-4:0 t4:3* ELEMENTS OF AL G E BR A. DESIGNED FOR SCHOOLS, ACADEMIES, AND COLLEGES. BY CHARLES D. LAWRENCE, PROF]lSOR OF MATNBMATICU IN CORTLAND ACADZXT. AUBURN: ALDEN, BEARDSLEY & CO. ROCHESTER: WANZER, BEARDSLEY & CO. 1853. Entered, according to Act of Congress, in the year 1852, BY ALDEN, BEARDSLEY & CO., In the Clerk's Office for the Northern District of New York. STEREOTYPED BY THOMAS B. SMITH, 216 William St. N. Y. PREFACE. PERHAPS every teacher of experience has formed in his own mind a conception of what he regarded as an unexceptionable text book on some favorite branch of study. The author of this treatise has had several years' experience as a teacher of mathematics, and during this time, he has given several of the more important works on algebra a fair trial in the recitation-room, the only proper place to test the merits of a school book. This trial has resulted in the belief; that a work on algebra, adapted to the wants of the mljority of those who make this science a study, is yet a desideratum. For this reason, the following treatise has been prepared. Great care has been taken to adopt such demonstrations of theorems, and discussions of principles, as are at once characterized by the greatest simplicity, rigor, and brevity. No effort has been made to simplify a demonstration, or discussion, by entering into mere matters of detail; for, it is believed, that such attempts to adapt everything to the comprehension of the meanest capacity, often fail in their object, and cause many an intelligent pupil to lose sight of important principles, as he is groping his way through a multiplicity of words. This work is, as its title indicates, an Elementary Treatise on Algebra; but, it must not therefore be inferred, that it may be studied without requiring any intellectual exertion. It has not been written for the purpose of pleasing " mental dyspeptics whose alim;tl; loathe vigorous thought as a diseased stomach loathes wlholesolle food." Such a work might suit the taste of some, but its puerilities would displease every intelligent scholar. Throughout thle whole work, it is hoped that the student will find opportunities for exercising his powers of thought and analysis. Our experience in teaching algebra has proved that the scholar can only acquire a familiarity with principles by applying these principles to the solution of practical questions. For this reason, a very large number of practical examples and problems will bie found iv PREFACE. in this work. Many of these examples are not contained in any other school book. In selecting these examples, it has been made a prominent object to select such as would be most likely to interest the student, and at the same time, to have them of such a nature, that their solutions would strengthen and increase his powers of analysis. On the Theory of Equations we have been concise, but it is hoped that that part of it of which we have treated will be found amply sufficient for most practical purposes. We will refer the reader to the table of contents for a brief enumeration of the topics discussed in this work. The utility of the study of Algebra cannot be questioned. Indeed, the study of any branch of mathematics is admirably calculated to give tone and vigor to the mental operations. Since the invention of representing by equations, lines, surfaces, and solids in geometry, the study of algebra has acquired a very great importance. It may be said to constitute the basis of sound mathematical learning. Should this volume be the means of facilitating the progress of the student in an important science, the wishes of the Author will be gratified. CORTLAND ACADEMY, Homer, Nov. 7, 1852. CONTENTS. CHAPTER I. PAGN Definitions and Notations........................ 9 CHAPTER II. GENERAL RULES AND PRINCIPLES. Addition.................................................... 13 Subtraction.................................................. 18 Multiplication.............................................. 21 Division................................................... 29 Greatest Common Measure............................39 Least Common Multiple........................ 44 CHAPTER IIL. FRACTIONS. To reduce a fraction to its lowest terms.................. 47 To reduce a mixed quantity to the form of a fraction...............49 To reduce a fraction to an entire or mixed quantity.......1........51 To add fractions............................................ 62 To subtract fractions.........................................55 To multiply one fraction by another............................57 To divide one fraction by another.............9............ 59 CHAPTER IV. SIMPLE EQUATIONS. Definitions................................................. 61 Axioms.....................................................62 Transposition............................................... 62 To clear an equation of fractions..................... 62 Examples................................................... 64 Solutions of questions which involve one unknown quantity........ 68 Problems.................6...........b.8 Vi COONTENTS. PAGE Equations which involve two or more unknown quantities......... 81 Elimination by addition and subtraction................. 81 Elimination by substitution.................................... 85 Elituinalion by comparison..................................... 86 Examples.................................................. 88 SoluLtion of problems which involve two or more unknown quantities. 96 Interpretation of negative results............................... 106 CtIAPTER V. INVOLUTION, EVOLUTION, RADICAL QUANTITIES AND IMAGINARY QUANTITIES. To involve a monomial......................... 111 To involve a polynomial.................................. 113 Binomial Theorem..................................... 114 Examples in Involution...................1.................. 116 Evol ution............................................... 118 To extract any root of a monomial............................. 118 Exatples.................................................. 119 To extract the square root of a polynomial...................... 120 Examples.................................................. 121 To extract the cube root of a polynomial........................ 123 Examples.................................................. 125 Radical Quantities............................................ 127 To reduce a rational quantity to the form of a radical............. 128 Examples.................................................. 128 To reduce radicals which have different exponents to equivalent ones having a common index.......1........1................. 129 Exampl les.................................................. 130 To redurce ra(lical quantities to tlheir silmplest firms.............. 131 Exatmples............................................. 132 Addition and subtraction of radicals............................ 133 Examples.................................................. 133 Mlultip)lication of radicals...................1..5.............. 1 5 ExsaIi ples................................................. 135 Division of radicals........................ 136 Exa sllll................................................ 136,Extralctionl of the square root of a binomial surd............... 137 Examples.................................................. 140 Imag-itary quantities.................................... 141 Examples.................................................. 142 CIIAPTER VI. rPUIRi' QUADI.RATICS. Definition of a pure quadratic.......................... 145 Examples........................................ 145 Problems in Pure Qtuadratics.................................. 154 CONTENTS. Vii CHAPTER VII. ADFECTED QUADRATICS. PAGE Definition of an adfected quadratic............................. 161 Method of finding the third term of the square of a binomial by means of the other two................................... 161 Examples................................................. 162 Three rules for finding the value of the unknown quantity in n adfected quadratic......................................... 163 In no quadratic equation, can the unknown quantity have more than two values.............................................. 166 Adfected quadratics containing more than one unknown quantity... 175 Examples.................................................. 176 Problems in adfected quadratics............................... 183 Discussion of an equation of the second degree................... 193 Application of the principles in the preceding discussion to the solution of problems......................................... 195 CHAPTER VIII. p!, P:;! SSIONS. Arithmetical Progression..................................... 201 Problem fir finding the last term.............................. 201 Problem for finding the sum of the terms........................ 202 Examples.................................................. 203 Geometrical Progression and Ratio............................ 207 Definitins.................................................. 207 Theorems in Proportion...................................... 208 Problem for finding the last term in a geometrical progression..... 211 Problem for finding the sum of all the terms..................... 211 Formnula for finding the suni of an infinite number of terms of a decreasing geometrical series................................ 212 Examples................................................ 313 CIIAPTERI IX. INDETER'MINATE CO-EFFICIENrTS, BINOMIAL THEOREM, SUMMATION OF SERIES, EXPONENITIAL THIEOREMN, AND LOGARITHMS. Indeterminate Co-efficients.................................... 220 Examples................................................. 221 BIinomial Theorem.......................................... 224 Investigation of the Binomial Theorem......................... 226 Examples.................................................. 228 Summation of Series......................................... 231 Differential Method..... 2.................................... 231 Formula for finding the nth term.............................. 232 viii CONTENTS. Formula for finding the sum of n terms of an increasing series.... 23.3 Examples................................................... 233 Summation of Infinite Series.................................... 235 Exponential Theorem............................ 237 Logarithms............................................. 239 Base of a system of logarithms................................ 239 Characteristic of logarithms-how determined................... 240 Logarithm of the product of two numbers equal to the sum of their logarithms............................................. 241 Logarithm of the quotient of two quantities equal to the difference of their logarithms....................................... 241 Logarithm of the nth root of a number equal to the nth part of its logarithm.............................................. 241 To find the logarithm of any number........................... 241 Modulus of the system....................................... 242 To find the base of the Naperian System........................ 245 Application of logarithms.................................... 245 Interest and annuities...................................... 247 Definitions and notation.................................. 247 Problems in interest and annuities............................ 249 Examples.................................................. 254 CHAPTER X. THEORY OF EQUATIONS. Propositions on theory of equations........................ 258 Transformation of equations................................... 264 Sturm's Theorem......................................... 283 Horner's method of resolving numerical equations of all degrees.... 295 Examples................................................... 296 ELEMENTS OF ALGEBRA. CHAPTER I. DEFINITIONS AND NOTATION. (1.) ALGEBRA is that method of computation in which letters and other symbols are employed. (2.) The first letters of the alphabet are generally employed to denote known quantities, and the last, those which are unknown. (3.) The symbol = denotes equality; as, a=b, which is read, a equals b. (4.) The symbol + denotes addition; as, a+b, or a added to b. Those quantities to which this sign is prefixed, are called positive quantities. (5.) The symbol - denotes subtraction; as, a-b, or b subtracted from a. Those quantities to which this sign is prefixed, are called negative quantities. (6.) The symbol X denotes multiplication; as, aX b, which is read, a multiplied by b. (7.) The symbol + denotes division; as, a+ b, which is read, a divided by b. (8.) The symbol > denotes inequality; as, a>b, which is read a is greater than b. The greater quantity is always placed next to the opening, as in the above example. I1 10 DEFINITIONS AND NOTATION. (9.) The use of the parenthesis, in Altgebra, may be illustrated by an example. If we wish to express, algebraically, the product of the sum of two quantities, a and b, by the difirence of two quantities, x and y, we may do it thus, (a+b)X(x-y). A straight line placed over a quantity is called a vinculuin. It is used for the same purpose as the parenthesis. Thus, a+-bXx-y is the same as the former expression. (10.) A co-eficient is a number or letter which is used to denote hovw many times a quantity is taken. Thus, in the expression 3x, 3 is the co-efficient of x. In the expression ax, a is the co-efficient of x, and it shows that z is to be taken as many times as there are units in a. (1 1.) An exponent is a number or letter which is used to denote how many times a quantity must be taken as a factor. Thus, the expression a' is the same as aX aXa. When an exponent is fractional, the denominator of the exponent shows the number of equal factors into which the quantity must be divided; and the numerator shows how many of these equal factors must be taken. Thus, in the expression 8, the denominator 3, shows that 8 is to be resolved into three equal factors, and the numerator 2, shows that two of these equal factors must be taken. Now 2 is one of the equal factors of 8; hence, 8 2 X 2 4. (1 2.) One of the two equal factors of a quantity is called the square root of that quantity. One of three equal factors of a quantity is called the cube root of that quantity. In general, the tllh root of a quantity is one of the n equal factors into which that quantity may be resolved. (13.) Roots are frequently denoted by fractional exponents; as a', which is read the cube root of a. 83 is read the cube root of the square of 8, or 64. The symbol V- denotes the square root. Thus, aC- and V/a are the same. (1 4.) When no exponent is expressed, the exponent 1 is always understood. The exponent or index of a quantity is written a little above and to the right of the letter; as, a'. DEFINITIONS AND NOTATION. 11 (15.) When a quantity is multiplied by itself, the product is called the square, or second power of that quantity. When a quantity is used as a factor three times, the product is called the cube of that quantity. And, in general, when a quantity is used as a factor n times, the product thus obtained is called the nth power of that quantity. (16.) The reciprocal of a quantity is a unit divided by that quantity; thus, - is the reciprocal of a. a (1 7.) The symbol.. signifies therefore or consequently. (1S.) An algebraical expression consisting of two or more terms connected by the si(rns + or -, is called a polynornial. An algebraical expression having only one term is called a monom;al. When a polynomial consists of two terms it is called a binomial; one consisting of three terms is called a trinomial. Thus, a-+b is a binomial, and a-b-c is a trinomial; a+b-3c +-4d+h is a polynomial. (1 9.) Every quantity written in algebraical characters is called an algebraical expression. Thus, 3a' is an algebraical expression for three tilnes the square of the number a. (20.),When each term of a polynomial contains the same number of literal factors, the polynomial is said to be homogeneous. a2+ax+y' is a homogeneous polynomial of the second deg~ree, because each term contains two factors. (21.) The product of several letters is indicated by writing the letters one after the other. Thus, the expression ab is the same as aX b. A period is sometimes used as a sign of multiplication; as a.b, which is read, a multiplied by b. (22.) Similar terms are those which contain the same letters, nf;acted with the same exponents. Dissimilar terms are those which do not contain the same letters, affected with the same exponents. 12 DEFINITIONS AND NOTATION. (23.) An axiom is a self-evident truth. (24.) A theorem is a truth which is made evident by means of a process of reasoning called a demonstration. A theorem is often called a proposition. (25.) A problem is a question which requires something to be ascertained. (26.) A corollary is an obvious consequence drawn from some proposition. (27.) A lemma is a proposition which is employed to establish some proposition which immediately follows it. CHAPTER II. ADDITION. (28.) ADDITION is the finding of an algebraical expression for the sum of several quantities. There are three cases in the addition of algebraical quantities. CASE I. (29.) When the quantities are similar, and have the same sign. If we wish to add 5ac2+4b to 7ac'+-6b, we can say that 5ac' and 7ac' are 12ac', and 4b and 6b are 10b,.. the sum is 1 2ac2+ ob. Operation. Sac'+ 4b 7ac2' 6b 12ac' + lOb sum. Whence we have the following RULE. Arrange the similar terms under each other, and then add the co-eficients of similar terms, and annex the common literal quantity. EXAMPLES. 1. What is the sum of 8b+ 7c, 4b+19c, and 1 7b. Ans. 29b+26c. 2. What is the sum of 4ax'+19c', 23ax'+4cS, and 17c2? Ans. 27ax'+40c2. 3, What is the sum of 16xy+17ay', 23xy+27ay2, and 47ay' 8 Ans. 39xy+91ay'. 14 ADDITION. 4, What is the sum of 14z+q-IlOab, 26z'2+17ab, and 13z2+ 5acb Ans. CASE II. (30.) When the quantities are similar, and have different signs. If we wish to add 10ab+14c' to 6ab-8c', we can say that 10ab and 6ab are 16ab. Now, the sign - before 8c2 indicates that it must be subtracted and not added. 8ca taken from 14c0 leaves +6c'. Hence the sum is 16ab+ 6c'. Operation. i Oab +14c' Gab- Sct 16ab+ 6cM Hence, when the quantities are similar, and have different signs, we have this RULE. Arrange the similar terms under each other, an)d then find the sum of the co-efficients of the positive quantities, and also the sum of the co-eficients of the negative quantities, in each set of similar terms; to the difference of these sums affix the common literal quantity, and prefix the sign of the greater sum. EXAM PLFS. 1. WNhat is the sum of x'-3x2y+3xyry-ys and x'+3xty+ 3xy' +y?.A fins. 2x'- +6xy2. 2. What is the sum of cx+10cx —l18ab+-4cx+ 10ab-4cx' +4ab? A ns. 11cx'-4ab. 3. What is the sum of 5a +6ay —15a 3+4ay —3a? -Ans. -13a3~+ 1 Oay.* 4. What is the sum of my' + 6.gx'+ 10mys- 4x3- 15mys. Ans. my2 +2x — 5my3. * NoTE. —5a is read five times the cube root of the square of a. See Art. 18. AUDITITON. 15 5. What is the sum of 16xy-4hnz+ 17xy+43hmk? Ans. 33xy+ 39hm. 6. What is the sum of 54' — 49an -10c 4+ l4an? AnS. 15c4-35an.. 7. What is the sum of 1Ox2y4 —17sk — 15X2y4 +-5sk —:xly 4 Ans. 21]xy.' 12 Ak. 8s What is the sum of 20a'cx-+15W -A- tF —h Ke-23ah? Ans. 5asc2x- 8ah. 9. What is the sum f 4x --?x+4, x-2x2-5. 1+-3x'-Sx, 2x-4+-7x2, 13-x2 —4x? Ans. 11x2-9x+-9. 10. What is tie sum of 4x —2x+y, 4x —y-x, 9y+7x'2-x, and 21x —2y+9x'? Ans. 19xS3+22x+7y. 11i What is the sum of 4 —3x, x-5, 2x- 6, -4x+ 13, and -5x+-1.Ans. 7 —9x. 12. WVhat is the sum of a —b3+3a2b-5ab2, 3a3 —4a2b+ 3b3- 3ab2, a -b3+ 3a2b, 2a3- 4b'- 5ab2, 6a2b +- 10ab2, and - G6a - 7a2b+4ab2+2b3? Ans. a9+ab-+ab-+b'. 13, What is the -sum of llbc+4ad-8ac+Scd, 8ac+7bc2ad+47mn, 2cd-3ab+5- ac+an, and 9an-2bc-2ad- 5cd 1 Ans. 16bc+5ac+12cd+4-mnr-3abr 10aXn. 1, *7What is the sum of x —4x'y+6xay2-4xy9+y', and x' -4xy +6x2y2 +4xy3 —4 Ans. 2x4+12x2y2+2yI. 15. What is the sum of x$- 54y +- Oxy2 - 10x2y+ 5y'-4+ y', and yl — 5y4x-+l 0oyx — 10Oy2x q+ 5yx' —x5? Ans. 2y+- 20y9x-+ 10yl. * NOTE.-The pupil must observe that the value of the expression 5x4y is the same as the value of 5yx4. The factors y and x4, are only differently arranged. Therefore the sum of these two expressions is 10z4y or 10y.4. 16 ADDITION. 16. What is the sum of 3(a+b)~+4a'x'-10, and 5Va+b 1 la2x2+6 Ans. 8(ab)2 +15ax2 — -4.* Or, 8V a+1 +15a2' — 4. 17. What is the sum of a2x2-2acx+c2 and a2 x2+2acx+c' Ans. 2a2x2-+ 2c. 18. What is the sum of x3 —ax2+ax-a' and x3+3ax2+3a'x +-a3? Ans. 2x3+2ax2+4a'x. (31.) In the preceding examples, the co-efficients have been numeral. WVhen the co-efficients are literal, find the sum of the co-efficients which have a common literal quantity, enclose this sum in a parenthesis, and prefix it to the common literal quantity. 19. What is the sum of ax+-16bx'+7cy —4ny2? In this example the common literal quantities are x2 and ye. The sum of the co-efficients of x2 is (a+ 16b), and the sum of the coefficients of y2 is (7c-4n). Htence, the answer to this example is (a+16b)x2+(7c-4n)y2. If, in this example, a=4, b=1, c=4, and n=6, the last expression becomes, by making the sub. stitution, (4 + 16)x'+(28-24)y2=20x'-4y2. In solving examples of this kind, it is convenient to arrange those terms which have a common literal quantity under each other. Operation. ax2+ 7cy' 16bx2- 4ny' (a+ 16b)x+ (7c- 4n)y' ans. 20. What is the sum of 5axy +10abz, and 15cz —2cxy? Ans. (5a-2c)xy + ( 1 Oab + 15c) z, 21. What is the sum of ax -+6bx, and mx'4+nx' 2 Ans. (a+b)x3 +((m+n)x. 22. What is the sum of ax2+by+-k, and dx2-hy+ck? Ans. (a+d)x2 +(b+h)y+(c+)k. - NoTE.-See Art. 13. ADDITION. 17 23. What is the sum of 5ca'+- 4ba9x+-mx'y2, and l0ca'z - 2bax' +- 6mx'y'.Ans. 15ca'x' +2ba'x'2+- 7mx'y2; or, it may be written, (1 5c - 2b)ax'+- 7mx'yS. 24. What is the sum of a'x'+b'x', and 9adx'-llb'x a Ans. (10a —10b')x'. If, in this example, we make a=3, b=2, and, consequently a'=9, b2=4, the expression a'x2+-b2x becomes 9xq+4x', and the expression 9a x2 —1 lb2x' becomes 81x'-44x'. Hence, (a'x'+ b2x) + (9ax-1 1 lbx) = (9x' + 4x') + (81x- 44x) =13x' + 37x' =50sx. The co-efficient 50, may be obtained by substituting the values of a and b in the co-efficient of x' in the answer. 10a'= 10X3'=90. 10b'=10X2'=40. Hence, lo0a-10b'=9040=50. This is only verifying the answer. CASE III. (32.) When the quantities to be added are all dissimilar, or when some are similar and some dissimilar. If we wish to add a to b we can only indicate the addition by connecting the two quantities by the sign of addition. Thus, a added to b is expressed, a+b. And in any case, where all the quantities to be added are dissimilar, we can only add them by writing them, one after the other, with their respective signs. Thus, ax+cy and 4nz are dissimilar quantities, and their sum is ax+cy+4nz. If we wish to add 4adx+-12cy, and 6cy+10z+3a'x', we can, obviously, add the similar quantities, and write the dissimilar quantity 10z after their sum. Operation. 4aax+ - 12ey 3a2'x'+ 6cy+- 0z 7a's +18&y+lQ10 sum. 18 SUBTRACTION. For convenience in adding we have arranged the similar terms under each other. Hence, to add quantities all or part of which are dissimilar, we have the following RULE. Add the similar terms according to the directions given in Case I. and Case II., and then write the dissimilar terms one after the other, with their respective signs. EXAMPLES. 1. What is the sum of ax' —6ab+y, and 4ax+10-lab-z? Ans. 5a2x2+4ab-y +z. 2. What is the sum of a'+2ab b', and a'+2ax+-x' Ans. 2a'+ 2ab + bl + 2ax +x'. 3. What is the sum of 4a" —4ab+4b', and a'9+2ab+b'+y? Ans. 5a-2-2ab + 5b' +y. 4. What is the sum of x'+9x+9, and -3x+10z? Ans. a +-6x 9 +-1 Oz. 5. What is the sum of lOm-+ 16ab+-10xy, and 15aclb + 7z -12ab+4-4m? Ans. 14m2+4ab+ lOxy+ 15acnb+7 z. SUBTRACTION, (33.) SUBTRACTION, in Algebra, is the finding of the difference of two algebraic quantities. If we subtract b from a the remainder is obviously a —b. Now, if we subtract (b-c) from a, it is plain that the remainder ought to be greater than the first remainder by c, since the minuend is the same, and the subtrahend is less by c. Therefore the remainder is, a-by+e. SUBTRACTION. 19 As another example, let it be required to subtract 20-5 from 50. If we subtract 20, the remainder will obviously be too small by 5. Therefore we must add 5 to the remainder, 50-20, in order to obtain the true remainder. Therefore the true remainder is, 50-20+5=35. Now, in each of these examples, we have simply changed the signs of the subtrahend, and added it, with the signs changed, to the minuend. Hence, for the subtraction of algebraic quantities, we have the following RULE. Change the signs of all the terms of the quantity to be subtracted, and then proceed as in addition. EXAMPLES. 1, From 8x' —3ax+5 take 5x2+2ax+5. Operation. The minuend. 8x-3ax+5 The subtrahend, with signs changed, -5x- 2ax- 5 3x- 5ax ans. 2. From 7xy —lOy+4x take 3xy+3y+3x. Ans. 4xy-13y+x. 3. From a+b+c take — a-b-c. Ans. 2a+2b+2c. 4, From a'x+6bx —9ax' —Tay take 11ay —4ax+3bx —2ax4. Ans. 5a'x+3bx — 7ax4- 18ay. 5, From 17cx+12px-7-ny" take 14cx —7px+3ny'-b. Ans. 3cx+19px — 10ny'+b.,6. From 13xy-14xy' + 17ay take 4xy + 7xy-10ay. Ans. 6xy —18xy' +27ay. 7. From 9ax' —16+ 10ay' take 4a'x2-.8+- ay+-y. Ans. ba'a'- 8+3ay* —y. ,20 SUBTRACTION. 8, From 24xy'-14my 1 8xSya —14+27xz' take 17xy' — Omy-4x2y2+ 20xz — 8. Ans. 7xy2 -4my+22x'y2 + 7xz2- 6. 9, From 17pmx2- 18n'3- 19m4-24 take'pmzx2-4nS + 10m4 -17. Ans. 10Omx —14n'+9m —7. 10. From a2+2ab+-b2 take a9-2ab+-b. Ans. 4ab. 11. From a9+3adb+-3ab-+b' take a'-3a'b+3ab2_-b. Ans. 6a2b+_2bs. 12. From x -+2x yt-y] take x —2x~y +y Ans. 4x y2. 13, From 6Vx+y+4adx2 take 3(x+y)2-4ax'2. Ans. 3 V+y+ 8a'x2. 14. From ax'+ay' take bx9z2+cy'. Ans. (a-b)x'+ (a-c)y2. 15. From axz —2cy+-nk take bx —4my+-pk. Ans. (a-b)x2+(4m- 2c)y+ (n-p)k. 16. From a'x9 —4axy+-4a'y2 take c'x'-8cxy+ 16x'y'. Ans. (a'-c2)x2+ (8c-4a)xy+(4a2- 16x')y2. 17. From 2x-y+(y —2x) —(x — 2y) take y-2x —(2y-x) +-x+ 2y. Ans. y —x. 18. From aex+bxy+cy' take dxs-hxy+hy'. Ans. (a-d)x2+(b+h)xy+(c-h)y2. (3 4.) If we wish to indicate the subtraction of one polynomial from another, without actually performing the operation, we must enclose the polynomial to be subtracted within the parenthesis. Thus, a3'+3a'b6+3abS +b'S-(a3-3a'b+-3ab' —b') signifies that the quantity within the parenthesis is to be subtracted from a's+-3a'b+ 3ab2+b'. If the subtraction be actually performed by the rule, the remainder will be 6a2b+2b'. * NOTE.-In this example, the difference of the co-efficients of xV2 is, (a - b), and the difference of the co-efficients of V2 is (a - c). Hence, ths~awx is, (4-b)*Ha cry. f see Art. 25. MULTIPLICATION. 21 On this principle polynomials may be made to undergo several useful transformations. Thus, a — 2ab+b6 may be written a2+ b2+(-2ab); a3-3a'b+3ab-2-b3 may be written a3-b3-(3a2b - 3ab) =-a- b + 3ab2 _ (3a2b)= a'- ('-3a 3ab'+3a2b). MULTIPLICATION. (3 5.) THE object of multiplication is to repeat one quantity as many times as there are units in the other. If it is required to multiply a by b, the product may be written ab. (Art. 21.) If a=5, b=4, then the expression ab, which represents the product, becomes 5X4=20. It is obvious that the product of any number of factors will be the same when they are arranged in one order as when they are arranged in any other order. Thus, ba is the same as ab, and 4 X 5 is the same as 5 X 4. In the multiplication of literal quantities, it is customary to arrange the letters in alphabetical order. (36.) In the multiplication of algebraical quantities, it is necessary to determine whether the positive or negative sign ought to be prefixed to the product of two monomial factors, or factors consisting of a single term. In order to do this, the following propositions must be established. PROPOSITION I. A positive quantity multiplied by a positive quantity gives a positive product. Let it be required to multiply +a by +b. Now, a is to be represented as many times as there are units in the multiplier, and as the sum of any number of positive quantities is positive, it follows that the product, ab, is positive. Hence the proposition is true. 22 MULTIPLICATION. PROPOSITION II. If a negative quantity be multiplied by a positive quantity, the product will be negative; and, if a positive quantity be multiplied by a negative quantity, the product will be negative. For, in the first place, let it be required to multiply -a by +b. Now, we must repeat — a as many times as there are units in b, and as the sum of any number of negative quantities is negative, it follows that the product of -a and +b is -ab. If b=5, then — a must be repeated five times, and the product will be, in this case, -a-a-a-a-a= —5a. In the second place, let it be required to multiply -a b.y — b. In this case, — b is to be repeated as many times as there are units in a, and as the sum of any number of negative quantities is negative, it follows that the product of +a and -b is -ab. Hence the proposition is true. PROPOSITION IIL If a negative quantity be multiplied by a negative quantity, the product will be,ositive. For, if +a bc multiplied by -+b, the product is +rab. (Prop. I.) And, if +a be multiplied by -b, the product is -ab. Hence we see that if a quantity be multiplied by a negative quantity, the sign of the product will be contrary to the sign of the product which was obtained by multiplying the same quantity by the same multiplier with a positive sign. Now, -a multiplied by +b gives — ab for the product; whence, -a multiplied by -b gives +ab for the product. Hence the proposition is true. This proposition may be established in this manner: let it be required to multiply a-b by c-d. In the first place, a-if multiply a-b by c, and we obtain for the )rOcduclt ac-bc. But as it is required to take a-b s manyll acas there are units in c-d, it follows that a-b has been taken d times too often. Hence, if we multiply Ia —b a-b by d, and subtract the product from the first product, the true product will be obtained. The prod MULTIPLICATION. 23 act of d and a —b is ad —bd; subtract this from ac —be, by changing its signs, and we obtain for the product of a-b by c-d, ac-bc-ad+bd. Hence, -b multiplied by -d gives +bd for the product. If, in this example, we make a=10, b=6, c=8, d=5, the product, ac-bc-ad+bd becomes 10X 8-6X 8-10X 5+ 6 X 5= 80 -48 —50+{30=12. This is obviously the true product, for a-b=10-6=4; and c-d=8-5=3, and 4X3=12. (37.) From the preceding propositions, the following rule for the signs is derived. RULE. The product of two quantities affected with the same sign, is positive, and the product of two quantities affected with different signs is negative. Multiplication may be divided into three cases. CASE I. (38.) When the multiplicand and the multiplier are both monomials. Let it be required to multiply 5a'x by 4xy. The product may be expressed thus, 5a'zX4xy. Now, the value of this product will not be changed by writing the factors in a different order, and therefore 5a'xX 4xy may be written 4 X 5 X a2xX xy= 20 X asx X xy=20Xa'xxy, since literal quantities may be multiplied by writing the factors which compose them, one after another. (Art. 21.) In the last expression the factor x is found twice, and x x =x2. (Art. 11.) Powers of the same quantity are multiplied by adding their exponents; for a=a X a X a, and a2=a X a. (Art. 11.) Therefore aXa'= (aXaxa)x (aXa) = aaaaa=ac. a"X=a-=a+. From what has been said, we derive the following rule for the multiplication of monomials. 24 I~MULTIPLICATION. RULE. To the product of the co-eficients affix that of the letters. EXAMPLES. 1, Multiply 5a'c by 4ab. Ans. 20a'bx. 2. Multiply 17a'x by 4a'x9z. Ans. 68a'x'z. 3. Multiply 19ay' by 7by'. Ans. 133aby. 4, Multiply 7ax by 41a'x'y'. Ans. 287a'x3y2. 5. Multiply 16ab by 17ax. Ans. 272a2bx. 6. Multiply 14ay' by 14ygz%. Ans. 196ay'z%. 7. Multiply 17'ah' by 1lah'. Ans. 18'7a'h. 8. Multiply — 6a~b by 14a'b'. Ans. - 84a4b'. 9, Multiply - 14c' by 1Oc'x. Ans. - 140c4x. 10. Multiply 21ay by 21by2. Ans. 441abys. 11, Multiply 15ay' by -1'7ax2. Ans. -255a'x'yg. 12. Multiply 23c'x by 40n'y. Ans. 920c'n'xy. CASE II. (39.) When the multiplicand is a polynomial, and the multiplier a monomial. In this case we have the following obvious RULE. Multiply each term in the multiplicand by the multiplier, and connect the partial products by their respective signs. EXAMPLES. Operation. 1. Multiply 4a —3b + c by 4a. 4a'- 3b-+c 4a An. 16a'- 12ab+4ac. MULTIPLICATION. 25 2. Multiply 7ax-ay+ 2c by 1la. Ans. 77ax —lla'y+22ac. 3. Multiply 1'7a'-2ac + 4 by 25c. Ans. 425a'c-50ac' + 100c. 4, Multiply a' + 2ab + b2 by 4b'. Ans. 4a'b& + 8ab' + 4b4. 5, Multiply 2ax + 5ax —4ac by 4c. Ans. 28acx- 1 6ac'.' 6. Multiply ay — 5ac + 3ac by 14a. Ans. 14a3y —28a'c. 7, Multiply a'- 2ab + b2 by -4a. Ans. -4a' + 8a'b-4ab'. 8. Multiply a2" —2a"bn + b'n by 4a"'. Ans. 4a"- 8a"b"- + 4a2'"b". 9, Multiply a' + 3a'b + 3ab' + b9 by 8aa. Ans. 8a' + 24a'b + 24a'ba + 8a'ba. 10. Multiply -4x' + 7xy-4ac by -4a'. Ans. 16ax' — 28a'xy + 16aSc. CASE III. (40.) When both multiplicand and multiplier are polynomials. Let it be required to multiply a + b by c + b. Now, we must repeat a + b as many times as there are units in c + b. Hence, a + b must be multiplied by c and then by b, and the two partial products added. Operation. a+b c+b ac + be - ab + b' ac +- be + ab + ba product. * NoTE.-In this example the similar terms 2az and 5ax may be added before performing the multiplication. 2 26 MULTIPLICATION. If the product of two polynomials contains similar terms, these must be arranged under each other in performing the multiplication, and their sum taken, as in the following example: Multiply 3a +b by 7a-2b. Operation. 3a + b 7a -2b 21a2+ 7ab -6ab-2b2 21a"'+ ab-2b' ans. In this example, we first multiply 3a + b by 7a, the first, or the left hand term, in the multiplier, and obtain for the product, 21a'+ 7ab; then we multiply 3a+b by the second term of the multiplier, - 2b, and obtain for the product - 6ab —2b'. If now, these two partial products be added, by placing the similar terms, 7ab, -Gab, under each other, the entire product, 21a' +ab-2b', is obtained. From the foregoing examples and illustrations, we derive the following RULE..Multiply each term in the multiplicand by each term in the multiplier, in succession, and add the partial products for the entire product. For the sake of system, it is better to multiply first by the left lianld term of the multiplier, and then by each one in order. 1. Multiply a' - 2ab + bV by a + b. Operation. a2 + 2ab - b6 a +b a3 + 2a2b + ab' =the prolduct by a. a'b+ 2ab +bs=the zproduct by b. as+- 3a'b + 3ab2 + b- the entire product. MULTIPLICATION. 27 2, Multiply x' +y' by x'-y'. Ans. x'-y'. 3, Multiply 2 + 2xy + y' by x —y. Ans. x'+xy-X —-y y 4. Multiply 5a'- 2a3b + 4a'b2 by a'- 4a2b + 2bs. Ans. 5a7 —22a6b + 12ab2 — 6a'b3-4a'Sb + 8a2b'. 5. Multiply x' + 2x' + 3x2 + 2x + 1 by x' —2x + 1. Ans. x'- 2x' — 1. 6. Multiply a' + 2ab + b2 by a-_ 2ab + b'. Ans. a4- 2ab' + b' = (a2 - b2)2. 7. Multiply x2+xy+y' by x2 —y 4-y'. Ans. X4 + x'y2+ y4 = (x y')'-x _ty. 8. Multiply x +x'y +xy' +y by x —y Ans.' -y'. 9. Multiply x' +y' +z-xy-xz —yz by x +y +z. Ans. x' + y' + z' - 3xyz. 10. Multiply aS+ 2ab + b by a —b'. Ans. a' + 2a'b —2ab _-b. 11. Multiply ab+cd by ab-cd. Ans. a'b'-c'd2. 12. Multiply a'+ 3a'b+3ab' +-b' by a'-3a'b+3ab' —b'. Ans. a' - 3a4b' + 3cL'b' -ba. 13. Multiply 6 —2 +4 by 9 —5..Ans. 54-27+ 36-30+- 15-20. 14. Multiply 1 —x+x' —x' by 1 +. Ans. 1 —x'. 15, Multiply a"+ bm by a' + b. Ans. a,' + a'b"- + ambn + b n 16, Multiply a+ b+c by a+ b+c. Ans. a +b ~c+- 2- ab + 2ac + 2bc. 17. Multiply a+b+c by a+b-c. Ans. a +2ab +- b'-c2. 18, Multiply a+b+c by a-b-c. Ans. a-b'-2bc —c'. 28 MULTIPLICATION. 19. Multiply 4xy +y' by 4y' + 1. Ans. 16xy' + 4y4 + 4xy + ye. 20. Multiply 3a2b2 +- 2a3b + 4ab' + 4a4 + b' by ab + a'.* Ans. 4a6+a Gab + 5a4ba + 7a3b3- + 5a2b4 + ab'. 21. Multiply 4x'-3x + x by 1 + x. Ans. 4x'4+x' —2xz' +x. 22. Multiply x —5 by x —7. Ans. xza-12x+35. 23. Multiply together (x-a), (x-b), (x-c). Ans. x'-(a +b+c)x 2 + (ab + ac + bc)x-abc. 24. Multiply aw + b3 by aw +b. 2 2. Ans. a + 2abx + b3. (41.) We will now establish some useful theorems. THEOREM I. The square of the sum of two quantities is equal to the square of the first, plus twice their product, plus the square of the second. For, let a-one of the quantities, and b=the other. Then, their sum is a + b, and the square of their sum is, (a + b) X (a + b) =a2 + 2ab-+b2, Hence the theorem is true. THEOREM II. The square of the diferences of two quantities is equal to the square of the first, minus twice their product, plus the square of the second. * NoTE.-In the preceding examples both multiplicand and multiplier have been arranged with reference to the powers of some letter; but in this example they are not so arranged. Before performing the multiplication, arrange them with reference to the powers of the letter a, that is, place the term which contains the highest power of a for the first, or left hand term, and the term which contains the next highest power of a for the second term, &c. DIVISION. 29 For, let a=one of the quantities, and b=the other. Then, a —b=their difference, and the square of their difference is, (a-b) X (a —b)=a2 —2ab+b'. Hence, the theorem is true. THEOREM III. The product of the sum and difference of two quantities is equal to the difference of their squares. For, let aCone of the quantities, and b=the other. Then, a+b-=their sum, and a-b=their difference. The product of their sum and difference is, (a +b)X (a-b)=a'-b'. Hence, the theorem is true. DIVISION. (42.) DIVISION has for its object the finding of one of two factors, when their product and one of the factors are given. The product of the two factors is the dividend, the given factor is the divisor, and the required factor is the quotient. Since the dividend is equal to the product of the divisor by the quotient, it follows from what has been said in regard to the signs in Art. 37, that when the signs of the divisor and dividend are alike, the quotient must be positive, and when they are unlike the quotient must be negative. Thus, +6 — -2, or -6- + 2 is equal to -3, because the signs are unlike. But, +6- - +2, or -6- — 2, is equal to +3, because the signs are alike. We have, then, to determine the sign of the quotient, this simple RULE. When the signs of the divisor and dividend are alike, the quotient will be positive, and when they are unlike, the quotient till be Regativw. 30 DIVISION. (43.) We have seen by Art. 38, that powers of the same quantity may be multiplied by adding their exponents. Now, division is the reverse of multiplication; hence, powers of the same quantity may be divided by subtracting their exponents. Thus, a' a- a'=a and a' —a'=a'= l, since a quantity divided by itself is equal to unity. Hence, we see that a quantity which has a cipher for its exponent is equal to unity. If we divide a' by a7, by subtracting the exponent of the divisor from that of the dividend, we obtain for a quotient a-'. But a' 1 1 a — a- -; whence, a —,, the reciprocal of a'. Similarly, a' as, a a-D-, the reciprocal of a'. In general, any quantity with a negative exponent, is equal to the reciprocal of that quantity with an equal positive exponent. Division may be divided into three cases. CASE I,. When the divisor and dividend are monomials. Let it be required to divide 24a'b'c by 4a2b'. Now, the quotient multiplied by the divisor must equal the dividend. Therefore, by the principles of multiplication, the quotient must be 6ab-'e. This quotient may be obtained by dividing the coefficient of the dividend by the co-efficient of the divisor, and the literal factors in the quotient are the same as those in the dividend with their exponents diminished by the exponents of the corresponding letters in the divisor. In this example, the factor c, which is found in the dividend, is not found in the divisor, and therefore the exponent of c in the divisor is nothing, since by placing the factor co in the divisor we do not change its value. (Art. 43.) Hence, there is nothing to subtract from the exponent of c in the dividend, which is understood, and c must be found in the quotient, ma a faetor. DIVISION. 31 If, in the example above, the letter c had been found in the divisor instead of the dividend, it is plain, from what has been said, that the exponent of c in the quotient must have been -1. Hence, we have the following rule for dividing one monomial by another. RULE. Divide the co-efficient of the dividend by the co-efficient of the divisor, and to the quotient affix the letters of the dividend with the exponent of each diminished by the exponent of the same letter in the divisor.'It must be observed, that in some examples, the same letters are not found in both dividend and divisor, but the dividend and divisor may always be made to contain the same letters, and then the rule may be applied. Thus, let it be required to divide 35a'3b by' 7abc'. By Art. 43, the dividend may be written 35a'b6c~ without changing its value. If the rule be now applied, we obtain for the quotient, 5a2bc 2. Whenever the exponent of any quantity in the quotient is a cipher, that quantity may be omitted, since any quantity which has a cipher for its exponent is equal to unity. (Art. 43.) ]EXAMPLES. 1. Divide 49a'b' by 7abc'. Ans. 7abc-2. 2. Divide — 72aSb'c by 9ab6. Ans. — 8aab2c. 3. Divide 64x9y2 by 16x2y'. Ans. 4xy-2. 4, Divide 625aab' by 25aS'b. Ans. 25a-'b2. 5, Divide 484a'm2 by 22am*. Ans. 22a2 6. Divide 73c'h' by 1 lchn2.* Ans. j-,ch2n-2. * NoTE.-In this example, 11 is not contained in 73, and the division is expressed by writing 11 under 73, in the form of a fraction. The same may be done in any similar case, and the fraction may be reduced to its lowest terms, as in arithmetic. 82 DIVISION. 7. Divide -48a'b' by — 24a~b'. Ans. 2ab". 8, Divide 96a'b' by -72a'b7. Ans. — 3acb2. 9. Divide 252e'k' by 364eky'..Ans. -&e8kry-2. 10. Divide 144m'n" by 240m'n', Ans. mmnr-. 1 11. Divide 192mn12' by 1536m'n'6 Ans. jm'.-. 12. Divide 246a"b'c' by 372a'bc. Ans..-a'bc'. CASE II. (45.) When the dividend is a polynomial and the divisor a monomial. Let it be required to divide 3a2b- 6ab by 3a, or, in other words, to find a quantity which multiplied by 3a, the divisor, will give 3a'b-6ab. Now, since each term in the required multiplicand, must be repeated 3a times, it follows that if each term of the quantity, 3a'b-6ab, be divided by 3a, the required quantity will be found. Hence, to divide a polynomial by a monomial, we have the following RULE. Divide each term in the dividend by the divisor, and the Jmn of these partial quotients will be the quotient required. EXAMPLES. 1. Divide 3a'9b-18a'b' + 6a2b by 3ab. Ans. a'b —6asb + 2, 2. Divide 24xy'-8- z4y' —24xy' by 8x. Ans. 3xy'-x'y'-3y'.. 3. Divide 2la'xs —7a'zx + 14ax by 7ax. Ans. 3ax2 —ax+2. 4, Divide 42a' —lla + 28a by 7a, Am. 8wia +4 DIVISION. 83 5. Divide 9k" —24kc + 48k6 by 3k'. Ans. 3kV~-.8k-'c + 16k. 6. Divide 72a'c'-48a'c —32aYc by 16aec. Ans. fa'c-1-3a'c- 2adc'. 