, a third quantity, which multiplied by the second shall produce the first. The first quantity is called the dividend, the second, the divisor, and the third, or the quantity sought, the quotient. 50. It was shown in multiplication that the product of two terms having the same sign, must have the sign +, and that the product of two terms having unlike signs must have the sign -. Now, since the quotient must have such a sign that when multiplied by the divisor the product will have the sign of the dividend, we have the following rule for signs in division, If the dividend is + and the divisor - the quotient is +; if the dividend is + and the divisor - the quotient is -; if the dividend is - and the divisor + the quotient is -; if the dividend is - and the divisor - the quotient is +. That is: TIhe quotient of terms having like signs is plus, ana the quotient of terms having unlike signs is mints. CtAP. II.] - DIVISION. 41 51. Let us first consider the case in which both dividend and divisor are monomials. Take 35a5b2c2 to be divided by 7a2bc; The operation may be indicated thus, 35a5b2c2 - 7a2c-; quotient, 5a3bc. Now. since the quotient must be such a quantity as multiplied by the divisor will produce thc dividend, the co-efficient of the quotient multiplied by 7 must give 35; hence, it is 5. Again, the exponent of each - etter in thie quotient must be such that when added to the exponent of the same letter in the divisor, the sum will be the exponent of that letter in the dividend. Hence, the exponent of a in the quotient is 3, the exponent of b is 1, that of c is 1, and the required quotient is 5a3bc. Since we may reason in a similar manner upon any two monomials, we have for the division of monomials the following RULE. 1. Divide the co-eficient of the dividend by the co-efficient of the divisor, for a newt co-efficient. 1[. Write after this co-eflcient, all the letters of the dividend and give to each an exponent equal to the excess of its expe nent in the dividend over that in the divisor. By this rule we find, 48a3b3c2d 4a2bd 150a%3cd3 5a23cd. 12a62c 30a3d2d. EXAMPLES. 1. Divid 16x2 by 8x. Ans. 2z. 2. Divide 15a2xy3 by 3ay. Ans. 5axy2 3. Divide 84ab3x by 12b2. Ans. 7abx. 4. Divide - 96a4b2c3 by 12a2bc. An_. - 8a2bc2. 5. Divide 144a9b8c7d5 by -36a4b6c6d. Ans. — 4a5b2cd1. 6. Divide - 256a3bc2x3 by - 16a2cx2. Ans. 16abcx. 7. Divide — 300a%4c3x2 by 30a4b3c2x. An. -lOabcx. a Divide — 400a866c4y5 by 25a8Bb5c3x. Ans. - 16bcx4 42 ELEMEINTS OF ALGEBRA.. [C'HAP. II. 5., It follows from the precediing rule that the exact division of monomials will be impossible: 1st. When the cc-efficient of the dividend is not divisible by that of the divisor. 2d. When the exponent of the same letter is greater in the divisor than in the dividend. This last exception includes. as we shall presently cee, the case in u hlichl the divisor has a letter which is not containxed in the dividend. Whlen either of these cases occurs, the quotient remlains under the form of a monomial fraction; that is, a monomial expression, necessarily containing the algebraic sign of division. Such expressions may frequently be reduced. 1 2a4b2cd 3ca2bd Take, for example, 2bc2 = 2c' Sa2bc2 2c Here, an entire monomial cannot be obtained for a quotient; for, 12 is not divisible by 8, and moreover, the exponent of c is less in the dividend than in the divisor. But the expression can be reduced, by dividing the numerator and denominator by the factors 4, a2, b, and c, which are common to both terms of the fraction. In general, to reduce a monomial fraction to its lowest terms: Suplress all the factors commnon to both numerator and dezomnz,nator. From this rule we find, 48a3b2cd3 4ad2 37ab3c5d.37b2c and, ta2b3Cc2tle - 3bce; 6aSbc4d2 -Ga2c 1 2ac816c* 3ab 7a2b 1 also, and. aalso, 9 1 - 4c2 a 14a3b2 2ab' In the last example, as all the factors of the dividend are found in the divisor, the numerator is: reduced to 1; for, in falct, both termns of the firaction are divisible by the numerator. 53, It often happens, that the exponents of certain lettcrs are, the same in the divide. d and divisor. 24a3b2 For examnple,.8..2. 8a%2'2 CHAP. Ii.] DIVISION. 43 is a case iN which the letter b is affected with the same exponent! in the dividend and divisor: hence, it will divide out, and will not appear in the quotient. But, if it is desirable to preserve the trace of this letter in the quotient, wve may apply to it the rule for exponents (Art. 51), which gives = b2-2 _ b0. b2 The symbol bh0, indicates that the letter b enters 0 times as a factor in the quotient (Art. 16); or what is the same thing: that it does not enter it at all. Still, the notation shows that b was in the dividend and divisor with the same exponent, and has disappeared by division. 15a2b3c2 In like manner, 2 = 5a02c = 5b2. 54. We will now show that the power of any quantity whose exponent is 0, is equal to 1. Let the quantity be represented by a, and let mn denote any exponent whatever. atm Then,: — m- = a, by the rule for division. ca" But, 1, since the n-umerator and denominator are equal: a,}m~~~~a nence, a- 1, since each is equal to -a~.We observe again, that the symbol a~ is only employed conventionally, to preserve in the calculation the trace of a letter which entered in the enunciation of a question, but which may disappear by division. 55, In the second place, if the dividend is a polynomial and the divisor is a monolilial, we divide each termn of the dividend bty the divisor, tancl connect the quotients by their respective signs. EXAMIPLES. Divide 6at2x4y6 - 12a3x3Y6 + 15a4x'5y3 by 3a2X2y2. Ans. 2x.2y4. 4axy4 + 5a2x3y. 44 ELEMENTS OF ALGEBRA. ICHAP. IL Divide 12a4y6 - 16a5y5 + 20a6y4 - 28a7y3 by - 4a4y3. Ans. - 3y3 + 4ay2 - 5a2y + 7X3. Divide 15a2bc - 20acy2 + 5cd2 by - 5abc. An - 4y2 d2 A4ns. - 3a + - - db ab 56. In the third place, when both dividend and divisor are polynomials. As an example, let it be required to divide 26a2b2 + 10a4 - 48a3b -+ 24ab3 by 4ab - 5a2 + 3b2. In order that we may follow the steps of the operation more easily, we will arrange the quantities with reference to the letter a. -Dividend..Divisor. 10a4 - 48a3b + 26ctW2b2 + 24ab3 - 5a2 ~ 4ab + 3b2 It follows firo'm the definition of division and the rule for the multiplication of polynomials (Art. 45), that the dividend is the sum of the products arising from multiplying each term of the divisor by each term of the quotient sought. Hence if we could discover a term in the dividend which was derived, without reduction, from the multiplication of a term of the divi sor by a term of the quotient, then, by dividing this term do the dividend by-that term of the divisor, we should obtain one term of the required quotient. Now, from the third remark of Art. 46, the term 10a4, con taining the highest power of the letter a, is derived, without reduction from the two terms of the divisor and quotient, containing the highest power of the same letter. Hlence, by dividing the term 10a4 by the term - 5a2, we shall have one term of the required quotient. -Dividend..Divisor. 10a4 - 48a3b + 26a2b2 + 24ab3 5a2 + 4ab + 3b2 10a4 - Sa3b - 6a2b2 - 2a2 + Sab - 40a3b + 32a2b2 + 24ab3 Quotient. - 40a3b + 32a2b2 + 24ab3. Since the terms 10a4 and - 5a2 are fibected with contrary signs, their quotient will have the sign -; hence, 10a4, divided by - 5a2, gives - 2a2 for a ternm of the required quotient. CHaAP. I.l] DIVISION. 45 After having written this term under the divisor, multiply each term of the divisor by it, and subtract the product, 10a4 - Sa3b + 6a2b2, from the dividend. The remainder after the first operation is - 40a3b + 32a2b2 + 24ab3. This result is composed of the products of each term of the divisor, by all the terms of the quotient which remain to be determined. We may then consider it as a new dividend, and reason upon it as upon the proposed dividend. We will therefore divide the term - 40a3b, which contains the highest power of a, by the term - 5a2 of the divisor. This gives + 8ab for a new term of the quotient, which is written on the right of the first. Multiplying each term of the divisor by this term of the quotient, and writing the products underneath the second dividend, and making the subtraction, we find that nothing remains. Hence, -2a2 + 8ab or 8ab - 2a2 is the required quotient, and if the divisor be multiplied by it, the product will be the given dividend. By considering the preceding reasoning, we see that, in each operation, we divide that term of the dividend which contains the highest power of one of the letters, by that term of the divisor containing the highest power of the same letter. Now, we avoid the trouble of looking out these terms by arranging both polynomials with reference to a certain letter (Art. 45), which is then called the leading letter. Since a similar course of reasoning may be had upon any two polynomials, we have for the division of polynomials the following RULE. 1. Arrange. the dividend and divisor with reference to a certain letter, and then divide the first term on the left of the dividend by the first term on the left of the divisor, for the first term of the quotient; multi21y the divisor by this term and subtract the pro. duct from the dividend. 46 ELEMENTS OF ALGEBRA. [CHAP. I., II. Then divide the first term of the remainder hy tI.e first term oj' the divisor, for the secold term of the quotient' multiply the divisor by this seccnd term, adcl subtract the prodLuct from the result of the first operatlon. Corntinue the same operation until a remainder is found equal to 0, or till the first term of th2e remainder is not exactly divisible by the first termz of tlhe divisor. In the first case, (that is, when the remainder is 0,) the division is said to be exact. In the second case the exact divi. sion cannot be performed, and the quotient is expressed by writing the entire part obtained, and after it the remainder with its proper sign, divided by the divisor. SECOND EXAMPLE. Divide 21 x3y2 + 25x2y3 + 68xy4 - 40y5 - 56x5 - lSx4y by 5y2 - Sx2 - G6xy. -. 4Oy5 - 68xy4 - 25X2y3 -- 21X3y2 - 1 - 56x5 2 X- zy —S82 — a Of5 + 48xy4 + 4zx2y3 _ 8y3 + 4xy2 - 3-x2y + 7x3 1st rein. 20xy4 - 39x2yf3 + 21x3f2 20xy4 - 24x2y3 3- 25~x3y2 2d rem. - 15 5x2y3 + 53x232 -- 1S4?y -- 1 52y3 + 1SX 3y2 + 24X4y 3d. rem. - 35x3y2 - 42x4y - 56x5 35x3y2 -4 2x4y - 56x5 Final remainder - - -O. 57, REMARIn. —In performing the division, it is not necessary to brinmg down all the terms of the dividend to form the first remainder, but they may be brought down in succession, as in the example. As it is important that beginners should render themselves.mf-iliar with algebraic operations, and acquire the habit of calcu. tlting promptly, we will trelt this last example in a different nalnner, at the same time, indicating the simplifications which should be introduced. These consist in subtracting each partial )roduct from the dividend as soon as this product is formed. CIAP. II.3 DIVISIOo. 47 - 40y5 + 68xy4 + 25X2fy ~- 21x3y2 - _ 61 t5y2 - G61W- 5.82 1st rein. 20xy4 - 39x2y3 + 21x.3y2 - 8y3-+ 4xy2 - _3x2y + 73J 2d rem. - - 15x2y3 + 53x3y2 _ 1Sxy 3d rein. - 35x3y2 - 42x4y - 565 Final remainder - - - 0. First, by dividing - 40y5 by 5y2, we obtain - 8y3 for 1tie quotient. Multiplying 5y2 by - Sy3, we have - 4Oy5, or, by changing the sign, + 40y5, which cancels the first term of the dividend. In like manner, - 6xy x - 8y3 gives + 48xy4, or, changing the sign, - 48Xy4, which reduced with + 68xy4, gives 20Xy4 for a remainder. Again, - 8x2 X - 8y3 gives +, and changing the sign, - 64x2y3, which reduced with 25x2y3, gives _ 39x2y3. Ience, the result of the first operati(n is 20xy4 - 39x2y3, fol lowed by those terms of the dividend which have not been reduced with the products already obtained. For the second part of the operation, it is. only necessary to bring down the next term of the dividend, to separate this new dividend from the primitive by a line, and to operate upon this new dividend ir the same manner as we operated upon the primitive, and so on. TIIIRD EXAMPLE. Divide 95a.- 73a2 + 56ca4 -- 25 - 59a3 by - 3aa + 5 - 1a + 7a3. 56a -- 59a3 - 73a2 a- 95a - 25 1 a3 - 3a2 - I la + 5 1st reml. - 35au3 4-'14' - 25 Sa 5. 24 remainder. 0. GENERAL EXAMPLES. 1, Divide lOab +- 15ac by 5a. Ans. 2b +- c. 2 Divide 30ax - 54x. by 6x. Ans. 5a- 9, 3, Divide 10x2y - 15y2 - 5 by 5y. Ans. 22 - 3y -1. 4. Divide 12a + 3ax - 1 Sax3 by 3a. Ans, 4 + x - -Cx2, t8 ELEMENTS OF ALGEBRA. LCHAP. 11 5. Divide 6ax2 + 9a2x + a2x2 by ax. Ans. 6x + 9a + ax, 6. Divide a2 + 2ax + x2 by a + x. Ans. a + x. 7. Divide a3 - 3a2y + 3ay2 - y3 by a -y. Ans. a2 - 2cay + y2. 8. Divide 24aib - 12a3cb2 - 6ab by - G6ab. Ans. -4a +- 2a2cb + 1. 9. Divide 6X4 - 96 by 3x — 6. Ans. 2X3 + 4X2 + 8S + 16. 10. Divide a5 - 5a4x +- 10a3x2- 10Ia2X3 + 5ax4 - XI by a2 - 2ax + x2. Ans. a3 - 3a2x + 3ax2 - x3. 1I. Divide 4Sx3 - 76ax2 - 64a2x + 105a3 by 2x - 3a. Ans. 24x2 - 2a - 35a2. 12. Divide y6 - 3y4X2 + 3y2x4 _ x6 by y3 - 3y2. + 3y - x3. Ans. y3 +_ 3y2x + 32yX2 + x3 13. Divide 64a4b6 - 25a2b8 by 8a2b3 + 5ab%. Ans. 8a2b3 - 5ab4. 14. Divide 6a3 - 23a2b - 22ab2 ~- 5b3 by 3a2 + 4ab -+ b2. Ans. 2a + 5b. 15. Divide 6ax6 + 6Gax2y6 + 42a2x2 by ax + 5ax. _Ans. X5 + xy6 + 7ax. 16. Divide - 15a4 + 37a2bd - 29a2cf - 20b2d2 - 44bcdf - 8c2 by 3a2 - 5bd + cf.. Ans. - 5a2 - 4bd — 8f 17. Divide x4+ 2y2 + y4 by x2 - xy + y2 Ans. x-2 + xy - y2. 18. Divide x4 - y4 by x -y. Ans. x3 + 2y + xy2+ y. 19. Divide 3a4 - 8a2b2 + 3a2c2 -- 5b4 - 3b2c2 by a2 - b2. Ans. 3a2- Sb2 - 3c2. 20. Divide x6 -6 5x5y2 - 6X4y4 + 6x3y2 +- 15l3y3 - 9X2y' - 1025 + lx y5 by 3X3 + 2x2y2 + 3y2 Ans. 2x3 - 3x2y2 4 5y3 CHAP. II.i D VISION. 49 REMARKS ON THE DIVISION OF POLYNOMIALS. 58. The exact division of one polynomial by another is impossible: 1st. When the first term of the arranged dividend or the first term of any of the remnainders, is not exatctly divisible by the first ternm of the arranged divisor. it may be added with respect to polynomials that we can often discover by mere inspection that they are not divisible. When the polynomials contain two or more letters, observe the two terms of the dividend and divisor, which contain the highest powers of each of the letters. If these terms do not give an exact quotient, we may conclude that the exact division is im possible. Take, for example, 12a3 - 5a2b + 7ab2 - 11b3 b 4a2 ~ Sab + 31". By considering only the letter C, the division would appear possible; but regarding the letter b, the exact division is imp(ssible, since - 11b3 is not divisible by 3b2. 2d. When the divisor contains a letter uwhich is not in the dividend. For, it is impossible that a third quantity, multiplied by one which contains a certain letter, should give a product independent of that letter. 3d. A monomial is never divisible by a polynomial. For, every polynomial multiplied by either a monomial or ia polynomial gives a product containing at least two terms which are not susceptible of reduction. 41A. If the letter, with reference to which the dividend is arranged, is nzot found in the divisor, the divisor is said to be independent ofy that letter; and in that case, the exact divisi(on is irmpossible, unless the divisor will divide separately the co-eftrients of the dliferent powers of the leading letter. For example, if the "dividend were 3ba4 -+ 9ba2 + 12b, arranged with reference to the letter a, and the divisor 3b, the divisor would be independent of the letter a; and it is evident 4 50 ELEMENTS OF' ALGEBRA. LCH AP, II. that the exact division could not be performed unless the coefficients of the different powers of a were exactly divisible by 3&. The exponents of the different' powers of the leading letter in the quotient wculd then be the same as in the dividend. EXAMPLES. 1. Divide 18a3x2 -- 36a2x3 — 12ax by 6x. Ans. 3a3x - 6a2x2 - a. 2. Divide 25a4b - 30a2b ~ 40ab by 5b. Ans. 5a4 - 6a2 + qSo From the 3d remark of Art. 46, it appears that the ters. of the dividend containing the highest power of the leading letter and the term containing the lowest power of the samt letter are both derived, without reduction, from the multiplication of a term of the divisor by a term of the quotient. Therefore, nothing prevents our commencing the operation at the right instead of the left, since it might be performed upon the terms containing the lowest power of the letter, with reference to which the arrangement has been made. Lastly, so independent are the partial operations required by'the process, that after having subtracted the product of the divi sor by the first term found in the quotient, we could obtain:another term of the quotient by arranging the remainder with:reference to some other letter and then proceeding as before. If the same letter is preserved, it is only because there is no,reason for changing it; and because the polynomials are already arranged with reference to it. OF FACTORING POLYNOMIALS., 59, When a polynomial is the product of two or more factors, it' is often desirable to resolve it into its component factors. This may often be done by inspection and by the aid of the:forlr las of Art. 47. WThen one factor is a monomial, the resolution may be effected iby writing the monomial for one factor, and the quotient arising from the division of the gi en polynomial oy this factor for the other factor. 1. Take, for example, the polynomial ab + ac; Ir. wvhich, it is plain, that a is a factor of both terms: hence ab + ac = a (b + c). 2. Take, for a second example, the polynomial ab2c + 5ab3 + ab2c2. It is plain that a and b2 are factors of all the terms: hence ab2c + 5ab3 + ab2c2 = ab2 (c + 5b + c2). 3. Take the polynomrial 25a4 - 30a3b + 15a2b2; it is evidentl that 5 and a2 are factors of each of the terms. We may, therefore, put the polynomial under the form 5a2 (5a2 - 6ab + 362). 4. Find the factors of 3a2b - 9a2c - 18a2x:y. Ans. 3a2 (b -- 3c + 6z8y) 5. Find the factors of 8a2cx - 18acx2 +- 2ac5y - 30a6c9x. Ans. 2ac (4ax - 9x2 + c4y - 15a5csx). 6. Find the factors of 24a2b2cx- 30a8b5c6y + 36a7b8cd - 6abe. Ans. 6abc (4abx - 5a7b4c5y - 6a6b7d + 1). By the aid of the formulas of Art. 48, polynomials having certain forms may be resolved into their binomial factors. 1. Find the factors of a2 + 2ab q- b2. Ans. (a d- b) X (a + 6) 2. 49x4 + 56x3y + 16x2y2 = (7x2 + 4xy) (7x2 + 4xy). 3. Find the factors of a2 - 2ab + b2. Ans. (a - b) X (a - b)., 64a2b2c2 - 4Sabc2d2 4- 9c2d' = (8abc - 3cd2) (8abc - 3cd2). F. Find the factors of a- - b2. Ans. (a + b) x (a - 6). ( 16a2c2Z-9d4 = (4ac - 3d2) (4ac - 3d2). 52 ELEMENTS OF ALGEBRA I CHAP. 1L GENERAL EXAMPLES. 1. Find the factors of the polynomial 6a3b 4- 8a2b5 - 10ab7 2ab. 2. Find the factors of the polynomial 15abc2 — bc2 9a3b5c6 12db6c2. 3. Find the factors of the polynomial 25a6bc6 - 30aSb4d 5ac4 - 60ac6. 4. Find the factors of the polynomial 42a2b2 - 7abcd + 7abd Ans. 7ab (6ab - cd + d). 5. Find the factors of the polynomial n3 + 2n2 + n. First, n3 + 2n2 + n = n (n2 + 2n + 1) =n(n ~-1) x (t + 1) =n(n + 1)2. 6. Find the factors of the polynomial 5a2bc + lOab2c + 15abc2. Ans. 5abc (a J 2b + 3c). 7. Find the factors of the polynomial a2x- x3. Ans. x (a + x) (a - xj. 00. Among the different principles of algebraic division, there is one remarkable for its applications. It is enunciated thus: The difference of the same powers of any two quantities is &xactly divisible by the difference of the quantities. Let the quantities be represented by a and b; and let m de note any positive whole number. Then, am - VI %will express the difference between the same powers of a ~d b, and it is to be proved that atm - bm is exactly divisible b a- b, If we begin the livision of we have am - bm a-b am - am-lb am-1 1st rem..... am'-lb - bm Or, by factoring - - - b(am-1 - bin-l). CHAP. II.] DIVISION. 53 Dividing am by a the quotient is aw-l, by the rule foi the exponents. The product of a - b by am-i lje:[ng subtracted fiom the dividend, the first remainder is am-lb- bm, which can be put under the form, b (am-l - bm-1). Now, if the factor (am- - bin-l) of the remainder, be divisible by a - b, b times (a"rl - bm-l), must be divisible by a —b, and consequently am" —bm nust also be divisible by a - b. HIence, If the difference of the same powers of two quantities is exactly divisible by the difference of the quantities, then, the difference oJ the powers of a degree greater by 1 is also divisible by it. But by the rules for division, we know that a2 - b2 is divis ible by a - b; hence, from what has just been proved, at3- b3 must be divisible by a - b, and from this result we conclude that a4 - b4 is divisible by a - b and so on indefinitely: hence the proposition is proved. 61. To determine the form of the quotient. If we continue the operation for division, we shall find am-2b for the second term of the quotient, and am-2b2 - bm for the second remainder; also, am-3b2 for the third term of the quotient, and am-bs3 - bm for the third remainder; and so on to the Mnth term cf the quo tient, which will be am'bm-l or bm-1; and the mth remainder will be amn-bmn -bm or bm bm = O. Since the operation ceases when the remainder becomes 0, we shal have m terms in the quotient, and the result may be wriW ten thus: am -a +- a -2b + a"n-3b2 +..... + ablm q- +Daa —b CHAPTER III. OF ALGEBRAIC iRACTIONS. fA, AN ALGEBRAIC FRACTION is a i expression of one or mnore equal parts of 1. One of these equal parts is called the fractional unit. Thus, b is an algebraic fraction, and expresses that 1 has been divided into b equal parts and that a such parts are taken. The quantity a, written above the line, is called the numer. atoro; the quantity b, written below the line, the denominator, and both are called terms of the fraction. One of the equal parts, as —, is called the fractional unit; and generally, the reciprocal of the denominator is the fraetional unit. The numerator always expresses the number of times that the fractional unit is taken; for example, in the given fraction, the!ractional unit - is taken a times. 63. An entire quantity is one which does not contain any [itactional terms; thus, a2b + cx is an entire quantity. A mzixed quantity is one which contains both entire and frao tional terms; thus, a2b +- is a mixed quantity. Every entire quantity can be reduced to a fractional form havirng a given fractional unit, by multiplying it by the deronoinator of the fracticnal unit and then writing the product over the ienominator; thus, the quantity c may be reduced to a fract ona] CHAP. III.] ALGEBRAIC FRACTIONS. 55 form with the fractional unit -b-, by multiplying by b and bc dividing the product by b, which gives —. 64. If the numerator is exactly divisible by the denominator, a fractional expression may be reduced to an entire one, by simply performing the division indicated; if the numerator is not exactly divisible, the application of the rule for division will sometimes reduce the fractional to a mixed quantity. 65. If the numerator a of the fraction? be multiplied by any quantity, q, the resulting fraction - will express q times as many fractional units as are expressed by -?-; hence: Multip lying the numerator of a fraction by any quanZtity is equivalent to multiplying the fraction by the same quantity. 66. If the denominator be multiplied by any quantity, q, the value of the fractional unit, will be diminished q times, and the a resulting fraction - will express a quantity q times less than the given fraction; hence: Mlltiplying the denominator of a fraction by any quan'tty, is equivalent to dividing the fraction by the same quantity. 67. Since we may multiply and divide an expression by the same quantity without altering its value, it follows from Arts, 65 and 66, that: Both numerator and denominator of a fraction, may be multiplied by thle same quantity, without changing the value of the fraction. In like manner it is evident that: Both numeratcr and denominator of c fraction mnay be divided by the sagme quanlity wuithout changing Mhe value of the fraction. 68. We shall now apply these principles in deducing rules For the transformation or reduction of fractions. 56 ELEMENTS OF ALGEBRA. LCHAP. III. I. A fractional is said to be in its simplest form when the numer: ator and denominator do not contain a common factor. Now, since both terms of a fraction may be divided by the same quantity without altering its value, we have for the reduction Of a fraction to its simplest form the following RULE. Resolve both numerator and denominator intc their simnple facw tors (Art. 59); then, suppress all the factore common to both terms, and the fraction wvill be in its simplest form. REMARK.-When the terms of the fraction cannot be resolved into their simple factors by the aid of the rules already given, resort must be had to the method of thea greatest common divi:or, yet to be explained. EXAMPLES. 3ab ~- 6ac i. Reduce the fraction 3ad + 12a t~'s simplest form. We see, by inspection, that 3 and a re- fpctoMr of th~ nu. merator, hence, 3ab + 6ac = 3a (b + 2c) We also see, that 3 and a are factors Ar the e r\omirn,.'s hence, 3ad + 12a = 3a (d +- 4). 3ab +- 6ac _ 3a ( 2c) b ~%- Pc Hence, 3ad + 12a 3a(d + 4) d — 4 6a2b - 3ac 2. Reduce a 3a to its simplest form. 9a b + 3a i 2AYb - C.?6 4- _ 3. Reduce 251bc 51f S. Reduce 356-2 + 156 to its simplest form. 7 + 3 4Reue 54abe 4, Reduce 45 + to its simplest Corm. 45a2e +- 9acd 6h aa + d CHAP. III.] ALGEBRAIC FRACTIONS. 57 36a2b ~' 12aSf b. Reduce 36a84 2a to its simplest furm. A3a +f 76 l2acd - 4cd2 7. Reduce;o its simplest form. 12caf + 4~2d 3a - d 3f+c 18a2c2 - 3acf "7. Reduce,o its simplest form. 27ac2 - fac3 Ans. f 9c -- 2C2' II. From what was shown in Art. 63, it follows that we may reduce the entire part of a mixed quantity to a fractional form with the same fractional. unit as the fractional part, by multiplying and dividing it by the denominator of the fractional part. The two parts having then the same fractional unit, may be reduced by adding their numerators and writing the sum obtained over the common denominator. HIence, to reduce a mixed quantity to a fractional form, we have the RULE. Multiply the entire part by the denominator of the fraction: hJen add the product to t/he numerator and write the sum over the danominator of the fractional part. EXAMPLES. 1. Reduce z- (a2 - x2) to the form of a fraction. a2 _ 2 X2 _ (a2 - X2) X2 - ca2liere, x- = _ = __ x x x, Reduce a - to the form of a fraction. 2a 2as 58 ELEMENTS OF ALGEBRA. [CRAP IL 2 — 7 3. Reduce 5 -i- to the form of a fraction. 3X 17z - 7 Ans. -<3 3z x-a-1 i Reduce 1 - to the form of a fraction,. 2a - x + 1 An8s. a 5. Reduce 1 + 2x - X to the form of a fraction. 5x 10x2 + 4r +- 3 Ans. 5x 6. Reduce 3x-1 - -x + to the form of a fraction. 3a -2 9ax — 4a - 7x + 2 3a - 2 REMARK.-We shall hereafter treat mixed quantities as though they were fractional, supposing them to have been reduced to a fractional form by the preceding rule. III. —From Art. 64, we deduce the following rule for reducing a fractional to an entire or mixed quantity. RULE. Divide the numerator by the denominator, and continue the oper ation so long as the first term of the remainder is divisible by the first term of the divisor: then the entire part of the quotientfound, added to the quotient of the remainder by the divisor, tuill be the mixed?uatitty required. If the Irmainder is 0, the division is exact, and the quotient is an entire quantity, equivalent to the given fractional expres. sion, EXAMPLES. 1. e-duce- + - to a mixed quantity. 2. Ans. - a + —.. CHAP, III] ALGEBRAIC FRACTIONS. 59 ax - X2 2. Reduce - to an enthie or mixed quantity. lan.. a -. ab - 2a2 3. Reduce to a mixed quantity. 2a2 Alzs, a-.. a2 - x2 4. Reduce a to an entire quantity. ct-x Ans. a + x. 5. Reduce - to an entire quantity. X-Y Ans. X2 - xy +- y2. 10x2 - 5x - 3 6. Reduce 5 to a mixed quantity. Anils. 2x - I + 5x IV. To reduce fractions having different denominators to equiv alent fractions having a common denominator. aX; C e Let b' d and f, be any three fractions whatever. b' 7 d It is evident that both terms of the first fraction may be mul adf tiplied by df giving b- f and that this operation does not change the value of the fraction (Art. 67). In like manner both terms of the second fraction may be multiplied by bf, giving b-f; also, both terms of the fraction e bcle - may be multiplied by bd, giving bf adf bcf bde If now we examine the three fractions df df and Wf we see that they have a common denominator, bclf, and that each numerator has been obtained by multiplying the numerator of the correspcnding fraction by the product of all the denominators except:its own. Since we may reason in a similar manner upon any fractions whatever, we have the following 60 ELE1MENTS OF ALGEBRA. [CHAP. III. RULE, Multiply each numerator into the pod:tcc t*f all the denomina. tors except its own, for new numerators, and all tl.e denominators together for a conmmon denominator. EXAMPLES. a b 1. Reduce b and - to equivalent fractions having a com C monl denominator. a X c = ac) b X b b2 a the new numerators. and. b X c = bc the common denominator. a a +a b 2. Reduce - and + to equivalent fractions having, coin b c ac ctb + b2 mon denominator. Ans. - and ab-62 bc be 3x 2b 3. Reduce 2T, 3- and d, to eouivalent fractions having a 9cx 4ab 6acd common denominator. Ans. - and 6ac'6ac 6ac 3 2x 2x 4. Reduce -, - and a +-, to equivalent fractions hav. 9a Sax 12a2 + 24x ing a common denominator. A.s. 12 and 1-a' f12a a 12a I a2 a2 + x2 5. Reduce -2-, and, to equivalent fractions hay ing a commcn denominator. 3a + 3x 2a3 + 2a2x 6a2 -1 6x2 Ans. and 6a d- 6x' 6a + 6x 6a +- Gx a c-b b 6. Reduce and -, to equivalent fractions hay a-b' ax C rig a common denominator. a2cx ac2 - abc - bc2 + cb2 a2bx - ab'x Ans. and x - ab a2cx - abex' a2cx - abcx a~cx - abcx tRAP. III.] ALGEBRAIC FRACTICN S. 61 V. To add fractions together. Quantities cannot be added together unless they have the same unit. Ience, the fractions must first be reduced to equivalent ones having the same fractional unit; then the sum of the numerators will designate the number of times this unit is to be taken. We have, therefore, for the addition of frao. d,1ns the following RULE. Reduce the fractions, if necessary, to a common denominator: then add the numerators together and place their szum over the oinmmon denominator. EXAMPLES. I. Find the sum of —, - and - V d Iere, - a x d xf = adf c X b X f = cbf. the new numerators. e X b x d-=ebd And - b X d X f =d bcf the common denominator. ence adf cf ebd adf + chf + ebd Hence, $bclf bd + b df bdf - the sum. 32. To 2ax 2abx- 3cx2 a T ---- add +-. Anls. a+b+ b c b x x x 3. Add - - and together. Ans. -+ 2' 3 4 12' x -2 4x 19x - 14 4. Add -2 ard 4- together. Ans. 214 x - 2 2x —3 1 lOx- 17 5. Add x + - to 3x + - -3. Ans. 4x + 12 5X2 x + a -6. It is required to add 4x a and 2+ together. 5Ans. 4 3 + ax + a2 2ax 2x 7x 2x~+ 1 7. It is required to add 2 7 and together. 3' 4 5 49x + 12 Ans.,2x -60 6O 62 ELEMENTS OF ALGEBRA. [CHAP. ITL 8. It is req~uired to add 4x, 7- and 2 + - together. 44x + 90 45 2w 8z 9. It is required to add 3x + 5- and x - together. 523x Ans. 3x + - 45' 10. What is the sum of aand a- b' a + b a + x' A a3 - ax2 + a2b - bx2 + a2c + acx - abc - bcx + a2d - b2d Ans. a3 _ b2a + a2x _ b2x a" + a2 (b - c ~d) - a (2 - cx bc) - b (X2 + C- + bd) a3 + a2x - ab2 - b2x VI. To subtract one fraction from ianother. Reduce the fractional quantities to equivalent ones, having the same fractional unit; the difference of their numerators will express how many times this unit is taken in one fraction more thanSin the other. Hence the following RULE. I. Reduc6 the frIactions to a common denominator. II. Subtract the numeratQr of the subtrahend from the nulmerator of the minuend, anzd place the dijerence over the common denominator. EXAMPLES. x —a 2a- 4 1. From - -subtract - 4-. 2b 3c Here, (xz- a) X 3c- = 3cx - 3ac t (2a - 4x) 2b - 4ab - bx the num And, 2b X 3c = 6bc the common denolinatch 3cX - 3ac {ab - 8bx 3cx - 3ac - 4ab + 8bx Henc be - bc 6be bc 12x 3x 399 2. From - - 7subtract-i A ns, -35 CHAP. III.] ALGEBRAIC FRACTIONS. 63 7iy 3. From - 5y subtract 8. Ans. 3x 2x 13x 4. From 3x subtract 2x A-Ls. -T VI'63 x + a -e dx ~ ad i-b c 5. From 6- a subtract c As. - b 3x -] a 2x+- 7 6. From - subtract 5b 8 24x 4- Sa - 1Obx - 35b 40b x x-a 7. Fiom - - 3x +- subtract x — c b c A + x + x - ab Ans. 2x -- bc VII. To multiply one fractional quantity by another. Let b represent any fraction, and c any other fraction; and 6 d any other fraction; and let it be required to find their product. If, in the first place, we multiply - by c, the product will be', obtained by multiplying the numerator by c, (Art. 65); but this product is d times too great, since we multiplied a g by a quantity d times too great. hIence, to obtain the true product we must divide by d, which is effected (Art. 66) by multiplying the denominator ])y d. We have then. O (' aC a & ac b. i hence d bd RULE. I. Cancel all factors co0mnmo to the numerator arnd de,1 smri: nartor. IIT Aflltiply the numerators together fior t/he numnerator of tle procdZe:t, and the denominators together for the denomnzator of the 64 ELEMENTS OF ALGEBRA. [CHAP. lIL E;XAiPLE5, I. ZMultiply a + - by d bx a2 4- bx First,. - - a+- + b a a a2 + bx c a2c + bcx Hence, - X- a d ad 3x 3a 9ct; 2. Required the product of andcl. Ans. 2x 3x2 3x3 3. Required the product of and 2a. Ans. 2x 3ab 3ac 4. Find the continued product of -, -- and Ans. 9ax bX a 5. It is required to find the product of b - and -. a 2 ab + bx Ans. X2 - 2 X2 -t- b2 6. Required the product of and bc b + c ns. b2C + bc5' 7. Required the product of x + + and x a a ~-b' a2 q- ab ax 2 - X2 8. Required the product of a + a -x and A 2 (a + x) x (1 + x) CHAP. III.3 ALGEBRAIC FRACTIONS. 65 VIII. To divide one fraction by another. Let b represent the first, and d the second fraction; then 6 d he division may be indicated thus. /a\ \b: d f now ve mfultioply both nwll nuerator and deno vinator of this ad fiaction (Art. 67), the new numerator will be bd, and the new ad denominator d- which is equal to I. (a\ lad\ a c ( b ) (abc) ad Hence, a -b a / c -) - bad 6 d C) I be TFhis last result we see might have been obtained by invertiIig, the terms of the divisor and multiplying the dividend by tile resulting fraction. IIence, for the division of fractions, we havethe following RULE. ~avert the terms of the divisor and miltiply the dividend by the. resulting firaction. EXAMPLES. b f 1. Divide a- by. 2c g b 2ac- b a -- - 2c 2c b f 2ac -b a 2aeg- bg ence, a2c g= 2c f = 2cf 7x 2. Let - be divided by 12 Ans, i 66 ELE MENTS OF ALGEBRA. [CHAP. III 4x2 4x 3. Let be divided by 5x. Ans. 17 35'. Let +1 2x x+As 1 4. Let be divided by -. Ans. 6 4x x x 2 5. Let be divided by.. Ans. 2 5x 2a 5bx 6. Let be divided by' Ans. 2 - 3b 2ab x - b CX -- bb 7. Let be be divide d by S.hLet s 4i - bo be divid ed by x2 h bx X2 - 2bxf tt b2 x b ii b2 Ans. x + —, 9. Divide ax- by a s. 1 - -x - X2' a' 10. Divide a- by - a Ans. - (1 + a). 69. if we have a fraction of the form b - we may observe that -a a -a -a -c, also =-c and - c; that is, The sign of the quotient will be changred by changing the sign cither of the numerator or denominator, but will not be afccted by changing the signs of both the terms. 70. We will add two propositions on the subject of fractions. I. If the same number be added to each of the terms of a proper fraction, the fractionz re'izuZing from lhese adclditionzs will be ygretle?'than the first; but if it be addled to the terms of an imiprofle? fraction, the resulting fraction will be less hian the first. Let the fraction be expressed by ab Let m represent the number to be added to each term: the7 ithe new fraction will be, a + m b + vm' CkHlAP III.] ALGEBRAIC FRACTIONS. 67 1k- order to compare the two fractions, they mni.st'Le reduced to the same denominator, which gives for a ab + am the first fraction, b + b2 + b}m a+ m ab +bm anld for the new fra + ct bm b + n- + bZn' Now, the denominators being the same, that fraction will bo the greater which has the greater numerator. But the two numerators have a common part ab, and the part bmi of the secornd is greater than the part amn of the first, when b > a: hence ab + — bm > ab + am; that is, when the fraction is proper, the second fraction is greater than the first. If the given fraction is improper, that is, if a > b, it is plain that the numerator of the second fraction will be less than that of the first, since bm would then be less than am. II. If the same number be subtracted frort each lern of a proper fraction, the value of the fraction will be diminished; but f' it be subtracted from the terms of an improper jf'action, the vatle of the fraction will be increased. Let the fraction be expressed by —, and denote the number to be subtracted by qnm. Then,. — will denote the new fraction. By reducing to the salne denominator. we have, a ab - am b b2 _- 6 bm2 a - m ab -- bmu and b —?n b2- brn Now, if we suppose a < b, then am < bin; and if am < bm then will ab - amn > ab - binz that is, the new fraction will be less than the first. If a > b, that is, if the fraction is improper, then Ua > bm, and.a5 - am < ab - bin, that is, the new fraction will Le greater than the first. 68 ELEMENTS OF ALGEBRA. LCHAP. 1Ii1 GENERAL EXAMPLES. 1+ X2 1-x2 2(1+x4) I Add 1 2 to Ans. 2. Add Z( -to Ans. ~a 4- b a-b 4ab 3. From a Z. take + b Ans. a- ba -— b a q-b'' -- 4. From - x take I - x' Ans. — _x 1-x2 12I-x -I X2 - 9x + 20 x2 - 13x +- 42 5. Multiply 2 ~ 2X by.2 -X X2 -- 5x Ans. - x s2 6. Multiply x4 - b2 by -+ bx Ans. x3 + b2xz X2 +fx + b2 b x-b a + x a -x a + x a — x 7. Divide -- ~t by a-x a-+-x ac-x a- x a2 + X2 Ans. 2ax n — n —1 8. Divide 1 n+ 1 oy 1- Ans.,n. EXAMPLES INDICATING ITUSEFUL FORMS OF IEDIt.'CTION. a c e adfx5 cbfx'4 ebd.3 bz d.*2 fx3 -bdf,6 bdfx6 bdfx16 adfx2 - b+fx + b~lde a c e g acldfx9 bcfhx8s becdxz bdffgdg bX dWx2 fX3 hX4 - bdfhx1 O bdfx10 - bdJf 10- bdfhx'1 adfhx3 + bceft, 2 -beJlL - e bdlf -bdflxlS liClAP. III. EXAMPLES IN FRACTIONS. 69 1 +,2 1 _- 2 (~ + X2)2 (1 -2)2 1 -- x2 1 + x2 -(1 -- X2)(1 + x (1 2) 2)(1 -+- 2) (1 + 2)2 + (1 -2)2 = (3. -- -) (1 + X2) 2 _ 1 1- x 1 — $(' + ) (- ) = 20. + b a -- (a 4 b)2- (a —b)2 a -b a + b (a + b) (a-b) 4ab a2 _ b' + 2 - 2 (1 + x2)2 ( - x2) - -' 1+ -x2 =( - x2)(1 +2) - (1 - x2)(1 + 2) (1 + X2)2 - (1 - x2)2 - ( - X2)(1 + x2) 4x2 lP 2 - x 1 + x x 1 + 2x ( 1 + x2)2 1-x2 1 +x2 1 - - x -- X 1-x2 (1 —x2)2' x4 - 64 x2 - x x4 -- b x — b x 2-b I ba z — b 6 22 —2x q- b2 X X2-bx (X - b4) (x -b) - (2 - 2b + b2) (2 + bx) (x2 - b2) (x2 + b2) (X - b) (- (-)2 x (v + b) (x + b) (x - b) (2 + 62) ( - b) ( - b) () - b) (x + b) 2 + 42 70 ELEMENTS OF ALGEBRA. LCHAP. IlL Of tAe Symbols 0, Xo cad 0 0O 71. The symbo. 0 is called zero, which significs in ordinary language, nothing. In Algebra, it signifies no quantity: it is also used to expres a quantity less than any assignable quaDntity. Tihe symbol oo is called the symbol for infiznity; that is, it is used to represent a quantity greater than aay assignable quat tily. If we take the fraction'a,' and suppose, whilst the value of a remains the same, that the value of b becomes greater and greater, it is evident that the value of the fraction will become less and less. When the value of b becomes very great, the value of the fraction becomes very small; and finally, when b becoames greater than any assignable quantity, or infinite, the value of the fraction becomes less than any assignable quantity, or zero. Hence, we say, that a finite quantity divided by infinity is eqtual to zero. We may therefore regard-, and 0, as equivalent symbols. If in the same fraction —, we suppose, whilst the value of a remains the same, that the value of b becomes less and less, it is plain that the value of the fraction becomes greater and greater; and finally, when b becomes less than any assignable quantity, or zero, the value of the fraction becomes greater than any assignable quantity, or infinite. Hence, we say, that a finite quantity divided by zero is equal to infinity. WVe may then regard 0- and c as equivalent symbols: Zers and infinity are reciprocals of each other. 0. The expression -- is a symbol of indetermznnatzon; that is, it is employed to designat-1 a quantity which admits of an infinite number of values. The origin of the symbol will be explained in the next chapter. C(HAP. III.] ALGEBRAIC FRATIgDNS. 71 0. It should be observed, however, that the expression - is not always a symbol of indetermination,, but frequently arises fromn the ezistence of a common factor, in both terlms of a fraction, which factor becomes zero, in consequence of a particular hypo-:thesis. 1. Let us consider the value of x in the expression a3 -_ 2 a2 _ b2 If. in this formula, a is made equal to b, there results 0. -. 2 But, - - -a3-b3 = -(a - b) (a2 ~ a -- b2) and - a2-6 b2=(a —b) (a + b), hence, we have, (a - 6) (a2 + ab b62) (a - b) ( + b) Now, if we suppress the common factor a - b, and then sup pose a- b, we shall have 3a 2 =. 2. Let us suppose that, in another example, we have a2 - 62 X = If' we suppose a = 6, we have 0 X —-- If, however, we suppress the factor common to the numerator and denominator, in the value of x, we have, (a )( - ) a + 6 (a - b) (a - b) - a - b' Tf now we make a b, the value of x becomes 2b 0, 72 ELEMENTS OF ALGEBRA. [CHA P. IlL 3. Let us suppose in another example, ((a - )2' 3 - -b3' 0 ill which the value of x becomes - when we make a 3. If we strike out the common factor a -- b, we shall find a-b a2 -+ ab + b2' If now we make a = b, the value of z becomes 0 0_ Therefore, before pronouncing upon the nature of the expres sion -, it is necessary to ascertain whether it does not arise from the existence of a common factor in both numerator and denominator, which becomes 0 under a particular hypothesis. If it does not arise from the existence of such a factor, we conclude that the expression is indeterlminate. If it does arise from the existence of such a factor, strike it out, and then make the particular supposition. If A and B represent finite quantities, the resulting value of the expression will assume one of the three f9rms; that is: A A 0 - or -; it will be either finite, infinite, or zero. This remark is of much use in the discussion of problems. CHAPTER IV. CQIATCNS OF THtE FIRST DEGREE IN70OLVING BUT ONE UNKNOWN QUANTITY! 72. A'EQUATION iS the algebraic expression of equality be, tween two quantities. Thus, = a + b, is an equation, and expresses that the quantity denoted by x is equal to the sum of the quantities represented by a and b. Every equation is composed of two parts, connected by the sign of equality. The part on the left of this sign is called the first 1?emtber, that on the right the second member. The second member of an equation is often 0. 73. An equation ma~ contain one unknow, quanltity only, or it may contain mnore tfhan one. Equations are also classified according to their degrees. Thie degrees are irzudicaled by the exponents of the uznknowvn quantities which enter thenm. In equations involving but one unknownu quantity, the degree is denoted by the exponent of'the highest power of that quantity in any termn. In equations involving more than one unknown quantity, the degree is denoted by the greatest sumn of the exponents of the unkcnorwn quantities in any term. For example: ax- b -cx -- d ) d Ix -,+ - b = c x + d = are equations of the first degree. ax)2 +'Thx + c = 0 ax2 ~ by + cy2 - are equations of the second degree. (,2X3 + 2(dgx2 = abx - Sc - are equations of the third degree, 4axy2 -- 2cy3 + abxy 3, and so on. 7: ELEM ENTS OF ALGEBRA. [CHAP. IV 74. Equations are likewise distinguished as numerical equations and literal eqaltions. The first are those which contain numbers only, with the exception of the unknown quantity, which iq always denoted by a letter. Thus. 4x - 3 2x- 5, 32 - -x-8 = are numerical equations. A literal equation is one in which a part, or all of the known quantities, are represented by letters. Thus, bx2 - a, - 3X = 5, and cx + dx2 = c +f, are literal equatiowns. 75. An identical equation is an equation in which one member is repeated in the other, or in which one member is the result of certain operations indicated in the otther. In either case, the equation is true for every possible vaz!u~ of the unknown quantities which enter it. Thus, 2 - Y2 x - + b = ax - b, (x + a)2 = X2 - 2ax - a2, - y - are identical equations. "76. From the nature of an equation, we perceive that it must possess the three following properties: 1st. The two members must be composed of quantities of the same kind. 2d. The two members must be equal to each other. 3d. The essential sign of the two members must be the same. 76.* An axiom is a self-evident proposition. We may here enumerate the following, which are employed in the transformationl and solution of equations: 1. If equal quantities be added to both members of an equation, the equality of the members will not be destroyed. 2. If equal quantities be subtracted from both members of an eqiUation, the equality will not be destroyed. 3. If both members of an equation be multiplied by equal quantities, the products will be equal. 4. If both members of an equation'be divided by equal quan itites, the quotients will be equal. 5. Like powers of the two members of an equation are equal 6. Like roots of the two members of an equation are equal, CHAP. IV.] EQUATIONS OF THE FIRST DEGREE. 70 Solution of Equations of the First Degree. 77. The solution of an equation is the operation of finding a value for the unknown quantity such, that when substituted for tne unknown quantity in the equation, it will satisfy it; that is, make the two members equal. This value is called a root of -the equation. In solving an equation, we make use of certain transJbrmationzs. A transformation of an equation is an operation by which we 3hange its form without destroying the equality of its members. First Transformation. 78. The object of the first transformation is, to reduce an equation, some of whose terms are fractional, to one in which all of the terms shall be entire. Take the equation, 2x 3 x o - x + = -11 4 6 First, reduce all the fractions to the same denominator, by the known rule; the equation then becomes 4Sx 54x 12x + t- -%- - 11. 72 72 + 72 If now, both members of this equation be multiplied by 72, the equality of the members will be preserved (axiom 3), and the common denominator will disappear; and we shall have 48x - 54x + 12x = 792; or by dividing both members by C, 8x — 9x + 2x = 132. The last equation could have been found in another manner by employing the least common multiple of the denominators. The common multiple of two or more numbers is any number which each will divide without a remainder; and the least common multiple, is the least number which can be so divided. The least common multiple of small numbers can be found by inspection. Thus, 24 is the least commcn multiple of 4,. 6 and 8; and 12 is the least common multiple of 3, 4 and 6. 76 ELEMENTS OF ALGEBRA. [CIHAP. IV. Take the last equation, 2x 3 x 3 4 6 We see that 12 is the least common multiple of the donomllinators, and if we multiply each term of the equation by 12, reducing at the same time tc entire te:ins, we obtain 8 x- 9x + 2x= 132, the same equation as before found. Hence, to transform an equation involving fractional terms to one involving only entire terms, we have the following RULE. Form the least common multiple of all the denominators, and then multiply both members of the equation by it, reducing fractiona, to entire termns. This operation is called clearing of fractions. EXAMPLES. 2 x 1. Reduce - - - 3 - 20, to an equation involving only entire terms. We see, at once, that the least common multiple is 20, by which each term of the equation is to be multiplied. ~X 20 Now, -x 20 =x x =- 4x, x S20 and x 20= x x 2 5.x; 41- 4 that is, we reduce the fractional to entire terms, by zmultiplyzn9 the nunmerator by the quotient of the common multiple divided by the denominator, and omitting the denominators. HEence, the transformed equation is 4x + 5z - 60 -- 400. 2. Rleduce - + -4 = 3 to an equation involving only i re terms. Ans. 7 entire terms. Ans. 7x + 5w.- I O = 105. UvAP, IV.] EQUATIONS OF THE FIRST DEGREE. 77 a c 3. Reduce -- - - -f= g to an equation involving only entire terms. Ans. ad - bc + bdf = bdg. 4. Reduce the equation ax 2c2x 4bc2x 5a3 2c2 b ab +4a= a3 - b2 -- to one involving only entire terms. Ans. a4bx - 2a2bc2x +- 4a4b2 = 4b3c2x - 5a6 + 2a2b2c2 -3-asb3a Sec;td Trfcansformaction. 79. The object of the second transformation is to change ony term fr'om one member *Yf an equation to the other. Let us take the equation ax -- b - d - cx. If we add cx to both membeis, the equality will not be do. stroyed (axiom 1), and we shall lhve ax + cx + b = d- cx + cx; or by reducing, ax + cx + b = d. Again, if we subtract b from both members, the equality will not be destroyed (axiom 2), and we shall have, after roluction, ax + cx = d - b. Since we may perform similar operations on any other equation, we have, for the change or transposition of terms, the following RULE. Any term of an equation may be transposed fromr oze member to the other by changing its sign. 80, We will now applf the preceding principles to,he solu. tion of equations of the first degree. For this purpose let us assume the equation a, - b a + d -- -— d bx —-- -- c a Clearing of fractions, we have, a (a + b) x - acd = abcx - c (a + d), 78 ELEMENTS OF ALTGEBRA. [CHAP. IV. If, now, we perfor:i the operatioun indicated in both members, we shall obtain the equation ax + abx - acd = abcx - ca - ed. Transposing all the terms containing x, to the first member, and all the known terms to the second member, we shall have, a2x - abx - abcz = ac(d -- ac - cd. Factoring the first member, we obtain (a2 + ab - abc) x:= acd - ac - cd: If we divide both members of this equation by the coefficient of x, we shall have acd - ac - cd a2 + ab - abc Any other equation of the first degree may be sclved in a similar manner: ience, in order to solve any equation of the first degree, we have the following RULE. I. Clear the equation of fractions, and peiform in both members ~~ll the algebraic operations indicated. II. Transpose all the terms containing the unknown quantity to the first member, and all the known terms to the second member, Cand reduce both members to ttheir simplest form. III. Resolve the first member into two factors, one of which shall be the unknown quantity; the other one will be the algebraic sumq of its several co-e flcients. IV. Divide both members by the co-efficient of the unknown quan. ity; the second mnember of the resulting equation will be the required value of the uznknown quantity. 1. Take the numerical example 5x 4x 7 13x - - 13- -- 12 3 8 6 Clyearing of fractions 10x - 32 - 312 = 21 - 52x; CHAP. IV.] EQUATIONS OF THE FIRST DEG-REE 7 transposing and recducing 30x - 533: W.AVhence, by dividing botl- members of the equation by 30, x- 11.1. if we substitute this value of x, for x, in the given equaticn, it will verify it, that is, make the two members equal to eacrh other. Find the value of x in each of the following EXAMPLES. 1. 3x - 2 4-24 = 31. Ans. x= 3. 2. x Jr 18 = 3x -5. Ans. x= 1i. 3. 6- 2 + 10 =20- 3x- 2. Ans. xz 2 4. x+ - - - -- 11 Ans. x = 6 2 3 1 6 5. 2x — x - 1 = 5x -2. Ans. x 2 a 6 -3a 6. 3ax + - 3 = bx-a. Ans. x -= 3ax ~C2 6a — 2b' x-3 x x —19 7. + -- - 20- 2 Ans. x = 238. 2 3 2 *= x + 3 x x — 5 3cax: 2b.* cdf + 4cd 81 - - = f. -Ans. Z = 3-. 2 x-2 13 3 2 33a —2b' ~2' -. A~.x --- O ?90 ELEMENTS OF ALGEBRA. [CHAP. IV x3 x A X abcdf 3 a { - - f. A2+. Ans. x=. b c d bcd - acd- abd - abe' 3x - 5 4x - 2 t4. 1 - l1=x-[1. Al4ns. x —6. x 8x x- 3 15. 122. Ans. x = 14. 7 9 5 -- 4x - 2 z I 16. 2;x - 2 3_ Ans. x = 3. 5 bx - d 3a - d 7. 3 +: x + a. Ans.= x + — 18. ( + ) (x — ) _a - 4ab-b2 a2- bac-b a + b b a4 +- 3a3b + 4a2b2 - 6ab3 + 2b4 Anl. rx = 2b (22 + ab - b2) Pro3nblcMs f2yving rise to iquations of the First )egree, involving but une Unlknown Quantity. 81. The solution of a problem, by means of algebra, consists of' two distillct parts — Ist. The statement of the problem; and 2d. The solution of the equation. We have already explained the methods of solving the equation; and it only iteirains to point out the best manner of making the statement. The stateme-nt (of a problem is the operation of expressing, algebraically, the relations between the known and unknown quantities which enter it. This part cllanot, like the second, be subjected to any welldefined rule. Soinetimes the enunciation of the problem furnishes the equation inlllediately; and sometimes it is necessary to disoover, from the cnlnciation, new conditions from which an equta tion nmay be foin._ed, cI-AP. I V. EQUATIONS OF THE FIRST DEGREE. 81 The, conditions enunciated are called explicit conditions, and those which alec deduced from them, implicit conditions. Il al]most all cases, however, we are enabled to discover the eqoa t4io. byS applying the following RULFE. Doloth The munczIownt quantity by olze of the final letters of the atlphabclt, C'and then iicndicate, by imeans of algebraic signs, the samne operations on' tihe known and'unkLJown quantities, as would be'necesscry to verify the value of the unlnkowl quantity, were such PROBLEMS. 1. Find a number such, that the sum of one half, one third and one fourth of it, augmented by 45, shall be equal to 448 Let the required number be denoted by - -. - Then, one half of it will be denoted by one, third of it - - - - - - by - - x 3' one fourth of it - - - - - - by and by the conditions -- + - - 45 = 448. Trn.tsposiing + 3 +- - 448 -45 = 403;:learillg of fiactions, - - - ~- 4x+4 - 3 4836; reducillng, - 13x 4836; hence, - - - - - - - a 372. ttt 1 08? -.a =372. Let ts see if this value will verify the equation. We. have, 37i2 O72 372 + -- -. - 45 5= 86 + 124 +- 93 ~- 45 _448. 0 t 4 82 ELEMENTS OF ALGEBRA. [CHAP. IV 2. What number is that whose third part exceeds its fourthl by 16? Let thu required aumber be denoted by x. Then, x X will denote the third part; anid z1 will denote the fourth part, By the conditions of the problem, 1 1.wx — Tx- _16. Cleai ing of fractions, - 4x - 3x = 192; reducing, - - -- 192. FVec ification. 192 192 - 16, 3'4 or, - - - 16 = 16. 3. Out of a cask of wine which had leaked away a third pari, 21 gallons were afterward drawn, and the cask was then half full: how much did it hold? Suppose the cask to have held x gallons. Then, - - - - will cdenote what leaked away; and - - - - + 21 will denote what leaked out and,also what was drawn out. By the conditions of the problem, x 1 3 + 21 = V. Clearing of fractions, - 2x + 126 = 3x; reducing x = — 126; dividing by — 1 - x 126. Verification. 126 126 1 + 21 = 23; 0r)..... 63 = 63. CHAP. IV.1 EQUATIONS OF THE FIRST DEGREE. 83 4. A fish was caught whose tai weighed 91b.; his head weighed as much as his tail and half his body; his body weighed as much as his head and tail together: what was the weight of the fish's Let - - 2x denote the weight of the body; then - - 9 k x will denote weight of the head; and since the body weighed as much as both head and tail, 2x =9+ 9+x or, 2x -- 18; whence x -18. Verfication. 2 x 1S8- 18 = 18; or. IS --- 18. Hence, the body weighed - - - - - - 36lbs; the head weighed - - - - - - 2'7lbs; the tail weighed - - - - - - - - - 9bs; and the whole fish - - - 721bs. 5. A person engaged a workman for 48 days. For each day that he labored he received 24 cents, and for each day that he was idle, he paid 12 cents for his board. At the end of the 48 days the account was settled, when the laborer received 504 cents. Required the number of working days, and the number o' days he was idle. If these two numbers were known, by multiplying them re. spectively by 24 and 12, then subtracting the last product fromn the first, the result would be 504. Let us indicate these operations by means of algebraic signs. Let - - x denote the number of working days; then 48 - x will denote the number of idle days; 24 x x = the amount earned, and 12 (48 - x)- the amount paid for his board. Then, fiom the conditions, 24x —12(48 — x) = 504 or, 24x - 576 - 12x - 504. Reducing 36x = 504 + 576 = 1080 whence, x- 30 the working days, A An Co G __1o I 84 ELEMENTS OF ALGEBRA. [CHAP. IV, T'7erification. Thirty days' labor, at 24 cents a day amounts to'-.......... 30 X 24 = 720 cts; and 18 days' board, at 12 cents a day, Jnuounts to - - - - - - - - 18 X 12 = 216 cts; and the amount received, is their difference, 504 cts. The preceding is but a particular case of a general problem which may be enunciated as follows. & person engaged a workman for n days. For each day that he labored, he was to receive a cents, and for each day that he was idle, he was to pay b cents for his board. At the end of the time agreed upon, he received c cents. Requirecl the number of working days, and the number of idle days. Let - - x denote the number of working days; then, n- -x will denote the number of idle days; ax will denote the number of cents he received; and b (n - x) will denote the number he paid out. From the conditions of the problem, ax - b (' - x) = c. Performing the indicated operations, transposing and factoring, we find, (a +- )x = c - bon, whence, x = + the number of working days; and aq - C n - x -—, the number of idle days. a + If we make n = 48, a = 24, e = 12 and c = 504, we obtain, 504~+ 576 5 = -- 76_= 30; and 48 —x= 18; as before found, 6. A fox, pursued by a greyhound, has a start of 60 leaps. lie makes 9 leaps while the greyhound makes but 6; but 3 leaps of the greyhound are equivalent to 7 of the fox. How many leaps must the greyhound make -o overtake the fox? CHAP. IV.] EQUATIONS OF THE FIRST DEGREE. 85 Let us take one of the fox leaps as the unit of distance; then, 3 leaps of the greyhound being equal to 7 leaps of the fox, one of the greyhound leaps will be equal to Let x denote the number of leaps the greyhound must make before overtaking the fox. Tihen, since the fox makes 9 leaps while the hound makes 6. 9 3 -x or -x will denote the number of leaps the fox makes in the same time. z will denote the whole distance passed over by the hound.; 3 xz will denote the whole distance passed over by the fox. Then, fiom the conditions of the problem, ~07 ~ X - 60 4- x. 2 Clearing of fractions, 14x - 360 4~ 9x, transposing and reducing, 5x = 360, whence, x 72; and -x- x 72 = 10, the nubIhcr of fox leaps. Verifica tion. 7 x 72 3 x 72 = GO J-2-; 3 2' or, -68 - 168. 7. A can do a piece of work alone in 10 days, and B in 13 days: in what time can they do it if they work tog'otl,? Denote the number of days by x, and tl:e work tG tlre Lty 1. Then, in 1 day A can do t- of the work; and in 1 day B can do of the work' Ihienl, hl x days A can do _ of the vork; and iiv x days B can do - of the work: 13 86 ELEM:ENTS OF ALGiEBRA. (EHAP. IV. Hence, by the conditions of the question,. X a,10 13; clearing of fractions, 13x + 10. = 130 helnce, X 5 —, the numlber of days. 8. I)ivide $1000 between A, B and C, so that A shall ha e $72 more than B, andcl (C 100 more thlan A. A;s. A's sharlce - $32~, IB's =- 252, C's - $4124. 9. A and B play togethel at cards. A sits down with $,84 and B with $48. Each loses and wins in turn, when it ap. pears that A has five times as much as B. How much did A win? Ans. $26. 10. A person dying, leaves half of his property to his wife, one sixth to each of two daughters, one twelfth to a servant, and the remaining $600 to the poor: what was the anlount of his property? Anzs. $7200. 11. A father leaves his property, amounting to $2520, to four sons, A, B, C and D. C is to have $360, B as much as C and D together, and A twice as much as B less $1000: how nuch do A, B and D receive? Ans. A $760, B $880, D $520. 12. An estate of $7500 is to be divided between a widow, twoT soilns, and three daughters, sc that each son shall receive twice as much as each daughter, and the widow herself $500 more than all the children- what w-as her share, and what the share of 0.ach chilcd? WVidow's share, $4000. Arts. Each son, $1000. Each daughter, $500. 13. A company of SO) persons consists of men, women and childirel. The men are 8 n3more in nlrm:be.r than the womlen, and.he children 20 more than thle m'nen and swomen together: how i,,any of each sort in the companly? Ans. 44 men, 36 women, 100 children. CHAP. IV.] EQUATIONS OF THiE FIRST DEGREE. 87 14. A father divides $2000 among five sons, so that each elder should receive $40 more than his next younger brother: what is the share of the youngest? Anr. $320. 15. A purse of $2850 is to be divided among three persons, A, B and C; A's share is to be, of B's share, and C is to have $300 more than A.anrd B together: what is each one's share? Alis. A's $450, B's $825, C's $1575. 16. Two pedestrians start from the same point; the first steps twice as far as the second, but the second makes 5 steps while the first makes but one. At the end of a certain time they are 300 feet apart. Now, allowing each of the longer paces to be 3 feet, how far will each have traveled? Ans. 1st, 200 feet; 2d, 500. 17. Two carpenters, 24 journeymen, and 8 apprentices, received at the end of a certain time $144. The carpenters received $1 per day, each journeyman half a dollar, and each apprentice 25 cents: how many days were they employed? Ans. 9 days. 18. A capitalist receives a yearly income of $2940: four fifths of his money bears an interest of 4 per cent., and the remainder of five per cent.: how much has he at interest? Ans. $70000. 19. A cistern containing 60 gallons of water has three unequal cocks for discharging it; the largest will empty it in one hour, the second in two hours, and the third in three: in what time will the cistern be emptied if they all run together? Ans. 32 1s min. 20. In a certain orchard - are apple-trees, a peach-trees, - plum-trees, 120 cherry-trees, and 80 pear-trees: how many trees in the orchard? Ans. 2400. 21. A farmer being asked how many sheep he had, answered that he had them in five fields, inl the 1st he had ~, in the 2d., in the 3d - in the 4th -, and in the 5th 450: how many had he? Ans. 1200. 88 ELEMENTS OF AIGEBRA.'CHAP. IV, 22. My horse and saddle tcgether are worth 2,$132, and the horse is worth ten times as -r.uch as the saddle: what is the value of the horse? Ans. $120. 23. The rent of an estate is this year 8 per cent. greater than it was last. This year it is $1890: what was it last year? Anis. $1750. 24. What number is that from which, if 5 be subtracted, Z- of tlhe remainder w-ill be 40? Anls. 65. 25. A post is - in the rnud, in the water, and ten feet above the water: what is the whole length of the post? Ans. 24 feet.'26. After paying - and 5 of my money, I had 66 guineas left in may purse: how many guineas were in it at first? Ilns. 120.'27. A person was desirous of giving 3 pence apiece to some beggars, but found he had not money enough in his pocket by 8 pence; he therefore gave them each two pence and had 3 pence temaining: required the number of beggars. Anlas. 11. 28. A person in play lost 1 of his money, and then won 3 shillings; after which he lost - of what he then had; and this done, found that he had but 12 shillings remaining: what had lie at first? Ans. 20s. 29. Two persons, A and B, lay out equal sums of money in trade; A gains $126, and B loses 887, and A's money is now double B's: what did each lay out? Ans. $300. 30. A person goes to a tavern with a, certain sum of money inl his picket, where he spends 2 shillings; he then borrows as much money as he had left, and going to another tavern, hle there spends 0 shillings also; then borrowing again as luch money as was left, he went to a third tavern, where, likewise, he spent') shillinugs and borrowed as much as he ihad left; and again spending 2 shillings at a fourth tavern, he then had aothing remaining. What had he at first?.Ans. 3s, 9d. CHAP. 111.] EQUATIONS OF THE FIRST DEIGREE. 89 31. A farmer bought a basket of eggs, and offered them at 7 cents a dozen. But'before he sold any, 5 dozen were brohen by a careless boy, for which he was paid. He then sold the remainder at 8 cents a dozen, and received as much as he would have got for the whole at the first price. HIow many eggs had he in his basket? 1s.~. 4(-,.zl.,qucations of the First Degree inzvolulvlg more thin~ o'e Unknown Quan2tity. 82. If we hatve an equation between two unknown quantities, we may find an expression for one of them in terms of the other and known quantities; but the vcalue of this unknown quantity could only be determined by assuming a value for the secorld. Thus, from the equatioln, x + 2y = 4, we may deduce - = 4 -2y, but cannot find a value for x without assuming one for y. If, however, we have another equation between the two un known quantities, the values of these quantities being the same in both, we may find, as before, an expression for x in terms of y, and this expression placed equal to the one already found, will give an equation containing but one unknown quantity. Let us take x + 3y -= 5, from which we find x = 5-3y. If we place this expression equal to that before found, wev deduce the equation 4 - 2y - 5 - 3/, from the solution of which we find, y = 1. This value of y, substituted in either of the given equlntiols, gives x - 2: hence, x - 2 and y = 1 satisfy both equations. We see that in order to find determinate values for two< unknown quantities, we must have two independent equations.,iniulltneous equations are those in which the values of the nlknown quantities ar-e the samee in them all at t.e same timne ELEMENTS )F ALGEBRA [CHAP. lV, In the same manner it may be shown that tc determine the values of three unknown quantities, we must have three equations; and generally, to determine the values il' n unknown quantities we must have n equations. Elin inlction. 63. Elimination is the operation (f combining several equtations involving several tunknowtv quanitities, and deducing therefrwomz a less 9,umber of equations involving a less number of unmknowl quantities. There are three principal methods of elimination: 1st. By addition or subtraction. 2d. By substitution. 3d. By comparison. We shall explain these methods separately. Elimination by Addition or Subtraction. 84, Let us take the two equations 4x-5y = 5, 3x + 2y = 21. If we multiply both members of the first equation by 2, the co-efficient'of y in the second, and both members of the second equation by 5, the co-efficient of y in the first, we obtain, 8x-1Oy = 10, 15x +- 10y = 105; in which the co-efficients of y are numerically the same in both. If, now, we add these equations member to member, we find 23x - 115. Ih this case y has been eliminated by addition. Again, let us take the equations 2x + 3y = 12, 3x + 4y = 17. If we multiply both members of the first equation by 8, the co-efficient of x in the second, and multiply both mern. bers of the second eequation by 2, the co-efficient of x in the first, we shall have, 6x + 9y = 36, 6x + 8y = 34; CHAP. IV.] EQUATIONS OF THE FIRST DEGREE. 91 n which the co-efficients of x are the sanle in both. If, now, we subtract the second equation fiom the first, member firom melnmber, we find, y =-2. HIere, x has been elimirated by subtraction. In. a simnilar manner we may elilminate one unknown quantity tetween any two equations of the first degree containing any nulnber of unknown quantities. The rule for el;imination by addition and subtraction may be simplifiedl by using the least common multiple. Hence, for elimination by addition or subtraction, We have the following RULE. Prepare the two equations in such a manner that the coefficients of the q-uantity we wish to eliminate shall be numerically equal in both': then, if thie two co-efficients have contrary signs, add the equations) nmember to member; if they have the same sign, subtract them m2We7mber from meember, and the resulting equation will be indedepedent of that quantity. Elimination by Substitution. 85. Let us take the equations, x+ 7y = 43, and 11x + 9y - 69. Find, fiom the first equation, the value of z in terms of y, which is, 43 - 7y 5 Substitute this value for x in the second equation, and we shall have 11 x (43 - 77y) - 9y 69; or, reducing, - - W - 3-77y + 45y 345. h1 a similar manner we may eliminate one unknown quartity between two equations of the first degree containing any nulmbez of unknown quantities. HIence, for eliminating by substitutior, we have the followina 92 ELEMENTS OF ALGEBRA. rCHAP. IV. RULE. Find 3'rons one equation the value of t1le unkzoi.:n quantity to be eliminated in term)ns of the others: substiltute this value in the other equation for the unklzo'wn quantily to be eliminlated, a.nd ths resulting equation, will be indepenlzdnt of that qutantity. Einminatior by Comp3arZison2. 86. Let us take the equations, 5x + 7y =- 43, 11x + 9y = 69. Finding the value of x in terms of y, from both equations we have, 43 - 7y 5 69 — 9y f, now, we place these values equal to each other, we shall have, 43- 7y 69 - 9y 5 11:educing, - - - 473 - 77y 345 - 45y. Here, x has been eliminated. Generally, if we have two equations of the first degree containing any number of unknown quantities, any one of them may be eliminated by the following RULE. F17ind the value of the quantity we wish to elinlinate, in terms of the others, fio'072 each equation, and then Tlace these values equal to each other: the resulting eqtlation will be inzdependent of the quantity whose values were foulnd. The new equations which'- arise, fi'om.the two last methodeof elimination, contain fractional terms. This inconvenience is avoided in the first method.: The method by substitttion, is, however, advantageously employed whenever the co-effieient of either of the unknown quantities in one of the equatiocns is equai to ), because then the inconvanience of which we have just CHAP. IV.] EQUATIONS OP ~HE FIRST DEGREE. 93 spoken doe. not o-cur. We shall sometimes have occasion to employ this method, but generally the method by addition and s',btraction is preferable. WVhen the co-efficients are not too great, the addition or subtraction may be performed at the same time with the multiplication that is made to rendler the t3o-efficients of the same unknown quantity equal to each other. The're is also a method of elimination by means of the greatest common divisor, which will be explained in its appropriate place. 87. Let us now consider the case of three equations involving three unknown quantities. 5x - 6F + 4z — 15, Take the equations, 7 +- 4y -3z- 19, 2x + y+6. - 46. To eliminate z from the first two equations, multiply the first equation by 3 and the second by 4; and since the co-efficients of z have contrary signs, add the two results together: this gives a new equation, 43x-2y = 121.. Multiplying both members of the second equation by 2, a factor of the co-efficient of z in the third equation, and adding them, member to member, we have - - 16 x- 9y _84.J The question is then reduced to finding the values of x and y, which will satisfy these new equations. Now, if the first be multiplied by 9, the second by 2, and the results be added together, we find 419x = 1257, whence x 3. By means of the two equations involving x and y, we may determine y as we have determined x; but the value of y may oe determined more simply, since by substituting for x its value found above, the last of the two equations becomes, 48 + 9y = 84, whence y -4. In the same manner, by substituting the values of x and y, the first of the three proposed equations becomes, 15- 24 4- = 15, whence z = 6. 94 ELEMENT'S OF ALtaEBRA. [CHAP. IV, If we have a group of mn simultaneous equations c(ntaining m unknown quantities, it is evident, from principles already ex.. plained, that the values of these unknown quantities may be found by the following RULE. I. Coombine one of the m equations with each of the m - 1 otherse separa tely, eliminating the same unknown quantity; there will result m —1 equations containing m - 1- unknown quantities. II. Combine one of these with each of the gn —2 others, sepa. rately, eliminating a second unknown quantity; there will result m- 2 equations containing n - 2 unknown quantities. -II. Continue this operation of, combination and elimination tilt we obtain, Jfinally, one equation cor Aining one unknozn, quantity. IV. Find the value of this unkrnxon quantity by the rule fol solving equations of the first deg-rce cotaining one unknown quarntity: substitute this value in either tf the two preceding equations containiny two unknown quantities, and determine the value oj' a second unknown quantity: substitute thesc tao values in either of the three equations involving three undknown quantities, and so on till we find the values of them all. It often happens that some of the proposed equations do ne contain all the unknown quantities. In this cause, with a littlf address, the elimination is very quickly perforrmed. Take the four equations involving four unknown quantities, 2x - 3y + 2z =13 - (1) 4y -F2z = 14 - (), 4u -2x= 30 - (2) 5y+-3u-32 (4). jBy examining these equations, we see that the elimination of z in equations (1) and (3), will give an equation involving z tand y; and if we eliminate u in' the equations (2) and (4), we shtll obtain a second equation, involving x and y. In the first placce, the elimination of z, in (1) and (3) gives 7y -. — 1. (5), that of u, in (2) and (4), gives - - 20y -+ z & 3E - (6). From (5) and (6) we readily deduce the values of y = I and -- 3; and by substitution in (2) and (s), we.also tbhd u, 9 lrnd z = 5. .CiHAP. 1V.] EQUATIONS OF THE FIRST DEGREE. 95 EXAMPLES. 1. Given 2x + 3y=16, and 3x -2y= 11 to find the values of x and y. Ans. x=5, y=2. 2x 3y 9 3x 2y 61 2. Given -2 +- and -- 1 to find the 5 4 20' 4 5 120 1 1 values of x and y. Ans. x = —, y = —. x y 3. Given x +7y 99, anzd -- 7x = 51 to find the values of x and y. Ans. x = 7, y=14. x y x y x 2 —x 4. Given - 12= - 8, and y -S - - - -27 2 +4 5 3 4 to find the values of x and y. Ans. x = 60, y = 40. r x y-+ z=29 x - 2y - 3z-62 5. Given + 2 } to find x, y, and z. 1 x yz=O Ans, x=8, y-9, z =12. 2x-+ 4y- 3z =220 6. Given 4x — 2y + 5z = 8 to hind x, y, and z. 6x 7y — z =63 Ans. x=3, y=7, z=4. I + + 1 32 7 Given 3-x y + -z= 15 to find x, y, and z., l. +t- -—, z -6 - 12.lns. x = 12, y=0, =30. I7x- 2z + 3u = 17 4y- 2z + t= 11 8. Given 5y — 3x — 2u = 8 t to findl, y, y, u, 4y — 3u - 2t = 9 and t. 3z Su = 33 J Ans. x=2, y=4,. z=3, u=3, t=]. 90 ELEMENTS OF ALGEBRA. [CHAP. IV. PROBLEMS GIVING RISE TO SIMULTANEOUS EQUATIONS OF THE FIRFST DEGREE. 1. WA'hat fraction is that, to the numerator of which, if I be added, its value will be one third, but if 1 be added to its lenominator, its value will be one fourth? Let x denote the numerator, and y thle denominator. From the conditions of the problem, x+1 1 Y 3' x I y +-w - I';Clearing of fractions, the first equation gives, 3x + 3 - y, and the 2d, 4x =y+ 1. WVhence, by eliminating y, x —3 = 1, and x - 4. Substituting, we find,' y= 15; 4 and the required fraction is 5. 2. To find two numbers such that their sum shall le equal, o a and their difference equal to b. Let x denote the greater number, and y the lesser number. From the conditions of the problem, x - y = a, x - y = b. Eliminating y by addition, 2 = -a - b, a b o0, - - -.By substitution, a b Y- 2-2' oHAL'. IV.j EQUATIONS OF THE FIRST IDEGREE 97 3. A person possessed a capital of 30000 dollars, for which he drew a certain interest per annum; but he owed tile sum of 20G00 dollars, for which he paid a certain interest. The interest that he received exceeded that which he paid by 800 dollars. Another person possessed 35000 dollars, for whlich he received interest at the second of the above rates; but he owed 24000 dollars, for which he paid interest at the first of the above rates. The interest that he received exceded that which he paid by 310 dollars. Required the twro rates of interest. Let x denote the first rate, and y the second rate. Then, the interest on $30000 at x per cent. for one year will be $30000x 100 or $300x. The interest on $20000 at y per cent. for one year will be $20000y $2000y or $200y. 100 Hcnce, from the first condition of the problem, 300z - 5OOy = 800; or, - - - - 3x- 2y = 8 - - - (1). In like manner from the second condition of the problem we find. 35y - 24x= 31 - - - (2). Combining equations (1) and (2) we find, y- = 5 and x= 6. Hence, the first rate is 6 per cent. and the second rate 5;per cent. Verification.;30000, placed at 6 per cent, gives $300 x 6 — 1800. $20000 do 5 do $200 x 5 - $1000. And we have 1800 - 1000 = 800. The second condition can be verified in the same manner. 4. There are three ingots formed by mixing together three. metals i.n different proportions. 98 ELEIMENTS OF ALGEBRA. [CIIP. IV 0(ne pound (of the first contains 1 ounces of silver, 3 ounces of copper, anld 6 ounces of pewter. One pound of the second contains 12 ounces of silver; 3 ounces of copper, and 1 ounce of pewter. 3)le pound of the third contains 4 ounces of silver, 7 ounces ef copper, and 5 ounces of pewter. It is reqlired to form from these three, 1 pound of a fourthi inlgot which shall contain 8 ounces of silver, 33- olunces of copper, and 44 ounces of pewter. let x denote the number of ounces taken from the first. y denote the number of ounces taken from the second z denote the number of ounces taken from the third. Now, since 1 pound or 16 ounces of the first ingot contains 7 oulnces of silver, one ounce will contain of 7 ounces: thfir is, ounces; and?x z ounces will contain L6 ounces of silver, y ounces will contain 1 ounces of silver, 16 z ounces will contain - ounces of silver. -But since 1 pound of the new ingot is to contain 8 ounces of silver. we have 7x 12y 4z - 16 1-6 ior, ciearing of f:ractions, we have, for the silver, 7x + 192y - 4z =- 128; for the cc-pper, 3 + 3y -- 7z - 60; nld for the pewter, t6x q- y - 5z — 68. Whenee, finding the values of X, y and z, we have.x- 8, the number of ounces taken from the first. y 5 " " " " "' second. -- 3 " " " " " " third. 5. What two numblers are they, whose sumn is 33 and whose di.'e..ence is 7 As. () an3 d it;. CHAP. IV.] EQUATIONS OF THE FIRST DEGREE. 99 6'. Divide the number 75 into two such parts, that three times the greater may exceed seven times the less by 15. AAns. 54 and 21. 7. In a mixture of wine and cider, 1 of the whole plus 25 gallons was wine, and -1 part minus 5 gallons, was cider; how many gallons were there of each? Ans. 85 of wine, and 35 of cider. 8. A bill of ~120 was paid in guineas and moidores, and the number of pieces of both sorts that were used was just 100; if the guinea were estimated at 21s., and the moidore at 27s., how many were there of each? Ans. 50. 9. Two travelers set out at the same time frolm London and York, whose distance apart is 150 miles; they travel toward each other; one of them goes 8 miles a day, and the other 7; in what time will they meet? Ans. In 10 days. 10. At a certain election, 375 persons voted for two candi dates, and the candidate chosen had a majority of 91; how many voted for each? zAns. 233 for one, and 142 for the other. 11. A's age is double B's, and B's is triple C's, and the sum of all their ages is 140; what is the age of each? Ans. A's = 84, B's - 42, and C's = 14. 12. A person bought a chaise, horse, and harness, for ~60; the horse came to twice the price of the harness, and the chaise to twice the price of the horse and harness; what did he give for each? ~13 6s. Sd. for the horse. Ans. X ~ 6 13s. 4d. for the harness. 4o0 for the chaise. 13. A person has two horses, and a saddle worth ~50; now, if the saddle be put on the back of the first horse, it will make his value double that of the second; but if it be put cn the back of the second, it will make his value triple that of the first what is the value of each horse? ~Ans. One ~30. and the other Y~4O. 100 ELEMENTS OF ALGEBRA. LCHAP. IV 14. Two pers4 as, A and B, have each the same income. A saves 1 of his yearly; but B, by spending ~50 per annulm more than A, at the end of 4 years finds himself ~100 in debt; what is the income of each? Ans. ~125. 15. To divide the number 36 into three such parts, that I of 0te first, I of the second, and 4 of the third, may be all equal to each other. Ans. 8, 12, and 16. 10, A footman agreed to serve his master for 8S a year and a livery, but was turned away at the end of 7 months, and re.eived only ~2 13s. 4d. and his livery; what was its value? Ans. ~4 16s. 17. To divide the number 90 into four such parts. that if the first be increased by 2, the second diminished by 2, the third multiplied by 2, and the fourth divided by 2, the sum, difference, product, and quotient, so obtained, will be all equal to each other, Ans. The parts are 18, 22, 10, and,40. 18. The hour and minute hands of a clock are exactly together at 12 o'clock; when are they next together? Ans. 1 h. 551i mill. 19. A man and his wife usually drank out a cask of beer in 12 days; but when the man was from home, it lasted the woman 30 days; how many days would the man be in drinking it ealone? A ns. 20 days. 20. If A and B together can perform a piece o)f work in 8 days, A and C together in 9 days, and B and C in 10 days; how many days would it take each person to perform the same work alone Ans. A. 14-~ days, B 1723, and C 23 r. 21. A laborer can do a certain work expressed by a, in a time expressed by 6; a second laborer, the work c in a time dl; a [bird, the work e in a time f. Required the time it would take thLe three laborers. workirg together, to perform the work g. Ans. bdfg cafa+bcf + 6de CHAP. IV.] EQUATIONS OF THE IFIRST ] EGREE. 10) 22. If 32 pounds of sea water contair I pound of salt, how much fresh water must be added to these 32 pounds, in order that the quantity of salt contained in 32 pounds of the new mixture shall be reduced to 2 ounces, or -L of a pound? Ans. 224 lbs. 23. A number is expressed by ilree figures; tile sum of these figures is 11; the figure in the place of units is double that in the place of hundreds; and when 297 is added to this number, the sum obtained is expressed by the figures of this number reversed. What is the number? Ans. 326. 241. A person who possessed 100000 dollars, placed the greater part of it out at 5 per cent. interest, and the other part at 4 per cent. The interest which he receivel for the whole amounted to 4640 dollars. Required the two parts.,, IAns. $64000 and $36000. 25. A person possessed a certain capital, which he placed out at a certain interest. Another person possessed 10000 dollars more than the first, and putting out his capital 1 per cent. more advantageously, had an income greater by 800 dollars. A third, possessed 15000 dollars more than the first, and putting out his capital 2 per cent. more advantageously, had an income greater by 1500 dollars. Required the capitals, and the three rates of interest. Sums at interest, $30000, $40000, $45000. lRates of interest, 4 5 6 per cent. 26. A cistern may be filled by three pipes, A, B, C. By the two first it can be filled in 70 minutes; by the firsG and third it can be filled in S4 miinutes; and by the second and tlhird in 140 minutes. What time will each pipe take to do it in? What time will ble require 1, if the three pipes run together! A in 105 minutes. 4ns. B in 210 minutes. C'n 420 minutes. All will fill it in one hour. 102 ELEMIENTS OF ALGEBRA. [CHAP. IV 27. A, has 3 purses, each containing a certain sum of money If $20 be taken out of the first and put into the second, it will' contain four times as much as remains in the first. If $60 be taken from the second and put into the third, then this will contain V1- times as much as there remains in the second. Again, if -40 be taken from the third and put into the first, then the third will contain 2-~ times as much as the first. W~hat were the contents of each purse? 1st. $120. Ans.!2d. $380. (3d. $500. 28. A banker has two kinds of money; it takes a pieces of the first to make a crown, and 6 of the second to make the same sum. Some one offers him a crown for 6 pieces. How many of each kind must the banker give him Ans. 1st kind, - 2dkind, a- b a- b 29. Find what each of three persons, A, B, C, is worth, knowing, 1st, that what A is worth added to 1 times what B and C are worth, is equal to p; 2d, that what B is worth added to m times what A and C are worth, is equal to 9; 3d, that what C is worth added to n times what A and B alre worth, is equal to r. If we denote by s what A, B, and C, are worth, we intro. duce an auxiliary quantity, and resolve the question in a very sinmple Inl allelr. 30. Find the values of the estates of six persons, A, B, C, D, E, ]F, fio-m the following conditions: 1st. The sum of the estates of A and B is equal to a; that of C and D is equal to b; andI that of E and F is equal to c. 2d. The estate of A is worth mn times thalt of C; the estate of D is worth it times that of E, and the estate of F is worth p tines that of B. This problemn lnay be solved by means of a single equatiotn involving batt one nikunown quantity. CHAP. IV.Ii EQUATIONS CF THE FIRST' DEGRiEE. 103 Of izdeterrnzinate Equations and Indeterminate Problems. 88, An equation is said to be indeterminate when it may be satisfied for an infinite number of sets of values of the unknown quantities which enter it. Every single equation containing two unknown quantities is indet terminate. For example, let us take the equation 5x - 3y = 12, 12 + 3y whence, o- x2- = Now, by making successively, y - 1, 2, 3, 4, 5, 6, &c., 18 21 24 27:~5 -) 5, 5, anrd any two corresponding values of x, y, being substituted in the given equation, 5x - 3y -= 12, will satisfy it: hence, there are an ifiXnile number of values for x and y which will satisfy the equation, and consequently it is idetermwinate; that is, it admits of an infinite number of solutions. If an equation contains more than two unknown quantities, we 1lay find an expression for one of them in terms of the others. if, then, we assume values at pleasure for these others, we can find from this equation the corresponding values of the first; and the assumed and deduced values, taken together, will satisfy the given equation. Hence, Every equation involzinyg more taan onle itnknozwn quanititu i. indeterminate. In general, if we have n equations involving more than n unknown quantities, these equations are indeterminate; for we may, by combination and elimination, reduce them to a single equation ovntainin g niore than one unknown quantity, which we }have already seen is indeterminate. If, on the contrary, we have a greater number of equations than we have unknown quantities, they cannot all be satisfied 10(4 ELEMENTS OF ALGEBRA. [CHAP. IV, unless some of them are dependent upon the others. If we combine them, we may eliminate all the unlinown quantities, and the resulting equations, which will then contain only known quantities, will be so many equtations of cov.dition, whiclh must be -satisfied in order that the given equations n.ay admit of solution. For example, if we have x + y a, X - y el xy -d; we may combine the first two, and find, a C a C x-= - and y- and by substituting these in the third, we shall find t2 C2 = d, 4 If which expresses the relation between a, c and d, that must exist, in order that the three equations may be simultaneous. o8". A Problem is indeterminate when it admits of an infinite number of solutions. This will always be the case whbzn its enunciation involves more unknown quantities than the —e are given conditions; since, in that case, the statement of the p:-oblem will give rise to a less number of equations than there are unknown quantities. Ist. Let it be required to find two numbers such that 5 times the first diminished by 3 times the second s.v1ll be equal to 12. If we denote the numbers by x and y, the conditional cf the problem will give the equation 5x - 3y -- 12, which we have seen is indeterminate: —Eence, the:-r,7l!em admits of an infinite number of solutions, or is indeter.::mlate. 2. Find a quantity such that if it be multiplied by a and the product increased by b, the result will be equal to c tim.A the quantity increased by dl. CHAP. IV.] EQUATIONS OF THE F IRST DEGREE. 105 Let x denote the required quantity. Then from the conditien, ax 4- b - cx + d, d-b whence, - - x= —d a - c If now we make the suppositions that d = b and a- c, the 0 value of x becomes 0, which is a sym'bol of indeterimination. If we make these substitutions in the first equatio%, it be comes ax + b = ax + b, an identical equation (Art. 75), which must be satisfied for all values of x. These suppositions also render the conditions of the problem so dependent upon each other, that any quantity whatever will fulfil thlem all. Hence, the result O indicates that the problem admits of an infinite number of solutions. 3. Find two quantities such that a times the first increased by b times the second shall be equal to c, and that d times the first increased by f times the second shall be equal to g. If we denote the quantities by x and y, we shall have from the conditions of the problem, ax - bjy c, - - (1) d:x +f g+, - - - - (2) cd g hg -cf whence - bl - a' and xz - f If now we make cc -ag, (3) and af= bd, (4) we shall find by multiplying these equations together, mcembex by member, cf= bg. 0 These suppositions, reduce the values of both x and y to Fiom (3) we find, d =, and from (4) f — da ELEMENTS OF ALGEBRA. [CHAP. IV whichl u'v ostit ted ill equation (2), reduce it to ax -r by - c, an eqatblor- wahich is the same as the first. Undsr th]i.3 supposition, we have in reality but one equatlik between two unknollo-wn quantities, both of which ought to be indeterminate. This supposition also renders the conditions of the problem so dependent upon each other, as to produce a less number of independent equations than there are unknown quantities. 0 Generally, the result -O-, with the exception of the case mentioned in Art. 71, arises from some supposition made upon the quantities entering a problem, which i.makes one or more conditions so dependent rupon the others as to give rise to one or 0 more indeterminate equations. In these cases the result -- is a true answer to the probilm, and is to be interpreted as indicating that the problem admits of an infinite number of solutions. Inte2:pretattion of Ncgative Results. 89. From the nature of thl signs q- and -, it is clear that the operations which they indicace are diametrically opposite to each other, and it is reasonable to infer that if a positive result, that is, one affected by the sign +, is to be interpreted:n a certain sense, that a negative r esult., or one affected by,he sign -, should be interpr(eted in exactly the contrary Se1se. To show that this infelrnce is correcet, wev sh.all discuss one or two problems giving rise to both positive and negativo results. 1. To find a number, which added to the number b, will give a sum equal to the number a. Let x denote the required number. Then from the tonditions x J- b = a, v hence, x - a - CHAP. IV.] EQUATIONS OF THE FIRST DEGREE. 107 This formvula w'll give the algebraic value of x in all the particular cases of the problem. For example, let a 47 and 6= 29; then, z = 47 - 29 18. Again, let a 24 and 6=31; then, x -24 - 31 - -7. This last value of x, is called a negative soclution. How is it to be interpreted? If we consider it as a purely arithmetical result, that is, as arising from a series of operations in which all the quantities are regarded as positive, and in which the terms add and s.ubtract imply, respectively, augmentation and diminution, the prob1erm will obviously be impossible for the last values attributed to a and b; for, the number b is already greater than 24. Considered, however, algebraically, it is not so; for we have found the value of x to be - 7, and this number added, in the algebraic sense, to 31, gives 24 for the algebraic sum, and therefore satisfies both the equation and enunciation. 2. A father has lived a number of years expressed by a; his son a number of years expressed by b. Find in how many years the age of the son will be one fourth the age of the faLther. Let x denote the required number of years. Then, a + x will denote the age of the father i at the end of the and b + x will denote the age of the son required time. Hence, from the conditions, a -F+ x a - 4b - b = -6 x; whence, x -= 3 54 — 36 19 Suppose a=54, and 6b9; then x 3 The father being 54 years old, and the son 9, in 6 years the bftiher will be 60 years old, and his son 15; now 15 is the i u'rth of 60; hence, = 6 satisfies the enunciation. Let us now suppoc'i a- =45, and b = 15; 45 - 6i then, Z - - 5, 108 ELEMENTS OF ALGEBRA. CIIHAP. IV. If we substitute this value of x in the ecupation, a + x 4- + A, 4 we obtain, = 15 5, Or, 10 = 10 IHence, — 5 substituted for x, verifies the equation, and therofore is a true answer. Now, the positive result which was obtained, shows that the age of the father will be four times that of the son at the expiration of 6 years from the time when their ages were considered; while the negative result, indicates that the age of the father was four times that of his son, 5 years previous to the time when their ages were compared. The question, taken in its general, or algebraic sense, demands the time, when the age of the father is four times that of tile son. In stating it, we supposed that the time was yet to come; and so it was by the first supposition. But the conditions imposed by the second supposition, required that the time should have already passed, and the algebraic result conformed to this condition, by appearing with a negative sign. HI-ad we wished the result, under the second supposition, to have a positive sign, we might have altered the enunciation by demanding, how many years since the age of the father was four times that of the son. If x denote the number of years, we shall ha e from the conditions, a —x 4b6-a = b —x' hence, = 4 3 If a = 45 and 6 -15, x will be equal to 5. From a careful consideration of the preceding discussion, we may deduce the following principles with regard to negative resul ts. 1st. Every negative valute fouand for the unknown quantity frona an equatio;b of the first degree, will, when taken with ~;s proper sign, satisfy the equation from which it was derived. CHAP IV.] EQUATIONS OF THE FIRST DEGREE. 109 2d. This negative value, taken with its proper sign, will also alftisft' the conditions of the problem, understood in, its alg braic sense. t3E Itf a positive result is interpreted in a certain sense, a negatire result must be interpreted in a directly contrary sense. 4th. The negative result, with its sign changed, may be regarded as the anszwer to a problem? of which the enunciation only clderrs from the one proposed in this: that certain quantities zowhich were additive have become subtractive, and the reverse. 90. As a further illustration of the extent and power of the algebraic language, let us resume the general problem of the laborer, already considered. Under the supposition that the laborer receives a sum c, we have the equations x + y =n bn + c an - c whence, x - b- y +ax -by-c ac-b' a b 1t; at the end of the time, the laborer, instead of receiving a sum c, owed for his board a sum equal to c, then,, by would Ibe greater than ax, and under this supposition, we should have the equations, x + y = n, and ax - by = -c. Now, since the last two equations differ from the preceding two given equations only in the sign of c, if we change the sign of c, in the values of x and y, found firom these equations, the results will be the values of x and y, in the last equations: this gives bn -c an t c ca +- b' a + b The results, for both enuncliations, may be conlprehended in the, same formulas, by writing bn~ -c an T c a+ b' a +b The double sign ~i, is read plus or,zinzus, and =, is read, menus or plus. The upper signs correspond to the case in which the laborer received, and the lower signs, to the case in 110 ELEMENTS OF ALGEBRA. LCHAP. IV. which he owed a sudii c. These torrn-ulas also comprehend the case in which, in a settlement between the laborer and his employer, their accounts balance. This supposes c-O0, which gives bn amc a+b a+b' Discussion of Problems. 91. The discussion of a problem consists in mak ng every possible supposition upon the arbitrary quantities which enter the equation of the problem, and interpreting itne results. An arbitrary quantity, is one to which we may assign a value, at pleasure. In every general problem there is always one or more arbi trary quantities, and it is by assigning particular values to these that we get the particular cases of the general problern. The discussion of the following problem presents nearly all the circumstances which are met with in problems giving rise to equations of the first degree. PROBLEM OF TIIlE COURIERS. Two couriers are traveling along the same right line and in the same direction from I' toward I. The number of miles traveled by one of them per hour is expressed by m, and the number of miles traveled by the other per hour, is expressed by n. Now, at a given time, say 12 o'clock, the distance be tween them is equal to a number of miles expressed by a: re quired the time when they are together. IR' A B FR. At 12 o'clock, suppose the forward courier to be at B, the,]ther at A, and R or Pt' to be the point at which they are together. Let a denote the distance AB, between the couriers at t2 o'clock,- and sup-pose that distances measured to the right, from A, are regarded as positive quantities. CHAP. IV.] EQUAXIONS OF THE FIRST DEGREE. 111 Let t denote the number of hours frt lrl 1 o'clock to tlhe time when they are together. Let x denote the distance traveled by the forward courier in t hours; Then, a + x will denote the distance trave!id by the other im the same time. Now, since the rate per hour, multiplied by the number of hours, gives the distance passed over by each, we have, t X =a+ - - - - (1) t Xn - - - - (2). Subtracting the second equation from the first, mneml.ber fiom member, we have, t (m - n) = a; whence, - - - - t - We will now discuss the value of t; a, z and n, being arbitrary quantities. Thirst, let us suppose m > n. The denominator in the value of t, is then positive, and since a is a positive quantity, the value of t is also positive. This result is interpreted as indicating that the time when they are together is after 12 o'clock. The conditions of the problem confirm this interpretation. For if im > mm, the courier from A will travel faster than the courier from B, and will therefore be continually gaining on lim: the interval which separates them will diminish more and more, until it becomles. and then the couriers will be found upon the same point of tle line. In this case, the time I, which elapses, must be added to 12?,'clock, to obtain the time when they are together. Second, suppose In < n. The denominator, mn - mm will then be negative, and the value cf t will also be negative. This result is interpreted in a sense exactly contrary to the interpretation of the positive result; that is, it indicates that the time of their being together was previous to 12 o'clock. 112 ELEMENTS OF ALGEBRA. [CHAP. IV. This interpretation is also confirmed by considering the circumstances of the problem. For, under the second supposition, the courier which is in advance travels the fastest, and therefore will continue to separate. himself from the other ourier. At 12 o'clock the distance between them was equal br) a: after 12 o'clock it is greater than a; and as the rate of travel has not been changed, it follows that previous to 12 o'clock the distance must have been less than a. At a certain holur, therefore, before 12, the distance between them must have bPen equal to nothing, or the couriers'vere together at some point R'. The precise hour is found by subtracting the value of t from 12 o'clock. Third, suppose mn = n. The denominator m - n will then become 0, and the value of i will reduce to 0, or A. This result indicates that the length of time that must elapse before they are together is greater than any assignable time, or in other words, that they will never be together. This interpretation is also confirmed by the conditions of thcl problem. For, at 12 o'clock they are separated by a distance a, and if m = n they must travel at the same rate, and we see, at once, that whatever time we allow, they can never come together; hence, the time that must elapse is infinite. Fourth, suppose a = 0 and mn > n or ns < n. The numerator being 0, the value of the fraction is 0 oi t — 0. This result indicates that they are together at 12 o'clock, or that there is no time to be added to or subtracted f-omn 12 o'clock. The conditions of the problem confirm this interpretation. Because, if a =- 0 the couriers are together at 12 o'clock; and since they travel at different rates, they could never have been together, nor can they be together after 12 o'clock: hence, I can have no other value than 0. Olt &P. IV.] OF INEQU ALITIES. 115 F;ifith, suppose a = 0 and m - n. The value of t becomes 0, an indeterminate result. This indicates that t may have any value whatever, or in other words, that the couriers are together at any time either before or after 12 o'clock: read this too is evident frown the cir curnlstances of the problem. For, if at = 0 the couriers are together at 12 o'clock; and since they travel at the same rate, they will always be together; hence, t ought to be indeterminate. The distances traveled by the couriers in the time v are, respectively, ma na and both of which will be plus when m > n, both min~us when?m < n, and infJinite when m- = n. In the first case t is positive; ih the second, negative; and in the third, ininite. When the couriers are together before 12 o'clock, the distances are negative, as they should be, since we have agreed to call distances estimated to the right positive, and from the rule for interpreting negative results, distances to the left ought to be r-egarded as negative. Of Inequalities. 92. An inequality is the expression of two unequal quantities connected by the sign of inequality. Thus, a > b is an inequality, expressing that the quantity a is greater than. the quantity b. The part on the left of the sign of inequality is called the first e.ncnvb6er that on the right the second mnember. The operations which may be performed upon equations, may in general be performed upon inequalities; but there are, never. theless, some exceptions. In order to be clearly understood, we will give examples of the different transformations to which ir equalities may 5'e sub 8 114 ELEMENTS OF ALGEBRA. [CHAiP. IV. jected, taking care to point out the exceptions to which these transfclrmations are liable. Two inequalities are said to subsist in the same sense, when the greater quantity is in the first member in both, or in the seo)nd member in both; and in a contrary sense, when the greater quantity is in the first member of one and in the sCeconl( member of the other. Thus, 25 > 20 and 18 > 10, or 6 < 8 ard 7 < 9, are inequalities which subsist in the same sense; and the in equalities 15 > 13 and 12 < 14, subsist in a contrary sense. 1. If we add the same quantity to both members of an inequality, or subtract the samne qzuantity from both members, the resulting,inequality will subsist inz the samze sense. Thus, take 8 > 6; by adlii;lg 5, we still have 5 > 6 + 5;.and subtracting 5, we have 8 - 5 > 6 - 5. When the two members of an inequality are both negative,'that one is the least, algebraically considered, which contains the -greatest number of units. Thus, - 25 < - 20; and if 30 be added to both members,'we have 5 < 10. This must be understood entirely in an alge. ibraic sense, and arises from the convention before established, to,consider all quantities preceded by the minus sign, as subtractive. The principle first enunciated serves to transpose certain terms'from one member of the inequality to the other. Take. fbr ex ample, the inequality a2 1- b2 > 3b2 _ -a2;'there will result, by transposing, a2 + 2a2 > 3b2 -b2, or 3a2 > 2b2. 2. If two inequalities subsist in the same sense, and we add themn;inembenr to member, the resulting inequality will also subsist in the,asme sense, CHAP. IV.] OF INEQUALITIES. 115 Thus, if we add a > b and c > d, member to member, there results a + c > b + d. But this is not always the case, -when we subtract, member from member, two inequalities established in the same sense. Let there be two inequalities 4 < 7 and 2 < 3, we have 4-2 or 2< 7 -3 or 4. But if we have the inequalities 9 < 10 and 6 < 8, by suttiactnlg, we have 9-6 or 3 > 10-8 or 2. We should then avoid this transformation as much as possible, or if we employ it, determine in whieh sense )the restilting inequality subsists. 3. If the two members of an inequality be multiplied by a positive quantity, the resulting inequality will exist in the same sense. Thus, - - - a< b, will give 3a< 3b; and, - - - a< -b, — 3a < -3b. This principle serves to make the denominators disappear. a2 _ b2 C2 -_ d2 From the inequality > 3a 2d' we deduce, by multiplying by 6ad, a(a2 - 7,2) > 2d (C2 d), and the same principle is true for division. But, When the two members of an inequality are multiplied or divided by a negative quantit?/ the resulting inequality will sub8ist in a contrary sense. Take, for example, 8 > 7; multiplying by - 3, we have -24< -21. 8 8 7 In like manner, 8 > 7, gives _-, or — < -. Therefore, when the two members of an inequality are multi. plied or divided by a quantity, it is necessary to ascertain whlother the multiplier or divisor is negative; for, in that case, the inequality will exist in a contrary sense. 116 ELEMENTS OF ALGEBRA. LCfIAP. IV 4. It tis V,oE permitted to change the signs of the two members of an inequality, unless we establish the resulting inequality in a contrary sense; for, this transformation is evidently the same as nultiplying the two members by - 1. 5. Both, members of an inequality between positive quantities aan be squared, and the inequality will exist in the samze sense. Thus, from 5 > 3, we deduce, 25 > 9; from a +. b > c, we find (a - b)2 > C2. 6. When the signs of both members of the inequality are not known, we cannot tell before the operation is performed, in which sense the resulting inequality will exist. For example, -2< 3 gives ( —2)2 or 4 < 9. But, 3 > — 5 gives, on the contrary, (3)2 or 9 < (-5)3 or 25. We must, then, before squaring, ascertain the signs of the two members. Let us apply these principles to the solution of the following examples. By the solution of an inequality is meant the oper ation of finding an inequality, one member of which is the unknown quantity, and the other a known expression. [EXAMPLES. i. 5x-6 > 19. Ans. x > 5. 14 2 3. + 3~ - -30 > 1O0. Ans. x > 4. x 1 x 13 17 3. -- + + >. x > G. 6 3 2+ 2 2! ax a A. + Jr - ab > 5 Ans. x > a. bx b2'5. y -ax + ab <'. Ans. z < b. CHAPTER V, EXTRACT'ION OF IHE SQUARE ROOT OF NUMBERS.- ORMATION OF TrSISQUARE AND EXTRACTION OF THE SQUARE ROOT OF ALGEBRAIC QUANTITIES. —TRANSFORMATION OF RADICALS OF THE SECOND DEGREE. 93. ThE square or second power of a number, is the product which arises from multiplying that number by itself once: for example, 49 is the square of 7, and 144 is the square of 12. The square root of a number, is that number which multiplied by itself once will produce the given number. Thus, 7 is the square root of 49, and 12 the square root of 144: for, 7 >< 7 = 49, and 12 x 12 = 144. Tihe square of a number, either entire or fractional, is easily found, being always obtained by multiplying the number by itself once. The extraction of the square root is, however, attended with some difficulty, and requires particular explanation. The first ten numbers are, 1, 2, 3,. 4, 5, 6, x7 8, 9, 10, and their squares, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100. Conversely, the numbers in the first line, are the square roots of the corresponding numbers in the second line. WVe see that the square of any number, expressed by one figure, will contain no unit of a higher order than tens. The numbers in the second line are perfect sqsares, and, generally, any number which results from multiplying a whole number by itself once, is a perfect square. If we wish to find the square root of any number less than 100, we look in the second line, above given, and if the number is there written, the, corresponding number in he first line 118 ELEMENTS OF ALGEBRA. [CHAP. V is its square root. If the number falls between anly two num1 bers in the seoond line, its square root will fall between the corresponding numbers in the first line. Thus, 55 falls between 49 and 64; hence, its square root is greater than 7 and less than S. Also, 91 falls between 81 and 100; hence, its square root is greater than 9 and less than 10. If now, we change the units of the first line, 1, 2, 3, 4, &c., into units of the second order, or tens, by annexing 0 to each, we shall have, 10, 20, 30, 40, 50, 60, 70, 80, 90, 100, and their corresponding squares will be, 100, 400, 900, 1600, 2500, 3600, 4900, 6400, 8100, 10000: Hence, the square of any number of tens will contain no unit of a less denomination than hundreds. 94. We may regard every number as composed of the sun, of its tens and units. -Now, if we represent any numnber by NW, and deuote the tens by a, and the units by b, we shall have, -IT = a + b; whence, by squaring both members, 2 = C2 2ab +- b2: HIence, the square of a number is equial to the square of the tens, plus twice the product of the tens by the units, plus the square of the units. For example, 78 = 70 + 8, hence, (78)2 - (70)2 + 2 X 70 X 8 + (8)2 -- 4(00 + 1120 + 64 =: 6084. 95. Let us now find the square root of 6084. Since this number is expressed by more than two figures, its root will be expressed by more than one. 6 30 84 But since it is less than 10000, which is the square of 100, the root will contain but two places of figures; that is, muits and tens. Now, the square of the number of tens must be found in the number expressed by the two left-hand figures, which we will separate from the other two, by placing a point over the placa CHaAP. V.] SQUARE ROOT OF NUMBERS. 119 of units, and another over the place of hundreds. These parts, of two figures each, are called periods. The part 60 is conmprised between the two squares 49 and 64, of which the roots are 7 and 8: hence, 7 is the number of tens sought; and the required root is composed of 7 tens plus a certain number of units. The number 7 being found, we set it on the right of the given number, from which we separate 49 it by a vertical line: then we 2-148 1184 subtract its square 49 from 60, 1184 which leaves a remainder of 11, 0 to which we bring down the two I next figures 84. The result of this operation is 1184, and this number is made up of twice the product of the tens by the units plus the square of the units. But since tens multiplied by units cannot give a product of a lower order than tens, it follows that the last number 4 canl form no part of double the product of the tellns by the units: this double product, is, therefore found in the part 118. Now, if we double the number of tens, which gives 14, and then divide 118 by 14, th quotient 8 is the nu?)?ber of units of the root, or a greater number. This quotient can never be too small, since the part 118 will be at least equal to twice the ptodu.ct of the number of tens by the units: but it may be too large; for the 118, besides the double product of the number of tens by the units, may likewise contain tens arising fronm the square of the units. To ascertain if the quotient 8 expresses the number of units, we place the 8 to the right of the 14, which gives 148, and then we multiply.148 by 8: Thus, we evidently form, 1st, the square of the units; and 2d, the double product of the tens by the units. This multiplication being affected, gives for a product 1184, tha same number as the result of the first operation. Having 120 ELEMENTS OF ALGEBRA. FCHAP. V, subtracted:he product, we find the remainder equal to 0: hence 78, is the root required. Indeed, in the operations, we have merely subtracted from the given nuinber 6084, 1st, the square of 7 tens or of 70; 2d, twice the product of 70 by 8; and 3Xd, the square of 8: that is, the three tarts which eter inlto te composition of the square of 78. II1 the same manner we may extract th.e square root of any number expressed by four figures. 95. Let us now extract the square root of a number expressed by more than four figures. Let 56821444 be the number. 56 82 14 44 75338 If we consider the root as the 49 sum of a certain number of tens 14 5 78 2 and a certain number of units, the 72 5 given number will, as before, be 150 31 5i 4 equal to the scluare of the tens plus 450 9 twice the product of the tens by r o0 8 12054 4 the units plus the square of the units. 12054 4 If then, as before, we point off a period of two figures, at the right, the square of the tens of the required root will be found in the number 568214, at the left; and the square root of the greatest perfect square in this number will express the tens of the root. But since this number, 568214, contains more than two figures, its root will contain more than one, (or hundreds), and the sqiare of the hundreds will be found in the figures 5682, at the left of 14; hence, if we point off a second period 14, the square root of tho greatest perfect square in 5682 will be the hundreds of the required root. But since 5682 contains more than two figures, its root will contain more than one, (or thousands), and the square of the thousands will be found in 56, at the left of 82: hence, if we point off a third peiiod 82, the square root of the greatest yerfect square in 56 will be the thousands of the required root. Hence, we place a point over 56. and then uroceed thus: ('.AP. V.j SQUARE ROOT OF NUMBERS. 121 Placing 7 on the right of the given number, and subtracting its square, 49, from the left hand period, we find 7 ibr a remainder, to which we annex the next period, 82. Separating the last figure at the right from the others by a point, and dividing the number at the left by twice 7, or 14, we have 5 for a quotient figure, which we place at the right of the figure already found, aidl also annex it to 14. Multiplying 145 by 5, and subtracting the product from 782, we find the remainder 57. Hence, 75 is the number of tens of tens, or hundreds, of the required square root. To find the number of tens, bring down the next period and annex it to the second remiainder, giving 5714, and divide 571 by douible 75, or by 150. The quotient 3 annexed to 75 gives 753 for the number of tens in the root sought. We may, as before, find the number of units, which in this case will be 8. Therefore, the required square root is 7538. A similar course of reasoiling may be applied to a number expressed by any number of figures. Hence, for the extraction of the square root of numbers, we have the following RULE. 1. Separate the given number into periods of two figures each, beginning at the right hand: the period on thl left will often cowtain but one f gzre. II. Find the greatest perfect square in the first period on the left, and 2place its root on the right after the manner of a quotient in division. Subtract the square of this root fr'on the first period, and to the remainzder bring clown the second period for a dividend. II. Dozuble th/e root already found and place it on0 the left obr a divisor. See how many times the divisor is contained inl the dividend, exclusive of the right hand figure, and place the quotien. ina the root and also at the right of the divisor. IV. Multiply the divisor, thus augmented, by thLe last,figur of thte root found, and subtract the product from the dividend 122 ELEMENTS OF ALGEBRA. CHAP. V, and to the remnac:-,der bring,owrn the next period for a new dividenid. V. Double the whole root already found, for a new divisor, and continzue the he peration as before, until all the p2eriods are 5rovght dlown. REMIARK I.-If; after all the periods are brought down, there is no remainder, the proposed number is a perfect square. Buit,f there is a remainder, we have only found the root of the greatest perfect square contained in the given number, or the6 entire part of the root soztght. For example, if it were required to extract the square root of 168, we should find 12 for the entire part of the root and a remainder of 24, which shows that 168 is not a perfect square. But is the square of 12 the greatest perfect square contained in 1680 That is, is 12 the entire part of the root? To prove this, we will first show that, the di'erence between the squares of two consecutive nubmbers, is equal to twice the less number autgmented by 1. Let a represent the less number, and a -t- 1, the greater. Then, (a + 1)2= a2 + 2a + 1, and (a)2 = a2 their difference is 2a + 1 as enunciated: hence, The entire part of the root cannot be augmented by 1, unless the remzainder is eual to, or exceeds twice the root found, plus 1. But, 12 X 2 - 1 -25; and since the remainder 2o4 is less than 25, it follows filat 12 cannot he augmented by a number as great as unity: hence, it is the entire part of the root. The principle demonstrated above, may be readily applied in findilng the squares of consecutive numbers. If the numbers are large, it will be much easier to apply tho albove principle than to square the numbers separately. CIAP. V.] SQUARE ROOT OF N UMBERS. 123 For examn:lple, if we have (651)2 -- 423801, and wlsh to find the square of 652, we have, (651)2 423801 +2 x 651 13029 +1 -- 1 and (652)2 - 425104. Also, (652)2 = 4-25104 +2 X 652 ~ 1304 +1 -= I and (653)2 _ 426409. REMARK Il1.-The number of places of figures in the root will always be equal to the number of periods into which the given number is separated. EXAMPLES. 1. Find the square root of 7225. 2. Find the square root of 17689. 3. Find the square root of 994009. 4. Find the square root of 85678973. 5. Find the square root of 67812675. 6. Find the square root of 2792401. 7. Find the square root of 37496042. 8. Find the square root of 3661097049. 