7. Divide 36n-48mt by 4mi. Ans. 9m 6 -- 12m'. 8. Divide mJ —mian by m+. Ans. m —ma-n. 9, Divide a4 — a6ba+a6 by a. Ans. a -b2+a 10. Divide 11a —33a3 by 11la. Ans. a-3a 9. 11. Divide 72mS — 6Omn by 24m6. Ans. 3m3-n3 12. Divide 2a' + 4a'b + 2a'b' by 2a'. Ans. a' + 2a + b'. CASE III. (46.) When both the dividend and divisor are polynomials. In order to give greater simplicity to the division of one polynomial by another, it will be found necessary to arrange both dividend and divisor according to the powers of the same letter. It will be understood, therefore, that in all cases they must be so arranged before performing the division. Let it be required to divide 2a'+ 12a + 22a' + 12a' by a'+3a. In this example, both dividend and divisor are already arranged with reference to the powers of the letter a, and the dividend being the product of the divisor and quotient, it follows that the quotient will be arranged with reference to the powers of the letter a Since the dividend is the sum of all the partial products that can be formed by multiplying each term of the divisor by each term of the quotient, it follows that the first term of the quotient is found by dividing the first term of the dividend by the first term of the divisor. Therefore the first term in the quotient is, 2a'+ —a' =2a'. If the divisor be now multiplied by the first term in the at 34 DIVISION. quotient, 2a', and the product, 2a+ 6a4, be subtracted from the dividend, as represented in the following operation, the remainder, 6a4+22a+ 12a 2, must be the sum of all the partial products that can be found by multiplying each term of the divisor by each of the remaining terms of the quotient. Therefore, the first remainder may be regarded as being a new dividend, and the second term of the quotient may be found in the same manner as the first term of the quotient was obtained, by dividing the first term of the new dividend by the first term of the divisor. Hence, 6a' -a'=6a', the second term of the quotient. Multiply the divisor by 6a', and subtract the product, 6a4+18a', from the first remainder, asnd the second remainder is, 4a' + 12a2. From what has already been stated, 4a.-a2= 4a, the third term in the quotient. Multiply the divisor by 4a, and subtract the product, 4a3 +12a', from the second remainder, and there is nothing remaining. Hence, (2a+ 12a4 + 22a' + 12a')- (a + 3a)=2a + 6a- +4a. Operation. 2ab + 12a 4+ 22a' + 12a'I a' + 3a =divisor. 2a5 + 6a4 2a + 6Ga' + 4a=quotient. 1st remainder= 6a4 + 22as + 12a' 6a4 + 18aa 2d remainder= 4a+- 12a' 4a2 + 12at 3d remainder= 0 From what has been said, we derive the following rule for the division of one polynomial by another. RULE. 1. Arrange the divisor and dividend with reference to the powers of the same letter. 2. Divide the first term of the dividend by the first term of the divisor, and the result will be the first term of the quotient, by which multiply the divisor, and subtract the product from the dividend. DIVISION. 35 3. iDivide the first term of the remainder by the first term of the divisor, and the result will be the second term in the quotient, by which multiply the divisor, and subtract the product from the first remainder, and proceed as before to find the remaining terms of the quotient. EXAMPLES. 1, Divide a' + 3aab + 3ab2 + b3 by a + b. Operation. a" +3a3ab + 3a b a + b =divisor. aS + a2b a + 2ab + b2=quotient. 2a2b + 3ab2* 2a2b + 2ab' ab' + b' ab~+b 2, Divide 6a' —96 by 3a-6. Operation. 6a' —96 3a-6 =divisor. 6a4- 12a' 2a' + 4a2 + 8a + 1 6 = quotient. 12a'- 96 12a'-24a' 24a'2-96 24a'- 48a 48a-96 48a-96 0 3. Divide a —2ab+ b by a-b. Ans. a-b. 4. Divide 12x4-192 by 3x-6.t Ans. 4x8+8x2+16x+32, * NOTE.-It will be observed that it is unnecessary to bring down the whole remainder, 2a2b+3ab2+b3. t NOTE.-In this example, and in many others, the dividend and divisor may be divided by a common factor before performing the division. 86 DIVISION. 5. Divide 6x —6y' by 2x2-2y2. Ans. 3x4 + 3x y2 + 3y4. 6. Divide x8 + 5xy + 5xy* + ya by zx' + 4xy + y..Ans. x+ y. 7. Divide a4'-b' by a —b. Ans. a' + ab+ab'+b3.8. Divide x8 —9' +27x —27 by x,3. AIns. xP-6 x+9. 9. Divide 48x-'76axz-64a'x+1lO5a by 2x-3a. Ans. 24x2 —2ax —35a2. 10. Divide x+x' x + tx + by Jx + 1. Ans. x2 +i. 11. Divide x4+y4 by x+y. Ans, x'- x'y + xy'-+ y' + 2 12. Divide x5-y' by x-y. Ans. x4 + xsy + xsys + xy +y4. 13. Divide a2' + 2amb" + b2" by am + b". Ans. a"+b". 14. Divide X4 + 2y2 +y' by x -xy +y, Ans. x + y +yq. 15, Divide s' +y' by x+y. Ans. x —xy + xy'-Xy + y*. 16. Divide 6c'+ 9c'-15c by 3c —3e. JAns. 2c + 2c+5. 17. Divide z —x'4+x —x2+ 2 —1 by x'+x-1. Ans. 4 —x +x —x+1. 18. Divide c6-ya by c3+2c'y+2cy'+y3. Ans. C3- 2c2y+2cy2-y3. Both dividend and divisor may also be multiplied by the same quantity before dividing. DIVISION. $7 19. Divide y' —I by y-1. Ans. y'+y+1. 20. Divide x'+y' by x+y. Ans. x —xy+yl. 21. Divide 1 +x by 1 —x.* 2x4 Ans. 1 +2x+2x2+2x'+ 1 — 22. Divide 1 by 1-2x+x'. Ans. 1+2x+3x+ —4x+a fraction. 23. Divide 2a"8 —6a"&b+ 6ab"'-_2bs' by a —bn. Ans. 2a'" —4ab"-+2b2". 24, Divide a' —b by a-b. Ans. a+b. PROPOSITION. (47.) The difference of the same powers of two quantities is divisible by their difference. Let a" and bn represent two quantities whose common exponent is n. Then an-b" is divisible by a-b. By commencing the division we have, an- b" i a-b divisor. a"-a'lb an- the first term of the quotient. Remainder= an'b - bn= b(a"-1 - b"-). Now since the dividend is equal to the quotient multiplied by the divisor plus the remainder, it follows that, an_ —b"= (a-b)a"-l+b(a"-'-b"-1) (A). In equation (A), the term (a-b)a"' is obviously divisible by a-b, the quotient being an"-. Therefore, whenever the term, b(an —-bn-l), or an —lbn-l, is divisible by a-b, the right hand * NOTE.-In this example the division does not terminate. In such cases, we can write the divisor under the last remainder, in the form of a fraction, and add this fraction to that part of the quotient which is obtained. 88 D)IVISIorN. member of equation (A) is divisible by a-b, and consequently its equal, an_-b, is also divisible by a-b. Hence, if the difference of the same powers of two quantities is divisible by the difference of those quantities, then the difference of the powers of the quantities will be divisible by their difference, when their exponents are increased by unity. Now, a' —b —a-b=a+b. Therefore, (a3-b') is divisible by (a-b), and in general a —b" is divisible by a-b. USEFUL FORMULAS. 1. (a'-b')=(a+b)(a-b). 2. (a'-b )=('+b')(a-b') = (a + b) (a b)(a — b). 3. (a +b')=(a'-ab+b)(a+ b). 4. (a'-b')= (a+ab+b2)(a-b). 5. (a~-br)= (a + b3) (a3'-b")=(a' +b9)(a's ab + b')(a- b). 6. (a'- bY)= (a' Pb3)(a-_ b) = ((b9-b) (a'- ab +b2) (a + bo). 7. (ad-b')= (a'+3 b')(a'- b) = (a'-b')(a +a'b'+ b). 8. (a' —b6)= (a'+ b)(a' —b)= (a-b)(a9'+ab + b)(9'-ab+ b')(a+b). 9. (a'-b')- (a+b)= a-b. 10. (a~+bS)- (a+b)=a'-ab+b2. 11. (a'-bs) (a-b)=a'+ab +'. 12. (a —b') (a'-b')= (a+b2). 13. (a"+b')+ (a+b)=a'-ab+a'b'-ab'+b'. 14. (ao-b ) *- (a-b)=a4+a'b+am +ab3+b'. 15. (a'-b') — (a'-b2)= c4+ ab'+ b'. 16. a~=1. GREATEST COMMON MEASURE. 39 THE GREATEST COMMON MEASURE. (48.) The greatest common measure of two or more quantities is the greatestfactor which is common to all of them. Thus, the greatest common measure of 49a'x' and 35a'x' is 7a2'x. Iln examples like this, the greatest common measure may be founl t by inspection, but in many others it will be necessary to apply the following rule for finding the common measure of two polynomials. RULE. Arrange the two polynomials with reference to the powers of some letter, and divide that which contains the highest power of this letter, by the other; then divide the last divisor by the last remainder, and continue th;.s process till there is no remainder; the last divisor will be the O; eatest common divisor. (49.) When the first term of any dividend is not divisible by the first term of the divisor, it may be made so by multiplying each term of the dividend by any quantity which will render it divisible. The reason of this is obvious. (50.) If either the dividend or divisor contain any simple factor which is not common to both, this factor may be rejected, before commencing the division, as this factor can form no part of the greatest common measure. (51.) If both polynomials contain a simple common measure, this common measure may be rejected from each before applying the rule; but as it forms a part of the greatest common divisor. it must be restored in the last divisor. DEMIONSTRATION OF THE RULE. Let P and p be any two polynomials, of which P is the greater, or the one which contains the highest power of that letter, with reference to which the two polynomials are arranged. If we now operate on these two polynomials, according to the rule, we shall have the following operation: 40 GREATEST COMMON MEASURE. p) P (q,* =first quotient. pq, 1st remainder= r,) p (q2 =second quotient. r,q2 2d remainder —= r,) r, (q, =third quotient. r2q3 3d remainder= 0 Now, since the dividend is equal to the divisor multiplied by the quotient plus the remainder, we have the following equations: P=pql +rl (1):p=rq2+r, (2) rl=rq3+0 (3) Since r, measures the right hand member of equation (3) it must also measure its left hand member, rl. Therefore, r, must measure each member of equation (2). Again, since ra measures rl and p, it must measure each member of.equation (1), and therefore measures P. Whence, rs is a common measure of P and p. We say, further, that it is their greatest common measure. For, equations (1) and (2) may be written as follows: P —pq, =r, (1)' p-rlq2=r2 (2)' Now, every common measure of P and p must measure each member of equation (1)', and consequently each member of equation (2)'. But it has been found that r, is a common measure of P and p, and therefore r2 is a common measure of each member of equation (2)'. The greatest common measure of r, is itself; therefore the greatest common measure of P and p is r2. Hence the rule is established. EXAMPLES. 1. What is the greatest common measure of a'-ab-2b' and -- 3ab+2b? * NOTE.-The numerals which are placed at the bottom of the letters q and r, are used to avoid the necessity of introducing too many letters into any calculation. They have other advantages which the pupil will soon learn to perceive. q, is read q sub one, q,, q sub two, &c. q, qa, &c., represent different quantities. GREATEST COMMON MEASURE. 41 Operation. a' — ab-2b2 a' —3ab+2b, divisor. a -- 3ab + 2b 1 1st remainder 2b)2ab-4b2 a -2b)a —3ab+ 2b'(a-b a2-2ab -ab+ 2b' — ab+2b' 0 In this example we first divide a'-ab-2b' by a'-3ab+2b' and find that it is contained once, with 2ab-4b2 for. a remainder By examining this remainder it may be seen that 2b is a factor of it, but that it is not a factor of the divisor, a —3ab+2b'. Therefore, by Art. 50, divide 2ab-4b2 by 2b, and we have a-2b, which is contained in the divisor a2- 3ab + 2b2, a —b times, with no remainder. Therefore, a-2b is the greatest common divisor required. 2. What is the greatest common measure of 12a'-8a-4, and 20a'c-1Oa'c-15ac+5c? Divide the first by 4, and the second by 5c, and we have 3a'-2a- 1 and 4a'-2a'- 3a- 1 Multiply by 3 (Art. 49) 3 12a3-6a — 9a+ 3 3a'-2a —1 divisor. 12a —8a2 — 4a 4a + 2 2a2 — 5a + 3 Multiply by 3 3 6a2 — 15a 9 6a2- 4a- 2 Divide by -11 (Art. 50) -11)-1 1a+-ll 1 a- 1)3a'-2a-1(3a+1 3a- 3a a-1 a-i Therefore at — is the greatest common measure. 42 GREATEST' COMMON MEASURE. 3, What is the greatest common measure of x'-xy' and x2- 2xy+-y2? Ans. x+ y. 4. What is the greatest common measure of 4 - 8x' + 2 1x — 20x+4 and 2x'-12x2+21x —10. Ans. x —2. 5. What is the greatest common measure of x-'7x'+ 14x- 8 and 3x4- 14x'+ 8x'?.Ans. x-4. 6. What is the greatest common measure of x'+4axq+ 5a'x + 2a3 and 3x -+7ax +2a' Ans. x+ 2a. 7. What is the greatest common measure of 5x'+13xS+2x' -8x and 2x' +x+12? Ans. x+2. 8. What is the greatest common measure of x'- 7x' + 16x12 and 3x2 —14x + 16? Ans. x —2. 9. What is the greatest common measure of 4x4- 1 OxS — x' +- 8x + 4 and 2xs +- 5x +- 4x+ 1 Ans. 2x + 1. 10. What is the greatest common measure of 12x2- 40x'+ 60x +32 and 3x' —6x2+43 1 Ans. x'-2x+2- 1. 11, What is the greatest common measure of x' - y' and x" - y13? Ans. x —y. 12. What is the greatest common measure of a' —ab + 3ab2 — 3bs and a'-5actb+4b 2? Ans. a-b. 13. What is the greatest common measure of x +yg and x2+ 2xy+y' 2 Ans. x+y. 14. What is the greatest common measure of a' —b4 and as + ab —ab —b' 2 Ans. a' —b2. (52.) In order to obtain the greatest common measure of three quantities, find the greatest common measure of two of them, then that of this common measure, and the third quantity will be the greatest common measure required. For, let a, b, c, be the three quantities, and m the greatest common measure of a and b, and n the greatest common measure of m and c. Since m is the greatest common measure of a and b, and n the greatest common GRIEATEST COerMMON M1EASURE. 43 measure of m and c, n must measure a, b, and e. Therefore n, which is the greatest common measure of m and c, is also the greatest common measure of a, b, and c. This reasoning may be applied to any number of quantities. 15. What is the greatest common measure of a4-b', a'-a'b +3a-3b, and a'+a-ab-b? In this example, first find the common measure of a' —b' and aS —ab + 3a-3b. a' -b' 4 (a2 - b')(a'-b'f)-=(a' + b')(a + b)(a-b) (mn) aa —aab + 3a-3b =a(a- 6) + 3(a - b)= (a' + 3)(a - b)*(n) By examining the two polynomials (m) and (n), it is obvious that a-b is their greatest common measure. 16. What is the common measure of x'+{ 5x'+6x', +3x' + 3x~- 2, and 3x' + 8x+ 5x - 2? Ans. x +2. In this example, find the common measure of x'+5x'+6x' and x'-+ 3x' + x + 2. Reject the factor x' from the first, before commencing the division. 17. What is the greatest common measure of a' + 3a'b + 3ab' +- b, 4a'b'+ 12ab'- 8b' and a'+ b' t Ans. a +b. 18. What is the greatest common measure of 5a'+ 10a'b+ 5a'b2, a'b + 2a'b'+ 2ab'6- b, and a' + b'? Ans. a + b. * NOTE.-M-uch of the pupil's success in algebra will depend on his ability to resolve algebraical expressions into their factors. To this end, he must carefully examine an algebraical quantity, in order to discover all the different forms in which it may be written, and thus cultivate a quickness of perception. It would be impossible to give rules for every operation in Algebra, and if we could, the labor would be unnecessary. Rules contain only general principles, and are chiefly designed to aid the beginner. 44 LEAST COMMON MULTIPLE. THE LEAST COMMON MULTIPLE. (53.) A COMMON MULTIPLE of any quantity is a quantity which contains it without a remainder. A common multiple of two or more quantities, is a quantity which contains each of them without a remainder. The least common multiple of two or more quantities is, therefore, the least quantity which contains each of them without a remainder. Let it be required to find the least common multiple of agx, cx, and abx. If these quantities be multiplied together, the product will obviously be their common multiple, but not their least common multiple; for the product is a'bcx', which contains as, b, c, and x'. But the highest powers of a and x, which are found in the three quantities, are a' and x, and therefore the least common multiple is a'bcx. Hence, in finding the least common multiple of two or more quantities, we must observe that when different powers of the same factor occur in the given quantities, the least common multiple will contain only the highest power of this factor. The least common multiple of two or more quantities, then, may be found by either of the following rules. RULE I. Resolve each quantity into its prime factors, and then, if different powers of the same quantity occur as factors, reject all but the highest power of this quantity. Of the remainingfactors, select all the different ones, and their product will be the least common multiple. RULE II. Arrange the quantities in a horizontal line, and then divide two or more of them by any prime factor which will measure them, and place the quotients and undivided terms in a horizontal line below. Proceed with this line as with the first, and so on, till all the quantities in the last line are prime with respect to each other. The continued product of the divisors and the quantities in7 the last line will be the least common multiple. LEAST COMMON MULTIPLE. 45 (54.) If no prime factor can be found by inspection which will measure two or more of the given quantities, the common divisors may be found by Art. 48 and Art. 52. 1. What is the least common multiple of 48a'b', 16a4b4, a'+ 2ab+-b, and af-'+b SOLUTION BY RULE I. 48a2bS= 24 X 3 X a' X b', the prime factors of 48a'b' 16ab4 =24 X a4X b, the prime factors of 16a4b4 a' + 2ab+ b2= (a + b)2 the prime factors of a' + 2ab+-b' a' + b'-= (a +b)(a' —ab+b2) the prime factor of a'S+b'.. X 24 X a4 X 64 X (a + b)'(a'-ab + b')= 4 8a4b4 X (a + b)2 (a'-ab +b') is the least common multiple required. 2. What is the least common multiple of 6a'x'(a-x), 8xS(a'-x2), and 12(a-x)'? SOLUTION BY RULE II. 2x' 6a'x'(a-x), 8x2(a'-x2), 12(a-x)' a-x 3a'(a-x), 4(a —x2), 12(a-x)' 3 3a', 4(a + x), 12(a-x) 4 a2', 4(a + x), 4(a-x) a, a+x a —x 4 X 3 X aX 2XX (a-x) X (a-x) X (a+ x)= 24a'22X (a'-x') X (a-x), the least common multiple. 3. What is the least common multiple of 15ca'b', 12ab', and 6a3b? Ans. 60a'b'. 4. WVhat is the least common multiple of (a+-b)', a'-b', (a-b)', and a+3a'b +3ab'+b'? Ans. (a+b)(a'-b'). Or, (a + b)(a - b)'(a + b)'. Or, (a + b)'(a- b)'. * NOTE.-By using 2x2 and 4 we do not rigidly follow the rule. because they are not prime factors, but it is easy to see that the result is the same as would have been obtained by dividing separately by the prime factors of 2z2 and 4. 46 LEAST COMMON MULTIPLE. 5. What is the least common multiple of x2+ 2xy+y2 and x3 —xy? Ans. x(x+y)(x2-y'). 6. What is the least common multiple of a —b2 and a+b31 Ans. (a'-b')(a'-ab + b2). 7. What is the least common multiple of y'-8y+7, and y - 7y-83 Ans. y'-57y2+ 56. 8. What is the least common multiple of a + b, a —b, a2 +ab +b2, and a'-ab+6?b Ans. as —b6. 9. What is the least common multiple of 16ab, 18a2b', 36ab4x, and 72abx I? Ans. CHAPTER III. FRACTIONS. (55.) ALGEBRAICAL FRACTIONS do nT;[ ditlfr iLi any respect from arithmetical fractions, and thercf.e t!., rules for operating upon them are the same as th:ee in c.)'~lnon arithmetic, and they are deduced in the same il.anner. (56.) Since the valc: of a fraction is the quotient which is obtained by dividc;g!l the numerator by the denominator, we infer the following principles, upon which the principal rules are founded: 1. That multiplying or dividing both numerator and denominator of a fraction by the same quantity does not change its value. 2. That multiplying the numerator, or dividing the denominator, of a fraction by any quantity, multiplies the fraction by that quantity. 3. That dividing the numerator, or multiplying the denominator, of any fraction by a quantity, divides the fraction by that quantity. CASE I. (57.) To reduce afraction to its lowest terms. By the first principle in the preceding article, we can divide both numerator and denominator of a fraction by a quantity without changing its value. If both terms of the fraction be divided by their greatest common divisor, it is obvious that the fraction will then be reduced to its lowest terms; whence the following 48 FRACTIONS. RULE. Divide both numerator and denominator by their greatest common measure. EXAMPLES. 2x- 16x — 6 1., Reduce to its lowest terms. 3x2-24x —9 By resolving both the numerator and denominator into factors, we readily discoyer the greatest common measure. Thus, 2xS — 16x —6 2 X (x3-Sx —3) x-24-16x-32X (x= x -8 3) — 3, the fraction required. 3x'-24x-9 3X (x3-8x-3)' d _a x4 2, Reduce 2 to its lowest terms. as a2 x- ax _+x By applying the rule for finding the greatest common measure, we find that it is a2 —x2. Divide both numerator and denominator by a-xA, and the fraction becomes -? 3. Reduce the fraction 22 to its lowest terms. x-y 4. Reduce the fraction to its lowest terms. a-b Ans. a'ab+ 5. Reduce the fraction to its lowest terms. a-b a4+a'b2 -b' Ans. a2+b2 4x2 + 3 7. Reduce the fraction to its lowest terms. A".s 482' $- ~~16 1 FRACTIONS. 49 8. Reduce the fraction 2 x4+ 2-o its lowest 25x4 +5x —x — 10x2- 1 4x2+ 1 terms. *n +x 9, Reduce the fraction,ab3 to its lowest terms. a2+ ab +b2 Ans. ns a2(a+b) a4- b4 10. Reduce the fraction b b to its lowest terms. a - a2b - ab+ toitslowesat b2 a +b. a-b aS-~_ 3aac q- 3ac2 — cs 11. Reduce the fraction a-3a'c+3a c to its lowest terms. a2 -2ac+c' Ans. -- a+c zs - ac2 12. Reduce the fraction to its lowest terms. a' +_2ac+c' a - ac a+c CASE II. (58.) To reduce a mixed quantity to the form of a fraction. RULE. IMultiply the integral part by the denominator of the fraction, and to the product add the numerator, if the sign of the fraction is positive, but if the sign is negative subtract the numerator f'on the product. Udder the result thus obtained, write the denominator. The pupil will readily understand this rule, if he comprehends the corresponding rule in arithmetic. EXAMPLES. a+ab a X (a +b) =a2+ab. The negative sign placed before the fraction shows that its numerator should be subtracted from this produ ct. Hence, +b — (a)-ab)_a' —a product. Hence, b a- is the fraction required. a +b a3 3 60 FRACTIONS. 2. Reduce 1 -a+ to the form of a fraction. 2a' Ans. 2 a2+x, 3, Reduce 1 -+ 2bc to the form of a fraction. 2be Ans. (b- c)-a 2bce 4. Reduce 1a 2abb2 to the form of a fraction. a2+b2 2ab Ans. a-+ b2 +c2 - a2 5. Reduce I - to the form of a fraction. 2bc a. - (b - c)2 Ans. 2bc a2 ~-ab q b2 6, Reduce a- to the form of a fraction. a+-b 2a2-2ab+ b.Ans. a+b 2x- 5 7. Reduce 5x- ~ to the form of a fraction. 13x+5 Ans. - -3 as-2a +bq8. Reduce a+ 2ab- - -- to the form of a fraction. aab a2~~ —2abns- bi 4ab x2~4x+4 10. Reduce x —2+- to the form of a fraction. x-2.2x: q 4x z —2 FRACTIONS. 51 aS — 3a2b 6-3ab2+-bT11. Reduce a -t 2ab+b-3a ab to the form of a fraction. A 6a26~Gab= Ans. = 6ab. a+b 12. Reduce x +1 X+ to the form of a fiaction. Ans. =x+2x+l (x+l)' x CASE III. (59.) To reduce afraction to an entire or mixed quantity. RULE. Divide the numerator by the denominator, the quotient will be the entire quantity, and under the remainder, if any, write the denominatorfor the fractional part. EXAMPLES. 1. Reduce ++14- to a mixed quantity. x+2 Operat-ion. x2+7x+ 14 x+2= —divisor. x2+ 2x x+ 5=quotient. 5x+14 5x+10 4 -remainder. 4 Therefore, the quantity required is, x+5- +2 x~2 a'+ 2abq-b2 q- 2c 2. Reduce + + to a mixed quantity. Ans. a+b+-.. Redu 6x' 2x10 to a mixed quantity. 3. Reduce to a mixed quantity. x+2 Ans. x2-4x~4+ 2 x+2' 52 FRACTIONS. aS + bs 4. Reduce +b to an entire quantity. a+b.Ans. a2 —ab+b2. 5. Reduce a+b to an entire quantity. a+b Ans. a4-aC6+a2b2- ab3t- b. 6. Reduce, _b2 to an entire quantity. An s. a' + a2b2 + b4. 7. Reduce -Y to a mixed quantity. x+y 2ys Ans. x2 —rcxy+y3+" x+y' a'2 - 4ab + 4b2 +2 c 8, Reduce + to a mixed quantity. a-2b.Ans. a +2b C - -a+2b+ as 9, Reduce a to a mixed quantity. a-Jrx Ans. a2-ax + x2a o ax x2 + 12x + 18 10. Reduce ---- to a mixed quantity. x+3 9 Ans. x + 9 — X CASE IV. (60.) To add fractions. RULE I. Reduce the fractions to a common denominrator, by multipl~ying the nulmerator anzd denominator of each fractcon by the product of the denominactors of the remcaiing fractions; then, add their numerators, and under their sum write the common denominator, which is the product of all the denlomilnators.# * NOTE:.-This rule may be explained by the 1st princille, Art. 56, for it is obvious that the terms of each fiaction are multiplied by the same quantity. Rule II. is, in effect, the application of the same principle. FRACTIONS. 53 RULE II. Fi;nd the least common rmultiple of the denominators, and it wll be the least common denominator of the fractions. Divide the least common denominator by the denominator of each fraction, and multiply the quotients by the corre.sponding numerators of the given fractions, and the products will be the numerators, uwnder each of which write the common denominator; then proceed as in Rule I. EXAMPLES. a in, C 1. What is the sum of —, and -? b' n x Here, a aX (n X) _anx m = nz X (b X x) bmx. c b b X (,n X x) bx'; n n X (b xx) b=nx' x cX (b X n) bce a m c alx bmx bcn _ Hence, +-k-+ —- + = X (b X n)- bnx' b n x b;u. bx bnx+= anx + bmx + bcn bn, the sum required. Let the student observe, that bnx by following the rule, the numerator and denominator of each of the given firactions are multiplied by the same quantity, and therefore the value of each fraction remains the same, as it should. a, -b a-b 2. What is the sunm of -- and -- a-b' a3 - b' Operation. a-b)a-b, as-b' 1, a2'+ab + b2. (a - b) X (a' + ab + b2) = a' - b = the least common denominator. ab- X a + b=(a2 +ab+ b2) X (a +b)=a' + 2a2b + 2ab- + b3, the numerator of first fiaction. a'-_b Xa —b=a —b, the numerator of second fraction. a + b a-b a3 +- 2ab + 2ab' + bS'+a- b W'hence, a-b ~3-b a3-b', the fraction required. 54 FRACTIONS. aq-b a —b 3. What is the sum of b and - a —naq- b' Ans. 2(a2+b2) a2 b2 a2- 2ab+ b a2-2ab+ b 4. What is the sum of a-b and a+ a —b a -b 2a3+6ab2' a2-b2 Ans. 21-x2 1-x' 5. What is the sum of IX and d 1- x2 1+x2x Ans. 2 (I +X4) 1 2x' 6. What is the sum of and - Ans. 1+x- 1_x 2 - _x+ a d C 7. What is the sum of and- a bx' dx-' d~ x adx - bdx+ bc Ans. bdx3 8, What is the sum of ~xand ~x a I —x' 1 —xl' 1 —xa An3 - 3+5x+- 6x2 +- 5x' 3x4 1 +x- 2 -x X4 a1q-b2. a'b2 9. What is the sum of -- and - d aq-b' Ans. 2(a'~b3)-2(a'-ab +b) a2 —ba a-b 1 c 1 10, What is the sum of and xy'yx8+/' x 3y Ans. 2(x-xyy) ~ NOTE.-If Rule I. be used the fraction obtained by it must be reduced to its lowest terms, in order to agree with the given answer. FRACTIONS. 55 3 4a n 5 11, What is the sum of 4a 4 and ax 48aa 64ax' 12a2x 12x2+12ax+80x 192a2x2 12. What is the sum of 6 7a and 2 6x2' ~ax' 21ax2 7a2+ 6cx 2n Ans 42ax' 13, What is the sum of 3_ 2' and 65'12' 60 36x+20x+7x x 60 20 14. What is the sum of _ and (? a2-62 (a - b)2 2a Ans. a3 -a2b -_ab2-+ b3 CASE V. (61.) To subtract one fraction from another. RULE. Reduce the fractions to a common denominator, if they have not one, and then subtract the numerator of the subtrahend from that of the minuend, an2d place the difference over the common denominator. EXAMPLES. 1. From take 3. c C3 2x 2x X c2 2C2X 2x 3x 2c2x 3x 2c2x-3x Here-= 2 =3. Hence, — - - C C c C' C C c3 C3 C~ C~ t the difference required. 2. From a2 take a- Ans. 2 2 56 FRACTIONS. Ox 2x 4x 3,. From 9 take.Ans. 7a 2a ]la 4, From - take -y. Ans. 6a 3y 4y' 6y 5. From 5x take -3Ans. 7x15 24 16 n 48 3x~ 2 C7ax-] Oa 6, From take — 2 a a2 Ans. 12a —4ax 4(3-x) a2 a 7. From 2x take 3x —2 Ans. 7 12 1836 8. From take a-b 4ab a-b a+. Ans. a.2 2 a+-b a+b 9. From a'b take a + a2_ b2 a'-22ab+b, Ans. — 2b(a-b) a'-s' a'-b'3a2b-(as 2+ b3 10. From ----- take b Ans. a +b) a —b a -+b ab —' 11. From 1 take a-b- a2 — 2ab + b2' a-b-1 a2-_ 2ab+ b' 12. From 1 take X2 - +2 x+-4 x2+10x+24'.Ans. (-,+2)a x2 + 10x+-24' I — x' 4x2 13. From 1-x2 take - Ans 4 From 2 1 +-x' 1.x4' 14, From 2x+5 take As3x2 As. X13 24 72 72 FRACTIONS. 57 1 1 15. From ab2 take a a 22ab 2+2a+ 2b Ans. 2b a'+ta2b-ab2-b3 ~ CASE VI. (62.) To multiply one fraction by another. RULE. Multiply all the numerators together for a new numerator, and all the denominators togetherfor a new denominator. a C Let it be required to find the product of the fractions - and d b _V In the first place, multiply the fraction a by c, the numerator of the second fraction. 13y the second principle, Art. 56, this productis aXc acb ct is b b Now, since it was required to multiply b by ac the dth part of c, it follows that the product, b must be d times ac too large. Therefore, in order to obtain the true product, b aC must be divided by d. By the third principle, Art. 56, d= ac ac baXd — b-d' Hence the rule is correct. (63.) By the first principle, Art. 56, we may, in the multiplication of fractions, cancel all the factors which are common to both numerators and denominators. EXAMPLES. a2 — x 3a2 + 3ac 1. Multiply by a+c 4ax+4a" By resolving a2-x2, 3a2+3ac, and 4ax-f-4a' into factors, and then multiplying, we have (a+x)(a-x) X 3a(a+c) = x a+c X 4a(a+x) — I 58 FRACTIONS. 2. Multiply 324 by 8n' Ans. 1 32nS 64a 16an' )7x 13 13 3. Multiply 7- by 18 Ans. 13 2 1 18x2' 54-~ 42m2 80aS 5a2nz2 4. Multiply 64 by 2lb Ans. 2b5an 5. Multiply am+m by-. Ans.. ax2 mc acx 6. Multiply y2+ by x n+Y acx4 3y2 An. 3acx"' 17a cc 7. Multiply 20b by 34a2. Ans.'Sb 8. Multiply a+- by c+ —.* ax ay ns. acyx + acx + acy + c ayx 9. Multiply a+ by 2a2-2ab~2b Ans.. 10. Multiply a2 + b2 by ab Ans. a+ xa-9x+20 x2-13x+42 11. Multiply _6- by 1-52 5 -'1z+28 Ans. 212. Multiply 2+3+ 2 by x.+5x+4 x2+2x+1 x2+7x+ 12' Ans. x2 x+3 * NOTE.-B efore multiplying, reduce the mixed quantities to improper fractions. FRACTIONS. 59 a-b62 a b2 -ab 13. Multiply b2- by a -+r a 2_b a'-2ab ab2' 14. Multiply a-~b- by a -ab+b2 Ans. (a+b)'. 15. Multiply +4 by 5ax+ x2 —+4x+4 27m 25ay 7x 7mx+ 14m' CASE VII. (64.) To divide one fraction by another. RULE. Invert the divisor, and then proceed as in multiplication. a C Let it be required to find the quotient of a divided by d. In the first place, divide - by c, the numerator of the divisor. By a a a the third principle, Art. 56, 6 -c= b Since the divisor, c, is d times as large as the true divisor, the quotient, a, must be one dth part of the true quotient. Hence, the true quotient is, a ad vC-X d —-.. This is the same result as may be obtained by the oc bc rule, and therefore the rule is correct. EXAMPLES. a+b aq —b 1. Divide ab by a-b. Ans. 1. a-b a~-2ab-+b' 2. Divide a+-b by a + a-b Ansb aa —b Ans....... 60 FRACTIONS. 3. Divide by 3x2 Ans. 5x+4 48 16 9x+6' 5x+10 x-+2 5x2 4. Divide by. Ans. +b a -pb x2 a -b 14x+-28 42x~-84 5. Divide 0+ by 72 Ans. 1. 96 72 4' a3.+ax a2 4ac+4cx+a+x 6,. Divide by Ans. 2c' 4c2-C' 2c.a2+2ab+ b a-2_b2 4a+4b 7. Divide ~ by -. As. a - b 4 a2- 2ab+bZ a4 -b4 a2+ab 8. Divide a2-2ab+b by a-b Ans. a2+b2 b' a a a 2 +b aqb a2~ b2 9. Divide a-_b2 by a-b. As. a2+2Lb+b-b X2- 9X+20 x- 5x 10. Divide 2 _ 6x by X2_ 6X x2__ 13x+42. x —11lx+28 Ans.,n — I n — 1 11. Divide 1-+ 1 by 1- 1. Ans. n. 12, Divide a - b y a b Ans. 1. a+b a-b bya- ab a2- b a3_b3 13. Divide 3b3 by b.Ahs. a —b6. $I a — b a- q+2ab-b2 14. Divide ab by a-b Ans a22ab+b' a-b ya+b' na'-2ab+b,c +cs 4c (a + C)2 15. Divide 24- y 12a+ 12c Ans. ( + c) a + x a-x a + x a-x a' + x 16 Divide a+ by a+ Ans.2ax a-x a+x ac-x a+x' 2ax~Z CHAPTER IV. SIMPLE EQUATIONS. (65a.) AcN equatit l is an all:Raic ex pression, consisting of two equal quantities, with the sign of equality )pliaced bcetw\een them. Thus, 5x+3:=14 is all equation. (66.) The two quantities which are separated by the sign of equality are called the members of the equation. That on the left of the sign of equality is called the first member, and that on the right, the second member. (67.) When one member of an equation is a repetition of, or the result of some operation indicated in, the other member, it is called an identical equation. Thus, (x2 —y2) — (-y)=x+y and 5x + 4= 5x + 4 are identical equations. (68.) Equations are divided into degrees. Those which contain only the first power of the unknown quantiy are simple equations, or equations of the first degree. Those which contain the square of the unknown quantity, and no higher power, are quadratic equations, or equations of the second degree; and, in general, those equations which contain the nth power of the unknown quantity, and no higher power, are equations of the nth degree. Thus, 5x + 2x +a 15 is an equation of the first degree, 3x2 + 4x= 20 is an equation of the second dceree, and x" + 34=a is an equation of the nth degree. (69.) Any value of the unknown quantity, which, when it is substituted for the unknown quantity, will satisfy tihe equation, that is, render the two members identical, is a root of that equation. Thus, 3 is the root of the equation, 2x+4=10, since, if in the place of 2x, 2 X 3, or 6, be substituted, the equation be 62 SIMPLE EQUATIONS. comes 6 +4=10, anjidentical equation. The solution of an equation consists in finding all of its roots. A simple equation has only one root. SOLUTION OF SIMPLE EQUATIONS. (70.) The rules for solving equations are founded upon the following axioms: 1. If the same quantity be added to equals, the sums will be equal. 2. If the same quantity be subtracted from equals, the renainders will be equal. 3. If equal quantities be multiplied by the same quantity, the products will be equal. 4. If equal quantities be divided by the same quantities, the quotients will be equal. TRANSPOSITION. (71.) By the term transposition is meant, the changing of terms in one member of an equation into the other member, without destroying the equality. Let it be required to find the value of x in the equation, x + 8 _12. Subtract 8 from each member, and we have x+8-8-= 12-8, or x= 4, the value required. This result may be obtained by transposing 8 into the second member, and changing its sign. If the equation had been x-8=12, it would have been necessary to add 8 to each member in order to have obtained the value of x, or the result might have been obtained by transposing 8 into the other member and changing its sign. Hence, we may observe that A quantity may be transposed from one member to the other by changing its sign. TO CLEAR AN EQUATION OF FRACTIONS. x x x (7 2.) Let it be required to clear the equation, - ++ + = =1, of fractions. This object may obviously be accomplished by SIMPLE EQUATIONS. 63 multiplying each member by any multiple of the denominators. Therefore, multiply each member by 24, the least common multiple of the denominators, and it becomes, 6x+ 8x+x=360. As any equation may be cleared of fractions in a similar manner, we have the following RULE. Multiply each member of the equation by any multiple of the denominators, or by the least common multiple of all the denominators. (7 3.) To find the value of the unknown quantity. Let it be required to find the.value of x in the equation, X X -+ —+7 (1) 6x + 4x= 3x+84 (2)=(1)X12, that is, the second equation is equal to the first, multiplied by 12, the least common multiple of the denominators. Ox+4x-3x=84 (3) The third equation is obtained by'transposing 3x, which is found in the second member of equation (2), into the first member, and changing its sign.'7x=84 (4) The fourth equation is obtained from the third, by uniting the terms in the first member of the third equation. Thus, 6x and 4x are 10x, and 3x taken from 10x leaves 7x. x=12 (5) = (4) 7 That is, the fifth equation is obtained by dividing each member of the fourth by 7. From what has been said, we derive the following general RULE. 1. Clear the equation of fractions, if it have any, and perform all the operations indicated. 64 SIMPLE EQUATIONS. 2. Transpose all the unknown quantities into the first meme ber, and all the known, quantities into the second member. 3. Unite the terms in each member, and divide by the coefficient of the unknown quantity. EXAMPLES. 3x —11 5x-5 97 —7x 1. Given 21+ = 8 + 2 to find the value of x. 16 8 2 Multiply each member by 16, the least common multiple of the denominators, and we have 336+3x-11=1Ox-10+ 776-56x (2) Bytransposingin (2),3x-Ox + 56x=- 10+ 776-336 + 11 (3) Uniting terms in (3), 49x= 441 (4) Whence, by dividing by 49, x=9 (5) 18x-19 11x+21 9x+15 2. Given 28 + 6 4 1 to find x. 28 6x+ 14 14 Multiply each member by 28, and the equation becomes 154x-294 18x —19+ - 3x 18x+ 30 3x{-7 By transposing 1-4x+29449 3x+ 7 22x+42 Dividing by 7 3x+7 3x+-7 Clearing of fractions, 22x + 42= 21x + 49 By transposing, x= 7 It may be observed that the solution of this example is abbreviated by partially clearing, at first, the equation of fractions. SIMPLE EQUATIONS. 65 X X X X X 3, Given + + + -+-=51, to find x. Mlultiply by 72, 24x+9x+8x+6x+4x=-5 1 X 72 Uniting terms, 51x= 3672 Dividing by 51, x= 72 Gx-7 7x-13 2x+4 4. Given to find x. 9 + 6x+3 3 Multiply each number by 9, and we have 6x+- 7 + — - 6x+- 12 63x — 1 17Subtract 6x+ t from each member, 3x — +7= Multiply by 6x+ 3, 63x-117 =30x15 By transposing, 63x-30x-132 Or, 33x=132 Whence, x=4.* 5. Given ( +x)(a-x)+2abx=2Aa2~b2_x2 to find x. Performing the multiplication indicated, we have a2- X2'+ 2Cabx= 2a2'- b'-x2 By transposing, 2abx=ca2 + b2 a2 + b2 Whence, x= - 2a 6. Given a+x= x+2ab to find x. a+x Multiply by a+x, a2+2ax +=x2I-+ 2ab. By transposing, 2ax=2ab-a2 2ab-a22 b-a WVhence, — 2 - 2 2a 2 * NOTE.-In the second and fourth examples, we have employed some artifices, in order to abbreviate the solutions. The student must learn to rely on his own ingenuity in solving examples and problzms, as no general rule can be given. 66 SIMPLE EQUATIONS. 7-x 6x-22 8x-t 15 7. Given +4= +- to find x. 2 8 68 Ans. x= 3. x+3 12x~26 8, Given 2x- -15= to find x. 3 5 Ans. x= —12. 2xGe 21 x~3 9. Given x_2+=..+. to find x. Ans. x=13. 10. G 3x-3 3x-3 15 27+4x to fid x. 10, Given 3to find. 4 3 3 9 Ans. x= 9. 10x-17 12x+2 5x- 4 11, Given td find x. 18 13x — 16 Ans. x=4. 12 Given 20x+36 5x+20 4x 86 12. Given 2+ = — +- to find x. 25 9X-116 5 25 Ans. x=4. 13, Given 9x+2 = 4x 2 + to find. 36 5x-4 4 Ans. x=8. Given z —1 23 —x 4~-x 14. Gien + =7- to find x. Ans. x=8. 3x —ll 5x-5 97 —7x 15. Given 21-+ -= + 2 to find x.!6 8 2 Ans. = —9. 16. Given23+ 5x —1 3x-2 llx-3 13x —15 8x-2 11 5 12 3 7 to find x..Ans. x= —9.? 3x-1f? J12+71s 9+5x 17. Given 4x+q$ - - -7s-33- 10 16 9 10 11x —17 l 1 to find x. Ans. =15. 8 SIMPLE EQUATIONS. 67 +x _9 10x2 —18 18. Given 5x+ x+ =9+ 2 3 to find x. 4xq-3 2xq-3 An s. x=3. a(b2+x2) Cax b 19. Given 6- =ac+ — to find x. Ans. C-. bx b c 5x-25 284-x 20. Given - + 6x=- to find x. 20 5 Ans. x=9. 21. Given -+-+-+-~ =40 to find x. Ans. x=63. 7 3 9 21 22. Given + - +- -=8S6 to find x. 3 5 7 11 Ans. x=1155. 23. Given -+ + +3-+ -+ =84 to find x. ( 7 4 12 42 3 Ans. x —= 84. a2 q- 2ax - x':3xIV 241 Given 22a+x2-abx + - to find x. 4 12 Ans. x 4b-2' 25, Given x'2V/3-ax+bx +cx to find x. Ans. xa - c'/3 26. Given ax=ab-bx to find x. Ans. x-: 3x+4 7x —3 tx-16 27, Given = ~ to find x. 25 2 4 Ans. x=2. 1 7-3x 4x+-2 7x+-14 28. Given.... -=5 —6x+.. to find x. As. x:= 4. 68 SIMPLE EQUATIONS. 0x-4 18-4x 29. Given - — 2= - +x to find x. 3 3 Ans. x=4. 7x 5 16 - 4x 3x~- 9 30. Given + - q- 2 — to find x. 3 5 2 Anzs. x —1. 5x+3 4z —10 31. Given -+ 7- -10 to fdl x. 4 10 25 Ans. x —. 17 X X X X 32. Given ----— +1 to find x. Ans. x-=12. SOLUTION OF QUESTIONS 1rWHICIh INVOLVE ONE UNKNOWN QUAXNTITY. (74.) The solution of a problem by algeblra consists in expressing the relations which exist between thie known and. unknown quantities by melans of cqu:1tionY, and thlcn obtaining the value of the utnknown quantities firom these equations. To obtain the equaitions is genra'ly thle mlost difficult part of the labor, but sometimes the solution of tllem p)resents the greater difficulty. In many problems, the enutncillion reatdily furnishes the necesmary equations, and then tlle conditions are said to be explicit. Imjplicit conditions are'those which are deduced fiom 1 xplicit col(litions., -?;a,mal rule can be given fr solving problems. We shall i, a Ie for the purpose of initiating the student, and then he must delend on his own ingenuity arid powers of analysis. He will find that much practice and reflection are necessary in order to become a good resolver of problems. PROBLEMS. 1. What two numbers are those whose difference is 7, and sum 33? SIMPLE EQUATIONS. 69 Let x=the less number; Then x+ 7=the greater number. x+x+7=33, by the conditions of the problem. By uniting terms and transposing, 2x= —26; Whence, x- 13 Ans. and x + 7=20 2. Out of a cask of wine which had leaked away A, 21 gallons were drawn, and then the cask was found to be half full; how much did it hold? Let x=number of gallons which it contained; Then — nuruber of gallons which leaked away; Whence, by the conditions of the question, 2 3 Multiplying by 6, we have 3x=- 2x - 126 By transposing, x=126 Ans. 3, A gentleman has stocks upon which he receives an annual dividend of 12 per cent. lie spends - of his yearly income for clothes, and 1 of it for the support of his family, and the rernainder, which is $360, he gives to the poor. What is the amount of his property? Let x=the amount of his property; 12 3x Then xX - his yearly income, 100 25 3x 1 3x 2X 8-20- =the cost of his clothing, 3x 1 -x and X - = - his family expenses. 25 2 50 Whence, by conditions of the question, 3x 3x 3x $&+-+3G0O — (1) 200 50 25 Or, 3x+12x+360X200=24x (2) By transposing, 9x=360x200 (3) ATWhence, *x=40 X 200= 8000 Ans. 70 SIMPLE EQUATIONS. 4. A merchant supports himself for three years for a dollars a year, and at the end of each year increases his stock, which was not thus expended by one third part of the remainder. At the end of the third year his original stock was doubled. What Awas his stock? Let x=hlis stock; 4 4x —4a'Thel, (x-a) X- = =his stock at the end of 1st year;* 4x —4 a) 4 161 — 6 1 = "4 " 2d year; a X — -- =" 8d year; ( 9 3 3 27 9 3 WVhence, by conditions of the question, 64x- -64a I a 4a - -- -=2x (1) 27 9 3 64x-64a-48a-36a=54x (2) By transposing and uniting terms in (2), \we have 10x=148c (3) 148a 74a Whence, x 1 =- Ans. 10 5 5. A hare pursued by a greyhound is 60 of her own leaps in advance of the dog; she makes 9 leaps during the time that the greyhound makes 6; but 3 leaps of the greyhound are equivalent to 7 of the hare. How many leaps must the greyhound take before he overtakes the hare? Let 6x=-the number of leaps that the dog takes, Then 9x= " " " hare takes after the dog starts. 7 of the hbarc'e leaps= 3 of the dog's leaps;. 1 " " = — of one of the dog's leaps; Whence 9x+-60 " -- X (9x+ 60) of the dog's leaps. Threfobre, by conditions of the question, 6x= 7(9x+6O) (1) Or, 42x=27x~ 180 (2) By transposing 1. 1I I180 (3).2= 12 (4) (6x= 72 Ans. (5) * NOTE.-A quantity may be increased by one third part of itself by multiplying it by 4. SIMPLE EQUATIONS. 71 In this example, it is necessary to express the leaps of the hare in terms of the leaps of the dog, in order that we may commit no absurdity by equating heterogeneous numbers, or numbers which are not related to the same unit of measure. 6. A can perform a piece of work in 8 days, and B can perform the same work in 24 days. In xhat time will they finish it if both work together? Let x=the time required. Since A can perform the work in 8 days, in one day he can perform - of it, and in x days he can perform x times {- of the x x work, or of it. In the same way we find that x is the part 8 24 of the work that B can do in x days. Now, the part that A can do, added to the part that B can do, must equal the whole work, or unity. Whence — t 1 (1) 24 Clearing of fracti-rns, x + x —24 (2) Or, 4x=24 (3) x=6 Ans. (4) 7. A laborer engaged to work for n days. For each day that he labored he received a cents, and for each day that he was idle he forfeited b cents. Now, at the end of the time he received c cents. It is required to find how many days he worked, and how many he was idle. Let x=the number of days which he worked; Then n —x= " " " was idle. ax= — what he earned. b X (n-x)= —what he forfeited. Whence, by conditions of the question, ax —b(.n-x)=-c (1) Or, ax-b+ +bx=c (2) Transposing, ax+bx=c-bn (3) x=- +-bl,the number of working days. c -. bn an-c.n —x=n- -- ab - a+b' the number of idle days. a-tb a+b' 72 SIMPLE EQUATIONS. If we make n=40 ] a=-50 cents We fnd that x=24 days, b= —0 cents and n-x=16 days. c-=400 cents J 8. A post is 4 in the mud, ~ in the water, and 11 feet out of water; what is the whole length? Ans. 20 feet. 9. After paying 4~ and - of my money, I had 66 guineas left in my purse; what was in it at first? Ans. 120 guineas. 10. Two persons, A and B, lay out equal sums of money in trade; A gains $126, and B loses $87, and A's money is now double that of B's; what did each lay out?.Atns. $300. 11. Divide the number 54 into three such parts, that ~ of the first, a of the second, and 4 of the third, may all equal each other. Ans. 12, 18, 24. 12. A person was desirous of giving 3 cents apiece to some beggars, but found that he had not money enough in his pocket by 8 cents, he therefore gave each one 2 cents, and lie then had 3 cents remaining; required the number of beggars. Ans. 11. 13. A person in play lost 1 of his money, and then won 3 shillings; after which he lost 1 of what he then had, and then won 2 shillings; lastly, he lost 1 of what he then had, and then he found that he had 12 shillings remaining; what had lie at first? Ans. 16 shillings. 14. Divide the number a into two parts which shall have to each other the ratio of m to n. A na na m +n' m +n' 15. 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, shall all be equal to each other. What are the parts Ans. 18, 22, 10, 40. SIMPLE EQUATIONS. 73 16. A man and his wife usually drank out a cask of beer in 12 days; but when the man was fiom home, it lasted the woman 30 days. In how many days would the man alone empty the cask? Ans. 20 days. 17. The hour and minute hand of a clock are exactly tcgether at 12 o'clock; when are they next together? Ans. I hour and 5-j1qm. 18. Divide the number a into three parts which shall be to each other as m, n, p. Ans. ma na pa nm +n +P-' mn +n+p' m+n +p 19. Two travellers set out at the same time fiom London and York, whose distance is 150 miles; one of them goes 8 miles a day, and the other 7; when will they meet? Ans. In 10 days. 20. The sum of $1320 is to be divided between A and B in such proportion, thalt as often as A receives an eagle B receives a dollar. What did each receive? Ans. A $1200, B $120. 21. Four places are situated in the order of the letters A, B, C, D. The distance fiom A to 1) is 34 miles, the distance from A to B is to the distance from C to D as 2 is to 3, and one fourth of the distance fiom A to B, added to one half of the distance from C to D, is three times the distance from B to C. What are the respective distances? Ans. A B —=12, B C=4, and C D=18 miles. 22. After paying 4 of my money, and then 4 of the remainder, I had $140 left. How much money had I at first? Ans. $180. 23. Two numbers are to each other as a to b, and if c be added to each, the first sum will be it times the second sum. What are the numbers? ac(n —1) bc(n —1) a-nb a-nb 4 74 SIMPLE EQUATIONS. 24. A cistern has two pipes; the first will fill it in 40 minutes, and the second in 60 minutes. In what time can the cistern be filled if both pipes are open? Ans. 24 minutes. 25. A cistern has three pipes; the first will fill it in a minutes, the second in b minutes, and the third in c minutes. In what tinne can the cistern be filled if the three pipes are open at the same time abc am b + ac + bc' 26. Twelve oxen can eat up 3 — acres of grass in 4 weeks, together with that which grows during the time that they are grazing; in the same manner, 21 oxen can eat 10 acres in 9 weeks. How many oxen can, in this way, eat up 24 acres in 18 weeks? Ans. 36 oxen. 27. A cask that held 146 gallons, was filled with a mixture of brandy, wine, and water; there were 15 gallons of wine more than there were of brandy, and the number of gallons of water was equal to the whole number of gallons of wine and brandy.'What was the number of gallons of each I Ans. 29 gallons brandy, 44 wine, 73 water. 28. The rent of an estate is 8 per cent. greater this year than it was last, and this year it is $1890. What was it last year? Ans. $1750. 29. A's age is double of B's, B's is triple of C's, and the sum of all their ages is 140 years. What is the age of each? Ans. A's=84, B's=42, and C's-14. 30, There is a fish whose tail weighs 9 lbs., his head weighs as much as his tail and half his body, and his body weighs as mluch as his head and his tail. What is the -whole weight of the fish? Ans. 72 lbs. 31. A field of 864 square rods is to be divided among three farmers, A, B, and C, so that A's part shall be to B's as 5 to 11, and C may receive as much as A and B together. How much does each receive? Ans. A 135, B 297, C 432 square rods. SIMPLE EQUATIONS. 75 32. On an approaching war, three towns, A, B, C, are to furnish 594 soldiers; the division is to be made in proportion to their population. Now, the population of A is to that of B as 3 is to 5, and the population of B is to that of C as 8 to 7. How many men must each toxwn furnish? Ans. A 144, B 240, C 210 men. 33, Divide the number a into three such parts that the first mav be to the second as m to n, and the s,2cond part to the third as p to q. Ans. mnpa npac nqa mnp+lp+nq' mp+n2+nq' m2p+n2+nq2i 31. A capital was put out for one year at 4. per cent. per annum; at the end of the year there was received $13167, in capital and interest. How much did the capital amount to? Ans. 12600. 35. A father leaves a number of children, and a certain sum, which they are to divide amongst them as follows: the first is to receive 8100, and the 10th part of the remainder, after this, the second is to receive $200, and the 10th part of what then remains, again, the third receives $300, and the 10th part of the remainder; and so on, each succeeding child receives $100 more than the- one immediately preceding, aild then the 10th part of the remainder. At last it is found that each child receives the same. What was the fortune left, and how many children were there? Ans. The fortune was $8100, and the number of children was 9. 36. Further, Xwhat must the fortune and the number of children be, when, in general, the first receives a dollars, together with the nth part of the remainder; and each succeeding child a dollars more, together with the nth part of the remainder, and it is found at last that each has received the same? Ans. The fortune=(n-l)'a; the number of children-n-1. 76 SIAMPLE EQUATIONS. 37. Find two such numbers that the one may be m tilnes as great as the other, and their sum=a. What are these two numbels a a ma Ans. a and m+l m+l 38. WVhat capital is that which being put at interest for 5 years at 4 per cent., amounts to $8208? Ans. $6840. 39. A person puts a capital of $5500 out at interest at 4 per cent., and 4- years after, another capital of $8000 at 5 per cent. If he leaves these two capitals constantly at interest, in how many years will he have drawn the same interest from both? Ans. In 10 years from the time he put out the first sum. 40. A merchant adds yearly to his capital one third, but takes from it, at the end of each year, $1000 for his expenses. At the end of the third year he finds that his original stock was doubled. What was his original capital? Ans. $11100. 41. A general, wishing to draw up his regiment in the form of a square, tried it in two ways; the first time he had 39 men over, the second time, having extended the side of the square by one man, he wanted 50 men to complete the square. What was the number of men in the regiment? Ans. 1975 men. 42. It is required to find a number, such, that if the two numbers, a and b, are added to it, the difference between the squares of these sums may be=d. What is the number? d-a2+ b2 Ans.2 (a-b) Does the solution of this problem include that of the precedin(r one? 43. Three merchants, A, B, C, enter into partnership. A advances $1200, B $800, C $600. A leaves his money in trade 8 months, B 10 months, and C 14 months. They gain $500. What is each one's share of the gain? Ans. A $184-,a-, B $153'1, C 161-'-7. 3 1 3, SIMPLE EQUATIONS. 77 44. Three merchants entered into trade; the first contributed $17000, the second $13000, and the third $10000. As they must have some person to conduct the business, the one who had furnished the least offered to undertake the management of it, on the condition that he shall receive 3 per cent. profit, besides what h:e is entitled to from his deposited capital. Now, it is found that they have gained $35262.50; how much is due to each? Ans. To the 1st $14875, to the 2d $11375, to the 3d $9012.50. 45. A person possesses a wagon with a mechanical contrivance by which the difference in the number of revolutions made by the fore and hind wheels may be determined. The fore wheel is a i.+,:t, and the hind wheel is b feet in circumference. What is the,istance gone over, when the fore wheel has made n revolutions Iore than the hind wheel? abn Ans. a feet. b-a 16. Two bombardiers threw different kinds of shells fiom a battery; the first had thrown 36 times before the second cominenced, and he throws 8 times while the second tihrows 7 times; but the second uses as much powder for three of his throws as the first does for four. HIow many throws must the second make in order to consume as much powder as the first? Ans. 189 throws. 47. Saltpetre and sulphur are mixed together in a mass of 80 lbs., and in such a prolportion that for every 7 parts of saltpetre there are three parts of sulphur. How much saltpetre mlust be added to the mass, so that the proportion of these substanie s may be such that for every 11 parts of saltpetre there may b,, 4 parts of sulphur? Arns. 10 lbs. 48. A courier who started fiom a certain place 10 days ago, is pursued by another firom the same place, and by the same way. The first goes 4 mniles every day, and the second 9. How many days will the second need to overtake the first? Ans. 8 days. 78 SIMPLE EQUATlIONS. 49. A courier left this place n days ago, and goes a miles each day. He is pursued by another going b miles daily. How many days will the second require to overtake the first? Ans. days. 50. In a full wine cask there are three faucets; by the first the wine can be drawn off in 2, by the second in 3, and by the third in 4 hours. What time will be required to empty the cask when all three are running at once? Ans. 55-i- minutes. 51. At 12 o'clock both hands of a clock are together. When, and how often will these hands be together during the next 12 hours? Ans. The hands will be together at 5-2 minutes past 1, 101- minutes past 2, 16-,-r minutes past 3, and so on, in each successive hour 5-A5 minutes later; and the hands will be together 11 times. 52. Two bodies move after one another in the circumference of a circle which measures a feet. At first they are distant fiom each other by an arc which measures b feet; the first moves c feet, and the second moves C feet in a second. When will these two bodies meet for the first time, second timle, and so on, supposing that they do not disturb each other's motion? b b-a 2a -b A ns. In -c' C —C C-' C — c' C-' 53. When will tllev meet if the first starts t seconds later than the second b-ct a+b-ct 2a+b-ct itns. C-c' C-c' C- &c. 51. When will they meet if tile fir- t!eginls to moNve t seconds sooner than the second? b+ct ac-b+ct 2a+b+ct C-A c' C-c' C-c' 55. When will they meet, if the first, instead of preceding the second, runs against it, and starts from the samle place t seconds sooners. b-ct a+b-ct 2a+b-ct CT-s ~' C+C' C+' SIIMPLE EQUATIONS. 79 56. When will they meet if the first starts from the same place as the second, and begins to move t seconds later? Ans. bct a-+b+ct 2a+b+ct As.(C+,c' C+c' C+c 57. A wine merchant has two kinds of wine; the one cost 9 shillings per gallon, the other 5. lIe wishes to mix l,,t1 sorts together, in such quantities, that he may have 50 galltons, and each gallon, without profit or loss, may b, suld f(r 8 shillings. Hlow much must he take of each sort nj reak.: up this mixture? Ans. 371 gallon!s,t' thc beost, 121 of the other. 58. Let the price (f the ),st wine in the preceding problem =a shillings, the pricn of utie poorest=b shillings, the number of gallons in the niix;.,tare=n, and the price of the mixture=c. How many gallons of each kind must he use? Ans. (a, —b gallons of the poorest, and (c-b)n of the other. a-b 59. A wine merchant has 40 bottles of wine, one of which he sells for 7 shillings; since he deems this price too high for his customers, he wishes to add as much water as will enable him to sell a bottle of the mixed wine for 6 shillings. How many bottles of water must he add to it? Ans. 6}2 bottles. 60. A father, who has three children, bequeaths his property by will in the following manner: To the eldest son he leaves a sum a, together with the nth part of what remains; to the second he leaves a sum 2a, together with the nth part of what remains after the portion of the eldest and 2a have been subtracted from the estate; to the third he leaves a sum 3a, together with the nth part of what remains after the portions of the two other sons and 3a have been subtracted. The property is found to be entirely disposed of by this arrangement. What was the amount of the property? An. n2-4n + )a (n- 1)9 80 SIMPLE EQUATIONS. 61. A person purchases goods for $4500, which he is to pay for at the expiration of a Pyar. He afterwards agrees to pay the seller $1500 cash, and the remaining $3000 he is to pay' in four equal payments of $750, at equal times. What period must be fixed upon for these equal payments to be made, so that neither party may be a loser? Ans. 7~ months. 62. During a panic there was a run on two bankers, A and B. B stoppied payment at the end of three days, in consequence of which the alarm increased, and the daily demand for cash on A being trebled, A failed at the end of two mrore clays. But if A and B had joined their capitals, they might both have stood the run, as it was at first, for 7 days, at the end of which time B would have been indebted to A $4000. What was the daily demand for cash on A's bank at first? Ans. $2000. 63. A waterman finds by experience that he can, with the advantage of a common tide, row down a river from A to B, which is 18 miles, in an hour and a half, and that to return from B to A, against an equal tide, though he rows back along the shore, where the stream is only three fifths as strong as in the middle, takes him just two hours and a quarter. From these data it is required to find at what rate per hour the tide runs in the middle, where it is strongest. Ans. At the rate of 2~ miles lel hour. 61. As A and B were going to school, A first shot an arrow in the direction in which they were going, which B took up and shot forward; and so on alternately till the arrow had passed exactly from one milestone to another; when it appeared that A had shot the arrow 8 times and B 7 times. Soine time afterwards, A and B were on the opposite banks of a rive r, the breadth of which they wished to ascertain; A first shot the arrow acrioss the river, and it flew 13 yards beyond the bank on wh1ch B stood; B then took it up, and from the plltce where it had fallen, shot it back across the river; it now fell 9-7 yards be.yond the bank on which A stood. Required the breadth of the river?.Ans. 100 yards. SIMPLE EQUATIONS. 81 EQUATIONS WHICH INVOLVE TWO OR MORE UNKNOWD QUANTITIES. (75.) In the solution of many problems in simple equations, it is necessary to employ two or more unknown quantities. In such cases, the number of equations must always equal the number of unknown quantities employed. For, in the equation, x=y + 4, any value may be assigned to y, which value augmented by 4 will furnish the corresponding value of x. Hence, in a simple equation, consisting of two unknown quantities, the unkno'kn quantities may have an infinite number of values. Such an equation is called an indeterminate equation, and x, the value of which depends on that of y, is said to be a function of y. But if we have two independent equations involving two unknown quantities, each of the unknown quantities admits of only one value, and these values may be found. The method of operating upon two or more simple equations, so as to obtain one equation, containing only one unknown quantity, is called elimination. There are three methods of elimination; namely, by addition and subtraction, by substitution, and by comnparison. FIRST METHOD. By Addition and Subtraction. (7 6.) Let it be required to find such values of x and y as will satisfy each of the following equations: 2x+3y=13 (1) 5x+4y=22 (2) Multiply equation (1) by 4, and equation (2) by 3, and they become 8x+12y=52 (3) and 15x+ 12y=66 (4) Eq. (4)-Eq. (3)* gives 7x=-14 (5) Whence x= 2 (6) 8x=-16 (7) * NOTE.-' Eq. (4) - eq. (3)" reads equation (4) minus equation (3). The abbreviation eq. stands for the word equation. 4* 82 SIMPLE EQUATIONS. By substituting 16 for 8x in equation (3), it becomes 16+12y=a2 (8). 12y=52-16 (9) Or 12y=36 (10).'. y- 3 (11) We might have first eliminated x; thus, multiply equation (1) by 5, and equation (2) by 2, and they become 10x+15y=65 (3) 10o+ Sy=44 (4) Eq. (3) —eq. (4) gives 7y=21 (5) y= 3 (6) and 8y=24 (7) Hence, eq. (4) may be written 10x+24=44 (8).'. 10x=20, and x=2 (9) Before commencing the process of elimination, unite all the similar terms, and all the unknown quantities in each of the equations into the first member. If any of the equations have fractions, they may be cleared of fiactions before transposing and uniting terms. It is, however, frequently advisable to transpose and unite some of the terms before clearing the equations of fractions. In reducing some equations it is better not to clear them of fractions. The student must rely on his own ingenuity. In the following equations transpose the 16 and the 18 before clearing the equation of fractions. 2x-y + 14=18 0,Y+14=18 (1) 2y + x 2y+ +16=19 (2) By transposing 14 and 16, these equations become, 2x-y 4 (3) x+2y x+2y= 3 (4) Eq. (3) multiplied by 4 gives 4x —2y=16 (5) Eq. (4) " 3 gives x~+2y= 9 (6) Eq. (5)+eq. (6) gives 5x=25 (7).'. x=5 (8) SIMPLE EQUATIONS. 83 In obtaining the values of x and y from the following equations, we do not clear them of fiactions. 147 14728 (1) -----— =28 (1) x y 17 56 41 t —- f- -- (2) x y 3 21 21 Eq. (1) divided by 7 gives ---— =4 (3) x y 17X21 56X21 Eq. (2) multiplied by 21 gives X + =287 (4) x y 17X21 17X21 Eq. (3) multiplied by 17 gives - =68 (5) x y Eq. (4) —eq. (5) gives 3X 219 (6) 7 Eq. (6) divided by 219 gives -=1 (7) y y=?7 (8) Substituting in (3) x=3 (9) (77.) From what has been done, we see that an unknown quantity may be eliminated from two equations by applying the following RULE. Make the co-eficients of the unknown quantity to be eliminated, the same in each of the equations; then, if the signs of the unknown quantity are alike, subtract one equatton fiSom the other, but if the signs are unlike, add one equations to the other. In order to obtain the values of three unknown quantities from three equations, we can eliminate one of the unknown quantities from the first and second equations, and then eliminate the same unknown quantity from the second and third equations, by the above rule. We shall thus have two equations, containing only two unknown quantities, fiom which we may obtain one equation containing only one unknown quantity, the value of which may be found. By retracing the several steps in the operation, it will be easy to find the values of the other two unknown quantities. 84 SIMPLE EQUATIONS. For the purpose of eliminating the unknown quantities, we may combine the equations in several ways, and the nature of the example should decide in what manner the unknown quantities should be eliminated. It is obvious that this method of eliminating may be applied, when we have m equations containing mn unknown quantities. The student will become familiar with this method of elimination by studying the solutions of the followinlg examples: (3iven +2y+ z=16 (1)) Given L2x + 2y + 2z= 18 (2) to find tl>, vOlve-.~ x. y, and z. 2x+2y+ z=14 (3) Eq.(2)-eq.(3)gives z= 4 (4) Eq.(1)-eq.(3)givesx= 2 (5) Whence 3x= 6 (6) Substituting these values of 3x and z, in eq. (1), and wq have 6+2y+4=16 (7) By transposing 2y- 6 (8) Whence y= 3 (9) (5x-6y+4z=15 (1)) Given 7x+4y-3z:=19 (2) tofind, y, 2d z. ( 2x y+6z=46 (3) Eq. (3) multiplied by 6 gives 12x+ 6y+36z= 276 (4) Eq. (l)+eq. (4) gives 17x+40z-= 291 (5) Eq. (3) multiplied by 4 gives 8x+ 4y+24z= 184 (6) Eq. (6)-eq. (2) gives x+27z= 165 (7) Eq. (7) multiplied by 17 gives 17x.+459z-2805 (8) Eq. (8) —eq. (5) gives 419z —2514 (9) Whence z=6 (x,?) Substitute this value of z in eq. (7), and we have x+162=165 (11) By transposing x= — 3 (12) By substituting the values of x and z in (3), we have 6 +y+36=46 (13) By transposing y= 4 (14) SIMPLE EQUATIONS. 85 SECOND MIETHOD. Elimination by Substitution. (7 S8.) Let it be required to find the values of x and y in the following equations: 2x+O3y=13 (1),5x+4y=22 (2) By transposing 3y in eq. (1), and dividing by 2, we have 13-3y'= 2 (3) 2 Whence 5x_- -- (4) By substituting this value of 5x in eq. (2), we have 65 —15y 2- +4y=22 (6) Or 65 —15y+8y2=44 (7) By transposing and uniting terms in (7), we have -7y=-21 (8) Whence y= 3 (9) Substituting this value in (3), x= 132 = 2 (10) Hence, to eliminate by this method we have the following RULE. ]Find the value of one of the unknown quantities in one of the equations, in terms of the other unknown quantities, and substitute this value for the unknown quantity in each of the remaining equations, and thus zake the number of equations and the number of unknown quantities one less. Proceed in this manner till a single equation is obtained containing but one unknown quantity, the value of which may then be found, and also the values of the other unknown quantities. 86 SIMPLE EQUATIONS. THIRD METHOD. Elimination by Comparison. (79.) Let it be required to find the values of x and y in the following equations: 2x.-3y=13 (1) 5x+4y=22 (2) By transposing 3y in (1) and dividing by 2, we have x 3-3y (3) By transposing 4y in (2), and dividing by 5, we have 22-4y (4) 5 Equating these two values of x, wve have 1 3-3y 22-4y 2 5 Eq. (5) multiplied by 10 gives 65-15y=44-Sy (6) By transposing -7y=-21 (7) Whence y=3 (8) Substituting tlhis value of y in (3), we have 13-9 x= 2 - 2 (9) As another example, take the equations 2xt-5y-3z= —3 (1) 3x-4y+ z= —2 (2) 5x- y+2z=9 (3) By transposing 5y and - 3z in (1), and dividing by 2, we have 3-5y+3z (4) 2 By transposing -4y and z in (2), and dividing by 3, we have 4y-z-2 X 3 ((.) By transposing -y and 2z in (3), and dividing by 5, we have y-2z + 9 SIMPLE EQUATIONS. 87 By equating these values of x, we have 4y-z-2 3-5 +3z (7) 3 2 And 4/-z —2 -2z+9 (8) 3 5 Eq. (7)multipliedby 6 gives 8y-2z- 4 =9 -15y+9z (9) Eq. (8) " 15 " 20y —5z —10 =3y- 6z+27 (10) By transposing in (9) 23y-llz-=13 (1 1) "s 4" (10) 17y+z=37 (12) From equations (11) and (12) we readily obtain 23y-13 (13) 11 And z=37-17y (14) 23y —13 = 37_ 17y Whence o-=0 17y (15) 11 Or 23y-13-=407-l87y (16) By transposing 210Qy=420~ (17) Hence y=2 (18) By substituting this value of y in (14), we have z=37-34=3 (19) By substituting the values of y and z in (5), we obtain 8-3-2 x = 3 1 (20) From what has been done, we see that we may eliminate by this method by the following RULE. Find the values of one qf the unknown quantities from each of the equations, as though the other unknown quantities were knower; and then take one of these values and equate it with each of the other values. We shall thus form as many equations less one us there were at first, containing as many unknown quantities as there are new equations. Proceed in this manner till a single equation is obtained containing but one unknown quantity,from which its value may be found. By retracing the steps in the operation, the values of all the unknown quantities may be successively obtained. 88 SIMPLE EQUATIONS. EXAMPLES. 2+4Given 7x+y=314 (21) } to find the values of x and y. 2x +-4y —14 (2) Eq. (1)+eq. (2) gives 9x+9y=45 (3) Whence x+ y= 5 (4) 2x+2y=10 (5) Eq. (2) —eq. (5) gives 2y= 4.. y=2 (6) Whence, by substituting, x= 3 (7) x+= — (1) I y 1 2. Given 1- (2) to find the values of x,y, and z.?;=~ (3) Eq. (l)+eq. (2)+eq. (3) gives 2 ++2-=8 (4) x y z 24 Eq. (4) divided by 2 gives -+-+- (5) x y z 24 Eq. (5)-eq. (1) gives -= 1 (6) z 24 Eq. (5)-eq. (2) gives y 8 (7) Eq.(5)-eq.(3) gives 1 5 (8) 24From(6),)and(8)wereadilyfindthat From (6), (7), and (8) we readily find that x=4A) y=8.Ans. and z=24) Given (y)=14 (1)) B. Given < y(z+z)=18 (2) to find the value of x, y, and z. z(x+y)=20 (3) SIMPLE EQUATIONS. 89 Perform the multiplications indicated, and then add the three equations together, and divide their sum by 2. xy+xz+ yz= 26 (4) Eq. (4) —eq. (1) gives yz= 12 (5) Eq. (4)-eq. (2) gives xz= 8 (6) Eq. (4)-eq. (3) gives xy= 6 (7) Eq. (5)Xeq. (6)Xeq. (7) gives x2y2z'=576 (8) Extract the square root of (8) xyz= 24 (9) Eq. (9) divided by eq. (5) gives x= 2 (10) Eq. (9) divided by eq. (6) gives y= 3 (11) Eq. (9) divided by eq. (7) gives z= 4 (12) x + ly + lz (1) ) to find the values of x, y, 4. Given mx+ y + mnz=q (2) and z, —, r, n, p, q, and nt-x+ny+z=r (3) r, are known quantities. Let x+y+z=s (A); then the equations may be inmde to assunle thlis form, by substituting s, adding and subtracting lx to the first member of (1), my to the first member of (2), and nz to the first member of (3). (1-l)x + ls=p. (4) ( —m)y +ms-=q (5) (1-n)z + ns=r (6) From (4), (5), and (6), we readily obtain (7), (8), and (9), x+ 8 _gs= _ (7) rn q y+~ s-8 Q (8)! -- m 1 -- m + - 1-s= (9) Eq. (7) + eq. (8)eq. (9) gives s+( I 1-m+1_ ) P+ + (10) 1-I m-1 1-11 Whence s= — I 1 n (11) 1+ +- + 1-I 1-m 1-n 90 SIM.IPLE EQUATION'S. Substituting this value of s in (), (87), (), and (9), and reducing, we obtain p= - 1 n{ I- -n J (12) 1. I -in 1-n 1 1 +1 —1-+ + ~ + — -(13) tions. The equations in example (3) are symmetrical equation~ 5 Given x+y () to fin d x and y. v= G i v e-8x-(21y 33 () 6. Gvn 5 —-177 (2) t fn -Ans. x=12, y=3.l 7. Given 17y- 2x: 1 ) 1 o -n 20y+6x=38 (2) to find x and y. Ans. y=l, x=3. 8. Given 18x+32y= 34 (1) to find x and y. z=y _x= (14) 7-n Giv -n ( tI rn 1, y. Sx 24y_254 to find and y. Ans. x_ 1, y= -l. SIMPLE EQUATIONS. 91 ( X(Z +y) 5a (1) ) 9. Given y(x+z)=7a (2) to find x, y, and z. z(x+y)=-4 (3) Ans. x* = V3a, y=2 /3a, and z =- Va. r[4x 5y' --- - (1 10. Given Y4 Y to find x and y. 5 4 7 3 -+-=-+- (2)j x y x2 Ans. x=-4, y=2. ru+x+y=13 (1) f!10 Given Iu+x-z=17 (2 ) 11. Given u+y+z=18 (3) to findx, y, z, and u. x+y+z -=21 (4) Ans. x-5, y=6, z=10, u=2. r2x=y+z+u (1) 12. Given 3yx+z (2) to find x, y, z, and u. 4z=x+y+u (3) L u=x-14 (4) Ans. x=40, u=2C, y=30, z=24. ( x+22=y+z (1)) 13. Given y+22=2x+2z (2). to find x, y, and z. z+22=3x+3y (3)) Ans. x=2, y=10, z=14. ( 4x-4y-4z=a (1)) 14. Given 6y-2x-2z=-a (2). to find x, y, and z. 7z-y-x =a (3)) Ans. x=l —a, y=a, z= —a. I ~12 — ~ (1) 15 Given 2 1 to find x and y. Y- = 41 -2 (2)I Ans. x= 2, and y= 7. * NOTE.-See Art., Ex. 92 SIMPLE EQUATIONS. 16. Given 3xy=1 0x+8y (1) to findand y. 2xy=26y-45x (2) Ans. x=4, y=10. 17. Given i xy3+-xy-2xy2=-4 (2) to find x and y. Ans. x=4, y=3. i 2x+2y+ 4z — 14 (1)) 18. Given 3x+-.y+42z= 63 (2) to find x, y andz. (5x~+3y —36z =-17 (3) Ans. x=2, y=3, z=l...~. 6x2-24y +130 19. Given 39xy- 110 151 —4 - 11-16,v to find x and y. yx= -3 4 4y- I Ans. x=9, y=2. fr 1 1 1 + I=- i (1) 20. Given q-+-z- (2) 5-to find x, y, and z. I I 1 I+ -- (3) 2abc 2abc 2abc Ans. -ac-ab- bc' Y=ab-ac+-bc' Z-acab-bc' 21. Giveun b+y 3a+-x ( to find x and y. ax+2by=c (2) 2b2 —6a2 c 3a2_-b ~-c.Ans. x= 3 X3 3b (x+ y2 =2a (1) ) 22. Given x 2b'y4 to find x and y. 2 =2b* (2) Ans. x= /a+b, y= /a-b. -~t~ ^ ~ ~ z,<.r SIMPLE EQUATIONS. 93 23. Given 17x+2 3y=103 () to find x and y. x3x+ 7y- 47 (2) Ans. x=2, y=3. 24. Given 2x+300=3y-150 (I1) tofind xandy. 9x —450=Sy+ 500 (2) tofi andy. Ans. x=300, y —350. 5. Given; bcx=cy-2b (1) -.to find x...... + c.~ (2) and y. a a~+2b -Ans. x —-, = — W6. Given {3xy -28-3 Y (2) to find x andy. Ans. x. =1, y=4. 7x-2z +3u=17 (1) 1 4y-2z + t —ll (2) to find x, yz, 27. Given 5y-3x-2u= 8 (3) d t. 4y-3u+2t= 9 (4), and t 3z +8u=33 (5) Ans. x=2, y=4, z=3, u=3, t=l. 28. Given' ~Y 140 (1) to find x.andy. x-_y2=40 (2) Ans. x=7, y=3. ( 5x+ 2y+z 2-=24 (1) ) 29. Given 4x+7y~+5z=49 (2) - to find x, y, and z. x+ y+3z —17 (3)) Ans. x:2, y=3, z=4. ELIMINATION BY INDETERAMINATE IMULTIPLIERS. (80.) Let it be required to find the values of x and y from he following equations: 3x-2y=10 (1) 4x+3y=30 (2) 94 SIMPLE EQUATIONS. Multiply the second equation by the indeterminate quantity, m, and we have 4mx+3my —36m (3) Add equation (3) to equation (1), and we have 4mx+3x+3my-2y= —36m+ 10 (4) Or, (4m+ 3)x+(3n —2)y=-36m -t 10 (5) Since mn may have any value, we may give to it such a value as will cause the co-efficient of y in equation (5) to become nothing. Hence, in order to eliminate y, we may assume 3m —2= —0 (6) Whence, m=- (7) Since 3m-2=0, equation (5) becomes (4m+3)x=-36m+ 10 (8) 36m + 10 9 Or, x'=4m +3 (9) Substitute the value of m in equation (7) in the place of m in equation (9), and we readily obtain x=6 (10) In a similar way we may find that y=4 (11) (81.) We will now, by the aid of this principle, determine the values of x, y, and z, from the following equations, in which a,, a2, a3, bl, b2, b3, cl, c2, c3, are the co-efficients of the unknown quantities, and mnz, m2, nm3, are the absolute terms. ax+ bly+clz=ml (1) ) a2x + b2y + c2z =m2 (2) ( ) a3x+bsy+cz=m3 (3) Multiply equation (2) by r, and equation (3) by s, and equations (A) become a, x+b, y+c, z=m1 (4)) a2,x + b2y + c21z = in, () (B) (asx + b.sy +~ casz =m- (6) J SIMPLE EQUATIONS. 95 — c3b, + b3c, Assume rbs=-b, Vhence sc - (D) and c2r +css= —c - bcl +bc C I The values of r and s may be found from the equations, b,2+b3s= —b,, and car+c3s= —cj, by applying the rules of either of the three common methods of elimination, or by the method of indeterminate multipliers. By taking the sum of equations (B), and recollecting that we have assumed that the sum of the co-efficient of y in equations (5) and (6), is equal to the co-efficients of y in equation (1), taken with a contrary sign, and that we have made a similar assumption in respect to the co-efficients of z in those equations, we have ax + aarx ~ a3sx=ml + m,2r + n1s (7) Whence, m1 + nr + ms — (8) a, +- f2r + as In the place of r and s in this last equation, put their values, as found in equations (D), and we have -blc3 + b6cl - b2 b-2c 2 bc2 m, + m.2 b. 2 2c2 a2b.c~23 ~ —- __blca q- b~Cl (9) - - blbc, + b cl _bcl - + blc2 a - a+2 al, b —b q + 3c b23 — -b3c2 Reducing this value of x to a more simple form by multiplying both numerator and denominator by b.2c-b,c,, and we have m6b.2c- - n b;c,-in-m2bc + m2,c, - nm;,h2c - mobc.2 (10) x: a b.c3 —a+bcq —a.2b ca + abc1 —a2b.2q +aAc1 x. b~c -a,b c-aablc, + a b,cl-asb.c, abc, By arranging the terms in the numerator and denominator in a different order, we have amb,2c+a2mbacl +ma:bIc2-amlb;c2-a2bbc3 —mtb2c1 (11) mac m3 a~hc, + m mjac2 - in ma c.2 - rnac,- - mranc,' albb2c3+a2bcl+ ~ctblc2-albC2-6c b c(1-2)ACE m3ab.2~mab3 + +.2ag.b. -n.2a b3- rn2b- mnabb~ a2- C + a. c,.c +- (t.b,.2 - a, b,, - aabc, — a nb.- a, 96 SIMPLE EQUATIONS. The values of y and z were obtained in the same manner. By carefully examining equations (E), and also equations (1), (2), (3), the intelligent student will find it easy to point out in what manner the values of x, y, and z may be obtained from the co-efficients and the absolute terms. It is obvious, that by giving particular values to the co-efficients and the absolute terms, we may, by the aid of equations (E), obtain the values of x, y, and z from any three simple equations involving these letters. ON THE SOLUTION OF PROBLEMS WHICHE REQUIRE TWO OR MIORE UNKNOWN QUANTITIES. 1. A banker has two kinds of money; it takes a pieces of the first kind to make a crown, and b pieces of the second kind to make the same. Some one wishes to have c pieces for a crown. How many pieces of each must the banker give him? Let x=the number of pieces that he uses of the 1st kind, And y= " " " " 2d " -=the part of a crown which 1 piece of the 1st kind makes,* a 1= 4 " " "i 2d " b -= " " " x pieces 1st kind make, a b Y " 2d " Whence x+y=c (1) x y (2) -+-=l2 ab * NOTE.-Since a pieces of the first kind make a crown, one piece of the first kind will make one ath part of a crown, or a part denoted by, and x pieces will make x times - part of a crown, or a part denoted by. In the same way we find that y pieces of 2d kind will make a part of the crown denoted by Y. These two parts added together make one crown. SIIMPLE EQUATIONS. 97 Eq. (2)X ab gives bx+ay=ab (3) Eq. (1)Xa gives ax+ay=ac (4) Eq. () X b gives bx +by=bc (5) Eq. (4)-eq. (3) gives ax-bx —=ac-ab (6) Eq. (3)-eq. (5) gives ay-by=ab-bc (7) ac-ab a(_ —b) (7) From (6) (c ) (8) Y a-b (a —=a b (9) 2. According to Vitruviu., the crown of Hiero, king of Syracuse, weighed 20 lbs., andc lost 1 lbs., nearly, wh(n weighed in water. Let it be assumled that it consisted of gold and silver only, and that 19.64 lbs. of gold lose 1 lb. in water, anld 10.5 lbs. of silelr, in 1,ke nmalner, lose 1 lb. I-How much gold, and how much silver, did this crown contain? If we let x-the number of pounds of gold, and y=the number of pounds of silver, and make 20=c, 19.64=a, 10.5=b, we shall find that the solution of this problem is similar to that of the preceding one. By reducing the equations which the problenm readily furnishes, we shall find that x=14.77, y=5.22. 3. Some hours after a courier had been sent from A to B, whiich are 147 imiles distant, a second waas sent, who wished to overtake the firt just as he entered B; in order to do this, he found that lie must perform the journey in 28 hours less time than the first did. Now, the time in wihich the first travels 17 mil s added to the timle in which thle second travels 56 miles is 183 houis. How mlny miles does ch;&l go per hour? Let x=number of imiles that the 1st travels per hour. And y= -- " " fd " " 147 -— number of hours which the 1st requires to travel 147 miles. x ]47' 17= "' s 7 I — = "" 1st 17 miles. 56 tr ar $c " -. 98 SIMPLE EQUATIONS. By the conditions of the problem we have, 147 147 -28...... =- ~ (1) xa y 17 56 — + — = la~~ =~x s (2) ax y These equations are the same as those found on page 83. Henlce. the first travels 3 miles per hour, and the second travels 7 miles per hour. 4, A certain number consists of three digits, the sum of which -is 9. If 198 be subtracted from the number, the remainder is a number consisting of the same digits as the one sought, but in an inverted order; and if the number be divided by the left digit,~ the quotient is 108. What is the number? Let x=the number of hundreds, y - " " tens, And z= " " units, Then, 100x + 1 Oy + z=the number. By the conditions of the problem, we have the following equations: x+y+z=9, (1) OOx+lOy+z —198= 100z+0ly+x, (2) 100x+lOy+z And =108 (3) By transposing, &c. in (2) 99x-99z= 198 (4) From (3) -8x-10y+z= 0 (5) Eq. (1)-eq. (5) gives 9x-9y= 9 (6) Eq. (6)-.9 gives x-y= 1 (7) Eq. (4) 99 gives x —z= 2 (8) Eq. (1)+eq. (7) gives 2x+z — 10 (9) Eq. (8)+eq. (9) gives 3x= 12 (10). = 4 (1 1) From (7), y=x-1=3 (12) From (8), z=x-2=2 (13) Hence, the number is 432. SIMPLE EQUATIONS. 99 5. Required two numbers whose sum is 70, and whose difference is 16. Anzs. 43 and 27. 6. Required two numbers whose sum is a, and whose difference is 0. a — b a —b -Ans.2 a and 2 2' 2 7. Two purses, together, contain $300. If ybu take 830 fiom the first, and put tllem into the. second, then there is the salme number of dollars in each. How many dollars does each contain? Ans. Thie first $180, the second $120. 8. A owes $1200, B $2500; but neither has enough to pay his debts. Lend me, said A to 1B, the 8th part of your fortune, and I shall be enabled to pay my debts. B answered, I can disch:arge my debts, if you will lend mie the 9th part of your!brtune. What was the fortune of each? Ans. A's 900, B's 2400. 9. Five gamllers, A, B, C, D, E, throw dice upon the co,ndition that he who has the lowest throw shall give all the rest the sun which they already have. Ecllh gamester loses in turl, conmencing' with A, and at the end of the fifth gamne, all lave the same sumn, viz., $32. How much had each at first? 10. A and B, jointly, have a fortune of $9800. A invests the 6th part of his l,roperty in business, and B the 5th part of hlis, and each has then the saeno sunm remailing. How imuch has each? Alzs. A has $4800, B 85000. 11. A capitalist borroxws $8000 on favorable conditions, li),cause lie has an opportunity of lending $23000 at such a rate per cent. that the interest which he receives exceeds that which he pays by $905. On the same conditions he borrows $9400, and then lends $17500, and the interest which he receives exceeds that which he pays by $539.50 yearly. At what rate per cent. did he borrow and lend money? Ans. At 4j and B5 per eent. 100 SIMPLE EQUATIONS. 12, Find two numbers of the following properties: WThen the one is multiplied by 2, the other by 5, and both products added together, the sum is 31; and if the first be multiplied by 7, and the second by 4, and the products added together, the sum is 68. Ans. The first is 8, and the second is 3. 13. A person has two kinds of goods. Now, 8 lbs. of tile first and 93 lbs. of the second cost $18.46; further, 20 lbs. of the first and 16 of the second cost $36.40.!low much does the pound of each article cost? Anis. c62 cents, and $1.50. 14. A person has two large pieces of iron, whose weiglht is required. It is known that # of the first piece weighs 96 lbs. less than 3- of the other piece; cand that - of the other piece weighs exactly as much as i of the first. IIow much did e-ach of these pieces weigh? Ans. The first weighs 720, the second 512 lbs. 15. A cistern containing 210 buckets, llay be filled by 2 pipes. By an experiment, in which the first pipe was open 4, and the second 5 hours, 90 buckets of water were obt,:ined. By another experiment, when the first was open 7, and the other 3~1 hours, 126 buckets were obtained. How many buckets does each pipe dischar(ge in an hour? And in whlt ti ne will the cistern be full, when the water flows from both pipes at,nce? Ans. The first pipe discblarges 15, and the second 6 buckets; it will require 10 hours for them to fill the cistern. 16. It is required to find a fraction such, that if 3 be subidtt fl,;,lfn the numerator and denominator, it is clangied into -,.! I.. i 5 be added to the numerator and denolinator, it becomes. What is the fiaction? Ans. 7. 17. A wine merchant has two kinds of wine. If he mix 3 gallons of the poorer wine with.5 of the better, the mixture is worth $1 per gallon; but if he mixes 3~ gallons of the poorer with 8a gallons of the better, the mixture is worth $1.03~ per gallon. What does each wine cost per gallon? Ains. The better $1.12, the poorer $0.80. SIMPLE EQUATIONS. 101 18. There is a fraction such that if 1 be added to the numerator, its value=j}, and if 1 be added to the denominator its value =-. What is the fraction a Ans. -4 19. A person has a sum of money which he places out at interest. Another person possesses $10000 more than the first, and placing his capital out at interest at a rate one per cent. higher than that of the first, he-had an income greater by $800. A third person possesses $15000 more than the first, and placing his capital out at interest at a rate 2 per cent. higher than that of the first, he has an income greater by $1500. Required the capital that each nad, and the rates per cent. Ans-. Captitals, $30000, $40000, $45000. s Rates of interest, 4, 5, 6, 20. It is found by experiment that 3'7 lbs. of tin lose 5 lbs. when weighed in water, and that 23 lbs.'of lead lose 2 lbs. in water. A composition of tin and lead, weighing 120 Ibs, loses 14 lbs. in water. How much does this composition contain of each metal? Ans. 74 lbs. tin, and 46 lbs. lead. 21. A given piece of metal, which weighs p pounds, loses a pounds in water. This piece, however, is composed of two other metals, A and B; of these we know that p pounds of A lose b pounds in water, and p pounds of B lose c pounds. How many pounds of each metal does this piece contain? Ans. (c-a)p lbs. of A, and (a-b)p lbs. of B. c-b c-b22. Three masons, A, B, C, alre to build a wall. A and B, jointly, caon build the wall in 12 days; B and C can accomplish,t in 20 days, and A and C in 15 days. How many days would each require to build the wall, and in what time will they finish it, if all three work together? Ans. A requires 20 days, B 30, and C 60; and all three require 10 days. 102 SIMPLE EQUATIONS. 23. Three laborers are employed on a certain work. A and B, jointly, can complete the work in a days; A and C require b days, B and C require c days. What time does each one, working alone, require to accomplish the wolrk, on the condition that each one, under all circumstances, does the same quantity of work? And in what time would they finish it, if they all three worked together? 2abe 2abce Ans. A requires + days, B days, bc ur-c-ab b +ab-ac 2abe and C 2abc ys. ab+ac-be Jointly, they require a bc days. ab-+ac + bc 21. A, B, C, together, possess $1820. If B gives A $200 of his money, then A will have $160 more than B; but if B receives $!70 from C, then both will have tlhe same sum. How much has each?' Ans. A $400, B $640, C $780. 25. A and B possess, together, a fortune of $570. If A's for tune were 3 times, and B's 5 times as great as each really is, they would have together $2350. What was each one's fortune? Ans. A's $250, B's $320. 26. B has lent out at interest $12600 more than A, and obtains 1 per cent. more for his money, and his yearly interest is $730 more than A's. C has lent $3000 more than A, at 2 per cent. hligher interest, and hlis yearly interest is $380 more than A's. How much has each lent? And at what interest? Ans. A 310000, B $22600, C $13000. A at 4 per cent., B at 5, and C at 6. 27. When the first of two numbers is increased by a, it becomes mn timhes as great as the second; but when the second is increased by b it becomes n times as great as the fir-t. How aro these numbers expressed? Ans. The first =the second = mn -1' mn- 1 SIMPLE EQUATIONS. 103 28. A work is to be printed, so that each page may contain a certain number of lines, and each line a certain number of letters. If we wished to have each page to contain 3 lines more, and each line 4 letters more, then there would be 224 letters more on each page; but if we wicshed to have 2 lines less in a page, and 3 letters less in each line, then each page would contain 145 letters less. How many lines are there in each page? And how many letters in each line? Ans. 29 lines in each page, and 32 letters in a line. 29. A father said to his son, "six years ago I was 3~ times as old as you; but three years hence I shall be 2k times as old as you." What is the age of each? Ans. The father 36, the son 15 years. 