9. Find the square root of 918741672704. REMARK III. —The square root of an imperfect square, is in commensurable with 1, that is, its value cannot be expressed in exact parts of 1. a To prove this, we shall first show that if - is an irreduci. a2 ble fraction, tts square -~ must also lbe an irreducible fraction. A number is said to be prime when it cannot be exactly divided by any other liumber, except 1. Thus 3, 5 and 7 are prime number F. 124 ELEMENTS OF ALGEBRA. [CCHAP. V, It is a fundamental principle, that every number Inay be re. solved into prime factors, and that any number thus resolved, is equal to the continued product Df all its prime factors. It often happens that some of these factors are equal to each other. For example, the number 50 = 2 x 5 x 5; and, 180 = 2 X 2 x 3 3 x 5. Now, from the rules for multiplication, it is evident that the square of any number is equal to the continued product of all the prime factors of that number, each taken twice. IIence, we see that, the square of a number cannot contain any prime facto? which is not contained in the number itse~; But, since ao, is, by hypothesis, an irreducible fraction, a and b can have no common factor: hence, it follows, fronm what has just been shown, that a2 and b2 cannollllt have a coma2 mon factor, that is, a- is are irreducible fraction, which was to be proved. a3 a4 an For like reasons, a-,, - - - are also irreducible fractions. Now, let c represent any whole number which is an imper fect square. If the'square root of c can be expressed by a firaction, we shall have Jc= bv in which - is an irreducible fraction. Squaring both members, gives, a2 or a whole number equal to arn irreducible fraction, whichl is absurd; hence, F -cannot be expressed by a fraction. We conclude, therefore, that the square root of,, ai imperfect square cannot be expressed in exact parts Jf 1. It may be Tqlown, in a similar manner, that any root -f an impenefect power of the degree indicated, cannot be expressed in exact parts of l. CHAP. V.] SQUARE ROOT OF FRACTIONS. 125 Extractacf of the Square Root of Fractions. 96. Since the second power of a fraction is obtained by squaring the numerator and denominator separately, it follows that the square root of a fraction will be equal to the square root of the numerator divided by the square root of the denominator. For example, b2 b' a a a2 since X - -. -- - But if the numerator and the denominator are not both perfect squares, the root of the fraction cannot be exactly found. We can, however, easily find the root to within less than the firactional unit. Thus, if we were required to extract the square root of the friction b to within less than -, multiply both terms of the fractions by b, and we have c6b Let r2 represent the greatest perfect square in ab, then will ab ab be contained between r2 and (r + 1)2, and!2 will be contained between r and (r + 1)2 and 62 ab a and the true square root of b2 - b-, will be contained be. tween r r 1 and - r r+1 1 but the difference between - and b s —; hence, either a i. will be the square root of, to within less than -. We ave then the following 126 ELEMENTS OF ALGEBRA. ICHAP. V. RULE. Mfultiply the numerator by the denominator, ordc extract the square root of the product to uwithin less than 1; divide the result by the denominator, and the quotient will be the apprcximate root. e3 For example, to extract the square root oft, we multiply 3 by 5, which gives 15; the perfect square nearest 15, is 16, 4 3 and its square root is 4; hence, 4 is the square root of 3 5 5 to within less than -. 97, If we wish to determine the square root of a whole number which is. an imperfect square, to within less than a 1 given fractional unit, as —, for example, we have only to place the number under a fractional form, having the given fractional unit (Art. 63), and then we may apply the preceding rule: or what is an equivalent operation, we may fiultiply the given number by the square of the denominator qft the fraction which determines the degree of approximation; then extract the square root of the product to the nearest unit, and divide this root by the denominator of the fryaction. EXAMPLES. 1. Let it be required to extract the square root of 59, to within less than First, (12)2 = 144; and 144 X 59 = 8496. Now, the square root of 8496 to the nearest unit, is 92: hcn.e 92 - 1 -= 7-8 — which is true to within less than 2. Find the 11 to within less than 4s' A 1s. A. 1. ind he to hn ns. 1 S. Find the ~223 to within`e.-s than:-6. Ans. 147-. CHAP. V.] SQUARE ROOT OF FRACTIONS. 127 97*o The n3anner of determining the approximate, root in de9% mals, is a consequence of the preceding rule. To obtain the square root of an entire number within 1 1 100' 1000' &C., it is only necessary, according to the preceding rule, to multiply the proposed number by (10)2, (100)2, (1000)2; or, which is the same thing, Annex to the number, two, four, six, &cc., ciphers: then extract the root of tze product to the nearest unit, and divide this root by 10, 100, 1000, &c., which is effected by pointing off oze, two, three, &c., decimal places from the right hand. EXAMPLES. 1. To find the square root of 7 to within less than 100' Having multiplied by (100)2, that is, having annexed four ciphers to the right 7 hand of 7, it becomes 70000, whose mot extracted to the nearest unit, is 264, 46 300 which being divided by 100 gives 2.64 1_276 for the answer, and this is true to within 524 2400 iL~~~ I ~~~2096'ess than'O0 304 Rem. 100 1 2. Find the E29 o within less than ~ Ans. 5.38. 1 3. Find the 027 to within less than Ans. 15.0665.REMARK.-Th e nwumler of ci phers to be annexed to the whole number, is always dcuble the number of decimal places required to be found in the root. 98, The manner of extracting the square root of a number containing an entire part and decimals, is deduced immediately from the preceding article. Let us take for example the number 3.425. This Ls equivas lent to 1425 Now, 1000 is not a perfect square, but the de. 100e o 128 ELEMAENTS OF ALGEBRA. [CHAP. V. nominator may be made such without altering the value of the fraction, by multiplying both terms by 10; this gives 34250 34250 or 1'0000 (100)2' Then, extracting the square root of 34250 to the nearest unit, 185 we find 185; hence, ~ or 1.85 is the required root to withinI less than 100' If greater exactness be required, it will be necessary to annex to the numlber 3.425 as many ciphers as shall make the number of periods of decimals equal to the number of decimal places to be found in the root. Hence, to extract the square root of a mixed decimal: Annex ciphers to the proposed number until the whole number of decimal places shall be egual to double the number required in the root. Then, exzract the root to the nearest unit, and point off, faomn the right hand, the required number of decimal places. EXAMPLES. 1. Find the 3/271.4707 to within less than.01. Ans. 57.19. 2. Find the 31.027 to within less than.01. Ans. 5.57. 3. Find the 0.01001 to within less than'.00001. Ans. 0.10004. 99. Finally, if it be required to find the square root of a vulgar fiaction in terms of decimals: Change the vulgar firaction into a decimal and continue the division until the number of decimnal places is double the'number required in, the root. Then, extract the root of the decimal by the last rule. EXAMArE., 11 1. Extract the square root of to within less than.001 This number, reduced to decimals, is 0.785714 to within less than 0.000001. The root of 0.785714, to the nearest unit, is CHAP. V.] SQUARE ROOT OF ALGEBRAIC QUANTITIES. 129 886: hence, 0.886 is the root of to within less than 001. 14 2. Find the 23 to within less than 0.0001. Ans. 1.6931. Extraction of the Square Root of Algebraic Quantitiea. 100, Let us first consider the case of a monomial. In order to discover the process for extracting the square root, let us see how the square of a monomial is formed. By the rule for the multiplication of monomials (Art. 42), we have (5a2b3c)2 - 5a2b3c X 5a2b3c = 25a4b6C2; that is, in order to square a monomial, it is necessary to square its co-effcient, and double the exponent of each letter. H-tence, to find the square root of a monomial, Exitract the square root of the co-eglcient for a new co-eficient, ~and write after this, each letter, with an exponent equal to its;ri.yinal exlponenit divided by two. Thus, I64ab4 -= 8a3b2; for, 8a3b2 X 8a3b2 = 64a6b4, atnd, I b - 25ab4c3; for, (25ab4c3)2 = 625a2b8c. From the preceding rule, it follows, that, when a monomial,s a perfect square, its numerical co-efficient is a perfect square,lnd every ex.2onent an even number. Thus, 25a4b2 is a perfect square, but 98ab4 is not a perfect:'LqLuare; for, 98 is not a perfect square, and a is affected with; [11 l uneven exponent. Of Polynomials. 101, Let us next consider the case of polynomials.. Let N5 denote any polynomial whatever, arranged with refer euce to a certain letter. Now the square of a polynomial is the product arising from multiplying the polynomial ly itself once: hence, the first term of the product, arranged with refer. ence to a particular letter, is the square of the first term of the polynomial, arranged with reference to the. same letter>,. 9 130 ELEMENTS OF ALGEBRA. r[CHAP. V. Therefore, the square root of the first term of such a product will be the first term of the required root. Denote this term by r, and the following terms of the root, arranged with reference to the leading letter of the polynomial, by r', r", r"', &c., and we shall have N= (r + r' + r" + r"' + &c.;)2 or, if we designate the sum of all the terms of the root, after the first, by s, N= (r + s)2 = 2.2 + 2rs + s2 = r.2 2r (r' + r" + r"' -+ &c.) + s2. If now we subtract r2 from N, and designate the remaindex by R, we shall have, N-r2 = R =2 (rT + r" + r"' + &c.) + s2, which remainder will evidently be arranged with reference to the leading letter of the given polynomial. If the indicated ~operations be performed, the first term 2rr' will contain a'higher power of the leading letter than either of the following terms, and cannot be reduced with any of them. Hence, If the first term of the first remainder be divided by twice the first term of the root, the quotient will be the second term of the root. If now, we place r + r' -- n,:and designate the sum of the remaining terms of the root, r'" r"', &c., by s', we shall have N = (It + SI)2 = 212 + 2;Zs' + S'2 If now we sul)tract a62 from A, and denote the remainder by 2',' we shall have, 1N- n2 _= _' 2ns' -+- s'2 - (r' - r) (r" + r"'/ J &c.) + s,"; in which, if we perform the mlultiplications indicated in the second member, the term 2'?r' will contain a higher power of the leading letter than either of the following terms, and can. lnot, consequently, be reduced with any of them. Hence, If the first term (f the second remainder be divided by tweice'the first term of the root, the quotient will be tie third term'of the root. CHAP. V.] SQUARE ROOT OF ALGEBRAIC QUANTITIES. 11. If' ve make r + r' + r" n', and r"' + r + &c. =s we shall have 2 aN (n' ~-s")2 = n'2 + 2n's" + s"12; and V 2 = " = 2 (r + r' + r") (r"' + r'v + &C.) -+ 5"2; in which, if we perform the operations indicated, the term 2rr"' will contain a higher power of the leading letter than any following term. Hence, If we divide the first term of the third remainder by twice the first term of the root, the quotient will be the fourth term of the root. If we continue the operation, we shall see, generally, that The first term. of any remainder, divided by twice the first term of the root, will give a new term of the required root. It should be observed, that instead of subtracting n2 from the given polynomial, in order to find the second remainder, that that remainder could be found by subtracting (2r + r')r' from the first remainder. So, the third remainder may be found by subtracting (2n1 -- r")r" from the second, and similarly for the remainders which follow. H!ence, for the extraction of the square root'of a polynomial, we have the following RULE. I. Arrange the polynomial with reference to one of its letters, znd then extract the square root of the first term, which will give the,first term of the root. Subtract the square of this termfrom the given polynomial. it. Divide the first term of the remainder by twice the first term of the root, and the quotient will be the second term of the root. III, From the firs, remainder subtract the product of twice the first term. of the root plus the second term, by the second term. IV. Divide the first term of the second remainder by twice the first term of the root, and the quotient will be the third term of the root. 132 ELEM3ENTS OF ALGEBRA. [CHAP. V. V. From the second remainder subtract the product cf twice the sum -of the first and second terms of the root, plus the thicWrd term, by the third term, and the result will be the third remanbder from which the fourth term of the root may be found as before. VI. Continue the operation till a remainder is found equal to O, or till the first term of some remainder is not divisible by twice the first term of the root. In the former case the root found is etact, and the polynomial is a perfect square; in the latter case, it is an imperfect square. EXAMPLES. i. Extract the square root of the polynomial 49a2b2 - 24ab3 + 25 a4 - 30a3b + 16b4. First arrange it with reference to the letter a. 625a4 - 30a3b + 49a2b2 - 24a43 -+ 1Gb4 5a2 - 3ab + 462 25a,4'10a2 - 3ab 1R =- -30a3b + 49a2b2 - 24ab3 + 16b4 - 3ab -30a3b +- 9a2b2 30a3b - 9a'62' = + 40a2b2 - 24ab3 +- 1664 10a2 - 6ab + 462 -+ 40a2b2 - 24ab3 + 16b4 4b2 = -......_.o_ 40a2b2 - 24ab3 + 16b4. 2. Find the square root of a4 + 4a3x + Ga'x2 + 4ax3 + x,. 3. Find the square root of a4 - 2a3x -- 3a2x2 - 2ax3 + x. 4. Find the square root of 4x6 + 12x'5 4 54 - 2x3 — 2-2 - 2, + I 5. Find the square root of 9~4 - 12a3b + 28a2b2 _ 16ab3 + 16b4. 6. Find the square root of 45a46 -- 40a3b2c + 76a262c2 - 48ab2c3 + 36b2c4._ 30a4bc + 24a.2boa - 36a2bc3 + 9a4c. CHAP. V.] RADICALS OF'THE SECOND DEGREE. 133 Remarks on the Extraction of the,Square Root of Polynomials. lst. A binomial can never be a perfect square. For, its root cannot be a monomial, since the square of a monomial will b)e a monomial; nor can its root be a polynomial, since the square of the simplest polynomial, viz., a binomial, will con tain at least three terms. Thus, an expression of the form a2 ~ b2 can never be a perfect square. 2d. A trinomial, however, may be a perfect square. If so when arranged, its two extreme terms must be squares, and the middle term double the product of the square roots of the other two. Therefore, to obtain the square root of a trinomial, when it is a perfect square, Extract the square roots of the tw'o extreme terms, and give these roots the same or contrary signs, according as the middle term is vositZve or negative. To verify it, see if the double product of the two roots is equal to the middle term of the trinomial. Thus, 9a6 - 48a4b2 -- 64a2b4 is a perfect square, for, 9a6 3a3; and, /64a2b4= — 8a2; also, 2 X 3as3 (- 8ab2)' — - 48ab2, the middle tenn. But 4a2 + 14ab + 9b2 is not a perfect square: for, although 4a2 and + 9b2 are per. feet squares, having for roots 2a and 3b, yet 2 x 2a X 3b is not equal to 14ab. Of Radical Quantities of the Second Degree. 102, A radical quantity is the indicated, root of an imperfeet power of the degree indicated. Radical quantities are sometimes called irrational quantities, sometimes sutrds, but more commonly, simply radicals. The indicated root of a perfect power of the degree indi eated, is a rational quantity expressed under a radical.form. 134 ELEMENTS OF ALGEBRA. LCHAP. V, An indil ated square root of an imperfect square, is called a radical of the second degree. An indicated cube root of an imperfect cube, is called a radi. m1 of the third dclqree. Generally, an indicated ntl' root of an imperfect nth power,. is called a radical of the nlt' degree. Thus, 2, e/ and //G, are radicals of the second degree; 3/4, 13 8 and 3 11, are radicals of the third degree; and /-4, I5 and a/ 1, are radicals of the nth degree. The degree of a radical is denoted by the index of the root. The index of the root is also called the index of the radical. 103. Since like signs in both factors give a plus sign in th.e product, the square of - a, as well as that of + a, will be a2: hence, the square root of a2 is either + a or -a. Also, the square root of 25a2b4 is either + 5ab2 or - 5ab2. Whence we may conclude, that if a monomial is positive, its square root mnay be affected either with the sign + or -; thus, 9aH4 -=:= 3a2, for, + 3-a2 or -3a2, squared, gives 9a4. The double sign ~ with which the root is affected, is read plus or minus. If the proposed monomial were negative, it would have ne square root, since it has just been shown that the square of every quantity, whether positive or negative, is essentially positive. Therefore, such expressions as, / — 9, /-4a2, -- 8a2b, are algebraic symbols which indicate operations that cannot be perftbrmed. They are called imaginary quantities, or rather, ianaginary expressions, and are frequently met with in the solution of equations of the second degree. Generally, Efvery indicated even root of a negative quantity is an zmmagiary cexpression. An odd root of a negative quantity may often be extrscte& Fo- example, -— 27 -= — 3, since ( —3) -— 27. CHAP. V.] hADPCALS.F THE SECOND DEGREE. 135 Radicals are similar when they are of the same degree and the quantity under the radical sign is the same in both. Thus, a a/ I and c V, are similar radicals of the second degree. Of the Sim;pification of Radicals of the Second Degree. 1040 Radicals of the second degree may often be simplified, and otherwise transformed, by the aid of the following principles. Ist. Let the a, and 6, denote any two radicals of the second degree, and denote their product by p; whence, x Jb=p - - - - (1). Squaring both members of equation (1), (axiom 5), we have, (,/)2 X ( )2 2 or, - ab =p2 _ - - - (2). Extracting the square root of both members of equation (2), (axiom 6), we have, but things which are equal to the same thing are equal to each other, whence, /fa-x /b —= ab; hence, T2he product of the square roots of twzo quantities is equtal to the square root of the product of those quantities. 2d. Denote the quotient of a by 6, by q; whence, - =q - - -- (1). Squaring both members of equation (1), we find, a (S/6)2 or, >= - - - - (2). Extrapting the. square root of both members of elquation (2), we have, a-'V -= 1]6 ELEMENTS OF ALGEBRA. [CHAP. V. Things which are equal to the same thing a e equal to each other, whence, V-=v a hence, The quotient of the square eroots of two quantities is equal to the square root of the quotient of the sanme quantities. 105. The square root of 98ab4 may be placed under the form 98_ab4 - V9(4 X 2a, which, from the 1st principle above, may be written, 49fr X 2a 7b622. In like nmanner, / 45a2b3c2C = aI 22 x 5bd 3abc/ 56. / 464 %'bi = —/1]= 144a%'c0c X (6bc- 12 ab cS/ 6bc. The quantity which stands without the radical sign is called the co-efficient of the radical. Thus, 7b2, 3a6c, and 12ab2c5, are co-efficients of the radicals. In general, to simplify a radical of the second degree: I. Resolve the quantity under the radical sign into two factors, one' of which shall be the greatest perfect square uwhich enters it as a factor. IT. Write the square root of thle perfect square before the radiica sign, under wahich place the other factlor. EXAMPLES. 1. Reduce 175a3bc to its simplest form. 2. I:educe -/125a6d2 to its simplest form. 8. Reduce V/ 2a9c to its simplest form. 4. Redlu.ce V/256ai4cs to its simplest form 5. Reduce 41024a967c5 to its simplest form, i6. Reduce /728a7b%6d to its simplest form, If the quantity under the radical sign is a polynomial, we may often simnplify the expression by the same rllle, CHAP. V.] RADICALS OF THE SECOND DEGREE. 137 Take, for exarmple, the expression 4 a3b + 4a2b2 - 4ab3. The quantity under the radical sign is not a perfect square: but it can be put under the form ab, (a2 ~- 4ab + 4b2). Now, the factor within the parenthesis is e ridently the square of a + 2b, whence we have /a3b +4a2b2 4ab3 (a + 2b) ab. 105*' Conversely, we may introduce a factor under the radical sign. Thus, a a2 b, which by article 104, is equal to a2b Hence, The co-efficient of a radical may be passed under the radical sigu?, as a factor, by squaring it. The principal use of this transformation, is to find an approximate value of any radical, which shall differ from its true value, by less than 1. For example, take the expression 64/13. Now, as 13 is not a perfect square, we can only find an approximate value for its square root; and when this approximate value is multiplied by 6, the product will differ materially from the true value of 6A/13. But if we write, GJ x13 - 6X =- V3x 3 I /46- 8 we find that the square root of 468 is the whole number 21, to within less than 1. Hence, 6/ 13 - 21, to within less than 1. In a similar manner we may find, 12/-7 =-31, to within less than 1. Addition and Subtraction. 106. In order to add or substract similar radicals: Add or subtract their co-efficients, and: the sum or difercsfwe an nex the common radical. 138 ELEMENTS OF ALGEBRA. LCHAP. V'hus, 3a - d- 5c/6 = (3at + 5c); and 3a, V ScV I = (3a - 5c). In like manner, 7 -2a + 3 + / = (7 + 3)v/- = ]o a; and 7 A2a-33 2a=(7-3)> 2a= 4 2a. Two radicals, which do not appear to be similar, may becomni so by simplification (Art. 104). For example, 48a2 + b~/75a =:4b 3- 5b 3a= 9b6/ 3-a; Also, 24 -3 = 6 - 3,- = 3. When the radicals are not similar, their addition or subtrac tion can only be indicated. Thus, to add 3, to 5/a, we write, ul/caton of Ra al Qantts of te Second Degree. ai~u7tipzlicat-o n of _Radical Ouantities of the econd.Degree. 107, Let aVb and c -d denote any two radicals of the second degree; their product will be denoted thus, aVb X cV(i; vwhich, since the order of the factors may be changed without altering the value of the product, may be written, aX< c x bx d The product oY the last factors from the 1st principle of Art. 104, is equal Jo fbfd; we have, therefore, a bxc d=acd Hence, t multiply one radical of the second degree Ly at.,tther, we have the following RULE. fAultiZly the co-e~tcients together for a zew co-eicient; after this write the radical sign, and under it the 1product of the zuanrtities under both radical signs. CHAP V.] RAElICALS OF THE SECOND DEGREE. 139 EXAMPLES. 1. 3 5ab >x 4 120a 2 1 O00a = 120ab. 2. 2abc X 3a -bc 6a2.a2c2 = a26c. 3. 2a x - 3a a2+ J b2 = - 6a(2 a(a. + 62) Div'ision of Radical Quantities of the Second Degree. 108. Let ao6- and c-d represent any two radicals of the second degree, and let it be required to find the quotient of the first by the second. This'quotient may be indicated thus, which is equal to - X but from the 2d principle of Art. 104, 6 a6/ a d V Cl hence c =-/ Hence, to divide one radical of the second degree by another, we have the following RULE. Divide the co-efficient of the dividenwd by the co-eficient of the divisor for a new co-eficient; after this, write the radical sign, placing under it the quotient obtained by dividing the quantity under, the radical sign in the dividend by that in the divisor. 5a For example, 5a/b. 2bV /c =; And, 12ac/6 bcq- 4c-2 -= 3a = 3a 3c. 109, The following transformation is of frequent application in finding an approximate value for a radical expression of a par. ticular forin. Having giren an expression of the form, a a or 140 ELEMENTS OF ALGEBRA. [CHAP. V in which e, and p are -any numbers whatever. and q not a per fect square, it is the object of the transformation to render the denominator a rational quantity. This object is attained by multiplying both terms of the fraeo tibon by p -- q, when the denominator is p - q_,, and W.y p + q, when the denominator is p —i/; and recollecting that the sum of two quantities, multiplied by their difTerence, is equal to the difference of their squares: hence, a ca(p-1 f) _,(p-V) a-a -ja p + gq (p+ yq)(pr - ) p2 q p2 -q a a(p~ + -i/) a(p + J) a_ a aJ p - q- (P p —) (p q2) P2 _ p2 - q in which the denominators are rational. As an example to illustrate the utility of this method of ap proximation, let it be required to find the approximate value of the expression - _. We write 3 —./5 7 7(3 + 5) 21 +~75 3 —5 9-5 4 But, 7 5= /49 x 5 245 = 15 to within less than 1. Therefore, 7 21 -F 15 to within less than 1 = 9 to within 3- -5 4 4 less than 4-; hence, 9 differs from the true value by less than onte fourth. If we wish a more exact value for this expression, etract the square root of 245 to a certain number of decimal places, add 2. to this root, ana divide the result,by 4. Take the expression, -7 1, and find its value to within less than 0.01. qHAP. V.] EXAMPLES IN THE CALCULUS OF RADICALS. 141 We have, 7JV 7AV _(v -11 V 7v- 7V 15 i~t+.j 11-3 8 Now, 7 55 =a /55 x 49 =2695 = 51.91, within less than 0.0, and 715=/5 X 49 =735 =-27.11; - therefore, 7 ~/i5 51.91 - 27.11 24.80 = - = - 3.10. V1 +~;3 8 8 HIence, we have 3.10 for the required result. This is true to 1 within less than By a similar process, it may be found, that, 3 + 2 2V — 2.123, is exact to within less than 9.001, 5V12-64J REMARPK.-The value of expressions similar to those above, may be calculated by approximating to the value of each of the radicals which enter the numerator and denominator. But as the value of the denominator would not be exact, we could not determine the degree of approximation which would be obtained, whereas by the method just indicated, the denominlator becomes rational, and we always know to what degree of accuracy the approximation is made. PROMISCUOUS EXAMPLES. 1. Sinlplify 125. Ans. 5/5. Z. Reduce V 5 7 to its simplest form. We observe that 25 will divide the numerator, and hencz 50 /25 x 2 2 1 47 147 147 Since the perfect square 49 will divide 147, /2 2 5 /~ 142 ELEMENTS OF ALGEIRA. [CIIAP. V. Divide the coefi-ci,;;t of the radical by 3, and mu tiply the num ber under the radical by 4the square of 3; then, 5 / 2 5/18 5 = 27 = -V6' 3 21" 3 21 3. Reduce 98a2x to its most simple form. Ans. 7a, /,;-. 4. Reduce (x3 - a2x2) to its most simple formn. 5. Required the sum of /72 and 1cl Ans. 14J i 6. Required the sum of /T and /147. Ans. 10 O. /'f2-2 V/ 27 7. Required the sum of - andcl -0 8. Required the sum of 2 ab and 3,/64bx. 9. Required the sum of 9 2a43 and 10 6 10. Required the difference of V / and 7-d. Ans. ~ 15 45 11. Required the product of 5t/- and 3/5. Ans. 30 /10. 12. Required the product of - - and -. 3O 4~4 Ans. 305i, 13~. Divide 6T10 by 3. i4. What is the sum Df V/48ab2 and bj/7a. - 15. What is the sumn cf 18/1aL 3 and /50a3b Aws. (3a2d +- 5ab)\ 2ab. C HAPTER V,. EQUATIONS OF THE SECCND lEGUEEg. 110. Equations of the second degree may involve bu onto unknown quantity, or they may involve more than one. We shall first consider the former class. 111. An equation containing but one unknown quantity is said to be of the second degree, when the highest power of the unknown quantity in any term, is the second. Let us assume the equation, aT b x2 - cz+ d =-Ce2 1 + 7x + a. Clearing of fractions, adx2 - bcdx + hd2 = bcdx,2 + b2- + abd; transposing, adx2 - bcdx2 - bcdx - b2x- abd - bd2; factoring, (ad - bcd)x2 - (bcd + 62) = ad - bd2; dividing both members by the co-efficient of x2, bcd - b2 abd- 6d2 Z2 s X - a/ bcd ad- bcd' If we now replace thle co-efficient of x by 2p, and the second member by q, we shall have x2 + 2px = q; and since every equation of the second degree may be reduced, in like manner, we concli;de that, every equation of the second degree, involving but one unknown quantty, can be reduced +a the form X2 + 2Vpx = q, by the following 144 ELEMENTS OF ALGEBRA. [CHAP. VI. RULE. 1. Clear the equation of fractions; II. Transpose all the known terms to the second member, and all thfe unknownr terms to the first. 1II. Reduce the terms involving the square of the unknown quantity to a single term of two factors, one of which is the square of the unknown quantity; IV. Then, divide both members by the co-efficient of the square ef the unknown quantity. 112. If 21p, the algebraic sum of the co-efficients of the first powers of x, becomes equal to 0, the equation will take the formi and this is called, an incomplete equation of the second degree. Hence, An incomplete equation of the second degree involves only the secondpower of the unknown quantity and known terms, and malt be reduced to the form x2. q. Solution of Incomplete Equations. 113. Having reduced the equation to the required form, we have simply to extract the square root of both members to find the value of the unknown quantity. Extracting the square root of both members of the equation x2 q, we have xz= - if q is a perfect square, the exact value of x can be found by extracting the square root of q, and the value of x will then be expressed either algebraically or in numbers. If q is an algebraic quantity, and not a perfect square, it must be reduced to its simplest form by the rules for reducing radicals of the second degree. If q is a number, and not a perfect square, its square root must be determined,' approximately, by the rules already given. CHAP. YIPj EQUATIONS OF THE SECOND DEGRE]. 145 But the square of any numbsr is +, whether the number itself have the + or - sign: hence, it follows that + and and therefore, the unknown quantity x is susceptible of two distiOcl v alumTs, viZ: x = +fq. and xz -; a;:d either of these values, being substituted for x, will satisfy the given equation. For, w2= +X +/;= and 2 - -- X - qT= q; hence, -Every incomplete equation of the second degree has two roots which are numerically equal to each other; one having the sign plus, and the other the sign minus (Art. 77). EXAMPLES. 1. Let us take the equation 1'5 7 299 X2 - 3 7 X2 -2 + 3 BC 12 24 + 24 which, by making the terms entire, becomes 8X2 - 72 - 10x2 = 7 - 24x2 + 299, and by transposing and reducing 378 42x2 = 378 and x2 9; 42 hence, ~= + 94 + 3; and x =-_2. As a second example, let us take the equation 3X2 - 5. Dividing both members by 3 and extracting the square root a which the values of x must be determined approximately 3. What are the values of x in the equation 11 (X2 -4) = 5(x2 A- 2). Apns. x -i 3 4. What are the values of x in the equation =n. Ans 2-= -. - z aI {'1A 146 ELEMENTS OF ALGEBRA. [CHAP. V1. Solution of Equations of the Second Degree. 114. Let us now solve the equation of the second de,,gree z2 + 2px =q. If we compare the first member with the square of + p, which is x2 +-2pz - p22 we see, that it needs but the square of p to render it a perfect square. If then, p2 be added to the first member, it will be come a perfect square; but in order to preserve the equality of the members, p2 must also be added to the second member. Making these additions, we have X2 ~- 2px +p -2 q + p2; this is called, completing the square, and is done, by adding tie square of half the co-efficient of x to both members of the equa tion. Now, if we extract the square root of both members, we have, xz +p -- - q,4: V/q -,:and by transposing p, we shall have x- — p q-/ q 2, and x p — _ 4q -+ p Either of these values, being substituted for x in the equation X2 +- 2px = q will satisfy it. For, substituting the first value, x2= ( -p +q+)2 = p2 2pl/2 + +qt p2. and 2p -2p x ( —p~ +V/q +p2) - - 2p q+p /+ by adding x2 + 2px = q. Substituting the second value of x, we find, 2 = ( - p — 2)2= p2 + 2p2q + X and 2x _=2p(-P- -/q +) =-2p v2pVq+p2; by adding x2 + 2px = q;:and consequently, both values found above, are roots of the Cequation. CHAP. VI.] EQUATIONS OF THE SECOND DEGREE. 147 In order to refer readily, to either of these valucs, we shall call the one which arises from using the + sign before the radical, the first value of x, or-the first root of the equation; and the other, the second value of x, or the second root of the equation. Iaving reduced a complete equation of the second degree to the form x2 + 2px = q, we can write immediately the two values of the unknown quaii tity by the following RULE. I. The first value of the unknown quantity is equal to half the co-eficient of x, taken with a contrary sign, plus the square root of the second member increased by the square of half this co-efficient. II. The second value is equal to half the co-efficient of A, taken with a contrary sign, minus the square root of the secofnd member increased by the square of half this co-eicient. EXAMPLES. 1. Let us take as an example, x2 - 7x + 10 = 0. Reducing to required form, x2 - 7x- - 10; 49 whence by the rule. z = 2 - A/ $ = 5; and, x - 10 - -. 2. As a second example, let us take the equation 5 x 1 3 = 2 273'~ -4z — x X2. 6 4 2 i i8 ELEMENTS OF ALGEBRA. [CueAP. VL Reducing tc: the required form, we have, 2 360 x2 ~q- X - -- 22 22' whence, z/=0 (-)2 and 22 + It often occurs, in the solution of equations, that p, and g are fractions, as in the above example. These fractions most generally arise from dividing by the co-efficient of x2 in the reduction of the equation to' the required form. When this is the case, we readily discover the quantity by which it is necessary to multiply the term q, in order to reduce it to the same denominator with p2; after which, the numerators may be added together and placed over the common denominator. After this operation, the denominator will be a perfect square, and may be brought from under the radical sign, and will l)ecome a divisor of the square root of the numerator. To apply these principles in reducing the radical part of the values of x, in the last example, we have /360 I1 2 /360 X 22 1 /720+1 4 — I — - ) --- - - -- 22 \22/ ( (22 (22)2 - (22))2 1 89 = 792 — ~and therefore, the two values of x become, 1 89 88 x -- +- = = 4' 22 22 22 1 89 90 45 and x = — 2 2 -- 22= t1 either of which being substituted for x in the given equation, will satisfy it. 3. What are the values of x in the equation ax2 - ac - cx - bX2 CHAP. VT.] EQUATIONS OF THE SECOND DEGREE. 149 Reducing to required form, we have, a+. ac a + b a + b -~C / CIC C2 hnd, x= + C c ~ 2 (a +-) b a b' 4 (a + b)2' Reducing the terms under the radical sign to a conaoron.denominator, we find, /cgc c2 4-a2c -''~+ C2 /42c +/ 4C1abc + ~c a+b 4 4(a~+b)2- 4 b)2 2(a + b) c + 4(/ 2c+ 4abc + c2 hence, x 2 (a: + b) 4. What are the values of x, in the equation, 6x2 - 37x = - 57. By reducing to the required form, we have, 37 57 2_ _ 6; 37 257 /37\2 whence, X= + -- — + D 12 / 6 \ 12/ Reducing the quantities under the radical sign to a common denominator, we have, 37 /- l14 x 12 (37)2 12 (12)2 (12)2' But, 114 x 12 = 1368; and (37)2 = 1369; 37 - 13(,:8 + 1369 37 1 hence,2 i \/ (12)2 + 1 22 37 1 19 or, x = - 12 - 37 1 and, = + -- = 3. 12 12 5. What are the values of x,' in the: equation, 4a2 - 2x"' + 2ax = lSab - liS6b. 150 ELEMENTS OF ALGEBRA. [CIAP. VI Reducing to the required form, we have, 2 - ax = 2a2 - 9ab + 92; a2 whence, = L /2a2 - 9ab + 9b2 +-. -~ -- --- 9ab _ a =k 9a _ 9ab + 9b2. 3a The radical part is equal to 3b; hence, a 3a ( - 2a-3b. 2 x=z = _ a + 3b. Find the values of x in the following EXAMPLES. X2 a b 2xz a b 1. — x —=1 —-. Ans. x=- x —-. 3 6 a 3 6 dx q 3X2 1 + c x2 x -'4 c 4 C~ 1 d Ans. x = -- X2 x 59x x 4 3 8 4 3' 3 5 Ans. = X 7 9, x 5 - 2' 3 2x - 10 x+3 4 X X-22 Ans. x - - 7, = x2'b -- I b b - b X5. -=. x = a, -- b ac a- b 3x2 a2 b - a x2 b2 7. ------- +... +...x C 2 C2 - c 2 b +-a b- a Ans, = 6 a 6 CHAP. VI.] EQUATIONS OF THE SECOND DEGREE 151. mx-l m4 n = 2m/Y -+r.zL Ans. x = -, z.Oa2 b2x ab - 262 3a2 9 abx1 +2 -= -- X. C2 c2 c 2a - b 3a -f 2b Ans. x- x --, ac bc 4X2 +2x 3. 2 58x Ans. x-9, x=-1. x - a a - x -2 11. + b -- Ans. x =- a 11 12. 2x + 2-24-5x -- 22. Ans. x- 2, x - 13. x2 - z - 40 = 171). Ans. x = 15, and x= -14. 14. 32 + 2x - 9 = 76. Ans. - 5, and x — 52-. 15. a2 + b2 - 2bx + X2 = Ans. x - (n - 2 + b - a2n). _ 2 7(b2 Problems giving rise to Equations of the Second lDegree involtt ing but one unknown quantity. 1. Find a nu-nber such that three times the number added to twice its square will be equal to 65. Let x denote the number. Then from the conditions, 2x2+3z=65 - - - (1) 3 (;5 9 Whence, = — - +; 4 S- 162 13 reducing x 5 and r —. 152 ELEMENTS OF ALGEBRA. [CHAtP IV Both of these roots verify the equation: for, 2 x (5)2 + 3 x 5=2 x 25 + 15- =;; (13\2 X 13 169 39 130 and (,~~ X The first root satisfies the concdtions of the problem as enuD ilated. The second root will also satisfy the conditions, if we regard its algebraic sign. I-lad we denoted the unknown quantity by — x, we should have found 2x2 —3x = 65 - - - (2) from which x and x -5. We see that the roots of this equation differ from those of tquation (1) only in their signs, a result which was to have been expected, since we canll change equation (1) into equation (2) by simply changing the sign of x, and the reverse. 2. A person purchased a number of yards of cloth for 240 cents. If lie had received three yards less, for the samle sum, it would have cost him 4 cents more per yard. How many yards dlid he purchase? Let x denote the number of yards purchased. 240 Then will denote the number of cents paid per yard. Had he received three yards less, x- 3, would have denoted the number of yards purchased, and 240 -24 would have denoted the number of cents he paid per yald. $-3' From the conditions of the problem, 240 240 X-3 x by reducing. x2 - 3x = 180 whence, x = 15 and x —12. The value x = 15 satisfies the conlitions (f the problem, anderstood in their arithmetical sense; for, -I yards for 240 CHAP. VI.] EQUATIONS OF THE SECOND DEGREE. 153 240 cents, gives -5, or 16 cents for the price of one yard, and 12 yards for 240 cents, gives 20 cents for the price of one yard, which exceeds 16 by 4. The value + x =-12, or — x —- 12, will satisfy the conditions of the following problem: A person sold a number of yards of cloth for 240 Menls: if he had c received the same sum for 3 yards morl, it woculd have brotught him 4 cents less _per yard. How manzy yards did he sell. If we denote the number of yards sold by x, the statement of this last problem, and the given one, both give rise to the same equation, x2 - 3x = 180, hence, the solution of this equation ought to give the answers to both problems, as we see that it does. Generally, when the solution of the equation of a problem gives two roots, if the problem does not admit of two solutions there is always another problem whose statement gives rise to the same equation as the given one, and in this case the two roots form answers to both problems. 3. A man bought a horse, which he sold for 24 dollars. At the sale, he lost as much per cent. on the price of his purchase, as the horse cost him. What did he pay for the horse? Let x denote the number of dollars that he paid for the horse: then, x -- 24 will denote the number of dollars that he lost. But as he lost x per cent. by the sale, he must have lost upon each dollar, and upon x dollars he lost a number 100 of dollars denoted by 00; we have then the equation x2 = =x - 24, whence X2 - 100x = - 2400; Thcrefore, x 60 and x - 40. Both of these values satisfy the conditions of the probleIn. 154 ELEMENTS OF ALGEBRA. LCHAP. VI. For, in the first place, suppose the man gave 60 dollars for the horse and sold him for 24, he then loses 36 dollars. But, from the enunciation, he should lose 60 per cent. of 60, that is, 660( 6 =60 x 60 of 60 = 1 =36; 100 100 therefore, 60 satisfies the problem. If he pays 40 dollars for the horse, he loses 16 by tile sale; for, he shonld lose 40 per cent. of 40, or 40 x 40;16 100 therefore, 40 satisfies the conditions of the, problem. 4. A grazier bought as many sheep as cost him ~60, and after reserving 15 out of the number, he sold the remainder for ~54, and gained 2s. a head on those he sold: how manydid he buy? Anls. 75. 5. A merchant bought cloth for which he paid ~33 15s., which he sold again at ~2 8s. per piece, and gained by the bargain as much as one piece cost him: how many pieces did he buy? Ans. 15. 6. What number is that, which, being divided by the product of its digits, the quotient will be 3; and if 18 be added to it, the order of its digits will be reversed? Ans. 24. 7. Find a number such that if you subtract it from 10, and multiply the remainder by the number itself, the product will be 21. Agns. 7 or 3. 8. Two persons, A and B, departed from different places at the same time, and traveled towards each other. On meeting, it appeared that A had traveled 18 miles more than B; and that A could have performed B's j,,urney in 153 days, but B would have been 28 da}s in performing A's journey. HIow far did each travel? Ai A 72 miles. Ans. B 54 miles. 9. A company at a tavern had ~8 15s. to pay for their reckoning; but before the bill was settled, two of them left CHAP. VI.] EQUATIONS OF TIlE SECOND DEGREE. 15o the room, and then those wlo rer-tmalned nad 10s. apiece n ore to pay than before: how mlany were there in the coimpany..Ans. 7. 10. What two numbers are those whose difference is 15, and of which the cube of the lesser is equal to half their prodltlet' Ans. 3 atd] 18. 11. Two partners, A and B, gained $140 in trade: A's money was 3 months in trade, and his gain was $60 less than his -tock: B's money was $50 mlore than A'fs'rd was in trace 5 months: what was A's stockl? Ans. $ 100. 12. Two persons, A and B, start from two difflerent points, and travel toward each other. When they meet, it appears that A has traveled 30 miles more than B. It also appears that it will take A 4 days to travel the road that B had comle, and B 9 days to travel the road that A had come. What was their distance apart when they set out? Ans. 150 miiles. Discussion of Equations of the Second Degree involving butt one unknown quantity. 115. It has been shown that every complete equation of the second degree can be reduced to the form (Art. 113) 2 2p2=q - - - (1), in which p and q are numerical or algebraic, entire ur frac. tional, and their signs plus or minus. If we make the first member a perfect square, by completing the square (Art. 112Q), we have x2 + 2pX + p2 = q p2, which may be put under the form (x +p )2 = q + Po2. Now, wha,'-.ver may be the value of q ~+p2, its square root may be represented by m, and the eouation put under the form (x z- p)2 =m2, and consequently, ( + p)2 _ m2 = 0. .156 ELEMENTS OF ALGEiBRA. LCHAPl. VI. But. as the first member of the last equation is the difference between two squares, it may be put under the form (x+p —m)(x+p +m)= O - - - (2), in which the first member is the product of t ro factors, and the second 0. No v, we can lmake this product eq sld to 0, and cornsequently satisfy equation (2) only in two different ways. viz., by making x+ p - m -0, whence, x = -p + m, or, by making x - p + m 0, whence, x = -p - m. Now, either of these values being substituted for x in equation (2), will satisfy that equation, and consequently, will satisfy equation (1), from which it was derived. Hence, we conclude, 1st. That every equation of the second degree has two roots, and only two. 2d. That the frst member of every equation of the second degree, whose second member is 0, can be resolved into two binomial fac. tors of the first degree with respect to the unknowon quantity, having the unknown quantity for a first term and the two roots, with their signs changed, for second terms. For example, the equation X2 + 3x - 28 = 0 being solved, gives x=4 and x=-7; either of which values will satisfy the equation. WNe also have (x - 4) ( - 7)= x2 +- 3x -28=0. If the roots of an equation are known, we can readily form the binomial factors and deduce the equation. EXAMPLES. 1. What are the factors, and what is the equatioi; of whirlc the roots are 8 and -9? Ans. x - 8 and x + 9 are the bincmial factors, and 2 + x -72- = O is /he equat on. CHAF, VI.] EQUATIO)NS OF THE SECOND DEGREE. 157 2. What are the factors, and what is the equation, of which the roots are -1 and + 1? x + 1 and x - 1 are the factors, and x2 - 1 = 0 is the equation. 3. What are the factors, and what is the equation, whose roots are 7+ /-Vo1039 and 7 - 0V/ — 039? 16 16 Ans. )7 +and x — ) are the factors, and 82x - 7x + 34 = 0 is the equation. 116. If we designate the two roots, found in the preceding article, by x' and x", we shall have, x" -p + -- m, or substituting for in its value V/q + p2, x' -= — p q p2, Adding these equations, member to member, we get x' +x" = -- p; and multiplying them, member by member, and -reducing, we find xI tl = _ q. Hence, after an equation has been reduced to the form of X2 + 2px = q, Ist. The algebraic sum of its two roots is equal to the co-efficient of the first power of the unknown quantity, with its sign dhanged. 2d. The product of the'wo roots is equal to the second membcr with its sign chapnSed. 158 ELEMENTS OF ALGEBRA. [CHAP. VI. If the sum of two quantities is given or known, their product will be the greatest possible when they are equal. Let 2p be the sum of two quantities, and denote their difference by 2d; then, p + d will denote the greater, and p - d the less quantity. If we represent their product by q, we shall have p2 - d2 = q. Now, it is plain that q will increase as d diminishes. and that it will be the greatest possible, when d = 0; that is, when the two quantities are equal to each other, in which case the product becomes equal to p2. Ilence, 3d. The greatest possible value of the product of the two roots, is equal to the square of half the co-efficient of the first power of the utnknown quantity. Of the Four Forms. 117. Thus far, we have regarded p and q as algebraic quantities, without considering the essential sign of either, nor have we at all regarded their relative values. If we first suppose p and q to be both essentially positive, then to become negative in succession, and after that, both to become negative together, we shall have all the combinations of signs which can arise. The complete equation of th.e second degree will, therefore, always be expressed under or e of the tb-4r following forms:x2 + 2pVx- q (1), 2 -2p x q (2), X2 + 2p =- q (3), x2 -2. = - q (4). These equations being solved, give x=-p-+/ q +p2 (1), x=~+p, q+p2 (2), x- -+- q 4 P2 (3), x= +p t +/q +p (4). CHAP. VI.] EQUTATIONS OF THE SECOND DEGREE. 159 in the first and second forms, the quantity under the radical sign will be positive, whatever be the relative values of p and q, since q and p2 are both positive; and therefore, both roots will be real. And since q +p2 >p2, it follows that, q +2 > 1p, and consequently, the roots in both these forms will have the same signs as the radicals. In the first form, the first root will be positive and the second negative, the negative root being numerically the greater. In the second form, the first root is positive and the second negative, the positive root being numerically the greater In the third and fourth forms, if P2 > q, the roots will 1 real, and since p >/ -q+ P2,1 they will have the same sign as the entire part of the root. Hence, both roots will be negative in the third form, and bote positive in the fourth. If p2 q, the quantity under the radical sign becomes 0, and the two values of x in both the third and fourth forms will be equal to each other; both equal to - p in the third form, and both equal to -- p in the fourth. If p2 < q, the quantity under the radical sign is negative, and all the roots in the third and fourth forms are imaginary. But from the third; prireil,l demonstrated in Art. 116, the greatest value of the product of the two roots is p2, and from the second principle in the satnle article, this product is equal to q; hence, the supposition of p2< q is absurd, and the values of the roots corresponding to the supposition ought to be im. possible or imaginary. When any particular supposition gives rise to imaginary reSults, we interpret these results as indicating that the suppo -ition is absurd or impossible. 160 ELEMENTS OF ALGEBRA. [CHAP. VL If p 0, the roots i;n each form become equal with contrary signs; real in the first and second forms, and imaginary in the third and fourth. If - 0, the first and third forms become the same, as also, the second and fourth. In the former case, the first root is equal to 0, and the second root is equal to - 2p; in the latter case, the first root is equal to + 2p, and the second to 0. If p -= 0 and q = 0, all the roots in the four forms reduce to 0. In the preceding discussion we have made p2> q, p2 b. The first value of x is positive; and since 6, we have #a-Va or, 2Va > ( /a-+ ), whence a > -; and consequently, -. - > a ~-tJ" 2 2 164 ELEMENTS OF ALGEBRA. LCHAP. VL The second value of x is also positive; but since V-> 1, it will be greater than c; and consequently, the secord point wil. be at some point C', on the prolongation of AB, and at the right of the two lights. This is as it should be; for, since the light at A is nmost intense, the point of equal illumination, between the lights, ought to be nearest'the light B; and also, the point on the prolongatioa of AB ought to be on the side of the lesser light B. Second, suppose a < b. The first value of x is positive; and since ~va _z 2A/ a, arnd consequently, c a < -/,/'1-a~ 6 V 2 The second value of x is essentially negative,'lince the numerator is positive, and the denominator essentially negative. We hate agreed to consider distances from A to the right positive; hence, in accordance with the rule already established for interpreting negative results, the second point of equal illuinination will be found at C", somewhere to the left of A. This is as it should be, since, under the supposition, the light at B is most intense; hence, the point of equal illumination, between the two lights, should be nearest A, and the pcint in the prolongation of AB, should be on the side nearest the fcebler light A. CHAP. VI.] EQUATIONS OF THE SECOND DEGREE. 165 Third, suppose a-b, and c> 0. The firs'. 7alue of x is then positive, and equal to 2 hences the first point is midway between the two lights. The second value of x becomes -= co, a result which in. dicates that there is no other point of illumination at a firite distance from A. This interpretation is evidently correct; for, under the supposition made, the lights are equally intense, and consequently, the point midway between them ought to be equally illuminated. It is also plain, that there can be no other point on the line which will enjoy that property. Fourklth, suppose b - a and c = 0. 0 The first value of x becomes, = 0, hence the first point 2 /a is at A. The second value of x becomes, -, a result which indicates that there are an infinite number of other pcints which are equally illuminated. These conclusions are confirmed by a consideration of the conditions of the problem. Under this supposition, the lights are equal in intensity, and coincide with each other at the point A. That point ought then to be equally illunminated by the lights, as ought, also, every other point of the line on which the lights are placed..]Afthl, suppose a > b, or a < b, and c = 0. Under these suppositions, both values of x reduce to 0, which shows that both points of equal illumination coincide with the point A. This is evidently the case, for, since a is not equa_' to 5, and the lights coincide at A, it is plain that no other point than A can be equally illuniniated by them. The preceding discussion presents a striking example of the precision with which the algebraic analysis respondi to all the relations which exist between the quant'ties that enter a pr(blm 166 ELEMENTS OF ALGEBRA. [CLAP. VI. EXAMPLES INVOLVING RADICALS OF TIHE SECOND DEGREE. 2a2 1. Given, x x =,a to find the values of x. By reducing to entire terms, we have, xVa i+ x2 + a2 + x2 = 2a2, oy transposing, x a2 = x2 - a2- x2, and by squaring both members, a2x2 + x4 = a4 - 2a2x2 + x4, shence, 3a2x2 = a4 and, X = tt2 2. Given, a + 2 - 2 =, to find the values of z, a2 a2 a a By transposing, + 2 - - b2 + b; a2 a22 squaring both members, -+ b2 = -z Jr 2b - b 2 q- b2a 4a2 2a squaring both members, b2 = 4a _ 462; and hence, 2 = 4a and x = 2a 5b2 b/V5 a C X2 X 13Z. Given, + = b to find the values of x. Ans. x -- 2ab- 6 x+a a x 4. Given, -+ 2 = 62, to find the xt-a x +a' vwlues of x. a Ans. x = 1 lb "- 1) 2' CHAP. VI.] EQUATIONS OF THE SECOND DEGREE.. 167 a- 1/ia2x-x 5. Given, a-=, to find the values of x. a a + Ans. x = + a 6+ Givena-=, to find the values of x. 1 Xb' VX-+ x-a I(1 + qn)2. Gien - + 1 / 27 Given, ~ --, to find the values of x. Anis. x = _ 2/ab- b)2 8. Given, + +/ + -b, to find the values of x. a + x aq-x Ans. x - /2b - b Of Trinomial Equations. 122. A trinomial equation is one which involves only terms colltainllig two different powers of the unknown quantity and a known terin or terms. 123. Every trinomial equation can be reduced to the form.m, +n = -. (1), ina which rn and n are positive whole numbers, and p and q known quantities, by means of a rule entirely similar to that given in article 111. If we suppose n = 2 and n - 1, equation (1) becomes x2 + 21X - q, a trinomial equation of the second degree. 124. The solution of tiinomial equations of the second degree, has already been explained. The methods, there explained, are, with some slight modifications, applicable to all trinomial equal tions in which m = 2-n, that is, to all equations of the form x28 + 2px'- = q. 168 ELEMENTS OF ALGEBRA. [CHAP, VI. To demonstrate a rule for the solution of equations of this form, let us place xn = y;. whence, x " = y2. These values of xn and x2', being Eubstituted in the giveD equation, reduce it to y2 + 2py = q, thence, y = — p + /q p2, or,,xn = -- p q ~/qXt- jp2. Now, the nth root, of the first member, is x (Art. 18), and although we have not yet explained ]low to extract the WtA root of an algebraic quantity, wre may indicate the 2nth root of the second member. Hence, (axioml 6), X -p V/ p- 2. Hence, to solve a trinomial equation which can be reduced to tho form 2 + 2px — q, we have the following RULE. Reduce the equation to the form of x2 +- 2px" = q; the values of the unknown quantity will then be found by extracting the rlth root of half the co-eficient of the lowest pozuer of the unknown quantity with its sign changed, plus or minus the square root of the second qmemnber increased by the square of half the co-egycient of the lowest power of the unknown quantity. If In = 2, the roots of the equation are of the form x = - p -- / q 2 We see that the unknown quantity has four val]es) since each of the signs + and -, which affect the first radical can be combined, in successicn, with each of the signs which affect the second; but these valees, taken two and utwo, are numiericully equal, and have contrary siqns. CHAP. VI.] TRINOMIAL EQUATIONS. 169 EXAMPLES. 1. Take the equation x4 - 25x2 -- — 144. This being of the required form, we have by appication of tit rule, x- 4 - — 144 65 25 7 whence, X 4 - 4- —; hence, the four roots are + 4, - 4, + 3, and — 3. 2. As a second example, take the equation X4 7X2 -. Whence, by the rule,'7 9 x = + ~ 4- + 9 —; hence, the four roots are, +2 2 -_22 -1 and -I-; the last two are imaginary. 3. x4 -(2bc -}- 4a2) x2 - 62c2. Ans. z = C Z-c + 2a2 4- 2a bc- a2. 4. 2x —7 —= 99. Ans. = —S1, =121 4 a C 2 5.- - bX4 + -x 2 = 0. Ans. X = C ~ V4ad2 +c2 125. The solution of trinomial equations of the fourth deg,. requires the extraction of the square root of expressions of tin form of a -- /; in which a and b %re positive or negative, numerical or algebraic. The expression / a - / can some. times be redluced to the form of a'- ~- or to the Form ~/-a" /b"; and when such transformation is possible, it is 170 ELEMENTS OF ALGEBRA. lCHAP. VI advantageouls to effect it, since, in this case, we have only to extract two simple square roots; whereas, the expression / requires the extraction of the square root of the square root. To deduce formulas for making the required transformation, let us assume p+q=~ a-~ 6 - - - - (1), P -- E -/m - (2); in which p and q are arbitrary quantities. It is now required to find such vattlies for p and q as will satisfy equations (1) and (2). By squaring both members of equations (1) and (2), we have p2 + 2pq + q2 a+ - - (3), p2 —2pq + q2= a-/ b- - - (4). Adding equations (3) and (4), member to member, we get p2+ q2= - - - - - - (5). Multiplying (1) and (2), member by member, we have, p _ 2 _ 2 - Let us now represent a2 - b by c. Substituting in the last equation, p2 /2 -- - (6). From (5) and (6) we readily cldeduce, - ac a-c p — + and q -42 2 these values slbstituted for p and q, in equations (1) and (2), give - -"2' 2'; — V 2 2 CHAP. VI.] TRINOMIAL EQUATIONS. 171 hence,,~~J~= 4 + ir 3- c a - c and -- a (c) - (8). Now,. if a2 - b is a perfect square, its square root, c, will Lo a rational quantity,.and the, application of one of the formrurlas (7) or (8) will reduce the given expression to the re. qmured form. If a2 - b is not a perfect square, the application of the formulas will not simplify the given expression, for, we shall still have to extract the square root of a square root. Therefore, in general, this transformation is not used, unless a2 - b is a perfect square. EXAMPLES. 1. Reduce / 94 + 42 = 94 + S 20, to its simplest form. We have, a = 94, b 8820, whence, c =a - b= /se36-S82 0- 4, a rational quantity; formula (7) is therefore applicable to this case, and we have 9 = / -4-(/94 +4 + 94-4 94+42V — 2 2 or, reducing, - ( 49 ~X 45) hence, / 94 + 2 + (7 + TV/ ). This may be verified; fcr, (7 + 3/5)2 = 49 + 45 + 421=- 94 + 42/5. 2. Reduce vnp9 q- 2mn2 - 2mnp q-m2, to its simplest form. vWe have a = p d — 2m2, and b - 4m2(np ~ m2), a2-5 —n2p2, and c- =/ -b=np; 172 ELEME~NTS OF ALGEBRA. LCHAP. VL and therefore, formula (7) is applicable. b! gives, ~('np- -F 22 -+ p -2') and, reducing, + -). 3. Reduce to its simplest ftorln. 16 + 30/- 1 6- 30~/-I. By applying the formulas, we fndc 16 30-1=5+ 3-1, and 1/6-0-1 = 5 - 3henGe, - 16+ 30V-l1,/ 16 I30V-1 = 10. This example shows that the transformation is applicable to imaginary expressions. 4. Reduce to its simplest form, /+ 3. Ans. 5 +J3. 5. Reduce to its simplest form, 1/+4-3. Ans. 2+ -3 6. Reduce to its simplest form, bc + 2 c c_ -2 bcb2. Ans. At 2b 7. leduce to its simplest form, ab + 4C2 - 2 -- 2 4a/cl2 — abd2. Ans. aa - t4c, - CMAP. VI.] EQUATIONS OF'THE SECOND DEGREE. 173 Equations of the Second Degree involving two or more unknown quantities. 126. Every equation of the second degree, containing two unknown quantities, is of the general form ay2 + bxy + cx2 + dy +fx + g = 0; or a particular case of that form. For, this equation contains terms involving the squares of both unknown quantities, theii product, their first powers, and a known term. In order to discuss, generally, equations of the second degree;nvolving two unknown quantities, let us take the two equations )f the most general form ay2 + b xy CX2+ dy - f x- g = 0, and a'y2 + b'xy + c'x2 + d'y +f'x - g' - 0. Arranging them with reference to x, they become cx2 + (by+f)x+ ay2+ dy+ g =0, cx2 +- (b'y + f') x + a'y2 + d'y - g' = 0; from which we may eliminate x2, after having made its co-efficient the same in both equations. By multiplying both members of the first equation by c', and both members of the second by c, they become, cc'x2 + (b y + J) c'x+ (y2 -- d y - g )c'= 0, cc' 2 + (b')y +f')cx + (a'y2 + d'y + y')c 0. Subtracting one from the other, member from member, we have [(bc' - cb')y +fc' - ~f']x + (ac' - ca')y2 + (d, - cd')y + gc' - cI'= 0, which gives (cC - (tC')y2 + (cd' - dc')y + g' - gc, (bc' - cb')y +/c' - Cf This value being substituted for x in one of the proposed equations, will give a final equation, involving only y. But without effecting the substitution, which would lead to a very complicated result, it is easy to perceive -hat the final equation involving y, will be of the fourth degree. For, the 174 ELEMENTS OF ALGEBRA. [CHAP. VI. numerator of the va;li of x being of tile form my2 + ny +, + its square will be of the fourth degree, an I this square forms one of the parts in the result cf the substitution. Therefore, in general, the solution of hto equations of the secona degree, involving two unknown quantities, depends upon that of an equation of the fourth degree, involving one unknown quantity. 127. Since we have not yet explained the manner of solving equations of the fourth degree, it follows that we cannot, as yet, solve the general case of two equations of the second degree involving two unknown quantities. There are, however, some particular cases that admit of solution, by the application of the rules already demonstrated. First. We can always solve two equations containing two unknown quantities, when one of the equations is of the second degree, and the other of the first. For, we can find the value of one of the unknown qua,: tities in terms of the other ahd known quantities, from the latter equation, and by substituting this in the former, we shall have a single equation of the second degree ccntaining but one unknown quantity, which can be solved. Thus, if we have the two equations x2 + 2y2 = 22 -. - (1), 2x - Y = - - - - (2), we can find from equation (2), l +y I 1+2y+y2~ x -- 2; whence, X2 —. 2Y ~ i f2' 7 4 aind by substituting this expression for x2 in equation (1), we find 1 +2y + y2 I +4 _ + 2y2 — 22; whlence we get the values of y: that is, 29 y = 3 and y — 29 and by substituting in equation (2) we finr,c 10 -=2 and x ,:,ttAr. VI.] EQUATIONS OF T'HE SECOND DEGREE. 175 Second. We can always solve two equations of the second degree containing two unknown quantities when they are bei h homogeneous with respect to these quantities. For, we can substitute for one of the unknown quantitiKs, an auxiliary unknown quantity multiplied into the second unknown quantity, and by combining the two resulting equations we can find an equation of the second degree, from which the value of the auxiliary unknown quantity mrnay be determined, and thence the values of the required quantities can easily be found. Take, for example, the equations x2 + xy — y2=5 - - - (1) 3x2 —2xy —2y2 6 - - - (2) Substitute for y, px, p being unknown, the given equat...-ls become x2 - pX2 _ p2x2 = 5 - - (3), 3X2 _ p2- 2p2X2 _ 6 - - - (4). Finding the values of x2 in terms of p, from equations (3) and (4), and placing them equal to each other, we deduce 5 6 1+pp — - 3 -2p — 2 9 or reducing, p2 + 4p; 1 9 whence, p 2. and p --. Considering the positive value of p, we have, by substituting it in equation (3), X2(I+ - 1 5 2 4 or, x2 - 4; whence, x = 2 and x =-2: and since y -px we lave y = - and Y —- Third. There are certain ether cases which admit of solution, but for wnich no fixed rule can be given. We shall illustrate the manner of treating these cases, ky the solution of the "ollowing [7E' ELEMENTS OF ALGEBRA. [CHAP. VL EXAMPLES. xy L. Given, 48, \/ y W to find the values of x and y. xy / = 24, Dividing the first by the second, member by member, we have 2/=2, or -y = 2; whence y 4; and by substituting in the second equation, we get /x= -6, and x=36. 2. Given, x +4 y+ y -Y 19, to find thevalues ofx andy. x2 + xy + y2 _ 133, to fin the values of Dividing the second by the first, member by member, we have x —y. yy= 7. But, x+ xy-t-y 19: adding these, member to member, and dividing by 2, we find x +y — 13, which substituted in the first equation, gives, 30 6, or xy =- 3G6, and x -. Substituting this expression for x, ir the preceding equation, we get, 36 -+ y = 13, or, y2 - 13y = -- 36; 13 169 13 5 lwhence, Y = o + 7 ~: and finally, y = 9, or y 4; and sinee x -+ y = 13, x = 4, or x 9. CE1A. VI.] EQUATIONS OF THE SECOND DEGREE. 177 3. Find the values of x and y, in the equations x2 + 3x + y 73 - 2xy y2 - 3y + x - 44. By transposition, the first equation becomes, x2 + 2xy + 3xw y = 73; o) which if the second be added, member to members tlere results, x2 + f2xy y2 + 4x + 4y = (x + y)2 + 4 (x + y) = 117. If, now, in the equation (x + y)2 + 4 (x - y) = 117, we regard x + y as a single unknown quantity, we shall lave x + y —2 /117+ 4; hence, x - y= - 2 -11- 9, and x + y = - 2- 11 — 13; whence, x = 9-y, and x= -13-y. Substituting these values of x in the second equation, we have y2 +2y=35, for x=9-y, and y2+_2y=_57, for x= —13 - y. The first equation gives, y-= 5, and y = -7, ad the second, y=- 1 + 5 and y -1 - /8. The corresponding values of x, are x =4, x = 16; = —12-58, and w=-12+~. 4. Find the values of x and y, in the equations x2Y2 + V2 + Xy = 600 - (y + 2)x'y3 x + y2 = 14 - y. From the first oqqation, we have X2y2 + (y + 2y) 2y2 + xy2 + xy = 600, or, 2y2 (1.4%y2 + 2y) + xy (1 + y) = 600, or, again, x2~y2 (1 + y) + xy (1 + Y) = 0oo; 12 17TS ELEMENTS OF ALGEBRA. LCHAP. VI. which is of the form of an equation of the second degree, re. garding xy (1 + y) as the unknown quantity. Hence, xy + y) = 2 io + _2 and if we discuss only the roots which belong to the + value of the radical, we have xy (4 + Y) -~- = 24; 24 and hence, x- -. y + y2 Substituting this value for x in the second equation, we ha ve (y2 + y,)2 - 14 (y2+ y) - -24; whence, y2 + y =12, and y2 + y = 2. From the first equation, we have Y -- 2+ 2 -- 3, or — 4; 2 2 and the corresponding values of x, from the equation 24 2 From the second equation, we have y -- 1, and y —2; which gives x = 12. 5. Given, x2y + xy2 = 6, and x3y2 + x2y3 = 12, to find the "alues of x and y. y-. or 2. 6.iven, 4 x"2 + X + y = 18 - y-2 to find the values of xy - x and y.' = 3, or 2; or — 3J- 3, {y=2, or 3; or -3 t'loblerms giving rise to Equations of the Second Degree con tainring two or more unknown qguantities. 1. Find two numbers such, that the sum of the respect'iv products of the first multiplied by a, and the second multiplied by b, shall be equal to 2s; and the product cf the one by athe other equal to p. CHAP. VI.] EQUATIONS OF THE SECOND DEGREE. 179 Let x and y denote the required numbers, and we have ax + by = 2s, and xy =p. From the first 2s - a ye=.b; wherce, by substituting in the second, and reducing, ax2 -sx =- bp. s 1 Therefore, X - - 2 - abp, a a s 1 and consequently, y = tb -s2 -abp. Let a = b = l; the values of x, and y, then reduce to x=s s-/2 -- p, and y =s s 2 —p; whence we see that, under this supposition, the two values of x are equal to those of y, taken in an inverse order; which shows, that if s' -2 -p represents the value of x, s — 4 2 -p will represent the co-responding value of y, and conversely. This relation is,:xlained by observing that, under the last supposition,'bf / on equations become x + y = 2s, and xy =-p; and'he t,,l~ tica is then reduced to finding two numbers of which the sav.. is 2s, and their product p; or in other words, to divide a ntmnb3cr 2s, into two such parts, that their product may be equal to a give Inumber 2. 2. To find four numbers, such that the sum of the first and fourth shall be equal to 2s, the sum of the second and third equal to 2s', the sum of their squares eq-nal to 4c2, and the product of the first and fourth equal to the product of the second and third. 180 ELEMENTS OF ALGEBRA. [CHAP. VLY Let u X, y, y, an z, denote the numbers, respectively. Then, from the colditions of the problem, we shall have t- + z = 2s 1st condition; x + y = 2s' 2d " u f + yx + z2 = 4C2 3d " uz _ xy 4th " At first sight, it may appear difficult to find the values of the unkncwn quantities, but by the aid of an auxiliary Uinknown yuantity, they are easily determined. Let p be the unknown product of the 1st and 4th, or 2d,nd 3d; we shall then have (t + z = 2s, 8, i u = s -S/S2 _ p, which give, UZ, = Z =S 8-o and xZG +y s', X= s'd which'give, X= p, s' -- p. Now, by substituting these values of u, x, y, z, in the third equation of the problem, it becomes (s + 8- p)2 + (S- p)2 + (3' + ~ / 2' p)2 - (s' -_s'2 p)2_4c2; and by developing and reducing, 4s2 4s2 - 4p = 4c2; hence, p = 2 + s2 -c2. Substituting this value for p, in the expressions for uz, x, 3,,re find U ~' = S'+ -' X- = S' These values evidently satisfy the last equation of she p-roblem; for uZ = (s 4 -s/2-,)(s -- 2 = f 2 _ -2 e - S12, Xy (,'+- ) (S -2 / - 2) = s'2-_ 2 + 2 CHAP. VI.] EQUATIONS OF THE SECOND DEGI'EE. 181 REMARK. —This problem shows how mulch the introduction of an unknowtn auxiliary often facilitates the determination of the principal unknown quantities. There are others problems of the same kind, which lead to equations of a degree superior to the second, and yet they may be resolved by the aid of equations of the first and second degrees, by introducing unknozwn auxiliaries. 3. Given the sum of two numbers equal to a, and the sun of their cubes equal to c, to find the numbers By the conditions = x3 + y3 = c. Putting x = s + z, and y =s —z, we have a=2s, -undl j x3 _ S3 + 3S2Z +~ 31s2 + Z3 and 3. y3 _ s3 3s2z + 3sz2 z3 hence, by addition, x3 +- 3 _ 2s3 + 6sz2 = c; c-2s3 c-Ss' whence, Z2 and z= 6s /c-2s3 c- 2s3 or, x = 6s and y =_s 6s and by substituting for s its value, a //c-3 a 4c - a3 2 / 3a ) 2 / 12a ada /c-&), a _ 4c -a3 n2 Y= 23a-J2-/ 12a 4. The sum of the squares of two numbers is expressed by a, and the differerce of their squares by b: what are the numbers ab -b Ans. a b2 - 5. What three numbers are they, which, multiplied two and two, and each product divid( I by the third number, give the quotients, a, 6, c Als..F/Yh, ac, Vv c. 182 ELEMET-YrS OF ALGEBRA. ICItAP. VI. 6. The sum of two numbers is 8, and the sum of their cubes is 152: what are the numbers! Al2s. 3 and 5. 7. Fiind two numbers, whose difference added to the diffcrence of their squares is 150, and whose sum added to the sum of their squares, is 330. Ans. 9 and 15. 8. There are two numbers whose difference is 15, and half their product is equal to the cube of the lesser number: what are the numbers? Ans. 3 and 18. 9. What two numbers are those, whose sum multiplied by the greater, is equal to 77; and whose difference, multiplied Jby the lesser, is equal to 12? AAns. 4 and 7, or 3 2 and'U!. 10. Divide 100 into two such parts, that the sum of their square roots may be 14. Ans. 64 and 36. 11. It is required to divide the number 24 into two such parts, that their product may be equal to 35 times their difference..Ans. 10 and 14. 12. What two numbers are they, whose product is 255, and tbe sum of whose squares is 514? A? s. 15 and 17. 13. There is a number expressed by two digits, which, when divided by the sum of the digits, gives a quotient greater by 2 than the first digit; but if the digits be inverted, and the resulting number be divided by a number greater by 1 than the sum of the digits, the quotient will exceed the former quotient by 2: what is the number?. Ans. 24. 14. A regiment, in garrison, consisting of a certain number of companlies, receives orders to send 216 men on duty, each company to hurnish an equal number. Before the order was executed, three of the companies were sent on another service, and it was ther_ found that each company that remained. would have to send 12 men additional, in order to make up the comrnplement, 216. HI-ow many companies were in the regiment, and what number of mren did each of the remaining companies send Ans. 9 companies: each that remained sent 36 men. CHEAP VI.J EQUATIONS OF THE SECOND DEGR:EE. 183 15. Find three numbers such, that their sum shall be 14, the sum of their squares equal to 84, and the product of the first and third equal to the square of the second. Ans. 2, 4 and 8. 16. It is required to find a number, expressed by three digits, such, that the sum of the squares of the digits shall be 104; the square of the middle digit to exceed twice the product of the other two by 4; and if 594 be subtracted from the number, the remainder will be expressed by the same figures, but with the extreme digits reversed. Ans. 862. 17. A person has three kinds of goods which togetner cost $230A-5. A pound of each article costs as many — _ dollars as there are pounds in that article: he has one-third more of the second than of the first, and 3~ timnes as much of the third as of the second: How many pounds has he of each article? Ans. 15 of the 1st, 20 of the 2d, 70 of the 3d. 18. Two merchants each sold the same kind of stuff: the second sold 3 yards more of it than the first, and together, they received 35 dollars. The first said to the second, "I would have received 24 dollars for your stuflf" The other replied, "'And I would have received 121 dollars for yours." How many yards did each of therl sell? Ans Ist merchant 15 5o Ans. 41 20 IS t or 2d s1 8. 19. A widow possessed 13000 dollars, which she divided into two parts, and placed them at interest, in such a manner, that the incomes from them were equal. If she had put out the first portion at the same rate as the second, she would have drawn for this part 360 dollars interest; and if she had placed the second out at the same rate as the first, she would have drawn for it 490 dollars interest. What wvere the tw( rates of interest' Ans. 7 and 6 per cent. CHAPTER VII. TORIMATION' OF POWERS-BINOMIAL TIHEOREM-EXTTRAClON OF P.RO'00 OP ANY DEGREE-OF RADICALS. 128. THE solution of equations of the second degree supposes the process for extracting the square root to be known. In like manner, the solution of equations of the third, fourth, &c., degrees, requires that we should know how to extract the third, fourth, &c., roots of any numerical or algebraic quantity. The power of a number can be obtained by the rules for multiplication, and this power is subject to a certain law of forInation, which it is necessary to know, in order to deduce the root from the power. Now, the law of formation of the square of a numerical or algebraic quantity, is deduced from the expression for the square of a binomial (Art. 47); so likewise, the law of a power of any degree, is deduced from the expression for the same power of a binomial. We shall therefore first determine the law for the formation of any power of a binomial. 129. By taking the binomial x + a several times, as a factor, the following results are obtained, by the rule for multiplication (+ a) =x + a, (x + a)2 = x2 + 2ax + a2, (;X ~ a)3 3 + 3ax2 + 3a2z -- a3, (. + a)4 X.4 + 4aX3 + 6a2x2 + 4a3x + a4, (x + a)5 - x5 + 5ax4 + 10a2x3 + 10a3X2 + 5a4x + al. By examining these powers of x + a, we readily discover thB law according to which the exponents of the powers of a ie CHAP. VII.] PERMUTATIONS AND COMBINATIONS. 185 crease, and those of the powers of a increase, in the successive terms. It is not, however, so easy to discover a law for the formation of the co-efficients. Newton discovered one, by means of which a binomial may be raised to any power, without per fbirning the multiplications. I-Ie did not, however, explain the course of reasoning which led him to the discovery; but the law has since been demonstrated in a rigorous -manner. Of' all the known demonstrations of it. the most elementary is that which is founded upon the theory of combinations. However, as the demonstration is rather complicated, we will, in order to simplify it, begin by demonstrating some propositions relative to permutations and combinations, on which the demonstration of the binomial theorem depends;. Of Permutations, Arracgements and Combinations. 130, Let it be proposed to determine the whole nlumber o/ ways in which several letters, a, b, c, d, &c., can be written. one after the "other. The result corresponding to each change in the position of any one of these letters, is called a p9er mutation. Thus, the two letters a and b furnish the two permutations, vb and ba. ach In like manner, the three letters, a, b, c, firnish cbec six permutations. cha PERMUTATIONS, are the results obtained by writing a certain 2zumber of letters one after the other, in every possible order, iln such a mannler thalt all the letters shall enter into each result, and each letter enter but once. To determine the number of permutations of which n letters are susceptible. Two letters, a and b, evidently give two per. ab nuutations. ba 186 ELEMENTS OF ALGEBRA. LCHAP. VI. Therefore, the number of perriutations of two letters is ex pressed by 1 x 2. Take the three letters, a, b, and c. Iteserve C either of the letters, as c, and permute the other ab two, gi ring b) Now, the third letter c may be placed before ab,?cab between a and b, and at the right of ab; and the ach same for ba: that is, in ONE of the f'rst permutations, the reserved letter c may have three different abe places, givinig three permutations. And, as the same cba may be shown for each one of the first permutations, it follows that the whole number of permutations of bea three letters will be expressed by, 1 x 2 X 3. bac If, now, a fourth letter d be introduced, it can have four places in each one of the six permutations of three letters: hence, the number of permutations of four letters will be ex. pressed by, 1 x 2 x 3 X 4. In general, let there be nt letters, a, b, c, &c., and suppose the total number of permutations of n - 1 letters to be known; and let Q denote that number. Now, in each one of the Q permutations, the reserved letter may have n places, giving n perrmutations: hence, when it is so placed in all of them, the entire number of permutations will be expressed by Q X n. If s- = 5, Q will denote the number of permutations of four quantities, or will be equal to 1 x 2 X 3 X 4; hence, the number of permutatins s of five quantities will be expressed by 2 3 4 5. If n =- 6, we shall have for the number of permutations of six quantities, 1 X 2 X 3 X 4 X 5 x 6, and so onl. HIence, if Y denote. the number of permutations of n letters, w; shall have Qx n = 1. 2. 3. 4.... (tL-1)n: that is, The number of permutations of n letters, is equal to the conttzued product of the nrtural numbers from 1 to n inclusively. CHAP. VI. PERMUTATIONS AND COMBINATIONS. 187 Arrangements. 131. Suppose we have a number m; of letters a, b, c, ed, &c. Ii they are written in sets of 2 and 2, or 3 and 3, or 4 aLid 4.. in every possible order in each set, such results are called sirrant?,geizentts. Thus, ab, ac, ad,... ba, be, bd,... ca, cb, cd,... are ar rangeements of m letters taken 2 and 2; or in sets of 2 each.'In like mnanner, abc, abd,... bcc, bad,... acb, acd... are wrran2geme ts taken in sets of 3. ARRANGEMENTS, are the results obltained by writi.ng a znumber m f letters, inz sets of 2 and 2, 3 and 3, 4 and 4,... n atid n; tle letters in each, set havizig every possible order, and m beinlg always greater than n. If we suppose m n, the arrangenzents, taken n and it, become permultations. IaEtving given a number m of letters a, b, c, d,... to determi-ze lk.e total number of arranyenments that may be formed of them by tainlg the,,m n in a set. Let it be proposed, in the first place, to arrange three letters, a, b and c, in sets of two each. First, arrange the letters in sets of one each, and a for each set so formed, there will be two letters reserved: the reserved letters for either arrangement, being those which do not enter it. Thus, with c reference to a, the reserved letters are b and c; with reference to b, the reserved letters are a and c; and with reference to c, thev are a and b. Now, to any one of the letters, as a, annex, in ad successibn, the reserved letters b and c: to the a second arrangement b6, annex the reserved letters a b and c and to the third arrangement, c, annex the ca reserved letters a and b. eb Since each of the first arrangerments gives as many new arrangements as there are reserved letcers, it follows, that the 188 ELEMENTS OF ALGEBRA. [CHAP. VIL number of arsrangements of three letters taken, roo in a set, will be equal to the number of arrangements of the saone letters taken one in, a set, multiplied by the number of reserved letters. Let it be required to form the arrangement of four letters, a, b, c and d, taken three in a set. First, arrange the four letters in sets of two: there fa b will then be for each arrangement, two reserved let- acs ters. Take one of the sets and write after it, in succession, each of the reserved letters: we shall thus b a form as many sets of three letters each as there are bd reserved letters; and these sets differ from each other 4 c by at least the last letter. Take another of the first t c b arrangements, and annex, in succession, the reserved cd letters; we shall again form as many different arrange- da ments as there are reserved letters. Do the sale for db all of the first arrangements, and it is plain, that the d c whole number of arrangements which will be formed, of four Letters, taken 3 and 3, will be equal to the cnumber of arrangements of the same letters, taken two in a set, rmultiplied by the number of reserved letters. In general, suppose the total number of arranyements of m letters, taken n - 1 in a set, to be known, and denote this number by P. Take any one of these arrangements, and annex to it, in succession, each of the reserved letters, of which the number is in - (n - 1), or m - n + 1. It is evident, that we shall thus form a number m - i + 1 of new arrangements of n letters, each differing from the others by the last letter. Now, take another of the first arrangements of n - 1 letters, and annex to it, in succession, each of the m - n - 1 letters which do not enter it; we again obtain a number n - n + 1 of arrangements of it letters, differing from each other, and from those obtained as above, by at least one of the n - 1 first letters. Now, as we may in the same manner, take all the P arrangeinents of the m letters, taken en - 1 in a set, and annex to them. CEAI'. VII.] PERMUTATIONS AND COMBIN ATIONS. 189 in succession, each of the mn - n + 1 other letters: it follows that the total number of arrangements of mn letters, taken n in a set, is expressed by P(Dn - n + 1). To apply this, in determining the number of arrangements of mn letters, taken 2 and 2, 3 and 3, 4 and 4, or 5 and 5 in a set, make s = 2; whence, mn-n + 1 m- = -; P in this case, will express the total number of arrangements, taken 2 - 1 and 2 - 1, or 1 and I; and is consequently equal to en; therefore, the expression P(m - n + 1) becomes m(m - 1). Let n= 3; whence, rm- n +r I = n-2; P will then ex. press the number of arrangements taken 2 and 2, and is equal to m(m - 1); therefore, the expression becomes m(?n - l)(m - 2). Again, take n = 4: whence, m —n r -l = m m-3: P will ex press the number of arrangements taken 3 and 3, and therefore the expression becomes m(m - 1) (m - 2) (n - -3), and so on. Hence, if we denote the number of arrangements of m let. ters, taken?, in a set by X, we shall have, X P (2n - n - 1) = m (nm —I) (m - 2). (m - n 1- ); that is, The number of arrangements of mn letters, taken n in a set, is equal to the continued product of the natural numbers from m down to mn - n + 1, inclusively. if in the preceding formula m be made equal to n, the ar rangements become permutations, and the formula reduces to X= n (n - 1) (n-2)..... 1; or, by reversing the order of the factors, and writing Y for X, Y= -1. 2. 3.... (n-l)n; the Dame formula as deduced in the last article. 190 ELEMENTS OF ALGEBRA. [CRAP. VTI combinations. 132, When the letters are disposed, as in the arrangements, 2 and 2, 3 and 3, 4 and 4, &c., and it is required that any two of the results, thus formed. shall differ by at least one letter. the products of the letters will be different. In this case, the results are called combinations. Thus ab, ac, bc,... ad, bd,.. are combinations of the let. ters a, b, c, and d, &c., taken 2 and 2. In like manner, abc, abd,.... acd, bcd,.. are combinations of the letters taken 3 and 3; hence, COMBINATIONS, are arrangements in which any two will differ fiom each other by at least one of the letters which enter them. To determine the total number of diferent combinations tlhai can be formed of sn letters, taken n in a set. Let X denote the total number of arrangements that can be formed of m letters, taken it and n: Y the number of per mutations of n letters, and Z the total number of different combinations taken n sold n. It is evident, that all the possible arrangements of m. letters taken n, in a set, can be obtained, by, subjecting the n letters of each of the Z combinations, to all the permutations of which these letters are susceptible. Now, a single combination of in letters gives, by hypothesis, Y permutations or arrangementstherefore Z combinations will give Y X Z arrangements; and as X denotes the total number of arrangements, it follows that X= rx Z; whence, Z- But we have (Art. 130), Y=- Q xn=l. 2. 3....n and (Art. 131), X = P (n n M 1 ) = m (m - 1) (m - 2).... (7n - n -+ I' therefore, P (7 - - l) t (vt - 1) (m -2).... - n + 1) Qxn 1 2........... that is, CHAP. VII.! BINOMIAL THEOREM. 191 The number cf combinations of m letters taken n in a set, is equal to the continued product of the natural numbers from m dowun to m —n +- 1 inclusively, divided by the continued product qf the natural numbers from I to n inclusively. 133. If Z denote the number of combinations of the in let ters taken n in a set, we- have just seen that m (m - -1) (m-2).... (n ~)-( +.2 3... If Z' denote the number of combinations of mn letters taken (m - -) in a set, we can find an expression for Z' by changing n into m - i in the second member of the above formula; whence'm (_ m,- 1)(mn 2).. (n - 1). 1. 2........ (-)'. If, now, we divide equation (1) by (2), member by member, and arrange the factors of both terms of the quotient, we shall have Z l. 2. 3.... (m —) X (m-n + 1)... (m- - )m Z' 1I 2. 3..... n X (n + I)...... (m- 1)m The numerator and denominator of the second member are equal to each other, since each contains the factors, 1, 2, 3, &c., to m; hence, Z - 1, or Z- Z'; therefore, The number of combinations of m letters, taken n.1 in a set, is equal to the number of combinations of m letters, taken m - in in a set. Binomi((al Theorem. 134. The object of this theorem is to show how to find any power of a binomial, without going through the process of con tnlued multiplication. 135, The algebraic equation which indicates the law of for. mation of any power of a biromial, is called the Binomiai F7tormula. 192 ELEMENTS OF ALGEBRA. [CHAP. VI. In order to discover this law for the mth power of the binomial x+ a, let us observe the law for the formation of the product of several binomial factors, x + a, x + b, x + c, x + d of which the first term is the same in all, and the second terms different. x - a x +b 1st product - x2 + a x + ab +b 2d.. 3 +- a x2 ab x +- abc + b + ac + c + bc X + d 3d. x4 +- at 3 + ab x2 + abe x Bbcd + b + ac + abd + c + ad +- acd +- d + be + bcd + bd + cd These products, obtained by the common rule for algebraic multiDlieartion, indicate the following laws: 1st. With respect to the exponents, we observe that the ex. ponent of x, in the first term, is equal to the number of bino-., mial factors employed. In each of the following terms to the right, this exponent is diminished by I to the last term, where it is 0. 2d. With respect to the co-efficients of the different powers of x, that of the first term is 1; the co-efficient of the second t rm is equal to the sum of, the second terms of the binomials; the co-efficient of the third term is equal to the sum of the products of the different second terms, taken two and two; CHAPf'. VII.1 BINOMIAL THEOREM. 193 the co-efficient of the fourth term is equal to the sum of their different products, taken three and three. Reasoning from analogy, we might conclude that, in the pro. duct of any number of binomial factors, the co-efficient of the term which has n terms before it, is equal to the sum of the diffieent products of the second terms of the binomials, taken n and n. The last term of the product is equal to the con. tinued product of the second terms of the binomials. In older to prove that this law of formation is general, suppose that it has been proved true for the product of m binomials. Let us see if it will continue lto be true when the product is multiplied by a new binomial factor of the same form. For this purpose, suppose x, + AXm — _ BXm- 2 + C~x-m-3 + M + 1 +Xm —n +. + U to be the product of mz binomial factors; _tAx"-'" representing the term which has n terms before it, and Afxmn-"+t the term which immediately precedes, Let x +- k be the new binomial factor by which we multiply; the product, when arranged: according to the powers of x, will be X.~L1+~A do x B t;x-1 +- C X-m-2 +... + N2 { -nyi+l. - Ac + Ak + Bk + Mk + U, freom which we perceive that the law of the exponents is ev,. dently the same. With respect to the co-eflicients, we observe; Ist. That the co-efficient of the first term is 1; and 2d. That A + k, or the co-efficient of xm, is the sum. of- tihe,e, owed terms of the n + I bionomials. 3d. Since, by hypothesis, B is the sum of the different- products of the second terms of the In binomials, taken two and two, and since A x k expresses the sum of the products of each of the.second terms of the first In binomials by the new second term k; therefore, B+ Ak is the szUm of the different products of the, second terms of the m + 1 binomials, taken two and two,. 13 194 ELEMENTS OF ALGEBRA. [CHAP. VIT. In general, sit!ce N1 expresses the satm of the products of the second terms of the m binomials, taken n and n, and AI the sum of their products, taken s - 1 and n - 1., therefore tN + M-I k, or the co-efficient of the term which has n terms before it, will be equal to the sum of the different products of the second termus of the m + 1 binornials, taken t and n. The last term is equal to the continued product of the second terms of the, m — 1 binomials. H1ence, the law of composition, supposed true for a number mr of binomial factors, is also true for a number-r denoted by m +- 1. But we have shown the law of composition for 4 factors, hence, the same law is true for 5; and being true for 5, it must be for 6, and so on; hence, it is general. 136. Let us take the equation, (x + a) (x + b) (x + c) X. =. -1 +'+B-2. + N2x"-"..... -L WI containing in the first member, m binomial factors. If we make a- b _ c = d.... &c., the first member becomes, (. + a)m. In the second member the co-efficient of xm will still be 1. The co-efficient of x,"', being a + b + c + d... will become -a taken m times; that is, ma.::he co-efficient of xn-2, being ab + ac + ad.... reduces to a2a - 2 - a2'that is, it becomes a2 taken as many times as there are corn binations of mn letters, taken two and two, and hence reduces (Art. 132), to 0 — 1 7ml.. 0-a2 The co-efficient of x'-3 reduces to the product of a3, multi plied by the number of different combinations of m lette.r, taken three and three; that is, to mn- i. - 2 m.. — a3. t4~e,. CHAP. VII.] BINOMIAL THEOREM. 195 Le t us denote the general term, that is iie one which has X terms before it, by 2Nxm-n. Then, the co-efficient 2V will denote the sum of the products of the second terms, taken n and n; and when all the second terms are supposed equal, it becomes equal to an multiplied by the number of combinations of mn letters, taken n and n. Therefore, the co-efficient of the general term (Art. 132), is P(m - n -Fl) + an. hence, we have, by makhing these substitutions, ( + a)m - xvn +,maxrn-4 ~ irnt.m 1 a2_2 m - 1 m-2 3 P(m - 1 + I) + 2 3 -a3ax-.. + Q which is the binomial formula. The termn P(m - n + l) is called the general term, because by making n - 2, 3, 4, &c., all the others can be deduced from it. The term which im mediately precedes it, is P P - a]lm? —-, 1, since Q Q evidently expresses the number of combinations of m letters taken n - I and n - 1. Hence, we see, that P (m7 -n + 1) which is called the numerical co-efficient of the general term, is equal to the numerical co-efficient - of the preceding term, multiplied by m - n + 1, the exponent of x in that term, and divided by n, the number of terms preceding the required term. The sizple lauw, demonstrated above, enables us to determine the numerical co-efficient of any term from that of the preceding term, by means of the follow;ng [96 ELEMENTS OF ALGEBRA. [CHLP. VII. RULE. The numerical co-eficient of any term after the first, is formed by mgultiplying that of the preceding term by the exponent of z in that term, and dividing the product by the number of terms which precede the required term. 137. Let it be required to develop (x + a)6. By applying the foregoing principles, we find, (x f a)6 = x6 ~- 6ax5 + 15a2x4 + 20a3X3 + 15a4x2 + 6a5x + al. Having written the first term x6, and the literal parts of the other terms, we find the numerical co-efficient of the second term by multiplying 1, the numerical co-efficient of the first term, by 6, the exponent of x in that term, and dividing by 1, the number of terms preceding the required term. To obtain the co-efficient of the third term, multiply 6 by 5 and divide the prouull uy 2; we get 15 for the required number. The other numerical co-efficients may be found in the same manner In like manner, we find (x +- a)~O = xl~ + l0ax9 + 45a2x8 + 120a3x7 + 210a4x6 + 252a5x5 + 210a6x4 + 120a7x3 + 45a8x2 + 10a9x + aG. 138. The operation of finding the numerical co-efficients may be much simplified by the aid of the following principle. We have seen that the development of (x + a)m, contains.n + 1 terms; consequently, the term which Ihs n terms after it, has m- n terms before it. Now, the numervical co-efficient of the term which has n terms before it is equal to the num. ber of combinations of m letters tkeaien n in a set, and the tlumenrical co-efficient of that term which has n terms after it, t rm - n before it, is equal to the number of combinations of r letters taken m —n in a set; but we have shown (Art. 133) that these numbers are equal. Hence, Ih the development of any power of a binomial of the forlm (x + a)-, the numerical co-eficients of terms at equal distances from the two extrenmes, are equal to each other. CHAP. VII.] BINOMIAL THEOREM. 197 We see that this is the case in both of the examples above given. In finding the development of any power of a binomial, we need find but half, or one more than half, of the numerical co-efficients, since the remaining ones may be written directly from those already found. 139. It frequently happens that the terms of the binomial, to which the formula is to be applied, contain co-efficients and exponents, as in the following example. Let it be required to raise the binomial 3a2c - 2bd to the fourth power. Placing 3a2c - x and - 2bd y, we have (X + y)4 = X4 ~4X3y + 6X2y2 + 4ry3 + y4; and substituting for x and y their values, we have (3a2c - 2bd)4 = (3a2c)4 + 4 (3a2c)3 (- 2bd) + 6 (3a2c)2 (- 26d)2 + 4 (3) (- 2sd)3 (- 2bd)4, or, by performing the operations indicated, (3a2c - 2bd)4 = 81ac4 - 216a6c3bd + 216a4c262d _ G96a2cb3d3 + 16b4d4. The terms of the development are alternately plus and minus, as they should be, since the second term is -. 140. A power of any polynomial may easily be found by means of the binomial formula, as in tle following example. Let it be required to find the third power of a + b + c. First, put b + c - d. Then (a+ b+ c) = (a +- d)3=a3 3a2d + 3acd + dB, and by substituting for the value of d, (a + 6 + C)3 = a3 + 3a2b + 3ab2 + bl 3a2e + 262c — + 66c + 3ac2 + 3bc2 4- c3 198 ELEMENTS OF ALGEBRA. [CHAP. VIL This developmei t is composed of the sum of the cubes of the three terms, plus the sum of the results obtained by multiplying three times the square of each term, by each of the other terms in succession, plus six times the product of the three terms. To apply the preceding formula to the development of the cube of a trinomial, in which the terms are affected with co efficierts and exponents, designate each term by a single lelttr, and peform the operations indicated; then replace the letters introduced, by their values. From this rule, we find that (2a2 - 4ab -p 3b2)3 - Sa6 - 48a5b 132a4b2 - 208a363 -+ 198a2b4 - 108ab5 + 27b6. The fourth, fifth, &C., powers of any polynomial can be developed in a similar manner. Extraction of the Cube Root of Numbers. 141. The cube root of a number, is such a number as being taken three times as a factor, will produce the given number. A number whose cube root can be exactly found, is called a perfect cube; all other numbers are imperfect cubes. The first ten numbers are, 1, 2, 3, i, 5, 6, 7, 8, 9, 10; and their cubes, 1, 8, 27, 64, 125, 216, 343, 512, 729, 1000. Conversely, the numbers in the first line are the cube roots of the corresponding numbers in the second. If we wish to tinld the cube root of any number less than 1000, we look for the number in the second line, and if 4t is there xvwritten, the corresponding number in the first line will be its cbt, root. If t1h number is not there written, it will fali betwenl two nulmbers in the second line, and its cube root will tlt!l between the corresponding numbers in the first line. In thlis case the culbe root cannot be expressed in exact parts of 1; ltence, the given nurnmber must be an imperfect cube (IRe. marlk Ill, Art. 85). CHAP. Vl1.j C JBE tOOT OF INUIBERS. 199 If the given number is greater than 1000, its cube root will be greater than 10; that is, it will contain a certain number of tens and a certain number of units. Let us designate any number by gN, and denote its tens by a, and its units by b; we shall have, 2V — a - b; whence, 3 =- a3 +- 3a2b ~- 3ab2 +- b3; that is, The cube of a nhmber is equal to the cube oJ the tens, plus three times the product of the square of the tens by the units, plus three times the product of the tens by the square of the units, plus the cube of the units. Thus (47)3 —(40) +~ 3 x (40)" X 7 + 3 X 40 X (7)2 ~- (7)3 = 103823. Let'us now reverse the operation, and find the cube root of I 38S 23. 103 823 47 48 47 64 8 48 47 42 x 3-48 1 39S'23 384 329 192 188 2304 2209 48 47 184')2 15463 9216 8836 110592 1 038O2 Since the number is greater than 1000, its root will contain tens and units. We will first find the number of tens in _the root. Now the cube of tens, giving at least thousands, we point off three places of fig ires on the right, and the cube of the number of tens will be i)und in the number 103, to the left of this pej iod. The cube root of the greatest cube contained in 103 being 4, this is the number of tens in the required root. Indeed, 103823 is evidently comprised between (40)' or 64,000, and (5(0)3 or 125,000; hence, the required root is comprised between 4 tens and 5 tens: that is, it is comnposed of 4 tens, plus a certa;u number of units less than ten. 200 ELEMENTS OF ALGEBRA. LCHAP. VII. Having found the number of tens, subtract its cube, 64, from 103, and there remains 39, to which bring down the part 823g and we have 39823, which contains three times the lproduct of tMe square of the tens by the units, plus three times the product o' the teas by the square of the units, plus the cube of the units, Now, as the square of tens gives at least hundreds, it follows that the product of three ti mes the square of the tens by the units, must be found in the part 389, to the left of 23, which is separated from it by a dash. Therefore, dividing 398 by 48, which is three times the square of the tens, the quotient 8 will be the units of the root, or something greater, since 398 is composed of three times the square of the tens by the units, and generally contains numbers coming friom the two other parts. We may ascertain whether the figure 8 is too great, by forming from the 4 tens and 8 units, the three parts which enter into 39823; but it is much easier to cube 48, as has been done in the above table. Now, the cube of 48 is 110592, which is greater than 103823; therefore, 8 is too great. By cabtng 47, we obtain 103823; hence the proposed number is a plerfect cube, and 47 is its cube root. By a course of reasoning entirely analogous to that prvrsucd in treating of the extraction of the square root, we may shew that, when the given number is expressed by more thsu six figures, we must point off the number into periods of three figlm es each, commencing at the right. Hence, for the extraction of the cube root of numbers, we have the following RULE i. Separate the given, number into periods of three figures each, beginning at the right hand; the left hand period will often con tail less than three places of figures. 11. Seek the greatest perfect cube in the fist period, on the left and set its root on the right, after the mannzer of a quotiert in division. Subtract the cube of this number from the first period, and to the remainder bring dowzn the first figure of the next period, and call this number the dividend. (t3AP. VII.J EXTRACTION OF ROOTS. 201 III. Take three times the square of the root just found Jbr a divisor, and see 7Low often it is contained in the ldividend, and place the quotient for a second figure of the root. Then cube the number thus found, and if its cube be greater than the first two periods of the given number, dinzinish the last figure by 1; but if it be less, subtract it from the first two periods, and to the r'emainder bring down the first figure of the next period, for a new dividend. IV. Ta/ce three times the square of the whole root for a new divisor, and seek how often it is contained in the new dividend; the quot;ient will be the third figure of the root. uabe the number thus found, and subtract the result from the first three periods of the given number, and proceed in a similar way for all the periods. If there is no remainder, the number is a perfect cube, and the root is exact: if there is a remainder, the number is an imvperfect cube, and the root is exact'to within less than 1. EXAMPLES. 1. 482528544 Ans. 364. 2. 327054008 Ans. 3002. 3. k/483249 Ans. 78, with a remainder 8697. 4. /916 32508641 Ans. 4508, with a remainder 20644129. 5. 3/32977340218432 Ans. 32068. Extractlion of the JATt: Root of Arunzbers, 142. The nzth root of a number is such a number as being taken n times as a fiactor will produce the given number, n being - ny positive whole number. When such a root can be exactly found, the given number is a perfect 1zth power; all other numbers are imperfect nth powers. Let N denote any number whaltever. If it is expressed by less than n? + I figures, and is a perfect n";i power, its 9tth root will be expressed by a single figure, and may be found by 202 ELEMENMT OF ALGEBRA. LCHAP. VII means of a tabl containing the nth powers of the first ten numbers. If the number is not a perfect nth power, it will fall between two n~th powers in the table, and its root will fall between the 1th roots of these powers. If the given number is expressed by more than n figures, its root will consist of a certain number of tens and a certain nlumber of units. If we designate the tens of the root by a, and the units by b, we shall have, by the binomial formula, N= (a bt)n =a +p nau1b + e 2 1-2b2 +, &c.; that is, the proposed number is equal to the nth power of the tens, plus n times the product of the n - 1 th power of the tens by the units, plus other parts which it is not necessary to consider. Now, as the nth power of the tens, cannot be less than 1 followed by n, ciphers, the last n figures on the right, cannot make a part of it. They must then be pointed off, and the nth root of the greatest nth power in the number on the left will be the number of tens of the required root. Subtract the nth power of the number of tens from the num ber on the left, and to the remainder bring down one figutre of the next period on the right. If we consider the number thus fuund as a dividend, and take n times the (a - 1)th power of the number of tens, as a divisor, the quotient will evidently be the number of units, or a greater number. If the part on the left should contain more than n figures, the n figures on the right of it, must be separated from the rest, and the root of the greatest flth power contained in the part on the left extracted, and so on. Hence the following RULE. I. Separate the nambher N iTnto periods of n figures each, be ginning at the right hcand; extract the nth root of the greatest perfect ntha power contained in the left hand period, it will be the first figure of the root. CItAP. VII.] EXTRACTLON OF ROOTS. 203 11. Subtract this nth power from the left hand period and bring down to the right 6f the remainder the first figure of t]he nex. period, and call this the dividend. [II. Form the n - 1 power of the first figure of the root, multipjl? it by n, and see how often the product is contained in the dividend: the quotient will be the second figure of the root, or'}omething greater. IV. Raise the number thus formed to the ntl power, then subtract this result from the twzo left-hand periods, and to the nere remainder bring down the first figure of the next period: then divide the number thus formned by n times the n - I power of the two figures of the root already found, and continue this operation until all the periods are brought down. EXAMPLES. 1. What is the fourth root of 531441 53 1441 127 24 - 16 4 X 23 = 32 371 (27)4 = 531441. We first point off, from the right hand, the period of four figures, and then find the greatest fourth root contained in 53, the first period to the left, which is 2. We next subtract the 4th power of 2, which is- 16, from 53, and to the remainder 37 we bring down the first figure of the next period. We then divide 371 by 4 times the cube of 2, which gives 11 for a quotient: but this we know is too large. By trying the numbers 9 and 8, we find them also too large: then trying 7, we find the exact root to be 27. 143, When the index of the root to be extracted is a multiple of two or more numbers, as 4, 6... &c., the root can be ob. tained by extracting roots of more siwp7e degrees, successively. To explain this, we will remark that, (a3)4 = a3 X a3 X a0 a3 = a3-++3+3 — a3X4 = a12 and, in general, from the definition of an exponent (am)- am x am X a"n X am... a"'cx' 204 ELEMENTS OF ALGEBRA. [CHAP. VII, hence, the nth power of the rnt:t power of a number is equal to the wnth power of this number. Let us see if the converse of this is alst true. Let -- b then raising both members to the nth power, we have, from the definition of the nth root, and by raising both members of the last equation to the mith power a = bmn Extracting the iznth root of both members of the last equation, we have, m a b; and hence, Ua - a, since each is equal to b. Therefore, the nth root of the mt'" root of any number, is equal to the tmnth root of that number. And in a similar manner, it might be proved that By this method we find that. l. dJi~ 2 5Gzzx'4 5G =; = =4. 3 2. 6/2985984 - 29S9S4 - 1728 = 12. 3 3. 61715 = 17771 = 11; 4. 89 =616 6,. REMARK.-Although the successive roots may be extracted in any order whatever, it is better to extract the roots of the lowest degree first, for then the extraction of the roots of the higher degrees, which is a more complicated operation, is effected upon numbers containing fewer figures than:he proposed number. CHAP. V1..] EXTRACTION OF ROOTS. 205 Extraction of Roots by Approximation. 144, When it is required to extract the nth root of a number which is not a perfect nth power, the method already explained, will give only the entire part of the root, or the root to within less than 1. As to the part which is to be added, in order to corn plate the root, it cannot be obtained exactly, but we can approximnate to it as near as we please. Let it be required to extract the nth root of a whole number, denoted by a, to within less than a fraction; that is, so near, teat the error shall be less than -. We observe, that we can write ap If we denote by r the root of the greatest perfect nth power in aXXp7n rCi apn, the number a, will be comprehended between - a- u (r 1); therefore, the Va will be comprised between the P two aunubers - and; and consequently, their difference -will be greater than the difierence between - and the true P P root. Hence, - is the required root to within less than the. P fraction -: hence, To extract the nth root of a whole number to within less than a fraction -, multiply the number by p"; extract the ntl root ef the prolduct to within less thacn 1, and divide the result by p. Extraction of the nt"h Root of Fractions. 145, Since the nth power of a fraction is formed by raising both terms of the fraction to the nth power, we can evidently find the nth root of a fraction by extracting the nth root of both terms. 206 ELEMENTS OF A TGE3RA. [CHAP. VIL. If both terms arhe aot perfect n'h powers, the exact nth root cannot be found, but we may find its approximate root to within less than the fractional unit, as follows:a Let - represent the given fraction. If we multiply both terms by bn-l it becomes, a a b b"' Let r denote the n"t root of the greatest nth power in a6b"' te abs- rn (r + l)' then bn will be comprised between and (; and consequently, - will be the it, root of a to within less than the fraction therefore, Multiply the numerator by the (n —l)th power of the denomi nator and extract the nth root of the product: D)ivide this rool by the denominator of the given fraction, and the quotient wili be the approximate root. When a greater degree of exactness is required than Tha.t indicatecl by -, extract the nth root of' ab"-l to withir,zny I' fraction -; and designate this root by _. Now, sinet is the root of the numerator to within less than, it fol ws, r' 1 that is the true root of the fraction to within less thar, EXAMPLES. 1. Suppose it were required to extract the cube root o 15I to within less than;> We have 15 X 12 -- 15 X 1728 = 25920. Now, the cube root of 25920, to within less thar. I is t9 hence, the required root is, 29 2 5 12- 12 CHAP. VII.] EXTRACTION OF ROOTS. 207 1 2. Extract the cube root of 47, to within less than 2 We have, 47 x 203 = 47 x 8000 = 376000. Now) the cube root of 376000, to within less than 1, is 72; 72 12 1 hence, 34 — 7 = 2 = 3-, to within less than 20 20' 3. Find the value of 3-25 to within less than.001. To do this, multiply 25 by the cube of 1000, or 1000000000: which gives 25000000000. Now, the cube root of this number. is 2920; hence, 3-25 2.920 to within less than.001. Hence, to extract ths cube root of a whole number to within less than a given decimal fraction, we have the following RULE. Annex three times as many ciphers to the number, as there are decimal places in the requzired root; extract the cube root of the number thus formed to within less than 1, ctnd point opT fSrom the right of this root the required number of decimal places. 146. We will now explain the method of extracting the cube root of a decimal fraction. Suppose it is required to extract the cube root of 3.1415. Since the denominator, 10000, of this fraction, is not a per feet cube, make it one, 1i. multiplying it by 100; this is equiva lent to annexing two ciphers to the proposed dcimal, which then,uomlnes, 3.141509. Extract the cube root of 3141500, that is, elf thew number considered independent of the decimal point to within less than 1; this -'yes 146. Then dividing by 100, o) / 1000000, and we find, v/3.1415 = 1.46 to within less than 0.01. HIence, to extract the cube root of a decimal fractions we have the following 208 ELEMENTS OF ALGEBRA. [CHAP. VII. RULE. Annex ciphers till the whole number of decimal places is equal to t.hree times the number of required decimal places in the root. Thzen extract the root as in whole numbers, and point off the reTvired number of decimal places. To extract the cube root of a vulgar fraction to within less than a given decimal fraction, the most simple method is, To reduce the proposed fractionz to a decimal fraction, continuing the division until the number of decimal places is equal to three times the number required in the root. The question is then reduced to extracting the cube root of a decimal fraction. Suppose it is required to find the sixth root of 23, to within less than 0.01. Applying the rule of Art. 144 to this example, we multiply 23 by (100)6, or annex twelve ciphers to 23: then extract the sixth root of the number thus formed to within less than 1, and divide this root by 100, or point off two decimal places on the right: we thus find, 6g/23 1.68, to within less than 0.01. EXAMPLES. i. Find the 7-3~ to within less than 1 Ans. 73 2. Find the 39 to within less than.0001. Ains. 4.2908. 3. Find the 6S13 to within less than.01. A/ns. 1.53. 4. Find the 3 3.00415 to within less than.0001. Ans. 1.4429. 5. Find the 3/0.00101 to within less than.01. Ans. 0.10. 6. Find the 3: to within less than.001. Ans. 0.824. C(IIAP. VII.] EXTRACTION OF ROOTS. 209 Extraction of Roots of Algebraic Quantities, 147. Let us first consider the case of monomials, and in order to dedlce a rule for extracting the n'-h root, let us examile the law for the formation of the nth1 power. From the definition of a power, it follows that each fitctor of the root will enter the power, as many times as there are uits in the exponent of the power. That is, to form the 7tn' power of a monomial, Wye form the n'4h ower of the co-efficient for a new co-efficient, and write after this, each letter affected with an exponent equal to n times its primitive exponent. Conversely, we have for the extraction of the n't root of a mnonomial, the following RULE. Extract the nt" root of the numerical co-efficient jor a new coeflicient, and after this write each letter affected with an exponent equal to I-th of its exponent inb the given monomial; the result n uill be the required root. Thus, 3/49 3c6 - -4a36C2 anlld V 16a8b12e4 = 2a2b3C. From this rule we perceive, that in order that a monomial may be a perfect nt" power: 1st. Its co-efficient must be a perfect nta power; and 2d. The exponent of each letter must be divisible by bn. It will be shown, hereafter, how the expression for the root of a quantity, which is not a perfect power, is reduced to its simplest form. 148. HIitherto, irn finding the power of a monomial, we have paid no attention to the sign with which the monomial may be affected. It has already been shown, that whatever be the sign of a monomial, its square is always positive. 14 210 ELEMENTS OF ALGEBRA. [CHAP. VIL Let n be any whole number; thenr every pc wer of an even degree, as 2n, can be considered as the ntk power of the square; that is, (a2)" = a2": hence, it follows, hiat every power of an even degree, will be essentia(lly posi yet, wlhether the quantity itself be positive.)r negative. ThIS, (42a263c)4 =+ G 6aSb12c4 Again, as every power of an uneven degree, 2n + l, is but the product of the power of an even degree, 2n, by the first power; it follows that, Every power of1 a monomial, of an uneven degree, has the same sign as the monomial itself. IHence, (+ 4a2b)3 = J- 64a63; and ( — 4a2b)a = 64a6b3. From the preceding reasoning, we conclude, 1st. That when the index of the root of a monomial is uneven,!the root will be affected with the same sign as the monomial. Thus, 3/+8 a= + 2a; a83 = a - 2a; V=-332aO1b 5 — _ 2a2b 2d. When the index of the root is even, and the monomial a,positive quantity, the root has both the signs ~ and -. Thus, 4 1a4b12 - 4- g3ab3; 3. 3d. TWhen the index of the root is even, and the monomial v.ega. tive, the root is impossible; For, there is no quantity which, being raised to a power of an even degree, will give a negative result. Therefore, V -a, 6 b, v:are symbols of operations which it is impossible to e secue'They are imaginary expressions. EXAMPLES. 1. What is the cube root of 8a6b3cT2? Ans. 2a4bc, -2. What is the 4th root of 81a4bSc'6 Ans. 3ab2c4. 3. What is the 5th root of - 32ab5c'ld'5? Ans. — 2ac2d3. 4. What is the cube root of - 125a9b6c3. Ans. — 5aC32c. CHAP. VII.] EXTRACTION OF ROOTS. 211 Extraction of the nt Root of PotlnomialTs. 148. Let 1N denote any polynomial whatever, arranged witg reference to a certain letter. Now, the nt, power of a polynomial is the continued product arising from taking the poly. nomial n times as a factor: hence, the first term of the product, wI;en arranged with reference to a certain letter, is the n'ri power of the first terra of the polynomial, arranged with reference to the same letter. Therefore, the nt" root of the first term of such a product, will be the first term of the nh root of the product. Let us denote the first term of the nt1' root of NT by r, and the following terms, arranged with reference to the leading letter of the polynomial, by r', r", r"', &c. We shall hav e, - = (r + r' + r " +-.. &c.)'; or, if we designate the sum of all the terms after the first oy s, N2 = (r + s)n yr + ~nr"-ls + &c., = r + nrfl-1(r' + r" + &c. ) + &c. if now, we subtract rn from A, and de signate the remainder by R; we shall have, R _N T - r -_ nrl- r' -f nr"l- r" + &c., which remainder will evidently be arranged with reference to the leading letter of the polynomial; therefore, the first term will contain a higher power of that letter than either of the succeeding terms, and cannot be reduced with any of them. Thence, if we divide the first term of the first remainder, by n times the (a - 1P)' power of the first term of the root, the quotient will be the second term of the root. If now, we place r + r' = ut, and denote the surm of the sub. maedng terms of the root by s', we shall have, N = (,u + s') - u t + Z-1S' + &M.. 212 ELEMENTS OF ALGEBRA, [CHAP. VIL If now, we subtract utn from N, and den:te the remainder by N', -e shall have, B' - N- a" = n(r + r2')n-is' - &c., = qr-l(r" ~ + " t &C. ) - &c., = n2~-lrt' -t- &Co h1' we divide the first term of this remainder by n times the (n —-l)t" power of the first term of the root, we shall have the third term of the root. If we continue the operation, we shll find that the first term of any new remainder, divided by n times the (n - l)th power of the first term of the root, will give a new term of the root. It [ may be remarked, that since the first term of the first remainder is the same as the second term of the given poly. rnomia], we can find the second term of the root, by dividing the seiond term of the given polynomial by n times tlhe (n-l)'h power of the first term. Hlence, for- the extriaetion of the nth root of a polynomial, we have the following RULE. L Arrange the given poolynomial with reference to one of its letters, wad extract the nth root of the first term; this will be the first term of the root. II. -Divide the second term by n times the (n - 1)t power of the first term of the root; the quotient will be the second term of the root. IM. Subtract the n`t power of the sum of the two terms already found from the given polynomial, and divide the first term of the remainder by n times the (n- 1)t power of the first term of the root; the quotient will be the third term of the root. IV. Continue this operation till a 1remainder is fiound equal to O, or, till oe is found whose first term is not divisible by n times the (n - 1)th power of the first term of the root: in the former case the root is exact, and the given polynomial a pefect rnh power; it the latter case, the polynomzial is an imp2eefect *th power. EIIAP. VII.] EXTRACTION OF ROOTS. 213 149. let us apply the foregoing rule to the following EXAMPLES. I. Extract the cube root of 6 — 6X5 -]- 5t'4-20x3+15xz —6 1. 