30. After A had won 4 shillings of B, he had only half'as many shillings as B had left. But had B won 6 shillings of A, then he would have hIad three times as many as A would have had left. How mav- hlld ct1ci? Ans. A 36, B 84. 31. A person has two snuff boxes. If he puts $8 into the first, then it is half as valuable as the second; but if he takes out $8 fiom the first, and puts them into the second, then the latter is three times as valuable as the former. What is the value of each. Ans. The first $48, second $112. 32. A, 13, C compare their fortunes. A says to B, give me $700 of your money, and I shall have twice as much as you retain; B says to C, give me $1400, and I shall have three times as much as you have remaining; C says to A, give me $420, and I shall then have five times as much as you retain. How much has each? Ans. A $980, B $1540, C $2380. 33. The sum of two numbers is 13, and the difference of their squares is 39. What are the numbers? Ans. 5 and 8. 34, The sum of two numbers is=a, and the greater number is n times the smaller. What are the numbers? a na Ans. -+l andn+l 11+17 L n +t- 104 SIMPLE EQUATIONS. 35. A merchant mixed tea which cost him 45 cents per pound, with some that was worth 75 cents per pound, and sold the mixture for $24.30, and by this means gained 20 per cent. How many pounds of each kind were there, the whole number of pounlds being 35? Ans. 20 lbs. at 45 cents, and 15 lbs. at 75 cents. 36. A certain number, which has two digits, is equal to 8 times the sum of its digits, and if 45 be subtracted from the number, its digits will be inverted. What is the number? Ans. 72. 37. Two persons, A and B, can perform a piece of work in 16 days. They work together for 4 days, when A being called off; B is left to finish it, which he does in 36 days more. In what time would each do it separately? Ans. A in 24 days, and B in 48 days. 38. There is a cistern, into which water is admitted by three pipes, two of which are exactly of the same dimensions. When they are all open, five-twelfths of the cistern is filled in four hours; and if one of the equal pipes be stopped, seven-ninths of the cistern is filled in 10 hours and 40 minutes. In how many hours would each pipe fill the cistern? Ans. Each of the equal ones in 32 hours, and the other in 24 hours. 39, A purse holds 19 crowns and 6 guineas. Now 4 crowns and 5 guineas fill 1 of it. How many will it hold of each? Ans. 21 crowns, and 63 guineas. 40. A rectangular bowling-green. having been measured, it was observed, that if it were 5 yards broader, and 4 yards longer, it would contain 116 yards more; but if it were 4 yards broader, and 5 yards longer, it would contain 113 yards more. Required the length and breadth? Ans. 12 yards long, and 9 yards wide. SIMPLE EQUATIONS. 105 41. A Corn-factor mixes wheat flour, which costs him 10 shillings a bushel, with barley flour, which costs him 4 shillings a bushel, in such proportion as to gain 433 per cent., by selling the mixture at 11 shillings per bushel. Required the proportion. Ans. 14 bushels of -vwheat flour to 9 of barlev. 42. Some smugglers discovered a cave which would exactly hold the cargo of their boat, viz., 13 bales of cotton, and 33 casks of rum. Whilst they were unloading, a cust,,m-house cutter coining in sight, they sailed away with 9 casks aind 5 bales, leaving the cave two-thirds full. How many bales or casks would it hold? Ans. 24 bales, or 72 casks. 43. A countryman, being employed to drive a flock of geese and turkeys to New York, in order to distinguish his own from any he might meet on the way, pulled three feathers out of the tail of each turkey, and one out of the tail of-each goose, and upon counting them, found that the number of turkey's feathers exceeded twice those of the geese by 15. Having bought 10 geese, and sold 15 turkeys by the way, he was surprised to find, as he drove them into his employer's yard, that the number of geese was equal to seven-thirds the number of turkeys. Required the number of each at first. Ans. 45 turkeys, and 60 geese. 44. Round two wheels, whose circumferences are as 5 to 3, two ropes are wrapped, whose difference exceeds the difference of the circumferences by 280 yards. Now the longer rot,e applied to the larger wheel, wraps around it a certain number of times, greater by 12 than the shorter round the smaller wheel; -and if the larger wheel turns round 3 times as quick as the other, the ropes will be discharged at the same time. Required the lengths of the ropes, and the circumferences of the wheels. Ans. The lengths of the ropes are 360 and 72 yards, and the circumferences of the wheels are 20 and 12 yards. 106 SIMPLE EQUATIONS. 45. A and B agree to reap a field of wheat in 12 days. The times in which they could severally reap an acre are as 2 to 3. After some time, finding themselves unable to finish it in the stipulated tine, they called in C to help them, whose rate of working was such, that if he had wrought with them from the beginnling, it would have been finished in 9 days. Also the times in which he could severally have reaped the field with A alone, and B alone, are in the proportion of 7 to 8. When was C called in? Ans. After 6 days. INTERPRETATION OF NEGATIVE RESULTS. (82.) Algebraical formulas can convey no distinct idea to the mind, unless they represent numerical calculations, which can be actually performed: Thus, the expression a-b indicates an absurdity, when b is greater than a, since it is impossible to subtract a greater quantity from a smaller one. As the solution of problems sometimes presents this difficulty, it will be necessary to show what signification ought to be attached to such a value of the unknown quantity. Every Equation of the first degree may be reduced to the following form, in which all the signs are positive, ax+n=cx+m (1)* Subtracting cx+n from each member, we have, ax-cx=m-n (2) Whence, x= nn (3) a-c This being premised, three cases present themselves: CASE I. When nz>n, and a>c. CASE II. Whent m>n, and ac; that is, when only one of the conditions in Case I. holds good. * NOTE.-This may be done by transposing the negative terms to the other members. SIMPLE EQUATIONS. 107 CASE III. When mc, for example, equation (1) must involve an absurdity, since, in this case, the two terms, ax and n, in the first member, are respectively greater than the two terms cx and m, in the second member. Now, as equation (1) may be regarded as truly representing the conditions of a problem, the conditions themselves must be absurd, whenever they give an equation, the solution of which presents this case. In the third case, both' of the subtractions, m —n and a-c, are impossible, but we must not, for this reason, conclude that equation (1) presents an absurdity. For, let it be observed, that in obtaining the value of x, we subtracted cx+n from each member, an operation which is impossible, since this quantity is greater than either the first or second member. To find the value of x, we will subtract axq+rn fiom each member of equation (1), and we then have, n —m=cx-ax, (4) Whence, x= —- (5) c-a The value of x in equation (5) may be derived from the value of x in equation (3), by changing the signs of both numerator and (enomninator of the fraction, which expresses the value of x in this equation. Now as the quotient of a negative quantity divided by a negative quantity is positive, we may change the signs of both numerator and denominator of the fraction, which expresses the value of x in equation (5), without changing its value from positive to negative. Hence equation (5) may be written, -(n-m) rn-n (6) -(c-a) a-c 108 SIMPLE EQUATIONS. Hence we see that the value of x in equation (6), which is the true value of x in Case III., is the same as the value of x in equation (3). One of the chief advantages which algebra presents, is to obtain formulas which shall include all the conditions of a problem. This end may be accomplished by establishing the convention, that the same operations shall be performed on isolated negative quantities, as if they were connected with other magnitudes. For example, in the expression a + b-d, we could not subtract d from b if d were greater than b, and we should write the expression in this form, a-(d-b); b may now be subtracted from d, and the difference taken from a. If' a does not exist, we may still write b-d in the form, -(d-b) when bn, and ac. Each of these values of x is negative. We conclude then, that Every negative solution denotes some absurdity zn the enunciation of the problem proposed. For the purpose of showing the use of a negative solution in pointing out the error in the enunciation of a problem, we will propose the following problem: A father is now 48 years old, and his son is 15. How many years must elapse, in order that the age of the father may be 4 times the age of the son a Let x=the number of years that must elapse. By the conditions of the problem, we have 48 +x=4X(15 +x) (1) Or, 48+x=60+4x (2) Whence, x= -4 (3) This negative result indicates that an error was made in the enunciation of the problem, We can correct the language in -the SIMPLE EQUATIONS. 109 problem by demanding how many years have elapsed since the age of thie father was 4 times that of his son. The equation to the problem may then be obtained fiom equation (1) by changing x into -x. (83.) Before closing this discussion, we will interpret the A 0 0 symbols -0 A' If in equation (2), in the preceding article, we make a-c, the m-n A value of x becomes, or in general, -. Now, the value of a fiaction increases as the denominator diminishes, the numerator rermainling the same. HJince, when the denominator is very small, the value of the fraction must be very great, and when the denomilltor is infiJnitely small, or nothing, the value of the fraction is infinitely great, or infinity. The symbol for infinity is o0. A Therefore, -= c. If in equation (2), we make m=n, the value of x becomes 0 0 — c, or in general, A. By a reasoning similar 4o that which has just been employed, we may show that =0. If in equation (2), we make m=n, and at the same time make a=c, the value of x becomes 0. In this case, equation (1) becomes ax+m=ax+-n, and it is plain that the value of r, O O. 0, may be represented by any number whatever. Hence, 0 is tiiu symbol of an indeterminate quantity. We must not, however, always conclude that when the solution of a problem is 0, the problem is indeterminate. Suppose that the solution of a certain problem is, a2 - b 110 SIMPLE EQUT1IT)NS. 0 If we make a=b, this value of x becomes B. But if we divide the numerator, a'-bs, and the denominator, a — b, by a-b, their greatest common measure, the value of x becomes, a2 +ab+b' a+b If we now make the hypothesis that a=b, we have, 3a- 3 X= —=-a 2a 2 This is the true value of x, when a=b., We therefore conclude that when the solution of a problem is a fraction-which reduces to 0 -, when a particular hypothesis is made, we must first determine whether the terms of this fraction have a common measure. If no common measure exists, then the value of x must be indeterminate, that is, x may have any value whatever. But if the terms of the fraction have a common measure, the fraction must be reduced by dividing both terms by this common measure. In the reduced fraction, we must make the same hypothesis, and it AA 0 will then assume one of the three forms, A O. B' O' O' In the first form, the value of x is determinate; in the second, it cannot be expressed in finite numbers; in the third, it is indeterminate. CHAPTER V. INVOLUTION, EVOLUTION, RADICA.L QUANTTITI ES, AND IMAGINARY Q"lANTITIIES. Ii VOLUTI ON. (84.) INVOLU:TION is the raising of a quantity to any required power. The involution of algebraical quantities may be divided into two cases. CASE I, To involve a monomial, (85.) Let it be required to raise 2a' to the third power. By definition 15, the third power of 2a3=2a3X2a3X2a3=8ae by the rules of multiplication. Let it be required to raise -5c to the third power. By definimion 1 5, the third power of - 5c= - 5cX — 5cX - 5c= - 125c3. In this example, there is an odd number of negative factors, and the product, — 125c3; is therefore negative. Let it be required to raise — 2n to the fourth power. By definition 15, the fourth power of - 2n= - 2nX - 2nX - 2n X — 2n= +1 6n'. In this example, there is an even number of negative factors, and the product, 16n4, is therefore positive. The fourth power of 3J= 3xX 3X X3JX 3xA=81x. The third power of 4c= 4.x X 42x X 4xi= 4x'. 112 INVOLUTION. In these examples, let it be observed that the exponents of the literal parts in the required power may be obtained by multiplying the exponents of the letters in the given quantities by the index of the required powers; and that the co-efficienlts of the literal parts in the required powers may be obtained by raising the numerical co-efficients in the given quantities to a power denoted by thile index of the required power. Thus, the fourth power of 3x~ is formed by multiplying the exponent ~ by 4, and raising 3, the co-efficient, to the fourth power. Hence, the foulth power of 3x3 is 81x'. The fourth power of 2a'b is 16a8b4. The exponent of b is 1 undeistood. It is often convenient to indicate powers of quantities by employing the exponent of the required power. Thus the third power of 2xi is expressed (2xA)'. The fifth power of 4a' is expressed (4a')5. Monomials may be involved by the aid of the following RULE. Multiply the exponents of the different letters by the index of the required power, and to this power of the letters prefix that of the co-egicient. EXAMPLES. 1. What is the fourth power of 5a'b? Ans. 625a'b4. 2. What is the fourth power of 7a'Sb?. Ans. 2401a'2b. 3, What is the fifth power of 3ac5? Ans. 243a6c'. 4, What is the sixth power of 2ac3? Ans. 64a6c2. 5,. What is the nth power of a'c'? Ans. a'"c". INVOLUTION. 118 6. What is the third power of — 3a'cB.Ans. -2 7a6c. a' a'lo 7. What is the fifth power of -2 Ans. 8. What is the sixth power of — 3acW Ans. 729aYc'. 9. What is the fourth power of 5a c'? Ans. 625ae'. 10. What is the third power of 2Ila ]331a2 Ans. 11.,What is the seventh power of -2a -? Ans. — 128a. 12. What is the second power of 21aic? Ans. 441ac3. CASE II. To involve a polynomial. (86.) A polynomial is involved to any required power by actual multiplication. Thus, the third power of a+b+c is found by multipllying it by itself, and that product by a+b+c. The quatntity to be involved must be used as a factor as many times as there are utits in the exponent of the required power. WThen a qu',ntity to be involved has an exponent, this exponent may be mulltil.,lied by the index of the required power, and the quanltity imay then be raised to the r(-quire(l power. Thus, to find the secoid power of (a +x), we can multiply the exponent 3 by 2, and then involve a-+- to the sixth power. This course will be found necessary when the exponent of the given quantity is a 114 INVOLUTION. fiaction. For example, let it be required to raise (a+X-z) to the fourth power. We multiply the exponent by 4, and we obtain 2. Hence, the fourth power of (a +x)2 is expressed (a + x)2=-a + 2ax+x2. BINOMIIAL THEOREZM FOR POSITIVE AND INTEGRAL EXPONENTS. (87.) It is obvious that po!ynomials may be raised to any required power by actual multil)lication; but in many cases,this labor would be very long anl d tedious. As itis often necessary to involve binomcals in solvinig problems in pure and quadratic equations, we will sh1ow in what way we can write doxwn any power of a binomial, when the exponent is a positive integer, without the labor of multiplicattion. When a binomial, as x —ta, is raised to any power, it is found that tile successive terms biar a certain relation to each other, and that this relation is the salme in all cases. This law, when expressed in algebraical characters, is called the " Binomial Theorem." It was discovered by Sir Isaac Newton, and it is very fiequently used in the higlelr blanches of mathematical analysis. For the present, we shall only show the application of the' Theorem in the case w-here the exponent is a positive integer. In a subsequent chapter, we shall give a rigorous demonstration of the Thlieorem. For the purpose of examining the powers of a V;nomial, we will find some of the powers of the binamial, x+a, and arrange them in a table. xpa x2+ xa xa ~ aa x2+ 2xc.- -a2= 2d power. x+-a xs'+ 2x2a+ xa2 X20 + Xa2 tsa' 3X2C 32 - 3 oer. x' ~3x2a + 3xa&+a'.-. 3d power. INVOLUTION. 115 s + 3x2a + 3xa2 + a' x+a' t+3xaa3x2a - xa3 x3a + 3xa2+ - 3,xa3 +a'4 x4 ~ 4x a + 6x(a 2 + 4xa3 + a4= 4th power. x5+4x4a+ 6x3a2+ 4x2a3 + xa4 x 4a I- 4xt3a2 - 6.,2a3 + 4xa4+ ~a5 5+5X a+ Ixxa O3S2 + 1 0X2a3 + ta4 + ac-= 5th power. TABLE OF THE POWERS OF x-a. x+a lx+a (x +a)' X2 +2xa+a2 (x +a) 3 + 3 x, +'I,,a s+_ a' (x +a)4 x4 + 4x'a +6.x'a2 + 4xa3+-a4 (x +a)5lx'5+ 5x4a+ I ox3a2-+ 1 ox2U — + 5zx4 +05 (x+ a)fl 6 q+ 6xa + 1 5. 44a2 2 Ox'3a3 + I 5x2a + 6,x(a5 +-6 (x A-a)'6 X7+ 7x'a + 2 xa 3 5x~a + 3 ox' ~1 5x a'+ + 7ua6 ~a' In this table, the quantities in the right hand column are called the developments or expansions of x+ce raised to the first, second, third, &c., power, and the corresponding quantities in the left hand columns are the expressions for these powers. By examining these expansions, we may make the following observations: I. In the first term of the expansion, the first term of the binomial is found raised to the given power, and the exponents of x decrease by unity in each successive term of the expansion. The first term of the binomial is found in each term of the expansion except the last. II. The second term of the binomial is found in each term of the expansion except the first, and its exponents increase by unity in each successive term. In the last term of the expansion, the exponent of a is the same as that of the given power. 116 INVOLUTION. III. The co-efficient of the first term in each expansion is unity and the co-efficient of the second term in any expansion is the exponent of the power to which the quantity is to be raised. Thus, in the fourth power of x+a, the co-efficient of the second term is 4, in the sixth power, 6; and so on. IV. The co-efficient of any term may be found by multiplying the coefficient of the preceding term by the exponent of x in that term, and dividing the product by the exponent of a increased by unity in thait term. Thus, to find the co-efficient of the fifth term in the expansion of the sixth power of x+a, we multiply 20, the co-efficient of the preceding term, by 3, the exponent of x in that term, and divide the product, 60, by 4, the exponent of a in that termn increased by unity, and obtain 15 for the co-efficient of the fifth term. V. All the signs of the terms in the expansion of the successive powers of x-a are positive. The development of any power of x-a is the same as the development of the same power of x ta, except that the signs are alternately positive and negative. Thus, (x —a)'= x-3x2a + 3xa -a'. EXAMPLES. 1. What is the third power of 3a+2c? In this exatnple; let m=3a, n=2c; then 3a+ 2c=m,+n, (3a + 2c)3 = (mn n)3'=ms +- 3m2n + 3mn2 n3. Since, m= 3a (1) And n= 2c (2) 7 e3=27' (3)=(1): -We have, 1 in 9 (4)-(l) sc ~CS )=(5 2) n2 4c0 (6) = (2) Substituting these values of min, i, ns, n, in the expansion of m+n, we filnd that, (m-n)3 = (3a+2c)3= 27ai+3X9a X 2c+3X3aX4c2 —S8c - 27a+54a2c + 36ac2 - 8c'. INVOLUTION. 117 2. What is the fourth power of a+c?.Ans. a4 4ac + 6a2c + 4ac -c4. 3. What is the square of 4ax+x+1? Ans. ] 6a2x2+8ax2+8ax+xZ+2x+ 1. 4. What is the sixth power of (a+ x)z?.Ans. a +- 3a2z +- 3axz + xS. 5. What is the square of a + b + c? Ans. a2 +- 2ab+ 2c + b2 +- 2bc +c2. 6. What is the cube of a — 2c? Ans. a3 -6a' c + 1 2cc2 - 8c. 7. What is the sum of the cubes of a +c, a-td a-c? Ans. 2tt + 6ac2. 8. What is the sum of the fourth powers of x + y, and x-y I Ans. 24 +. 12x2y 2+2y4. 9. What is the sum of the fifth powers of a +c, and (a-c)? Aszs. 2a5 - 20a3ca +- 10ac'. 10. What is the sixth power of (a + c)? Arns. (a +c)2=a~ + 2ac + c. 11. What is the second power of 9x +- As. Slxq+18 +~2. 12. What is the second power of 4 + x? Ans. 16 + 8xZIx. 13, What is the square of x —? a Ans. x2 — - (4_ 4)-+ -a -14. What is the sqdare of y4+ 3+x Ans. Vi -4X + 9X. 118 EVOLUTION. 15. What is the cube of a +ci. Ans. a + 3a3c~ + 3acI+c. 16. What is the fourth power of ac+?~ Ans. a2 + 4a c2+ 6ac + 4a2c2 + C2. EVOLUTION. (88.) Evolution is the extracting of roots. It is the reverse of Involution. The square root of any quantity may be denoted by the index ~. For, the square root of any quantity, as a, is a quantity which multiplied by itself will produce the given quantity. Now, a2 X a= —a, by the principle of multiplication. Hence, the square root of a may be written au. In general, the nth root of a is a ", since the product of n factors, each of which is a", is a. We see, thereflre, th:t ioots, a:s well as powers, may be correctly represented by exll-o(- its. On this principle we may denote a root of any quantity, by enclosing this quantity within the parenthesis, and then regarding g it as a single quantity. Thus, the square root of a'22ac- c2 mlay be wlritten, (a2+2ac+c2) Evolution may be divided into three cases. CASE I. To extract any root of a monomial. RULE. Extract the root of the numeral co-eficient; and then extract the root of the literal part by dividing the exponent of each letter by the denominator of the fraction, which is the index of the required root. EVOLUTION. 119 When the root of a quantity cannot be accurately found, we may indicate it by a fractional exponent. Thus, the cube root of 4as may be indicated by enclosing it in a parenthesis, and then treating it as a single quantity. It is written (4a')~. (89.) Since all even powers of a negative quantity are positive, it follows that all even roots of a positive quantity may be either positive or negative. All odd roots of a negative quantity are negative. An even root of a negative quantity is impossible. For example, we cannot take the square root of — a, since no quantity multiplied by itself will produce — a. EXAMPLES. 1. What is the cube root of 64a6b3'? In this example, the cube root of the co-efficient is 4, and dividing the exponents of the letters by 3, the denominator of the fractional index which denotes the cube root, we have a'b for the cube root of the literal part. Hence the root is 4a2b. 2. What is the cube root of 27a'c9. Ans. 3ac' 3. What is the square root of 625a"ac 8 Ans. 25a'c. 4. What-is the seventh root of 128al'4s 2 Ans. 2a'x7. 5. What is the fourth root of:-1/Ve2 Ans. c6. 6. What is the square root of 64a~. Ans. 8ak. 7, What is the fourth root of 16a"6c? Ans. 2a2c. 8. What is the fifth root of 32a'5c"? Ans. 2a'c. 9, W hat is the sixth root of 729a"c's? Ans. 3a2 Vc. 120 EVOLUTION. 10. What is the square root of 49m~n~ l Ans. 7mn6..1. 11. What is the cube root of alra'b2 8 Ans. iab6. 12. What is the square root of 100a4x5. Ans. 1Oa'xk. CASE II. To extract the root of a polynomial. We shall only show in this place how to extract the square and cube roots of polynomials. To extract the square root of a polynomial. (90.) Since the square of a+b is a2+2ab +b, it follows that a +b is the square root of a2' + 2ab + b'. Now the sqluare root of the first term a2, is a, the first term of the root. If wre subtract a2, the square of the first term of the root, friom Root. as + 2ab + b2, we have for a remainder, 2ab ac2 + 2ab + b2(a + b +b'=(2a+b)Xb. This remainder is a2 composed of two filctors, one of which, b, 2a+b)2ab+ bz the second term of the root, may be ob- 2ab +6 tained by dividing the first term of the re- O mainder, 2ab, by twice the first term of the root. If we add b to 2a we obtain 2a +b, the complete divisor of 2ab + bP. This complete divisor, multiplied by b, the second term in the root, and the product subtracted fiom the remainder, finishes the operation. If the root has three terms, it may be represented by a+b+c, and its square is (a + b)'+ 2c(a+b) + c. Now, we can find the root of the first part, (a-tb)'=a'+2ab+b', as above, and then find c from 2c(a+b) in the same way that b was derived fiom 2ab, in the last example. It may be shown that, in a similar manner, the root of a polynomial may be found, which has four or a greater number of terms. EVOLUTION. 121 Hence, the square root of a polynomial may be found by the followinll RULE. I. Arrange the polynomial according to the pouers of one of its letters, as in d&vision. II. Extract the squajre root of the first term, and subtract its square from the polynomial. Bring down the next two termns of the polynornial for a dividend, aid divide the first te m of the dividend by twice the first term of the root ju t jbund. The quotient will be the second tem maf the root. Add the second term of the root to twice the first term of the root, and multiply the sunm by the'ecotld te, m of the 3 oot, and subtract the product from the dividend. To the remainder add two or more terms of the polynomial, and proceed as before. EXAMPLES,. 1. What is the squa;re root of a4'+ 4a3c + 6a2b2 + 4acs + c4? Operation. Root. a4 + 4a3c + 6a2c2 + 4ac3 +- c4(al -+ 2ac + c2 a4 2a2 + 2ac)4a'c + 6a2c2 4 ac + 4a'2c' 2a2 + 4ac + c2)2a2c +2- 4ac3 +c4 2aL2C2 + 4ac9 + c4 0 2. What is the square root of 16a2xa + 8ax' + 8ax + x2 + 2x +1? Operationl. Root. I 6aax2 + 8ax2 + 8ax+xa 2x + 1(4ax+x+ I 1 6ax2 8ax +x)8ax2 + 8ax + x2 8aX2 X2 8ax+ 2x-+ 1)8ax +2x+ 1 8ax +2x+ 1 0 122 EVOLUTION. 3. What is the square root of x4-ax3+a2 2? Ans. x2-6ax. 4. What is the square root of 9x'- 30ax-3a2x- + 25as a4 a' +1? Ans. 3x-5a — 5. What is the square root of 4x4+ 8ax3+ 4ax2+ - I 662x2+ 16ab2x+ 16b4' Ans. 2x2+- 2ax+ 4b2. 6. What is the square root of 49x4-28x3" — 7x+6x+-.-? Ans. 7x —2x —,. 7, What is the square root of 4x4 —16x3+24'- 16x+4? Ans. 2x2 —4x+2. 8. What is the square root of 16x4 +24x3 + 89x2 + 60x+ 100? Ans. 4x2 +3x+10. 9. What is the square root of 9x8- 12x5 + 10Ox4- 28x3 + 17x' -8x +- 16 a Ans. 3x3- 22+x-4. 10, What is the square root of 4a4 —4a3 +- 5a — 2a 1? Ans. 2a2 —a+1. 11.'What is the square root of 4aa- 4am - 4an + m2 + 2mn +?n a Ans. 2a-m-n. 12. What is the square root of 4a+4an n2- 4a —2n +-1 8 Ans. 2a+n-1. 13. What is the square root of 9a2 - 24ac + 6a + 16c2-8c + 1? Ans. 3a-4c+1. 11. What is the square root of 4x4 + 4ax + 4x2+a +2a + 1? Ans. 2x+a+1. EVOLUTION. 123 To extract the cube root of a polynomial. (91.) Since the cube of a+b is a'+3a'b+3ab'+b', it follows that the cube root of as +3a'b + 3ab' + b3 is a + b. Now, the first term of this root, is obtained by extracting the cube root of the first term of the polynomial, a3 + 3a26b+ 3ab2 + b3. Subtract the cube of the first term of the root from the polynomial, and the remainder is 3a'b+3ab2+b3. By resolving this remainder into its factors, we find that, 3a2b + 3ab' + b3= (3a + 3ab + b) Xb If, then, we divide the remainder by 3a + 3ab +b', we shall obtain the second term of the root. We may call 3a +3ab6+b' the true or complete divisor of the remainder. We may also observe that the second term of the root may be obtained by dividing the first term of the remainder, 3a'b, by three times the square of the first term of the root, that is, by 3a. We will call 3a' the trial divisor, since it may be used for determining the next term in the ioot. For convenience in calculation, the trial and complete divisors may be formed in the following manner: Write the first term of the root in a column, designated Column I., and its square in a column designated Column II. Add the first term of the root to the term in Column- I., multiply the sum by the first term in the root, and add the product to the term in the second column. The sum thus obtained is the trial divisor. The second term in the root is obtained by dividing the first term of the remainder by the trial divisor. Now add the first term in the root to the last term in Column I., and to this sum add the second term in the root, and the latter sum is the next term in Column I., and if this term be multiplied by the second term in the root, and the product be added to the trial divisor, the sum Awill be the true or complete divisor. Multiply the true divisor by the second term in.the root, and subtract the product from the first remainder, and the work is finished. We have, then, for extracting the cube root of a' + 3a'b + 3ab2 + b3, the following 124 EVOLUTION. Operation. COL. I. COL. II. a3+3a2b+3ab+b'a a+ b a ac as a 2a2 3a2b + 3ab' + b =Remainder. 2a 3a= —Trial divisor. 3a2b + 3ab' + b3 a 3ab+b' 0 3a 3a2 + 3ab + b 2=True div. +b 3a+b If the ioot has three terms, it may be represented by a +b +c, and its cube may be written (a+b)3 +3(a+b)3c+3(a+b)c2+cS =a' + 3a2b + 3ab2+ b'+3-(a + b)c + 3(a + b)c' + c'. Now, we may return to the root of this polynomial by finding the first part of the root, a+b, as before, and then deriving c by means of a+ b, in the same way that b was derived by means of a. In a simniar manner, we may find the cube root of a polynomial, when its root has four, or any number of terms. From what has been said, we conclude that the cube root of any polynomial may be found by the following RULE. I. Arrange the terms of the polynomial with reference to the powvers of oine of its letters. II. Extract the cube root of the first, or left hand term, for the first term of the required root. Write the first term of the root in a column, designated Column I., and its square in a column, designated C(;lumn II. Ml~ultiply the term in Column II. by the first term in the root, and subtract the product from the given polynomial, and take three terms of the remainder for the FIRST DIVIDEND. III. Add the first term in the root to the term in Column I., multiply the sum by the first term in the root, and add the product to the term in Column II. The sum thus obtained is the FIRST TRIAL DIVISOR. Add the first term in the root to the last term in Column I., and the sumn is the next term in this column. EVOLUTION. 125 IV. Divide the first term of the first dividend by the trial divisor, and the quotient is the next, or second teim of the root. Add the second term of the root to the last te) m in Column I., for the next term in this column, and then multiply this te; m in Column I. by the secoJnd ter m in the root, and add the product to the first trial divisor. The sum thus obtained is the FIRSTr TRUE DIVISOR. Multiply the true divisor by the second te m in. the root, and subtract the product from the first dividend. To the rema.:'nder annex the next thJee teirm., of the polynomial for' the SECOND DIVIDEND. TO the last te,? c, CUolumn I., crld the last term in the root for the next term in Column I., (and,,lt p.ly this term of Column I. by tie lust teirm in the root, (cd add the pproduct to the Jfirst t'rue divisor. The sum thus obtained is thle SECOND TRIAL DIVISOR. Then, with the second tr'ial divisor, and the second dividend, proceed as before.* EXAMPLES. i, What is the cube root of 8a + 1 2a2c + 6ac' + c 2 Solution. Root. CoL. I. COL. II. 8a' + 12a2c+-6ac2 -c3'12af c 2a 4aa 8a3 2a 8a2 1 2a2C + - ac2 +c' 4a 12a2=First trial divisor. 1 2a2c + 6ac2+c' 2a 6ac+c2 0 6a 12a 2+ 6ac + c=First true div. +c 6a + c If, in obtaining the several terms in Column 1. and Column II., we perform the additions mentally, these columns will assume a more compact form, as represented in the following operations. NOTE.-From the rule here given for extracting the cube root of a polynomial. we might easily deduce a rule for extracting the cube root of any number. See Hutton's Mathematics, late edition, and Perkins' Higher Arithmetic. Operation. I -4 COL. I COL. IL 27c'-54cB+63c'-44c3+21c2-6c+ 1(3c- -2c+1 + 3cs 9c4 2 7c 6c2 27c' -- 54c5 + 63c — 44cs 9c2 27c4 — 18cS + 4c — 54c5- 36c4 — 8c3 9c —2c 27c4-36c3 +- 12c2 27c4 — 36c3 + 21c2-6c+1 9c'2-4c 27c-86c3+21c2-6c+ 1 27c4-36c- 21c2-6c+1 9c —6c o 9c2 — 6c+1 1o O0 -3. 3. What is the cube root of c6+6c —40c+- 96c-64? - COL. I. COL. IL c +6c~-40c+ 96c-64(c2+2c —4 c C2 C Ca 2c2 3c4 6c6- 40c3 + 96c + 3c2 3c4 + 6c3+ 4c' 6c5+ 12c4 8c3 3C2 2c 3c4 +- 12c3 + 12c -- 2c4-48c3 +- 96c —64 3c2+4c 3c'+ 12c3-24c 16 -12c —48c + 96c-64 3c + 6c 0 + 3c+- 6c-4 t~ RADICAL QUANTITIES. 127 4. What is the cube root of 8a6-12a5+18a4 —13as+9ae -3a + 1 Ans. 2a2 —a+1. 5, What is the cube root of 2 7c' + 2W7c + 9c + 1? Ans. 3c+1. 6. What is the cube root of 8x + 36x2+54x+27? Ans. 2x+ 3. 7. What is the cube root of xe + 6x —40x'+ 96x-64? Ans. 22 +2x-4. 8. What is the cube root of x6-6x5+15x'-203x+15x2-6x +1 Ans. x2-2x+-1. 9. What is the cube root of a'+-6a2c + 12ac2 + 8Sc3 Ans. a +2c. 10, What is the cube root of 8c'-36c'+ 54c-27? Ans. 2c-3. 11. What is the cube root of x3 + 6x- + 12x + 8? Ans. x+2. 12. What is the cube root of 8a'- 84a2x - 294ax2 —343x.9 2 Ans. 2a-7x. 13. What is the cube root of x —6cx'+ 12c2x4-8c'x? Ans. x -- 2cx. RADICAL QUANTITIES. (92.) A RADICAL QUANTITY is one that is affected with a radical sign or fractional exponent, without which it cannot be accurately expressed. Thus, the expressions, /3, /2a, (2a + c")j n4, c5, are radical quantities. They are also called surds and irrational quantities. The square root of a quantity is a radical'of the second degree, the cube root of a quantity is a radical of the third degree, and, in general, the nth root of a quantity is a radical of the nth degree. 128 RADICAL QUANTITIES. (93.) Radical quantities are similar when each is affected with the same radical sign, and the quantity under the radical is the same in each. Thus, 5a V, 3c V-2b, are similar radical quantities. The expressions, (3a + c)T, 4c(3a + c) are dissimilar radicals. REDUCTION OF RADICALS. (94.) The reduction of radicals consists in changing their form without altering their value. There are three cases. CASE I. To reduce a rational quantity to theform of a radical. RULE. Raise the quantity to a power denoted by the degree of the radical. and then affect it with the sign of the required radical. EXAMPLES. 1. Reduce 7c to the form of cube root. In this example, the radical is of the third degree. Hence, we must raise 7c to the third power, and affect this power with the sign of the cube root. (7a)3=243a'. By extracting the cube root we have 7a-(243a')*; or it.may be written, 7a-=243a. 2. Reduce 3a to the form of the fourth root. Ans. (81a'4)4 3. Reduce 5a} to the form of the square root. Ans. v.d 4. Reduce 4cy to the form of the cube root. Ans. (642)3. 5. Reduce 45c0 to the form of the cube root. Ans. (91125c3). RADICAL QUANTITIES. 129 6. Reduce 4c0 to the form of the nth root. Ans. (4nc2n)n. 7. Reduce 16cs to the form of a radical whose exponent is 3. Ans. (164c4 = ~a(8C 8. Reduce l 25c' to the form of a radical whose exponent is a. ia 9. Reduce 32ab 9 to the form of a radical whose exponent is. Ans. (32ab 5]) 6 = (8a5b3) 110. Reduce 64a 5 to the form of a radical whose exponent is -. Ans. (32a2)5. 11. Reduce a 5x, to the form of the nth root. Ans. (a252. _) n. 12. Reduce 8c3x to the form of a radical whose exponent is i. Ans. ( 312ce) CASE II. (95.) To reduce radicals which have different exponents to equivalent ones having a common index. RULE. Reduce the exponents to fractions having a common denominator, and then raise each quantity to a power denoted by the numerator of its exponent, and the reciprocal of the common denominator will be the common index of each. This rule depends upon the obvious principle, that the values of the exponents are not altered, and therefore the value of the radicals themselves are not changed. 6*~~~~~ 130 RADICAL QUANTITIES. EXAMPLES. 1. Reduce 2~ and 41 to radicals having a common index. In this example, the exponents reduced to fractions having a common denominator become -- and 13. Hence, 23 =2- =16-; and 4441 2 —6412 2. Reduce 22 and 3~ to radicals having a common index. 1~.1. Ans. 86, and 96 3. Reduce 23 and 34 to radicals having a common index. Ans. (256)-, and (27)"h. 4. Reduce 3- and 21 to radicals having a common index. Ans. 38, and 48 5. Reduce 6~ and 41 to radicals having a common index. Ans. (36), and (64). 6. Reduce 3(3)4 and 212 to radicals having a common index. Ans. 3(27)-2, and 21. 7, Reduce a1 and bn to radicals having a common index. Ans. (a'n)m, and (bm)M. 8. Reduce m' and n6 to radicals having a common index. Ans. (Mn4)_f, and (n3) 19, Reduce 45 and 38 to radicals having a common index. Ans. (48) 5, and (35)T-. 10. Reduce a + and (a-ec)~ to radicals having a common in4ex. Ans. (a'+2ac+ )", and (a-c)+. RAI)ICAL QUANTITIES. 131 11. Reduce (a+c)' and (a-c)2 to radicals having a common index. Ans. [(a +c)'], and [(a-c)]3. 1 1 12. Reduce 22 and 35 to radicals having a common index. Ans. (32)1T, and (9)TI CASE III. (96.) To reduce radical quantities to their simplest formqs. Before giving the rule for this case, we will establish the following proposition on which it depends. PROPOSITION. The product of the nth roots of any number of quantities is equal to the Math root of their product. Take two quantities, a and b, and the demonstration will be the same for any number. Then we are to prove that, VaXq/b — v,b; or, what is the same, a' Xb'=(ab)n. Let P relpresent the product of their nth roots. Then P =Va X V. (1) By raising each member to the nth power, we have P=naXb —ab (2) By extracting the tntlh root of equation (2), we have p =v-b (3) By equating iight-hand mernbers of equations (1) and (3), we have V/aX Vb_=Vab; hence, the proposition is established. We will now give the rule for reducing radicals to their simpllest forms. RULE. Resolve the radical into ttco such factors that the root of one of them may be accurately taken; then find this root, and multi)ly it by the other factor. When the radical cannot be resolved into two such factors, it cannot be made to assume a more simple form. 132 RADICAL QUANTITIES. EXAMPLES. 1. Reduce V147a2x3 to its simplest form. We can resolve 147ax23 into the two factors, 49ax2 X 3x, one of which is a perfect square; therefore, /147a2x3=- V49a2 2X V3x'=-7ax/3x, since the square root 49a2z2 is 7ax. 2. Reduce V/l08a2c to its simplest form. Ans. G6aS/3c. 3. Reduce V/192a4c3 to its simplest form. Ans. 8a'c / 3c. a, Reduce V/48a2x to its simplest form. Ans. 4a V/3x. 5, Reduce va2'a'x to its simplest form. Ains. as /l-ax. 6. Reduce V243c2n3 to its simplest form. ilns. OcnV V3n. 7, Reduce V 64a 5c to its simplest form. Ans. 4ac2Va c-2. 8. Reduce V/54a2+9a2c to its simplest form. Ans. 3aV6 +c. 9. Reduce /V64 a'+32x6 to its simplest form. Ans. 2V/2a7 +x6. 10. Reduce /J- to its simplest form.* Anls. ~/6. Am. A * NoTE.-When the radical is a fraction, multiply both terms of the fraction by some quantity that will render the denominator such a power that the required root may be taken. Thus, in this example, 3X33 6-s RADICAL QUANTITIES. 133 11. Reduce /-2 to its simplest form. Ans. 1/15. 12. Reduce ( to its simplest form. Ans. 1 (a-2-x2)~ a+x ADDITION AND SUBTRACTION OF RADICALS. (97.) The addition and subtraction of radical quantities may be performed by the following RULE. Reduce the radicals to their simplest form; then, if the radicals are similar, add or subtract their rational parts, and to the sum or difference annex the cbmmon radical part; but if the radicals are dissimilar after the reduction is made, they can only be added or subtracted, by waiting them one (ffter another, with their proper signs. EXAMPLES. 1. What is the sum of /147a2c and V/48Ca'c? By reducing, v/147a2c= /49a2X V/3c= 7aV/3c And, / 48a2c= — /1aX V/3c= 4a /3c their sum = 1 la a/ 3c. 2, What is the sum of tV27atc and V/12a2c? Ans. 5a V/3cc. 3, What is the sum of V/243ab2 and'/192ab'? Ans. 1 7b /3a. 4, What is the sum of V500 and V/108? Ans. 8V/4. 5. WThat is the difference of V/jx and Vb2x? Ans. (a-b) Vx, 134 RADICAL QUANTITIES. 6. What is the difference of 5 V/20 and 3 V45? Alns. V5. 7. What is the difference of V108ax2 and /V48axr2 2 Ins. 2x V3a. 8. 1W1hat is the difference of nm4/1 + )2 and 0 2hat ithsm 3 TV/l A+ns ins. (-' // +) xt 9. AWhat is the sum of 3 V- and 2 v-L;-? Ans. AV10. 10. What is the sum of 3V/32 ancd 2V/ 2 a? Ans. 6V4 ~ 0V2. 11. What is the sum of v/27 and 3 V/75? Ans. 18 /3. 12. WThat is the sum of V/2ax2" 4act x~-4.2a aind 2ct6x2~4ax~2at? inzs. 2xcV 9a. 13. Wlhat is the sum of Vz and V4,-? 11. What is the sum of a +Q-) ( and * NOTE.-Tllis is the length of "the longest straight pole that can be put up a chinmney, when the heighlt from the floor to the mantel=a, and the depth from the front to the back=b." See Hutton's Mathematics, Dif. Cal. page 726. RADICAL QUANTITIES. 135 MULTIPLICATION OF RADICALS. (98.) The multiplication of radical quantities may be performed by the following rule, which is founded on the proposition that was demonstrated in Art. 96. RULE. Reduce the radicals so that they shall be of the same degree (Art. 95); and then annex the product of the rational parts to that of the radical parts. EXAMPLES. i. Multiply Vf8 by V48. I 3- I1I I I Here, /8=82-=86-(83)6 =(512) =(64X8) — 2(8). And, V48:(48)3- =(48)t —(482) =(48 X 48)6 — (64 X 36)6' =-2(36)6. Therefore, 8 X, 3 —- 2(8)' X 2(3G6) — 4V288. 2. Multiply V/18 by /48. Ans. 12/6. 3. Multiply V'24a2x by V/12x. Ans. 12a /2. 4. Multiply aa by b2. Ans. Va2b3. 5, Multiply 3V/4 by 4/3. Ans. 12V4-432. 6. Multiply aL" by b". Ans. (a"lbf")... 7, Multiply 2V/3 by V 72. Ans. 2V'6. 8. Multiply (a + /c) L bya — Vc) Ans. (a2 —c)j 9, Multiply 4a2 by 13a3. Ans. 52Va*. 10. Multiply the cube root of 3 by the square root of 2. Ans. V72. 136 RADICAL QUANTITIES. 11. Multiply Va —'a —x by Va+ Va-x, Anr. x. 12. Multiply (a+b)l by (a+b)~. Ans. V(a + b). DIVISION OF RADICALS. (99.) Before giving the rule for the division of radicals, it will be necessary to establish the following proposition, on which the rule depends. PROPOSITION. The quotient of the nth roots of two quantities is equal to the nth root of their quotient. Let a and b represent the two quantities; then will Va —n V =Va b. Let P=their quotient, Then, p-=a (1) By raising each member to the nth power, we have p.= a (2) By extracting the nth root of each member of equation (2), we have n p=_/a b; hence the proposition is b established. RULE. Reduce the radicals so that they shall be of the same degree; and then annex the quotient of the rational parts to the quotient of the radical parts. EXAMPLES.. 1, Divide 4 v3 by 33V4. Here, 4V3=4X3-4X36=4X(27)' 3V4=3 X 4=3X 46=3 x (16) Whence, 4V3 3- V4(a = )'1 41/ 35~4 = le RADICAL QUANTITIES. 137 2. Divide V/a2x —a2x2 by a. VX-x2. 3. Divide V96a, by 4a. Ans. V6. 4. Divide V1024a2b by 8a. Ans. 4 Vb. 5, Divide V324a2 by V/9a, Ans.'144a. 6. Divide ( 7-2+ V32-4) by V8. Ans. 5 — 2. 7. Divide /ab2 —b2c by Va —c. Ans. b. 8. Divide /64 by 2. Ans. V2. 9, Divide 4/V12 by 2 V3. Ans. 2VI-6. 10. Divide V/12 by V24. Ans. V3. 11. Divide (144a32a)I~ by (6ax)~. Ans. (4a)T-'. 