6-6X5+ 15x4 -20X3+z 15X2- Ox+ l x2- 2x+ (2 —2x)3- 6-_X5 t+1 2,4- 8x3 3x4 Ist rem. 3x4 — 1~3+ &c. (2 — 2x 1- )3- 6 _6-X5 + 1 5x4-20x3 + 15x2 -O+ 1. In this example, we first extract the cube root of r6, which ives x2, for the first term of the root. Squaring x2, and multiplying by 3, we obtain the divisor 3x4: this is contained in the second term - 6x5, -2x tinmes. Then cubing the part of the root foundl, and subtracting, we find that the first term of the remainder 3x4, contains the divisor once. Cubing the vwhole root found, we find the cube equal to the given polynomial. Hence, x2 - 2x + 1, is the exact cube root. 2. JFind the cube root of x6 - 6x5 - 40x3 + 96x - 64. 3. Find the cube root of 8X6 -- 12x5 + 30x4 - 25x3 + 30x2 - 12x -- S. 4. Find the 4th root of 16a4 - 96a3x + 216a2x2 - 216ax3 + 8sx4 la-96a3x+ a2216-2 GaX- 3 +Sl 81X4 1 a_ (2c - 3X)4 — l a4 -9Oc3x+2 1a2X2-210aX3+8X4 S 4 X (2(a)3- 32a3. We first extract the 4th root of 16a4, which is 2a.: We then raise 2a to the third power, and multiply by 4, the index of the root; this gives the divisor 32a3. This divisor is contained in the second term -96a3, - 3x times, which is the second term of the root.% IRaising the whole root found to the 4th rower we find the power equal to the given polynomial. 5. What is the 4th root of the polynomial, 81ac4 + 1Gb4d4 -- 96a2cb3d3.. 216a6C3bd + 216atc2%2d2. C;. Find the 5th ).oot of 33q5 -- 8Ox4 t+ S0`_ - 40x2 + 10O - 1. 214 ELEMESI'TS OF ALGEBRA. LCHAP. VIL Transformation of Radicals of any Degree. 150. The principles demonstrated in Art. 104, are general. For, let'Va- and' 6, be any two radicals of the n-th degree, ard dienote their product by p. WX e shall have, Sx V=r p - - - - (1). By raising both mcmbers of this equation to the n' power, we find (nV,)7L x (x' b)n =Jpn) or ab = p; Ywhence, by extracting the n0" root of both members, "a-p - - - (2). Since the second members of equations (1) and (2) are the same, their first members are equal, whence, a! X / -n a: hence, 1st. The poclduct of the nt' r'oots of two quantities, is equal to the n"e root of the product qf the quantities. Denote the quotient of the given radicals by q, we shall have n"a V = q.. (1); and by raising both members to the,'th power, Q' a) - or - "; whence, by extracting the nt' root of the two members, we ha ye, n / -- -.. (2).'The second members of equations (1) and (2) being the same, their first members are equal, giving'V~a ~ 7-hence, 2d. The quotient of the n"h roots of two quantities, is equal to thc ut' root of the quotient of the quantities. CEHAP. VII.1 TRANSFOPrMATION OF RADICALS. 215 151. Let us apply the first principle of article!50, to the simpliiication of the radicals in the following RXAM PLES. 1. Tale the radical Y3/54ab'Lc2. This may be NNitten: 54-c2 -_ 73 a2 - 3a 2. 2. In like manner, = 8a2 -. 2 V 2 and ~ 8 4a%58cI6 2ab26 c I4C2; 3. Also, 6 9 2 4c12 X - 3a b a2c26 6 In the expressions, 3SO ab 2, 2 a2, 2ab2c4 V 2, each quantity placed before the radical, is called a co-efficient of the radical. Since we may simplify any radical in a similar manner, we have, for the simplification of a radical of the nh' degree, the following RULE. Resolve the.quantity under the radical sifgr into tzoo factors, one of which shall be the greatest pe/ffect nth powter which enters it; extract the'nth root of this factor, and write the root without the radical sign,' under which, leave the other fctor. Conversely, a co-ef!cient nmay be introduced tunder the radical sion, by simply raising it to the n"t power, and writing it as a factor under thte radical sign. Thus, 3ab 3:22 -- — 73b3 X -2 - _ac2 -:.a4b C2. 152. By the aid of the principles demonstrated in article 1.43, vwe are enabled to mlie another kind of simplifi.:ation. Take, for example, the radical 6/4a2; from the principles re, ferred to, we hale, s /4a2 2 /v4a, 216 ELEMENTS OF ALGEBRA. [CHAP. VII. find as the quantity under the radical sign of the second degree is a perfect square, its root can le extracted: hence, 6 4a2 2a. hi like manner, In general, r —-- z =/ that is, when the index of a radical is a multiple of any ni-mber 7, and the quantity under the radical sign is an exact at" power, ITe can, withoTut chaningi the vczlue of the radical, divide its izdex by n,, and extract the nt"' root of the qzantity under the sign. 153. Conversely, The icldex of a racdical may be multipllecd by any number, _provided we raise the q.antity under the sign to a power of which this msember is the exponent. For, since a is the same thing as "a" we have, 154. The last principles enable us to reduce two or more radicals of different degrees, to equivalent radicals having a coinnon index. For example, let it be required to reduce the two radicals,2a and,(a+ b) to the same index. By multiplying the index of the first by 4, the index of the second, and raising the quantity 2a to the fourth power, the1n multiplying the index of the second by 3, the index of the first, ancl cubing a 4- 6, the value of neither radical..will bp changed, and the expressions will become 2a -a =1 24a-11)a4; anld + (a -+,) and similarly for other radicals: hence, to reduce radicals to a common index, we have the following CHAP. VIi.] TRANSFORMATION OF ZRADICALS. 217 RULE. [iultl2ply the index of each, radical by the Zproduct of the indices of all the other radicals, and raise the quantity under each radicag 3ign to a power denoted by this product. This rule, which is analogous to that giver: for the reduction of fractions to a common denominator, is susceptible of similar modifications. For example, reduce the radicals to a common index. Since 24 is the least common multiple of the indices, 4, 6, and 8, it is only necessary to multiply the first by 6, the second by 4, and the third by 3, and to raise the quantities under each rad ical sign to the 6th, 4th, and 3d powers, respectively, which gives 4F a 24; 6 - 24 8 2-b2 (a2+62)3 Acldition and Subtraction, of Radicals of any Degree. 155. We first reduce the radicals to their simplest form by the aid of the preceding rules, and then if they are similar, in order to add them together, we acid their co-egficients, and after this sum write the common radical; if they are not similar, tho addition can only be indicated. Thus, 33 / + 2 /- = 5. EXAMPLES. 1. Find the sum of /48b2 and b -75a. Ans. 9b'a, 2. Find the sum of 3 42 and 23. Ans. 5 3 2a 3. Find the sum of 2 /45 and 3. Ans. 9'155'. In order to subtract one radical from another wh(e they are similar, Subhtract the co-efficient o the subtrahendfrom the co-efficient ra e minulendl, aznd write tI is di/Terence before the common radela4 218 ELEMEN-TS OF ALGEBRA.,C.1HA.P. VII, SThus, Oa 46 - 2c44-(3a- 2c) 4; but, 2ab d - 5ab are irredacible. 1. From VSac3b +- NO14 subtract 3b4 + 2b3. Ans. (2a - b) 3/ h5-)2 2, From 3 4a2 subtract 2V 2a. Ans. 1'a. lffultZlztcation of ]Radicals of aCy Degree. 156. We have shown that all radicals may be reduced to equivalent ones having a common index; we therefore suppose this transformation made. Now, let a J/ and c" /- denote any two radicals of the same degree. Their product may be denoted thus, a V x cx; or since the order of the factors may be changed without affecting the value of the product, we may write it, acx X >< V- or (Art. 150), since nd X n =nd'; we have finally, a( b X c d=ac bd; hence, for the multiplication of radicals of any degree, we have the following RULE. 1. 1Reduce the radicals to equivalent ones having a common index. II mluliplyV the co-ec'ienlts together for a new co-egicient; after this write the radical sign with, the cozmmzon, index, placing u2nder it the prodact of the quantities unzder the radical signs in tle two fcitors; thle result is the produzct reqaired. EXAMPLES. 1. The product 2 a 3 aI2 + (a2 + b2)2 -a3 (a2 + 6( 3 2a / X-3 -- Cja c d cd _a2 (a2 + 62) 3q /v d CHItAP. VII.] TRANSFORMATION OF RADICALS. 219 2. Th-e product 3a /8a x 2b 4uC Gab 4J32;a4c - 1226 1/2. 3. The product 3 3 / 3 1 13 _ X 4- 7 16 21 4. The product 3a 6 b X 5b 2 = 152- X 24 b4c. _ 3 5. Multiply 2x by -x xs 6. Multiply 2f 15 by 33 /10. Ans. G 6 337500. 7. Multiply 4 byv 2. A,2S. 8 V 8. Multiply.2, 3, and 4, together. _Ans. 12 64800G.?'4 3 1 D. Multiply, ad 1'6, together. 42 /2 Ans. V. 27@ 10. Multiply (4 +. 5 -) by (\-/ 2- ). Anzs. + 4' Division of Radicals of any Degree. 157. We will suppose, as in the last article, that the radieala have been reduced t'o equivalent ones having a common inlex. Let a b- and cf-d represent any two radicals of tho Ua, degree. The quotient of the first by the second may be written, a f_ Xa cVd C d 220 ELEMENTS OF ALGEBRA. [CHAP. VIL or, since V - - (ANt. 150), h-e have, ca ac n b I d C Hience, to divide one radical by another: we Tave the Ail lowing RULE. I. Reduce the radicals to equivalent ones having a common index, II. Divide the co-egicient of the dividend by that of the divisor for a new co-efficient; cfter this write the radical sign with the common index, and p2lace underl it the quotient obtained by dividing the quantity under the radical sign in the d(ividend by that in the divisor; the result will be the. quotient required. EXAMPLES. i. What is the quotient of c V3a2b2 + b4 divided by d - C b2 3 +b~ C 3 /8ba(;L22 + b4) 2cb3 /(t2 + b2 d 3 2t - d a2 - b2 -d a2 -b_2 Sb 2. Divide 2 X 3 4 by 2 4 X3 3. Anls. 4 /,28. 3, Divide 3 by 4 1 12 4. Divide - -/- by (+ 3 ). Ans. 10 5. Divide 1 by 4 -+4 b. Ans. - a- CHAP VI..I TRANSFORMtATION OF RADICALS 221 f. Divide 4a, +4 46 by 4a-11b. izs. -+ b 2 ab + 2 a a- b Ibormnation of Powers of Radicals of any Degree. 158. Let a n/1-b; represent any radical of the nIth degree. then we may raise this radical to the nth power, by taking it m times as a factor; thus, a it lb X a n 1 b - - a - c t a But, by the rule for multiplication, this continued product is equal to am bn/; whence, (a tb )m = an n; (1). We have then, to raise a radical to any power, the following RULE. Raise the co-efficient to the required power for a new co-effcient; after this write the radical sign with its primitive index, placing under it the required power of the quantity under the radical sign in the given expression; the result will be the power required. EXAMPLES. 1. ( /4a3)2-=,/(4a3)2 — 4iOa6- 2a4 a- 2a /5-7,. 2. ( 3.2) -= 35 (2a)5 24 3'2a = 496a3 When the index of the radical is a multiple of the exponent of the power to which it is to be raised, the result can be simplified. For, -2a - /.a/ (Art. 152): hence, in order to square ~/2a, we have only to omit the first radical sign, which gives (4>)2 = 2a Again, to square 69 6 we have 6 36 3: hence, (6 36)2 = 36; herce, 222 ELEMENTS OF ALGEBRA. [CHAP. VII, When the indes of the radical is divisible by the exponent of the power to which it is to be raised, perform the division, leaving the quantity under the radical sign unchanged. Extraction of Roots of Radicals of any Degree. 159. By extracting the mth root of both members of equation (1), of the preceding article, we find, a tm X /6tlt = a Uc; Whence we see, that to extract any root of a radical of any degree, we have the followingn RULE. Extract the required root of the co-efficient for a new co-efficient; after this write the radical sign with its primitive index, under which place the required root of the quantity under the radical sign in the given expression; the result will be the root required. EXAMPLES. 1. Find the cube root of 8 4s. Ans. 2 43. 2. Find the fourth root of 6253 5ns. - 4. 159*. If, however, the required root of the quantity under the radical sign cannot be exactly found, we maly proceed in the following manner. If it be required to find the'mth root of cA/ the operation may be indicated thus, t mn whence, by substituting i the previous-d butmt "4 -d I-'"'5/ whence, by substituting in the previouz equation, Consequently, when we cannot extract the required root of the quantity under the radical sign, CHAP VII.] TRANSFORMATION OF RADICALS. 223 Extract the required root of the co-eicient for x new co-eficient; after this, write the radccal signr, with an index equal to the pro. duct of its primitive index by the index of the required root, leaving the quantity under the radical sign unchanged. EXAMPLES. 1 59 =c' 32c; and, l c 6 When the quantity under the radical is a perfect power, of ihe adegree of either of the roots to be extracted, the result can be simplified. Thus, 2 St. 8a3 4 a In like manner, 2/ 9= 9a 5a- _ 2. Find the cube root of Ails.. 3. Find the cube root of -b- b Ans. - 62ab DIsfferent Roots of the same Power. 160, The rules just demonstrated depend upon the principle, that if two quantities are equal, the like roots of those quantities are aleso ecual. This principle is true iso long as we regard the term root in its general sense. but whel the term is used in a restricted sense, it requires sorlme modification. This modification i particulirly necessary in operating Lupon imaginary expressions, w-lic}h are lnot roots, strictly speaking, but mere indications Qf operations which it is impossible to perform. Before pointing out these modifications, it will be shown, that every quantity has more than one cube root, fourth root, &c, It has already been shown, that every quantity has two square roots, equal, with contrary signs. 224 ELEMENTS OF ALGEBRA. [CHAP. VII. 1. Let x denote the general expression for the cube root of a, and let p denote the numerical value of this root; we have the equations x3= a, and X3=p3. The last equation is satisfied by making x =p. Observing that the equation x3 = p3 can be put under the form ~ -- p3 - 0, and that the expression x3-p3 is divisible by -2p, giving the quotient, x2 + px + p2, the -above equation can be placed under the form (X -p) (X2 +p + p2) - = 0 Now, every value of x that will satisfy this equation, will satisfy the first equation. But this equation can be satisfied by supposing — p = 0, whence, x =p; or by supposing x +~ px 2 = 0, from which we have, X -- - — 3, or x ( -2 hence, we see, that there are three dijerent algebraic expressions for the cube root of a, viz: -1+ /-) and (, Y x 2' 2 2. Again, solve the equation x4 = p4 in which p denotes the arithmetical value of V/a~ This equation can be put under the form x4 - 4 = 0; which reduces to (x2 - 2) (2 + p2) = 0; tmrd this equation can be satisfied, by supposing x2 -p2 0; whence, xI= p; or by supposinglo Xz l p2 -0 whence, z = =_ p2_p. CHAP. VII.] TRANS]EoRMATION OF RADICALS. 225 We therefore obtain four different algebraic expressions for the fourth root of a. 3. As another example, solve the equation X6 - p6 = 0. This equation can be put under the form (X3 - 3) (x3 + p3) =0; which may be satisfied by making either of the factors equal to zero. But, x3 - p3 = O, gives x=p, and x = I_ 2 And if in the equation x3 +p3 -0, we make p -- p', it becomes x3 _'3 = 0, from whici we deduce x = p and x = ( I a:i or, substituting for p' its value -p, x=-p, and x-p 2 ) Therefore, x in the equation X6 _p6 = 0, aund consequently, the 6th root of a, admits of six diferent alg& braic expressionzs. If we make a -+ and a - 2' 2 these expressions become p, ap, at, -p,-ap,-atp. It may be demonstrated, generally, that there are as many different expressions for the nth root of a quantity as there are units in n. If n is an even number, and the quantity, is positive, two of the expressions will be real, and equal, with contrary signs; all the rest will be i.?ginary: if the quantity is, negative, they will all be imaginary. 15 226 ELEMENTS OF ALGEBRA. LCH1 P. VIL If is is oddl, mle of the expressions will be real, and all the rest will be imaginary. 161. If in the preceding article we make a = 1, we shall find the expressions for the second, third, fourth, &c., roots of 1. Thus, + 1 and — 1 are the square roots of 1. Also, + 1, - and are. the cube roots of 1: And + 1, 1, 1, / — and -— 1, are the fourth roots of 1) &c., &c. -Rutes for Imaginary.Exyressions. 162. We shall now explain the modification of the rules for operating upon radicals when applied to imaginary expressions. The product of V-a by V/ —, by the rule of Art. 156, would be /+ a. Now, +/ a2 is equal to +- a, whence there is an apparent uncertainty as to the sign of a. The true pro-.duct, hcwever, is - a, since, from the definition of the square root of a quantity, we have only to omit the radical sign, to -olbtain the quantity. Again, let it be required to form the product,By the rule of Art. 156, we shall have V-a — X /- -- +;':but the true result is — a/, so loing as both the radical3 — a and / —b are afficted with the sign +. -For, la-a= a.V-\l; and -= v. i x -a. 1 (6 x - z -: / -6 CHAP. VII.] TRANSFORMATION OF RADICALS. 227 i! a similar manner, we treat all other imaginary expressions of' Lhe second degree; that is, we first reduce them to the form of a V/'-l1, in which the co-efficient of - 1 is real, and then proceed as indicated in the last article. 162*. For convenience, in the application cf the preceding F!iilciple, we deduce the different powers of A/-1, as follows: k V_3 ( -1)X - I -, (.V ( - 1)4 - )2 X ) 1) = 2 + 1 The fifth power is evidently the same as the first power; the sixth power the same as the second; the- seventh the same as the third, and so on, indefinitely. 163. If it is required to find the product of 4/ —a and 4/ -b, we should get, by applying the rule of Art. 156. 4/ —a 4 X — b -- + ab, but this is not the true result, Fpor, placing the quantities under the form 4 XV -1 and 4 X4-, and proceeding to form the product, we find 4 -a- X b)2.T a X' —*=4.X bX (V/ —1>2 V —>>< V —1, since, (' l-J)2= ( / —) =~ —1 from the definition of a root. HIence, generally, when we have to apply the rules for radieals to imaginary expressions of the fourth degree, transform thems, so that the only factor under the radical sign shall be - 1, and then proceed as in the above example. -+ -— 3 Let us illustrate this remark, by showing that 2 is an expression for the cube root of 1, or that, ir the restricted sense, it is a cube root of 1. 228 ELEMENTS OF ALGEBRA. I CRAP. VI!. We have \ 2,/\ 2 (-) +3. (-).. +3. (-1 2). (2 2), + (I 3). (i) =_ 8-.1. I + 3.-3 3 X - - 3 V3 - In like manner, we may show, that is another expression for the cube root of 1, when understood in the restricted sense. It may be remarked that either of these expressions is equal to the square of the other, as may easily be shown. Of Fractional and Negative Exponents. 164. We have yet to explain a system of notation by means of which operations upon radical quantities may be greatly simplified. We have seen, in order to extract the nth root of the quantity am, that when ILm is a multiple of n, we have simply to divide the exponent of the power, by the index of the root to be extracted, thus, a am = a. When m is not a multiple of n, it has been agreed o retain the notation. am a, these two being regarded as equivalent expressions, and both indicating the nth root of the mth power of a, or what is the same thing, the mth pc ver of the nth root of a; and generally, When. any quantity is written with a fractional exponent, the;n.imerator of the fraction denotes the power to which the quantity is to be raised, and the denominator indicates the root of this xower which is to be extracted, CHAP. VII.] THEORY OF EXPONENTS. 229 165. We have also seen that am may be divided by an, when ra, and n are whole numbers, by simply subtracting it from rn, giving - a"" = aP; a" in which we have designated the excess of m over n by p. Now, if n exceeds m, p becomes negative, and the exact division is impossible; but it has been agreed to retain the notation - a"-" -= a-p. But when m < n, in the fraction, am a, we may divide both terms by am', and we have a"n a —- = ap; hence, a-P is equivalent to -, and both denote the recipro. cal of aP. We have, then, from these principles, the following equivalent expressions, viz.:?ij/ah equivalent to an. (v(t~f or; ( )" a" a.m. an 1 " /_ or / a /am a 166. It has been shown above that a'; if now we divide 1 by both members of this equation, we shall have, e= =-: heune we conclude that, 230 ELEMENTS OF ALGEBRA. [CHAP. VII.Any factor mnay be transferred from the numrerator to the denominator, or from the denominator to the numerator, by changing the sign of its exponent. 167. It may easily be shown that the rules for operatiirg upon quantities when the expolents are positive whole lunblors, are equally applicable when they are firactional or negative. In the first place, it is plain that both nuineratoi and denominator of the fractional exponent may be multiplied by the same quantity without altering the value of the expression, since by definition the mth power of the mnth root of a quantity is equal to the quantity itself. This principle enables us to reduce quantities, having fi'rtional exponents, to equivalent cnes having a common denominator. Let it be required to find the product of a' and a,, m r ms n.r1 We have, a~ X as = an X a"s' or (Art. 164), n -a2is- X n" -an7 = f/afiSs + n7ims -+ 7r This last result is equivalent to a I'S' hence, mn r ms-7ns r an X a, -s a ns the same result that would have been obtained by the application of the rule for the'multiplication of mlonomials, when the exponents are positive whole numibers. If both exponents are negative. we shall have, _2 _ L 1 1 1 _ m+ "r a X)( X -= -- - = a _& - in r S11+ j11r a1' as a ns if one of the exponents is positive, and the other negati-e Ye shall have, m r m 1 rns 1 a X a, 8= a" X - ans Xas ac" _?aS 6XS~~1~ CHAP VIIi.] THEORY OP EXOIONENTS. 231 n / S M m8- nr whence, ns m X n asnr - a n"S Xa a 7F — a,, — aa'0m rn r ils - 71r and finally, aX a sa "a We have) therefore, for the multiplication of quantities when the exponents are negative or fractional, the same rule as when they are positive whole numbers, and consequently, the same rule for the formation of powers. EXAMPLES. 3 _1 23 I 1 12 81. a4ab 2C- ( X a2b3c' = a 4 b6-c' 2. 3a-2bz X 2a 562c2 = 6a 5 b6c2 1- 1 b13. 2b4c-m X 5a3b-_cn = 30a 6b6-cn-. 4. Find the square of -a3.." ~ X2 2 7yWe have, (a (2)2 X a 9a 5. Find the cube of 3a2. ns. a 168. Let it be required to divide all by a3. W\e shall have, a: m. - m- nan a - =al X a s or (Art. 167), - a ns as as If both exponl its are negative, m m r rn- ms -- = a " X a' =a "s a by the last artice. r a ~ one exponent is negative, a m = a" X i' = a ns' by the preceding article. a R 232 ELEMENTS OF ALGEBRA. [CHAP. VII. Hence, we see that the rule for the division of quantities, with fractional exponents, is the same as though.he exponents were positive whole numbers; and consequently we.,have the same rule for the extraction of roots, as whert the exponents are positive whole numbers. EXAMPLES. 2 _ 7 1a. a a -a _ 4 A 4 1 2. ad.5 a4 = a25- 3.a X b~ — 1a - a 1 5 ] 3 4. Divide 32a'b%6c2 by 8a6b-c 2. Ans. 4aJbc. 3 3 3 5. Div'de 64a9b6c 5 by 32a- 9b 2c. Ans. 2a'sb6. 3. X 4 8 a 2 a3=a9; 7. a al. 3 _ 3 3 8. X4 =a 8 9. a a5b. 169. We see from the preceding discussion, that operations to b)e performed upon radicals, require no other rules than those previously established for quantities in which the exponents are entire. These operations are, therefore, reduced to simple oper ations upon fractions, with which we are already familiar. GENERAL EXAMPLES. 1. Reduce 22Jf/x(3)~ 1. Rredv/uce - to its simplest terms. Acns. 4 3,. 2. Reduce ~ (2) to its simplest terms. Ans. 384 CHAP VII.] THEORY OF EXPONENTS. 233 a3. RIeduce (e d1\/ to its simplest terms. 2/4 (;)2__ __ Ans. 4 4+I 4.'What is the product of. ~35 1. 32 b1 4 5 a2 + a2b8 + a2bb3 - b ab + a- b3, by a2-b Ans. a3 - b2. 5. Divide a3 —a2b - ab -6 bx, by au -b Ans. a2 - b. 170. If we have an exponent which is a decimal fraction, as. for example, in the expression 10' 301 from what has gone be 301 fore the quantity is equal to (10)i~~~' or to 1000~(1)3'01), the value of which it would be impossible to compute, by any process yet given, but which will hereafter be shown to be nearly equal to 2. In like manner, if the exponent is a radical, as 3, 11-, &e., we may treat trhe expression as though the exponents were fractional since its values may be determined, to any degree of exactness, in decimal terms. CHAPTER VIL. OF SERE —-AR THMETICAL PROGRESSION-GEOMETRICAL PROPO1IION AN]D PROGRESSION —RECURRING SERIES-BINOMIAL FORMULA —SUMMATION OF SERIES-PILING SHOT AND SHELLS. 171. A SERIES, in algebra, consists of an infinite number of terms following one another, each of which is derived from one or more of the preceding ones,by a fixed law. This law is called the low of the series. Arithmetical.Progression. 172. An ARITHMETICAL PROGRESSION is a series, in which each term is derived from the preceding one by the addition of a constant quantity called the common, dierenzce. If the common difference is positive, each term will be greater than the preceding one, and the progression is said to be in creasing. If the common difference is negative, each term will be less than the preceding one, and the progression is said to be decreasing. Thus,... 1, 3, 5, 7,... &c., is an increaszng arithmetical progressioz, in which the common difference is 2; and.... 19, 16, 13, 10, 7,... is a decreasing arithmetical przgression, in which the common difference is - 3. 1.73. When a certain number of ternms of an arithmetical progression are considered, the first of these is called the first terem of the progression, the last is called the last term of the progression, and both together are called the extremes. All the terms between the extremes are called arithmetical means. An arithmetical progression is often called a progression by differences. CHAP. VIII.] ARITHMETICAL PROGRESSION. 2385 174. Let d represent the common difference of the arithmetical progression, a. b. c. e. f. g. h. k, &c., which is written by placing a period between each two of the terms. From the definition of a progression, it follows that, b:= a +- d, c b + d = a + 2d, e = c + d = a +- 3; sid, in general, any term of the series, is equal to the first ter-ml plies as many timnes the cozmmon difference as there are pre.ceding terms. Thus, let I be any term, and n the number which marks the place of it. Then, the number of preceding terms will be deuoted by n - 1, and the expression for this general term, will be z = a + (n -l) d. If d is positive, the progression will be increasing; hence, In an increasing arithmetical progression, any term is equal to the first term, plus the product of the common difference by the number of preceding terms. If we make n = 1, we have I — a; that is, there will be but one term. If we make n- = 2, we have I = a - d; that is, there will be two terms, and the second term is equal to the first plus the common difference. EXAMPLES, 1. If a= 3 and d- 2, what is the 3d term? Ans. 7. 2. If a=5 and d 4, what is the 6th termll? Ais, 25. 3. If a = 7 and d- 5. what is the 9th term? Ans. 47. The formulae I = a + (n - ) d, serves to find any term whatever, without determining those which precede it. 236 ELEMENTS OF ALGEBRA. [CHAP. III. Thus to find the 50th term of the progressi(n, 1. 4. 7. 10. 13. 16. 19, we have, 1= 1 +49 X 3 = 148. And for the 60th term of the progression, 1. 5. 9. 13. 17. 21. 25,... we have, 1= 1 + 59 x 4 = 237. 174', If d is negative, the progression is decreasing,, and thle formula becomes I = a-(n - 1)d; that is, Any term of a decreasing arithmetical progression, is equal to the first term plus the product of the common dziereznce by the?number of preceding terms. EXAMPLES. 1. The first term of a decreasing progression is 60, and the common difference - 3: what is the 20th term? I = a-( - ) d gives = 60-(20 - 1)3 = 60 -57 = a. 2. The first term is 90, the common difference — 4: what is the 15th term. Ans. 34. 3. The first term is 100, and the common difference — 2' what is the 40th term? Ans. 22. 175. If we take an arithmetical progression, a. c..b.. i. k. 1, having n terms, and the common difference d, and designate the term which has p terms before it, by t, we shall have t-=+pd - - - - - (1) If we revert the order of terms of the progressicn, considering I as the first term, we shall have a new progression whose common difference is -d. The term of this new progression which has p terms before it, will evidently be the same as that which has p terms after it in the given progression, and if we represent that term by t', we shall have, i, -pd -. 2). CHAP. VIIIj ARITHMETICAL PROGRESSION. 237 Adding equations (1) and (2), member to member, we find t + t' = a + I; hence, The sum of any two terms, at equal distances from the extremes of an arithmetical progression, is equal to the sum of the extremes. 176. If the sum of the terms of a progression be represented by S, and a new progression be formed, by reversing tha order of the terms, we shall have S== a - b - c.... - i - ~ + 1, S= I + k i+.... + c b + a. Adding these equations, member to member, we get 2 ~(a + I)+ (b k) (c+ i) +(i + c) + (k + b)+ (I+a); and, since all the sums, a + 1, b + k, c +- i.... are equal to each other, and their number equal to n, the number of terms in the progression, we have 2S-(a + 1)n, or S= (a l) I; that is; The sum of the terms of an arithmetical progression is equal to hayf the sum of the two extremes multiplied by the number of terms. EXAMPLES. 1. The extremes are 2 and 16, and the number of terms 8: what is the sum of the series? S=(a) X ), gives S- 2 X 8 —72. 2. The extremes are 3 and 27, and the number of terms 12a what is the sum of the series? Ains. 180. 3. The extremes are 4 and 20, and the number of terms 10: what is the sum cf the series? Ans. 120. 4. The extremes are 8 and 80, and the number of terms 10: what is the sum of the series? Ans. 440. The formulas I = a + (n-1) d and S = a n, 2-/~ 2 238 ELEMENTS OF ALG(EBRA. [CHAP. VIII, contain five quantiticn, a, d, n, 1, and S, and consequently give rise to the following general probleml, viz.: Any three of these five quantities being gzven, to determine the other two. This general problem gives rise to the ten following cases - No. Given. Unknown. Values of the Unknown Quantities. 1 a, d,, S 7 = a (n-1) dC; S- n [2a + (n -l1) d]. I-a (I + a) (i-a -+ d) 2, d, S n, S 12- ( -; 2s- d —2a zi//V(d - a)2 +SdS 3.a, dS n, d n-,Sn, nladI; I n —a (n- ()d. 4 1 n, 1 S a, S= a (+ — ); d — 12 ( - -,an) l, 1a, S a- I - (n - 1) d; 1 I - 1. 8 om the se o a, a —ind 9 d, 1, S`,1 a n= 2 1 d - -,. 2 S id, —( 2 (nI1- S) 10 nl a), d a-a=- -; d= F77, Fro m the formula w ke have. c1-dt- - )d; that is. Thie, first term of a -, increasing arithmetical progression, is eqaua to any fbllowi~,# term, mninus the produ.ct of the common dilt'erenee by the num7ber of.preceding terms. 178. From the same formrul, we al.so find d.... that is, CHAP. VIII. ARITHMETICAL PROGRESSION. 239 In any arithmetical progression, the common di ference is eqztal to the last term minus the first ternm, divided by the nutmber of ternms less one. If the last term is less than the first, the common difference Hivl be negative, as it should be. EXAMPLES. 1. The first term of a progression is 4. the last term 16, and the numnlber of terms considered 5: what is the common difference? The formula I-a 16-4 d= — gives d= = 3. n- 1 4 2. The first term of a progression is 22, the last term 4, and the number of terms considered 10: what is the common difference? Ans. - 2. 179. By the aid of the last principle deduced, we can solve the following problem, viz.: To find a number m of arithmetical means between two fgiven numrbers a and b. To solve this problem, it is first necessary to find the common difference. Now, we may regard a as the first term of an arithmetical progression, b as the last term, and the required means as intermediate ti:-ns. The number of terms considered, of this progression, will be expressed by mn + 2. Now, by substituting in the above formula, b for 1, and m -- 2 fBr n, it becomes b —a b-a d or dthat is, the common difference of the required progression Is obtained by dividing the difference between the last and first terms by one more than the required number of means.* 240 ELEMENTS OF ALGEBRA. [CHAP. VIII. Having obtained the common difference, form the second term of the progression, or the first arithmetical mean, by adding d, or b-a + to the first term a. The second mean is obtained by augmenting the first by d, &c. EXAMPLES. 1. Find 3 arithmetical means between 2 and 18. The formula b -a 182 — d_- gives d_ 4; m+ 1' 4 -4 hence, the progression is 2. 6. 10. 14. 18. 2. Find 12 arithmetical means between 77 and 12. The formula b -a 12 - 77 m + gives d 13 -; hence, the progression is 77.72. 67. 2.....22.17. 12. 3. Find 9 arithmetical means and the series, between 75 and 5. Ans. Progression 75. 68. 61.....26. 19. 12. 5. 180. If the same number of arithmetical means be inserted between the terms of a progression, taken two and two, these terms, and the arithmetical means together, will form one and the same progression. For, let a. b. c. e. f.... be the proposed progression, wd m the number of means to be inserted between a and b, a and c, c and e.... From what has just been said, the common difference of each partial progression will be expressed by b-a c -b e - c mz+ 1' Mn+ 1' mV - 1 which are equal to each other, since, a, b, c,... are in pro gression: therefore, the common difference is the same in eall ClAP. VIII.J ARITHMETICAL PROGRESSION. 241 of the partial progressions; and since the last term of t'le first, forms the first term of the second, &c., we may conclude that all of lhese partial progressions form a single progression. GENERAL EXAMPLES. I, Find the sum of the first fifty terms of the progression 2. 9. 16. 23;For the 50th term, we have = 2 - 49 X 7 = 345. 50 Hence, S- (2 +- 345) X -347 X 25 = 8675. 2. Find the 100th term of the series 2. 9. 16. 23 Ans. 695. 3. Find the sum of 100 terms of the series 1.3.5. 3 7. 9... Ans. 10000. 4. The greatest term considered is 70, the common difference 3, and the number of terms 21: what is the least term and the sum of the terms. Ans. Least term 10; sum of terms 840. 5. The first term of a decreasing arithmetical progression is 10, the common difference is - -, and the number of terms 21: required the sum of the terms. Ans. 140. 6. Inl a progression by differences, having given the common difference 6, the last term 185, and the sum of the terms 2945: find the first term, and the number of terms. Ans. First term — 5; number of terms 31, 7 Find 9 arithmetical means between each antecedent and cnsequent of the progression 2. 5.. 11. 14... Ans. d = 0.3. 8. Find the number of men contained in a triangular battalion, the first rank containing 1 man, the second 2, the third 3, and so on to the rnth, which contains n. Ill other words, 16 242 ELEMENTS OF ALGE BRA. LCHAP, VIII, find the expression for the sum of the natural numbers 1, 2, 8,... from 1 to 9n, inclusively.s. S ( 9, Find the sum of the first n term s of the progression of uneven numbers 1, 3; 5 7, 5... Ans. S = O. 10. One hundred stones being placed on the ground, in a straight line, at the distance of two yards from each other, how far will a person travel who shall bring them one by one, to a basket, placed at two yards from the first stone? Ans. 11 miles 840 yards. Of Ratio and Geometrical Proportion. 181. The RATIO of one quantity to another, is the quotient which arises from dividing the second by the first. Thus,'the ratio of a to b, is -. a 182. Two quantities are said to be proportional, or in proportion, when their ratio remains the same, while the quantities'themselves undergo changes of value. Thus, if the ratio of a to b remains the same, while a and b undergo changes of value, then a is said to be proportional to b. 183. Four quantities are in proportion, when the ratio of the first to the second, is equal to the ratio of the third to the fourth. Thus, if b d a C the quantities a, b, c and d, are said to be in proportion. VWe generally express that these quantities are proportional by writing them as follows: a: b:: c d. This algebraic expression is read, a is to b, as lb to d,:and is called a proportion. CHAP. VIII.] GEOMETRICAL PROGRESSION 243 184. The quantities compared, are called terms of the pro. portion. The first and fourth terrils are called the extremes, the secon& and third are called the means; the first and third are called antecedents, the second and fourth are called consequents, and the fourth is said to be a fourth proportional to the other three. If the second and third terms are the same, either of these is said to be a mean proportional between the other two. Thus, in the proportion a:::b:c, b is a mean proportional between a and c, and c is said to be a third proportional to a and b. 185. Two quantities are reciprocally proportional when one is proportional to the reciprocal of the other. Geometrical Progression. 186. A GEOMETRICAL PROGRESSION is a series of terms, each of which is derived from the preceding one, by multipyy'lg it by a constant quantity, called the ratio of the progression. If the ratio is greater than 1, each term is greater than heo preceding one, and the progression is said to be increasing.. If the ratio is less than 1, each term is less than the p.ri ceding one, and the progression is said to be decreasing. Thus,... 3, 6, 12, 24,... &c., is an increasing progression. 1 1.., 8, 4, 2, 1, 2-14, -.... is a decreasing progressifn It may be observed that a geometrical progression is a cons tinued proportion in which each term is a mean proportions between the preceding and succeeding terms. 187. Let r designate the ratio of a geometrical progression, a: b: c: d,... &c. W\e deduce from the definition of a progression the follow ing equations: b-ar, c=r ar, d-= cr = ar3, e = dr=a4..; 244 ELEMENTS OF ALGEBRA., [CHAP. VIIL and, ill geleral, any term n, that is, one which has n- I terms before it, is expressed by ar"-1. Let I be this term; we have the formula I= arn-= by means of which we can obtain any term without being obliged to find all the terms which precede it. That is, Any term of a geometrical progression is equal to the first term multiplied by the ratio raised to a power whose exponent denotes the number of preceding terms. EXAMPLES. 1. Find the 5th term of the progression 2: 4: 8: 16, &c., in which the first term is 2, and the common ratio 2, 5th term = 2 24 = 2 >< 16 = 32. 2. Find the 8th term of the progression 2: 6: 18: 54 8th term = 2 x 37 = 2 X 2187 = 4374o 3. Find the 12th term of the progression 64: 16 4: 1 4 1I43 1 1 12th term = 64 () - 43 = 4 65536' 188. We will now explain the method of determining the sum of n terms of the progression a: b: c: d: e: f:..: i: 1, of which the ratio is r. If we denote the sum of the series by S, and the nth term By 4, we shall have S = a + ar + ar2.... + arn-2 + arn-1 If we multiply both members by r, we have Sr = ar + art + ar3... + atr-s +- art; CHAP. VIII.] GEOMETRICAL PROGRESSION. 245 and by subtracting the first equation frcm the second, member from member, ar - a -Sr-S = ar - a, whence, S substituting for arn, its value Ir, we have Ir - a S....; that is, r —1 To obtain the sum of any number of terms of a progression by quotients, lulet1ply the last term by the ratio, subtract the frst term from this product, and divide the remainder by the ratio diminished by 1. EXAMPLES. 1. Find the sum of eight terms of the progression 2: 6: 18: 54: 162.. 4374. lr - a 13122 - 2 S — = -=6560. r —1 f2 2. Find the sum of five terms of the progression 2:4: 8: 16: 32; S=r -a 64 - 2 S —- = 1 =62. 3. Find the sum of ten terms of the progression 2: 6: 18: 54: 162... 2 X 39 = 39366. Ans. 59048. 4. What debt may be discharged in a year, or twelve months, by paying $1 the first month, $2 the second month, $4 the third month, and so on, each succeeding payment being double the last; and what will be the last payment? Ans. Debt, $4095; last payment, $2048. 5. A gentleman married his daughter on New-Year's day, and gave her husband is. toward her portion, and was to double it on the first day of every month during the year: what was hei pojrtion 2 Ans. ~204 15s. 246 ELEMIENTS OF ALGEBRA. [CIHAP. VIII. 6. A man bought 10 bushels of wheat on the condition that lhe should pay 1 cent for the first bushel, 3 for the second, 9 foi the third, and so on to the last: what did he pay for the last bushel, and for the ten bushels? Ans. Last bushel, $196 83; total cost, $295, 24. 189. When the progression is decreasing, we have r < 1 and I< a; the above formula for the sum is then written under the form =a - Ir 1 -r in order that both terms of the fraction may be positive. By substituting arn-l for 1, in the expression for S, arn -a a - art.S,,= - 1 or S - r-l' 11 —r EXAMPLES. 1. Find the sum of the first five terms of the progression 32: 16: 8: 4: 2. a -- i 31 S- = = = --- =62. 1-r 1 T162 2. Find the sum of the first twelve terms of the progression I 1 64: 16: 4: 1: I...:. 4 65536' a64 —x 256 - I - a- ll. 65536 4 65536 65535 s= 3 - = 85 + I- r 33 t-196608' 4 We perceive that the principal difficulty consists in obtaining the numerical value of the last term, a tedious operation, even whon the number of terms is not very great. 190. If in the formula o a(rn- 1).-1. CHAP. VIII.] GEOMETRICAL PROGRESSION. 247 we make r == 1- it reduces to 0 This result sometimes indicates indetermination; but it often arises fiom the existence of a co-rmmon factor in both numerator and denominator of the fraction, which factor becomes 0, in con. sequence of a particular supposition. Such is the fact in the present case, since both terms of the fraction contain the factor r- 1, which becomes 0, for the particular supposition r = 1. If we divide both terms of the fraction by this common factor, we shall find (Art. 60), S = r-1 + ar -2 + arn3 +... J ~r + a, in which, if we make r- 1, we get S = a+-a + a.-..... + a - na. We ought to have obtained this result; for, under the supposition made, each term of the progression became equal to a, ard since there are n of them, their sum should be na. 191. From tho two formulas = ar"-1, and S -a several properties may be deduced. We shall consider only some of the most important. The first formula gives nl —, - whence ra) — a a The expression furnishes the means for resolving the following problem, viz. To find mn geometrical means between two given numbers a and b; that is, to find a nvumber m of means, which will form with a and 6, considered as extremes, a geometrical progression. 248 ELEMfENTS OF ALGEBRA. [tHAP. VIIL. To find this series, it is only necessary' to know the ratio. Now, the required number of means being?n, the total number of terms considered, will be equal to rn + 2. M]oreover, we have I = b; therefore, the value of r becomes r = /-; that is, 7'o find the ratio, divide the second of the given numbers by the first; then extract that root of the quotient whose index is one /realter than the required number of neans: H]ence the progression is,ii+1 b m+ /b "2 +l /3 a: a -: a V -: a a 3t' EXAMPLES. 1. To insert six geometrical means between the numbers 3 a:d 384, we make m = 6, whence from the formula, r = __4= W 2; hence, we deduce the progression 3: 6 12: 24: 48: 96: 192: 384. 2. Insert four geometrical means between the numbers 2 and 486. The progression is 2: 6: 18: 54- 162 486. REMARK. —When the same number of geometrical means are inserted between each two of the terms of a geomet-ical pro. gressioln, all the progressions thus formed will, when taken together, constitute a single progression. Progressions having an infinite number of terms, 192o Let there be the decreasing progression a: c: d: e:f,.., tonltaniing an infinite number of terms. The formula r, - arn 1 —r' CHAP. VIII.] GEOMETRICAL PROGRESSION. 249 which expresses the sum of n terms, can be put under the form a a ar' — r I -r Now, since the progression is decreasing, r is a proper fraction, and rn is also a fraction, which diminishes as n increases. Therefore, the greater the number of teims we take, the more will a X r79 diminish, and consequently, the nearer will the sum of these terms approximate to an equality with the first part of S; that is, to 1 Finally, when n is taken greater than any assignable number, or when n==o, then X rn 1 -r will be less than any assignable number, or will become equal to 0; and the expression 1 - will represent the true value of the sum of all the terms of the series. Hence, Th/e sum of the terms of a decreasirng progression, in which the number of terms is infinite, is a 1-r This is, properly speaking, the limit to which the partial su..ms approach, as we take a greater number of terms of the progression. The number of terms may be taken so great as to make the difference between the sum, and 1 as small as we please, and the difference will Dnly become zero >'hen the number of terms taken is infinite. EXAMPLES. 1. Finld the sum of I 1 1 1111 3 27 (r&c. 250 ~ ELEMENTS OF ALGEBRA. [CHAP. VIIL We have foi: the sum of the ternls, a 1 3 1 —r 1~'12, Again, take the progression l 1 I 1 1 { T2: -—. a 6 c &.... We have S a We have S- 2 1-r 1 What is the error, in each example for b = 4, n-r,, n=-6 Indeterrminate Co-efficients. 193. An IDENTICAL EQUATION is one which is satisfied for any values that may be assigned to one or more of the quantities whichl enter it. It differs materially from an ordinary equation. The latter, when it contains but one unknown quantity, can only be satisfied for a limited number of values of that quantity, whilst the former is satisfied for any value whatever of the indeterminate quantity which enters it. It differs also from the indeterminate equation. Thus, if in the ordinary equation ax + by - cz + d = 0 values be assigned to x and y at pleasure, and corresponding values of z be deduced from the equation, these values taken together will satisfy the equation, and an infinite number of sets of values may be found which will satisfy it (Art. 88). But if in the equation ax ~- by + cz + d = 0, we impose the condition that it shall be satisfied for any values of x, y and z, taken at pleasure, it is then called an idenztical equation. 194. A quantity is indeterminate when it admits of an infinite number of values. Let us assume the identical equati n, A+Bx-. C X3 +D &=+ &C - -.(1), CHAP. VIII.] GEOMIETRICAL PROGRESSION. 251 in which the co-efficients. A, B, C, D, &c., are entirely ilde, pend.ent of x. If we make x = 0 in equation (1) all the termr: containing z reduce to 0, and we find -A —O. Substituting this value of A in equation (1), and factoring, it becomes, x(B + + DAx + &c.,)= - - - - (2), which may be satisfied by placing x O, or by placing B + C +D 2+. = - - - - (3). The first supposition gives a common equation, satisfied only for x = 0. Hence, equation (2) can only be an identical equa. tion under a supposition which makes equation (3) an identical equation. If, now, we make x 0 in equation (3), all the terms containing x will reduce to 0, and we find B = 0. Substituting this value of B in equation (3), and factoring, we get (C + D + &c.) 0 - - - - - - - (4). In the same manner as before, we may show that C(7 0, and so we may prove in succession that each of the co-efficients D, E, &c., is separately equal to 0: hence, In every identical equation, either member of which is 0, involving a single indeterminate quantity, the co-effcients of the diferent powers of this quantity are separately equal to 0. 195, Let us next assume the identical equation a + bx + cx2 + &c. = a' + b'x + c'C2 + &e. By transposing all the terms into the first member, it may be piacr-d under the form (a - a') + (6 -') x + (c-') x2+ &O. = 0. Now, from the principle just demonstrated, a -a a - O - b'- c - c' =-0, &c., &c.; whence a = a', b = b= C =', &C., &c.; that is, 252 ELEMENTS OF ALG EBRA. [CHAP. VIII. In an identical equation containing but one indeternminate quantity, the co-eaticients of the like powers of that quantity in the two nmermcrs, a-e equal to each other. 196. We may extend the principles just deduced to'dentical equations containing any number of indeterminate quantities. For, let us assume that the equation a + 7 )t b'y + b"z + &c. + cx2 + c'y2 + c"z2 + &c. -+ dz3 + d'y3 + &c.-o - - (1), is satisfied independently of any values that may be assigned to r, y, z, &c. If we make all the indeterminate quantities xcept x equal to 0, equation (1), will reduce to a + bz - cx2 +- dx3 + &c. 0; whence, from the principle of article 194, a-0, b = 0, c = 0 d = 0, &c. If, now, we make all the arbitrary quantities except y equal to 0, equation (1) reduces to, a + b'y + c'y2 + d'y3 + &c. = 0; whence, as before, a- = 0 6'-, c' = 0 d' = 0, &c. and similarly we have b"=O, c" _0 &c. The principle here developed is called the principle of inde. terminate co-eficients, not because the co-efficients are really indeterminate, for we have shown that they are separately equal to 0, but because they are co-efficients of indeterminate quantities. 197. The principle of Indeterminate Co-efficients is much used in developing algebraic expressions into serf es. For example, let us endeavor to develop the expression, a a' + b'x' into a series arranged according to the ascending powers of x. CHTAP. VIII.] GEOMETRICAL PROGRESSION. 253 Let us assume a development of the proposed form, a _ P+ Qx+ - + Sx3+&c. - - - (1), a' -4- b'x in which P, Q, B, &c., are independent of x, and depend upon a, ai and b' for their values. It is now required to find such values for P, Q, R, &c., as will make the development a true one for all values of x. By clearing of fractions and transposing all the terms into she first member, we have Pa' + Qa' x + Ra2' X2 + &c. = 0. -a + Pb' +- Qb'[ &c. Since this equation is true for all values of x, it is identical, and from the principle of Art. 194, we have Pa' - a -- 0, Qa' + Pb' O, Rat' Qb' = 0, &c., &c.; whence, a Pb' ab' Qb' ab'2 &c., &C. P, Q- a, = ag, - -— aa a/ a/ a'sa a3' & Substituting these values of P, Q, R, &c., in equation (1), it becomes a a ab' ab'2 ab'3 a' + a' a2Z a3 aX43 & - - (2) Since we may pursue the same course of reasoning upon any like expression, we have for developing an algebraic expression into a series, the following RULE. I. Place the given expression equal to a development of the form P + Qx + Bx2 + -&c., clear the resulting equation of fractions, and transpose all of the terms into the first member of the equation. IT. Then place the co-efiicients of the dif'erent powers cf the let. ter, with reference to which the series is arranged, separately equal to 0, and from these equations find the values of P, Q, R, &c. III..Having found these values, substitute them for P, Q, R, &c., in the assumed development, aad the result will be the develop, ment required. 254 ELEMENTS OF ALGEBRA. [CHAP. VIII. EXAMPLES. 1. Develop a into a series. a-x X 2 x3 Ans. 1 + 2' - + a a a 2. Develop 1 into a series. (a Vx)2 1 +x 3z2 4x3 Ans. a a3+ a+ a5 + 1 q —2x 3. Develop 1 +, into a series. Ans. - + 5x + 15x2 +- 45x3 + 135x4 + &c. 198. We have hitherto supposed the series to be arranged according to the ascending powers of the unknown quantity, commencing with the 0 power, but all expressions cannot bE: developed according to this law. In such cases, the application of the rule gives rise to some absurdity. For example, if we apply the rule to develop 3 —' we shall have, =P + Qx + Rx2 + &c. - - (1). 3x - X2 Clearing of fractions, and transposing, -1I + 3 Px~ + 3Q 2 -+ &c. =0; P Whence, by the rule, -1-0 = 0, 3Q - P =0, &c. Now, the first equation is absurd, since — 1 cannot equal O. HeIce, we conclude that the expression cannot be developed ao cording to the ascending powers of x, beginning at x~. We imay, however, write the expression under tile forrn 1 1 -X, and by the application of the rule, develop the factc x 3 --' - which gives 1 1 X 1 1 + &C.; x-4 —x - t- -- -+- -+&c.; 3- z 3 9 271 81 CHAP. VIII.] RECURRING SERIES. 255 whence, by substitution, ___ _+1 1 1 1 3 - + x - + X + X+2 + &c. Since -i is equal to 3x-1 (Art. 166), we see that the true level. opmcnt contains a term with a negative exponent, and the supposition made in equation (1)' ought to have failed. Recurring Series. 199. The development of fractions of the form a' b &c., gives rise to the consideration of a kind of series, called recurring series. A RECURRING SERIES. is one in which any term is equal to the algebraic sum of the products obtained by multiplying one or more of the preceding terms by certain fixed quantities. These fixed quantities, taken in their proper order, constitute what is called the scale of the series. 200. If we examine the development a a cab' ab'2 ab'3 a' + b'x a a /'2 3 4X3 we shall se, that each term is formed by multiplying the preo 6' ceding one by -, x. This is called a recurring series of the first order, because the scale of the series contains but one term. The expression - — x is the scale of the series, and the ex pression --- is called the sca le of the co-efficients. It may be remarked, that a geometrical progression is a recur ring series of the first order. 201. Let it be required to develop the expression a ~- bx a + b'x into a series.:2' + Y'x + c'x2 256 ELEMENTS OF ALGEBRA. [CHAP. VIII, Assume a=. P +- Qx A- Rx2 + SQ3 -3+ &C. a' + b'x + c'x2 Clearing of fractions, and transposing, we get P' - Qa' x + - Ra' x2 + Sa' x3 + &c. =0. -a |- Pb' - Qb' + Rb' -b - Pc' + Qc' Therefore, we have a Pa' - a = 0 or, P=a a.a' - Pb' - b = O, or, Q - a'P + b D/ C/ a/' — Qb/ + Pc'= O, or, R -- 7 Q- a P; Sa' + _Rb' + Qc' = 0, or, S =- - Q; a' a &c., &c., &c., &c.; fiom which we see that, commencing at the third, each co-effi. cient is formed by multiplying the two which precede it, re. seteyby''ci spectively, by -- and — C' viz., that which immediately precedes the required co-efficient by — i, that which precedes C' it two terms by - -, and taking the algebraic sum of the pro ducts. Hence, a/" a" is the scale of the co-efficients, From this law of formation of the co-efficients, it follows that the third term, and every succeeding one, is formed by multipl) ing the one that next precedes it by - -7x, and the second pr'eding one by --- 2, and then taking the algebraic sum of these products: hence, ( b/ ~2 )2 is the scales of the series. is the scales qo the series. CHAP. V111.j BINOMIAL THEOREM. 257 This scale contains two terms, and the series is called a recurring series of the second order. In general, the order of a recurring series is denoted by the number of terms in the scale of the series. The development of the fraction a + bx + cx2 a' - byx + Cx2 + dd X3' gives rise to a recurring series of the third order, the scale of which is, ( b' c'a2 _, and, in general, the development of a bx + cx2 +... kx"a' + b'x + c'x2 +... kxn' gives a recurring series of the nith order, the scale of which is b( $ / 2 - n General demonstrcation of the Binomial Theorem. 202. It has been shown (Art. 60), that any expression of the form zm - yn, is exactly divisible by z - y, when m is a pos;t;v. whole number, giving, Zn- yn= zm'-1 + zm-2y + zn-3y2 - +.y.m-1 The number of terms in the quotient is equal to m, and if we suppose z = y, each term will become zm-l; hence, z -- y /y=z The notation employed in the first member, simply indicates what the quantity within the parenthesis becomes when we make We now propose to show that this form is true when, m: isfractional and when it is negative. 17 258 ELEMENTS OF ALGEBRA. LCHAP. VIII. Jr~'gt, suppose nm fractional, and equal to 1 P Make zq -v, whence zq =vP and z-=vq; L..L aud yq -u, whence y q -up and y = u. p p VP -?tP zq -yq VP -up V - it hence, z-y vq- U? Vqq- q' V — f If now, we suppose y -z, we have v = u, and since p and q are positive whole numbers, we have P p v z z -, Z-y Vq a u qVq —1 q a x Q — y)(vq- 1q " Y. 7J V- ~t ]v —u v - u = Second, suppose mn negative, and either' entire or fraclional. By observing that - rn X (m?m) — _ ym we have, zr —M ng- 7y — X m' Y — y _ z-n z-ym If, now, we make the supposition that y = z, the first factoi of the second member reduces to -— 21', and the second feitor, from the principles just demonstrated, reduces to mz"'-;n hence, \ -XY y=z''We conclude, therefore, that the Jbrnm is general.:203. By the aid of the principles demonstrated in the, laf. article, we are able to deduce a formula for the- develop mnent of (x + a)m, wvhen the exponent m is positive or negative, entire or fr actional Let u-s assume the equation,.(1 + z), - P + Qz + Rz2 4- S3 +&. -& (1), CHAP. VIII.] BINOMIAL THEOREUI. 259 il which, P, Q, 3R, &c., are independent of z, and depend upon I and mn for their values. It is required to find such values for them as will make the assumed development true for every possible value of z. If, in equation (1) we make z- 0, we have P = 1. Substituting this value for P, equation (1) becomes, (1 + z)m = 1 + Qz +R z2 + S3 + &C. - - - (2). Equation (2) being true for all values of z, let us make z y; whence, (1 +-/) = 1 + Qy+ y2 +Sy3+&c. - - (3). Subtracting equation (3) from (2), member from member, and dividing the first member by (1 + z)- (1 + y), and the second member by its equal z- y, we have, (1 -L+- z)m — (I y)m y _ 22 - y2 3 _ Y3 + =Q -t Y - z y S -Y- &c -(4). If, now, we make 1 + z = 1 + y, whence z = y, the first member of equation (4), from previous principles, becomes nz (1 + z)m-l, and the quotients in the second membcr become respectively, (x /=) 1,( /- ) =, 2z( ) = 32, &. &c. n(1 ~z)m-1 = + 2Rz 3Sz2+4Tz3+&c. - - - (5). Mlultiplying both members of equation (5) by (1 + z), we find, \z (1 + zm = Q 2R lz + 3 S 2 + 4T 3 + &C.- - +2 -- 3S Sr we multiply both members of equation (2) by w ne we have ti (1 + - )m = n + m Qz + nmRz2 + nSz3 ~ z TTz + &e. - - (7). Tne second members of equations (6) and (7) are equal to each other, since the first members are the same; hence, we have the equation, m Qz -V-nRz2-tJ-mS3+& c- Q+92Kz+3f S 2 +4T 1z3+ c - (& ) 1 + 0A.l'R + 3 5 260 ELEMENTS OF ALGEBRA. LCHAP. ViHl This equation being identical, we have, (Art. 195), Q =, - - or, - Q = 2+ Q=mQ,- or, - - - = - S+ 2 = m, - or, - - (l)(. 2. 3 - 4T- 3S=mS, - or, T ( -1) (m - 2) (- 3) 1. 2~.3 4 &c., &c. &c. Substituting these values in equation (2), we obtain (1 + )m- + Mz (m-]) 2 m (n — ) ( m —2) 1.2 1. 2. 3'* m (m - )(m - 2) ( - 3) +. (9) ~ 1. 2. 3. 4 If now, in the last equation, we write - for z, and theln Imun tiply both members by xm we shall have, (x a)m =x + maxm-+ m(2n-1) a2x-2 + m(Mn-l) (m-2) a3n9.- &c... (19). Hence, we conclude, since this formula is identical with that deduced in Art. 136, that the form of the devcl>ormnent of (x+ a)" will be the same, whether rn is positive or negfative, entire or fractional. It is plain that the number of terms of the development, when m, is either fractional or negative, will be infinite. Ajplica.Jions of the B]inomial Formula. 04,. If in the formula (x + a) = a m -1 a2 m — i m-2 a3: m. + 2' 2 * 3+'M CHAP. YIII.] BINOMIAL THEOREAL 261 1 we make mn =-, it becomes (x + a)n or za-_ 1 1 1 2 a a - 1 2 1 n - a3 x 1 + —.+ —..- - n — n x 12 2 2 3 + or, reducing, n - = 1 a 1 n- a l a1 l 2n - 1 3 a a' n 2,r *2 +l 2n 3mr x3 The fifth term, within the parenthesis, can be found by multiplying the fourth by 4- and by -, then changing the sign of the result, and so on. 205. The formula just deduced may be used to find an approximate root of a number. Let it be required to find, by means of it, the cube root of 31. The greatest perfect cube in 31 is 27. Let x = 27 and a = 4: making these substitutions in the formula, and putting 3 in the place of n, it becomes 3 1 +- 3'3'29 3+ 3 9' 19683 1 1 5 2 256 3 3 9 3'531441- &c. or, by reducing, 4 16 320 2560 /31= 3 + 27 2187 531441 43046721 + Whence, V31 - 3. 