12. Divide (a+b)~ by (a' —b2)~. Ans. (. +b )a+ EXTRACTION OF THE SQUARE ROOT OF A BINOMIAL SURD. (100.) A binomial surd is a binomial, one or both of whose terms are surds. Thus, a4- V6, V8a — V/b, 5 - V/7, aare binomial surds. The square root of a binomial surd may sometimes be found to be another binomial surd. For example, the square root of 6+ V/20 is 1 +2 5. Before we can obtain a formula for extracting the square root of a binomial surd, it will be necessary to establish the following lemmas. 138 RADICAL QUANTITIES. LEMM5A I. The square root of a quanttity cannot consist of the sum or difterence of two quantities, oze of which is rational and the other irratilonal. For, if it be possible, let us have the equation vc'=b+ f'c (1) By squaring, a-=b2-2b /-cfc (2) By transposing, &c., 4- /c; that is, an irra2b tional quantity equal to a rational quantity, which is absutd. LEMMAA IT. If we have any equation, which has a rational and an irrational quantity in each member, the rational and the irrational quantities are equal ecch to each. If we have the equation ai Vb=.x l /y; (1) Then will a=x (2) And ~L V/ b=:- Vy (3) If a be not equal to x, make x=a+c (3) Substitutin, this value of x in equation (1), we have a+ V/b=a +c~+ /y (3) ~.. i Vb:- c V_'y, which, by Lemma I., is impossible; whence, a=x, and -vb- / Vy. LEMMIAA III. If x/a + V/b:x+y; thlen /a — b — x-y; where x and y are both supposed to be radicals of the second degree, or only one of thetm. For by squaring the equation, V y /a + b=x+y,We have a+ V/b=x2 + 2xy+ y2; Whence V/b = -x2 +2xy + y2 — a RADICAL QUANTITIES. 139 In this last equation, x2, y2 and a are rational quantities, and therefore 2xy must be an irrational quantity; for, if it were not, we should have an irrationcal quantity equal to a rational quantity. But, by Lemma I., there can be no such equality.. Hence, 2xy is an irrational quantity, acnd by Lemma II., we have Cl-x~+y2 +(1) an d, vb= 2xy (2) Eq. (1) —Eq. (2) gives a — /b=2 —2;yy2 (3) Whence x//a- Vb-x-y (4) (101.) We will now obtain a formula for extracting the square root of' a bilnomial surd. Assume.+y-/ + Vb (1) T'lhen, Iby L, mm IIT., ax-y=/ l- b- (2) Eq. (1) sqtuared gives x 2+Or2+ y2= —a+ /b (3) Eq. (2) squared gives Xd_2 _- _2 = a- xb (4) By adding eq. (3) to eq. (4), ai.d dividing by 2, - +y2-ca (5) Eq. (i)x eq. (2) gives x2- VI _-.b=c* (6) Wlhence, from equlations (5) and (6) x -= -c (- +) audil ZS=8/ 2' (8) and NV/-c (s) ncl - - 2-0 - C 22 V/"c _ aWVhence \/C A -ba+ (A) and 2 )2 2 * We makc V/Oa-b=-c for the sake of brevity. 140 RADICAL QUANTITIES. If the extraction of the square root of a binomial surd is possible, c must be rational, and therefore a —b must be a perfect square. We will now apply formulas (A) and (B), in extracting the square roots of binomial surds. EXAMPLES. 1. What is the square root of 11 + v/72, or 11 +6 /2 8 Iere a=11, b=-72; andc-= /a2 —b= v121 —72='7. Therefore, \/11+ V/72 _ a - c a -- 2. 2 2 2. What is the square root of 7 7+4 V3 Ans. 2 + V3. 3, What is the square root of 43 —15 V/8? Ans. 5-3V/2. 4. What is the square root of 5- V24? Ans. V3- V2. 5, What is the square root of 3 —2 V2? Ans. V/2- 1. 6, What is the square root of 28 + 5 V12? Ans. 5+V-3. 7. What is the square root of 2m + 2 Vm2-n2 2 Ans. Vmn+n +'m —n. 8, WVhat is the square root of x —2 — 1 Ans. x-1 — 1. 9. What is the square root of 28 + 10 /3? Ans. 5 + 3. 10. What is the square root of 3 —4/- V? Ans. 2 - V-1. IMAGINARY QUANTITIES. 141 11. What is the square root of 18~i2 /77. Ans. /7i V/11. 12. What is the square root of 94+42 V/5? Ans. 7 + 3 V5. 13. What is the square root of m2+ n + 2mn /n? Ans. m + Vn. 11. What is the square root of 42 +3 174? Ans.:/28 + V4. IMAGINARY QUANTITIES. (102.) Any expression which represents an even root of a negative quantity is called an imaginary quantity. (103.) We have seen that an even root of a negative quantity is impossible, and therefore imaginary quantities have no real value; notwithstanding this circumstance, they are of much use in many parts of mathematical analysis, since, when they are employed according to the rules for operating upon them, they lead to important and possible results. (104.) The addition and subtraction of imaginary quantities are performed by the ordinary rules for adding and subtracting radicals; but in the multiplication and division of them, we must pay attention to some important particulars which do not belong to other quantities. (105.) In the first place, it is plain that V/-aX V-a= -a, since the square root of any quantity multiplied by itself must produce the quantity under the radical. But by the rule for the multiplication of quantities, -aX —a=a2. Therefore, -/aX Ja-a= V a2; whence, it follows that Va2= -a. If we say that /a2= +a, then we shall have +a= -a, as the conse 142 IMAGINARY Q UANTITlES. quence. But this reasoning is incorrect, since the square root of a2 cannot be +a and -a at the same time. In this case, we know that the square root of a2 is -a, since a2 was produced by multiplying -a by -a. (106.) If we have two imaginary quantities which are different, as V/-a and V —b, we do not at once perceive whether their product, V'ab, is to be talen neyatively or p2ositively. But, by observing that V/-a= / X V — 1, and that V- V b= v/bX -.1!, their product is equal to VtaX V/- X - bX /- - Vab X ( Vt-1)= -/bX — 1= - /ab. Hence, it appears that the proper sign of the product, /ab, is minzzs. (107.) From what has been said, and firom the principle in regard to the signs in multiplication, we may form the following table for the multiplication of imaginary quantities. (+ -a) +,x( ~-c) —- (- (1) ( - V-i) x ( /-V() -a (2) ( -t I/-ca) x (+ V/-)- - - a b (3) (- /-C)x(- "/-i)= —,/ (4) (4) + V/ —) X (- V',) + V'Z~ ~) I.:X.\ B, LES. 1, Multiply 3o V _.i 1,Y 4 29 v, 12 —4= -o'2. Multiply 2 V3 -V' 5 by 4 V3' 2 V/-. Ans. 14-8 —15. IMAGINARY QUANTITIES. 143 3. Multiply 9+Gv'/ — by 3+7 v/-1. Ans. -15+81 V-1. 4. Multiply 4 / —5 by 5 V-1. Ans. -20 /5. 5. Multiply -5 /-3 by — 3 — 4. Ans. — 1J3V2. 6. Multiply 8 + — 5 by V —3" lj)S s,/3- 3 /T15. 7. Multiply 2 + N/- 3,:, — +- "'5. As. 8+ 4 -3+ 2 v/5-'15. (108.) Thle qu.cticnt of two radicals which have the same sign, and which arise from taking the square root of a negative quantity, is equal to the square root of their quotient. That is, +Vz-c- acX1 /; _ Vb b X./-a_- V'; _ =,/ And V -/ — / If the two imaginaries have not the same sign, it is plain that their quotient will be minus the square root of their quotient. EXAMPLES. I. Divide 10 / —1i4 by 2 V/ —7. Ans. 5 2..2 Divide 6 + / — 2 by 6 — /-2. Operation.,d+ v-_2 6 —2 Multiply both the numerator and the denominator by the numeator,; and we have 36+12v'-2-2 17+6V2 — 36+2 19 144 IMAGINARY QUANTITIES. 3. Divide 3+2/-1 by 3-2 / —-1. Ans. -13(5 + 1 /-). 4, Divide 4 + v —2 by 2- / —2. Ans. 1+ -/2. 5, Divide 1 + V-1 by 1-/-1. Ans. V-i1. 6. Divide - 1- IV by -6 /-3. Ans. +6v/3 CHAPTER VI. ON THE SOLUTION OF PURE QUADRATICS AND OTHERS, WHICH MAY BE SOLVED WITHOUT COMPLETING THE SQUARE. (109.) AN equation of the second degree, which contains only the second power of the unknown quantity, is called a pure quadratic. Thus, ax2 —6x'=c+10 is a pure quadratic. (1 10.) When the terms of an equation involve only the square of the unknown quantity, its value may be readily found, since we may so reduce the equation, that the square of the unknown quantity may constitute one member, and known quantities the other; then, by extracting the square root of each member, the value of the unknown quantity is determined. Since the square root of a quantity is positive or negative, it follows that the unknown quantity has two values in a pure quadratic. These two values are numerically equal, but they have contrary signs. In practical problems, it can generally be determined, which of these values or roots should be taken. (1 1 1.) No general rule can be given for solving such examples as will be found in this chapter. The student must rely- on his own ingenuity, and a thorough acquaintance with the preceding chapters. If he will carefully study the solutions which are given of the following examples, he will get an idea of the methods to be employed in reducing such equations. EXAMPLES. 1, Given V.x + 16-=2 + V/x to find.x. 7 146 QUADRATICS. By squaring each member, we have, x+16=4+4 V-+x (1) By transposing in (1), 4'x=12 (2) Or, Vx= 3 (3) Whence, x= 9 (4) Vx+ /b a 2. Given to find x. VX-,- b By clearing of fractions, we have, b Vxq+b Vb'=a /'x-a Vb (1) By transposing in (1), (a- b).V /= V/b.(a + b). (2) By squaring (2), (a —b).x=b.(a+b)2; (3) Whence, x= (a -b)2 (4) (a - b)2 I1 / /4 9 3. Given -+-=V -+2/a- + to find x. x a a ax x By squaring each member, we have, 1 2 1 I 9 -ta~+~=~+ + $- 2 33; (1) Whence 1 +a 2 —=; (2) a2 ax a"x2 x 1 4 4 4 9 Squaring(2), -+ - 3 + a a a () q4 ax3 a2=2 a2(2 4 8 By transposing in (3),,x —=x- (4) Whence x-2ac (5) I x2-ny=54 (1) 4. Given xy-y2=18 (2) to find the values of x and y. Eq. (1)-eq. (2) gives x2-2xy+ y= 36 (3) Sq. root of eq. (3) gives x-y= 6 (4) Eq. (1)+eq. (2) gives x2-_y2=72 (5) Eq. (5)-eq. (4) gives xz y-12 (6) Eq. (4) + eq. (6) gives 2x —18 (7) Eq. (6)-eq. (4) gives 2y= 6 (8) Whence 9, and y=3 QUADRATICS. 147 y 2 ien{ x (2)' to find the values of x and y. 1 2 1 1 Eq. (1) squared gives T +- -~=- (3) y 4 4 2 Eq. (2)X 2 gives - (4) Xy 9 Eq. (3)-eq. (4) gives 1 2 (5) XT _y y2 36 I 1 1 Sq. root of eq. (5) gives 1 (6) x y 6 2 2 Eq. (1)+eq. (6) gives (7) x 3 2 1 Eq. (1)-eq. (6) gives -2- (8) y 3 WVhence x=3, and y=6 6. Given x'+y+xy(x+y) —68 (1) to find the values tXs+y'f-3- ax y'~ - 12 (2) of x and y. From eq. (1) 2 +y +2y + y2= 68 (3) From eq. (2) x'+-y3-3(x2+y2)= 12 (4) Eq. (3) X 3 gives 3(x2 +y2) + 3x2y + 3xy2= 204 (5) Eq. (4)+(5) x3+3x2y+33y2 + y3= 216 (6) Cube root of eq. (6) gives x+y= 6 (7) Eq. (7) squared gives 2+ 2xy +y2= 36 (8) The value of x + y put in eq. (1) gives x2 +y2+6xy= 68 (9) Eq. (9)-eq. (8) gives 4xy= 32 (10) Eq. (8) —eq. (10) gives x —2_2xyz-y2 4 (11) Sq. root of eq. (11) gives x-y- 2 (12) Eq. (7)+eq. (12) gives 2x= 8 (13) Ex. (7)-eq. (12) gives 2y- 4 (14). =4, and y=2 148 QUADRATICS. x'2~xy y=7 (1) 7. Given +Y y () to find the values of x X — Y+y (2) y and y. Multiply both terms of the first fraction by x -y, and both terms of the second fraction x+y, and equations (1) and (2) become X Y/ (3) x3 -y3__ and 9 (4) X2 -y2 2x'2 Eq. (3) +eq. (4) gives -2 =16 (5) 2y3_ Eq. (4)-eq. (3) gives 2 2 () Eq. (5)-.-(6) gives 3= 8 (7) Whence, -= 2 (8) Y Whence, x= 2y, x2 =4y2, and x'= 8y3. By substituting these values of x, x2, and x3, in (5), we have 16y3 4y_ —y2= 16; WVhence, y= 3, and x= 6. Va-1 V/a —x 8,. Given --,-a to find the value of x. V/a+ Va —x Multiply the numerator and denominator of the fraction by the numerator, and the equation becomes =(8t-a-x) -a, (1) 2 Or, (,a-/ a-x)/a —ax. (2) Extracting sq. root, Va- CV-x V/('. (3) By transposing, V-_ = v'c-z. (4) QUADRATICS. 149 By squaring (4), a-2a Vx+ax=a —; (5) Whence, ax- 2a /x — -x. (6) Dividing by.z, aV /x-2a= — x. (7) Or, (a + 1) /x= 2a; (s) - 2a WVhence, X =-2a (9) 4a2 By squaring (9), C' (10) 9. Given V9x —4 + 6- 8, to find the value of x. Ans. x=4. 10. Given V/x-32=16-V/x, to find the value of x. Ans. x=81. 11. Given V4x+21= —2x+1, to find the value of x. Ans. x=25. V6x —2 4 Vx-9 12. Given, to find the value of x. 6- x+2 4V/ x+ 6 Ans. x=6. 13. Given - - - — 2, to find the value of x. /Sx+3 2 Ans. x= 5. 9x-4 15+ V/9x 14, Given _ 2=15+49, to find the value of x. /x + 2 V'x+40 Ans. x=4. 15. Given Va +x= -2Vx+ 5ax+b2, to find the value of x. a2_ b2 Ans. x —3a 36 16. Given Vx+ Vx-9=, to find the value of x. Ans. x=25. 9 150 QUADRATICS. 17. Given 2+z-x+ v/x, to find the value of x. Ans. x= —. 15 18, Given V/5 V-+'/x --, to find the value of x. V/5+x Ans. x-4. 19. Given V/x~- X- /x- V ~/ X_ to find the value of x. Als. x —-. X. 20. Given ax. c --- -., to find the vallue of x. a c-1 vx A2a x/x+ 4a 21. Given V -: ~-, to find the value of x. Vx + b VAx+ 2 Ans. (a-b 22. Given.. +.-, to find the valIue of x../.- + 4 V x+ Ans. x= 4. a t'c- x'L x 23, Given - + a = — to find thie value of x. X x b Ans. x= — I 2ab — b. 2a2 21. Given x+ V/a+x —-2 —- 27 to find the value of x. An s. x4 — ak 3/' 25. Given x 2 xxy124 to find the values of x and y. ns., y=2xy24.Ans. x=-~-2, y= —-4. QUADRATICS. 151 26, Given x 3 to find the values of x and y. Ans. x. 16, or -4-, y=4, or 1-5. 27. Given 3+yz 13 1 to fial the values of x and y. +y_3= 5 Ans. x- 27, and y=8. 28. Given + y3189} to find the values of x and y. Ans. x=5, or 4; y=4, or 5. 29, Given I + x- 2-4(x +Y)xy to find the values of x and y. x xy x- xy = 4xy Ans. x —2, y=2. 30. Given ~" t Y -I -Y 9 to find the values of x and y.., Ans. x-il, y-=:4. 31. Given {Y y -216 to find the values of x and v. xy _-y2 - 6 Ans. x=3, or -2, y=2, or -3. 32 Givren I x2y2(zx +y2)= 3600 to find the values ofx and y. Ans. x=4, or 3, y=3, or 4. 33, Given y' q- yx2y- 1053 Ans. x=~:k8, y=~4-27. ( Giy56 ) 34. Given { 16 to find the values of x and y. x-y-2 Xy Ans. x=4, or -2, y=2, or -4. 152 QUADRATICS. 35. Givenx fxi + y= 6 35, Given +Y =72 to find the values of x and y.* Ans. x=64, or 8, y=8, or 64. 36. Given (2x2+3 ~(x3\ (429 R2x3 2x- 3 13 4x2-9 to find the value of x. 4Ars. x= —. (V4x~I +* -4x 37. Given ~ + - -+=9 to find the value of x: las. x=-. 18 38. Given x2+3x —7=x+2+ —, to find the values of x. Ans. x= —-3. vx+a 2 Va 2'.x 39. Given + Gi b2._ to find the values of x. VX Vx+a /x+a a~. - a v'x+ V/x-a n1a 40. Given,to find the values of x.' x- a x -a A.2-a a( Ian)2 Ans. x=- 1-t-2n Va+x+x a —x 41. Given va+__ =b, to find the values of x. Va +x- Va-x 2ab Ans. X=_2 b2 +1 42. Given a+x+ X+2ax=b, to find the values of a. a+x 1i /26 -b2 Ans. x —a.'2b,/ b-b' NOTE.-In this and in some of the preceding examples, we may avoid fractional exponents, by substitution. Thus, let m-=xz, and n=y3; whence, m3=z, n3=y. QUADRATICS. 153 43. Given { 3xy-3x- -Y3- y} to find the values of x and y. Ans. x= 1i, y= /3. 44. Given (x4_Y4) X(x2_y2) _45x2y2 to find the vallues of x and y. Ans. x=4, or 2, y=2, or 4. 45. Given {,+ y233 to find the values of x andy. Ans. x=9, or 4, y=4, or 9. x + Vx2 — 1,2 x 46. Given, to find the real value of x. x-V Xa X2_ a la Ans. x=a." 4 2(x2~yX)X(X+y) =l XY Gi 4(-y4) X(x2 +y2)-75x2y to find the values of x andy. Ans. x= 2, y= 1. 48. Given1 ((x 2+Y) X(x -+Y)= 6 } to find the values of x and y. Ans. x=11, or 5, y=5, or 11. 49. Given x2_ 2y2=25/ y 6 to find the values of x and y. Ans. x= -16, y=4. 9( +y)' 9 ( +y)_s 8] 8 x l~ 8 y (8 Gx 7 Ito find the values 7(x-yy) 7 (x-y)~ 1 ox acyl. T4 y 4' x 9 Ans. x=a, y=A. * There are two other roots of x, both of which are imaginary. In Hutton's Mathematics the imaginary roots only are given. 7*k 154 QUADRATICS. PROBLEMS IN PURE QUADRATICS. (1 1 2.) In solving problelms in pure quadratics, it will be necessary to recollect thllat, in many cases, the product of two or more factors must only be expressed. Much also depends on the notation which is adopted. To show the necessity of these observations, we will give the solutions of a few problems. 1. Two workmen, A and B, were engaged to work for a certain number of days at different rates. At the end of the time, A, who had played 4 of tlhose days, had 75 shillings to receive; but B, who had played 7 of those days, received only 48 shillings. Now, had B only played 4 days, and A played 7 days, each would have received the same sum. For how many days were they engaged; and how many did each work, and what had each per day? Let x=the number of days for which they were engaged;.. x-4=tlhe number A worked, And x-7 — = the number B worked, 75 And ~=the number of shillings A received per day. 48 And 48~:the number of shillinYs B received per day. x-7 75(-74 )=the number of shillings A received in (x —7) days. x-4 And 48(x —4)-=the number of shillings B rec'd in (x —4) days; Whence, 75(x- 7)__ 48(x- 4) (1) x-4 x-7 ( Or,.75(x-7)2-=48(x-4)2 (2) Dividing by 3, 25(x-7)2=-16(x-4)' (3) Extracting sq. root, 5(x-7) =4(x-4) (4). x=19; hence they were engaged to work 19 days, A worked 15, and B 7 days; B received 4 shillings per day, and A 5 shillings per day. If we had actually performed the multiplications indicated in equation (1), then cleared it of fr'actions, trL.ullp,.)-lA a:d u,!ited its terms, we QUADRATICS. 155 should have obtained an equation which could not be reduced by any method yet considered. 2. It is required to find two numbers such, that the product of the greater an(d square root of the less may be equal to 48, and the product of the less and square root of the greater may be 36. Let x2 and y2 be the two numbers... 2=-48 (1) and xy12= 36 (2) Eq. (l) —eq. (2) gives -:= (3) y —x. (4) fiom (3) Substitute this vlue of y in (1), and it becomes, ~-14- 8 (5).- 4 (6) 1y= 3 (7) Hence the numbers are 16 and 9. 3, A vintner draws a certain quantity of wine out of a full vessel that holds 256 gallons; and then filling the vessel with NNwater, draws off the same quantity of liquor as before, and so on0 for four draughts, when there were only 81 gallons of pure wine left. How much wine did he draw at each time? Let a=256, and x=nurnber of gallons drawn the first time..'. a —x " " left in the cask. The cask is now filled with water, and in a gallons of the mixture thus formed, there are (a-x) gallons of pure wine. In one gallon of the mixture there is one ath of this quantity, and in x gallons of the mixture, the number drawn at each draught, there are x times one ath of a —.x gallons. lIenc e, (a-.1) - — the number of gallons drawn the a a 2d time; ax-x2 2aX -x2 And.x~-Z a+ a — a =the number drawn the 1st and 2d a time. time. 156 QUADRATICS. 2ax-x2 a2 — 2ax x2 Whence a — -- =the number of gallons a a left after the 2d draught. Reasoning as before, we find that, a2 —2axtx2 x a2x-2ax2+x3 X -= 2 =the number of gallons a a a drawn the 3d time. Whence, a C.X-x2 ax2ax'X2 + a'-3a2X +3ax2-X3 a-x x+ a- -2 a aS a =the number of gallons left in the cask after the third draught. a3- 3ax2+3ax2 —x2 x a3x- 3ax2~+3axs-X4 Therefore, - X a -=the a Ca a number of gallons drawn the 4th time. Whence, ax-x2 a2x-2ax2 x3 a'x-3a2x2 + 3ax9-x4 a —- a + 2 -- 3 +a a a a,4 —4a3x ~- 6a'x2 —4ax3 - x4 a4 —4a + 6a2-4the number of gallons left after the 4th draught. Therefore, by the question we have, a4 — 4a'x - 6a2x -- 4ax3 x- x4, -=81 (1) a4- 4a3x + 6a2x-4ax3+4-=81a' (2) Extracting 4th root of eq. (2), a-x=3ac (3) Or, 256 —x=3.256-= 3.64=192 (4) Whence, x=64, and the quantities drawn off at each time were 64, 48, 36, and 27 gallons. 4. What two numbers are those, whose sum is to the greatel as 10 to 7; and whose sum multiplied by the less produces 270. Ans. I21, and ~4i9. 5. What two numbers are those, whose difference is to tle greater as 2 to 9, and the difference of whose squares is 128? Arns. *: i 8, and ~E 14. 6. There are two numbers in the proportickl of 4 to 5, the diI ference of whose squares is 81. What are the numbers? Ans. ~-12, and 15. QUA)DRATICS. 157 7. A mercer bought a piece of silk for 324 shillings; and the number of shillings which he paid for a.yard was to the number of yards as 4 to 9. How many yards did he buy, and what was the price per yard? Ans. 27 yds. at 12s. per yd. 8. It is required to divide the number 18 into two such parts, that 25 times the square of less part may be equal to 16 times the square of the greater part. What are the parts? Ans. 10 and 8. 9. A number of b;ys:et out to rob ar orchard, each Carrying as many bags as there were boys. in all, and each bag cap,;tle of containing 4 times as many apples as there were boys. They filled their bags and found that the number of apples was 2916. How many boys were there.? Ans. 9 boys. 10. From two towns, C and D, which were at the distance of 396 miles, two, persons, A and B, set out at the same time, and met each other, after travelling as many days as are equal to the difference of the number of miles they travelled per day; when it appears that A has travelled 216 miles. Iow many miles did each travel per day? Aas. A went 36, B 30. 11. There is a number consisting of two digits, which being multiplied by the digit on the left hand, the product is 46; but if the sum of the digits be multiplied by the same digit, the product is only 10. Required the number' Ans. 23. 12. There are two rectangular vats, the greater of which contains 20 solid feet more than the other. Their capacities are in the ratio of 4 to 5; and their bases are squares, the side of each of which is equal to the depth of the other. What are the depths? Ans. 5 feet, and 4 feet. 13. Bought two square carpets for ~62 ls.; for each of which I paid as many shillings per yard as there were yards in its side. Now had each of them cost as many shillings per yard as there were yards in the side of the other, I should have paid 17s. less. What was the size of each? Ans. One contained 81, the other 64 square yards. 158 QUADRATICS. 14. A and B are two towns, situated on the bank of a river which runs at the rate of 4 miles per hour. A waterman rows from A to B and back again, and finds that he is 39 minutes longer upon the water than he would have been had there been no stream. The next day he repeats his voyage with another waterman, with whose assistance he can row half as fast again; and they find that they are only 8 minutes longer in performing their voyage than they would have been had there been no stream. Determine the rate at which the waterman would row by himself. Ans. 6 miles per hour. 15. A person bought a number of apples and pears amounting together to 80. Now the apples cost twice as much as the peals: but had he bought as many apples as he did pears, and as many pears as he did apples, his apples would have cost 10 cents, and his pears 45 cents. How many did he buy of each? Ans. 60 apples and 20 pears. 16. A gentleman exchanged a quantity of brandy for a quantity of rum and ~11 5s.; the brandy and rum being each valued at as many shillings per gallon as there were gallons of that liquor. Now had the rum been worth as many shillings per gallon as the brandy was, the whole value of the rum and brandy would have been ~56 5s. How many gallons were there of each? Ans. 25 gallons of brandy, and 20 of rum. 17. A and B carried 100 eggs to mralket, and each received the same sum. If A had carried as many as B, he would have received 18 pence for them, and if B had taken as many as A, he would have received only 8 pence for themn. How many had each? Ans. A had 40, B 60. 18. A farmer has two cubical stacks of hay. The side of one is 3 yards long'er than the side of the other; and the difference of their contents is 117 solid yards. Required the side of each. n/s. lThcl side of oo is 5 yVds, that of the other 2. QUADRATICS. 159 19. What two numbers are those, whose difference multiplied by the greater produces 40, and by the less 15? Ans. 4-8 and 4-3. 20. What two numbers are those, whose difference multiplied by the less produces 42, and by their sum 133? Ans. 4-13 and — 6. 21. Find two numbers which are to each other as 5 to 8, and whose product is 360. Ans. 4-15 and ~-24. 22. There is a field in the form of a rectangular parallelogram, whose length is to its breadth as 6 to 5. A part of this, equal to one sixth of the whole, being planted, there remain for ploughing 625 square yards. What are the dimensions of the field? Ans. The sides are 30, and 25 yds. 23. The paving of two square yards cost ~205; a yard of each costing one fourth of as many shillings as there were yards in the side of the other. And a side of the greater and less together measures 41 yards. Required the length of a side of each? Ans. 25, and 16 yards. 24. Two traders, A and B, set out to meet each other, A leaving the town C at the same time that B left D. They travelled the direct road C D, and on meeting it appeared that A had travelled 18 miles more than B; and that A could have gone B's journey in 154 days, but B would have been 28 days in performing A's journey. What was the distance between C and D? Ans. 126 miles. 25. A and B lay out some money on speculation. A disposes of his bargain for $11, and gains as much per cent. as B lays out; B's gain is $36, and it appears that A gains four times as much per cent. as B. Requlired the capitll of each. ul1 s. B's capital.1 20, A's PI5. 160 QUADRATICS. 26. The captain of a privateer descrying a trading vessel 7 miles ahead, sailed 20 miles in direct pursuit of her, and then observing the trader steering in a direction perpendicular to her former course, changed his own course so as to overtake her without making another tack. On comparing their reckonings, it was found that the privateer had run at the rate of 10 knots in an hour, and the trading vessel at the rate of 8 knots in the same time. Required the distance sailed by the privateer. Ans. 25 miles. 27. A merchant laid out a certain sum on speculation, and found at the end of a year that he had gained a dollars. This he added to his stock, and at the end of the second year found that he had gained exactly as much per cent. as in the year preceding. Proceeding in the same manner, and each year adding to his stock the gain of the preceding year, he found that at the end of the fourth year his stock was equal to eighty-one sixteenths of his original stock. What was his original stock? Ans. 2a dollars. CHAPTER VII. ADFECTED QUADRATICS WHICH INVOLVE ONE UNKNOWN QUANTITY. (1 13.) An Adlfected Quadratic is one that contains only the square and the first power of the unknown quantity..Thus, ax-2+bx-c is an adfected quadratic equation. Every adfected quadratic equation may be reduced to the form of ax2+bx=c. (1 1 4.) Since (x+ta)2=-x22ax + a', we may infer the following propositions, by the aid of which we may obtain one of the three terms of the square of a binomial by means of the other two. I. The first term of the square of a binomial may be obtained by dividing the second term by twice the square root of the third, and then squaring the quotient. II. The second term may be obtained by taking twice the product of the square roots of the first and second terms, and prefixing to it the double sign. III. The third term may be obtained by dividing the second term by twice the square root of the first term, and then squaring the quotient. (1 15.) If, in the expression x2I2ax, we regard 2a as the co-efficient of x, we see that x24-2ax may be rendered a complete square by adding to it the square of one half the co-efficient of x. If we have the quadratic equation x2+8x= 20, we can render the first member a complete square by adding to it 16; but if we add 16 to one member, we must add the same quantity to the other member, in order to preserve the equality. Hence, we may have, 2+8x + 16=36. By extracting the square root of this 162 ADFECTED QUADRATICS. equation, we have x+ 4=6, or x=-2. Hence, it appears that in order to solve an adfvcted quadratic equation, the tirst n illbe' must be rendered a complete square, and then, by extracting the square root of each member of the resu!t;ng equation, a simlle equatiot is obtained from which the value of the unknown quantity is readily found. As the preceding propositions will be of use in reducing some (equations, we will add a few examples for the purpose of mlaking the student familiar with them. We shall write a cipher in the ulace of that term which is to be found, in order that we may have the square of a binomial. 1. Render 81x2+0+-2 a complete square. Here the second term is wanting. By Prop. II. the second 1 1 term=9xX-X2=18. Therefore, 81X-l2+18+- is a complete x square. I2 Render x+1 2x+0 a complete square. Ans. x'+12x+36. 3. Render 4x2+ 4x+0 a complete square..Ans. 4x2+4x+1. 17 4, Render x4 +-7 X +0 a complete square. Ans. X4+1-7x3+ 28 x. 2 16 5. Render 49x + 98+0 a complete square. 49.Ans. 49x+ 98 +-. 6. Render y4 + 12xy2+ 0 a complete square. Ans. y4 +12xy2 + 36x2. 7. Render 25x2+30x + 0 a complete square..Ans. 25xt + 30x+ 9. ADFECTED QUADRATICS. 163 8. Render 81 +39x+O0 a complete square. 169x2 Ans. 81 +39x+- 6 36 9. Rendel x2y2 + 2xy+-0 a complete square. Ans. x2!2 + 2xy + 1. 10. Render 4x2 72x + 0 a complete square. Ails. 4x2+72x- 324. (1 1 6.) If we represent the sum of all thle known quantities in an adfected quadratic equation by c, tlte sum of all the co-efficients of the first power of the unknown quantity by b, and the sunm of all thle co-efficients of the second power by a, the equation will then take this form, ax2+bx=c (A) In order to avoid fractions, we will multiply equation (A) by 4a, and it becomes 4a2x2 + 4abx= 4 ac (B) For the piurplose of rendering the first member a perfect square, we find by Prop. I., in the preceeling article, that we must add b2 to each member. Equation (B) then becomes 4a2x2 + 4abxb2 = 4ac + b2. (C) * By extracting the squatre root of equation (C), we have, 2ax b= - V 4ac~2: (D) Whence, = 2 - +. (E) From formula (E) we may derive the following rule for writing out the value of the unknown quantity. RULE I. Reduce the equation to the form of equzation (A). Th e;n tlte value of the ui.known quantity is equal to the co-efficient of the first power of the unkniown quantity taken with a contrary sign, plus or minus the square root of four times the preoduct of the co-efficient of the highest power of the unknown quantity and the * NOTE.-The process of rendering the first mermber a perfect square is called " oompleting the square." 164 ADFECTED QUADRATICS. term inlependent of the unknown quantity, increased by the square of the co-efticient of the first power of the u),mnowun quantity, and the whole divided by twice the co-egficient of the second polwer of the unknown quantity. (1 17.) If we divide equation (A) in the preceding article by b c a, and then substitute for -, 2p, and fol -, q, it becomes a a x +'21pZ=q (F) Add p2 to each member of tilis equation, and it becomes x2 ~- 29X +,p2I =)p2 + q By extracting the square root, we have, x +p= -- V'2 ~q; NWhence, x= — p- Vp + q (G) Formula (G) furnishes, by translating it into common language, the following rule. RULE II. Reduce the equation to the form of equation (F). Then the value of the unkntownz quantity is equal to one half of the co-eFicient of its first power, taken with a contrary sign, plus or min.uts the square 1root of the square of one half the co-egicient of the first power of the u1knowz qulantity, increased by the term which is independent of the unknown quantity. (1 1 8.) If we substitute 2p for b, in equation (A), Art. 1 16, i'd then mIlltiply each mlember by a, and add p' to-each member;i' t t;i':,,t equ~ition, it becomes a2x'2 q+ 2apx +p2 = ac + p2 By extracting the square root of this equation, we have, ax+p=: /Vac +p2; Whence, x= ac +p ([I) a By translating formula (H) into common language, we have the following rule. ADFECTED QUADRATICS. 165 RULE III. The value of the unknown quantity is equal to one half of the co-efJcie)tt of the first power of the unknown quantity taken with a contrary sgyn, plus or minus the square root of the product of the co-efficient of the highest power of the unknown quantity and the te'rm independent of the unknown quantity, increased by the square of one half the co-efficient of the first power of the unknrown quantity, and the whole divided by the co-eyhicient of the second power of the unklnown quantity. (1 19.) Before applying Rule I. or Rule III., we should divide each member of the equation by the greatest common divisor of the co-efficient of the first and second powers of the unknown quantily, and the term which is independent of the unknown quantity. Tlihu%, in the equation 9x + 15x-66, we should, before applying the rule, divide each member by 3, the greatest common measure of 9, 15, and 66. If, after an equation is reduced in this manner, the co-efficient of the first power of the unknown quantity is an even number, Rule III. should be employed, if it is not an even number, Rule I. should be employed. Rule II. should be used when the co-efficient of the second power is unity, and that of the first power is an even number. (120.) By examining formulas (E), (G), and (H), we find, that by using the plus sign before the radical. we shall obtain one value of the unknown quantity, and that by using the mints sign we shall obtain another value. IHence, In every quadratic equation the unknown quantity tmay have two values. Either of these values will, when substituted for the unknown quantity, satisfy the equation from which it was obtained. In a quadratic equation the unknown quantity cannot have more than two values. For, if possible, suppose that the equation ax'+pxz —q has three distinct values, and represent these values by r, s, and t. Then each of these values must satisfy the equation. Hence, we have 166 ADFECTED QUADRATICS.,a2~ pr=~q, (1) as,+ ps=q, (2) anzd at2 +pt =q. (3) (Eq. (1)-eq. (2) gives a('2 - 2) ~p('-s)= O (4) Eq. (l)-eq. (3) gives a(2'-t2) +p(r —t) =O (5) Eq. (4) divided by r-s, gives at(r s) p= O (6) Eq. (5) divided by r —t, gives a(r+-t)-=O (7) Eq. (6)-eq. (7) gives a(s-t)=O (8) Now the last equation can only be satisfied by making a-O, or s-t=O. But a cannot equal 0, for then the proposed equation would not be a quadratic; hence, s-t=O, or s=t. Therefore, in a quadratic equation the unknown quantity cannot have three distinct values. (121.) Thle rules -which have been given for solving quadratics, will also enable us to solve any equation which can be reduced to the form X2n4-2ax-n=C; that is, any equation which conta;ins the unknown quantity in two terms, and having the index of the unknown quantity in one term, double its index in the other. It must be observed, however, that we can seldom obtain all the roots in this rnannerl.* EXAMIPLES. 1. Given 6.+- = —44, to find the values of x. Clearinlg of firactions, 6x2. - 35 — 3 _r _44,x By trar!sposing,, 6:2 —47x- -- 35 4 7V472' 4 X 35 47-4-37 5 Therefore.(Ruie I.)... 1 7 or-. " 1'2 12 -- 1 6 * NoTE.-Every ecquatioa whlich contains only one unknown quantity has as many roots as there are units in the highest power of the unknown quantity. Such equations will be treated of in another part of this work. ADFECTED QUADRATICS. 167 2. Given 9x —4x'2 /4x2_ 9x + 11 = 5, to find the values of x. Add -11 to each member of the equation, and then change its signs, and it becomes 4x2-9+11- V4x2_9x+ 11=6 (1) Let y= V 4x2 _ 9x + 1 1; whence, y2= 4x2-9x + 11. By substituting y and y2 in place of their values in equation (1), we have'-8y=c (2) Therefore, (Rule I.) y=- 2 =3 or -2 (3) Y2=9, or 4 (4).. 42-Ox+ 11 —9 (5) And 4x2 —9x+11=4 (6) From eq. (5), we find tlat.v=2, or 4; 9q- / —31 From eq. (6), v:_: V-;: V= 813x? 3, Given x4+ —-39x-81, to find the values of x. 2 1 3x By transposition, + —3 =81 + 39 (1) By Prop. I., Art. 114, we find that we can render each mem169x2 ber of equation (I) a perfect square, by adding ---- to each member. Hence, we have 133 169 169 69x2 + 1 81+ 39x +- (2) x+3 36 1 36 By extracting the square root of equation (2), 13x 13x Z+ -=9+- (3).'. x —-3 8'7 4, Given /x —, to find the values of x. x /x-2 168 ADFECTED QUADRATICS. Let y= ~'x, then y2=x. Substitute y2 and y in the place of their values in the given equation, and it becomes 8 7 Y-2=y_2/ ~(1) y-2 By clearing this equation of fractions and transposing, it becomes y4-2y' — 7y2 —Sg +16= —O (2) If we proceed to extract the square root of equation (2), we shall discover that we can render the first member a perfect square by adding to it 16y2. Therefore, add 16y2 to each member of equation (2), and we have y4 —2y+ 92 - 8y + 1- 6 = 6 (3) By extracting square root, y2 —y+ 4-=-4y (4) y2-y=-4; (5) Whence, y=4 or 1 (() y2=16 or 1 (7) And x=y2-16 or 1 (8) By using -4y instead of +4y, we shall find that the other! +3V/ —7 1 —3V —7 roots of x are, and, both being imaginary. 2 2 b 5. Given + - =-, to find the values of x. x- V z2_ a2 a Multiply the numerator and denominator of the first mnember by the numerator, and the equation becomes 2 (x + Vx2 - a)_z a2 a (1) Multiply each member by a2, and then extract the square root, and we have x2+ Vx2- 2= + ax (2) By transposing, Vx2 — a= -_/( —x (3) By squaring each member, x2-a'"=ax-i2xVaX+X2 (4) By transposing and using the negative sign of 2x /arx, we have -ax+2xVax + x2= 2 - a (5) ADFECTED QUADRATICS. 169 Add 2ax to each member of eq. (5), axf 2xV/ux+x2 —x2+2ax+a' (6) Extracting sq. root of (6), Va-x + x =x+a (7).'. vax=c; (8) Or, ax=a2, (9) And, x —a (10) If we take the negative sign of 2xVax, equation (6) may be written, ax — 2x V/ax ~-2 =x2 + 2ax + a2 (11) Extracting sq. root of(11), V'ax —x=x+a (12) Or, v/ax-2x+a (13) By squaring (13), ax= 4x2 -+4ax-a2 (14) 4x2 + 3ax —-a2 (15) e(k:/ — 7 —3)*1 Whence, (Rule I.), x= ( 8 --- (16) 6. Given x3 — 8x2 19x —12-=O, to find the values of x. _Multiply each member by x, in order to make the first term, X, a perfect square, and the equation becomes x4-8x3 + 19x2- 12x=0 (1) If we endeavor to extract the square root of the first member, we shall find that the first two terms of the root are x2 and -4x; and that the remainderl 32 — 12x, may be written 3(2 —4x). HIence, equation (1) may take this form, (X2-4x)2 + 3(X2 -4x)=0 (2) Let y=zx-4x, acnd then y2-(X —4x)2. Substitute y and y" in the place of their values in equation (2), and we have y2+3y=0, (3) Or, y2= - 3y; (4) WThence y=- -3.. x' 4-4x=- 3; Whence, x=2~-1-=3, or 1. * NOTE.-In Hutton's Mathematics, from which this example is taken, only the imaginary roots are given. 8 170 ADFECTED QUADRATICS. If we divide each member of the given equation by x —l, we shall obtain an equation from which the other root, 4, may be found. y — 10y2 + 1 7. Given:Y2 +9 y —3, to find the values of y. By clearing of fractions, we have, y-10oy2p + 1 —y3' y + 27y-27. (1) By transposition, y2 + 27y —28 —27+1 (2) 27-i- v'272 +4X 27 +4 By Rule I., 2: 2(3)4X +4 Or, Y-27i(27 + 2)=1, or -28. 2 8. Given 45x2+64x=308, to find the values of x. Let a=45., and c=-26. Then 2a-c —64, and 8a —2c=308. Substitute a, 2a-c, and 8a-2c in the place of their values, and the equation becomes ax2+(2a-c)x= 8a-c. (2) - 2a + ci 36,,?,2- 12ac-c2 By Rule I., - 2a (3) 2a - 2a+c4- (Ga-c). Or, x= 2a (4) 109 Whence, x=3, or (5) 45 We may generally avoid a great deal of numerical calculation by judiciously substituting letters for numerals in those examples in which the value of the unknown quantity is an integer. The student should exercise his ingenuity in this way. 9. Given 133x —218x —543, to find the values of x. Let a=133, and c —48. Then, 2a-c-218, and 3a-+3c543. Substitute a, 2a-c, and 3a+3c in the place of their values in the given equation, and it becomes ax2 -(2a-c)x= 3+3c (1) * Let the student observe that the quantity under the radical sign is the square of a binomial. ADFECTED QUADRATICS. 171 ~ 2a —c~4- V/16a -8uc-c2 By 1ule I., x- + 2a (2) 2a Or,x - = 2a- c~ -(4a + c) ~Or, x — ~ P2a; (3) 1Si Whence, x= 3, o1 - 3 (4) 1 3 3 10. Given 3x + 37x= 876, to find the values of x. Let a-37. Then 23a 25=876. Substitute a, and 23a 25:i. the place of their values, and the equation becomes 3x2+Cax=23a + 25. (1) Multiply (1) by 3, 9x2 + 3ax= -69a+75. (2) Since a-37, 75=2a+-1. Therefore 69a+75 —69a+2a+-l= 71a+-1. Observe that 74a-2X37Xa=2a2. Now, if fiom 7 la + 1, we slltlrct 74a, and add to thle remainder its equal, 2a2, we sihall have, 71a t- 1 = 2a-2 3a+ 1. Therefore, 69a + 7 = 2a2 — 3a+l. HIence, equation (2) inma be written 9x2 + 3cx= 2a2 - 3a+-1. (3) a2 We find by Prop. I., Art. 114, that we must add - to each member of equation (3), in order to render each a perfect square. Therefore, we have, a2 O9aC2 9x2 + 3ctx+r- = —3a+1 (4) Extracting square root, 3x+- = (r -i)I (5) 3x —=a-1 —36, or -2a+1=-73; Whence, x= 12, or -24~. 11. Given x2-.-4x=140, to find the values of x. Ans. x —10, or -14. 12. Given x2 —6x + 8 S80, to find the values of x. Ans. x —12 or -6. 13. Given x2+6x=27, to find the values of x. Ans. x-3, or -9. 172 ADFECTED QUADRATICS. 14, Given 3x —2x=65, to find the values of x. Ans. x=5, or — 4. 15. Given x2-37x= —252, to find the values of x. Ans. x=28, or 9. 16. Given x2- 1x+ 17 = I, to find the values of x. Ans. x= —8, or 2. 17. Given 7x2-21x+13=293, to find the values of x. Ans. x= 8, or -5. 18. Given x2-x-40=170, to find the values of x. Ans. x — 15, or - 14. 19. Given x2-26x+105 —=0, to find the values of x. Alns. x-5, or 21. 20. Given 11x2 —59x 78=0, to find the values of x. Ans. x-? 21. Given 9x —7x=116, to find the values of x. Ans. x=4, or — 3-. 22. Given 9x2 —x=140, to find the values of x. Ans. x-4, or -3-3-. 36-x 23. Given 4x — 36 =46, to find the values of x. Ans. x=-12 or - x+3 16-2x 26 21, Given +2 + 2 6- 5 - to find the values of x. Ans. x=5, or -91. _x+ 9-4x 25. Given 14+4x- -- 3x+,to find the values of x x-7 3 Ans. x =9, or 28. x+4 7-x 4x+7 26, Given 3 9-1, to find the values of x. Ans. x=21, or 5. ADFECTED QUADRATICS. 173 x 7 27. Given = to find values of x. x+60 3x —5 Ans. x=14, or -10. 8x 20 28. Given - -6 — to find the values of x. x+2 3x' Ans. x —=10, or -. x+ll 9+4x 29. Given - + -=7, to find the values of x. X X Als. x= 3, or —-. 2x+9 4x —3 3x+38 30. Given +, to find the values of x. 9 4x-3 18 Ans. x=6, or -1-. x~12 x 78 31. Given - 124+ x12 1_ to find the values of x. Ans. x=3, or -15. 32, Given (V4x+5) X(/7x+ ])=30, to find the values of x. Ans. x=5, or 1r - -2. xVx x+2-x - 33. Given -- -, to find the values of x. x- x x 4' Ans. x=4, or 1, or — /+ 1 5 31. Given - to find the values of x. Ans. x=8, or —. x + 2x — 9 35, Given = ( —2)2, to find the values of x. Ans. x=5, or 3. 36. Given 2x2 + 3x-5 v/2x2 + 3x + 9 =-3, to find the values of x. Ans. x- 3, or — ~. * NOTE.-In this, and some other examples, only the real roots are given. 174 ADFECTED QUADRAlTI1C3. 37. Given Vx+12+V x+12=6, to finld the values of x. Ans. x=4, or 69. 38. Given x+-16 —7Vx/+16=10 —4/4x+16, to find the.values of x. Ans. x= 9, or -1.2. 39, Given (x+6)'+2x(x + 6)=138+ +-x, to find the values of x. Ans. x=4, or 9. 40, Given x- 1-2 +-,, to find the values of x. Ans. x=4, or 1. 41. Given x4 —2xs +x=132, to find the values of x. Ans. x=4, or -3. 12~-8x~ 42. Given x=- 12, to find the values of x. x —5 Ans. x=9, or 4. 49x2 48 6 43. Given 4-9 +i —49=9+-, to find the values of x. Ans. x=2, or —. 44. Given x- — = 51 +-, to find the values of x. Ans. 9, or 4. 15. Given 4x'-+-=4x3s + 33, to find t!he values of x. 46. Given -3 - to find the values of x. Ans. x=1, or -_2. 47. Given x( /x-+ 1)2=102(x+ Vx)- 2576, to find the values of x. Ans. x —49, or 64. ADFECTED QUADRATiCS. 175 48. Givell _ Qa Q-),~+( J - a ), to find the values.... 2 a' of X.. )49. Given X4-lOx'+3Sx` —50.+-214 0 to find the vi;'laou. of x. Anl s. x= 1, 2.;, (4. -. 