14138, which, as we shall show presently, is exact to within less than.00001. We may, in like manner, treat all similar cases: hence, for extracting any root, approximatively, by the binomial formula, we have the following RULE. Find the perfect power of the deg/ree indicated, which is nearest to the given number, and place this in the formula for x. Sub. tract this power from the given number, and substituzte this difference, which will often be negative, in the formula for a. Perform the operations indicated, and the result will be the required root. 262 ELEMENT1S OF ALGEBRA. LCHAP. VIII.:EXAMPLES. L. 3F_ 278(1 4 27) = 3.0366. 2. 3-0 (32-2)5 = 325(1 1- )= 1.9744.. 39 = (32 + 7)5+ -325= 2.0807. 4. [1 08- (128 - 20) 128(1- = 1.95204. 206. When the terms of a series go on decreasing in value, the series is called a decreasing series; and when they go on increasing in value, it is called an increasing series. A converging series is one in which the greater the numbey of terms taken, the nearer will their sum approximate to a fixed value, which is the true sum of the series. When the terms of a decreasing and converging series are alternately positive and negative, as in the fir-;t example above, we can determine the degree of approximation when we take the sum of a limited number of terms for the true sum of the series. For, let a — b - c - d + e-f.., &c., be a decreasing series, b, c, d,... being positive quantities, and let x denote the true sum of this series. Then, if n denote the number of terms taken, the value of x will be found between the sums of n and -n + 1 terms. For, take any two consecutive sums, c - b - c-d -t-e-f, and a-bqe-d -e —f g. In the first, the terms which follow -f, are +g-h, + k —+..; but, si.ice the series is decreasing, the terms g -- 4, k - 1 &c., are positive; therefore, in order to obtain the complete valtic of x. a -)(ositive number must be added to the sura a -- b t — d J e-f. IIence, we have a - b + c - d - e - f< x. CRAP. VIIIJ. BINOMIAL FORMULA. 263 In the second sum, the terms which follow + g, are - h + k- -l+m,.... Now, -h +, -- I + m.. &c., are negative; therefore, in order to obtain the sum of the series, a negative quantity must be added tc a - b ~ c -d + e -f g; or, in other words, it is necessary to diminish it. Consequently, a —b + c - d+ e —f+g> x. Therefore, x is comprehended between the sums of the first,n and the first n + 1 terms. But the difference between these two sums is equal to g; and since x is comprised between them, g must be greater than the difference between x and either of them; hence, the error comnmitted by taking the sum of n terms, a - b + c - d + e -f, of the series, for -the sum of the series is numerically less than the Jfllow'ing term. 207. The binomial formula serves also to develop algebraic eqx)1ressions into series. EXAMPLES. 1 1. To develop the expression 1- z' we have, -Z(1 - Z)-. [n the binomial formula, make = - 1, x = 1, and a - a and it becomes (1 - -z)1- 1. (-Z) - z)2 2-1- -- or,'performing the operations indicated, we find for the de velopment, — =(1 - z)-= 1 + Z + 3 +4 + &Ce. 1 z We might have obtained this result, by applying the rule f)r division. 264 ELEMENTS OF ALGEBRA. -CHAP. VUJ 2. Again take the expression, (1-_ or 2(1 Z)-3. Substituting in the binomial formula — 3 for, I Rr %$ and -- z for a, it becomes, Z) - (1 -z)-3 = 1 -3.(-z) —3. _2 (- x)z 2 3 _33 —3 -.. —S2. (- ~ - Z)3&o. rerforming the indicated operations and multiplying by 2, we find (_-3 _: 2 (1 + 3z + 6z2 + 10z3 + 15z4 + &c.). 3. To develop the expression z- z2 we first place it under the form 3, x (1 - I By the application of the binomial formula, we find (12) 3 2( ) 3 2 (- - 2' 1 I 5 = 1 —-~z- _3_ Z2 _- - _ Z3'''; hence, 3 _ /1 1_ 1 —,&. 4. Develop the expression (-+ b) (a + b)-2 into a series 5. Develop r + into a series. r + -x x2 x3 X4 Ans. r-x+ - — +-, & c. r r2 r q (. Develop the square root of 2 - 2 into a series. it 2 2 2. X2 x32 X4 36 Ans. 1 +_- + ~ + x c. a2 7, Develop the cube root of + into a series. 1 / 22 54 40a series. Anws.2X 1 + — + 2 X C),32 4 e. O3a2 9a4 $1 ~ CHAlP. VIII.J SUMMATION OF SERIES. 265 Summation of Series. 208. The Summaton of a Series, is the operation of finding an expression for the sum of any number of terms. Many useful series may be summed by the aid of two auxiliary series. Let there be a given series, whose terms may be derived from the expression - by giving to p a fixed value, and then attributing suitable values to q and n. Let there be two auxiliary series formed from the expressions q q? and +, so that the values of p, q, and n, shall be the same as in the corresponding terms of the first series. It can easily be shown that any term of the first series is equal to - multiplied by the excess of the corresponding term in the second series, over that in the third. For, if we take the expression and n n Jrindicated, we shall get the expression, and perform the operations indicated, we shall get the expression, (- p); hence, we have n(n+ p)' q _1( qq\ n(n+P) A- p n nJ-r ); which was to be proved. It follows, therefore, that the sum of any number of terms oj the first series, is equal to - multiplied by the excess of the sum of the corresponding terms in the second series, over that of tMe corresponding termrs in the third series. Whenever, therefore, we can find this last difference, it is always possible to sum the given ser'es. 266 ELEMIENTS OF ALGEBRA. LCHAP. VIIL EXAMPLES. I RequireL the sum of n terms of the series 1 1 1 1 1.2 + 2.3 +3.4 + 4-.5 + & Comparing the terms of this series withf the expression q n(n + P)' we see that making p=l, q = 1, and n = 1, 2, 3, 4, &c., in succession, will produce the given series. The two corresponding auxiliary series, to n terms, are i 1 1 1 1 I 1 1 1 Band - + +. - n 2 3 4 n n + 1 The difference between the sums of n terms of the first and second auxiliary series is 1 or, if we denote the sui n+-~' of n terms of the given series by S, we have, S = 1 If the number of terms is infinite n = - and S — =1. 2. Required the sum of n terms of the series 1 1 I I 1 - + + + -+ + &c., 1.3 3.5 5.7 7.9 9.11 If we compare the terms of this series with the expression q n(n + p)' we see that r 2, q-1, and n7=1, 3 5, 7, &c., in sue 2esai on. CHAP. VIII.J SUMMATION OF SERIES. 267 The two auxiliary series, to n terms, are, 1 1 1 1 I+ +-+ +d-+......+ -' 1 1 1 + 1 and - - +- 1. + 3 5 -7 2n-1 2- -.+ hen ce, as before, - S= -(1- 2 + ) if n= Go, we find S=-. 2 3. Required the sum of n terms of the series 1 1 1 1 -+ - 4- +- &ec 1.4 2.5 3.6 4.7 Here p = 3, q 1 = n=1, 2, 3, 4, &c. The two auxiliary series, to n terms, are, 1 1 1 1 1 1I 1 1 1 5 a n+1 n 4 5 7 +?+ 1 n I 2 n+9s hence, S- +1 1+ 3 nq- n +- 12 -3' If t n= oo, S -i 4. Required the sum of the series 4 4 4 4 4 1.5 -? 5. 9 9.13 13. 17 17.21 Ans. 1. 5. Find the sum of n terms of the series, 2 3 4 5 6 _ &C - + 3.5 5.7 7.9 9,11 11.13 Here p = 2, q -2, — 3, +4, — 5, — 6, &e n — 3, 5, 7, 9, 11, &c. 268 ELEMIENTS OF ALGEBRA. LCHAP. VIII. The two auxiliary series are, 2 3 4 5 n+ I 3 5 7 9+ 2n + 1 2 3 4 I -n + 1 5 7 V3. 2.. -.. I 2'n F 3 ah:noe, S - 4 - + --- +]) hein'e gS- 2 ( 3 f 2n + 2 If n is even, the upper sign is used, and the quaitity in ithe last parenthesis becomes + 1, in which case S _ 1 (2 +r -+- _ _1 I I + z + 1) If n is odd, the lower sign is used, and the quantity in Jt last parenthesis becomes 0, in which case 1S —-- 2 n + 3 S - ( - n ~ 1,, If in either formula we make:n + 1 f -- 1 = 2n + becomes and S -. 2 +6. Find the sum of n terms of the series, l 1 1 1 __ - +- Ac. 1.3 2.4 3.5 4.6' HIere, p =2, q= - 1, +1, -1,, -1,, -1, &c. n=1?, 2, 3, 4, &c. The two auxiliary series are, 1 1 1 I 1 1 2 3 4 5 6 1 1 1 1 1 1 I o t4 +.5 - --— 6 +... whence, S -+2 F l n d- 1 + If n-c-, we find S -. 4~ CHAP. VIII.] METHOD BY. DIFFERENCES. 269 Of thee Method by Difzerences. 209 Let a, b, c, d.... &c., represent the successive terms of a series formed according to any fixed law; then if each term be subtracted from the succeeding one, the several re. mainders will form a new series called the first order of differences. If we subtract each teram of this series from the succeeding one, we shall form another series called the secondi order of differences, and so on, as exhibited in the annexed table. a, b, c, d, e, b-a, c-b, d-c, e - d, &c., 1st. c-2b+a, d-2c. + b, e -2d+ c, &c., 2d. d-3c+3b-a, e - 3d + 3c - b, &c., 3d. e-4d + 6c - 4b + a, &c. 4th. If, now, we designate the first terms of the first, second. third, &c. orders of differences, by d,, d2, d3, d4, &c., we shall have, dl= b- a, whence b= a+ d,, d = c- 2b + a, whence c = a + 2d1 d- d2, ds = d - 3c + 3b - a, whence d — a - 3 d- + 3da + d3, d,= e - 4d + 6c - 46b + a, whence e = a + 4dcl, + 6d2+ 4d3+ d4, &c. &C. &c. &c. And if we designate the termn of the series which has. a terms before it, by T, we shall find, by a continuation of the above process, T- a + d ) +'(- 1) ( 1-) (n 2- )-13 1. 2 1 2. 3 n. (n - I). (n - 2) (l ) - -. (&'- -. (1). This formula enables us to find the (n-t l)th term of a series when we know the first terms of the successive orders of differences. 270 ELEMENTS OF A LGEBRA. LCHAP. VIII. 210. To find an texpression fir the stun of n terms of the series a, b, c, &c., let us take the series 0, a, a+b, a+ b +c, a +6 c+d, &c. - - -. (2) The first order of differences is evidently a, b, c, d,..... &c- - - - -. (3) Now, it is obvious that the sum of n terms of the series (3), is equal to the (n l1)th term of the series (2). But the first term of the first order of differences in series (2) is a; the first term of the second order of differences is the samne as d, in equation (1). The first term of the third order of differences is equal to d., and so on. HTence, making these changes in formnula (1), and denoting the sum of n terms by S, we have, s=~~a~n(~nn = H )(n2) n"(n -])(n-2)(n-3)J, na ( d, + 2 +1.2.3 1.2.3.4 - &c. - - (4). When all of the terms of any orcder of differences become equal, the terms of all succeeding orcdrs cf differences are 0, and formulas (1) and (4) give exact results. When there are no orders of differences, whose terms become equal, then formnulas do not give exact results, but approximations more or less exact according to the number of terrns used. EXAMILEiS. 1. Find the sum of n terms of the series 1.2, 2.3, 3.4, 1. 5, &c. Series, 1.21, 2.3, 3.4, 4.5. 5. 6, &c. Ist order of differences, 4, 6, 8, 10, &c. 2d order of differences, 2, 2, 2, &c. 3d order of differences, 0, 0. Hncee, we have, a=2, d-43, dI2= c4d, d4, &c., equal to O. CHAP. VIII.. METHOD BY DIFFERENCES. 271 Substituting these values for a, dl,, d2, &c., in formlula (4), we find, 2=n + (n - 1) X4+n(n - 1) ( - 2)x; 1.2 X 1.2.a waeane,,. _ n(n + l) (n + 2) whence, 2. Find the sum of n terms of the series 1.2.3, 2.3.4, 3.4.5, 4.5.6, &c. 1st order of differences, 18, 26, 607 90, 126, &c. 2d order of differences, 18, 24, 30, 36, &c. 3d order of differences, 6, 6, 6, &c. 4th order of differences, 0, 0. &c. We find a = 6, d,= 18, d = 18, d3 = 6, d — O &e. Substituting in equation (4), and reducing, we find, n(n + 1) (n -+ 2) (a 3) 3. Find the sum of n terms of the series 1, 1 + 2, 1+2+3, 1 4-2+3+4, &c. Series, 1, 3, 6, 10, 15, 21. 1st order of differences, 2, 3, 4, 5, 6. 2d order of differences, 1, 1, 1, 1. 3d order of differences, 0, 0, 0, a- =, d,=-2, 2 C12, =d3= 0, d-, &c.; hence, S =n + n(- ) n( —1) (2) na3 ~ 3n2 + -2 1.2 1.2.3 1.2. 3 Or, reducing, S- n(n + 1) (n + 2) 1.2.3 4. Find the sum of n te-rms of the series 12, 22, 32, 42, 52, &c. We find, a = 1, d,= 3, d = 2, ds 0, d4 = 0, &.,&e. Substituting these values in formula (4), and reducing, we find, S n(n + 1)(2n + 1) 1.2 3 272 ELEMENTS OF ALGEBRA. [CHAP. VIII. 5. Find the sum of n terms of the series, 1. ( + 1), 2 (n + 2), 3 (n3 + ), 4 (m + 4), &c. We find, a = rn + 1, d = m + 3, d2= 2, dc = 0, &c.; whence, S =n (m - 1) 4.( 1) (mnz 3) n( 1)(n2) x2.'n. (n + 1). (1 + 2n + 3m) or, 1. 2. 3 Of Piling Balls. The last teree formulas deduced, are of practical application in determining the number of balls in different shaped piles. First, in the Triangular Pile. 211. A triangular pile is formed of succcessive triangular layers, such that the number of shot in each side of the layers, decreases continuously by 1 to the single shot at the top. The number of balls in a complete triangular pile is evidently equal to the sum of the series 1, 1~-2, 1l- 2 -3, 1 - 2 +3 + 4, &c. to 1 + 2-... - n, n denoting the number of balls on one side of the base. But from example 3d, last article, we find the sum of n terms of the series, _ (n2 + I) (n 2)- ) 1. 2.3 - (1). Second, in the Square Pile. 212. The square pile is formed, as shown in the figure. The nlumber of balls in the top layer is 1; the number in, the second layer is denoted by 22; in the next, by 32, and so on. Hence, the number of balls in a pile of ni layers, is equal, to the sum of the series, 12, 22 32, ,EIAP. VIII.] PILING BALLS. 273 zc., n2, which we see, from example 4th of the last article, is n.+ (n +- 1). ).. (2). Third, in the Oblong Pile. 213. The complete oblong pile has (m + 1) balls in the. upper layer, 2. (m + 2) in the next layer, 3 (m + 3) in the third, and so on: hence, the number of balls in the complete pile, is given by the formula deduced in example 5th of the preceding article, n. (n - 1). (I + 2n -f 3m) (3 S4- *. 2. 3 - (3). 214. If any of these piles is incomplete, compute the lumber of balls that it would contain if complete, and the number that would be required to complete it; the excess of the forz iner over the latter, will be the number of balls in the pile. The formulas (1), (2) and (3) may be written, Triangular, S + (~- ) - - -.... (1); square, S 2 + (n- + +1) - (2); 3 2 ( Now since ( -+) is the number of balls in the triaingular face of each pile, and the next factor, the number of balls in the longest line of the base, plus the number in the side of the base opposite, plus the parallel top row, we have the following 18 274 ELEMENTS OF ALGEBRA. LCHAP. VII1. RULE. Add to the number of balls in the longest line oj tlhe base the number in the parallel side opposite, and also the numnber in the top parallel row; then mnultiply this sum by one-third the ) numnber in the triangular face; the product will be the number of balls ia the pile. EXAMPLES. 1. How many balls in a triangular pile of 15 courses? Ans. 680. 2. How many balls in a square pile of 14 courses? and how many will remain after 5 courses are removed? Ans. 1015 and 960. 3. In an oblong pile, the length and breadth at bottom are respectively 60 and 30: how many balls does it contain? Ans. 23405. 4. In an incomplete oblong pile, the length and breadth at bottom are respectively 46 and 20, and the length and'breadth at top 35 and 9: how many balls does it contain? Ans. 7190.'5. How many balls in an incomplete triangular pile, the num ber of balls in each side of the lower course being 20, and;ill each side of the upper, 10? 6. How many balls in an incomplete square pile, the number nhi each side of the lower course being 15, and in each side of the upper course 6? 7. How many balls in an incomplete oblong pile, the numbers in the lower courses being 92 and 40; and the numbers in the Corresponding top courses being 70 and 18I CHAP ElER IX. CONTINUED FRACTIONS-EXPONENTIAL QUANTITIES-LOGARI FHAIS, AND FORMULAS FOR INTEREST. 215. Every expression of the form a "+ 1 a-a +1 a +1 b -+ -1 b c+I c C + i in which a, b, c, d, &c., are positive whole numbers, is called a continured fraction: hence,.A CONTINUED FRACTION has 1 for its numerator, and for its denominator, a whole znutmber plus a fraction, which has 1 for its nunerator and for its denominator a whole number plus a frac. liion, and so on. 216. The resolution of equations of the form ax _ b, gives rise to continued fractions. Suppose, for example, a = 8, b -- 32. We then have 8 = 32, it which x> 1 and less than 2. Make x= 1 +-, 276 ELEMENTS OF ALGEBRA. [CHAP. ILX in which 9 > 1, and the proposed equation becomes 1+ 32 = 8 Y =8 X 8 Y; whence, 1 8Y = 4, and consequently, 8 = 4. It is plain, that the value of y lies between 1 and 2. Suppose y= 1 ~ y —l 3. z hence, 4z = 2, and 4- = 2, or z=2. 1 1 2 5 Blut, y = — 1 _1 + 1 _ o z 2 2 land z = - 1 = 1 - - - Y 1 3 3 and this value will satisfy the proposed equation. 5 5 For, &3 - = 38-623) = 5)3=2 =32. 217. If we apply a similar process to the equation (10)X = 200, we shall find x= 2 + -; y= 3 + —: z 3 +-. y U Since 200 is not an exact power, x cannot be exactly expressed either by a whole number or a fraction: hence, the value of x will be incommensurable with 1, and the continued fraction will not terminate, but will be of the form x-2 + 1 1 2 1 1 1 -t- 3 I+ 3+ 1 + + &o. CHAP. IX.I CONTINUED FRACTIONS. 277 218, Vulgar fractions may also be placed unlder the form of continued fractions. Let us take, for example, the fraction 14 and divide both its terms by the numerator 65, the value of the fraction will not be changed, and we shall have 65 1 149 149' 65 65 1 or efficting the division, 149 65 1_ 19 Now, if we neglect the fractional part, -5, of the denomina 1 toI, we shall obtain - for an approximate value of the given fraction. But this value will be too large, since the denominator used is too ssmall. If, on the contrary, instead of neglecting the part 6-5 we were to replace it by 1, the approximate value would be, which would be too small, since the denominator 3 is too large. Hence, 1 65 a 65 49 and3 1. Substituting this value in the given equation, it becomes, 2+1 1 2 x' -69 or 22 x 2x'-6; hence, 6 3 2' = 4 f2 284 ELE MENTS OF ALGEBRA. [CHAP. IX. and by changing the order of the members, and raising both to the x' power, (X/ To determine x'. male x' successively equal to 1 a i 2; we find, -(3) = -<2; and - 4 >2; therefore, x' is comprised between 1 and 2. Make, x' = -- I,I in which x" > 1.'3 )~ By substituting this value in the equation 2- = 2, we find, " 2; hence, - x = 2, and consequently, 4 3 The supposition, x" = 1, gives < and // = 2, gives >, therefore, x" is comprised between 1 and 2. Let x"// + 1,,; then, (4)1+ A x ( whence, (=+ ) 3' If we make x"' 2, we have 9 \2 81< 4 \8J-6 ~<4 3' and if we make x/"' 3, we have (9\3 729 4 \s- =512> -; CHAP. IX.] EXPONENTIAL QUANTITIES. 285 therefore, x'/ is comprised between 2 and 3. ~M~ake "x/// =" -2 q — ) and we have 1'V (9\2+- 4 81 /9\! 9 4 ( 8 =i; hence, - Xlv — 64 \8/ (256 \xlv 9 ftnd consequently, \ 24 -. Operating upon this exponential equation in the same manner as upon the preceding equations, we shall find two entire num bers, 2 and 3, between which xIv will be comprised. Making XZlV 2 + - x can be determined in the same manner as xWV, and so on. Making the necessary substitutions in the equations 1 1 1 1.X -- =2t + x x'"=X X x X/we obtain the value of under the for of a whole n we obtain the value of x under the, form of a whole numbe'r7 plus a continued fraction. x + -- 1+ 2+ XV, hence, we find the first three approximating fractions to be 1 1 3 1' 2' 5' and the fourth is equal to 3x2~-1 7 5 x 2+2 (Art. 220), 5 x 2 + 2 12 which is tne true value of the fractional part of x to within less than (1 )2' or 44 (Art. 222). (12)27 144 286 ELEMENTS OF ALGEBRA. [CHAP. IX. Therefore, 7 31 1 z= = 12 =+ _- 2.58333 + to within less than 144' 12= -- 12 144' and if a greater degree of exactness is required, we must take a greater number of integral fractions. EXAMPLES, 3 = 15 - - x= 2.46 to within less than 0.01. (10) 3 - - - x = 0.477 " " 0.001. 2 5 x= - 0.25 " 0.01. Of Logarithms. 227. If we suppose a to preserve a constant value in the equation ax; -- \T whilst N is made, in succession, equal to every possible number, it is plain that x will undergo changes corresponding to those made in NV. By the method explained in the last article, we can determine, for each value of N, the corresponding. value of x, either exactly or approximatively. The value of x, corresponding to any assumed value of the number N, is called the logarithm of that number; and a is called the base of the system in which the logarithm is taken. Hence, The logarithm of a number is the exponent of the power to which it is necessary to raise the base, in, order to produce the given number. The logarithms of all numbers corresponding to a given base conslitute a system of logarithms. Any positive number except 1 may be taken as the base of a system of logarithms, and if for that particular base, we suppose the logarithms of all numbers to be computed, they will constitute what is called a system of logarithms. IIence, we see that there is an infinite nw mber?f systems of ]oga rithrlls. CHAP. IX.] THEORY OF LOGARITHMS. 287 228. The base of the common system of logarithms is 10, and if we designate the logarithm of ally number taken in that system by log, we shall have, (10)~ = 1; whence, log 1 = 0; (10)1 = 10; whence, log 10 = 1; (10)2 = 100; whence, log 100 = 2; (10)3 -- 1000; whence, log 1000 = 3; &c., &c. -We see, that in the common system, the logarithm of any number between 1 and 10, is. found between 0 and 1. The logarithm of any number between 10 and 100, is between 1 and 2; the logarithm of any number between 100 and 1000, is be. tween 2 and 3; and so on. The logarithm of any number, which is not a perfect power of the base, will be equal to a whole number, plus a fraction, the value of which is generally expressed decimally. The entire part is called the characteristic, and sometimes the index. By examining the several powers of 10, we see, that if a number is expressed by a single figure, the characteristic of its logarithm will be 0; if it is expressed by two figures, the characteristic of its logarithm will be 1; if it is expressed by three figures, the chamacteristic will be 2; and if it is expressed by n places of figures, the characteristic will, be n- 1. If the number is less than 1, its logarithm will be negative, and by considering the powers of 10, which are denoted by negative exponents, we shall have, 1 -(10)=.1; whence, log.1 = —1. (10) - =.01; whence, log.01 = - 2. 100 (1) =. 10 — =.001; whence, log.001 = - 3. 100o &e., &e. &c., &C. Here we see that the logarithm of every number between 1 and.1 will be found between 0 and - 1; that is, it will oe equal to - 1, plus a fraction less than 1. The logarithm of any number 288 ELEMENTS OF ALGEBRA. [CHAP. IX. between.1 and.01 will be between — 1 and - 2; that is, it will be equal to - 0, plus a fraction. The logarithm of any number between.01 and.001, will be between - 2 and - 3, or will be equal to - 3, pluhs a fraction, and so on. In the first case, the characteristic is - 1, in the second - 2, im the third - 3, and in general, the chcaracteristic of the logarilthms of a decimtal fr'action is negative, and nznumerically 1 greater than the numzlber of O's which imnmediately follow the decinzal poinl. The deciimal part is always positive, and to indicate that the negative sign extends only to the characteristic, it is generally written over it; thus, log 0.012 = 2.079181, which is equivalent to - 2 +.079181. 2238", A table of logarithms, is a table containing a, set of numbers, and their logarithms so arranged that we may, by its aid, find the logarithm of any number from I to a given number, generally 10,000. The following table shows the logarithms of the numbers, from I to 100. N. o. N. Log. Log. N. og. 1 0.000000 26 1.414973 51 1.707570 16 1.880814 2 0.301030 27 1.431364 52 1.116003 17 1.886491 3 0 477121 28 1.4-17158 53 1.724276 18 1.892095 4 0.602000 29 1.462398 54 1.732394 19 1.897627 5 0.698970 30 1.477121 55 1.740363 S0 1.903090 6 0.78151 31 1.491362 56 1.748188 81 1.908485 7 0.845098 32 1.505150 57 1.155815 82 1.913814 8 0.903090 33 1.518514 58 1.763428 83 1.919018 9 0.954243 34 1.531479 59 1.770852 84 1.924279 10 1.000000 35 1.544068 60 1.718151 85 1.929419 11 1.041393 36 1.556303 61 1.785330 86 1.934498 12 1.079181 37, 1.568202 62 1.792392 87 1.939519 13 1.113943 38 1.579784.63 1.799341 88 1.944483 14 1.146128 39 1.591065 64 1.806180 89 1.949390 15 1.176091 40 1.602060 65 1.812913 90 1.954243 16 1,204120 4 1 1.612184 66 1.819544 91 1.959041 17 1.230449 42 1.623249 67 1.826075 92 1.963788 1 8 1.255273 43 1.633468 68 1.832509 93 1.968483 19 1.278754 444 1 G13453 69 1.838849 94 1.973128 20 1.301030 45 1.6532300 13 70 1.845098 95 1.917724 21 1. 32221I9 46 1.662758 71.1.851258 96 1.982271 22 1.342123 41 1.672098 12 1.851333 97 1.986772 23 1.361128 48 1.681241 3 1.863323 98 1.991226 24 1,380211 49 1.690196 14 1.869232 99 1.995635'25 1. 391'i940 5 0 1.698910 7 15 1.875061 100 2.000000 <~~~ _ _ _ _ _ CHAP. IX.] THEORY OF LOGARITHMS. 289 When the number exceeds 100, the characteristic of its logae rithm is not written in the table, but is always known, since it is 1 less than the number of places of figures of the given number. Thus, in searching for the logarithm of 2970, in a table of logarithms, we should find opposite 2970, the decimal part.4.72756. But since the number is expressed by four figures, the characteristic of the logarithm is 3. Hence, log 2970 = 3.472756, and by the definition of a logarithm, the equation ax = \, gives 103'172d56 -- 2970. General Pro2perties of Logarithms. 229. The general properties of logarithms are entirely inde, pendent of the value of the base of the system in which they are taken. In order to deduce these properties, let us resume the equation, ax = V, in which we may suppose a to have any positive value ex(ept 1. 230. If, now, we denote any two numbers by N' and ".3 and their logarithms, taken in the system whose base is a,,. by x' and x", we shall have, from the definition of a logarithm, ax' ='..(1), and, ax" N- - (2). If we multiply equations (1) and (2) together, member by, member, we get, C.'+x"- =,N'. " -. - (3). But since a is the base of the system, we have from theo definition,'V + /'- = log (N1 x N"); that is, The logayrithm of the product of two numbers is equal to the, sum of their logarithms. 19 290 ELEMENTS OF ALGEBRA. [CHEAP. X.m 231. If we divide equation (1) by equation (2), member by member we lave) NI -N' (4) Ilut from the definition, X.' —X=" log\~-; (v; N that is, The logarithm of the quotient which arises from dividing one number by another is equal to the logarithm of the dividend mnzinus the logarithm of the divisor. 232. If we raise both members of equation (1) to the nth power, we have, -an = N1V" (5). But froml the definition, we have, nx' -- log (N"') ); that is, The logarithmn of any power of a number is equal to the logarithm of the number m9ultiplied by the exponent of the power. 233. If we extract the nth' root of both members of equation,(1), we shall have, 1 n= (N1')" Nv - - (6). iBut from the definition, -- log (/fN'); that is,;The logarithnm Of any root of a zn6mber is equal to the loqarithm of the number divided by the index of the root. "23~4. From the prinlcii)s dcitlio-trated ir. the four precedillg articles, we deduce the following practical rules - First, To multiply quantities by means of their logarithm~s. Find from a table, the logarithms of the given factors, take the sum of these logarithms, and look'in the table fior the -or. res"pondin#g nvher;: his will be the p'ro0d!ct 3required, CHAP. IX.] THEORY OF LOGARITHMS. 291 Thus, log 7 - - 0.845098 log 8 - - 0.903090 log 56 - - - - - - 1.748188; hence, 7 x 8 56. Second. To divide quantities by means of their logarithms. Find the logarithm of the dividend and the logarithm of the divisor, from a table; subtract the later from the former, and look for the number corresponding to this difference; this will be tIe quotient required. Thus, log 84 - - 1.924279 log 21 - - - - 1.322219 log 4 - - - - - 0.602060; 84 hence, = 4= Third, To raise a number to any power. Find from a table the logarithm of the number, and multiply it by the exponent of the required power; find the number corresponding to this product, and it will be the required power. Thus, log 4 - - - - - 0.602060 log 64 - - - 1.806180; hence, (4)3 = 64.'Fourth, To extract any root of a number. Find from a table the logarithm of the number, and divide this by the index of the. root; find the number corresponding to this quotient, and it will be the root required. Thus. log 64 - - 1.806180(6 log 2.301030; hence, 6 644 = 2. By the aid of these principles, we may write the following eqpuivalent expressions: 292 ELEMENTS OF ALGEBRA. [CHAP. IX. Log (a.b..d....)- log a +log b logc.... Log ( a log a -l logb + log c-log d - log s Log (a". bn. cp....) =-n log a + nlog b - p log c -.... Log (a,2 - x2) = log (a + x) + log (a - ). Log (a — x2) =2 log (a - x) -) 2 log (a - x). Log (a3 X 4a) = 3- log a. -234. We have already* explained the method of determining the characteristic of the logarithm of a decimal fraction, in the common system, and by the aid of the principle demonstrated in Art. 231, we can show That the decimal part of the logarithmn is the same as the decimal part of the logarithm of the numerator, regarded as a whole number. For, let a denote the numerator of the decimal fraction, and let m denote the number of decimal places in the fraction, then will the fraction be equal to a and its logarithm may be expressed as follows: log = log a -- log (1o0)n= log a -- mlog10 - log a —m, but m is a whole number, hence the decimal part of the loga rithm of the given fraction is equal to the decimal part of log a, or of the logarithm of the numerator of the given fraction. I-Ience, to find the logarithm of a decimal fraction from the common table,Loik for the logarithm of the number, neglecting the decimal point, and then prefix to the decimal part found a negative characteristic equal to 1 more than the number of zeros which immediately follow the decimal point in the given decimal. The rules given for finding the characteristic of the logarithms taken in the common system, will not apply in any other systern, nor could we find the logarithm of decimal fractions CHGAP. IX-.' THEORY OF LOGARITHMS. 293 directly from the tables in any other system than that whose base is 10. These are some of the advantages which the common system possesses over every other system. 235. Let us again resume the equation = N. 1st. If we make N= 1, x must be equal to 0, since ac I; that is, T/he loyarithm of I ian any system is 0. 2d. If we make N - a, z must be equal to 1, since a' a. that is,?VWhatever be the base of a system, its logarithm, taken in that systems is equal to 1. Let us, in the equation, ax —, First, suppose a > 1. Then, when N= 1, x 0= 0; when 2N> 1, x > 0; when N< 19 W < 0, or negative; that is, In any system whose base is greater than 1, the logarithms of all numbers greater than I are positive, those of all numbers less than I are negative. If we consider the case in which _A < 1, we shall have a-X-X = or -= a. ax Now, if N diminishes, the corresponding values of x must increase, and when N becomes less than any assignable quantity, or 0, the value of x must be cm: that is, Th/e logarithmn of 0, in a system whose base is greater than I, is equal to - w. Second, suppose a < 1. Then, when N- 1, x - 0; when N < 1, x > 0; when N >l, z < 0, or negative: that is, 294 ELEMENTS OF ALGEBRA. LCHAP. IX In any system whose base is less than 1, the logarithms of all numbers greater than 1 are negative, and those of all numbers less than I are positive. If we consider the case ill which N< 1, we shall have ax _ N, in which, if gN be diminished, the value of x must be. increased; and finally, when N = 0, we shall have x = oo: that is, The logarithm of 0, in a system whose base is less than 1, is equal to - ca. Finally, whatever values we give to x, the value of ax or N will always be positive; whence we conclude that negative numbers have no logarithms. Logarithmic Series. 236, The method of resolving the equation, ax = be explained in Art. 226, gives an idea of the construction of loga. rithmic tables; but this method is laborious when it is necessary to approximate very near the value of x. Analysts have discovered much more expeditious methods for constructing new. tables, or for verifying those already calculated. These methods consist in the development of logarithms into series. If we take the equation, ax - y and regard a as the base of a system of logarithms, we shall have, log y = x. The logarithm of y will depend upon the value of y, and also upon a, the base of the system in which the logarithms are taken. Let it be required to develop log y into a series arranged according to the ascending powers of y, with co-efficients that are independent of y and dependent upon a, the base of the sy stem. CHAP. IX.3 LOGARITHMIC SERIES. 295 Let,is first assume a development of the required form, log y=M+Ny+P+ y ~3 + &c., in which lt, IV, P. &c. are independent of y, and dependent upon a. It is now required to find such values for these cox efficients as will make the development true for every va:lue of y. Now, if we make y = 0, log y becomes infinite, and is either negative or positive, according as the base a is greater or l.ess than 1, (Arts. 234 and 235). But the second member under this supposition, reduces to M, a finite number: hence, the development cannot be made under that form. Again, assume, log y = il]y + Ny2 + Py3 + &c. If we make y = 0, we have log 0 = O that is, -4- c = 0, which is absurd, and therefore the development cannot be made under the last form. Hence, we conclude that, The loyarithm of a number cannot be developed according to the ascending powers of that number. Let us write (1 + y), for y in the first member of the assumIed development; we shall have, log (1 + y) = My + ~n 2 + py3 Qy4 _+ &e. - - (1), making y = 0, the equation is reduced to log 1 = 0, which does lot present any absurdity. Since equation (1) is true for any value of y, we may write z for y; whence, log (1 + z) = Mz + l Y2 + PZ3 + Q4 + ac. -. - (2). Subtracting equation (2) from equation (1), meamber from memDer, we obtain, log (1 + y) - log (1 + z) = M(y - z) + N(y2 -- ) -- P(y3.- 3) + Q(4 - z) -. - (3). 296 ELEMENTS OF ArJGEBRA. [CHAP. IX, The second member of this equation is divisible by (y -z), let us endeavor to place the first member under such a form that it shall also be divisible by (y - z). We have, log (I + y) -1g ( Z) -_ log( {)= + - But since can be regarded as a single quantity, we may I + zr substitute it for y in equation (1), which gives, lo (1.+ 1; ( z + +2 Substituting this development for its equal, in the first member of equation (3), and dividing both members of. the resulting equation by (y - z), and we have,' + (Y - z) ( + -) + & -M + N(Y + ) +{2 + + )3 + P(y2 + yz + z2) - &c. Since this equation is true for all values of y and z, nrake -=y, and there will result = i q- 2Vy ~- 3Py2 + 4Qy.f + 5Ry4 + &c. + y Clearing of fractions, and transposing, we obtain, +J —f +2N y + 3P y2 + 4Q y3+ 5R ly4 + &c. = O, -~M + 21 + 2N + 3P + 4Q and since this equation is identical, we have, ~f — i= =O; whence, if-1f; 2N~- if —0; whence, V =3P + 2N=O; whence, _ 2N - ~I 3P 3 4Q + 3P; whence, Q = = - &C. CHAP. IX.] LOGARITHMIC SERIES. 297 The law of the co-efficients in the development. is evident; the co-efficient of y' is t- -, according as n is even or odd. Substituting these values for N, P, Q, &c., ir. equation (1), we find for the development of log (1 + y); 2 a +y3 log (I + y) =- My -- y y3 _ y 4. ( y2 + + 5. =~?2 y4 7o which is called the loyaritAmic series. Hence, we see that the logarithm of a number may be developed into a series, according to the ascending powers of a number less than it by 1. In the above development, the co-efficients have all been determined in terms of Al. This should be so, since M depends upon the base of the system, and to the base any value may be assigned. By examining equation (4), we see that, The expression for the logarithm of any number is composed of two factors, one deperdent on the number, and the other on the base of the system in which the logarithm is taken. The factor which depends on the base, is called the modulus of the system of logarithms. 237. If we take the logarithm of 1 + y in a new system and denote it by 1(1 + y), we shall have, "('1+y)-?y( 2 y3 y4 y5 +( ~y) (Y- q t- - ac. - (5), a 3 4: in which M' is the modulus of the new system,. If we suppose y to have the same value in equations (4) and (5), alud divide the former by the latter, member by member, we have log (I + y) _= e; whence, (Art. 183,) 1(1 - y): log (l + y):::,I':.I; hence, The logarithms of the same number, taken in two different systeme, are to each, other as the mol.ui of those systems. 298 ELEMENTS OF ALGEBRA. ICHAP. IX. 238. Having shown that the modulus and base of a system of logarithms are mutually dependent on each other, it follows, that if a value be assigned to one of them, the corresponding value of the other must be determined from it. If then, we make the modulus I'Z = 1, the base of the system will assume a fixed value. The system of logarithms resulting from such a modulus, and such a base, is called the Naperian System. This was the first system known, and was invented by Baron Napier, a Scotch mathematician. If we designate the Naperian logarithm by 1, and the logarithm in any other system by log, the above proportion becomes, I (1 +y): log(l +y):: 1 M; whence, 2 X I (1 + y) = log (1 + y). Hence, we see that, The lNaperiacn logarithm of any numbeer, multiplied by the modutus of any other system, will give the logarithm of the same number in that system. The modulus of the Naperian System being 1, it is found most convenient to compare all other systems with the Naperian; and hence, the -modulus of any system of logarithms, is The nu.mber by which if the Nacverian, logarithm of any number be multiplied, the product will be the logaritlhm of the same number in thmat system. 239. Again, M x (l1 + y) = log (1 + y), gives 1(l -y)= log(1 +Y); that is, T'he logarith!m of any number divided by the modulus of its system, is equal to the Naperian logarithm of the same number. 240. If we take the Naperian logarithm and make y-= equation (5) becomes, 1 1 1 1 12 3 4 5 CHAP. IX.] LOGARITHMIC SERIES. 299 a series which does not converge rapidly, and in which it would be necessary to take a great number of terms to obtain a near approxmnation. In general, this series will not serve for deterainling the logarithms of entire numbers, since for every number greater than 2 we should obtain a series in which the terms would go on increasing continually. 241. In order to deduce a logarithmic series sufficiently con verging to be of use in computing the Naperian logarithms of numbers, let us take the logarithmic series and make M'= 1. Designating, as before, the Naperian logarithm by Al we shall have, 2 y3 ~ y5 1(1 +y)- y + Y -- v-1+ I &c.. — (1). If now, we write in equation (1), -y for y, it becomes, 2 3 4 5 Subtracting equation (2) from (1), member from member, we have, (1 + y)- I(- )= (y ++ 3 5 7 + ) 75 9 But,.(1 + )-1 (- ~; whence, I1 ~ )=2 (i +- + -. 7 + y + &c.)- (4). 1 +-y z ~- 1 If now we make 1+ - z, wee shall have, 1 (1 + y)z = (1 — y) (z+ l), whence, y =ZF I Substituting these values in equatlor (4), and observing that + 1 l (z + 1) - wa we find, zZ 300 ELEMENTS O' ALGEBRA. [CHAP. IX. l(z + 1)- lz = 2(2 3(2 5( &c (5) 2z + 3(2+1 )3 5(2z+1)5 or, by transposition,'1 1 1 +a) 1 1(z +1) = lz + 2 + 3 (2+1) + 51) + (2t+l) & ) Let us make use of formula (6) to explain the method of computing a table of Naperian logarithms. It may be remarked, that it is only necessary to compute from the formula the logarithms of prime numbers; thoze of other numbers may be found by taking the sum of the logarithms of their factors. The logarithm of 1 is 0. If now we make z = 1, we canl find the logarithm of 2; and by means of' this, if we mako z = 2, we can find the logarithm of 3, and so on, as exhibited below. 11 = - - - - - - - = 0.000000; 12 2,( = ) - - - = 0.693147; 3' 3.oa 50. 3'' 4 = 2 X I2 - - - - - - - - - - - -=1.386294; /1 1 1 1 16 ~ 1~+13 - - - - - - - - - - -- -. 1 5-1.386294~ 2 ( -3+ 5- 93- 5 —5 + 7... =1.69437; 1I6 = 12+ 1 -- = 1.791759; 17 = 1.791759+2 (+ 3(13)3 2 5.(13)5 t...) 1.945910; 18 =14+12- = 2.079441; 1 9 =2 X 13 -.- - 2.197224; 110() 15 + 12. - -. — = 2.302585; &c. &c. In like manner, we may compute the Naperian logarithnls of all numbers. Other formulas may be deduced, which are CHIAP. IX.] LOGARITHMIC SERIES. 301 more rapidly zonverging than the one tAbove given, but this serves to shom the facility with which logarithms may be compated, 241*. We have already observed, that the base of the common system of logarithms is 10. We will now find its modulus. We have, 1(1 + y): log (1 + y): 1:21 (Art. 238). If we -make y= 9, we shall have, 110: log10::: lM. But the 110 = 2.302585093, and log 10 = 1 (Art. 22G); hence, 1 = 2.302585093 0.434294482 = the modulus of the common system. If now, we multiply the Naperian logarithms before found, by this modulus, we shall obtain a table of common logarithms (Art. 238). All that now remains to be done, is to find the base of the Naperian system. If we designate that base by e, we shall have (Art. 237), le: loge:: 1: 0.434294482. But le = (Art. 235): hence, I: loge:: 1: 0.434294482; hence, log e 0.434294482. But as we have already explained the method of calculating the common tables, we may use them to find the number whose logarithm is 0.431294482, which we shall find to be 2.718-281828; hence, e = 2.7182S1828...... We see frc m the last equation but one, that The modulus of the common system is equal to the common loga Tithm of the Naperian base. 302 ELEMENTS OF ALGEBRA. [CHAP, IX Of Interpolation. 242. When the law of a series is given, and several terms taken at equal distances are known, we may, by means of the formula, T a d + (n 1) n(n -- () -- (2) already deduced, (Art. - 209), introduce other terms between them, which terms shall conform to the law of the series This operation is called inzterpolation. In most cases, the law of the series is not given, but only uumerical values of certain terms of the series, taken at fixed intervals; in this case we can only approximate to the law of the series, or to the value of any intermediate term, by the aid of formula (1). To illustrate the use of formula (1) in interpolatirng a terl,, in a tabulated series of numbers, let us suppose that we have the logarithms of 12, 13, 14, 15, and that it is required to find the logarithm of 12~. ForIning the orders of differences from the logarithms of 12, 13, 14 and 15 respectively, and taking the first terms of each, 12 13 14 15 1.079181, 1.113943, 1.146128, 1.176091, 0.034762, 0.032185, 0.029963, - 0.002577, - 0.002222, + 0.000355, we find d -= 0.034762, de - 0.002577, ds = 0.000355. If we consider log 12 as the first term, we have also ac — 1.079181 and eMaking these several substitutions in the formula, and ne gleeting the terms after the fourth, since they are inappreclable, we find, T= (+ d,- + -1d = log 12l CHAP. IX.J FORMULAS FOR INTEREST. 303 or, by substituting for d,, d2, &c., their values, and tbr a its value, a...... - 1.079181 d -... 0.017381 -d2 - - - 0.000322 d3 - - -. - 0.000022 Log 121.. 1.096906 Had it been recquired to find the logarithm of 12.39, we should have made n =.39, and the process would have been the same as above. In like manner we may interpolate terms between the tabulated terms of any mathematical table. INTEREST. 243. The solution of all problems relating to interest, may be greatly simplified by employing algebraic formulas. In treating of this subject, we shall employ the following notation: Let p denote the amount bearing interest, called the principal; r " the part of $1, which expresses its interest for one year, called the rate per cent.; t " the time, in years, that p draws interest; i the int(,rest of p dollars for t years; S " p + the interest which accrues in the time t. This sum is called the amount. Simple Interest. To find the interest of a sum p for t years, at the rate r, and the amount thein due. Since r denotes the part of a dollar which expresses its interest for a single year, the interest of p dollars for the same 304 ELEMENTS OF ALGEBRA. [CHAP. IX. time will be expressed by pr; and for t years it will be t times as much: hence, i =ptr.... (1); and for the amount due, =p + ptr =p ( + tr) (2). EXAMPLES. 1. What is the interest, and what the amount of $365 for three years and a half, at the rate of 4 per cent. per annum. HIere, p -= $365; r - - 0.04; 100 t -3.5; i =ptr = 365 X 3.5 X 0.04 = $51,10: hence, S = 365 + 51,10 = $416,10. Present Value and Discount at Simple Interest. The present value of any sum X, due t years hence, is the prin cipal p, which put at interest for the time t, will produce the amount S. The discount on any sum due t years hence, is the difference between that sum and the present value. To find the present value of a sum of dollars denoted by S, due t years hence, at simple interest, at the rate r; also, the discount. We have, from formula (2), S =p + ptr;. and since p is the principal which in t years will produce the lum n we have, P _ + t a... (3); 1 + tr CHAP'. IX.I FORMULAS FOR INTEREST. O05 and for the discount, which we will denote by D, we have Str D)= ___ - tr - -- (4). 1 +t 1 I tr I 1. RequiTed the discount on $100, due 3 months hence, at thre rate of 5~ per cent. per annum. S- = $100 = $100, t = 3 months = 0.25. 5.5 r = 100 =.055. Hence, the present value p is 8 100 = =1+ $98,643. 1 + tr 1 +.01375 hence, D =S —p = 100- 98,64' = $1,357. Compound Interest. Compound interest is when the interest on a sum of money becoming due, and not paid, is added to the principal, and the interest then calculated on this amount as on a new principal, To find the amount of a sum p placed at interest for t years, compound interest being allowed annually at the rate r. At the end of one year the amount will be, =p + pr = p(l + r). Since compound interest is allowed, this sum now beccrnes the principal, and hence, at the end of the second year, the amount will be, s, = p( -+ r) + pr(l + r) p(1 + r)2. Regard p(l + r)2 as a new principal; we have, at the end of the third y ear, " =p( 1+ r)2+ pr(1 + r)2= p(1 + r)3; 20 306 ELEMENTS OF ALGEBRA. IOHAP. IX, and at the end of t years, S _=p( + (I) - -. (5). And from Articles 230 and 232, we have, log S= logp + t log (1 + r); and if any three of the four quantities S, p, t, and r, are giver. the remaining one can be determined. Let it be required to find the time in which a sum p will double itself at compound interest, the rate being 4 per cent. per annum. We have, from equation (5), S =p(l + r)t. But by the conditions of the question, S= 2p =p(1 + r)t: hence, 2 (1 + r)t. log 2 0.301030 log (1 +- r) 0.017033' = 17.673 years, = 17 years, 8 months, 2 days. lo find the Discount. The discount being the difference between the.lomn S and, we have, D= — (1 _ -- = S -(1. r CHAPTER X. GENERAL THEORY OF EQUATION. 244. Every equation containing but one uninown q lantity which is of the mth degree, in being any positive whole number, may, by transposing all its terms to the first member and dividing by the co-efficient of xm, be reduced to the form Xm + P.lm-1 + QXm —2 +....+ Tx + U- =0. In this equation P, Q...., U, are co-efficients in the most general sense of the term; that is, they may be positive or negative, entire or fractional, real or imaginary. The last term U is the co-efficient of xO~, and is called the absolute term. If none of these co-efficients are 0, the equation is said to be comnplete; if any of them are 0, the equation is said to be incomplete. In discussing the properties of equations of the mnth degree, involving but one unknown quantity, we shall hereafter suppose them to have been reduced to the form just given. 245. We have already defined the root of anr, equation (Art. 77) to be any expression, which, when slbstitzuted for the tcnkcnown qFf'etity in the equation, will satisfy it. XN e have shown that every equation of the first degree has one root, that e-very equation of the second degree has two roots; and in general, if the two members of an equation are equal, they must be so for at least some one value of the 308 ELEMENTS OF ALGEBRA. [CHAP. X. unknown quantity, either real or imaginary. Such value of the unknown quantity is a root of the epqation: hence, we infer, that ezery equation, of whatever degree, Ihas at least one root. We shall now demonstrate some of the principal properties of equations of any degree whatever. First Property. 246. In every equation of the form xm + Px'mI~ + Qxm-2' +.... + T + U= O, i' a is a root, the first member is divisible by x - a; and con versely, if' the first member is divisible by x - a, a is a root of the equation. Let us apply the rule for the division of the first member by x - a, and continue the operation till a remainder is found which is independent of x; that is, which does not contain x. Denote this remainder by R and represent the quotient found by Q', and we shall have, m + pxm-1.. + Tx + U Q'(x - a) + P. Now, since by hypothesis, a is a root of the equation, if we substitute a for x, the first member of the equation will reduce to zero; the term Q'(x - a) will also reduce to 0, and consequently, we shall have R - 0. But since 12 does not contain., its value will not be affected by attributing to x the particular value a: hence, the remainder P is equal to 0O whatever may be the value of x, and consequently, the first member of the equation xnm - Pxm-1 + Qxm-2.... ~ TX+ U 0, is exactly divisible by x - a. Conversely, if x - a is an exact divisor of the first member of the equation, the quotient Q' will be exact, and we shall have R = 0: hence, xmn + x.. + Tx + U= Q'(x- a). CHAP. X.] THEORY OF EQUATIONS. 309 If now, we suppose x = a, the second member will reduce to zero, consequently, the first will reduce to zero, and hence a will be a root of the equation (Art. 245). It is evident, fronm the nature of division, that the quotient Q' will be of the forim X2m- + P'_xm-2.+...... + + = 0. 247. It follows from what has preceded, that in order to di.s cover whether any polynomial is exactly divisible by the bino. mial x- a, it is sufficient to see if the substitution of a for a will reduce the polynomial to zero. Conversely, if' any polynomial is exactly divisible by x-ca, then we know, that if the polynomnial be placed equal to zero, a will be a root of the resulting equation. The property which we have demonstrated above, enables us to ditninish the degree of an equation by I when we ki -)w one of its roots, by a simple division; and if two or ra,re roots are known, the degree of the equation may be still fur,mer diminished by successive divisions. EXAMPLES. 1. A root of the equation, X4 - 25x2 + 60 - 36 - 0, is 3: what does the equation become when fireed of this c,,t? x4 - 25X2 + 60x 3 x - 3 - 3X x3 + 3X2- 16x - 12. -- 3x3 _ 25X2 3X3- 9,r2 - 16x2 -- 60x - 16x2 + 48x 12x - 36 12x - 36 Ans. x3 +- 3.2 - 16x -r 12 2. Two roots of the equation, x4 — 12.3 +48x2- 68x+15 = 0, are 3 and 5: what does the equation become when fred f4 theim. Aqns. x2 - 4x + I - C 310 ELEMIENTS OF ALGEBRA. [CtHAP. X. 3. A root of the equation, X3-6x2 + li- X-6 - 0 is 1: what is the reduced equation? Ans. x2 - 5x l- 6=0. 4. Two roots of the equation, 4x4- 14x3 - 5x2 + 31x + 6 =0, are 2 and 3: find the reduced equation. Ans. 4x2 + 6x + 1 = 0. Second Property. 248. ELvery equation inrvlvinjq but one unknbuwn quantity, has 8s msany roots as there are units in the exponent which denotes its deyree, and no, more. Let the proposed equation be xm - Pxm- - + Qxn"-2 -.... Tx+ U —0. Since every equation is known to have at least one root (Art. 245), if we denote that root by a, the first member will be divisible by x - a, and we shall have the equation, m + pxl-.. - (X - a) (m- - P'Xm-'...)- - - -( But if we place, x"l-1 -- P/xm-2 +..' 09 we obtain a new equation, which has at least one root. Denote this root by b, and we. have (Art. 246),,1-I + PXM-2 +... = (x - ) (.,,m-2 _ P//xm-3 +.. ) Substituting the second member, for its value, in equation (1), we have, 7m+ Px.-1 +... = (x - a) (, - b) (xm-2 -+ P"/ rm-3 -+.) * (2. Rleasoning upon the polynomial, xm-2 + P'xm-3a. as1 upon the preceding polynomial, we have x~-2 + Px,-3 +.= ( - C) ( m-3 + P//Xm-4 +..) anld by substitution, _4p.- pxm —t -,= (x -- a) (x -- 6) (x - c) (vm-3+P///xm-4) -. (3) CHAP. X. 1 THEORY OF EQUATIONS. 311 By continuing this operation, we see that for each binomial factor of the first degree with reference to x, that we separate, the degree of the polynomial factor is reduced by 1; therefore, after m- 2 binomial factors have been separated, the polynomial factor will become of the second degree with reference to x, which can be decomposed into two factors of the first degree (Art. 115), of the form x - k, x - 1. Now, supposing the mn - 2 factors of the first degree to have already been indicated, we shall have the identical equation, xm + Pxm: -.. - (x - a) (x - b) (x - c).. (x - k)( - I) = 0; from which we see, that the first member of the proposed eqaltion may be decomposed into m binomial factors of the first degree. As there is a root corresponding to each binomial factor of the first degree (Art. 246), it follows that the m binomial factors of the first degree, x- a, x- b, x- c....., give the m roots, a, b, c..., of the proposed equation. But the equation can have no other roots than a, b6 c... k, 1. For, if it had a root a', different from a, b, c.... 1, it would have a divisor x- a', different from x - a, x - b, x - c... x —, which is impossible; therefore, Every equation of the mth degree has mn roots, and can have no more. 249. In equations which arise from the multiplication of equal factors, such as (X - a)4 (x - b) (x - c)2 (x - d) = 0, the number of roots is apparently less than the number of units in the exponent which denotes the degree of the equation. But this is not really so; for the above equation actually has ten roots, four of which are equal to a, three to b, two to c, and one to d. It is evident that no quantity a', different from a, b, c, d, can verify the equation; for, if it had a root a/, the first mnrcn ber would be divisible by x —a', which is impossible. 312 ELEME NTS OF ALGEBRA. I CHAP. X, Consequence of the Second Property. 250. It has been slown that the first member of every equa. tion of the rnt-' degree, has m binomial divisors of the first degree, of the form x-a, x- b, - c,.. x - c, x- 1. If we multiply these divisors together, two and two, three asnd three, &c., we shall obtain as many divisors of the second, third, &c. degree, with reference to x, as we can form different combinations of m quantities, taken two and two, three and three, &c. Now, the number of these combinations is expressed by m-i mr-1 m —2 m 2. 2 m. 2 (Art. 132); 2 3 hence, the proposed equation has 2on divisors of the second degree; -- i Mn-2 2 3 divisors of the third degree; Mn-I rn-2 m- 3 2 3 4 divisors of the fourth degree; and so on. Composition of Equations. 251, If we resume the identical equation of Art. 248, xm+ Pxm-1 + Qxm-2... +. U =(x-a) (x — b)(x - c)... ( — )... and suppose the multiplications indicated in the second member to be performed, we shall have, from the law demonstrated in article 135, the following relations: P =- a —b —c -...-k-1, or -P = a+bc+.+-tl, Q = ab +-ac +- hc.... ak + -kl, X1= -- abc -abd -bcd... - ikl, cr -R abe + abd 4-...-4 ikl, U = +- abcd.... ikl, or i UT= abc... ikl. EHAP. X.] COMPOSITION OF EQUATIONS. 213 The double sign has been placed before the product of a, b, c, &c. in the last equation, since the product - a x - b X - c.. X - -, will be plius when the degree of the equation is even, and milnus when it is odd. By considering these relations, we derive the following conclusions with reference to the values of the co-efficients: 1st. The co-efficient qf the second term, withl its sign changed,?3 equal to the algebraic srnum of the roots of the equation. 2d. The co-efficient of the third term is equal to the sum of the different produtcts of the roots, taken, two in a set. 3d. The co-eticient of the fourth term, with its sign changed, is equal to the sum of the different products of the roots, taken three in a set, and so on. 4th. The absolute term, with its sign chang/ed whken the equation ts qof an odd degree, is equal to the continued product of all the roots of the equation. Consequences. 1. If one of the roots of an equation is 0, there will be no absolute term; and conversely, if there is no absolute term,:ne of the roots must be 0. 2. If the co-efficient of the second term is 0, the numerical sum of the positive roots is equal to that of the negative roots. P). Every root will exactly divide the absolute term. It will be observed that the properties of equations of the second degree, already demonstrated, conform in all respects to the princ ples demonstrated in this article. EXAMPLES OF THE COMPOSITION OF EQUATJONS. 1. Find the equation whose roots are 2, 3, 5, and - 6. We have, from the principles already established, the equation (x- 2) (x- 3) (x -5 ) (x + 6) = 0; whence, by the application of the preceding principles, we obtain the equation, 4 - 4x-3 29x2 + 156x - 10 = 0. 314 ELEMENTS OF ALGEBRA. LCHAP. X, 2. WN hat is the equation whose roots are 1, 2, and.- 3? A its. 3- 7x +6=0... WAhat is the equation whose rlts are 3, -4, 2 - nd ~2 -3 Ans. x4 - 3' - 15V2 + 49x - 12 -0. 4. What is the equation whose roots are 3 5, 3- 5, and - 6 ills. v3-3-2x+24 —-= 0. 5. W~hat is the equation whose roots are 1, - 2, 3, — 4, 5, and - 6? AIs. x6 + 3.5 - 41x4 - 87.3 + 4002 + 444x - 720 = 0. 6. What is the equation whose roots are.... 2 + -F 1, 2 --— 1, and — 3? Aits. x3 - 2-7x 1- 15 =0 Greatest Common Divisor. 252. The principle of the greatest common divisor is of frequent application in discussing the nature and properties of equations, and before proceeding fuirther, it is necessary to investigate a rllle for determining the greatest common divisor of two or niore polynomials. The greatest common divisor of two or more polynomials is the greatest algebraic expression, with respect both to co-efticients and exponents, that will exactly div\ide them. A polynomial is prime, when no other expression except l will exactly divide it. Two polynomials are primze with respect lo eacl other, when they have no common factor except 1. 253. Let A and B designate anly two polynomials arranged nwith reference to the same leading letter, and suppose' tha polynomial A to contain the highest exponent of the leading letter. Denote the greatest common divisor of A and B by 1) and let the quotients found by dividing each polynomLial by i CHAP. X.] GREATEST COMMON DIVISOR. 315 be represented by A' and B' respectively. We shall then have the equations, D -=-A', and -B'; w henlce A. A' x D and B —B' x D. Now, D contaiS all the factors colmmon to A aind B. For, if it does not, lt t us SlpI)poSe thalt A arid B have a co(,illinon ftlictor d which does not enter ),:and let us designlate the iuo. tients of A' and B', by this ftlteto, by A" and B". W\e shall then have, A -- A".d.D and B B".d.D; or, by division, A B -- A" and -- -" d. ad d. D B'd Since A"// and B" are entire, both A and B are divisible, by d. D, which must be greater than D, either with respect to its co-efficients or its exponents; but this is absurd, since, by hypothesis, D is the greatest common divisor of A and B. Therefore,.D contains'all the factol's comlmon to A and B. Nor can D contain any factor which is not common io A and B. For, suppose D to have a factor d' which is not con tained in A and B, and designate the other factor of D by D'; we shall have the equations, = A/. d. D' and B=B'.d'.D/; or, dividing both members of these equations by d', A B = A'. D' and -= B'. D'. d' Now, the second members of these two equations being en. tire, the first members must also be entire; that is, both A and B are divisible by d' and therefore the supposition that d' is not a common factor if A and B is absurd. HIence, 1st. The greatest common, divisor of two polynomials contains all the factors common to the polynonials, and does not contain any other factors. 