50. Given 2x2('+a~)2' 2x(zd-2, ) +,' —,,) to find the values of x. ~ a - or -a. (1 22.) In some examl?.,-. e Cay obtain values of the unknown quantity, which' rc' e:i: to be taken with some limitation. If, in example w.q, wre -- ibstitute 9 in the place of x, the equation becomes (9+6) -+6(9+6)-138+3, or 225+90=141, which is not correct. But in solving this example it will be found that xz is equal to -3, and therefore in substituting 9 for x in the equation, we should place for x, —3 instead of + 3. Observing this, the equation becomes by substituting, (9 + C)2 —6(9 + 6)= 138-3, or, 225-90=135, a true result.. ADFECTED QUADRATIC EQUATIONS CONTAINING TWO UNKNOWN QUANTITIES. (I 23.) An equation which contains two unknown quantities, and hlas terms in which the sum of the exponents of the unknown quantities equals, but never exceeds 2, is said to be of the second degree. Thus, xy + +yx =a, xy + y' + h=a, are equations of the second degree. (124.) If from two equations of the second degree, which contain two unknown quantities, and which are of the general form ax + bxy + cy + dx + ey +f= 0, we eliminate one of the unknown quantities, we shall obtain an equation of the fourth degree, which contains only one of the unknown quantities. Hence, in general, the solution of two equations containing two unknown quantities gives rise to the solution of an euation of the fourth 176 ADFECTED QUADRATICS. degree. There are, however, many examples which may be solved by the aid of rules already given. No general rule can be given for the solution of such examples. The learner must depend on his own ingenuity. EXAMIPLES. 1 y4-432 —12?xy2 (1) t to find the values of x y2=12~ +2xy (2) and y. By transposing in (1) and (2) J y' —12xy2=432 (3) we hlave, y2_ 2xy — 12 (4) } By completing the square { y4- 12xy2 + 36x2=432 + 362 (5) in (3) and (4) we have, y2 —2xy+x'2=12+x2 (6) y4 -_1 2X'2 ~ 362X Dividing eq. (5) by eq. (6), y'- 2xy + =36 (7) y" — 2xy + 2- x Extracting sq. root of eq. (7), Y — =6 (8) y-x Clearing (8) of fractions, y2'-6x=6y-6x; (9) y2= 6y, (10) and y=6 (11) Substitute this value of y in equation (2), and it becomes 36=12+1i'x (12) Whence, x= 2 2. Given { xy=6 (1) to find the values of z x4+y4 =272 (2) and y. + 4x'3y + 6xy2y + 4xy3 y4 = 1 2 96 (3) Eq. (3) is the 4th power of eq. (1) Eq. (3)-eq. (2) gives 4x3y+4xy3 + 6xy2= 1024 (4) E1. (1) squared gives x2 +y2 + 2xy=36 (5) By transposing, x2 + y2=36- 2xy (6) From (4), we have, 4xy(x2+y2) + 6x2y2= 1024 (7) For x2+y2 in (7), substitute its value, as found in (6), and we have, 144xy- 2x2y2= 024 (8) Whence, x2y2_ 72xy= - 512 (9) If we regard xy as the unknown quantity, equation (9) is a quadratic, and by Rule II., we find that, xy-=8, or 64 (10) ADFECTED QUADRATICS. 177 From the nature of the example we know that the second value should be rejected. Eq. (5) is x + 2xy +y2=6 (11) Multiply eq. (10) by 4, 4xy=32 (12) Eq. (11)-eq. (12) gives xa —2xy+y'=4 (13) Extracting sq. root x-y=2 (14) Eq. (1) +eq. (1 4) gives 2x=8 (15) and, x=4 (16). -. y=2 (17) 3. Given x + +y_-2y + 2y'+x+1 ) to find the values of x and y. By transposing 2x'ya in eq. (1), we have, X4-2x2y2 +y-_= 1+ 2xy+x y2, (3) Whence, x2- y= 1 + xy, (4) Or, x'-xy-y2=1 (5) From (2) by transposing, x' +y'- 2y'x- 2y2= x+ 1 (6) Let x=-ny; then'=n'y', and x'=n3y3. In the place of x, x2, and x4, in equations (5) and (6), substitute their values, and we have n2/2 _ —y2 -y2= 1 (7) and, ny +Sy3'- 2ny- 2y2=ny + 1, (8) n3y3-y 3 — 2ny3 —2y2 Eq. (8) -eq. (7) gives n y2'~y22n =ny+ (9) Divide the numerator and denominator of the first member of the last equation by y2, and then clear the equation of fractions, and we have, n~y +y-2ny-2 - n3y +n-n2y-n-ny-1 (10) By transposing, n'2y-ny+y=n2-n +l, (11) Or, by factoring, y(n —n + )=n2n + 1; (12) y=l. (13) By substituting this value of y in equation (7), it becomes n2_ n-1-1, (14) or, n2_ -n=2 (15) By, Rule I., n=2; (16) x=ny=2 (17) 178 ADFECTED QUADRATICS. 4. Given { +2 12 (2) to find the values ofx and y. xy-2y2- 1 (2) In this example each of the literal quantities is of the second degree; that is, the proposed quadratics are homogeneous. Such examples may always be reduced by substituting for one of the unknown quantities an unknown multiple of the other; because by this substitution we sllall introduce the square of this other into every term, and therefore it may be eliminated fiom the two equations, and the resulting equation wll be a quadratic involving' the unknown multiple, alnd this being found, the values of the two unknown quantities are readily obtained. Therefo)re, let xz=izy; then, x2=n2y2. Substitute for x and x' in the given equations, their values, and they become n2y2y2-= 12. (3) ny -2y2= I. (4) From (3) we have, Y= (5) Y'- + n' From (4) we have, y=-2; (6) 12 1 n2 +n n-2 (7) Clearing (7) of fiactions, n2+n=12n-24, (8)1 or, n2_11n —= - 24 (9) By Rule I., n-3, or 8; (10) 2 n=9, or 64. (11) By substituting these values of n and n' in equation (3), we have, 9y + 3y2 = 12, (12) and, 64y2 + 8y2= 12; (13) Whence, y= t 1, or 1 — and, x= ny= 4- 3, or - _v 5/ ADFECTED QUADRATICS. 179 x+ V x2_ y 2 17 x- VX2_- ~/~ 2 5. Given x - V/x-_ 2 4 x+ z-,2_to fi2nd.+x(x +y)=52- V/x2 +xy+4 (2) j the values of x and y. X+' V2_ y2 1 x_ /2_y2 Let n= ___ then -= Substitute, i x- V2_ —y2 n + 2_ 2 equation (1), n and - in the place of their values, and it becomes 1 17:n+ -=, (3) or, 42a —17n= —4; (4) from (3) by Rule I., n=4, or ~; (5) x + vx' — y Whence 4, or (6) x-,/x2-yClearing (6) of fractions, z+ Vx/2_ -y=4x- 4 V-2-y/, (7) x — g x~__?/' or, x+ Vx2- y2 = - (8) By transposing in (7), 5 VX2 y —2 3x. (9) By squaring eq. (9), 25x2-_25y2=9x2, (10) or, 1 6x2=- 25y (11) Whence, 4x= 5y, and x= 5 (12) We shall obtain the same result if we employ equation (8). By adding 4 to each member of equation (2), and transposing, it may be written x2 +xy+4 + VX2+xy+4= —56 (13) Let s='x2 ++xy 4; then x'+xy+4=s'. Substitute s' and s for their values, and equation (13) becomes s'+s=56, (14).'. s-:t 2-21=7 or -8 (15) 2 2 and s2=49, or 64 (16) X'+xy+4=s'=49, or 64. (17) 180 ADFECTED QUADRATICS. By substituting in the place of x, in equation (1I7), its value as found in (12), it becomes, by reduction, y=-4-4, or'and x=s =~5, or 4- ~. 4 7/3 6. Given { 3x+y=73-2xy. to find the values y 2+ 3y+x=44. (2) of x and y. By transposing in (1), x2+ 2xy + 3x + y=73. (3) Eq. (2)+eq. (3) x2+2xy+y2+4x+4y-=117, (4) or, (xq+y)2+4(x+y)=-117. (5) If we regard x+y as the unknown quantity, equation (5) is a quadratic, and by Rule II., we find that x+y —ll- -2=9 or -13 (6) x=O —y, or — 13-y (7) By substituting these values of y in equation (2), we have y2+2y=35, (8) And7, y2 +2y5= 7. (9) From (8), y=5, or — 7 (1 0) From (9,), y= —I+ /5, or — 1 — v' ( i) Hence, x=9 —y=4, or 16, (12) And, x= —13 —y= —12- -V58, or -12+ V58 (13) The values of x in (12) correspond to the values of y in (10) and the values of x in (13), to the values of y in (11). 7. Given S x2+x y=18-y2 (1) } to find the values ot.~Xy= (2) x and y. As. s. — 3, y —2.....'+ {x~d_2xy+y2+ 2zq_ 2y __120 8. Given {x +2XYZY2 x+2= 120 (1) to find thQ s of x n Z= ol, 4orl(2) values of z and y. Alns. x=-6, or 9, y-=4, or 1. ADFECTED QUADRATICS. 181 9. Given +y —x — y —-78 (1) to find the values of xy+x+y-39 (2):.. and y. Ans. x=9, or 3, y=3, or 9.* 10. Given {xY _ ry2 =1710 (1) } to find the values Xy-y= 12 (2) of x and y. Ans. x=5, or, y=3, or -15. 11. Given X:y+xy23-=12 (22) to find thle values of x Y+Xy3=18 (2) and y. - Ans. y —2, or -, =2, or 16. 12. Given { j~2+'33 -1- t42 (1) } to find the values of x and y. As. x-3, or -7, y=2, or 7. 13. Given 3y2_ =2=-39 (2) to find the values of x and y. -Ans. x= —k-, or:102, y —=~5, or 4 —59. 11. Given (1) y to find the values of x (2) and y. Ans. x-=4, or 2, y=2, or 4. 15. Given (x + l)y- xyl 2 44 (2) to find the values of xandy. Ans. x-5=, orl, y-6, or 150. 16. Given xfy+ /x+y~=2 (2) to find the values - x+p3=8ls9 (2) of x and y. 1An s. 5, or 4, y=4, or 5. 17. Given { x2y2~x_-y_132 (1) }tofind the values' (x2+fy).(x-y)='22o (2), of x and y. Ans. x=11, or -1, y=I, or -11. * NoTE. —In this and some of the following examples, the rational roots only are given. 182 ADFECTED QUADRATICS. 18. Given g 1Y"=P ( ) =to find the values of x and y. Ans. x=4, or 1, y=S. x' - ++2 19. Given t (1) to find the values y +: y2 + 4y (2) of x and y. xAns. x=4, or 1, y=, or -2. I 2x+ Vy 10 2x- V' T 20. Given 2x-/ 15 2x+to find 2xy -267N/2x+y+Y- (2) the values of x and y. Anzs. x=2, or — 10, y=l, or 25. 21. Given (1) to find the values (:-+-2) (3 + y:) 455 (2) of x and y.( Ans. x=. 3, or2; y-2, or 3. 22. (Give'n ( -4y) to find the values of x and y. _Ans.-x3=3, r 2, y 2, or 3. 23. Given (2 +; - 2'=5 (1) to find the values of x and?. A.. -:, _ ~, y=l, or. values of x and y. S ns. x-6, y1o. i 2xy-24( +,y)=240 (1) to find the 25. Given 12 y o=1o (2) )y values of x and y. Ans. x=8, or 6, yG, or 8. (xy -y=~r=2O+ (2) and y. Ans. Xz=5, y=2. ADFECTED QUADRATICS. 183 27. Given ( 2) +( -2 ), (1) x2- 1 8= -4xy —9 (2) to find the values of x and y. Ans. x=6, y=3. 28. Given x+y=6 (1) y to find the values of x ~X5+ yr=105 6 (2) S and y., Ans. x=2, or 4, y=4, or 2.. -,......... [xvx+y -N/x+x4y 89 1 29. Given I- -V -y xj -y 40 to find y _ /y 9=4 (2) the values of x and y. Ans. x=9, y=4. xy x-y 24 (1) x-y x+y 5 39. Given xY 1 to find the x Vx)_y values of x and y. ns. x= -3, or A; or oa, or { y=2, or -_]; or o, or _1~. PROBLEMS IN ADFECTED QUADRATICS. (1 25.) If a proper notat;on be employed, some of the following examples nare susccllti1)l1, of solutions which do not involve adfected quadratic equationls. The student should endeavor to obtain different solutions of the same problem by atlopting different notations, and thus strengthen his powers of analysis. 1. There are two numbers, one of which is greater than the other by 8, and whose product is 240. What are the numbers? 184 ADFECTED QUADRATICS. First Solution. Let x=the less number;* Then, + 8 = the greater number. By the question, X (x + 8)= 240, Or, x2 + 8= — 240; Whence, by Rule II., x= —4-1'256=12, or -20, And,' x+ 8 = 20, or -12; hence, the numbers are 12 and 20, or -12 and -20. Second Solution. Let x+4=tlhe greater; Then, x —4 =the less. *. (x-+ 4)X (x —4)= 240, Or, 2 -- 16=240;.'. x2_256, And x=- 1; Whence, x + 4 = 20, or -12, x- 4 = 12, or - 20. 2. A person dies, leaving children, and a filrtune of:;48600, which, by the will, is to be divided equally anmongst them. It happens, however, that immediately after the death of the farther, two of the children also die. If, in consequence of this, each remaining child received $1950 more than he or she was entitled to by the will, how many children were there? Let x+ 1 =the whole number of children; Then x- 1 =the number after two had died. Put a=1950 Then 24a=48600=24X 1950. * NOTE.-The greater of two numbers is equal to half their sum increased by half their difference, and the less is equal to half their sum diminished by half their difference. For, let x the greater, and y the less number; and let 2s=their sum, and 2d1=their difference. Then x —?=2s, And x-y=2d; Whence, %=sl+d, and y=s —d. Q. E. D. ADFECTED QUADRATICS. 185 24a Now, -=what each child would have rec'd by the will, x+ 1. And, - -" " "6 did receive. x —1 24a 24a By the question, -+a= x-L Cle'tring of fractions, 24ax- 24a + ax- -a= 24ax+ 24a By transposing, ax'= 49a, or, x2=49.- x —7, and, x+1=8, the number of children. If we had represented the number of children by x, instead of x —1, the solution of the problem would have involved an adfected quadratic equation. 3, A surveyor is requested to lay out a piece of land in a rectanlgular forll, so that its perimeter may be 100 rods, and its area, 589 square rods. What is its length, and width? Since its perimeter is 100 rods, its length and width, together, must - 50 rods. Let x= the length; Then 50-x= tile width. By the question, x X (50 —x) = 59,* 9 Or, x — 50x= —589 By Rule II., x= +25 /625 — 589= 31, or 19 50-x=19'3, or 31. Hence, the length is 31 rods, and the width 19. 4. A set out from C towards D, and travelled 7 miles a dlay. After he had gone 32 miles, B set out from D towards C, and went every day -l-th of the whole journey; and after he had travelled as many days as he went miles in one day, lie met A. Required the distance fiom C to D. * NOTE.-The area of any rectangle is equal to the product of its length and width. 186 ADFECTED QUADRATICS. Let 19x= tlle required distance; Then, x=tlie number of miles B travelled per day, And, x= " " of daoys lie travelled before he met A.'7x+32= " " of miles A travelled, And, x'=t " " B " Whence, 2 t 7x +- 3 2 = 19x, Or, x2 —12X= —32 By Rule II., x=6~ V/'- -32=- 8, or 4..'. 19x=152, ol 76. Both of these values will answer the conditions of the question. The distance, then, of C fiom D was 152, or 76 miles. 5. A grazier boullht as many lambs as cost him $60; out of which he reserved 15, and sold the remainder for $54, gaining 10 cents a head by them. How many lambs did he buy, and what was the price of each? Let x=the number, Then, -=the price of each. By the question, (x —15)X -l- -) 54 (1) Or (x-15) X (600 +x)= 540x (2) And x2 + 585x- 9000-= 540x (3) By transposing, x2' + 45x- 9000 (4) By Rule I., x= -45 /9(4000+225)_ -4 5+43 4225 2 2 -45+3-X65 Or, x= 4 X — -=75, or -120 From the nature of the problem, we know that the negative value of x should be rejected. -=-=0.80. Therefore he bought 75 lambs, and paid 80 cents a head. 6, Bacehus caught Silenus asleep by the side of a full cask, nnd seized tlhe opportunity of drinking, lwhich he continued for two tirds, (!f the time that Silenus woulll have taken to empty ADFECTED QUADRATICS. 187 the whole cask. After that S lenus awoke, and drank what Bacchus had left. Had they drank both togetler, it would have been emptied two hours sooner, and Bacchus would have drank only half what he left Silenus. Required the time in which they would empty the cask separately. Let x=the number of hours, in which Bacchus would drink it, And y= " " " Silenus " " Since Bacchus would empty the cask in x hours, in 1 hour he' would empty -th part of it, and by the same reasoning, Silenus would drink -th part of it in 1 hour; hence, both together Y would drink, ini 1 hour, -th part plus -th part of the cask, or a x y 1 1 x y part denoted by -+ — -. Now, as the whole cask may x y xy be represented by 1, the time in which both would have emptied the cashk, if they had drank together, may be found, by dividing ulnity by the part of the cask which they both drink in 1 hour. ilence 1 + = =Y the time in which both would have xy x+y drank the cask, by drinking together. By the question-X 2y x 3 -= =the part which Bacchus drinks. Therefore, 1-2y3x 3x.-2y- =the part which was left for Silenus to drink. There3x 3x-2y 1 3xy-2yq fore, 3 — y 3 - =the time which Silenus required 3x * 3x to drink hlis parlt. Again, by the question, 3- 2yX x- te part which 3 2 6x Bacchus would have drank, had they both drank together. The time which Bacchus occupied in drinking this quantity may be found by dividing the whole quantity by the quantity which he 188 ADFECTED QUADRATICS. drank in one hour. Therefore,'3 -- 2y the time 6' x 6 that Bacchus would have drank if they had both drank together, or the time in which both, by drinking together, would have drank it. Therefore, 3x- 2y x (1) 6 x+y By another condition of the question, we have, 2y 3xy- - 2y xy (2) 3+ 3=, (2) 3 + 3x -TClearing (1) of fractions, 3x2 + xy — 2y = Gxy. (3) Let x-ny. Substitute this value of x in equation (3), and it becomes 3n'2y2+ny — 2y2= G-Ony. (4) Dividing (4) by y', 3' + n-2 = — 6n By transposing, 3n2-5n2 = 2; Whence, n=2, and, x-=ny=2y Substitute 2y for x in equation (2), and it becomes, 2y 6yy2- 2y2 2y2. — 2=3 6y 3Y By reducing, y=3 x= 2y= 6 7, What two numbers are those whose sum is 19, and whose difference multiplied by the greater is 60? Ans. 12 and 7. 8. There is a field in the form of a rectangular parallelogram, whose length exceeds its breadth by 16 yards; and it contains 960. Required the length and breadth. Ans. 40, and 24 yards. 9. A man travelled 105 miles, and then found that if he had not travelled so fast by 2 miles an hour, he should have been 6 hours longer in performing the journey. How many miles did he go per hour I Ans. 7 miles. ADFECTED QUADRATICS. 189 10. From two places, at the distance of 320 miles, two persons, A and B, set out at the same time to meet each other. A trlavelled 8 miles a day more than B, and the number of days in which they met was equal to half the number of miles B went in a day. I-ow many miles did each travel per day, and horw far did each travel? Ans. A went 24, B 16 miles per day; A went 192, and B 128 miles. 11. A tailor boulllt a iieCo of cloth forl $147, from which he cut off 12 yards for hi; own use, and sold the ircntiAdelr for $120.25, gaining 25 cents per yard. How many yards Awere there, and what did it cost him per yard? Ans. 49 yards, at 83 per yard. 12. A regiment of soldiers, consisting of 1066 men, is formed into two squares, one of which has 4 men more in a side than the other. What number of men are in a side of each of the squares? Ads. 21, and 25. 13. What number is that, to which if 24 be added, and the square root of the sum extracted, this root shall be less than the original quantity by 18. Ans. 25. 14. A poulterer bouoght 15 ducks, and 12 turkeys for 105 shillings. He had 2 ducks moie for 18 shillings than he had of turkeys for 20 shillings. What was the price of each? Ans. The price of a (luck was 3s., and of a turkey 5s. 15. The joint stock of two partners, A and B, was $416. A's money was in trade 9 months, and B's 6 months; when they shared stock and gain. A received $228, and B ~9252. What was each man's stock? Ans. A's 192, and B's g224. 16. Three merchants, A, B13, and C, made a joint stock, by which they gained a sum less than that stock by $8 A. A's share of the gain was $60; and his contribution to the stock was 8$17 more tihan B's. Also 13 and C together contribute *325. I-ow much did each contribute? Ans. A S75, B,$58, C $267. 190 ADFECTED QUADRATICS. 17. A and B hired a pasture into which A put 4 horses, and B as many as cost him 18 shillings a week. Afterwards B put in two additional horses, and found that he must pay 20 shillings a week. At what rate was the pasture hired? Ans. 30 shillings per week. 18. A and B engage to reap a field for ~4 10s.; and as A alone could reap it in 9 days, they promise to complete it in 5 days. They found, however, that they were obliged to call in C, an inferior workman, to assist them for the two last days, in consequence of which B received 3s. 9d. less than lie otherwise would have done. In what time could B or C alone reap the field? Ans. B in 15, C in 18 days. 19. What two numbers are they whose product is 255, and the sum of whose squares is 514? Ans. 15 and 17. 20. What two numbers are they, whose difference is 8, and the sum of whose squares is 544 Anls. 12 and 20. 21. What two numbers are they, whose sum is 41, and the sum of whose squares is 901. Ans. 15 and 26. 22. Divide the number 16 into two such parts, so that tile products of the two parts added to the sum of their squares may be 208. Ans. The parts are 4 and 12. 23. Find two numbers, whose difference, multiplied by the difference of their squares=160; and whose sum, multiplied by the sum of their squares=580. Ans. 3 and 7. 124. The fore wheel of a carriage makes 6 revolutions more than the hlind wheel in going 120 yards; but if the circumference of each wheel be increased one yard, it will make only 4 revolutions more than the hind wheel in going the same space. Required the circumference of each? Ans. 5 and 4 yards. ADFECTED QUADRATICS. 191 25. Find two numbers such, that their sum and product together may equal 34, and the sum of their squares exceed the sum of the numbers themselves by 42. What numnbers are they? Ans. 4 and 6. 26. A and B were going to market, the first with cucumbers and the second with three times as many eggs; and they find that if B gave all his eggs for the cucumbers, A would lose 10 cents, according to the rate at which they were then selling. A therefore reserves two fifthss of his cucumbers; by which B would lose 6 cents, according to the same rate. But B, selling the cucumbers at 6 cents apiece, gains upon the whole the price of six eggs. Required the number of eggs and cucumbers, and their price. An s. 30 eggs and 10 cucumbers, and the price of an egg 1 cent, cucumber 4 cents. 27. A person bought two cubical stacks of hay for ~41, each of which cost as many shillings per solid yard as there were yards in a side of the other, and the greater stood on more ground than the less by 9 square yards. What was the price of each? Ans. ~25, and ~16. 28. A cask, whose contents is 20 gallons, is filled with wine, a certain quantity of which is then drawn off into another cask of equal size; this last cask is then filled Ivith water; after which the first cask is filled with the mixture, and it appears that if 63 gallons of the mixture be drawn off from the first into the second cask, there will be equal quantities of wine in each. Required the quantity of wine first drawn off. Ans. 10 gallons. 29. It is required to find two numbers such, that their sum, product, and difference of their squares, shall all be equal..An~s. ~1V5, and 2~ 1 /a. 30, A person buys a horse, and pays a certain sum for it; he then sells it for $144, and gains as much per cent. as the horse cost him. How much did the horse cost him? Ans. 880. 192 ADFECTED QUADRATICS. 31. A merchant buys a certain quantity of cloth, and pays a certain sum for it, besides 4 per cent. for carriage. He sells it fur $390, and gains as much per centum as the twelfth part of the purchase money amounted to. What did he buy it four? Ans. $300. 32. A rectangular garden, which is 8 rods long and 6 rods wide, is to be surrounded with a gravel walk of uniform width, so that the number of square rods in the walk may be equal to one half the whole number of square rods in the gardt-n. What must be the width of the walk? Ans. 1 rod. 33. In the preceding problem, if the walk had extended half way around the garden, what would have been its width? Ans. 2 rods. 31. A carpenter has a square board, a side of which is a inches, and he wishes to form a regular octagon out of the square, by cutting off from its four corners four equal right-angled triangles. What is the length of a side of one of the triangles, which is adjacent to the right-angle? Ans. a(1i-V /2)= aX0.293. 35, The alea of a rectangular garden is 560 square rods, and its length exceeds its width by 8 rods. RPtquired its length and width. Ans. 28, and 20 rods. 36. Divide 100 into two such parts, that the sum of their square roots may be 14. Ares. 64 and 36. 37. A poulterer going to market to buy turkeys, met with four flocks. In the second were 6 more than three times the square root of double the number in the first. The third contained three times as many as the firslt and second; and the fourth contained 6 more than the square of one third the number in the third; and the whole number was 1938. How many were there in each flock? Ans. The numbers were 18, 24, 126, and 1V70. ADFECTED QUADRATICS. 193 38. There are two square buildings, that are paved with stones a foot square each. The side, of one building exceeds that of the other bty 12 feet, and both their pavements taken together contain 2120 stones. What are the lengths of them separately? Ans. 26 and 38 feet. DISCUSSION OF AN EQUATION OF THE SECOND DEGREE. (1 26.) We have seen that all quadratic equations may be reduced to the form x2+ 2ax=c; (A) Whence, x= -a:h Va2+c By using the positive sign before the radical, and then the negative sign, we find that the two values of x are, x= -a+ a2+ c, (1) and, x= -a- /a' + c. (2) Now, whether a is positive or negative, its square must be positive, and therefore the quantity under the radical will be positive when a2>c, and negative when a2c. The first value of x is positive and less than a, since a is multiplied by a proper fraction, H'. Hence, this value of x shows that there is a point P1, between A and B, which is equally illuminated by the two lights. And this point must be nearer to B than to A. For, since b>c, V >, and Vb aVb a therefore aX o or ->. Now, this is the conb+ /c' vb + Vc 2 clusion to which we ought to arrive, since we here suppose that the intensity of the light A is greater than that of B. The corresponding value of a-x is positive and less than 2. The second value of x is positive and greater than a, since a is'Vb multiplied by an improper fraction, H_ Hence, this value of x shows that there is a point P2, in the prolongation of AB, and to the right of the light B, which is equally illuminated by the two lighlts. In fact, since the two lights emit rays in all directions, there may be a point in the prolongation of A B which is equally illuminated by the two lights. But this point must be to the right of the light B, so that it may be nearer to the weaker light. 198 ADFECTED QUADRATICS. We may readily perceive why these two values are derived from the same equation.. If we had taken AlP2 for the unknown quantity, instead of AP,, the equation would have been c x2 (x —a)"' Now, as (a-x)' is identical with (x-a)', the new equation is the same as the first equation, the solution of which ought, therefore, to give AP2 as well as AP1. The second value of a-x,, is negrative, as it should 2/b- /C be, since x>a. By changing the signs of each member of the ac/ c a__ _c equation a-x= _, it becomes x —a=- V n- th b-/ Cb-c value of x-a expresses the real length of B P2. CASE II. ]WVhen by Art. 176 thle fifdth part of thLe logarithm of any number is equal to the 1olgarithm (f the fifth root of th'it humbler. The looaritlln of 385 is 2.585461. Therefore, the loarithm of V/385 is 2.58461 -- 5 —0.51 7092. The inumber answerino to this logairitllhm is 3.289 nearly, which is the fiftlh root of 385. 3, What is the cube root of 551? Ans. 8.198 nearly., N. \Vlat is the cube', I4o-ot of 267? Ad~s. 6.439., YAt.V;: is the i: tltth root of 245? An:. 1.733. 6, Wlhat is the v:tlue of x in thle equation 3x=45? 7, C1alcullate tl:e colirmmo.n l%';-l'ih.,r of 7. l1;ns. 1.2304489. 80 Given the logarithm of 2 to find the logarithml of 25. A7s. Log. 25= — -2 log. 2. 9. Given the logarithms of 2 and 3 to find th.e fl u,, ritlnm of 22.5. nzs. Log. 22.5-1~-2 I 1 o. II 10. Find the value of x in the equation 1-. log'. (de —c) -- log. a 1,;. b LOGARIT3HMS. 247 11. Show that log. ~ -Ilog. (a-x)-3 log. (a+x)]. 12. Show that log. va2 —_x2= loo. (a +x)~+ log. (a —x). 13. Show that log. (3-3) — log. log. (a' + ax + x2). INTEREST AND ANNUITIES. (1 8 1.) Interest is the sum which is paid by the borrower to the lender for the use of money. It is computed at a certain per cent. per annum; that is, the borrower pays the lender so many dollars for the use of $100 for a year, the amount of interest paid for the use of $100 for one year, is called the Rate of Interest, or the Rate per cent. Thus, when $7 is paid for the use of $100 for one year, the rate per cent. is 7. The money lent is called the principal, and the interest added to the principal is called the amount. (182.) When interest is regularly received at stated periods, it is called Simple Interest; but when it is not so received, and the interest which accrues on the principal during the first of these periods, is added to the principal, and interest is calculated on this amount for the next stated period, and added to it, and so on for each period, the amount which is thus obtained, is called the amount at Compound Interest of the principal for the given time. If from this amount, the principal be subtracted, the remainder is the Coompouncd Interest. (183.) The preseot worth of any sum of money which is due at some future period, without interest, is obviously such a sum as will amount to the given sum, in the given time, at the given interest. (1 84.) Discount is an allowance which is made for the payment of money before it is due. It is obtained by subtracting the present worth from the given sum. 248 LOGARITHMS. (18 5.) The interest which is obtained by regarding the interest as being due at the end of every instant, and then finding the amount at compound interest, is called Instantaneous Compound Interest.* (186.) An Annuity is a sum of money which is paid regularly once a year, for a certain number of years. Payments which are made semi-annually, quarterly, monthly, &c., are also called annuities. (1 87.) The present worth of an annuity which is to continue forever, is a sum, the interest of which is equal to the annuity. The present worth of an annuity which is to continue for a given number of years, is a sum, which would, if it were placed out at compound interest during the time for which the annuity was to continue, amount to the same sum, which the payments would amount to, if each was placed out at compound interest from the time it became due till the termination of the annuity. In treating of interest and annuities, we shall employ the following notation: Let P=the given principal. r=the interest of $1 for one year. n=the number of years. R=the amount of $1 for one year. I=the interest of P dollars for n years. S=the amount of P dollars for n years at r per cent. a=an annuity. A=the amount of an annuity for n years. p=the present worth of annuity due in n years. K=the present worth of a given principal, due in n years, at compound interest. * NOTE.-For a very interesting article on the subject of Instantaneous Compound Interest, by Prof. Geo. R. Perkins, see American Journal of Science, vol. xlvii., No. 1. See also Perkins' Higher Arithmetic, late edition, and Young's Algebra. LOGARITHMS. 249 SIMPLE INTEREST. PROBLEM I. (1 88.) Find the interest on P dollars for n years at r per cent. Since the interest on 1 dollar for 1 year is r, the interest on P dollars for the same time is P times r, or Pr; and the interest for n years is n times Pr, or nPr. Therefore we have I=-nPr, (1) or, log. I=log. n+log. P+log. r. Cor. Since the amount is equal to the principal increased by the interest on that prillcil)al for the given time, we have S —=P+nPr. (2) PROBLEM II. (I 89.) To find the present worth of any sum of money, which is due in any given time. By formula (2) we find that the amount of a sum P due in n years is S=P+-nPr. Therefore, P must be the present worth of the sum S due in n, years, and we have Ynr- (3) that is, the present worth of any sum is equal to the quotient obtained by dividing the given sum by the amount of 1 dollar for the given, time. Cor. Therefore, by Art. 184, the discount on a sum S dlue in n years at r per cent. is s- _ -= - —; (4) that is, the discount on any sum is equal to the quotient obtained by dividing the interest on that sum for the given time, by the amount of 1 dollar for the given time. 11* 250 LOGARIT1IMS. COMPOUND INTEREST. PROBLEM III. (190.) To find the amount of any sumn for a given number of years, and at a given rate per cent., compound interest being allowed. By Problem I., the amount of P dollars for one year is P+Pr=P(l +r). Since compound interest is allowed, we must find the amount of P(1 + r) dollars for the next year. Now, the amount of 1 dollar for 1 year is +r, and the amount of P(l+r) dollars is, then, P(1+-r) times 1 +r, or P(1 +r)2. By similar reasoning, we can show, that the amount of P dollars for three years is P(1 +r)3, and hence we conclude, that the amount of P dollars for n years is P(1 +n)n-PRn; and, if we denote this amount by S, we have S=PRn.; (5) that is, the amount of any sum at comipound interest for a given time, is equal to the product of the principal, by the amount of 1 dollar for 1 year, raised to a power denoted by the number of year)s. PROBLEM IV. (1 9 1.) To find the present worth of any sum due in a given number of years, compound interest being allowed at a given rate per cent. By Problem III., we find that the amount of P dollars due in xn years is S=PRn. Therefore, P must be the present worth of a sum S due in n years, and we have S (6) that is, the present worth of any sum S is equal to the quotient obtained by dividing this sum by the ameougnt of I dollaer for 4he qi;ven time. LOGARITHMS. 251 ANNUITIES AT COMPOUND INTEREST. PROBLE-M V. (192.) To find the amount of an annu'ity which is to contillue for a given numnber of years, compound interest being allowed at a given ro lte per cent. The annuity which is due at the end of the first year, will be on interest during the n —1 remaining years, and therefore, by Probem l111I., its amount is aR-1; and the amount of the annuity which is due at the end of the second year is acRn-2, and so on. Hence, the amount of an annuity a, which is to continue for ni years, is A=aR.'- *+ aRl - t-R 3 - - - - - +a. The last equation may be written A= —(1 +R+R 2 - +-Rn-2~ Rn-1) Thle quantities within the parenthesis form a geometrical series, tile first termn of which is 1, the ratio R, and the last term Rn-'. T'e sum of this series is Rn-I R-1 Hence,'the last equation may be written A=1(nn-1) A= R-1 ) (7) PROBLEM VI. (193.) To jfind tfhe 2present worth of an annuity which is to conatiuce tfcr a givenz izeZmbe' of yeears, comp2ound interest being If we let IK represent the present value of the annuity, then, by Art. 187, the amount of K for the given number of years, at compound interest, must equal the amount of the annuity. Therefore, we shall have, by Probs. III. and V., 252 LOGARITHMS. K.Rn = a(R —1) R-1 K= (R 1) 8) PROBLEmI VII. (194.) To find the present worth of an annuity which is to commence after a given number of years, and continue for a given number of years. Let N denote the number of years that will elapse before the annuity commences, and let n represent the number of years for which the annuity is to continue. It is plain, that if we subtract the present worth of the annuity a for N years, from its present worth for N+n years, we shall have the present value required. By Prob. VI., the present worth of a for N+V n y'rs is a(R ) B~(R- l) -, I', "e "V _" is R(R —I) Therefore, if we denote the present worth required by K, we have K- a(R"+ln- 1 ) a(R —1) (1 ) If we multiply both terms of the last fraction by RTn, the two fractions will have a common denominator, and by subtracting their numerators, we have += "(R- - 1) R~h (P-1) Cor. 1. Equation (9) may be written in this form, by making a slight reduction, LOGARITHAIS. 253 a K=R —"(R — 1 - Now, if n=oo -0, and the last equation will become a K= Rx(- - 1(10) Cor. 2. By making N —0, equation (9) becomes Kt- ((R 1) () Equation (11) agrees with equation (8), and, under this supposition, it is plain that these equations should be identical. Cor. 3. If, in equation (10), we make -V=O, it will become a a R — r' that is, the present worth of an annuity, which is to commence immediately, and continue forever, is a sum, the yearly interest of which is equal to the annuity. This result agrees with what was said in Art. 187. PROBLE3M VIII. If the interest on $1 for the xth part of a year is I, ulhat is the amount of $1 for 1 year, when x=oo. Denote the amount by A, then we have, by Problem IIt., A=n(1 tI b Expanding the righlt-hand member of this equation by the binomial'theorem, we have x( + -l) x2- x(x —1)( —2) r3 A=1+r+ 1.2 xx 1.2.3 X &C. 254 LOGARITHMS. Since x is equal to infinity, x-1, x —2, &c., cannot differ, essentially, from x, and thlerefore we may substitute x for x-1, x —2, &c., and then'the equation will readily become* r2 r3 A 1 = r + 2 + r 2- + &c. 1.2 1.2.3 By the exponential theorem, we know that the right-hand member of this equation is equal to Er. Therefore, we have A= sr, Or, log. A=rXlog. e=rX log. 2.7182818= —rX0.434294-18. If we make r=0.07, then we have Log. A=0.030400~' Whence, A= 1.0725, nearly. If we wish to ascertain what must be the rate per cent. when the interest is compounded every instant, in order that one dollar may amount to $1.07, we must make A=$1.07, and then the equation above will become Log. 1.07=rX0.43429448, log. 1.07 0.0293838 nearly. 0.43429448 0.43429448 EXAMPLES. 1. Required the present worth of $100 due in 18 months hence, at the rate of 7 per cent. per annum. In formula (3) Problem II., make S=100, r —0.07, n=1 =1.5, and we have for the present worth required: 00 100 P1 + 1.5 X 0.07 1.105' Or, Log. P= —log. 100-log. 1.105=2-0.043362=1.956638. The number answering to thislogarithm is 90,497. Therefore, P=$90.497, the present worth required. * NOTE,-See Pierce's Differential Calculus, page 176, paragraph 42. We may find the value of A by a different process of reasoning. LOGARITHMS. 255 2. What is the amount of $800 for 12 years,.at 7 per cent., compound interest? In formula (5), Problem MII., make R=-1.07, P=800, and n=12, and we have S= soo x (1.07)12, Or, Log. S=log. 800+12 log. 1.07=3.255698. The number answering to this logarithm is 1801.76, nearly. Therefore S=$1801.76, the amount required. 3. In what time will $275 amount to $1000, at 6 per cent. compound interest? In formula 5, Problem III., make R=1.06, S=1000, and P=275, and we have 1000=275 X (1.06). Or, Log. 1000=log. 275+n log. 1.06. log. 1000-log. 275 3-2.439333,Whence, n= 22.16 years, log. 1.06 0.025306 the time required. 4, At what rate per cent., compound interest, will $100 amount to $112.48, in 3 years? In formula (5), Problem III., make S=112.48, P=100, and n=3, and we have 112.48=100 X Rs, Or, BR3=1.1248, Or, 3log. R=log. 1.1248, log. 1.1248 Or, log. R= — 1 =12480.017025, And, R= 1.04, nearly, and the rate per cent. is 4. 5. What is the present worth of an annuity of $500, to last for 40 years, at the rate of 2- per cent. per annum? In formula (8), Problem VI., make a=-$500, R= 1.025, n= 40, and we have 5oo0 X [(1.025)4~-1] (1.025)40X 0.025 256 LOGARITHMS. Now, Log. (1.025)4~=40 log. 1.025, =40X 0.0107239=0.4289560, =log. 2.685072, (1.025)40=2.685072. 500 Also, 0 20000. 0.025 1.685072 K.. _20000 X 2.685072' = 20000 X 0.62757, =-12551.4 dollars. 6. What is the amount of $1000 placed out at compound in terest of 5 per cent. for 10 years? Ans. $1628.9. 7. What sum must be placed out at compound interest, at 4 per cent., to amount to $2000 in 15 years? Ans. $1110.5. 8, In how many years will $200 amount to $318.80, at 6 per cent. compound interest? Ans. 8 years, nearly. 9, At what rate of compound interest must $518.30 be placed out, to amount to $600 in 3 years? Ans. 5 per cent. 10. In how many years will a given principal double itself, at 4 per cent. compound interest? Ans. 17.67 years. H11 In what time will $100, at 7 per cent. compound interest, amount to the same sum as $1000 would at 4 per cent. compound interest? Ans. 12. Find the amount of $1200, placed out at compound interest, at 6 per cent. for 10 years, the interest being converted into principal every half year. Ans. $2167.3. 13. What ought to be given for the lease of an estate for 20 years, of the clear annual rental of $100, in order that the purchaser may make 8 per cent. of his money? Ans. $981.40, nearly. LOGARITHMS. 257 14. A usurer lent a person $600, and drew up for the amount a bond of $800, payable in 3 years, bearing no interest. What did he take per cent., if compound interest be taken into consideration. Ans. 10.064 per cent. 15. How long must a capital a remain at interest, the rate per cent. being p, to become as large as a capital a,, at p, per cent., in n' years? A Log. a, + n log. p, —log. a A~s. 4 7 lyears. log. p 16, The present value of a freehold estate of $100 per annum, subject to the payment of a certain sum (Y) at the end of every two years, is $1000, allowing 5 per cent. compound interest. Find the sum (Y). Ans. Y=$162.50. 17. What will a capital of $12000 amount to in 10 years, if it bear 6 per cent. compound interest, the interest being paid half yearly? Ans. $21673.333. 18. A capital of $800 increased, in the space of 6 years, to $3600. What did the capital gain per cent.? Ans. 28.5, nearly. 19. A capital a is put out at p per cent. interest, and at the expiration of each year the interest is added to the principal, and at the same time it is increased or diminished yearly by the sum b. What will this capital amount to in n years hence? Ans. Apn(i]? — ); where the upper sign must be used when b is added, and the lower sign when 6 is subtracted.20. What will a capital of 83740 amount to in 8 years, if at the end of each year $450 be added to it, reckoning 4 per cent. compound itterest?, 4ns. $9264.833, nearly. CHAPTER X. THEORY OF EQUATIONS. (195.) A function of a quantity is any expression involving that quantity. Thus, ax2+cx' + 4x7, and 4 +5x + 2x+4 are functions of x. When an expression involves several quantities, it is a function of those quantities. Thus, ax'+5y'+ 2xy +y, and 2x3-=x4 —7y'+14 are functions of x and y. Such expressions are generally represented by the symbols F(x), F(x,y), which are read function of x, functions of x and y. PROPOSITION I. (1 96.) If any function of the form xn+Ax-' + Bxn- - - -.- -+P be divided by x-a, the remainder will be the same function of a that the given polynomial is of x. Let F(x)=xn+axn- + bxn- - - - - - +p. Now divide F(x) by x-a, and represent the quotient, whatever it may be, by Q; and let R represent the remainder which does not involve x. Since the dividend is equal to the divisor multiplied by the quotient plus the remainder, we have F(x) — Q(x-a)+R This equation is true for all values of x; hence, we may let x=a, and the equation then becomes F(a)=O+R But R is independent of x, and therefore the remainder is the same function of a, that the given polynomial is of x. Cor. It is obvious that whenever we wish to divide a polynomial of the form of x"+Ai1'q+Bx ~- -+B by xs-a, TIEORY OF EQUATIONS. 259 we can find the remainder without performing the labor of division. Thus, if we divide the polynomial x3 —6x+7 by X-2, we have for the rernainder, R=22- 6 X 2 + 7 — 1. PROPOSITION II. (1 9 7.) If a is a root of the equcation x"~Bx~- + Cn- 2 - - -2 +P —O, then the equat,:on is divisible by x-a. if the given equation be divided by x-a, and the division is carricd on till a remaind:r is obtained which does not involve x, this remainder must be the sam l fuinction of a, that the given polynomial is of x. (Piop. I.) H(nee, the relnail:der is an+ BaC."-'+ Ca -2 - - P. But, since a is a root of the given equation, it must verify that equation, and we have a-n+Ba"- + 6Cx-2 - P - _, P=0; h(-nce, the remainder is equal to zero, and therefore the given equation is exactly divisible by x-a. Cor. If the first member of any equation of the form xn+,B.n-I + C6c-2 - - P=O is divisible by x-a, then a is a root of that equation. For, since the given equation is exactly divisible by x-a, the remainder, which is the same function of a that the given equation is of x, must be equal to zero; that is, a nr+ Bxn- + Cxn-2 - - - - -P=O. Hence, it appears that if a, be substituted for x in the given equation, the equation is verified, and therefore a is a root of that equation. PROPOSITION III. (198.) Any equation which contains only one unknown quantity has as many roots as there are units in the highest poewer of the unknown quantity.* Let F(x) represent any equation of the nth degree, and let a represent a root of this equation. Then, by Proposition II., F(x) is exactly divisible by x-a, and if we represent the quotient, which is a function of x, by F/(x), we shall have, by a principle in division, F(x)=(z(-a,) X F2(x)=O0 (l) * NoTE.-In this demonstration it is assumed that every equation has at least one root. 260 THEORY OF EQUATIONS. Now, the equation (x-a,) X F(x) —= O may be satisfied by making either of its factors equal to zero. Hence, we may have F,(x)=O. If we represent a root of this equation by a2, we shall have, by reasoning as before, F(x) = (z-a.2) X E3(x). Substitute this value of F,(x) in equation (1), and it becomes F(x) = (x-aa,).