316 ELEMENTS OF ALGEBRA. lCHAP. X 254. If, now, we apply the rule for dividing A by B, and continue the process till the greatest exponent of the leading letter in the remainder is at least one less than it is in the polynomial B, and if we designate the remainder bh R, and the quotient found, by Q, we shall have, A-Bx Q -R - - (1). If, as before, we designate the greatest comnllon divisor of. and B by D, and divide both nlembers of' the last equation by it, we shall have, A B B,=X Q+ ~ Now, the first member of this equation is an entire quantity, and so is the first term of the second member; hence must be entire; which proves that the greatest common divisor of A and B also divides R. If we designate the greatest common divisor of B and R by D', and divide both members of equation (1) by it, we shall have. A B R @,= _>x Q +. Now, since by hypothesis D' is a common divisor of B and R, both terms of the second member of this equation are entire; hence, the first member must be entire; which proves that the greatest common divisor of B and R, also divides A. We see that D', the greatest common divisor of B and R cannot be less than D, since D divides both B and R; nor can D, the greatest common divisor of A and B, be less than D', because D' divides both A and B; and since neither can be less than the other, they must be equal; that is, D = D'. Ilelce., 2d. Thle greatest common divisor of two polynomials. is the same as that between the second polynomial and their remainder after division. From the principle demonstrated in Art. 253, we see that we may multiply or divide one polynomial by any factor that is, C:HAP. X.1 GREATEST COMMON DIVISOR. 317 not coltained in the other, without affecting their greatest common divisor. 255, From the principles of the two preceding articles, we dedulce, for finding the greatest common divisor of two polynomials, the following RULE. I. Suppress thle mononmial factors common to all the terms of the first polynomial; do the same with the second polynomial; and if the factors so suppressed have a common divisor, set it aside, as forminyg a factor of the common divisor sought. II. Prepare the first polynomial in such a manner that its first term shall be divisible by the first term of the second polynozbial, both being arranged with reference to the same letter: Apply fthe rule for division, and continute the process till the greatest exponent of the leadinyg letter in the remainder is at least one less tian it is i, the second pol?/nomial. Sutppress, inz this remainder, all the.factors t/at are common to the co-efficients of the di /~reut powers of the leading letter; then take this result as a divisor and the second polynomial as a dividend, and proceed as before. III. Co~ntinue the operation until a remainder is obtainzed which will exactly divide the precediing divisor; tl.Jast remainder, mzultiplied by the factor set aside, will be the greatest common7 dczisor sought; if ano remainder is found which will exactcly divide the preceding divisor, then the fictor set aside is the greatest common divisor so'ught. EXAMPLES. 1. Find the greatest common divisor of the polynomials a3 - a2b - 3b3, and a2 - 5ab + 4b2. First Operation, AS~econd Operation. a3 —a2b - 3ab2 _ 3b3 a2 5ab + 4b2 a2 - 5ab - 4b2 — b 4a2b- ab2 - 3b3 a 4b -4ab - 4b2 a - 4 1st rem. 19ab2 - 19b3 0. or, 19b2 (a - b). tHence, a - b is the greatest common divisor. $18 ELEMKNTS OF ALGEBRA. [CHAP. X We begin by dividing the polynomial of the highest degree by lhat of the lowest; the quotient is, as we see in the above table, a + 4b, and the remainder 19ab2 - 19b3. But, 19ab2 - 19b3 = 1962 (a - b), Now, the factor 19b2, will divide this remainder without dividing a2 - 5ab + 462: hence, thb factor must be suppressed, and the question is reduced to finding the greatest common divisor between a2 - 5ab q- 462 and a - b. Dividing the first of these two polynomials by the second, there is an exact quotient, a - 4b, hence, a - b is the greatest coinmon divisor of the two given polynomials. To verify this, let each be divided by a - b. 2. Find the greatest common divisor of the polynomials, 3a5 - 5a3b2 + 2ab4 and 2a4 - 3a2b2 + b4. We first suppress a, which is a factor of each term of the first polynomial: we then have, 3a4 - 50262 -+ 2b4 4i 2a4 - 3a2a2 + b6. We now find that the first term of the dividend will not contain the first term of the divisor. We therefore multiply the dividend by 2, which merely introduces into the dividend a factor not common to the divisor, and hence does not affect the common divisor sou(ght. We then have, 6a4 - 10a2b2 + 4b4 1-a4 - 3a2b2 + b4 6a4 - 9a262 + 3b4 3 - a2b2 + b6 - b2 (a2 - b2). We find after division, the remainder - a2b2 +- b which we put under the formn - 2 (a2 - b2). We then suppress - 62 nld dio ide, 2a4 - 3a2b2 +,4 a2 -_ b2 2a4 - 2,262 2a2 - b2 - 2b2 + b4 _ a2b2 + b4. HIence, a2 —b2 is the greatest crnmon divisor. CHAP. X.J GREATEST COMMON DIVISOR. 319 3. Let it be required to find the, greatest common divisor between the two polynomials, -313 3ab2 - a2b + a3, and 4b2 - 5ab -+ -2. First Operation. - 12b3 + 12ab2 - 4a2b + 4a3 A14b2 -.jab + a2 st reml. _ - 3ab2 - a2b +- 4a3 -3, - 3a - 12ab2 - 4a2b - 16a3 2d rem. - -1 9a2b +- 19a3 or, 19a2 ( —b + a). Second Operation. 4b2 -- 5tb + a2 -- b + a,- b 4- a-2 -4b -+ a O. Hence, - b + a, or a - b, is the greatest common divisor In the first operation we meet with a difficulty in dividing the two polynomials, because the first term of the dividend is not exactly divisible by the first term of the divisor. But if we observe that the co-efficient 4, is not a factor of all the terms of the polynomial 412 - 5ab + a2,,uad therefore, by the first principle, that 4 cannot form a part of the greatest common divisor, we can, without affecting this common divisor, introduce this factor into the dividend. This gives, - 1263 + 12a62 - 4ia2 + 4a3, and then the division of the terms is possible. Efflcting this division, the quotient is - 3b, and the re nmainder. is, - Oab2 - a2b + 4a3. As the exponent of b in this remainder is still equal to that of b in the divisor, the division may be continued, by multiplying this remainder by 4, in order to render the division of the first term possible. This done, the remainder becomes - 12u.b2 - 4a2b + 16Ga3; 820 ELEMENTS OF ALGEBRA. [CHAP, X. which, divided by 4b2 — 5a6b ~+ 2, gives the quotient — 3a, which should be separated from the first by a comma, having no cnlnexion with it. The remainder after this division, is - 19a2b + 19a3. Placing this last remainder under the form 19a2 (-b + ca), and suppressing the factor 19a2, las forming no part of the common divisor, the question is reduced to finding the greatest common divisor between 4b2 - 5ab + a2 and -b +- a. Dividing the first of thess polynomials by the second, we obtain an exact quotient, - 4b - a: hence, — b + a, or a -b, is the greatest common divisor sought. 256. In the above example, as in all those in which the exponent of the leading letter is greater by I in the dividend than in the divisor, we can abridge the operation by first multiplying every term of the dividend by the square of the coefficieitt of the first term of the divisor. We can easily see that by this means, the first term of the quotient obtained will contain the first power of this co-efficient. Multiplying the divisor by the quotient, and making the reductions with the dividend thus prepared, the result will still contain the co-efficient as a factor, and the division can be continued until a remainder is obtained of a lower degree than the divisor, with reference to the leading letter. Take the same example as before, viz.: - 3b3 + 3ab2 - a2b + a+ and 4b2 - 5ab + a2, and multiply the dividend by 42 = 16; and we have First Operation. - 4Sb3 + 48ab2 - 16a2b + 16a3 4b2 - 5ab - a2 - 12ab2 - 4a2b + 16a3 12b - 3a Ist remainder, - 19a2b - 19a3 or, 19a2 (-b +- a). CHAP. X.] GREATEST CO0MMION DIVISOR. 321 Second Operation. 4b2 —5ab + a2 -- b a - ab + a2 - 4b + a 2d remainder, - 0. W.hen the exponent of the leading letter in the dividend exceeds that of the same letter in the divisor by two, three, &c., multiply the dividend by the third, fourth, &e. power of the co-efficient of the first term of the divisor. It is easy to bee the reason of this. 257. It may be asked if the suppression of the factors, cornl mon to all the terms of one of the remainders, is absolutely eecessary, or whether the object is merely to render the operations hmore simple. It will easily be perceived that the suppression of these factors is necessary; for, if the factor 19a2 was not suppressed in the preceding example, it would be necessary to Mlultiply the whole dividend by this factor, in order to render its first term divisible by the first term of the divisor; but, then, a factor would be introduced into the dividend which is also contained in the divisor; and, consequently, the require&d goreatest common divisor would contain the factor 19a2 whichl, should form no part of it. 258. For another example, let it be required to find the7 greatest common divisor of the two polynomials, a4 +- 3a3b + 4a2b2 - 6ab3 + 2b and 4a2- + 2ab2 - 2b3, vr simply o0 a4 + 3a3b + 4a2b2 - 6ab3 - 2b4 and 2a2 + ab -b2, since the factor 2b can be suppressed, being a factor of th'e second polynomial and not of the first..Fizst Operation. 8a" -4- 21a3b +J 32a2b2 - 48ab3 + 16b4 l 2a2 + ab - b2 -+ 20a3b + 36a2b2 - 48ab3 H- 16b4 4a2 - 10ab -- 03l/ + 26ab2 - 38ab3 -j- 16b4 lst remainder, - 51ab3 + 29b4 or, -- b3(51a - 29b). 322 ELEMENTS OF ALGEBRA. I CHAP. so Second Operation. Multiply by 2601, the square of 51. 5202a2 + 2601ab - 260162 1 5ia - 291 5202a2 - 2958ab 102a + 1096 1 st remainder, + 5559ab - 2601 b6 5559ab - 3161 b2 2d remainder, + 5606b2 The exponent of the letter a in the dividend, exceeding that eof the same letter in the divisor, by two, the whole dividend is multiplied by 23 = 8. This done, we perform the division, and obtain for the first remainder, - 51ab3 + 2,94. Suppressing - b3, this remainder becomes 51a - 29b; and the new dividend is 2a2 + ab - b62. Multiplying the dividend by (51)2 =2601, then effecting the division, we obtain for the second remainder + 560b2. Now, it results from the second principle (Art. 254), that the greatest icommon divisor must be a factor of the remainder after each division; therefore it should divide the remainder 560b2. But this remainder is ilzdependent of the leading letter a: hence, if'the two polynomials have a common divisor, it must be inde-:pendent of a, and will consequently be found as a factor in the Aco-efficients of the different powers of this letter, in each of the,proposed polynomials. But it is evident that the co-efficients of these powers have not a conimo n eactor. tence, the two gir,nC polynomials are primze with resperct to each other. 259, The rule for-finding the greatest common divisor of two,polynomials, may readily be extended to three or nlore pily lnomials. For, having the polynomials A, B, C, I), &c., if we find the greatest common divisor of A and B, and then th}e Lreatest common divisor of this result and C, the divisor so ob CHAP. X.] GREATEST COMMON DIVISOR. 323 t..inled will evidently be the greatest common divisor of A, B, and C; and the same process may be applied to the remaining polynomials. 260. It often happens, after suppressing the monomial factors common to all the terms of the given polynomials, and arranging the remaining polynomials with reference to a particular letter, that there are polynomial factors common to the co-efficients of the different powers of the leading letter in one or both polynomials. In that case we suppress those factors in both, and if the suppressed factors have a common divisor, we set it aside, as forming a factor of the common divisor sought. EXAMPLE. Let it be required to find the greatest common divisor of the two polynomials a2d2 - c2d - a2c2 -+ C4 and 4a2d' 2ac2 +r 2c3 - 4acd. The second contains a monomial factor 2. Suppressing it, acp.d arranging the polynomials with reference to d, we have (C2 - 2) d - a2c2 - c4, and (2c - 2ac) d - ac + C3. By considering the co-efficients, a2 - c2 and - ac2 c4, in the first polynomial, it will be seen that - a2rc +- c4 can be put under the form — c2(a2 - c2): hence, a2 - C2 is a common factor of the co-efficients in the first polynomial. In like manner, the co-efficients in the second, 2a2 - 2ac and -ac2 + C3, can be reduced to 2a(a -c) and — c2(a - c); therefore, a -c is a common factor of these co-efficients. Comparing the two factors a2 - c2 and a - c, we see that the ias'; will divide the first; hence, it follows that a- c is a common factor of the proposed polynomials, and it is therefore a factor of the greatest common divisor'. Suppressing a2 - c2 in the first polynomial, and a - c in the second, we obtain the two polynomials, d2 - c2 and 2ad —-, 324 ELEMENTS OF ALGEBRA. L[CAP. X. to wnich the ordinary process may be applied. d2 - c2 2ad c2 4a2d2 - 4a2c2 2ad + c2 +- 2ac2d - 4a2c2 - 4a2c2 + C4. After having multiplied the dividend by 4a", and performed the division, we obtain a remainder - 4a2c2 + c4, independent of the letter a': hence, the two polynomials, d2 - c2 and 2ad - c2, are prime with respect to each other. Therefore, the greatest common divisor of the proposed polynomials is a - c. 261. It sometimes happens that one of the polynomials coil tains a letter which is not contained in the other. In this case, it is evident that the greatest common divisor is independent of this letter. Hence, by arranging the polynomial which contains it, with reference to this letter, the required comNmon divisor will be the same as that which exists between the coefficients of the dfi'erent powers of the'principal letter and the second polynomial. By this method we are led, it is true, to determine the great est common divisor between three or more polynomials. But they will be more simple than the proposed polynomials. It often happens, that some of the co-efficients of the arranged polynomial are monomials, or, that we can discover by simple inspection that they are prime with respect to ea(ch other; and, it, this case, we are certain that the proposed l(,lcynomials are p[rime with respect to each other. Thus, in the example of the last article, after having suppressed the, common factor a - c, which gives the results, d2 - c2 and 2ad - c2 we know immediately that these two polynomials are prime witil resplect to each other; for, since the letter a is contained in the second and not in the first, it follows from what has just been said, that the common divisor must be contained in the co-efficients 2,2 CHAP. X.] GREATEST COOMMON DIVISOR. 325 and - C2; but these are prime with respect to each other, and consequently, the expressions d2 - c2 and 2ad - c2, are also prime with respect to each other. Let it be required to find the greatest common divisor of the two polynwmials, 36c-q + 30top + 18bc + 5rnpq, and, 4adq - 4'2fg + 24ad - 7fgq. Now, the letter b is found in the first polynomial and not in the second. If then, we arrange the first with reference to b, we have, (3cq + 18c) b + 30tnp + 5npgq, and the required greatest common divisor will be the same as that which exists between the second polynomial and the two co-efficients of b, which are, 3cq + 1Sc and 30mop 5mpq. Now, the first of these co-eficients can be put under the form 3c(q -- 6), and the other becomes 5mnp(q + 6); hence, q + 6 is a common factor of these co-efficients. It will therefore be sufficient to ascertain whether q + 6 is a factor of the second polynomial. Arranging this polynomial with reference to q, it becomes (4ad - 7f)q - 42fg + 24ad; and as the second part, 24ad - 42fg = 6(4ad - 7fg), it follows that this polynomial is divisible by q + 6, and gives the quotient 4ad - 7. Therefore, q + 6 is the greatest common divisor of the proposed polynomials. EXAMPLES. 1. Find the greatest common divisor of the two polynontia!s 6x5 - 4;4 - 113 - 3x2 - 3 - 1, and 4z4 + 2.x3 - 18x2 4- 3x- 5. Ans 2x3 _-,2._L - 326 ELEMENTS OF ALGEBRA. I CHAP. X. 2. Fiid the greatest common divisor of the polynomials 20x6 - 12x0 + 16X4 - 15x3 + 14X2 - 15x + 4, and 15x4 - 9x3 + 472 - 21x -+ 28. Ans. 5x2 - 3x r 4, 3, Find the greatest common divisor of the two pdlynorliais 5ca42 + 2a3b3 -F ca2 - 3a2b4 + bca, nid a5 + 5aSd -as3b2 + 5a24d. Ans a2 + ab. Transformation of Equations. 262. The object of a transformation, is to change anl equation frorn a given form to another, from which we can more readily determine the value of the unknown quantity. First. To change a given equation involving fractional co-eficients to anothe? of the same generalform, but having the co-efficients of all its terms entire If we have an equation of the form xm + pxm-1 + Qxn-2 +... Tx + U =' 0, y and make x = k in which y is a new unknown quantity, and k entirely arbitrary; we shall have, after substituting this value for x, and multiplying every tei m by km, ym + p.ypm-l + Qlc2ym-2 + Rks3yn-3 -... Tkm-ly + Ukm U, an equallltion in which the co-efficients of the different powers of y are equal to those of the same powers of x in the given equation, multiplied respectively by ~O, k1, )2, k3, k4, &c. It is now requtred to assign such a value to k as will make thie coefficients of the different powers of y entire. To ilI;:trate, let us take, as a general example, the equation a x3 c e g -0, b d~ f T CHEAP X.] TRANSFORMATlONT OF EQUATIONS. 327 which beccmes, after substituting - for x, and multiplying by AP, ak ck/2 ek3 gk4 Now, thlere may be two caseslst. Where the denominators b, d, f, h, are prime with respect to each other. In this case, as k is altogether arbitrary, take -h b dfh, the product of the denominators, the equation will then become, p4 ~ adfA. y3 + Cb2df2h2. y2 e13d3f2h3. y + gb4d4Pf4h3 =0, in which the co-efficients of y are entire, and that of the first term is 1. 2d. When the denominators contain common factors, we shall evidently render the co-efficients entire, by making k equal to the least comnmon multiple of all the denominators. But we can silmplify still more, by giving to k such a value that k1, k2, 3,... sbhall contain the pririme factors which compose b, d, f, A, raised to powers at least equal to those which are found in the denominators. Thus, the equation 5 5 7 13 x4 - X3 ~- 3 - = 0 X 12 x 9000 becomes 5/k 5k2 7c3 7 13,k4 y4 - y3 + ~ — -- y 900 =0, Y4- 6 f ~ Y3 + l2 Y 150 9000 after making z = Y and reducing the terms. First, if we make /c = 9000, which is a multiple of all the other denominators, it is clear that the co-efficients become entire nulmbers. But if we decomnpose 6, 12, 150, and 9000, into their prilme factors, we find, 6= 2 x 3, 12=22 3, 150 =:2 3 2, 900 = 23 32 X 5' and by mraking = 2 > 3 x 5, 328 ELEMENTS OF ALGEBRA. [ CHAP. X the product of the different prime factors, we obtain k2 = 2 X 32 X 52, c3 =:3 33 X 53, c4 =- 24 X 3' X 54; N hence we see that the values of k, k2, k3, k4, contain the prime factors of 2, 3, 5, raised to powers at least equal to those which enter into (6, 12, 150, and 9000. IHence, making k -2 x 3 x 5, is sufficient to make the denomir ators disappear. Substituting this value, the equation becomes 5.2.3.5 5.22.32 52 7.23.3.3 13.2434.54.3 2.3.52 Y 2.32.53 0 which reduces to yQ - 5.5y3 + 5.3.52y2 _ 7.22.32.5y - 13.2.32.5 _ 0; or, y4- 25y3 + 375y2 - 1260y - 1170 -= 0. Hence, we perceive the necessity of taking k as small a number as possible: otherwise, we should obtain a transformed equation, having its co-efficients very great, as may be seen by reducing the transformed equation resulting from the supposi tion k = 9000. Having solved the transformed equation, and found the values of y, the corresponding values of x may be found from the equation, x by substituting for y and kI their proper values. EXAMPLES. 3 7 2 11 25 3 36 72 Making x =,Y and we have, 3 - 14y2 + Ily - 75 = 0. 13 x4 21 32 43 1 2., x'" —x4+ —— x2- X -—,0 2.-012 40 25 600 800 Mauking, x 22Y = Y and we have, -, 65y4 + 1890y3 - 30720y2 - 928800y - 972000 = 0. CHAP. X.] TRANSFORMATION OF EQUATrIONS. 329 Second. To make the second or any other terrn disappe:r from an equation. 263. The difficulty of solving an equation generally dilLinishes with the number of terms involving the unknown quantity. Thus the equation x2 = q, gives immediately, x =- + while the complete equation x2 + 2px + q = 0, requires preparation before it can be solved. Now, any given equation can always be transformed into ans incomplete equation, in which the second term shall be wanting. For, let there be the general equation, xm~P i xzm- + Qxm-2 +.. Tx + U- O. Suppose x -- + --, u being a new unknown quantity, and x' entirely arbitrary. By substituting n + x' for x, we obtain (U +- X)m + P (~b 2')m1 - Q ( + X')M-2... + T(it + )') -- U= 0. Developing by the binomial formula, and arranging with reference to u, we have U~+ mxj | fum~- 1 X+' 2 tm —2 4- X/m + P + (m - 1) Px' + Px'n-I1 + Q + Q,/n2,-2 -0. + Tx' + U Since x' is entirely arbit:rary, we may dispose of it in such way that we shall have mn'+ P=O; whence, x' 330 ELEMENTS OF ALGEBRA. [CHAP X. Substituting this value of x' in the last equation, we shall obtain an incomplete equation of the form, um +~ Q'um-2 + R'z,'-3 +... T'u +- U' = 0, in, which the second term is wanting. If this equation were solved, we could obtain any value of a corresponding to that of u, friom the equation = u + X/, since x = -- -. We have, then, in order to make the second term of all equation disappear, the following RULE. Substitute for the unknown quantity a new unknown quantity mninus the co-efficient of the second term divided by the exponent which expresses the degree of the equation. Let us apply this rule to the equation,,2 + 2px = q. If we make x = U —, we have (U - p)~+2p (U -p)=q; and by performing the indicated operations and transposing, we find,2 p2 + q. 263*. Instead of mlaking the second term disappear, it may be required to find an equation which shall be deprived of its third, fourth, or any other term. This is done, by making the co-efficient of u, corresponding to that term, equal to 0.'For example, to make the third term disappear, we make, in the transformed equation, (Art. 263), 2-x/2 +- (i2 -- 1) P' ~- Q = 0, fromn which we obtain two values for x', which substituted in the transformed equation, reduce it to the form, Urm + P'um —1 A- R'um... T'u + U, = 0. CHAP. X.] OF DERIVED POLYNOMIALS. 331 Beyond the third term it will be necessary to solve an equation of a degree. superior to the second, to obtain the value of x'; and to cause the last term to disappear, it will be necessary to solve the equation, x'm +- px/rm-1.. 4- Tx' + U r O, which is what the given equation becomes when x' is sub~ stituted for x. It may happen that the value, P m= which makes the second term disappear, causes also the disap pearance of the third or sonle other term. For example, in order that the third term may disappear at the same time with the second, it is only necessary that the value of x, which results from the equation, P x = - -, shall also satisfy the equation, — m 2 + (n - 1) Px' + Q = 0. P Now, if in this last equation, we replace xo by —, we have m-1P2 P2 2 -(m-1)- - + Q =O, or (m-1 )P2-2Q= O; and, consequently, if rn- 1' the disappearance of the second term will also involve that of the third. Formationm of Derived JPolynomials. 264. That transformation of an equation which consists in substituting u + x' for x, is of frequent use in the discussion of equatiIons. In practice, there is a very simple method of obtainirg the transformed equation which results from this sub 86.2 tfutiol ^ 332 ELEMENTS OF ALGEBRA. LCHAP. X, To show this, let us substitute for x, u + x' in the equation x.m + pm-1 + Qxm-2 + RXm-3 +.. Tx - U=O; then, by developing, and arranging the te; ms according to the ascending powers of ut, we have Xm + mX'm-l u + -t 1 X/m-2 U2 + t. m 2 ~ Qxl"'-2 + (m-2) Q —3 +Qm +(m2)" Qxm- O. + Tx' + T By examining and comparing the co-efficients of the different powers of ut, we see that the co-efficient of u~, is what the first mnember of the given equation becomes when x' is substituted in place of x; we shall denote this expression by X'. The co-efficient of u1 is formled from the preceding terml X', by multiplying each term of X/ by the exponent of x/' in that term, and then diminishing this exponent by 1; we shall denote this co-efficient by Y'. The co-efficient of u2 is formed from YF, by multiplying each term of Y' by the exponent of x' in that term, dividing the product by 2, and then diminishing each exponent by 1. Repre. senting this co-efficient by 2, we see that Z' is fornmed from F. in the same manner that Y/ is formzed from X~. In general, the co-efficient of any power of u, in the abo-v transformed equation, may be found from the preceding co-efficient in the following manner, viz.:Mtltiply each term of the preceding co-e icient by the exponent of x' in that term, and diminish the exponent of x/ by I; then divide the algebraic sum of These expzressions by the number of y re. ceding co-eficients. CHAP. X.] OF DERIVED POLYNOMIALS. 333 The law by which the co-efficients, Z' V' XI, y/, 1.2' 1.2.83' are derived from each other, is evidently the same as that which governs the formation of the numerical co-efficients of the terms in the binomial formula. The expressions, Y, 1 Y V, W',, &c., are called successive derived polynomials of X', because each is derived from the preceding one by the same law that Y' is derived from AY'. Generally, any polynomial which is derived from another by the law just explained, is called a derived polynomial. Recollect that X' is what the -given polynomial becomes when ~x is substituted for x. Y' is called the first-derived polynomial; Z' is called the second-derived polynomial; V' is called the third-derived polynomial; &ec., &Gc. We should also remember that, if we make u = 0, we shall have x' x= whence X' will become the given polynomial, from which the derived polynomials will then be obtained. 2650 Let us now apply the above principles in the following EXAMPLES. 1. Let it be required to find the derived polynomials of the first member of the equation 3X4 + 6x3 - 3x2 + 2x + 1 = 0. Now, u being zero, and x' = x, we have from the law of form'.Lg the derived polynomials, X' 3x4 + 6X3 - 3x2 + 2 + 1; Y/ = 12x3 + 18"-6x -_ 2; ZA 26X2 + 16,X 6 Z' =-36x2~36x -6; V' -72x + 36; W' = 72. 334 ELEMENTS OF ALGEBRA. LCHAP. X. It should be remalrkcd that the exponent of x, in the terms 1, 2, — 6, 36, and 72, is equal to 0; hence, each of those terms disappears in the following derived polynomial. 2. Let it be required to cause the second term to disappear in the equation 4 - 12x3 + 17x2 - 9x + 7 = 0. 12 Mfake (Art. 263), x = u +- = u t +3; whence, x'=- 3. The transformed equation will be of the form Z/ T~ 2 + V 3 + U4 = 0, X' + Y'u + U 2 x 3 2 2x3 and the operation is reduced to finding the values of the co. efficients X', Yr', L -, 2.3' Now, it follows from the preceding law, for derived poly. nomials, that I' - (3)4-12.(3)3+17.(3)2-9. (3)+7, or X' = -110; / =4. (3)3 —36. (3)2+-34.(3)1 —9, or - - -- 13; =6.(3)236. (3)1~17, or - 37; 2 =4(3)112 - - = 0. Therefore, the transformed equation becomes u4 -37u2-123z -1 110 0. 3. Transform the equation 4x3 - 5x2 + 7 - 9 -- 0 Into another equation, the roots of which shall exceed those of the given equation by 1. Make, x u —1; whence x'=- 1 and the transformed equation will be of the form Y 4,, -z12 Jr- 12 u4Y 1,.2,3 CHAP. x.] DERIVED POLYNOMIALS. 335 We have, from the principles established, X' -_ 4.( —1)3- 5.( —1)2+7.(-1)19, or e = -72,; Y = 12. (-1)-10 - -. 11- + 7 - 29; 2 = 12.(- 1)- 5 I) e = 17; 2' 23 --- =....-.+ 4. _. = l~2. 3 Therefore, the transformed equatior is, 4u3 - 17u2 + 29 - 25 = 0. 4. What is the transformed equation, if the second term be made to disappear from the equation x5 — 10x4 + 7x3 + 4x — 9 = 0? Ans. u5 - 33u3 - 118u2 -152u - 73 = 0. 5. What is the transformed equtation, if the second term tb made to disappear from the equation 3x3 - 15x2 + 25x - 3 0 152 An3s. U3 - 0. 27 O. Transform the equation 3x4- 13x3 3+ -7x2- 8 9 _ 0 into another, the roots of which shall be less than the roots of the given equation by 3. ns 34 93 tt2 65 34 Properties of Derived Polynomials. 206. We will now develop some of the properties of derived polynomials. Let xm + px- + Qx2-2... Tx + U = O0 be a given equation, and a, b, c, d, &c., its in roots. We shall then have (Art. 248), m+ Pm-1 + Qx2 -.. = (x - a) (v -,) (x c)... (X - ) 336 ELEMENTS OF ALGEBRA. [CHAP. X, Making x = x' + u, or omitting the accents, and substituting x + u for x, and we have (v + u)tt + P(x u)-l - ~ 1 =+ (x + — a) (x -ub)...; lr, changing the order of x and v, in the second member, and regarding z —a, - b,... each as a single quantity, (X + U)m +rP(x -' U)m-1.. =-(u + x -a) (z+x-b-)... (u+x —). Now, by performing the operations indicated in the two members, we shall, by the preceding article, obtain for the first mnember, z X+ YIu ~-U2 +.. tm; X being the first member of the proposed equation, and Y, Z, &c., the derived polynomials of this member. With respect to the second member, it follows from Art. 251: 1st. That tha term involving,u0, or the last term, is equal to the product (z - a) (x - b)... (x - ) of the factors of the proposed equation. 2d. The co-efficient of u is equal to the sum of the products of these m factors, taken m - 1 and m - 1. 3d. The co-efficient of It2 is equal to the sum of the products of these m factors, taken m- 2 and m- 2; and so on. Moreover, since the two members of the last equation are identical, the co-efficients of the same powers of u in the two members are equal. Hence, X = (x - a) (x - b) (x - c).. ( -1), which was already shown. HTence, also, Y, or the first derived polynomial, is equal to the sutm of the produlcts of the in factors of the first degree in the proposed equatiosn, taken m- 1 and in - 1; or eqztal to the alyebraic sum of all the quotients kthat can be obtained by dividing X by each of the m factors of the first degree in the proposed eqluation, tlhat is, X X X X -a x —b' -c +- -. Xs-a x -bb xc- c X CHA P.] EQUAL ROOT'S. 337 Als-o,, that is, the second derived polynomial, dividled by 2, is equlC to the suOm of the products of the mi faclors of the first me- mber of the proposed equation, tacenz m- 2 and la — 2; or equzl to the sun, of tie auotients obtained by dividing X by each of th6:lbfiletet factor's of Athe secondc degree; that is, ~ — (,~ -. ~) (~ -- ~) + (x --,) (x -- )... (x — h.) ( — _' aid Lo on. Of Equal.Roots. 267. An equation is said to contain equal roots, when its first member contains equal factors of the first degree with respect to the unknown quantity. Wllen this is the case, the derived polynomial, which is the sum of the products of the m factors takenl mn - 1 and m - 1, contains a factor in its different parts, which is two or more tilmes a factor of the first member of the proposed equation (Art. 066): hence, There muls be a commono divisor between, the first member of the proposed eq'uation, and its first derived polynomial. It remains to ascertain ihe relation between this common divisor and the equal fhetors. 268~ HavIing given an equation, it is required to discover whether it has equal roots, and to determine these roots if possible. Let us make X =m +xP- QxPv'f m-2 + -.. + Tx + U=Oq and suppose that the second member contains n factors equal to z- a, n' factors equal to x - b, n" factors equal to x - c.., and also, the simple factors.-p, x - q, x - r...; we shall then have, X:- (-a)"t (x - b) /(x -c)".. (x - p) (x - q) (x - r) (1). We have seen that YI or the derived polyn.)mial of X, is the sum of the quotients obtained by dividing X by each of the in factors of the first degree in the proposed equation (Art. 266). 22 338 ELEMENTS OF ALGEBRA. ICHIAP, X. Now, since X contains n factors equal to x - a, we shall X have n partial quotients equal to; and the same reason -ig applies to each of the repeated factors, x - b, x.- c.... Moreover, we can form but one quotient for each simple factor, which is of the form, X X X x_ —p — q' x -r therefore, the' first derived polynomial is of the form, nY In'kx n"X X X X y= Jr X _+ + + + +. (', x —a x-b x-c x-p x-q x-r By examining the form of the value of X in equation (1), it is plain that (X - a)n-1, (X - b)n'-1, (X -C)ff-l... are factors common to all the terms of the polynomial Y; Ience the product, (x - a)n-l X (x - b)n'-l X (X -- c)1"-.'is a divisor of Y. Moreover, it is evident that it will alsc divide X: it is therefore a common divisor of X and Y; and it is their greatest common divisor. For, the prime factors of X, are z -a, x - b, x -c..., and x:-p, x-q, x —r...; now, X-p, x-q, x-r, cannot divide Y, since some one of them will be wanting in some of:the parts of Y, while it will be a factor of all the other parts. Hence, the greatest common divisor of X and Y, is D (x - a)"n-l (z - b)n/-1 (X -c)n/-1...; that is, The greatest common divisor is composed of the product of those factors which enter two or more times in the given equation, ecch raised to a power less by 1 than in the primitive equation. 269. From the above, we deduce the following method fol finding the equal roots. To discover whether an equation, XY= 0, conrtains any equal roots: C'HAP. IX.] EQUAL ROOTS. 339 Ist. Form n, or the derived polynomzial of X; then seek Jbr C}ie greatest commzon divisor between X and Y. 2d. If one cannot be obtained, the equation has no equal rcots, or equal factors. If uwe find a common divisor D, and it is of the first degree, or of the form - h, make x - h = O, whence = h. We then conclude, that the equation has two roots equal to h, and has but one species of equal roots, from which it may be freed by dividing X by (x - h)2. If 2D is of the second degree with reference to x, solve tht equation D = O. There may be two cases; the two roots will be equal, or they will be unequal. 1st. When we find D = (x - h)2, the equation has three roots equal to h, and has bult one species of equal roots, from which it can be freed by dividing X by (x - h)3. 2d. WNahen D is of the form (x - h) (x - h'), the proposed equation has two roots equal to h, and two equal to h', fronm which it may be freed by dividing X by (x - h)2 (z -h')2, or by _D2. Suppose now that D is of any degree whatever; it is necessary, in order to know the species of equal roots, and the number of roots of each species, to solve completely the equation, D = 0.'hzen, every sim2ple root of the equation D -0 will be twice a root of thbe given equation; every double root of the equtation D = 0 will be three times a root of the given equation; and so on. As to the simple roots of X= O0 we begin by freeing this equation of the equal factors contained in it, and the resulting equation, X' = O, will make known the simple roots. 340 ELEMENTS OF ALGEBRA. [CHAP. X. EXAMPLES. 1. Determine whether the equation, 2x4 -- 12x3 + 19x2 -- + 9 = 0, contains equal roots. We nave for the first derived polynomial, SX3 - 36x2 + 3Sx - 6. Now, seeking for the greatest common divisor of these poly. nomials, we find D x — 3-0, whence x = 3: hence, the given equation has two roots equal to 3. Dividing its first nlember by (x - 3)2, we obtain 2x2 + 1 =0; whence, x- 2 2. The equation, therefore, is completely solved, and its roots are 3 3 +- a-2 nd -2 7 7 + 2 2I- T 22. For a second example, take x5 - 2x4 + 3x3 - 7X2 + 8 - 3 = 0. The first derived polynomial is 5x4 - 8x3 + 9x2 - 14x +- 8; and the common divisor, X2 - 2x I = (x - 1)2 hence, the proposed equation has three roots equal to lo Dividing its first member by (x- 1)3 = X3_ -22 + 3x - 1, the quotient is x2 +- x 3 0; whence, x — thus, the equation is completely solved. CAP. x.J EQUAL ROOTS. 341 3. For a third example, tak,~ the equation X7 - - 5X6 53 6 x — _x ]53 3X2 + SX - 4- = 0. The first derived polynomial is 7x6 + 30x5 +- 304 - 24x3 - 4a-2 - 6- + 8; and the common divisor is X4 +- 33 + x2 -_X - 2. The equation, 24 -F- 32I3 - x2 _- 3X - 2 = 09 cannot be solved directly, but by applying the method -f equal roots to it, that is, by seekinll fuor a coinnion divisor between 4ts first member and its derived polynolnial, 4x3 + 9xZ - 2x - 3: we find a common divisor, x f- I; which proves that the squcare of x +- 1 is a fictor of x4 - 3x3 + X2 3 — 2, and the cutbe of x +- 1, a factor of the first member of the given equation. Dividing x2 -- 3X3 -x2 -3x-2 by (x + 1)2 = x2 + 2. + 1, we have -2 - x-2 2 which being placed equal to zero, gives the two roots x= -, I x — 2, or the two factors, x - I andl z + 2. Ihence, we have 4 + 3x3 J.2 _ -- _3 -2 )= (x + )2( — )(~ + 2). Therefore, the first menlber of the proposed equation is equal to (x + l) (.X - )2 ( + 2)2; that is, the proposed equation has three roots equal to - 1, two equal to + 1, and twvo equal to - 2. 4. What is the product of the equal faictors of the equation x" - 7x6 -- 10x5 + 22.c4 - 43x - -352 + 48x -P- 36 = 0 As. (.e - 2)2 (X - 3)2 ( 4- 1)3. 5. What is the product of the equal factors in the equation, x7 3Z + -9x5- 19x4 + 273 -- 332 27x-9 = Ans. (x- 1)3 (x2 ~- 3)2. 342 ELEINENTS OF ALGEBRA. [CHAP. I. Elimnination. 270, We have already explained the methods of eliminating oml unlknown quantity fiom two equations, when these equations are of the first degree with respect to the unknown quantities. AVShen the equations are of a higher degree than the first, tile methods explained are not in general applicable. In this cas:, t]he method of the greatest comzmonr divisor is considered the best, and it is this method that we now propose to investigate. One quantity is said to be a function. of another when it depends upon that other for its value; that is, when the quantities are so connected, that the value of the latter cannibt be changed without producing a corresponding change in the former. 271. If two equations, containing two unknown quantities, be combined, so as to produce a single equation containing but one unknown quantity, the resulting equation is called a final equation; and the roots of this equation are called compatible values of the unknown quantity which enters it. Let us assume the equations, P -0 and Q = 0, in which P and Q are functions of x and y of any degree whatever; it is required to combine these equations in such a manner as to eliminate one of the unknown quantities. If'we suppose the final equation inyolving y to be found, and that y = a is a root of this equation, it is plain that this value of y, in connection with some value of x, will satisfy both equations. If then, we substitute this value of y in both equations, there mill result two equations containing only x, and these equations will have at least one root in common, and consequently, their first members will have a common diviscr involving x (Art. 246). This common divisor will be of the first, or of a higher degree with respect to x, according as the particular value of y a cor respunds to one or more values of x. CHAP. XI.I ELIMINATION. g43 Conversely, every value of y which, being substituted in the. two equations, gives a common divisor involving x, is necessarily a comnlpatible value, for it then satisfies the two equations at the same time with the value or values of x found from this common divisor when put equal to 0. 272. We will remark, that, before the substitution, the frst members of the equations cannot, in general, have a common divisor which is a function of one' or both of the unknown quantities. For, let us suppose, for a moment, that the equations P =O and Q = 0, are of the form P' x = 0 and Q' X R = 0, R being a function of both x and y. Placing - =0, we obtain a single equation involving two unknown quantities, which can be satisfied with an infinite nunmber of systens of values. Moreover, every system which renders 1 equal to 0, would at the same time cause P'.R and Q'. R to become 0, and consequently, would satisfy the equations P = and Q = 0. Thus, the hypothesis of a common divisor of the two poly. nomials P and Q, containing x and y, brings with it, as a consequence, that the proposed equations are indeterminate. Therefore, if there exists a common divisor, involving x and y, of the two polynomials P and Q, the proposed equations will be indeterminate, that is, they may be satisfied by an infinite number of systems of values of x and y. Then there is no data to determine a final equation in y, since the number of values of y is ilqfinite. S.r Again, let us suppose that 12 is a function of x only. Placing R =0, we shall, if the equation be solved with reference to x, obtain one or more values for this unknown quarltity. Each of thboq values, substituted in the equations P'. -= O0 and Q'. R = 0, 344 ELEMENTS OF ALGEBRA. [CHAP. X will satisfy them, whatever value we may attribi te to y, sinco these values of x would reduce R to 0, independently of y. Therefore, it, this case, the proposed equations admit of a finite number of values for x, but of an infinite number of valucos for y and then, therefore, there cannot exist a final equation in y. IHence, when the equations P =, Q -O are determinate, that is, when they admit only of a limited nzmber of systems of values for x and y, their first memlbers cann:,t have for a common divisor a fumiction of these uicnowz quan tilies, unless a particular substitution has been made for one of these quantities. 273. Froml this it is easy to deduce a process for obtaining the final equation involving y. Since the characteristic property of every compatible value of y is, that being substituted in the first nm-embers of the two equations, it gives them a common divisor involving.r, xwhich they had not before, it follows, that if to the two proposed polynomials, arranged with reference to x, we apply the process for finding the greatest common divisor, we shall generally not find one. But, by continuing the operation properly, we shall arrive at a remainder independent of x, but which is a1 function of y, and which, placed equal to 0, will give the required final equatioz. For, every value of y found from this equation, reduces to zero the last remainder in the ooperation for finding the common divisor; it is. then, such that being substituted in the preceding remainder, it will render this remainder a common divicr of the first members P and Q. Therefore, each of the roots of the equation thus formed, isa compatible value of y. 274. Admitting that the final equation may be completely solved, which would give all the compatible values, it would tfterward be necessary to obtain the corresponding values of x. Now, it is evident that it would be sufficient for this, to snlb. stituate the different values of y in the remainder preceding the CHAP. X.j IELIMINATION. 345 last, put the( polynomial involving x which results from it, equal to 0, and find firom it the values of x; for these polynomials are nothing more than the divisors involving x, which become common to A and B. But as the final equation is generally of a degree superior to the second, we cannot here explain the methods of finding the values of y. Indeed, our design was principally to show that, two equations of any degree being given, we can, without supposing the resolution of any equation, arrive at another equation, contanbing only one of the unknown quantities which enter into the pro.?posed equations. EXAMPLES. 1. IIaving given the equations X2 + xy + y2 - 1 = 0, x3 + y3 0, to find the final equation in y. First Operation. X3 + y3 X2- x2 y 4 y2__ x3+yX/2+(y2 l)X - -y - yx2 - (y2 - 1)x + y3 - yx2 - y2- y3 + y z + 2y3 - y = Ist remainder. Second Operati:n. x2 + yX +y2 —1 l- 2y3- y 2 f - (~3 x -- Y _XX 2y3 - (yf2y) - (2y3 - 2y) + y2 - - (2y3 - 2y) x - 4y6 - 6y4 - 2yg2 4y - _6y4 + 3y2 - 1. Henco, the final equation in y, is 4y6 - 6y4 + 3y -- 1 0. 346 ELEMENTS OF ALGEBRA. I CA P. X. If it were required to finJd the final equation in x, we observe that x and y enter into the primitive equations under the same fborlms; hence, x may be changed into y and y into x, without destroying the equality of the members. Therefore, 4xs-6x4 -3x2 —1 =0 is the final equation in x. 2. Find the final equation in y, firom the equations x3 - 3yx2 +- (3y2 - y + 1) x - y3 + y2 -2y =, x2 —2yx t- y2_y=0. First Op)era tion. 3 3yz2 + (3y2 - y + 1) X - y3 + y2 - 2yllx2 - 2zy + y2 - 3 -2yx2 + (y2 )x x - - yx2+- (2y2 - 1)x -y3 -y2 -2y - yx2 + 2y2x -_ 3 3+ y2 x -2y Second Operation. x2 - 2xy + y2 -_ yI - 2 X2 - 2x.ry y2 _ y. Hence, y2 _ y _ 0, is the final equation in y. This equation givws y —1 and y = O. Placing the preceding remainder equal to zero, and substYi tuting therein lle values of y, y = 1 and O = 0; we find for the sorresponding values of x, x = 2 and x = 0; fronm w3hich Ptu: is-en eqlations may be entirely solved. 1HAPTERl XI. BOI,1UTION OF N'ME1R AL EQUATIONS CONTAINING BUT ONE UNKI{WWN QUANTITY.-SI UIIM'S THEOREM.-CARDAN S RULE. —HORNERS 5 M.ETHIOD. 275. TIE principles established in the preceding chapter, are applicable to all equations, whether the co.efficients are nunmerical or algebraic. These principles are the elements which are employed in the solution of all equations of higher degrees. Algebraists have hitherto been unable to solve equations of a higher degree than the fourth. The formulas which have beer deduced for the solution of algebraic equations of the higher degrees, are so complicated and inconvenient, even when they can be applied, that we may regard the general solution of an algebraic equation, of any degree whatever, as a problem more curious than useful. Methods have, however, been found for determining, to any degree of exactness, the values of the roots of all numerical equations; that is, of those equations which, besides the unknown quantity, involve only numbers. It is proposed to develop these methods in this chapter. 276. To render the reasoning general, we will take the equation, X= xxm + Pxm~-1 + Qxm- ~... U- =. in which P, Q... denote particular numbers which are real, antd either positive or negative. If we substitute for x a number a, and denote byt A what, becomes under this suppositi )n; and again substitute a -- u fbr x. and denote the new polynomial by.A': then, u may be taken s) small, that the difference between A" and A shall be less than any assignable quantity. 348 ELEMENTS OPF LGEBRA. L[CIAP. XI. If, now, we denote by B, C,.... what the co-efficients z V Y) 2' 2.3 (Art. 264), become, when we make x - a, we shall have, A' -A Bu CU2 -DU + i. t *+ - - (1); whence, A' — A Bu + CU2 + DU3+... + U - - - (2). It is rnow required to show that this difference may ble ren. dered less than any assignable qulantity, by attributing a value sufficiently small to u. If it be required to make the difference \'tween A' and A less than the number 1V, we must assign a value to u which will satisfy the inequality b'+u d — C D2- 3 +....um < n - - - (3). Let us take the most unfavorable case that can occur, viz,, let us suppose that every co-efficient is positive, and that each is equal to the largest, which we will designate by K. Then any value of u which will satisfy the inequality K(u + u2 + 3 +....m) < N (4), will evidently satisfy inequality (3). Now, the expression within the parenthesis is a geometrical progression, whose first term is u, whose last term is un, and whose ratio is u; hence (Art. 188), ~ - I m + 1 u+U+3+.. - u.. rn= - + z2 J lt,3 + m== - - - - X (1 -- ~). u - 1 1 — u 1- u Substituting this value in inequality (4), we have, 1u (1I uv) pxm,-1 + Qxm-2 +.. ~ + Tx + U. Let k denote the greatest numerical co-efficient, and substitute it in place of each of the co-efficients; the inequality will then brcome xm > kx-1 ix-2 +... + +kx+k. It is evident that every number substituted for x which will satisft this condition, will satisfy the preceding one. Now, dividing both members of this inequality by xm, it becomes 1 x + —', + 3 + X-., Making x k, the second member reduces to 1 plus the suln of several fractions. The number k will not therefore satisfy the inequality; but if we make x = k + 1, we obtain ft:,r the second member the expression, k k + k + k k + ( + - (le+ q)2 (k + )3 + (k + l)M- + (k + 1)' This is a geometrical progression, the first term of which is' +- 1' the -last term, (+ 1) and the ratio, k 1; hence the expression reduces to k k (khl)m-)l I A+ _ 1,+ 1 1 1 which is evidently less than 1. Now, any number > (Ak + 1), put in place of x, will render. the sum of the fractions - + -... still less: therefore, a; The greatest co-eficient plus 1, or any greater number, being substituted for x, will render the first term Xm greater than the. arithmetical sum of all the other terms. 23 834 ELEMENTS OF ALGEBRA. [CHA P. XI, 283. Every number which exceeds the greatest of the positive roots of an equation, is called a superior limit of the positive roots. From this definition, it follows, that this limit is susceptible )f an infinite number of values. For, when a number is found to exceed the greatest positive root, every number greater than this, is also a superior limit. The term, however, is generally lapplied to that value nearest the value of the root. Since the greatest of the positive roots will, when substituted for x, merely reduce the first member to zero, it follows, that we shall be sure of obtaining a superior limit of the positive roots by finding a number, which substiluted iz place of x, renders the first member positive, and which at the' same time is such, thaa every greater number will also give a positive result; hence, The greatest co-efficient of x plus 1, is a superior limnit of the positive roots. Ordinary Limit of the Positive Roots. 284. The limit of the positive roots obtained in the last article,'is commonly much too great, because, in general, the equation contains several positive terms. We will, therefore, seek for a limit suitable to all equations. Let Xm-n denote that power of x that enters the first negative term which follows x", and let us consider the most unfivoratble case, viz., that in which all the succeeding terms are negative, and the co-efficient of each is equal to the greatest of the negative co-efficients in the equation. Let S denote this co-efficient. What conditions will render *.m > sj7?7-m -n +.,n-n-i. D,V @ S? Dividing both members of this inequality by xi', we have S S S S Now, by supposing - = S/-+ 17, or for simplicity, mn king n - S. whtkih gives, S' S'", and x = S' 4- 1, CHAP. XI.] LIMITS OF POSITIVE ROOTS. 855 the second member of the inequality will become, S'X S' S, S/ (8s + J)n (St + W)++ (S'+ I)M-1 (- + )~ which is a geometrical progression, of which (is + )n first term, and the ratio. Hence, the expression for the sumn of all the terms is (Art. 188), (S' + 1)n+ - (S' + )' So/n-i SI/-i 1 - = (S' +-)n- (5' ~ 1) < S'+1 - Moreover, every number > S + 1 or "/S- 1, will, when substituted for x, render the sumn of the fractions S S Xn xn+l still smaller, since the numerators remain the same, while the denominators are increased. Hence, this sum will also be less. Hence, /S q — 1, and every greater number, being substituted for x, will render the first term xm greater than the arithmetical sum of all the negative terms of the equation, and will conse quently give a positive result for the first member. Therefore, That root of the numerical value of the greatest negative co-efftcient whose index is equal to the number of terms which precede thle first negative term, increased by 1, is a superior limit of the positive roots of the equation. If the co-efficient of -z term is 0, the term must still be counted. Make n = 1 in which case the first negative term is the second termn of the equation; the limit becomes S~Sq + 1 = S + 1; that is, the greatest negative co-eficient plus 1. Let n = 2; then, the limit is 2 —S+ 1. Whe n = 3. tfhe limit is aS + 1. 856 ELEMENTS OF ALGEBRA. rCOlAP. XL EXAMPLES. 1. What is the superior limit of the positive roots of the eshquation x4- 5x3 + 37x2 — 3x 39-= 0 Ans." s+1=I + 1 =6. 2. What is the superior limit of the positive roots of the equation X5 + 7x4 - 12x3- 49x2 + 522x - 13 — 02 Ans. VS- 1 =-49+-1 = 8. 3. What is the superior limit of the positive roots of the equation 4 +- 11x2 - 25x — 67 = 0? In this example, we see that the second term is wanting, that is, its co-efficient is zero; but the term must still be counted in fixing the value of n. We also see, that the largest negative co-efficient of x is found in the last term where the exponent of'is zero. THence, and therefore, 6 is the least whole number that will certainly fulfil the cDnditions. Smallest Limit in Entire it2Thebs. 285. In Art. 282, it was shown that the greatest co-efficient of x plus 1, is a superior limit of the positive roots. In the last article we found a limit still less; and we now propose to find the smallest limit, in whole numbers. Let X= O te the proposed equation. If in this equation we make x = + 4- S' being arbitrary, we shall obtain (Art. 264), Z CHAP. XI.] LIMIITS Ou POSITIVE ROOTS. 357 Let us suppose, that after successive trials we have determined a number for x', which substituted in Z X', Y, - -.... renders, at the same time, all these co-efficients positive, this num. ber will in general be greater than the greatest positive root of the equation X -0. For, if the co efficients of equation (1) are all positive, no positive value of u can satisfy it; therefore. ali the real values of u must be negative. But from the equation z - X' q-u, we have = Zx —x'; and in order that every value of u, corresponding to each of the values of x and x/, may be negative, it is necessary that the greatest positive value of x should be less than the value of x'. Hence, this value of x' is a superior limit otf the positive roots. If we now substitute in succession for x in X the values a' - 1, x'- 2, x' - 3, &c., until a value is found which will make X negative, then the last number which rendered it positive will be the least superior limit of the positive roots in whole numbers. EXAMPLE. Let x-_ - 5x3 - 6X2 - 19x + 7 - 0. As x/ is indeterminate, we may, to avoid the inconvenience of vwriting the primes, retain the letter x in the formation of the derived polynomials; and we have, X - xI — 5x3 — 6x2 —19x+ 7, Y = 4X3 — 15x2 - 12x -19, -= G6x2 - 15x — 6, 3=4 — 5. 2.3 The question is now reduced to finding the smallest entire number which, substituted;n place of x, will render all of these polynomials positive. 358 ELEMENTS OF ALGEBRA. CRHAP. XI. It is plain that 2 and every number > 2, will render the polynomial of the first degree positive. Ba-t 2, substituted in the polynomial of the second degree, gives a negative result; and 3, or any number > 3, gi ves a positive result. Now, 3 and 4, substituted in succession in the polynomial of the third degree, give negative results; but 5, and any greater number, gives a positive result. Lastly, 5 substituted in X, gives a negative result, and so does 6; for the first three terms, x4- 5X3 - 6X2, are equivalent to the expression X3 (x- 5)- 6x2, which reduces to 0 when =- 6; but x = 7 evidently gives a positive result. Hence 7, is the least limit in entire numbers. We see that 7 is a superior limit, and that 6 is not; hence, 7 is the least limit, as above shown. 2 Applying this method to the equation, x- 3x4- 8x3 - 25x2 + 4x - 39 - 0, the superior limit is found to be 6. 3. We find 7 to be the superior limit of the positive roots of the equation, x5 - 5x4 - 13x3 + 17x2 - 69 = 0. This method is seldom used, except in finding incommensurable roots.'Superior Limit of Negative oots. —Inferior Limit of Posi live and ANegative Roots. 286. HIaving found the superior limit of the positive roots, it remains to find the inferior limit, and the superior and inferior limits of the negative roots, numerically considered. First, If, in any equation, X = 0, we make x = — we shall have a new equation Y = 0. 1 Since we know, from the relation x-=., that the greatest g CHAP. XI. CONSEQUENCES OF PRINCIPLES. 359 positive value of y in the new equation corresponds to the least positive value of x in the given equation, it follows, that If we determine te the perior limnit of the positive roots of the equation Y = 0, its reciprocal will be the inferior limit of the positive roots of the given equation. HIence, if we designate the superior limit of the positive roots of the equation Y = 0 by L', we shall have for the inf'rior limit of the positive roots of the given equation, tSecond, If in the equation X = 0, we make - x=-y, which gives the transformed equation Y'= 0, it is clear that the positive roots of this new equation, taken with the sign _, will give the negative roots of the given equation; therefore, determining by known methods, the superior limit of the positive roots of the new equation Y' = 0, and designating this limit by L//, we shall have - L" for the superior limit, (numerically), of the negative roots of the given equation. Third, If in the equation X- O, we make x _ —, we shall have the derived equation Y" = 0. The greatest positive value of y in this equation will correspond to the least negative value (numerically) of x in the given equation. If, then, we find the superior limit of the positive roots of the equation Y" -— 0, and designate it by L"', we shall have the inferior limit of the negative roots (numnerically) equal to - // Consequtenoss deduced from the precedil'g Princip7es First. 287. Every equation in which there are no variations in the signs, that is, in which all the ternms are positive, must have all of its real roots negative; for, every positive number substituted for x, will render the first member essentially positive. 360 ELEMENTS OF ALGEBRA. [CHAP. XL Second. 288. Every (cmplete equation, having its terms alternately posi tive and negative, must have its real rocds all positive; for, every negative number substituted for x in the proposed equation, would render all the terms positive, if the equation be of an even de gree, and all of them negative, if it be of an odd degree. IHence, their sum could not be equal to zero in either case. This principle is also true for every incomplete equation, in which there results, by substituting — y for x, an equation having all its terms affected with the same sign. Third. 289. Every equation of an odd degree, the co-efficients of which are real, has at least one real root affected wit'h a sign contrary to that of its last term. For, let xm -P- p n-... Tx ~o U= 0, be the proposed equation; and first consider the case in whicl the last term is negative. By making x -= 0, the first member becomes - U. But by giving a value to x equal to the greatest co-efficient plus'l, or (K+ 1), the first term xm will become greater than the arithinetical sum of all the others (Art. 282), the result of this substitution will therefore be positive; hence, there is at least one real root com2prehended between cand It A — 1, which root is positive, and consequently affected with a sign contrary to that of tho last term (277). Suppose now, that the last term is positive Making x = 0 in the first member, we obtain + U for the result; but by putting - (K 4 1) in place of x, we shall obtain a neyga tivee result, since the first term becomes negattive by this sub stitution; hence, the equation has at least one real root corn prehendled between 0 and — (K+ 1), which is negative. ox eafected with a sign contrary to that of the last;erm. CHAP. XI. I CONSEQUENCES OF PRINCIPLES, 361 F0ourth. 290, Every equation of an even degree, which involves only rea. o-efficients, and of which the last term is negative, has at least two real roots, one positive and the other negative. For, let - U be the last term; making x - 0, there results - U Now, substitute either AK~+ 1, or - (K+ 1), At being the greatest co-efficient in the equation. As mn is an even number, the first term xm will remain positive; besides, by these substitutions, it becomes greater than the sum of all the others; there. fore, the results obtained by these substitutions are both positive, or aflected with a sign contrary to that given by the hypothesis x = 0; hence, the equation has at least two real roots, one positive, and comprehended between 0 and _K]- 1, the other negalive, and comprehended between 0 and - (K 1) (277) 291. If an equation, involving only real co-epicients, contains imagynary roots, the number of such roots must be even. For, conceive that the first member has been divided by all the simple factors corresponding to the real roots; the co-efficients of the quotient will be real (Art. 246); and the quotient must alsc be of an even degree; for, if it was uneven. by placing it equal to zero, we should obtain an equation that would contain at least one real root (289); hence, the imaginary roots must enter by pairs. RIEMARK.-There is a property of the above polynomial quotient which belongs exclusively to equations containing only imaginary roots; viz., every such equation always remains positive for any real value substituted for x. For, by substituting for x, fK l —1, the greatest co-efficient plus 1, we could always obtain a positive result; hence, if the polynomial could become negative, it would follow that when placed equal to zero, thelre -wrould be at least one real root corn 362 ELEMENTS OF ALGEBRA. [CHAP. XI. prehended between K + I and the number which would give a negative result (Art. 277). It also follows, that the last term of this polynomial must be positive, otherwise x 0 would give a negative result. SixtJh. 29.2 When the last term of an equation is positive, the number of its real positive roots is even; and when it is negative, the number of such roots is uneven. For, first suppose that the last term is + U, or positive. Since by making x -= 0 there will result + U, and by making x = K + 1, the result will also be positive, it follows that 0 and + I1 give two results affected with the same sign, and consequently (Art. 279), the number of real root.s, if any, comprehended between them, is even. When the last term is - U, then 0 and K + 1 give two results affected with contrary signs, and consequently, they comprehend either a single root, or an odd number of them. The converse of this proposition is evidently true. Descartes' Rule. 293. An equation of any degree whatever, cannot have a greater number of positive roots than there are variations in the signs of its terms, nor a greater number of negative roots than there are permanences of these signs. A variation is a change of sign in passing along the terms. A permanence is when two consecutive terms have the same sign. In the equation x - a = 0, there is one variation, and one positive root, x — a. And in the equation x + b = 0, there is one permanence, and one negative root, x =-b. If these equations be multiplied together, member by member, there will resul5 an equation of the second degree, x2 — a x-ab) 0 I 7 b CHAP. XI.j DESCARTES' RULE 363 If a is less Lhan 6, the equation will be of the first form (Art. 117); ard if a > 6b, the equation will be of file sacond form; that is, a < b gives x2 2+px —q= 0, and a > b " x2 -- 2px - q = O. In the first case, there is one permanence and one variation, and in the second, one variation and one permanence. Since in either form, one root is positive and one negative, it follows that there are as many positive roots as there are variations, and as many negative roots as there are permaz nences. The proposition will evidently be demonstrated in a general manner, if it be shown that the multiplication of the first mem. ber of any equation by a factor x - a, corresponding to a positive root, introduces at least one variation, and that the multiplication by a factor x + a, corresponding to a negative root, introduces at least one permanence. Take the equation, xm - _Axm-1 q- Bxm-2 - Cxm —3 -... +- TX T U- 0, Mn which the signs succeed each other in any manner whatever. By multiplying by r- a, we have XM+-1 -+ A x" m- B I Xm-l q- C XmI-2 ~... U Ix - a TA a:: Ba - TTa - Ua O. The co-efficients which form the first horizontal line of this product, are those of the given equation, taken with the same signs; and the co-efficients of the second line are formed from those of the first, by multiplying by a, changing the sig ns, and advancing each one place to the right. Now, so long as each co-efficient in the upper line is greater than the corresponding one in the lower, it will determine the sign of the total co-efficient; hence, in this case there will be, from the first term to that preceding the last, inclusively, the same variations and the same permanences as in the proposed equation; but the last term A: Ua having a sign contrary to that which immediately precedes it, there mlust ce one more varia. tic: than in the proposed equation. 064 ELEMENTS OF ALGEBRA. [CIA P. Xi. When a co-efficient in the lower line;s affected with a sign contrary to the one corresponding to it in the upper, and is also greater than this last, there is a change from a perina nence of sign to a variation; for the sign of the term in wihiuet this happens, being the same as that of the inferior co-efieient, nlust be contrary to that of the preceding term, which has been supposed to be the same as that of the superior co-efficient. Hence, each time we descend from the upper to the lower line, in order to determine the sign, there is a vatriation which is not found in the proposed equation n; and it; after passing into the lower line, we continue in it throughout, we shall find for the remaining terms the same variations and the same permanences as in the given equation, since the co-eflicients of this line are all affected with signs contrary to those of the primitive co-efficients. This su pposition would therefore give us one variation for each positive root. But if we ascend fi-om the lower to the uppei line, there may be either a variation or a permanence. But even by supposing that this passage, Froduces permanences in all cases, since the last term - Ua be'-ris a part of the lower line, it will be necessary to go once -.nore from the upper line to the lower, than from the lower t,) the apper. Hence, the new equation must /have at least onie more variation than the proposed; and it will be the same for each positive root introduced into it. It may be demonstrated, in an analogous manner, that the multlijlication of the first member by a factor x + a, correspond. ing to a negative root, would iintroduce one permanence mnore. Hence, in any equation, the number of positive roots cannot be greater than the number of VARIATIONS of signs, nor the number elt negative roots gieater than the number of PERMANENCES. Consequence. 291. When the roots of an equation are all real, the number of positive roots is equal to the number of variations, and the nume ber of tzegative roots to the nurmber of pe'manences.~ CHAP. XI.] DESCARTES' RULE. 365 For, let m denote the degree of the eqaation, n the number of variations of the signs, p the number of permanences; then, m - n +-p. Moreover, let n/ denote the number of positive roots, and p' the number of negative roots, we shall have wn = as' +- p; whelce, n p =' +- p', or, n - n' =p' -p. Now, we have just seen that n' cannot be > n, nor can it be less, since p' cannot be >_p; therefore, we must have n/- = n, and p' = p. REMARK.-When an equation wants some of its terms, we can often discover the presence of imaginary roots, by means of the above rule. For example, take the equation x3 + px + q = 0, p and q being essentially positive; introducing the term which is wanting, by affecting it with the co-efficient - O0; it becomes x3 -~- 0. x2 + p + q = 0. By considering only the superior sign, we should obtain only permanences, whereas the inferior sign gives two variations. This proves that the equation has some imaginary roots; for, if they were all three real, it would be necessary, by virtue of the superior sign, that they should be all negative, and, by virtue of the inferior sign, that two of them should be positive and one nega. tive, which are contradictory resulls. We can conclude nothing from an equation of the form x3 -px+q= 0; for, introducing the term -- 0. x2, it becomes x3 0. z2 -px qX - =- 0, whicn contains one permanence and two variations, whether we take the superior or inferior sign. Therefore, this equation may have its three roots real, viz., two positive and one negative; or, two of its roots may be imaginary and one'negative, since its last term is positive (Art. 292). 366 ELEMENTS OF ALGEBRA. LCHAP. XL Of the cosnme'a~rable Roots of Numnericait Equations. 295, Every equation in which the co-efficients aie whole numbers, that of the first term being 1, will have whole numbers only for its commensurable roots. For, let there be the equation xm- + pxm-l + Qxm-2+,. — + T. +- U= O; inl which P, Q... T, U, are whole numbers, and suppose that It were possible for one root to be an irreducible fraction a. Substituting this fraction for x, the equation becomes am am-1 am —2 a am Pb -__+Qm+ +. T+ + T U= O; whence, multiplying both members by b'n-l, and transposing, am - Pam-' - Qam-2b... -- Tabm-2 - Ubm-1. But the second member of this equation is composed if the sum of entire numbers, while the first is essentially fractional, for a and b being prime with respect to each other, am and b will also be prime with respect to each other (Art. 95), and hence this equality cannot exist; for, an irreducible fraction cannot be equal to a whole number. Therefore, it is impossible for any irreducible fraction to satisfy the equation. Now, it has been shown (Art. 262), that an equation containing rational, but fractional co-efficients, can be transformed into another in which the co-efficients are whole numbers, that of the first term being 1. Hence, the search for commenisurable roots, either entire or fractional, can. always be reduced to that for entire roots. 296. This being the case. take the general equation xm +- P-1 + Qx-2... - RX33+ Sx2 + Tx + U'= O, and let a denote any entire number, positive or negative, whiclh will satisfy it. Since a is a oot, we shall have the equation am4 Pa" P-F.. -Ra3 4-.a2 + Ta U= O (). CHAP. XI.] COMMENSURABLE ROOTS OF EQUATIONS. 367 Now replace a by all the entire numbers, positive and negative, between I and the limit +L, and between -1 and -L": those which verify the above equality will be roots of the equation. But these trials being long and troublesome, we will deduce from equation (1), other conditions equivalent to this, and more easily applied. Transposing in equation (1) all the terms except the last, and dividing by a, we have, - am- - pa... -Ra _ Sa - T (2). Now, the second member of this equation is an entire number; U hence, - must be an entire number; therefore, the entire roots of the equation are comprised among the divisors of the last term. Transposing T- in equation (2), dividing by a, and making — + T - T', we have, a — a' —-Pa3 -Ra-.. - a-S (3). a T' The second member of this equation being entire, -, that is, the quotient of - + T by a, fs an entire number. Transposing the term — S and dividing oy a, we have, by supposing - + S = S', a -- = a-3 - PaM —-,,. (4), a'rIe second member of this equatinm being entire. - that is, the quotient of — +S by a, a an entire number. 368 ELEMENTS OF ALGEBRA. [CHAP, XI. By continuing to transpose the terms of the second member into the first, we shall, after m - 1 transformations, obtain an equation of the form, Q, -a-P. a Then, transposing the term - P, dividing by a, and making Q' PP P1 - + P P/, we have -— 1, or P- 1 = 0. a a a This equation, which results from the continued transformas tions of equation (1), expresses the last conditions which it is requisite for the entire number a to fulfil, in order that it may be known to be a root of the equation. 297. From the preceding conditions we conclude that, when an entire number a, positive or negative, is a root of the given equation, the quotient of the last term, divided by a, is an entire number. Adding to this quotient the co-efficient of x1, the sum will be exactly divisible by a. Adding the co-efficient of x2 to this last quotient, and again dividing by a, the new quotient must also be entire; and so on. Finally, adding the co-efficient of the second term, that is, of x7-l, to the preceding quotient, the quotient of this sum divided by a, must be equal to - 1; hence, the result of the addition of 1, which is the co-efficient of Xm, to the preceding quotient, must be equal to 0. Every number which will satisfy these conditions will be a root, and those which do not satisfy them should be rejected. All the entire roots may be determined at the same time, by the following RULE. iAfter having determined all the divisors of the last term, write those which are comprehended between the limits + L and - Le upon the same horizontal line; then underneathl these divisors write the quotients of the last term by each of them. CHAP. XI.] COMMENSURABLE ROOTS. 369 1Add the co-efficient of x1 to each of these quotients, and write the sums underneath the quotients which correspond to them. Then divide these sums by each of the divisors, and write the guotients underneath thIe corresponding sums, talcing care to reject the fractionzal quotients and. the divisors which produce them; and so Oll. WVhen there are terms wantin g in the proposed equtation, their co-ericients, which are to' be regarded as equal to 0, must be taken in to consideration. EXAMPLES. 1. What are the entire roots of the equation, - X3 - 13x2 + -16x- 48 =0 A superior limit of the positive roots of this equation (Art. 284), is 13 q- 1 - 14. The co-efficient 48 need not be considered, since the last two terms can be put under the form 16 (x - 3); hence, when x > 3, this part is essentially positive. A superior limit of the negative roots (Art. 286), is -(1+ 48), or — 8. Therefore, tle divisors of the last term which may be roots, are 1, 2, 3, 4, 6, 8, 12; moreover, neither + 1, nor - 1, will satisfy the equation, because the co-efficient — 48 is itself greater than the sum of all the others: we should therefore try only the positive divisors from 2 to 12, and the negative divisors from — 2 to — 6 inclusively. By observing the rule given above, we have 12, 8, 6, 4, 3, 2, - 2, - 3,- 4, - 6 - 4, - 6, -8, -12, -16, —24, ~-24, + 16, + -12, + 8 -- 12, +- 10, + 8, + 4, 0,- 8, - 40, + 32, +28, 2 - 24 1,.... 1, O.- 4, -20,.., 7- 7,.- 4 -12,..,.. 2 -12, -o3, -17, -33,.. -20, -17 -1,..,..,- 3,..~....,.. -2 54 - 2,..., — 4,.......,, J- 4.,.... - 1,..,,..,.. - 1,. 24 370 ELEMENTS OF ALGEBRA. [CHAP. XI The first line contains the divisors, the second contains tlie quotients arising from the division of the last term - 4S, by each of the divisors. The third line contains these quotients, each augmented by the co-efficient + 16; and the fourth, the quotients of these sums by each of the divisors; this second condition excludes the divisors + 8, + 6, and - 3. The fiftYh contains the preceding line of quotients, each au=g mented by the co-efficient - 13, and the sixth contains the quo tients of these sums by each of the divisors; the third condition excludes the divisors 3, 2, — 2, and — 6. Finally, the seventh is the third line of quotients, each aug mented by the co-efficient - 1, and the eighth contains the quotients of these sums by each of the divisors. The divisors - 4 and — 4 are the only ones which give - I; hence, + 4 and — 4 are the only entire roots of the equation. In fact, if we divide -- 3 _ 132 - 16x - 48, by the product (x - 4) (x + 4), or X2 - 16, the quotient wiL be x2 - x + 3, which placed equal to zero, gives 1 1 x=~ ~~ -11' 2 2 V gtherefore, the four roots are 1 1 1 1 4) — 4,- 2 +2-V/-i- and 2 2 -1' 2. What are the entire roots of the equation x4 - 5x3 - 25x -21 = O? 8. What are the entire roots of the equation 155 - 19x4 + 6x3 - 152 - 19 + 6 = O? 4. What are the entire roots of the equat-on 9x6 -+ 30x5 + 22x4 -- lOx3 + 17W2 -20x + 4 = O? CHAP. XI.] STURM S THEOREM. 871 Stuizm's Theorem. 298. The object of this-;theorem is to explain a method of do termining the number and places of the real roots of equations involving but one unknown quantity. Let X=-O -. (1), represent an equation containing the single unknown quantity x; X being a polynomial of the mzth degree with respect to x, the co-efiicients of which are all real. If this equation should have equal roots, they may be found and divided out as in Art. 269, and the reasoning be applied to the equation which would result. We will therefore suppose -= 0 to have no equal roots. *299. Let us denote the first derived polynomial of X by X,, and then apply to X and X1 a process similar to that for find.. ing their greatest common divisor, differing only in this respect, that instead of using the successive remainders as at first obtained, we change their signs, and take care also, in preparing for the division, neither to introduce nor reject any factor except a positive one. If we denote the several remainders, in order, after their signs have been changed, by X.2, X3... X, which are read X second, X third, &c., and denote the corresponding quotients by Q,, Q...e Q_, we may then form the equations X=XQ1-X2 X - - (2). X1 =X2Q2-X, { x;1, X.Q 1-+ - - - (3)o kr;* 2 = x Ql - X,, Since by hypothesis, X- 0 has no equal roots, no common divi6 or can exist between X and X, (Art. 267). The last re. mainder - X,, will therefore be different fioom zero, and inde, pendent of x. 372 ELEMENTS OF ALGEBRA. LCHAP. XI. 300. Now, let us suppose that a number p has been substi tuted for x in each of the expressions X, X,, X... X,_,; and that the signs of the results, together with the sign of X,, are arranged in a line one after the other: also that another number q, greater than p, has been substituted for x, and the signs of the results arranged in like manner. Then will the number of variations tin the signs of the first arrangemnent, diminished by the number of variations in those of the second, denote the exact number of real roots comprised between p crand q. 301. The demonstration of this truth mainly depends upoll the three following properties of the expressions X, X,.. X,, &c. I..If any number be- substituted for x in these expressions, it is impossible that any two consecutive ones can become zero at the same time. For, let X,,,, X., X,+il, be any three consecutive expressions. Then among equations (3), we shall find..-,-..X Q. X'n fl1.... (4), from which it appears that, if X,,_ and Xi, should both become 0 for a value of x, X,,Y+ would be 0 for the same value; and since the equation which follows (4) must be X,, X.+, Q.+1 -X.+, we shall have X+,,. =0 for the same value, and so on until we should find X -= 0, which cannot be; hence, X,, and X, cannot both become 0 for the same value of x. 11, By an examination of equation (4), we see that if X, b&. comes 0 for a value of x, X,_ and Y,.+, must have contrary signs, that is, oIf any one of the expressions is reduced to 0 by the substitution of a value for x, the preceding and followinVg ones?ill have contrary signs for the same value. CHAP. XI.] STURMI'S THEOREM. 373 Ill. Let us substitute a + u for x in the expressions X and. X,, and designate by U and U, what tney respectively become under this supposition. Then (Art. 264),'we nave U = A + A'u - A" + &c. U, Al+Al + au + A", — +&c. 3 in which A, Al' Al, &c., are the results obtained by the sub stitution of a. for x, in X and its derived polynomials; and A1, A/,, &c., are similar results derived from X1. If, now, a be a root of the proposed equation X= 0, then A = 0, ard since A' and Al, are each derived from X1, by the substitution of a for x,' we have A' = A,, and equations (5) become U- =A + A'u -. A ( &. 2 - - - (6). U = A + A',u + &c. Now, the arbitrary quantity u may be taken so small that the signs of the values of U and U, will depend upon the signs of their first terms (Art. 276); that is, they will be alike when u is positive, or when a + u is substituted for x, and un. like when u is negative or when a —zu is substituted for x. Hence, If a nunmber insensibly less than one of the real roots of X = 0 be substituted for x in X and X,, the results will have contrary signs; and if a number insensibly greater than this root be substituted, the results will have the same sign. 302. Now, let any number as k, algebraically less, that is, nearer equal to - o, than any of the real roots of the several equations x- O... X_.-o = 0, be substituted for x in the expressionsX X,, X1, &ec., and the signs of the several results arranged in order; then, let x be increased by insensible degrees, until it becomes equal to h, the least of all the roots of the equations. As thel e is no 374 ELEMIENTS OF ALGEBRA. LCHAP. XI. root of either of the equations between k and h, none of the signs call change while x is less than Ih (Art. 277), and the numnber of variations and. permanences in the several sets of results, will remain the same as in those obtained by the first suDstitution. When x becomes equal to A, one or more of the expressions Ad, X-, &c., will reduce to 0. Suppose X~ becomes 0. Then, as by the first and second properties above, explained, neither X,,, nor XVF+ can become 0 at the same time, but must have contrary signs, it follows that in passing from one to the other (omitting X- 0), there will be one and ontly one variation; and since their signs have not changed, one must be the sazme as, and the other contrary to, that of Xn, both before and after it becomes 0; hence, in passing over the three, either just before X, becomes 0 or just after, there is one and only one variation. Therefore, the reduction of X, to 0 neither increases nor ditninishes the number of variations; and this will evidently be the case, although several of the expressions X1, X2, &c., should become 0 at the same time. If x= h should reduce X to 0, then h is the least real root of the proposed equation, which root we denote by a; and since by' the third property, just before x becomes equal to a, the signs of X and X, are contrary, giving a variation, and just after passing it (before x becomles equal to a root of X, = 0), the signs are the same, giving a permanence instead, it follows that in passing this root a vtriation is lost. In the same way, increasing x by insensible degrees from z = a - u until we reach the root of X- 0 next in order, it is plain that no variation will be lost or gained in passing any of the roots of the other equations, but that in passing this root, for the same reason as before, another variation will be lost, and so on for each real root between k and the numlber last substituted, as g, a variation will be lost until x has,beon incr'eased beyond the greatest real root, when no more can be lost or gained. Hence, the excess of the number of variations CflAP. XI.] STURM'S THEOREM. 375 obtained by the substitution of k over those oltained by the substitution of g, will be equal to the number of real roots comprised between kc and g: It is evident that the sane course of reasoning will apply when we commence with any number p, whether less than all the loots or not, and gradually increase x until it equals any other number T. The fact enunciated in Art. 299 is therefore established. 303. In seeking the number of roots comprised between p and q, should either p or q reduce any of the expressions X,, XL, &O., to 0, the result, will not be affected by their omission, since the number of variations will be the same. Should p reduce X to 0, then p is a root, but not one of those sought; and as the substitution of p + u will give X and X, the same sign, the number of variations to be counted will not be affected by the omission of X= 0. Should q reduce X to 0, then q is also a root, but not one of those sought; and as the substitution of q -, will give X *and X1 contrary signs, one variation aust be counted in passing from X to X1. 304. If in the application of the preceding principles, we observe that any one of the expressions,, X,... &c., X,, for instance, will preserve the same sign for all values of x in passing from p to g, inclusively, it will be unnecessary to use the succeeding expressions, or even to deduce them. For, as XY preserves the same sign during the successive substitutions, it is plain that the same number of variations will be lost among the expressions X, X1, &c... ending with X,, as among all inlcluding X. Whenever then, in the course of the division, it is found that by placing any of the remainders equal to 0, an equation is obtained with imr ginary roots only (Art. 291), it will be useless to obtain any of the succeeding remainders. This principle will be found very useful in the sol. tion of numerical examples. 376 ELEM[ENTS OF ALGEBRA. [CIAP. XI 305. As all the real roots of the proposed equation are, feces. sarily included between - oo and -oc, we may, by ascertaining the number of variations lost by the substitution of these, in succession, in the expressions X, X1... X,,,.. &c., readily determine the total number of such roots. It should be observed, that it will be only necessary to make these substitutions in the first terms of each of the expressions, as in thlis case the sign of the term will determine that of the entire expression (Art. 282). Hlaving found the number of real roots, if we subtract this number from the highest exponent of the unknown quantity, the remainder will be the number of imaginary roots (Art.'248). 306, Having thus obtained the total number of real roots, we may ascertain their places by substituting for x, in succession, the values 0, 1, 2, 3, &c., until we find an entire number which gives the same number of variations as + om. This will be the smallest superior limit of the positive roots in entire numbers. Then substitute - 1, — 2, &c., until a negative number is obtained which gives the same number of variations as -)o. This will be, numerically, the least superior limit of the negative roots in entire numbers. Now, by commencing with this limit and observing the number of variations lost in passing from each number to the next in order, we shall discover how many roots are included between each two of the consecutive numbers used, and thus, of course, know the entire part of each root. The decimal part may then be sought by som e of the known methods of approximation. EXAMPLES. 1. Let 8s3 —6x- 1I = 0 —'The first derived polynomial (Art. 264), i; 24,2 - 6, CHAP. XI.] STURt'S THEOREM. 377 and since we may omit the positive factor 6 wvithout affecting the sign, we may write 4x2 — 1 = X1. Dividing X by XTY, we obtain for the first remainder, — 4x —. Changing its sign, we have 4x 1 =-X2. Multiplying X, by the positive number 4, and then dividing by Xi, we obtain the second remainder — 3; and by changing its sign + 3 = XY. The expressions to be be used are then X = Sx -- 6x -1, X —= 4x2 -1, X2,= 4x + 1, X= - 3. Substituting - o and then + Go, we obtain the two following arrangements of signs: - + -..... 3 variations, +++......0 there are then three real roots. Hf, now, in the same expressions we substitute 0 and + 1, and then 0 and - 1, for x, we shall obtain the three following arrangements: For x- -1 + -F + + 0 variations. X z 0 -- l "c " 1x —1 -- - + - + 3 "6 As x -+ 1 gives the same number of variations as -l w, and x = -1 gives the same as — G, -+ and — 1 are the smallest limits in entire numbers. In passing from - 1 to 0, two variations are lost, and in passing from 0 to -- 1, one, variation is lost; hence, there are two negative roots between — 1 and 0, and one positive root between 0 and.- 1. 2. Let 2x* - 13x2 - 10x - 19 = 0. 878 ELEMENTS OF ALG EBRA. [CHIAP. xI If we deduce X, X,, anid X2,, we have the three expressions X = 2x4 - 13X2 10x-19, = — 4x3 -13. + 5, 2= — 13x2-15x +-38. If we place X, = 0, we shall find that both of the roots of the resulting equation are imaginary; hence, Xi will be positive for all values of x (Art. 290). It is then useless to seek for X3 and X4. By the substitution of - o and + oo in X, X1, and YX, we obtain for the first, two variations, and for the second, none; hence, there are two real and two imaginary roots in the proposed equation. 3. Let 3 -5x2 8x — I1 = 0. 4. X4 - x3 _ 32 +x2 -x- 3 = 0. 5. xs - 2X3 + I = 0. Discuss each of the above equations. 307. In the preceding discussions we have supposed the equations to be given, and from the relations existing between the co-efficients of the different powers of the unknown quantity, have determined the number and places of the real roots; and, consequently, the number of imaginary roots. In the equation of the second degree, we pointed out the relations whllich exist'between the co-efficients of the different powers of the unknown quantity when the roots are real, and when they are imaginary (Art. 116). Let us see if we can indicate corresponding relations among the co efficients of an equation of the. third degree. Let us take the equation, x3+ Px2 + Q' + U=- and by causing the second term to disappear (Art. 263), it will take the form, x3 -px + q -= 0. CHAP. XI CARDAIN'S RULE. 379 I-Tence we have IXr = o3 + pt _ qx 1X = 3xZ -- p, X~ = - 21.xr- 3q, X3 = -4ps -- 27q2. In order that all the roots be real, the substitution of co for z in the'above expressions must give three permaniences; and the substitution of - o for x must give three variations. But the first supposition can only give three permanences when -423 - 27q2 > 0; hat is, a positive quantity, a condition which requires that p2 be negative. If, then, p be negative, we have, for x co, 4p3 27q2 > 0; that is, positive:. or, 4p3 + 27q2 < 0; that is, negative: 2 3 hence, q +- < 0, which requires that p be negative, and that o-? > -; conditions which indicate that the.27 4 roots are all real. Cardan's Rude for Solving Cubic Equations. 308. First, free the equation of its second term, and we have the form, x3+p g= - (1). Take x = y + z; then x3 yS + z3 _ 3yz (y - z); or, by transposing, and substituting x for y + z, we have X3-3yr.. -(y3 + Z3)=O - - - (2); and by comparing this with equation (1), we have - 3yz -- p; and y3 z3 = - q. 380 ELEMENTS OF ALGEBRA. [CHAP. XL From the 1st, we have p3 27y3$ which, being substituted in the second, gives p3' 27y3 -q; Gr clearing of fractions, and reducing y6 - qy3a - p Solving this trinomial equation (Art. 124), we have 3 q 2 p3 Y= A - 4 27' and the corresponding value of z is 3 q2 p3 = -2 - 4 27 But since x = y + z, we have X q + q + q This is called Cardant's fornmula. By examining, the above formula, it, will be seen, tha. it is inapplicabie to the case, when the quantity 92 I 3 q7 p3 4 27' under the radical of the second degree, is negative; and hence is applicable only to the case where two of the roots are imag inary (Art. 307). IIaving found the real root, divide both members of the giveL equation by the unknown quantity, minus this root (Art. 247); the result will be an equation of the second degree, the roots of which may be readily found. CHAP. XI.1 HORNER'S METHOD. 381 EXAMPLES. 1. What are the roots of the equation x3 - 6x2 10x- = 8? Ans. 4,. 1+ l -- 2. WhIat are the roots of the equation x3 - 9x2 + 2Sx = 30? Ans. 3, 3+ -1, 3- / -. 3. What are the roots of the equation X3 - 7x2 +- 14x = 20? Ans. 5, 1+~-3, 1-1- a3 PrelfInirnaries to Horrer's 2Method. 309. Before applying the method of Horner to the solution af numerical equations, it will be necessary to explain, 1st. A modification of the method of multiplication, called the method by Detached U(o-e cients: 2d. A modification of the method of division, called, also, the method by Detached Co-ejicients: 3d. A second modification of the method of division, called Synthetical Division: and, 4th. The application of these methods of Division in the Transformation of Equations. Mfultipliccttion by Detached Co-efficients. 310, When the multiplicand and multiplier are both homogeneous (Art. 26), and contain but two letters, if each be arranged according to the same letter, the literal part, in the several terms of the product, may be written immediately, since the exponent of the leading letter will go on decreasing from left to right by a constant number, and the sum of the exponents.~f both letters will be the same, in each of the terms. 382 ELEMENTS OF ALGEBRA, [CHAP. Xl. EXAMPLES. 1 Let it be required to multiply x3+x2y-+xy2+y3 by x-y. Since x3 X = X4, the terms of the product will oe of the 4th degree, and since the exponents of x decreasP by 1, and those of y increase by 1, we may write the literal parts thus, x4, x3y x2y2, xy3, y4. In regard to the co-efficients, we have, Co-efficients of multiplicand, - - - 1 1 + 1 1.'" "L multiplier - 1 —1 -1- 1 —1co-efficieints of the product, - 1 + 0 + 0 + 0 - 1; aikd writing these co-efficients before the literal parts to which they belong, we have x4 + 0. Xq y ~ O. 3- O. 22 -. y3 -y y4. 2. Multiply 2a3 - 3ab2 + 5b3 by 2a2 — 5b2. In this example, the term a2b in the multiplicand, and ab in the multiplier, are both wanting; that is, their co-efficients are 0. Supplying these co-efficients, and we have, Co-efficients of multiplicand, - 2 - 0 - 3 -+ 5 " "L nmultiplier, - - 2 + 0- 5 4q-0- 6+-10 - 10 — 0+15 —25 co-efficients of the product, - - 4 + 0 - 16 +- 10 + 15 - 25. Hence, the product is, 4a5 — 16a3b2 ~- 10a2b3 + 15ab4 - 25b5 3. Multiply x3 —3X2+ 3x-1 by x2 - 2x + 1. 1 1 4. Multiply y2 ya- a2 by y2 + ya z. 4 4 REPMAR.-T-Th e method by detached co-efficients -s also appli. cable to the case, in wlich the multiplicand anld multiplier con. tain but a single letter. The terms whose co-efficients are zero must be supplied, when,vanting, as in the previous exanmples. CHAP'. XI. DETACHED CO EFFICIENTS. 383 EXAMPLES. 1. What is the product of a4 + 3a2 l1 by a2 — 3 2, What is the product of 2 -I-1 by b + 2? Division by.Detached Co-efcients, 3110 When the dividend and divisor are both homogeneous and. contain but two letters, the division may be performed by means of detached co-efficients, in the following manner: I. Arrange the terms of the dividend and divisor according to a common letter. 2. Subtract the highest exponent of the leading letter of the divisor from the exponent of the leading letter of the dividend, and the remainder will be the exponent of the leading letter of the quotient. 3. The exponents of the letters in the other terms follow the same law of increase or decrease as the exponents in the corresponding terms of the dividend. 4. Write down for division the co-efficients of the different terms of the dividend and divisor, with their respective signs, supplying the deficiency of the absent terms with zeros. 5. Then divide the co-efficients of the dividend by those of the divisor, after the manner of algebraic division, and prefix the several quotients to their corresponding literal parts. EXAMPLES. 2. Divide 8a5 —4 ax — 2~'z2 + a2x3 by 4a2 —. The literal part will be a3, adx, ax2 x3 i and fir the numerical co-efflicients, 8-4-2+ 11 4+0-1; 8+0 —2 2 —1 -4 +1 -4 +1 0 0 384 ELEMENTS OF ALGEBRA. LCHAP. XI. hence, the true quotient is 2a3 - a2x; the co-efficlents after -1, being each equal to zero. 3. Divide x4 —3ax3 — Sa2x2 + 18a3x —8a4 by x2-2ax —2a2. 4. Divide 10a4 - 27a3x + 34a2X2 - 18ax3 - 8x4 by 2a2 3ax + 4x2. REMARK.-The method by detached co-efficients is also appli. eable to all cases in which the dividend and divisor contain but a single letter. The terms whose co-efficients are zero, must be supplied, when wanting, as in the pirevious examples. EXAMPLES. 1. Let it be required to divide 6a4 —96 by 3a —6. The dividend, in this example, may be written under the form, Ga4 0. a3 +- 0. a2 + 0. a - 96a~. Dividing a4 by a, we have a3 for the literal part of the first term of the quotient; hence, the form of the quotient is a3, a2, a. For the co efficients, we have, 6 + 0 + 0-+0 - 96 113 -6 6 —12 2 + 4 + 8 + 16 quotient; hence, the true quotient is, 2a3 +- 4a2 + 8a + 16. ASynthetical Division. 312. In the common method of division, each term of the divisor is multiplied by the first term of the quotient, and the products subtracted from the dividend; but the subtractions are Ierformed by first changing the sign of each product, and then.adding. If, therefore, the signs of the divisor were first changed, we should obtain the same result by adding the products, instead of subtracting as before, and the same for any subsequent oper. ation. CHAP'XI.] SYNTHETICAL DIVISIOIn. 385 By this process, the second dividend would be the same as by the common method. But since the second term of the quo. tient is found by dividing the first term of the second dividend by the first term of the divisor; and since the sign of the latter has been changed, it follows, that the sign of the second term of the quotient will also be changed. To avoid this change of sign, the sign of the first term of the divisor is left unchanged, and the products of all the terms iof the quotient by the first terim of the divisor, are omitted; be cause, in the usual method, the first terms. in each successive dihidend are cancelled by these products. H1aving made the first term of the divisor 1 before commene ing the operation, and omitting these several products, the co-efficient of the first term of any dividend will be the co-efficient of the succeeding term of the quotient. Hence, the co-efficients in the qu)tient are, respectively, the co-effcients of the first terms of thei successive dividends. The operation, thus simplified, may be fa.rther abridged by. omitting the successive additions, except so much only as may be necessary to show the first term of each dividend; and also,. by writing the products of the several terms of the quotient by the modified divisor, diagonally, instead of horizontally, the first product falling under the second term of the dividend. Hence, the following RULE. 1. Divide the divisor and dividend by the co-fficient of the first; term of the divisor, when that co-eficient is not 1. it. Wfrite, in a horizontal line, the co-efficints of the dividendc4 aw ii their prcoper signs, and place the co-eliceezts of the dizisor,: with all their signs changed, e.xcept the first, on the right. [Il. Divide as in the method by detached co-eficients, except thda no term of the quotient is multiplied by the bfist term of the divisol, and that all the products are written diagonally to the, rigias4. under the terms of the dividend to which they correspond. 25 386 ELESMENTS OF ALGEBRA. [CHAP. XI, IV. The f rst tern of the quot:i!nt is the same as that of the dividend; tihe second term is the sum of the numbers in the second column; the third term, the sum of the numbers in third column, and so on, to the right. V. FWhen the division can be exactiy made, columns will lie found at the right, whose sums will be zero: when the division is not exact, continue the operation until a sufficient degree of approximation is attained. tHaving found the co-efficients, annex to them the literal Tarts. EXAMPLES. 1. Divide a65- 5a4x +- 10a3x2 — 10a23 +5ax - -z by a2- 2x x + x. - 5 10 - 10 5-1 I1+ 2-1 2 - 6 63+3- 1 + 3 —3+1 1-3+ 3 1 0 0. Hence, the quotient is a3 - 3a2x -F 3ax2' -'3 PREMARK. — The first term of the divisor being always 1, need:not be written. The first term of the quotient is the same as:that of the dividend. 2. Divide - -5x5~+ 1 5x4 -24x3+27x2-1 3-+5 by 4- 3.-4x2 —-2x- 1 - 5 +Q 15-24 J+27 — 13 +51[ 1+ -4- 2- i +2,- 6+10 I- 3 + 5 - 4+12 - 20 + 2 — 6+10 - 1 _L 3 5 *1 —_3+ 5 <0 0 0 0. H'ence, the quotient'is x2 - 3x + 5. $. Divide ae +.-a4b + 3(b2 - 23 —.2ab -- 3b5 by, -+- 2ab 4 3ba..Ans. -a + 0. a2b 4+ 0a. ab3 -- b3- a3 - b3.; CHAP. XI.] SYNTHETICAL DIVISION. 387 4 Divide 1 —x by 1 + x. Ans. I -2x -t xa -- 2X3+ &C. 5. Divide I by 1 -X. Ans. l + x + xe + x3 + &C. 6. Divide x7 - y7 by x - y. Ans. x6 + xS5y - x4y2 + xzy3 + x2y + xy5 q+ y6. 7.' Divide a6 - 3a4x2 + 3a2x4 - x6 by a3 3a2x + 3a-2 2 — 3 Ans. a3 +' 3a2x + 3ax2 + x3. 313. To transform an equation into another iwhose roots shall be the roots of the proposed equation, increased or diminished by a given 7uantity. A method of solving this problem has already been explained (Art. 264); but the process is tedious. We shall now explain a more simple method of finding the transformed equation. Let it be required to transform the equation axn + P2"-1 + Qxm-2.... T + U = 0 into another whose roots shall be less than the roots of this equation by r. If we write y q- r for x, and develop, and arrange the terms with reference to y, we shall have ay, + P/,y-_ + Q'y~2.... +- T' +;,U' - - (1) But since y = - r, equation (1), may take the form a(X-_.r)n+P p(X-_r)m-1 + Q'(x - r.)2... T'(x -r)+'=0 (2), which, when developed, must be identical with the given equa. tion. For, since y q- r was substituted for x in the proposed equation, and then x - r for y in the transformed equation, we must necessarily have returned to the given equation. Henee, we have a(X -,)n + P,(X - I.)-1 + Q'(x - r)~-2... TI (x - r) + U( = ax + P'xn-I + Qx —2... Tx + U = 0. If now we divide the first member by.z —r, the quotient will be a(x - r)mn-l + PI(x - r)"-2 + Q'(x - r).v. and the remainder P7. 388 ELEMENTS OF ALGEBRA. lCHAP. I. But since the second member is identical with the first, the very same quotient and the same remainder would arise, if the second member were divided by x- r: hence, If the first member of the given equation be dividclcd by!he unknoewn "antity minus the number which expresses the difertence between th t1s, the remainder will be the absolute term of the transformed equation. Again, if we divide the quotient thus obtained: viz., a(x - r) —l + P'( -,.)n —2 + Q(,- r)m-3..T by z - r, the remainder will be T', the co-efficient of the term last but one of the transformed equation; and a similar result would be obtained by again dividing the resulting quotient by s —r. Hence, by successive divisions of tile polynomial in the first member of the given equation and the quotients which result, by x - r, we shall obtain all the,v-efficients of the transformed equation, in an inverse order. REMARK.-When there is an absent term in the givev equation, its place must be supplied by a 0. EXAMPLES. Transform the equation 5X4 - 12x3 + 3x2 + 4x - 5 = 0 into ano,:er whose roots shall each be less than those o' the gives equation ",y 2. First Operation. 5,4 12x3 -3x2 4x-5 jx -2 5x4 - 10x3 5x3 -- 2X2 - x + 2 - 2x3 + 3x2 - 2X3 + 4x2 -- x + 2x 2w - 5 2w -4 - 1 1st remainder CHAP. XI.l SYNTHETICAL DIVISION. 389 Second Operation. 5x' - 2x2 - x + 2 11 -2 5X3 - 10X2 5X2 - 8 - +15 8x2 - Sx2 - 16x 15x + 2 15x - 30 32 2d remainder, Third Operation. Fourth Operati o. 5X2 + 8+ 15r x- 2 5x+ 18! -2 5x2 - 10x 5x.- 18 5x - 10 5 18x q- 15 28 4th remainder. 18x - 36 51 3d remainder. Therefore, the transformed equation is 5y4 + 28y3 - 51y2 + 32y -1 = 0. This laborious operation can be avoided by the synthetieal method of division (Art. 312). Taking the same example, and recollecting that in the synthetical method, the first term of the divisor not being used, may be omitted, and that the first term of the quotient, by which the modified divisor is to be multiplied for the first term of the product, is always the first term of the dividend; the whole of the work may be thus arranged: 5- 12 + 3 + 4 5 112 10 -4 -2 4 -2 -1 2 -1.'. UT'= 1 10 16 30 8 15 32.'. 32 10 36 28 51. Q' = 51 28.'. P = 28 - 890 ELENMENITS OF ALGEBRA. LCHAP. XL for it is plain that the first remainder will fall under the absolute term, the second under the term next to the left, and so on. Ience, the transformed equation is 5y4 + 28y +- 51y2 + 32y — 1 = 0. 2. Find the equation whose roots are less by 1.7 than those of the equation x3 - 2x2 + 3x- 4 = 0. First, find an equation whose roots are less by 1. 1-2 + 3 — 4111 1 — 1 2 -1 2 -2 1 0 0 2 1 We have thus found the co-efficients of the terms of an etfuation whose roots are less by 1 than those of the given equation: the equation is x3 + x2 2x - 2 = 0; and now by finding a new equation whose roots are less than those of the last by.7, we shall have the required equation: thus, 1+1 +2 -2f1.7.7 1.19 2.233 1.7 3.19.233.7 1.68 2.4 4.87 3.1 henee, the required equation is y3 + 3.1y5 + 4.87y +.233 = 0. Thils latter operation can be continued from the former, witl. out arranging the co-efficients anew. The operations have been explained sep3ratoi.y, merely to indicate the several steps in the CHAP'. XI.] SYNTHE'ICAL DIVISION. 891 transformation1 and to point out the equations, at each step resulting from the successive diminution of the roots. Conbining the two operations, we have the following arrangement: 1 -' +3 — 4(1.7; or, 1 -2 +3 -4(1.7 I - 1 2 1.7 -.51 4.233 - 1 2 -2 -.3 2.49.233 1 0 2.233 1.7 2.38 0 2.233 1.4 4.87 I 1.19 1.7 1.7 3.19 3.1.7 1.68 2.4 4.87.7 3.1 We see, by comparison, that the above results are the same as those obtained by the preceding operations. 3. Find the equation whose roots shall be less by 1 than the roots of x3 - 7x:- 7 = 0. Ans. y3 +4_ y2 4y -} 1 = 0. 4. Find the equation whose roots shall be less by 3 than the roots of the equation Xz- 3x3 - 15x2 + 49x - 12 = 0. Ans. y4 +- 9y3 + 12y2 - 14y -- 0. 5. Find the equation whose roots shall be less by 10 than the roots of the equation x4 + 2x3 + 3x2 + 4x - 12340 = 0. Ans. y4 + 42y3 1 663y2 + 4664y = 0. 6. Find the equation whose roots shall be less by 2 than the roots of the equation x5 +- 2x3 - Ox2 - 10x _ O. A s. y5 I OQy4 -I — 42f 86- y2 + 70y + 1 O=O 892 ELEMENTS OF ALGEBRA.'CHAP. XI Horner's Method of approximratina to the Real -Roots of Numerical Equawions. 314, The nethod of approximating to the roots of a nLmeri cal equation of any degree, discovered by thp English nmathe, matician bW. G. Horner, Esq., of Bath, is a,rocess of very remarkable simplicity and elegance. The process consists, simply, in a succession of transforma tions of one equation to another, each transformed equation, as it arises, having its roots equal to the difference between the tlrue value of the roots of the given equation, and the part of the root expressed by the figures already foiund. Such figures of the root are called the inilial figyurs.. Let V = xZm + pxtn-1 Qx,2n-2.... + Tx + U 0- - - (1) be any equation, and let us suppose that we have foun'r a part of one of the roots, which we will. denote by mn, and denote the remaining part of the root by r. Let us now transform the given equation into another, wh,ose roots shall be less by rm, and we have (Art. 313), V = rm {_+ P1/m-1 + Q'rm-2.... + T'r + U = 0 (2). Now, when r is a very small firaction, all the terms of the second member, except the last two, may be neglected, and the first figure, in the value of r, may be found from the equlatlon U/ U' T'r + U' = 0; giing -r -; l' r =-; hence, Thle first figure of r is the first figure of the quotient obtained by lividing the absolute terme of the trainsformzed equation by the penu.ti. fnate co-efficient. If, now, we transform equation (2) into another, whose roots shall be less than those of the previous equation by the first figure of r, and designate the remaining part by s, we shall hav e, VT"';m ~ P'/os- L Q/'sm —2.. + ~Tns + to ^ = 0, CHAP. XI. I HORNER'S METHOD. 393 the roots of which will be less than those of the given equa. tion by inm the first figure of r. The first figure in the value of s is found froll the equation, T"'s + U" 0, giving S =Ti,. xWe may thus continue the transformations at pleasure, and each one will evolve a new figure of the root. Hence, to find the roots of numerical equations. 1. Find the mtnzber and places of the real roots by Sturms' theorem, and set the negative roots aside. 11. Transform the given equation into another whose roots shall be less thal those of the given equation, by the initial figure or figures al}ready found: then, by Slturs' theorenm, find the places of the roots of this new equation, and the first figure of each will be the first decimal place in each of the required roots. III. Transform the equation again so that the roots shall be less than those of the given equation, and divide the absolute term of the transformzed equation by the penullimate co-efficient, which is called the trial divisor, and the first figure of the quotient wtill be the next figure of the root. IV. Trancsfor-m the last equation into another whose roots shall be less than those of the previous equation by the figure last found, and proceed in a similar nzmacnner until the root be fo.und to the required degree of accuracy. REMAR I. —This method is one of approximation, and it may happen that the rejection of the terms preceding the penultimate term will affect the quotient figure of the root, To avoid this source of error, find the first decimal places of Lhe root, also, bhy the theorem of Sturm, as in example 4, page 899, and when the results coincide for two consecutive places of decimals, those subsequently obtained by the divisors may be relied on. REMARK II. —When,he decimal portion of a negative root is to be found, first transform the given equation into another by changing the signs of the alternate terms (Ar'.. 280), and then find the decimal part of the corresponding positive root of this new equation. 394 ELEMENTS OF' ALGEBRA. [CMAY. Xl. I]1. When several decimal places are found in the root, the operation may be shortened according to the method of con. tractions indicated in the examples. 314, Let us now work one example in full. Let us take the. equation of the third degree, 3 - 7x + 7 = 0. By Sturm's rule, we have the functions (Art. 299), X - x3-7 +7 XA1 = 3x2 - 7 X=-2x -3 X3 = + 1. Hence, for x = o, we have + + + + no variation, X - C -GO + - + three variations; therefore, the equation has three real roots, two positive and one negative. To determine the initial figures of these roots, we have for x = O. + —-- forx= 0.. -.+- -+ Z = 1. + - - — 1... -+ -=2....+ + + x= -2... + — + x 3... ++ + z= —4... —+ -+ hence there are two roots between 1 and 2, and one between - 3 and -4. In order to ascertain the first figures 1 0 - 7 + 7 ( in the decimal parts of the two roots I 1 - situated between 1 and 2, we shall trans- I - 6 + 1 form the preceding functions into others, I + 2 in which the value of x is diminished by 2 - 4 1. Thus, for the function X, we have 1 this operation: 3 And traniforrring the others in by y3 3y2 - 4y 1;.he same wayr, we obtain the Y, 3Y2 + 6y - 4; thncticns = - 1; Y, —- 1. CHAP. XI.] HORNER'S METHOD. 395 Let y =.1 we have + — - + two variations, y=.2 + ---- y-.3 " + _- _ + - y.4 " - - + one variation, y=.5 " -- y-, " -- + + Y- =.7 " +'- + no variation.'therefore the initial figures of the two positive roots are 1.3, 1.6, i et us now find the decimal part of the first root. I o - 7 + 7 (1.356S95856 1 1 -G — 6 1 1 2 -.903 2 *-4 **.097 1.99 -.086625 *3.3 - 3.0 1 *~.010375 3 1. 08 --.009048984 3.6 ** —. 9 3 "***.001326016 3.1975 -.001184430 4*3.9 5 -1.7 3 2 5.000141586 5.2 000 -.000132923 4.0 0 * — 1. 5 3 2 5.000008663 5.024336 -.000007382 <**4.0 56 - I. 5 08 1 6 4.000001281 6.024372 -.000001181 4.0 62 * -- 1.4 83 7 9 2.000000100 6. 0 3 2 5 4 -.000000089'*4.0168 8 - 1.4 8 0 5 38.000000011 8.00325i4 -.000000010 14.0169C — 1.4 77 218 1 00036 0 0 0 316 — II 414171 0,r 396 ELEMENTS OF ALGEBRA. [CHAP. XI The operations in the example are performed as follows: 1st. We find the places and the initial figures of tle positive roots, to include the first decimal place by Sturms' theorem. 2d. Then to find the decimal part of the first positive roc.t, we arrange the co-efficients, and perform a succession of trans formations by Synthetical Division, which must begin Fwith the initial figures already known. We first transform the given equation into another whose roots shall be less by 1. The co-efficients of this new equation are, 1, 3, - 4 and 1, and are all, except the first, marked by a star. The root of this transformed equation, corresponding to the root sought of the given equation, is a decimal frae tion of which we know the first figure 3. We next transform the last equation into another whose roots are less by three-tenths, and the co-efficients of the new equation are each marked by two stars. The process here changes, and we find the next figure of the root by dividing the absolute term.097 by the penultimate co-efficient -1.93, giving.05 for the next figure of the root. We again transform the equation into another whose roots shall be less by.05, and the coefficients of the new equation are marked by three stars. We then divide the absolute term,.010375 by the penultimate co-efficient, - 1.5325, and obtain.006, the next figure of the root: and so on for other figures. In regard to the contractions, we may observe that. having decided on the number of decimal places to which the figures in the root are to be carried, we need not take notice of figures which fall to the right of that number in any of the dividends. In the example under consideration, we propose to carry the operations to the 9th decimal place of the root; hence, we may reject all the decimal places of the dividends after the 9th. The fourth dividend, marked by four stars, contains nine decimal places. and the next dividend is to contain no more. CHAP. XI.] HORNER'S METHOD. 397 But the corresponding quotient figure 8, is the fourth figure from the decimal point; hence, at this- stage of the operation, all the places of the divisor, after the 5th, may be omitted, since the 5th, multiplied by the 4th, will give the 9th order of decinals. Again: since each new figure of the root is removed one place to the right, one additional figure, in each subsequent divisor, may be omitted. The contractions, therefore, begin by striking off the 2 in the 4th divisor. Yn passing from the first column to the second, in the next operation, we multiply by.0008; but since the product is to be limited to five decimal places, we need take notice of but one decimal place in the first column; that is, in the first operation of contraction, we strike off, in the first column, the two figures 68: and, generally, for each figure omitted in the second column, we omit two in the first. It should be observed, that when places are omitted in either column, whatever would have been carried to the last figure retained, had no figures been omitted, is always to be added to that figure. Having found the figure 8 of the root, we need not annex it in the first column, nor need we annex any subsequent figures of the root, since they would all fall at the right, among the rejected figures. Hence, neither 8, nor any subsequent figures of the root, will change the available part of the first column. In the next operation, we divide.000141586 by 1.4772, omitting the figure 8 of the divisor: this gives the figure 9 of the root. We then strike off the figures 4.0, in the first column and multiplying by.00009, we form the next divisor in t0l second column, — 1.4769, and the next dividend in the column,.000008663. Striking off 5 in this divisor, we the next figure of the root, which is 5. It is now evident that the products from the first column, will fall in the second, among the rejected figures at the right; we need, therefore, in future, take no notice of them. Omitting the right hand figure, the next divisor will be 1.476, and the next figure of the root 8. Then omitting 6 in the 898 ELEMENTS OF ALTGEBRA. LICAPr, X divisor, we obtain tLbe quotient figure 8: omitting 7 we obtain 6, and omitting 4 we ubtait 7, the last figare to be foundI We have thus found the root x =.356895876....; and all sim'ila examples are wrought after the same manner. The next operation is to find the root whose initial figures are ].6, to nine decimal places. The operations are entirely similar to those just explained.'We find for the second root x - 1.69202141. For the negative root, change the signs of the secocnd und Ifurth terms (Art. 280), and we have, 1- - 7 7 (3.04891733,96 3 9 + 6 3 2 — 1 3 1 8.814464 6 20 -.185536 3.3 6 1 6.166382592 9.0 4 2 0.36 6 1 6 19153a408 4.3 6 3 2 18791228 9.0 8 2 0. 7 24 8 — 362180 4 7 3 0 2 4 208875 9.1 28 2 0.79782 4 -153305 8 73088 146212 9.1 36 20.8 70912 -70193 8 8 2 3 0 6266 1 |9.1 144 20.879 1 42 -827 8 2 30 626 20.88737 -201 19 188 2 0. 8 874 6 -13 9 12 210.18181715 I ~P Find t;he:roots of the equation I3 * 112 - 102x - 181 =0. E[AP. XI.] HORNER'S METHOD. 399 The functions are X X3 $+ 11x2,-102x+1 1 X- -32 + 22x - 102 X- 122x - 393 ad the signs of the leading Jerins are all +; hence, the sub titutien cf - - and + oo must gibe three real roots. To discover the situation of the roots, we make the substitu; ions z _0 which gives + - -+ two variations x= 1 " - - -- " X 2 " + - - + s x 3 + - + - x 4 + + + + no variation; lence the two positive roots are between 3 and 4, and we must;herefore transform the several functions into others, in whicl, x Ihall be diminished by 3. Thus we have (Art. 314), Y -- y3 + 20y2 -9y 1 Y = 3y2+ 40y — 9 Y, = 122y - 27 = - + Make the following substitutions in these functions, Niz.: y = 0 signs +- + two variations y=.1 " + ----- + =.2 " -- -- + y-.3 " + + + + no variation; hemce, the two positive roots are between 3.2 and 3.3, and we mli-t again transform the last functions into others, in which y h al1 be diminished by.2. Effecting -this transformation, we have Z - z3 -{ 20.6z2 -.88 +.008 Z, = 3z2 + 41.2z -.88 Z, = 122z - 2.6 Z3= +. 400 ELEMENTS OF ALGEBRA. [CHAP. XI. Let z 0 then signs are + - - + two variations, z.01 " " + - - + z =.02 " " - - + one variation, z -.03 " " ~ + + + no variation; nornce we have 3.21 and 3.22 for the positive roots, and the sum of the roots is - 11; therefore, - 11 - 3.21 - 3.22 - 17.4. is the negative root, nearly. For the positive root, whose initial figures are 3.21, we have x = 3.21312775; and for the root whose initial figures are 3.22, we have x = 3.229522121; and for the negative root, x -- 17.44264896. EXAMPLES. 1. Find a root of the equation x3 + x2 + x - 100 = 0. Ans. 4.2644299731. 2. Find the roots of the equation x4- 12x2 + 12 - 3 = 0. ~+ 2.858083308163 -A..606018306917 +.443276939605 - 3.907378554685. 3. Find the roots of the equation 4 - 8x3 + 14x2 + 4x - 8= —0, F+ 5.2360679775 I +.7639320225 + 2.7320508075, --.7320508075. k Find the roots of the equation x5 -103 + 6.X+ 1 — 0. 3- 3.0653157912983 -.6915762804900 4n3s, -.1756747992883 +.8795087084144 + 4- 3.0530581626622.