(x-a2) X F3() (2) It is plain that we may continue this process of reasoning till we have resolved F(x) into n fhctors, and no more; hence, the equation, F(x)=O, mnay assume this form F()= (x —(a-).(x —a,).(x-a-a3 --- (.X-an)=0 (3) Whence, it appears that there are as many roots a, factors, that is, n roots; for equation (3) may be satisfied by making any one of the n quantities, a,, a, a3, - - - - - a, eqal to x. Therefore the proposition is established. PROPOSITIONT IV. (199.) If x represents the unknowe quantity in any equation whose roots are a,, a2, a,- an, the-e the equation is equal to the coatinued product of x-a,, x —a,, x-a,,- - - x-a,. For, if F(x) represents the required equaclion, we have, by the last Proposition, F(x) = (x-,). (-,).( -a,)- - - - -(x- a) - TIence the proposition is true.? w..,erfoi;m tihe multiplications indicated in the last equation, we shall hlave, by making n equal 1, 2, 3, 4, &c., the following equations: When n=2. F(x)=x —a2 x + ala — 0 -a, aa,) When n=3. F(x)=x' —a, aa,.'+aa x+aa2a3=O -a + aa, a, THEORY OF EQUATIONS. 261 +a,a2-a,) q- ala3 When n=4. F(z) =X4 a2 Iz +a2a, -x-a a3 1 a2aa4 -a:J +aa4 | aa2a4 O, &c. - 4 +- ada4 I -aa 3a4 -aaa4 -a2a3a4 J By examining these equations, we discover the following properties: 1. The co-efficient of the second term in either of the equations is the sum of all the roots with their signs changed. 2. The co-efficient of the third term in either of the equations is the sum of all the different products that can be obtained fiom the roots, taken two and two, with their signs changed. 3. The co-efficient of the fourth term in either of the equations is the sum of all the different products that can be obtained fiom the roots, taken three and three, with their signs changed; and, by deduction, we conclude that the co-efficient of the nth term is the sum of all the different products that can be obtained from the roots, taken n — and n-1, with their signs changed. 4. The last, or absolute term, is the product of all the roots with their signs chauged. COR. 1. If the signs of the terms of an equation are alternately positive and negative, the roots are all positive, and if the signs are all positive, then the roots are all negative. CoR. 2. If the co-efficient of the second term of any equation is 0, that is. if the second term is wanting, the sum of the positive roots is equal to the sum of the negative roots. Con. 3. Every root of the equation is a divisor of the last or absolute term. PROPOSITION V. (200.) No equation has an odd number of surd, or an odd number of imaginary roots. 262 THEORY OF EQUATIONS. For, let nx + ~Ax~-+Bx-2 - - - - - - P=O be an equation, one lroot of whl;ch is of the form a + b -- l; then will a-b V -1 be a root of this equation. Since a + b V —1 is a root of the equation, it must, when substituted for x, satisfy the equation, and we shall have (a++b )/-l)n+A(a+b V'_)n- +B(a+bV /_)n-2 P=- (1) Now, as all even powers of an imaginary quantity give real quantities, and all odd powers imaginary quantities, it follows that if we expand the terms in the last equation, the resulting equation will contain some terms which are real, and some that are inmaginary. If we represent the sum of the real quantities by P, and the sum of the imaginary quantities by Qv' —1,' we shall have P~+ Q/ —1=0 (2) It is obvious that this last equation can only be satisfied by making P and Q equal to 0 at the same time. If we now substitute for x in the given equation a —b/ —l, and then expand the different terms as before, an expression will be obtained which will only differ from the former expanded result in the signs of the odd powers of b /-1. HIence, if we represent the former result by P+ Q V/- 1, we may represent the latter by P-Q 9 V-1. But it has been shown that P=O, and Q=0. Therefore P —Q V' —-=. Hence the substitute of a -b /- I for x verifies the equation, and therefore it is a root of that equation. In a similar manner it may be sliown that if a~- /b is a root of an equation, the a — Vb is also a root of that equation. Con. 1. All thle roots of an equation of an even degree may be imaginary, bnt if they are not all imaginary, there must be at least two real roots. THEORY OF EQUATIONS. 263 Con. 2. The product of two imaginary roots, a+bV -—, a-b — 1, is a'+b', a positive quantity; and, therefore, when all the roots are imaginary, the absolute term must be positive. COR. 3. An equation of an odd degree must have at least one real root, and the sign of this root is contrary to that of the al!solute term. COR. 4. An equation of an even degree, whose last or absolute term is negative, must have at least two real roots. PROPOSITION VI. (201.) If the signs of the alternate terms in any equation be changed, the corresponding roots of the resulting equation will have contrary signs. Let x +A "-'+-Bx -2+ Cx'-. — - P=O, be any equation. If we change the signs of the alternate terms, we shall have x" —Ax" —A 1Be-2-'_ Cxn-3 _ _ P = O; (1) or, -x+Ax"-'-Bn- + C-3 - - - - =FP=O (2) Equations (1) and (2) are identical, for by transposition in equation (1), we have xn+Bx - - - IP= Axn-1 + Cx'n- + &c. (3) By transposition in equation (2), we have fn + Bn"-2 _ - P = Axn- + Cx-1' +&c. (4) Now, as equations (3) and (4) are identical, equations (1) and (2) from which they were derived, must also be identical. Now let a be a root of the given equation, and let a be substituted for x in the given equation, and -a for x in equation (1), and the results in each case will be the same. But since a is a root of the given equation, it must verify it, and hence -a must verify equation (1), and therefore equation (2), equations (1) and (2) being identical. Therefore, -a is a root of equations (1) and (2). Hence the proposition is established. 264 TRANSFORMATION OF EQUATIONS. TRANSFORMATION OF EQUATIONS. PROBLEM I. (202.) To transform any equation into another, so as to remove the second term, Let n + Axn — +Bx" - - - - + P= 0 represent any equation Let x=r+s, and then substitute this value of x in the given equation, and we have (r + s)n (r +)n- B(r J S)n-2 +P=O By expanding the terms in this equation by the Binomial Theorem, we have + sO, + n(n-1) a, 2sn-1 + 1)s. +.As" + ns 1 1.2 r-2__ +Bs"-2 =O t( +( -l) As Now, if we regard s as being an indeterminate quantity, we can attribute to it such a value as will make the co-efficient of rn-l in the last equation equal to 0. Hence, we may have ns +A=0; -A ~. 8= —._ n But x=s+r; A ~', X —-r ——. n Hence, the second term of any equation may be made to disappear, by substituting for the unknown quantity another unknownr quantity, connected with the co-efficient of the second term, with c contrary sign, and divided by the exponent of the highest power of the unknown quantity in the given equation. TRANSFORMATION OF EQUATIONS. 265 (203.) If we have the quadratic equation x'+2ax=b, we can find the values of x by the aid of the preceding article. Thus, let x=y-a, and then substitute this value of x in the given equation, and we have (y-a)'- 2ac(y-a) =b, or y — 2ay +a 2+ 2ay- 2a2= b, or y -a2 - Whence, y= 8Va2 +-b; x= —a -- a'+b. PROBLEM 11. (204.) To translform any equat'on into another whose roots shall be equal to the roots of the given equation increased or diminished by a given quantity. L, t x"+ Ax' —+ -.Bx" - - - - - - -+P=O X be any equation. Substitute for x in' this equation s+r, and it becomes (s+ r)+ + (s r)- + B(s + r)n- ----- +P= 0 X. By expanding the terms in this equation by the Binomial Theorem, and arranging them according to the ascending powers of r, we have s'h ~ns1"-' ~n.!~2 - +A —s~+( n-l)As. ~( —n-'1)fl122As3 r2n I-sn-1n- (A + Bs- +-(1n-2B-3 - (n-2) 12 Bs " (A) 1.2,- + q,I IS — If, in this equation, we represent the sum of the terms which do not involve r, or what is the same thing, the co-efficient of r~, X3 by X, the co-efficient of r by X2, the co-efficient of rA by -, the co-efficient of r' by 2 3' and so on, we shall have the following equations: 12 266 TRANSFOR{MATION OF EQUATIONS. X,=sn+Asn-'+Bsn-,- +-P, (1) X,=ns"-' + (n- l)Asn-2 + (n- 2)Bsn-3 + 0, (2) (B) x=-n(n-1 )Sn-2+(n- l)(n- 2)Asn-3+(n- 2)(n- 3) Bs-3 - -...N, &. (3) By examining these equations, we find that the first may be derived from the given equation by simply changing x into s; the second may be derived from the first by multiplying each term by the exponent of s in that term, and then diminishing the exponent of s by a unit in each term. The third may be derived from the second in the' same manner that the second was derived from the first, and so on. Equations (1), (2), and (3), are called derived polynomials. If we substitute for the co-efficients of r in equation (A), the quantities which represent them, we shall have X,+X2r+-r + -. +rn=0 (C) 2 2.3 To show the application of equations (B) and equation (C), let it be required to find an equation whose roots shall be less by 3 than the roots of the equation X4-3x3- 15x+49x-12=0. Let x=r+ 3. It is evident that if we substitute this value of x in the given equation, the roots of s in the resulting equation will be less by 3 than the roots of x in the given equation. The transformed equation will be of the form Xi X + xr+ 3 r2 +X4 r X5r =o (nz) 2 2.3 2.3.4 In this example, s= 3, and by the aid of equations (B), we find that x,= (3)4- 3(3)3- 15(3)2 + 49(3)- 12 =0 X2=4(3)3 —9(3)2- 30(3)+ 49 = —14 =- =6(3)2- 9(3)- 15 =+12 x4 — =4(3)-3 =+9 X,. -=1 =+1 2.3.4 TRANSFORMATION OF EQUATIONS. 267 By substituting these values of the co-efficients in equation (m), it becomes r4 + 9r3 + 12r2-1 4r=O (205.) We may also transform an equation into another, whose roots shall be less or greater than the roots of the given equation by a given quantity, by division. Let x"+Axn-l+'BxSn — - - ~Ox+P=O (1) be any equation. If we make x=s + r, and then substitute this value of x in the proposed equation, the transformed equation will obviously be of the form 81n +?1, -I -- +i.n-A. B.... — 01 + OsP1-O (2) If, in this equation, we substitute fdr s its value, x —r, it will become (x-r)n +A,(x-r)n- +B,(x-r)n- -- - - + 01(x-r) P,=O (3) Equation (3), when developed, must be identical with equation (1). For, equation (2) was obtained by substituting s+r for x in equation (1), and then by substituting (x-r) for s in equation (2) we must needs have returned to the original equation. Hence, -we have, (X -r)n +A,(x-r)n- +B1(x-r)n- - - + 01(x-r) +Pl=x + Ax" -2 _ - OX + P If we divide the first member of this equation by x —r, the remainder will obviously be P1, and the quotient will be (x-r)n — + ~A1(x -r)n —+B1.B(x- r)n-3 --- Ni(x-r) + 01 If we divide the second member by x-r, the quotient and remainder must be the same, since the second member is identical with the first. Therefore, if the first member of equation (1) be divided by x-r, the remainder is the absolute term in equation ('2), which is the required transformed equation. Again, if we divide the quotient alleady obtained by x-r, the remainder will be 0, the co-efficient of s in the transformed equation. By proceeding in this way, it is plain that we may determine all of the co-efficients of s in the transformed equation. 268 TRANSFORMATION OF EQUATIONS. Hence, we have the following rule for transforming an equa tion into another, whose roots adr equal to the roots of. the giver equation, increased or diminished by a given quantity. RULE. Transpose all the terms of the given equation into the firs. rmenlber, and then divide the given equation by the unknown quan tity increased or diminished by a given quantity, according as the roots of the propo.ed equation are to be increased or diminished, and then div'de the quot'ent list obtained by the same divisor, and so on till a quotient is obtained which is independent of the unkn own quantity. Continue the divisi on in each case till the remainder is independent of the unknown quantity. The co-efficient of the highest power of the unknown quantity in the transformed equation will be the same as the co-efficient of the highest power of the unknown quantity in the proposed equation, and the co-eficients of the following terms in the transformed equation will be the several remainders, taken in a reverse order. Wherever there is an absent term in the equation, supply its place with a cipher. We will now apply the above rule in transforming the equation x" —'x -7-=0 into another whose roots are less by 1 than those of the given equation. We shall designate the first remainder by RB,, the second by R,, and so on. Operation. x- l)x'- 7x+ 7(x-2+x-6 = 1st quotient. 3 2 r2 — X -6x+ 7 -6x+6 _t- l - /~ TRANSFORMATION OF EQUATIONS. 269 x-l)x2+x-6(x+2=2d quotient. 2x-6 2x- 2 x —1)x2(1 =3d quotient. x-1 +3=R3 Hence, the required transformed equation is xs + 3x2- 4x + 1= 0. This method of transforming equations will be of no practical value, on account of its length, unless we have some means of iminihishing the tabor. To this end we shall here introduce HIORNERXS SYNTHIETIC METHOD OF DIVISION. (206.) We shall first show how division may be performed by means of detached co-eficients, when the divisor and dividend are homogeneous, and contain only two letters. Let it be required to divide a'3ax+ 3a 3ax +x3 by a +x. By examining the divisor and dividend, it will appear evident that the first terim in the quotient is a2, and that the exponent of a is one less in any term than in the term ~which immediately p)recedes it; and that the exponent of x in the second term of tlhe quotient is 1, and that the exponent of x is greater by 1 in any term than in the term which immediately precedes it. H-ie!c-, the literal parts of the quotient are a', ax, x2. The literal p):ults of the successive terms in the quotient follow the same law of increase and decrease as those in the dividend. Now, the co-efficients of the terms in the quotient may be determined by writing down the co-efficients of the divisor and dividend, and then proceeding to divide as though the literal parts were attached to them, 270 TRANSFORMATION OF EQUATIONS. Operation. 1+-1)1+3+3+-1(1.+2+1, co-efficients. 1+1 2+3 2+2 1+1 1+1 0 Therefore (a3 + 3ax+ 3ax2 + as) -- (a +x) = a +- 2ax+x2. As another example, we will divide x4 —3ax's-8a'x'+18a~x -8a' by xs + 2ax-2a'. Operation. 1 +2-2)1-3-8 + 18-8(1-5+4 11+2-2 -5- 6+18 -5-10+10 4+ 8 — 4-+ 8-8 In the following example, there are no terms wlici n'ai "3, a2, a, or each of the co-efficients of these term., ir 0. Divide 6a4 —96 by 3a —O. Operation. 3-6)6 — 0+0+0-96(2+4-+8+16 6-12 12 12-24 24 24-48 48-96 48-96 Since a'-. a=a', the literal parts of the quotient are a', a', a, therefore, the quotient is 2a' + 4a' + 8a + 16. TRANSFORMATION OF EQUATIONS. 271 (207.) 13y the common rule for division, each term in the divisor is multiplied by the first term in the quotient, and the several terms of the product are subtracted from the dividend. Now, if we change the signs of the terms in the divisor, and then multiply each term of the divisor by the first term of the quotient, the terms of this product must be added to the dividend::i;;tl since we can do the same for each term in the quotien:, tilh st-veral terms of the products of the terms in the divisor by the successive terms in the quotient would all bhc'-!Iiac a;> ive. By this process, the first r-?K..il c. I is the same as would be obtained by the ordinmry uile, i s:t since the sign of the first term in the divisor has been cli:;.,ged, it is plain that if we divide the first term of the reilainder by the first term of the divisor we shall obtain the next term in the quotient with its sign changed. Now, to avoid the liability of error, incident to this change of sign in the quotient, we may let the first term of the divisor remain unchanged, and then omit altogether the products of the first term of the divisor by the successive terms in the quotient, since by the usual method of division, the first term of the dividend, and the first terms of the several remainders are cancelled by these products. Let it be required to divide a'- 5ax + 10a3x'- 10a'x3+s5ax' -x' by a —2ax+x2. By detaching the co-efficients, we shall have the following Operation. 1+2-1)1-56+10-10+5-1(1-3+3-1 4+2- 1 (-3+ 9-10)=1st remainder. -- 6- 3 (+ 3- 77+15)=2d remainder. *+ 6-3 (- 1 + 2-1)=3d remainder. 272 TRANSFORMATION OF EQUATIONS. This operation may be abbreviated by omitting all the additions except those which must be performed for the purpose of obtaining the first terms in the remainders, which are the successive terms in the quotient. For example, instead of adding -1 to 10, and then adding to their sum -6, we may omit the first addition, and find the sum of +10-1-6, at once. For the sake of convenience in performing the additions, multiply the changed terms in the divisor by each term in the quotient, as it is found, and then arrange the terms of each series of products in a diagonal line passing downwards. The first terms of the several series of products are arranged immediately under the terms of the dividend. The last operation may be Inodified as folluws: 1 + 2 -1)1-5+ 10-10+5-1(1 +2- 6+ 6-2 - 1+ 3-3+1 1-3-+ 3- 1+0+0 In this operation, +2 and -1, the terms in the divisor whose signs are changed, are multiplied by 1, the first term of the quotient, and the products are arranged in a diagonal line, as represented. By adding -5 and +2, the terms in the second column, we obtain -3, the second term in the quotient. Now multiply +2 and -1 by the second term of the quotient, and arrange the products, -6 and +3, in a diagonal line, as before. By adding + 10, -6 and — 1, the terms in the third column, we obtain +3, with which we can proceed as before. By supplying the literal parts of the quotient, we find that the required quotient is a'-3a'x+3ax'2- x From what has been said, we may derive the following RULE. I. Divide the dividend and divisor by the co-efficient of the first term of the divisor, after having arranged the dividend and divisor with reference to the powers of the same letter. The first term of the quotient is the same as that of the dividend. TRANSFORMATION OF EQUATIONS. 273 II. Change the signs of all the terms in the divisor except the first, acnd then multiply the terms with changed signs, by the first term of the quotient, and arrange the series of products thus obtained in a diagonal line passing downwards, so that the first term of the series may be directly under the second term of the dividend. III. Find the sum of the terms in the second column for the second term of the quotient, and multiply the terms in the divisor with changed signs by the second term in the quotient, and place the series of products in a diagonal line, so that the first term of the series may be directly under the third term of the dividend. IV. Find the sum of the terms in the third column for the third term of the quotient, with which proceed as before. In some examples, it will be observed that there are absent terms in the divisor or dividend, and in such cases, supply the places of absent terms with ciphers. For the purpose of illustrating this point, take the following example. Divide a —3a4$x+3ax'4-x6 by as-3ax+ 3ax'-xs. Here there are no terms in the dividend which contain a', as, a, x, xs and x5. That is, the second, fourth, and sixth terms are absent, or their co-efficients are ciphers. Operation. 1 +3-3 + 1)1 +0-3+0+3+0-1 +3+9+99+3 -3 —9-9-3 +1+3+ +1 1 +3 +3 + 1 +0+0 +0 Hence, a' + 3a'x + 3ax2 + x3=the required quotient. (207.) We can now transform any equation into another whose roots are less or greater than the roots of the given equa12* 274 TRANSFORMATION OF EQUATIONS. tion by a given quantity, by the aid of Synthetic Division, and thus avoid much labor. We will first take the equation x-'7x$7-=O, which we have already transformed by division. The roots in the transformed equation are to be less by unity than those in the proposed equation. The several divisions are exhibited in the following Operation. 1 +1)1 +0-7+7 1+1-6 1+1-6+1.'. R —1 1+2 1+2-4..~ R-=-4 +1 1+3.~. R3=3 In this example, the co-efficient of x2 is 0. We see that the co-efficients of the terms in the quotient are 1, 1, -6, and the remainder is 1. The other quotients are obtained in the same manner. We may omit all the terms in the first column except the first, and write the changed term in the divisor at the right. The operation may then be written as follows: +o0 -7 +-7(1 1 1 -6 1 — 6.'. R,=1 1 2 2 -4.R. R2=-4 1 3.'. R3=3 Hence, xs + 3x — 4x + 1 =0 is the transformed equation. As another example, transform the equation x4 — 3xs -- 15x + 49x-12=0 into another whose roots shall be less by 3 than those of the given equation., TRANSFORMATION OF EQUATIONS. 275 Operation. 1-3 - 15 ~49-12(3 3 0 -45 12 0 -15 4 0.'. R1=0 3 9 -18 3 - 6 -14.'. R —-14 3 18 6 12'. R2-12 3 9.'. R3=9 Hence, the transformed equation is x4+ 9x3 122- 14x= 0. PROPOSITION I. (208.) If a, a, a - - - an represent the n roots of an equation of the nth degree, taken in the order of their magnitudes, so that a, is less than a1, a3 less than a,, and so on; and if a series of numbers, b,, b2, b3, b4 - - - bn, in which b, is greater than a,, b, a number between a, and a,, b3 a number between a, and a,, and so on, be substituted for the unknown quantity in the given equation, the results thus obtained will be ctlternately positive and negative. For, let xn - A.-1 + Bx- -2 - - Ox + P=O represent the given equation whose roots are a,l, a, a3 - - - a,,. Then, by what has already been shown, this equation may be written (x-al).(x — a).(x —a3) - - -- (x-a)=O If, in this last equation, we substitute b,, b,, b3, - - - bn in succession, we have the following series of results: (1) (b,-a,).(b,-a,).(b, -a3) ---- (b,1 —a,) = a positive quantity. (2). (b,-a,).(b,-a,).(b,-a,) - - - - (b,-a,)=a negative quantity. (3) (b,-a,).(b —a,).(b3-a3) ---- (b,-a.)=a positive quantity. (4) (b4-a,).(b4-a,).(b4-a,) -- -- (b4 —an)= a negative quantity. (n) (b, —al).(bn-a a).(bn-a,) - - - - (bn-a)=a negative or positive quantity according as n is even or odd. 276 TRANSFORMATION OF EQUATIONS. In the first series, the product is positive, since all the factors are positive, b, being great;-r than any of the roots. In the second series, one of the factors b2-al is negative, and all the others are positive; hence the product is negative. By examiing each series of factors, it may be seen that the several products are alternately positive and negative, which was the thing to be demonstrated. COR. 1. If two.numbers be substituted for x, and the results obtained by these two substitutions have like signs, then there must be some even number of roots between the two numbers, or no root between the two numbers. COR. 2. If two numbers be substituted for x, and the results obtained have contrary signs, then there must be some odd number of roots between the two numbers. CoR. 3. If any number p, and every number greater than p, gives a positive result, when substituted for x, then p is greater than the greatest root. Therefore, if the signs of the alternate terms in any equation be changed, and p and every number greater than p lenders the result positive, then -p is less than the least root of the given equation. As an example, take the equation x' —7x+ 7=0. If we substitute 1 for x we obtain for the result + 1. If we substitute 2 for x, the result is also + 1. Hence, by Cor. 1, there must be an even number of roots between 1 and 2, and as the equation has onlv three roots, there are two roots between 1 and 2; that is, each root is equal to 1 increased by a proper fraction. If we now transform the given equation into another whose roots are less by unity, we shall obtain for the transformed equation x9+ 3x2-4x-+l=0. If we substitute in succession, for x, 0.1, 0.2, 0.3. 0.4, we shall find that there is one root between 0.3 and 0.4. Hence, the first two figures of one of the roots in the proposed equation are 1.3. TRANSFORMATION OF EQUATIONS. 277 PROBLEM. (21 0.) -Hawving given an equartion of the nth degree, it is required to find another of the (n — 1)th degree, such that the roots of the former sha/l be limits to the roots of the latter. Let xn +Ax + Ax - - A- - 1n-x + A= -O be the given equation, whose roots are a,, a,, a, - - - an, taken int the order of their magnitude. If we transform this equation into another whose roots are less by r, we have the following operation: 1+A1+A - - - - An_2+ A_, 1 +An( r B1r Bn,_r Bn-2r B,,-r B, B2 Bn_ Bn_1 B r Cir Cn-_3r Cn 2r C1 C2 C,_2 - C.-, In this transformation, we represent A,4+r by B1, A2+B,r by B,, and so on. The co-efficient of x in tile transformed equation is Cn1. By retracing the steps we have Cn-1= An_ + rB_2 + r. Cn_2 (a) =An1_,+ (A_ B+'n_3) + 2(An-2+B_, r + Cn_,3r) (b) = A_1 + 2rA_,+~ 21'Bn_-3 +~.2 C._3 (C) = An,_ + 2roAn_ + _2r2(A-,_ + -B,-4) + r'2(A,,_3 + rB,,_4 +? Cn,_,) (d) -An_, + 2r.A_, 3rAn_3 +3r3B_4+r3'C_4 (e) =An_1 + 2rAn2+ 3r'2An-+4r'An-4 - - - (n - 1)r n-2Al +~ n"-'. (f) Or, Cn,_ =n n- + (n -1 )Arn-2 + (n -- 1)(n - 2)A2rn" — + 2An,_,-2A,_, (1) Now, since the roots of the transformed equation are less by r than those of the given equation, they are a,1-r, a2-r, a —r - - - - a-r. Since Cn,_ is the co-efficient of the first power of the unknown quantity in the transformed equation, it is, by a preceding proposition, equal to the suln of all the different products that can be formed from the n roots, taken n —1 at a time. Hence, we have 278 ThANSFOMi'1 tATION OF EQUATIONS. (2-a).(r-a)(r-a3) - - - - -to (n-1) factors - | J(r-a,).(r-a).(r-a4) " " C i-= +(r-a). (r-a3).(r-a4) 4 ( ( +(r-a,).(r-a3).(r-a4) 4 iJ If we substitute a, for r in the second member of equation (A), it is plain that all the terms will become zero which contain al, and the equation will become C = (a, -a).(a, — a)(a, - a4) - - - - to (n —l) terms. Since a, is the greatest root of the proposed equation, each of the factors in the second member of the last equation is positive, and therefore the value of Cn_, is positive. Again, if a, be substituted fur r in equation (A), it will become Cn —= (a.2-al).(a2- a3).(a2-a - - - to (n —l) terms. In this equation, all the factors in the right-hand member are positive except the first, a2-a,, which is negative, a, being greater than a.. EItlnce, this vallue of Cn_, is negative. By continuing this proc'lss of realsolling, it will be found that if a,, a2, a3, --- -on be succesively substituted for r in equation (A), the results will be alternately positive and negative. But, by a preceding proposition, if a se;ries of numbers a,, a2, a3.- - - an, be substituted for the unknown quantity in any equation, and the results are alternately positive and negative, the real roots of the equation are situated between these numbers, the numbers and the roots being arranged in the order of their magnitude. But the value of C,_l in equation (A) is equal to the value of C,n_ in equation (1). Therefore, if the series of numbers a,, a2, as, a4 - - - - an, be likewise substituted in equation (1) for r, the results will be alternately positive and negative. Hence, the real roots of equation (1) are situated between the roots of the given equation, and it is therefore the equation required. If we change r into x in equation (1), it becomes nx"-'+ TRANSFORMATION OF EQUATIONS. 279 (n —1)Axt-2 + (n —1)(n —2)A2x"-3 - 2A,,x+An-_= O, by making Cn_,=O. This equation may be derived' from the given equation by multiplying each term by the exponent of x in that term, and then diminishing the exponent of x in each term by a unit. This equation is the limiting equation. Con. 1. If we make al=a,, then r —a will be a factor of each term in the right-hand member of equation (A), which is equal to the right-hand member of equation (1), which is the limiting equation. By a preceding proposition x-a, is an exact divisor of the given equation. Hence, if any equation and its limiting equation have a common divisor, that equation must have equal roots. CoR. 2. If al=a2=a3, then (x-a,)2 will be found in each series of factors in equation (A), and the given equation and the limiting equation will have a common divisor of the form (x-a,)'. We can, then, always determine whether an equation has equal roots by ascertaining whether the proposed equation and its limiting equation have a common measure. As an example, find the equal roots of the equation x'-5x-8x+48=0O. The limiting equation is 3x'-1 Ox-8 = O. For finding the common measure of these two polynomials, we have the following Operation. X3 — 5x- 8x +48 3x —10x —8)3x3 —15x2 — 24x+144(x 3x'-1 Ox2- 8x 5x2 —_16x +144 3 3x2- 1x-8)15x2a - +48x-432(5 15x2 50Ox- 40 98)98x —392 x-4)3x'-10X-8(3x+2 3x2-]2x 2x- 8 2x-8 280 TRANSFORMATION OF EQUATIONS. Therefore x-4 is the common measure required, and there are two roots equal to 4. If we divide the given equation by x-4, we obtain the quadratic equation, x —x-12=0, or, x-x=-12; whence, x= 4, or - 3, Therefore, the three roots of the given equation are 4, 4, -3. The solutions of the following examples will test the pupil's knowledge of the principles which have been developed in this chapter, and at the same time they will serve to fix these principles in his mind. EXAMPLES. 1. One root of the cubic equation x' —6x+1 lx —6=0 is 1; find the quadratic which contains tile other two roots, and employ synthetic division in the operation. A1nis. x'-5x+6=0. 2. One root of the cubic equation x + 3x - I 6x+ 12=-0 is 1; find the remaining roots, and employ synthetic division in the operation. Ans. 2 and -6. 3. Two roots of the biquadratic equation 4x4- 14x-5x2- +31x +6=0 are 2 and 3; find the quadratic equation which contains the remaining roots, and employ synthetic division in the operation. Ans. 4x2+6.+1=0. 4. Two roots of the biquadratic equation x4 —6x + 24x- 16 =0 are 2 and -2; find the other two roots. Ans. 3t: /5. 5. Transform the equation x' + 2xs + 3x + 4x- 12340 =0 into another whose roots shall be less by 10 than the roots of the given equation. Ans. x4+ 2x3 + 663x2+4664x=0. 6. Transform the equation x'+2x'-6x — 0x-8=0 — into another whose roots are less by 2 than the roots of the given equation. Ans. +10x -x4 +42x3+86x2+70x-4=0. TRANSFORMATION OF EQUATIONS. 281 7. Transform the equation 5x4 —1 2x3 + 3x2 + 4x- 5=0 into another whose roots shall be less by 2 than the roots of the given equation, and employ synthetic division. Ans. 5x4+28x3 + 51x2 + 32x —1=0. 8. Transform the equation x'+x2-10x+4=0 into aiioth-lCwhose roots shall be greater by 4 than the roots of the givenl equation. Ans. 9, Transform the equation x'-4x' + 6x2 — 12=0 into one whose roots shall be greater by 5 than the roots of the given equation. Ans. 10. Transform the equation x'- 6x2+9x-12=O into another whose roots shall be less than the roots by 6. Ans. 11. Form the equation whose roots are 1, 2 and -3. Ans. X3- -7x+ 60. 12. Form the equation whose roots are 3 + V/5, 3-'5, and -6. Ans. x —32x+24=0. 13. One root of the equation x4-7x's+15x2-llx+2=0 is 2 + V3; what are the other roots of the equation. Ans. 2 — V3, 1, and 2. 14. The equation x — 15x2+66x-80 —0, has two roots whose sum is 13; find all the roots. Ans. 8, 5, and 2. 15. The roots of the eqhation x — 15x2 + 66x —80 =0, have a common difference; determine them. Ans. 2, 5, and 8. 16. In the equation x —6x + 11x-6 =0, one root is double another; determine all the roots. Ans. 1, 2, and 3. 17. The product of two roots of the equation x4 +x —62x2 -80x+1200=0, is 30; it is required to determine the roots. Ans. 5, 6, — 6+2/-1, and -6 —2V'-1. 282 TRANSFORMATION OF EQUATIONS. 18. Determine the roots of the equation x — 1 7x' + 94x- 168 =0, two of them being in the proportion of 2: 3. Ans. 4, 6, 7. 19. One root of the equation x4-5x3-x+5=0 is 5; find the other roots. Ans. 1 and 3(-l: V- - 3). 20. The equation x' — x2 +i7x —1=0, has two roots of the form a, -; determine all the roots. Ans. 1, 2, and i. 21. Determine the roots of the equation x4- 1Ox3 + 35x' —50Ox +24=0, they being of the form a+l1, a-i, b+l1, b-1. Ans. 1, 2, 3, and 4. 22, Solve the following equations, whose roots are in arithmetical progression:* 1. 9- 6x2 —4x+ 24=0. Ans. -2, 2, 6. 2. x93-9x2a 23x-1 6=0. 3. x - 6x2+11 Ix — 6 = 0. 4. x3 - 3x2 + 6x + 8 = 0. 23. Solve the following equations, whose roots are in geometrical progression: 1. x3 —7x2+14x-8=0. Ans. 1, 2, 4. 2. x —13x- + 39x — 27=0. 3. x' —14x2+56x-64=0. 4. x3-26x2 + 156x-216=-0. 24. Determine whether the equation x4-9x2+4x+12=0 has equal roots, and find the roots of the equation. Ans. 2, 2, -3, and — 1. * NOTE. —If the roots of the equation xn —p.r-1+qx-2 —&c.=O be in arithmetical progression, very neat formulas may be obtained for finding the least root. and the conlmon diffrencoe of tHie ro(ots. STURM'S THEOREM. 283 25. Having given the equation 2x4- 12x'+19x2-6x-9 =0, determine whether it has equal roots. Ans. It has. 26. Find the equation whose roots are less by 1.7 than the roots of the equation y'- 2y2+~3y-4=0. Ans. y +3.1y +4.87y+0.233=O. 27. One root of the equation x'-11x2+37x-35=-O being 3 + V 2; find all the roots. Ans. 3- /2, 3 + V2, and 5. 28, Transform the equation x —px2+O.x —r into another whose roots are less by p than those of the proposed equation. Ans. x' + 2p2+p'x —r= O. STURM'S THEOREM. (21 1.) Sturm's Theorem is a Theorem for finding the number -f real and imaginary roots in any proposed equation. This was a problem of great difficulty, but the ingenious mathematician, M. Sturm, has given its solution. THEOREM OF STURM. Let x"n +Axn- + Bxn-2 - -- Px + Q=O be any equation which has no equal roots, and let n-'1 + (n —1)Ax-2 + (n —2)Bx" - - - P be its first derived polynomial, formed by multiplying each term by the exponent of x in that term, and then diminishing the exponent of x by 1 in each term. Divide the given equation by the derived polynomial, and continue the division till a remainder is obtained whose degree is less by unity than that of the divisor, and change the sign of this remainder. Now, divide the last divisor by the last remainder with its sign changed, and continue the division till a remainder is obtained whose degree is less by unity than that of the divisor, and 284 STURM'S THEOREM. change the sign of this remainder. Proceed in this manner till a remainder is obtained which is independent of x. If we denote the given equation by X, its derived polynomial by X,, and the successive remainders, with their signs changed, by R2, R, R4, - - - - -, we shall have a series of functions of x, the degrees of which regularly decrease, the last, or Rm.+, is indepenldent of x. If we now substitute any number, m, for x in this series of functions, and note the number of variations of signs* in the results, and then substitute any other number, n, for x in the same series, and note the number of variations of signs in the results, then the exact number of real roots between m and n is found by taking the difference of the two numbers which express the number of variations of sign that arise from the two substitutions. This is the point to be established. For the purpose.of facilitating the demonstration of this theorem, we shall state the following lemsnas. LEMMA I. Tvo consecutive functions cannot reduce to zero for the same value of x. If we represent the successive quotients by Q,, Q2, Q- - - - QM, we shall have, firom the process which has been described for obtaining the series of functions, X, X,, R., B3 - - -, the following equations: x=x Q, -R2 (1) X,=R2Q2-R3 (2) R2=R = Q3-R4 (3) R3=R4Q4 —R6 (4) Ri _= Rm Q — R4m+ (m) * NOTE.-By "the number of variations of signs" is meant the number of changes that can be made in passing from plus to mintus, and from minus to plus in a series of signs. Thus if the series is + — + —- + ++-, the number of variations is 4. STURM'S THEOREM. 285 Let us suppose that R=-0, and R3=O. Hence, each of the terms R2, R3Q,, in equation (3) must become 0, and therefore R4=O. Since R3=O, and R,=0, we find from equation (4), that R5=0, and finally, we shall find that Rml=-0. But the equation X=O has, by hypothesis, no equal roots, and consequently X an4 X1 have no common mneasure, and therefore there must be a final remainder independent of x. lHlnce, we cannot suppose that two consecutive functions become zero for the same value of x. LEMZMiA 1I. If one of the derived fuzct o'ns reduces to zero for any part'cular value of x, the adjacentfunctions have contrary signs. For, let us suppose that R3=O, then equation (3) becomes R, = —R4. But R, and R, are the adjacent functions, and as they have contrary signs, the lemma is true. DEMONSTRATION. Let p be less than the least root of the equations, X=O, X,=O, R=o-~ R — - =, and let q be greater than the greatest root of these equations. Now, so long as p remains less than the least root of any of these equations, the several factors into which each of the equations may be divided* must be negative, and therefore the signs of the results obtained by the substitutions of any numbers for x in the above equations, which are greater than p, and less than the least root of any of the equations, must be the same for each of these numbers. Now, as p increases, it must pass successively over the roots of the given equation, and also over those of the derived equations. As p passes over one of the roots of the derived equations, that * NOTE.-It has been shown in Prop. III., page 259, that any equation may be resolved into as many factors as it has roots. 286 STURM'S THEOREM. equation will vanish; but, by Lemma I., neither of the adjacent functions can vanish for the same value.of x, and, by Lemma II., these two adjacent functions must have contrary signs. Hence, when one of the derived equations vanishes, the number of variations of signs is not altered. Now, let p gradually increase till it becomes very nearly equal to one of the roots of the equation X-O, and let this value of p be substituted. for x in the equations X=O, X1=0, and the results will have contrary signs. Now as p increases till it becomes a very little greater than this root of the equation X=O, the sign of X must have changed, but the difference between these two values of p may be made so small, that in passing fiom that value of p which is a little less than a root of the equation X=O, to the one which is a little greater than this root, gp could not have passed over any of the roots of the other functions, and consequently their signs remain unchanged. Before this small increase in the value of p the signs of X and X, were contrary,* and after * NOTE.-Let it be observed that if we substitute for x in any equation a numbel a very little less than one of its roots, and also substitute the same number for x in its limiting equation, the two results will have contrary signs. For example, take the cubic equation.r3+2x-2-6.x+4=0, and its limiting equation 3x2-+4x-6=0, and represent the roots of the former by as, a2, as, and the roots of the latter by bl, b2. Then the cubic equation may be written (x-a1).(x-a).(.-a3)=0, and the limiting equation may be written ( —bO).(x —b2) = If the roots of the two equations be arranged in the order of their magnitude, regarding al as being the least root, the arrangement will be al, bl, 02. b2. a3. If we substitute c, a number a very little less than a2, for x in the two equations, they will become (c — al).(c-as).(c-a3)=a positive quantity, and, (c-bl).(c —b)=a negative quantity. The first result is positive, since (c-a2) is a negative quantity, and (c-as) STURM'S THEOREM. 287 it, they became alike. Hence, whenever p passes over a root of the equation X=O there is a loss of one variation in sign. Suppose, that in substituting that value of p which is a very little less than a root of the equation X=O, for x in each of the equations, we obtain results which are affected with the following series of signs: X XI RB R3 R - + - + 3 variations. Now, if we increase p till it becomes a very little greater than this root, the sign of X will become changed, while the signs of the other functions will not be changed, and we: shall have the following series of signs: X X, R, R R4 + + - - + 2 variations. In substituting any number which is a very little less than a root of the equation X=O, we may have any order of signs in which the signs of X and -, are contrary. The last term, being independent of x, its sign is not changed by any of the substitutions. We have'seen that when p, by its continual increase, reaches and passes a root of the equation X —O, there is a loss of one variation. Suppose now that it increases from a number a very little greater than this root, till it reaches a number a very little less than that of the next root of this equation. During this increase ofp it may pass over some of the roots of the other equations, thus changing the order of their signs, but the number of is negative, aa being greater than a2. The factor (c-al) is positive, since c is very nearly equal to a2, and al is less than a2. Hence, two of the factors being negative, and the other positive, the result is positive. The second result is negative, since (c-bl) is a positive quantity for the reason that. a2 is greater than bl and is very nearly equal to c..It is clear that the other factor, (c-b2) is negative. Hence, the product of the two factors is a negative quantity. If we take an equation of any other degree, and reason in a similar manner, we shall arrive at the same result. 288 STURM'S THEOREM. variations is neither increased nor diminished. (Lemma II.) If we substitute this value of p, then, for x in each of the equations, we shall obtain the same number of variations of signs as we would obtain if we substituted any other value of p which is comprised between the limits of these two roots. But for this value of p, X and X, must have contrary signs, as has been shown in the note. Now let p increase till' it becomes a very little greater than the next root of the equation X=O, and the sign of X will have changed, but the signs of the other functions will not have changed, for a reason already given. Hence, the signs of X and X1 are now alike, and consequently another variation of sign has been lost as p passed another root of the equation X=0. From what has now been shown, we know that if we substitute a number p which is less than a root of the equation X= — 0, for x in each of the equations X=-O, X1=0, R=O0 Rm_-=O, and note the number of variations of sign, and then substitute a number q which is greater than the next root of the equation X=-O for x in the same equations, and notr the number of variations of signs, the difference in. the number of variations is 2, which is the number of roots between p and q. In the same manner it may be shown, that if there are more than two roots between the limits of p and q, this number may always be determined by substituting p and q for x in each of the above equations, and then taking the difference of the two numbers which express the number of variations'of signs corresponding to the two substitutions. APPLICATION OF STURM'S THEOREM. EXAMPLES. 1. Find the number and situation of the roots of the equation x' -x'- 2x+8=0. In this example we have X-=xS- x'- 2x + 1 X=3x — 2x-2 STURM'S THEOREM. 289'or finding the other functions we have the following operations: x- - 2x + 1 3* 3x — 2x-2)3-3x2 — 6x + 3(x-1 3x'-2 r2-2x - xs-4x+3 3 Multiply by 3 to avoid fractions. - 3x2- 1.2x + 9 — 3x2+ 2x+2 7)- 14x+7 - 2x+1.'. Rs=2x-1 3x-2x- 2' 2 2x-1)6x'-4x-4(3x-1 6x'- 3x - x-4 2 -2x-8 - 2x+ 1 -9.'. R3=+9 Hence, we have the following series of.functions: X xSx- x- 2x+ 1 X,- -;x — 2x- 2 R2=2x-1 R3=9 Now, as all the real roots of any equation are comprisod between' the limits of -c+ and -0o, we can readily find the number of roots by substituting +00 and — o for x in the first * NOTEr.-In the application of this Theorem. we may multiply or di-ide any of the quantities by a positive factor. but we cannot do the same with a negative factor, and the reason of this is plain. 13 290 STURM'S THEOREM. terms of the several functions; for if x be equal to infinity, any power of x is greater than the sum of all the following terms in the function, and therefore the sign of the first term is the same as that of the whole function. Let m=the number of variations of sign which is obtained by substituting + c for x, and n the number for — A; then m-n=the number of real roots in the equation. If we substitute +0O for x in the above series of functions, we have the following series of signs + + + + giving 0 variations. If we substitute — c0 for x in tho same series of functions, we have the following series of signs - + - + giving 3 variations. Therefore, 3 —0=3 is the whole number of real roots, and as the equation has only 3 roots, it follows that they are all real. We will now find the situations of the roots. For this purpose we must employ narrower limits than + oo and -Xo. Let us commence at 0, in the series -3,-2, -1, 0, 1, 2, 3, &c. and extend the limits in both directions. SIGNS. For x=1 we have - - + + giving 1 variation. x=2 " + +. + + giving 0 variations. =0O " + - - + giving 2 variations. =-1 " + + - -+ giving 2 variations. x=-2 " - -+ - + giving 3 variations. It is unnecessary to proceed any farther, since, by inspecting the column of variations, we discover that there is one root between 1 and 2, one between 0 and 1, and one between — 1 and.-2. Hence, the initial figures of two of the roots are 1 and — 1. STURM'S THEOREM[. 291 To find the initial figure of that root which is between 0 and 1, we might transform the given equation into another whose roots are less by 1 than those of the given equation, and then apply the theorem to the transformed equation. 2. Find the number and situation of the roots of the equation x — 4x2-G6x + 8=-0. In this example we find that the functions are X= -3 4x- 6x+ 8 X,=3x2 -8x — 6 R2= 17x —12 BR3=514 Substitute +o- and — O for x in the first terms of the sevsral functions, and we have the following series of signs: SIGNS. For x=o- we have -- + + + giving 0 variation. x=-o - - - - + giving 3 variations. m-n=3-0=3, the number of real roots in the given equation. For x=O we have + — - + giving 2 variations. x=1 i" — + + " 1 " x=2 " -- + " 1 ".:C- 3 " ----- q- "' 1 " x=4 " - +- " 1 " x= 5 " -++ " 1 " x=6 " + +++ " 0 " x= — " ~-+-+- " 2 " x=-2 " -- " 3.3 By examining the column of variations, we discover that the three roots of the proposed cubic equation are between the limits of +6 and -2, and that there is one root between 0 and 1, one between 5 and 6, and one between -1 and -2. Hence, the initial figures of the root are 0, 5, and -1. It may be observed 292 STURM'S THEORE[M. that 1 very nearly satisfies the given equation, when substituted for x, and therefore the root, which is between 0 and 1, nmust be very nearly equal to 1. Henc%, in order to determine the first decimal figure of this root, substitute I for x in the series of functions, then substitute 0.9, 0.8, 0.7, &c. for x in the same series, and continue the substitution till the situation of the root is determined. Thus for x= I we have — + + giving 1 variation. x=0.9 " + - + - " 2 variations. Hence, the initial figure of this root is 0.9. 3. Find the conditions that all the roots of the general cubic equation xs3ax2+bx+c=0, may be real. In this example we have the following functions: X=x-3+ax' +- bx +c X =32 + 2ax+b (A) -R,= 2(a2- 3b))xab- 9c R3= — 4a3c + a2b'- 1 8abc- 4b'- 2 7c In order that all the roots may be real, there must be no variations of sign when + Go is substituted for x in the several functions, and three variations of sign when -so is substituted for x. Hence, we must have for the conditions required (a2-3b)>o (1) aild (ab —4a'c- I 8abc-4b3 - 2 7C2) > O (2) When a-0, we have the following functions: X=x' + bx + c X1 =3x2 +b 1R=-2bx-3c (B) -R3 -4bs-27c0 In order that all the roots of the equation x'+bx+c=O may be real, the first terms of the several functions must be positive, and therefore -2bx, and — 4b3-27c0 must be positive. As -27c' is negative for all values of c, whether positive or negative, STURM'S THEOREM. 293 b must be negative, in order that -4b' and -2bx may be positive, and at the same time, 46b must be greater than 27c', or ( 3> (c )2 Functions (A) and (B) may be used in applying Sturm's Theorem to particular examples in cubic equations. In order to obtain the several functions of x, we only substitute for a and b, their values in the given example. 4. Find the number and situation of the roots of the equation xs + 1 lx2-102x+ 1 81 =0. The functions are X=x' + lx'-102x+181 X,=3x'+22x-102 R2-= 122x-393 R3= + For x-=+C-, we have + + + + giving 0 variations. x=-00 " " - " — + "- 3 " m —n=3-0=3, the number of real roots. For x-0, we have + --- + giving 2 variations. x=1 " + —- " 2 " x-=2 " + — " 2 " x=3 " v —+ " 2 " x=4 " - + - - " 0 Therefore two of the roots are between 3 and 4. In order to determine these roots within narrower limits, we will transform the given equation into another, whose roots are less by 3. The tranf4orned equation is x3+ 20x —9x + 1 = 0, and for this equation we have the following functions: X=rx3 + 20x2- 9x+ 1 X — 3x2 + 40x- 9 _R=122x-27;,= + 294 STURM'S THEOREM. For x=0 we have + — + giving 2 variations. x=0.1 " + —+- " 2 " x=0.2 " -— + " 2 " x=0.3 " +++ "- 0 " Therefore there are two positive roots between 0.2 and 0.3, belonging to this transformed equation, and hence the given equation has two roots between 3.2 and 3.3. If we transform x3+ 20x2-9x'+1 into an equation whose roots are less by 0.2, we shall obtain x3+20.6x2-0.88x+0.008=0 for the transformed equation. The functions of this equation are X==x3 + 20.6x2-0.88x+-t 0.008 X= 3x2 + 41.2x-0.88 R2 —122x-2.6 For x=0 we have + —— + giving 2 variations. x=0.01" + + —+ " 2 " x=0.02 " -— + " 1 variation. x=0.03 " + +++ " 0 " Hence, this transformed equation has one root between 0.01 and 0.02, and one root between -0.02 and 0.03. Therefore two roots of the proposed equation are 3.21 and 3.22. Since the sum of the three. roots of the equation is -11, the co-efficient of the second term taken with a contrary sign, the other root is -11 -3.21-3.22= - 17.4. NUMERICAL SOLUTION OF EQUATIONS. 295 HORNER'S METHOD OF RESOLVING NUMERICAL EQUATIONS OF ALL DEGREES. THERE are several methods for finding the roots of numeric;:i equations of all degrees, but in point of simplicity and elegantc, the method of W. G. EHorner, of Bathl, England, is superior to arty that has yet been given. Teil initial figtures of the roots of any equation may be found by Sturm's Theorem, and then the equation may be transformed into another whose roots are less than those of the proposed equation by one of these initial figures. The initial figures of the roots of this transformed equation may now be determined by Sturm's Theorem, and this process may be continued till a transformed equation of the form x + Ax"-'-Bx"- Ox + P=O is obtained, in which x i, -.. sm::tii quantity, as 0.08, for example. Now, as x is a small qualtity in the above transformed equation, the higher powers of x must be smaller still, so that we shall obtain a near value of x, in the transformed equation, if we neglect the terms which contain these powers. By omitting these terms the transformed equation becomes Ox+ P= O, or,; = —; that is, the next figure of the root may be obtained by employing the last, or penultinmate co-efficient, of each transformation, as a trial divisor of the absolute term, and with this figure of the root,:ve nmay proceed as before. Hence, by successive transformations, we may obtain the roots of a numerical equation to any required degree of exactness. (212.) It is advisable to first find the positive roots of the proposed equation. To find the negative roots, change the signs of the alternate terms, and the positive roots of the resulting 296 NUMERICAL SOLUTION OF EQUATIONS. equation, taken with a negative sign, will be the negative roots of the given equation. EXAMPLES. I. Find all the roots of the equation xs —3x-1 =0. By Sturm's Theorem, the several functions are X=xS'-3x-=I O X, —3x'- 3 R2 =6x+3 (A) R3 r 9 For x=0, we have -- + + giving 1 variation. x= 1 " — O ++ " 1 " x=2 " +- + " 0 " Hence, there is one root between 1 and 2, and the initial figure of this root is therefore 1. In order to obtain the first decimal figure of this root, we must transform the preceding functions into others, whose roots are less by unity. Thus, for the function X, we have the following Operation. 1+o0 - 3 - 1(1 1 1 -2 1 -2 -3 1 2 2 0 And transforming the others in the same way, we have the fol lowing series of transformed functions: X=x~+ 3x'-, 3 X,=3x+ 6x R,=6x+9 (B) R, _a NUMERICAL SOLUTION OF EQUATIONS. 297 For x=l, we have + + + + giving 0 variation. x=0.9 " + + + + " 0 " x —=0.8 " -++ " 1 " lHence, one root of the transformed equation is between 0.8 and 0.9, and therefore the initial figure of this root, or the first decimal figure of the corresponding root in the proposed equation, is 0.8. To obtain the next decimal figure in the root, we must transform functions (B) into others, whose roots are less by 0 8. For transforming the first of these functions, we have the following Operation. 1 3 0 -3 (0.8 0.8 3.04 2.432 3.8 3.04 -0.568 o.8 3.68 4.6 6.'72 0.8 5.4 And transforn'ng the others in the same way we have X=x'+ 5.4x2 + 6.72x- 0.568 Xl= 3x2+ 10.8+6.72 R26x + 13.8 (C) R3 9 If we were to apply Sturm's Theorem to this series of functions, we should find that the function X had one root between 0.07 and 0.08, and therefore the next decimal figure of the corresponding root in the proposed equation is 0.07, and the toot itself, as thus far determined, is 1.87. Now, according to what has been shown, the value of x in the first equation of functions (C), may be nearly found by dividing the absolute term, 0.568 by 6.72, the co-efficient of x; and this division gives for this value of x 0.08, nearly. Now, if we transform this equation into another whose roots are less by 0.08, we obtain for the transformed equation x' + 5.64x2 + 7.6032x + 0.004672=0. 298 NUMERICATL SOLUTION OF EQUATIONS. As all the terms of this equation are positive, it follows that all its roots are negative. IIence, in diminishing a positive root of the equation, xS+-5.4x2 + 6.72.x- 0.5G8=0, by 0.08, we must have rendered it negative, and hence this positive root is less than 0.08. It will be observed that the last tlansformration caused the absolute term to change its sign. Now, it is plain that the absolute term cannot clhange its sign, unless the co-efficient of x chlanlges its sign at the same time. In obtaining, thlen, a positive root of an equation, we must recollect, that whenever a figure of the root is obtained by means of the trial divisor, which causes the sign of the absolute term to change, but not that of the co-efficient of x, this figure must be diminished, till one is obtained which does not cause such a change of signs. Since the first figure of the root of the equation X3 + 5.4x2 + 6.72x-0.568-=0, is 0.07, we will, in order to obtain the next figure in the root, transform it into another, whose roots are less by 0.07. For this transformation, we hive the following Operation. 1 5.4 6.72 -0.568 (.07 0.07 0.3829 0.497203 5.47 7.1029 -0.070797 0.07 0.3878 5.54 7.4907 0.07 5.61 Hence, the transformed equation is x+- 5.6 1x2a -.4907. — 0.070797 = 0 0.070797.'. -- -- =0.009, nearly. 7We can contnue this process of successive transformations4907 We can continue this process of successive transformations, till NlUMERICAL SOLUTION OF EQUATIONS. 299 the root of x is obtained to any required degree of exactness. For the sake of brevity, these transformations may be condensed into one operation, as follows: Operaction. COL. I COL. II. COL. III. ROOT. 1 0 -- 3 - 1 1.87938 1. 1 - 2 1 - 2 *3 1 2 2.43'2 2 *0 -*-0.568 1 3.04 0.497 203 *3.8 3.04 -*0.070797 0.8 3.63 0.067871439 4.6 *6.72 -*t.002925561 0.8 0.s 29 0.002278084257 *5.47 7.1029 — 0.000647476743 0.07 0.;8878 o.54' 7i.4907 0.07 0.050571 *5.619 7.541271 0.009 0.050652 5.628`7.591923 0.009 0.001691 19 *5.6373 7.59361419 0.0003 0.00169128 5.6376 7.59560547 0.0003 *5.63798 By the application of Sturm's Theorem, we find that the initial * The co-efficients of the successive transformed equations are indicated by asterisks. 300 NUMERICAL SOLUTION OF EQUATIONS. figure of one of the negative roots is -0.3, and for the successive transformations, we have the following Operation. COL. I. COL. II. COL. IIT. ROOT. 1 0 -3 — 1 1-0.3473 - 0.3 0.09 0.873 - 0.3 2.91 -*0.127 - 0.6 0.18 0.107696 - 0.3 -'2.73 — 0.u19304 — *0.94 0.0376 0.018522077 - 0.04 - 2.6924 — *0.00)781923 - 0.98 0.0392 - 0.04 - 2.6.532 — *1.027 0.0071 89 - 0.007 - 2.646011 - 0.034 0.007'238 - 0.007 — 2.6838 3 3 *0.0413 Since the algebraical sum of the three roots is equal to the coefficient of x2, the other root is equal to 0-(1.8793-0.3473)= — 1.5320. (21 3.) Let it be observed that each successive figure in the decimal pairt of the root extends Col. I. one decimal place, Col. II. two, and Col. III. three decimnal places. We may also observe, tlat the co-efficients of x' and x in the successive transformed equations change in value very slowly after the third or fourth transformation. These co-eficients are indicated by the asterisks in-eacll column. In the operation for finding the positive value of x, in the preceding example, it will be noticed that the co-efficient of x inl the last transforlned equation is 7.59 nearly, and fhat it is 7.59 nearly, in the last transformed equation but one. He-ice, after three or four decimals of the root have been found, we may find several of the remaining decimals of the root by dividing the NUMERICAL SOLUTION OF EQUATIONS. 301 absolute term in the last transformed equation, by the co-efficient of x in that equation. Now, since the last decimal figure ill the root extended this absolute term three decimal places, and the coefficient of x only two de cimnal places, we may omit, in per forming this division, a number of decimals in the dividend, and a number less by one, in the divisor. The accuracy of the result will not be much affected. The division mrayjbe perfornled by the abridged method' for the division of decimnals. For finding three or four additional figures of the positive root of x, in the last example, we have the following Operation. 7.59560)0.0006474767 (0.000085243 6076480 398287 379780 18507 15191 3316 3038 278 227 Hence, x=1.879385243. The last figure should be 2 instead of 3. 2. Find the roots of the equation x+ 1 x2- 102x+181=0. This equation is one of the examples in the application of Sturm's Theorem, and we then found that the roots were nearly 3.21, 3.22, and - 17. * See Perkins's Higher Arithmetic, and Davies' University Arithmetice 02 NUMERICAL SOLUTION OF EriQUATIONS. Operation. COL. I. COL IL COL. III. 1 11. -102 181 13.213127 3 42 -180 14 - 60 1 3. 51 -0.992 17 - 9 0.008 3 4.04 -0.006739 20.2 -4.96 0.001261 0.2 4.08 -0.001217403 20.4 -0.88 0.000043597 0.2 0.2061 20.61 -0.6739 0.01 0.2062 20.62 -0.4677 0.01 0.061899 20.633 -0.405801 0.003 0.061908 20.636 -0.343893 0.003 20.639 343893)0.000043597(0.000127 34389 ~q208 6878 2330 2406 The other roots may be found in the same way. They ar. 3.229521, and -17.44265. 3. Find a root of the equation x-+xo +x,2+3x-100=0. NUMERICAL SOLUTION OF EQUATIONS. 303 By Sturm's Theorem, we have the following functions: X= -x4 +x'+ x3x — 100 Xl — 4x'3 3x2 2x + 3 22-=- 5x2-34x 1603 B3- -1132x+ 6059 R4 — If in these functions, we substitute -3 for x, and then -4, and note the number of variations of sign in encli case, we shall find that there is a loss of one va,'i.'iotl (1:f sign, and hence the initial figure of one of the roots is — 3. To find this root we have the following Operation. -1 + 1 -3 -100(3.43 3 6 21 54 2'7 18 46 3 15 66 41.6896 5 22 84 4.3104 3 24 20.224 3.84456501 8 46 104.224.46583499 3 4.56 22.112 11.4 50.56 126.336 0.4 4.72 1.816167 11.8 55.28 128.152167 0.4 4.88 12.2 60.16 0.4 0.3789 12.63 60.5389 The student may find three or four more decimals of this root. The root is -3.433577, and the other root is 2.8028. Two of the roots are imaginary. 4, Find one root of the equation x9 -22x-24=0. Ans. x=5.16227. 804 NUMERICAL SOLUTION OF EQUATIONS. 5, Find a root of the equation x3 x — 500=0. Ans. x='7.617279. 6. Find a root of the equation x3s +x -x-100=0. Ans. x=4.264429. 7, Find a root of the equation 2x +3x —4x-10 —0. Ans. x= —1.624819. 8, Find a root of the equation x4- 12x2+ 12x- 3=0. Ans. x=2.85808. 9. Find a root of the equation x- -9=0. Ans. x=2.0800. 10f. Find a root of the equation xs +2x2 + 3x=- 13089030. Ans. x=235. 11. Find a root of the equation x3- 17x2+54x=350. Ans. x — 14.05406. 12, Find a root of the equation 1 53 x9-18-x +29- 0. 12 or 208 Ans. x=2.333, or 24. APPENDIX. APPLICATION OF ALGEBRA TO GEOMETRY.' (214.) No general rules can be given for the solution of geometrical protlens, but if the student is familiar with the principles of algebra and geometry, he will be enabled, by practice, to solve such problems as follow, without much difficulty. PROBLEM I. Having given, in a right-angled triangle, the sum of the base and perpendicular, and the hypothenuse, it is required to determine the base and perpendicular. Let A B C represent the right- C angled triangle, right-angled at A. We shall represent quantities which are known, by the first letters of the al)habet, and those which are not known, by the last letters of the alphabet. Let x=A C, the perpendicular, y=A B. the base, x+y=a, the sunm of the base and perpendicular, and c=C B, the hypothenuse. e NoTe.-The Geometry referred to is Davies' Legendre, new edition. 806 APPENDIX. Then, x + y=a, (1) And, x'+-y2=c. (2) (Geom. Bk. IV., Prop. XT.) Eq. (1) squared, gives x2+2xy+y-=a2 (3) Eq. (3) —Eq. (2), gives 2xy-=a —c' (4) Eq. (2)-Eq. (4) gives x'-2xy+y2=2c2-a2 (5) The sq. root of eq. (5) gives x —y=- -2'2c2 —a2 (6) Eq. (6)+eq. (1) gives 2x —Za4-/ 2 —a2 (7) Eq. (1)-Eq. (6) gives 2y'-=a::V/2c2 —a (8) x=-a= 2c2-a2 (9) And, y=~a:ai2c20 —a2 (10) If we let a= 7, c=5, and then substitute these values for a and c in equations (9) and (10), we shall find that x-4, or 3, and that y=3, or 4. We will now give the geometrical solution of this problem, in order that the student Inay coinpare the twvo imethods of solution. A Let the line A represent the suim of tlle base and B perpendicular, and let the line 13 represent the given h\kvpothenuse. Take F C equal to A, and from the point C, as a centre, with a radius equal to B, describe F~/ 4"05~ [~,!' ~Ic the indefinite arc E n. n D From the point F draw the line F E, making the angle E F D equal to half a right-algle, and intersecting the arc E n at E. From the point of intersection, E, draw E D perpendicular to F C, and draw E C. E D C is the triangle required. For, since the angle E F C is equal to half a right-anole, the other acute angle F E D, in the right-angled triangle E F D, must also be half a right-angle. Hence, the triangr!e E F D is issseel:s, and ED equals FD. Therefore, ED+L)D C=FC=A. By construction, C E=B, the given hypothenuse. Hence, the triangle ED C is the one required. ALGEBRA. APPLIED TO GEOMETRY. 807 PROBLEMI II. In a right-angled triangle, having given the perimeter, and the peipendicular let fall fjoin the right-angle on the hypothenuse; to determine its sides. Let A BC represent the required triangle, right-angled D at A, ald AD the perpendicular let fill on the bhypothenuse. Let x + y=-A C, A. x-y=A B, Whence p-2x=B C. p — the perimeter, ) and c= the p,erpendicular A 1). (xz+y)2 + (x-y)2, (p- 2)2. (1) (Geom. Bk. IV., Prop. XL) 2- y2_cp-2cx (-) Equation (2) is obtained by finding two expressions for the area of the trianfflo B A C, and then equating themn. Thu4, (x + y).(x —y) — x'-2 _ -doubllc the area of the triangle A B C, AndcX(p-2x)=-cp-'-cx= " " " By expmanding', and':liting terms,in equation (1), we have,2 2:+,2y -2 4_ p+ 4x2 (3) By transposing, - 2x2 2y =p2 — 4px (4) Eq. (2)X2, gYi-es 2x'-2?/ _2cp-4cx (5) Eq. (4)+Eq. (5) gi\es' 0=-p2 + 2cp-4px-4cx (6) Transposing, 4cx + 4px =-p2 + 2cp; WThence, x- 2c= b, by putting b for this 4c+4p- value of x. Substitute the value of x for x in equation (5), and it beconmes 2b2 -2y2= 2cp-4bc, (7) Or, y2 2bc-cp + b2, (8) Whence, y- V 2/'bc-cp + b. (9) 08 APPE-NDIX. GEOMETRICAL SOLUTION. G Upon D E; the given perilmeter, describe a D / - se monent such that A -B every angcrle inscrlibed in it mmy be eqlual to 135~. Dlralv E G perpendicular to D E, and equal to the given perlendicularl. IDraw G F parallel to D E, inters, cting the are D C E inl the points C and F. Froim one of these p),ilnts, as C, draw the chords D C and C E, andll bisect tllese chords by tile perlendicliclars A P and C B, and fhorn the points A and B where tlhese perpendiculars meet the line DE, draw A C and C B. A C B is the triangle required. For, since D P=P C, by construction, and A P is common, the two rigrlt-argled trian(rles, A I' D and A P C, are equ:!l. (Georm., Book I., prop. V.) HIence, AD= A C. In a similhar manner, it may be shown, that C B —B E. Tfilerefol'e, the perimeter of the triangle A C ) is equal to D E, the given perimeter. Furtherl, the angle CAB equals the, sumr of the an gles AD C, and A( D, or twice the aingle AD C. (Ge(,m., Book I., prop. XXV., Cor. 6.) In the s:une matnner, it Imay be shlown, tlat the angle A B C is equal to twice the anole A E C. Tlme angle C D E is measured by one half of the are E C. (Geoni., Book llI., plrop. XVIII.) Hence, the angle C A B is mIeasured by the wlhole are C E. For a like reason, the angle C B A is Inmsured by the arc D C. Hence, the two angles A B C and C A B are togetler measured by the arc D C E. But, by constlucetion,* D C E is an arc of 90~. IIence, the sum of the two angles C A B and AB C is equal to one right-angle, since an arc of 90~ is the measure of a right-angle. Therefore, the angle A C B is a right-angle, (Geom., Book I., prop. XXV.) As the perpendicular,lt fall * By construction, the angle D C E=1350. Hence, the are which measures it must be an are of 2X1350=2700, and as this arc, together with the are D C E, make up the whole circumference, or an are of 360~, it follows that D C E is an are of 90~. ALGEBRA A!'I'LIED1 To) GEOMETRY. 309 from the right-angle AC B upon thle hypothenuse AB, is obviously equal -to E G, whicn, by construction, is equal to the given perpendicular, it follows that the triangle A BC answers all the conditions of the probleml, and it is, therefore, the triangle required. PROBLEM IT. Having given the Lase. and perpendicular of a right-angled triangle, it is required to determine the hypothenuse. Let AB C repr(csent the right- C angle ti'iangle, right-angled I:t A. From the point B, ats a centre, with a radius A1B, deserilbe t:he circumference of a circle, intersecting the hypothenuse C B in D. AC is a tangent to the circumference. (Geom., Book II., prop. IX.) Let AC=:., AB Ob. and, C B3-x; %\ i-lwc,, C l-CB- B=C B —A B=x-b, C E-=C B+B E=C B + A B-=x+b. C E: A C:: A C: C D, (Geom., Book IV., prop. XXX.) Or, x+b:a::a:x-b, Or, x2 —b2=a2-' (1) Whence, x2= a2 + b6, (2) And, x= Vt2+b2, (3) From equation (2), we see that the square described on the hypothenuse of a right-angyled triangle is equal to the sum of the squares described upon the base and perpendicular, a theorem which is, perhaps, the most important of any in elementary geometry. 310 APPENDIX. PROBLEM IV. 7Iaving given the base and perpendicular of a plane triangle, it is required to bisect it by a line drawn parallel to the base. C LJet A B C represent the triangle, of whichrthe base A B, and thle perpendicular C D, are given. Let A B= b, CGD =p, and CG=x. The line FE is drawn parallel to the base. By similar triangles, we have CD:A B:: C G': F E, Or, p:b:: x:FE;.'. FE=-. ~bp= the artea of tri-iangle A B C, bx' and - =the area of tria,,lle C F E. (Geom., Book IV., prop. VI.) Therefrte, \by tlie rol'(,!eln, we have' — _2;b' (1) Thcretbluf!',~, if ve ttk, t}k l line (C G equal to? 22, and draw, through the point G, FE pa'allel to thie bIse A B, the'triangle A B C is divided inito two eqiial lats. The form whicll the value,f x as-;,t,ines sur('gests the following GEOMETRICAL SOLUTION. C Draw C D perpendicular to tlie base A B antld then draw C K e.'l t, C 1). anld,7 (: >\making witil; t:ill i anle A B ~of4a~. 14.5* v lea KF parallel D to A B, and the triangle ACB is divided in the manner required, that is, the triangle F C E is equal to one half of the triangle A B C. ALGEBRA APPLIED TO GEOMETRY. 311 For, since the angle G C K=450, by construction, it follows that the angle C K G must also equal 45~. Hence, the triangle G C K is isosceles, and C G=G K.. Hence, (C K)'=(C G)2+ (G K)'=2(G C)2=(C D)2. By silnillar tl-ia)lgles, we have ABC: FCE:: (B C)': (CE)f:: (C D)': (G C)':: 2: 1. Hence, the triangle F E C equals one half of the triangle A B C, which was the t;hing to be hlown., PROBLEM V. Having given the three sides of a plane triangle, it is required,to determine its area. Let A B C represent the triwangle, whose sides are AB -c, A C=b,;nd B1 C —a. Let C fhill the p)erpetelicilarl B 1), anlld let D C= -x. c =a' + b — -l2?; ((Geom., BVolk IV., prop. XII.) i 2 + 2_ _c2 Whence, x 2b D4a'b2-(a2 + b'_c')2 bF(a'+2tb' c').(c' - (ac - 2a.b-). 4b'2 2b _r( + b)'- C).(c,- (a- b)2)]it-= [(a +-+ c)(- + b-c)(c+a-b)(a-a+-b)] 2b If we multiply the value of B D in its last form, by one half of A C, or ib, the product will be the area of the triangle A B C, and calling this area S, we have s=I (aJb+c).(aJ+b —c)(c + a —b).(c-a + b)l 812 APPENDIX. The several changes in the form of the value of B D were made by the application of the theorem, that the difference of the squares of two quantities is equal to the product of their sum and difference. The quantities within the parentheses were regardecd as being single quantities. PROBLEM VI. How high above the surface of the earth must a person be raised to see one-third of its surface 2 Let A F B G represent a plane section of the earth madle by a plane passing through its centre D. Let C represent the required point of elevation. Draw CD, and thle two tanzeults A C and B C. A atl iB wit b.,oilts in that circu.L i'.rICHe,,tJ thie -uriftce of tie eartll, twhichI bounds the person's vi.sio. Draw AD, B D, and A B. Let x=FC the required altitude, antd 3D=the radius of the earth. Since the surface of the zone.hose altitude is F E is equal to i of the surface of the whole sphere, or i of the surface of the hemi-sphere, and since the surfaces of zones are to each other as their altitudes, (Geomr., Book VI1I., prop. X., cor. 4,) it'follows that D F is to FE as 3:2. Hence, F E-= D, and E D=D. Now,'in the right-angled triangles A E C, A C D, and A E D, we hWave (A E)'= (A D)'y- (E D)'= 92 -- D: -= 8D (A C)'=(A E)' + (C E)= 8D'+(x + 2D)=12D' + 4Dx +, Aud (A C)'= (C )' —(A D)'= (x+ 3D)'- 9D0=x$2 + Dx. Hence, x2 + 4Dx+ - 12D2' + 6.Dx; Whence, xm6D. ALGEBRA APPLIED TO GEOMETRY. 318 Therefore, the person must be elevated to the height of the diameter of the earth, in order to see one-third of its surface. In this solution we have made no allowance for the atmospheric refraction of light, and we have-assumed that the earth is a sl:-,l'e. PROBLEM VII. Having given the area of a rectangle inscribed in a triangle, it is required to determine the trianigle. PROBLEM VIII. Having given. the perimeter of a right-angled triangle, and the radius of its inscribed circle, it is required.to determine the triangle. PROBLEM IX. JIuily yiV!c,, the hypothenuse of a right-angled triangle, and the side of the insclribedsquare, it is required to determine the triangle. PR(')BLEM I. Havingy y ve the side of an inscribed square, in a right-angled triangle, and the radius ofjits is inscribed circle, it is required to determine the triangle. PROBLEM XI. Find the area of a regular octagon, whose side is a.' PROBLEM XII. Find the area of a regular hexagon, whose side is a. PROBLEM XIII, Find the area of a regular polygon of twelve sides, each side being equal to a. 314 APP EN-I)IX. PROBLEM XIV. Determine the radii of three equzal circles, described in a given circle, which are tangent to euch other, and to the given circle. PROBLEA XV. The distances from a point P to two parallels, are denoted by a and b. It is required to draw a line through the point P intersecting the two parallels, so that the part of the line between the two parallels mry be equcal to a given line c. PROBLEM XVI. Draw a circle througy to given points, and tangent to a given line. PROBLEM, XVII. Four equal b-alls, 2a inches ii diameter, are placed so as to form a trria, qalar pyramid, and they are ta. ngent to each other. Find the caltittude of this pyramid. PROBLEIM XVIII. If R and r alre the radii of two spheres inscribed in a cone, so that the greater may touch the less, and also the b:ase of the cone then will the solidity of the cone be r(R ) 3r(R' —r) THE FARM AND THE FIRESIDE; ORn, THE ROMANCE OF AGRICULTURE. BEING HALF HOURS OF LIFE IN T1IE COUNTRY, FOR RAINY DAYS AND WINTER EVENINGS. BY REV. JOIIN L. BLAKE, D.D. AUTHOR OF FARMER'S EVERY-DAY BOOK; THE FARMER AT HOME; AND A GENERAL BIOGRAPHICAL DICTIONARY. COMMENDATIONS OF THE PERIODICAL PRESS. From the Ohio Farmer. Di. BLAKE is justly regarded as one of the best agricultural writers in the country, and the work before us is one of the most interesting productions of his pen. Its peculiar merit, as a work for the fireside, consists in the variety of its topics, its plain and simple, yet attractive style, its fine engravings, and the interesting romance which the author hlas thrown around Rural and Agricultural Life. In this respect, "' The Farm and the Fireside" is a work well adapted to the youthful mind. We hope it may be extensively read, as it cannot fail to improve the taste and promote inquiry in the most useful and practical of all departments of science. From the New-York Evangelist. The aim of the author has been to throw over labor, home and agricultural life, their true dignity and charm; to introduce the farmer to the delights and privileges of his lot; to embellish the cares of toil with those kindly sentiments so naturally associated with the country and its employments. It is a pleasant book-one that will enliven the 2 COMMENDATIONS OF THE PRESS. fireside, elevate and purify the thoughts, and, at the same time, impart a great deal of valuable agricultural knowledge. We know not how the natural trains of thought of the farmer could be more aptly met or more safely and agreeably led, than they are by these brief ans varied discussions. The range is as wide as life itself-morals, religion business, recreation, education, home, wife and daughters-every rela tion and duty is touched upon, genially and instinctively. From the New-York Tribune. We have here another highly instructive and entertaining volume from an author, who had laid the community under large obligations by the enterprise and tact with which he has so frequently catered to the popular taste for descriptions of rural life. Its contents are of a very miscellaneous character, embracing sketches of natural history, accounts of successful farming operations, anecdotes of distinguished characters, singular personal reminiscences, pithy moral reflections, and numerous picturesof household life in the country. No family can add this volume to their collection of books without increasing their sources ol pleasure and profit. From the Northern Christian Advocate. The venerable author of this work is entitled to the warmest thanks of the public for his numerous and valuable contributions to our literature. He is truly an American classic. We have been conversant with his writings for the last twenty years, and have always founa them both useful and entertaining in a high degree. His writings on Agriculture contain much real science, with numerous illustrative inci dents, anecdotes, and aphorisms, all in the most lively and pleasing manner. By this means the dry details of farming business are made to possess all the interest of romance. The style is clear, easy, ant dignified; the matter instructive, philosophical, and persuasive. This work is an eloquent plea for the noble and independent pursuit ot Agriculture. From the National Magazine. We return our thanks for the new volume of Dr. Blake, " The Farm and the Fireside, or the Romance of Agriculture, being Half Hours and COMMENDATIONS OF THE PRESS. 8 Sketches of Life in the Country," a charming title, certainly, and one that smacks of the man as well as of the country. Eschewing the dryness of scientific forms and erudite details, the'author presents detached, but most entertaining, and often very suggestive Articles on a great variety of topics-from the " Wild Goose" to " Conscience in the Cow,"-from the " Value of Lawyers in a Community " to the "Objec tions to early Marriages." The book is, in fine, quite unique, and just such a one as the farmer would like to pore over at his fireside on long winter evenings. From t-he New-York Recorder. "The Farm and the Fireside," is a most interesting and valuable work, being a series of Sketches relating to Agriculture and the numerous kindred arts and sciences, interspersed with miscellaneous moral instruction, adapted to the life of the farmer. From the Germantown Telegraph. We have looked through this work and read some of the " Sketches," and feel a degree of satisfaction in saying that it possesses decided merit, and will commend itself, wherever known, as a volume of much social interest and entertainment. The sketches comprise "'Country Life" generally-some of them are just sufficiently touched with romance to give them additional zest; while others are purely practical, and relate to the farmer's pursuit. We regard it as a valuable book, and are sorry our limits will not admit of bestowing upon it such a notice as it really deserves. From Harper's New Monthly Magazine. This work is a collection of miscellaneous sketches on the Ronmance of Agriculture and Rural Life. Matters of fact, however, are not excluded from the volume, which is well adapted for reading in the snatches of leisure enjoyed at the farmer's fireside. From the True Democrat. Dr. Blake's publications are all of a high order, and are doing a most important work towards refining the taste, improving the intellect, and 4 COMMENDATIONS OF THE PRESS. rendering attractive the various branches of Agricultural science. Indeed we know no author who has so successfully blend d the romantic, the rural and beautiful with the poetical, the uneful, and true, as has Dr. Blake. This is a peculiar feature of all his works His style is plain, simple, and perspicuous; and, with unusual tact and judgment, he so manages to insinuate himself upon you, that you are at once amused, delighted, and instructed with the subject he is discussing. In this respect he relieves the study of agricultural science from the abstruseness of technical science, and thus renders himself easily comprehended by all classes of readers. From the New-York Evening Post. The author's object is to improve the soil through the mind-not so much to place in the hands of farmers the best methods of raising large crops-for these he refers them to Leibig's Agricultural Chemistry, and to treatises of the like description —but to make them feel how usefill, agreeable, and ennobling, is the profession of agriculture, and, above all, how profitable the business must become when skilfully and economically carried on. These money-making considerations are, we suspect, the best moral guano that can be applied to the farmer s spilitual soil. The author writes well of the countryman's independence, the good effect of fresh salubrious air upon his health, and the moral influence of his every-day intimacy with nature upon his mind. "The Farm and the Fireside " is a kind of Bucolical annual-to be read in seasons of leisure-intended for the Phyllises and Chloes, as well as for the Strephons and Lindors. Dr. Blake has enriched it with curious anecdotes of domestic animals, and of the best way of raising and selling them. He describes model-farms, and the large incomes made from them. He expatiates on the advantages of matrimony in rural life, expounds the true theory of choosing a helpmate, discusses the advantages of Sunday-Schools, ar.d recommlends neatness of attire and punctuality in bathing. In shoi t, this volume is as diversified in its aspect as the small garden of a. judicious cultivator, where, in a limited space, useful cabbages, potatoes, and all the solid esculent greens, grow side by side with choice fruits and pleasant flowers. , P. WP. LLIS'S SELECT VIRKS, IN UNIFORM 12M,, VOLS. RURAL LETTERS, AND OTHER RECORDS OF THOUGHTS AT LEISURE, embracing Letters from under a Bridge, Open Air Musings in the City, "Invalid Ramble in Gerusany," "Letters from Watering Places," &c., &c. I voL Fourth Edlition.'There is scarcely a page in it in which the reader will not remember, and turn to again with a fresh sense of delight. It bears the imprint of nature in her purest and most joyous forms, and under her most cheering and inspiring inltuences." —. Y. Tribz,e. " If we would show how a modern could write with the ease of Cowley, most gentle lover of nature's gardens, and their fitting accessaries from life, we would offer this volume as the best proof that the secret has not yet died out."-Literary World, PEOPLE I HA\VE MET, or Pictures of Society and People of Mark —drawn under a thin vell of fiction.:By N. P. WILL&s 1 voL, 12mo. Third Edition. It is a collection of twenty or more of the stories which have blossomed out from the Sunmmer soil oftthe author's thoughts within the Inst few years. Each word in some of them the author seems to have picked as daintily, for its richness or grace, or its fine fitness to his purpose, as if a humming-bird were picking upon his quivering wing the flower whose sweets he would lovingly rifle, or a belle were culling the stones for her bridal necklace."-N. Y. Independesnt. "The bhnrk-mhraces a great variety of personal and social sketches in the Old World, and concludes with some thrilling reminiscences of distinguished ladies, including the Belles of New York, e:". —:r ae.ccpucv. LIFE HERE AND THERE, or Sketches of Society and Adventure'at far-apart timer and places. By N. P. WILLIS. 1 vol., 12mo. "This very agreeable volume consists of sketches of life and adventure. all of them, the author assures us, having a foundation strictly historical, and to a great extent auttobiogra. phical. Such of these sketches as we have read, are in MIr. Willis's happiest vein-a vein, by the way, in which he is unsurpassed" —Sartain'a M lagazine. " Few readers who take up this pleasant volume will lay it aside until they have perused every line of its contents." —Jersey Journacl HURRYGRAPHS9 or.Sketches of Scenery, Celebrities, and Society, taken from Life By N. P. WILLIS. 1 vol, l2mo. Third Edition. "S' ome of the best specimens of Mr. Wlllis's prose, we think, are hereto actained," 2. Y. Evanegelist. "In the prcsent volume, which is filled with all sore 4 enticements, we prefet the descriptions of nature to the sketches of character, and the dusty road-side grows delightful under the touches of Willis's blossoming-dropping pen; aed when we come u, the mountain and lake, it is like revelling in all the fragrant odors -i PatlA"-dBw-soat, At"a THE FRUIT GARDEN. Suer)"b EDMnONw. A Treatise Intenled to nilustrte end explain the Physiology of Fruit Trees, the Theory and Practice of all operations connected with the Propagation, Transolanting, Pruning antd Training ol Orchard and Garden Trees, as Standards, Dwarfs, Pyramids, Eslaliers, &c., the laying out and arranging different kinds of Orchards and Gardens. the selection of suitable varieties for different purposes and l'calities, gathering and. preserving Fruits, Treatment of Discase, Destruction of Insects. Descriptions and Uses of Implements, &c., illustrated with upward of one hundred ard fifty figures, represent. tng different parts of Trees, all Practical Operations, Forms of Trees, Desi-gns for Plantations, Implements, &c. By P. Barry, of the Mount Hope Nurseries, Ilochester, IN'ew York. 1 vol. 12mo. U It is one of the most thoronugh worlks of the kind we have ever seen, dealing in particular as well as gene-alities, and imparting many valuable hints relative to soil, manures, pruning and transplanting.'"-Boston Gazette. "A mass of useful information is collected, which will give the-work a value even to those who possess the best works on the cultivation of fruit yet published."-Evening Post. "Itls work is one of the completest, and, as we have every reason for believing, most accurate to be obtained ou the ubjuct,."- —. V. uvangyelit. " A concise Manual of the kind here presented has long been wanted, and we will venture to say that, should this volume be carefully studied and acted upon by olr industrious farmers, the quantity of fruit in the State would be doubledl in five years. and the quality, too, greatly improve(L Here may be found alvice suited to all emergencies, and the gentleman farmer may find direction for the rimplest matters, as well as those which trouble oldler heads. The book, we think, will be found valuable."-Newark:Dailg.Ad/sertiser. " It is full of directions as to the management of trees, and buds, and fruits, and is a valuable and pleasant Book-"-Albany Evenisng Jouernal. " The work is prepared with great judgment, and founded on the practical experience of the Authl,)r-is of far greater value to the cultivator than most of the popular compilations on the subject,"-N. Y. Tributn, This Book swuplies a place in fruit culture, and that is saying a great deal, while we have the popular works of Downing, Thomas, and Cole. Mr. Barry has then a field to himself which he occupies with decided skill and ability.-Prairie Farmer. Among the many works which within a few years have been brought before the publio designed to give impulse and shape to practical husbandry and horticulture, this is among the best, and in many respects, the very best. It ought to be in every family in the United States.-As8htabula Sentinel. It is a manual that ought to be in the poasseeiou of every mua that owns a foot of land, ~-N. Y. Observer Both to the active fruit grower and the novice in Pomology, this book will be found Invaluable.- Arthur's lonms Gazette.