WILLIAM E. DEAN, PRINTER AND PUBLISHER, N1o. 2 ANN STREET, NEW YORK, OFFERS TO THE TRADE IN QUANTITIES ~AE FOLLOWING CLASSICAL AND SCHOOL BOOKS, INFANT TIHERAPEUTIES; by John B. Beck, M. D. NELIGAN'S MATERIA MEDICA; by Prof. McCready, M. D. RYAN'S ALGEBRA. 12mo. SCOTT'S COMMENTARIES ON THE BIBLE, 3 vols. Royal Octavo. LEMPRIERE'S CLASSICAL DICTIONARY; containing the principal Names and Terms relating to the Geography, Topography, History, Literature, and Mythology, of the Ancients. Revised, corrected, and arranged in a new form, by Lorenzo L. da Ponte and John D. Ogilby. 8vo. ADAMS' ROMAN ANTIQUITIES; a new Edition, from a late English Copy, illustrated with upwards of 100 Engravings on steel and wood, with notes and improvements, by L. L. da Ponte, Editor of the Seventh, Eighth, Ninth, and Tenth American Editions of Lempriere's Classical Dictionary. LATIN READER; Parts I. & II. by Frederick Jacobs and Frederick William Doring; with Notes and Illustrations, partly translated from the German and partly drawn from other sources. By John D. Ogilby. 12mo. MAIR'S INTRODUCTION TO LATIN SYNTAX; from the Edinburgh Stereotype Edition; revised and corrected by A. R. Carson, Rector of the High School of Edinburgh. To which is added, Copious Exercises upon the Declinable Parts of Speech, and an Exemplification of the several Moods and Tenses. By David Patterson, A. M. 12mo. ADAMS' LATIN GRAMMAR; with numerous expansions and Additions, designed to make the work more elementary and complete, and to facilitate the acquisition of a thorough knowledge of the Latin Language. By Jamas D. Johnson, A. M. 12mo. SALLUST; with English Notes. By Henry R. Cleveland, A. M. 12mo. PLAYFAIR'S EUCLID; a New Edition, revised and corrected; for the use of Schools and Colleges in the United States. By James Ryan. LEE'S PHYSIOLOGY. 12mo. COFFIN'S NATURAL PHILOSOPHY, with Questions for Practice, Experiments and Questions for Recitation. BONNYCASTLE'S ALGEBRA; with Notes and Observations, designed for the use of Schools. To which is added, an Appendix on the Application of Algebra to Geometry. By James Ryan. Also, a large collection of Problems for exercise, original and selected. By John F. Jenkins, A. M. 12mo. KEY TO BONNYCASTLE'S ALGEBRA; containing correct So. lutions of all the Questions. By James Ryan. 18mo. JACOB'S GREEK READER; Corrected and Improved, with nu. merous Notes, Additions, and Alterations, not in any former edition, alsc a copious Lexicon. By Patrick S. Casserly, T. C. D. 8vo. CASSERLY'S TRANSLATION TO JACOB'S GREEK READER; for the use of Schools, Colleges, and private lessons, with copious notes, and a complete Parsing Index. 12mo. LEUSDEN'S GREEK AND LATIN TESTAMENT. 12mo. GR2ECA MINORA; with;extensive English Notes and a Lexicon. VALPY'S GREEK P GRPAMMAR; greatly enlarged and improved, by Charles Anthon, LL. D. 12mo. BECK'S CHEMISTRY; a new and improved edition. THE SCHOOL FRIENO,. ByMiss Robbins. 18mo. LEVIZAC'S FRENCH GRAMMAR; revised and corrected by Mr. Stephen Pasquier, M. A. With the Voltarian Orthography, according to the Dictionary of the French Academy. 12mo. CHRESTOMATHE DE LA LITTERATURE FRANCAISE, &c. By C. Ladreyt. 12mo. RECUEIL CHOISI deo Traits Historiques et de Contes Moraux; with the signification of Words in English at the bottom of each page; for the use of Young Persons of both Sexes, by N. Wanostrocht. Correctedl and enlarged, with the Voltarian Orthography, according to the Dictionary of the French Academy, by Paul Moules. 12mo. HISTORY OF CHARLES XII., in French, by Voltaire. 18mo. LE BRETHON'S FRENCH GRAMMAR; especially designed for persons wvho wish to study the elements of that language. First American from the seventh London edition, corrected, enlarged and improved; by P. Bekeart. I vol. 12mo. SIMPLE AND EASY GUIDE TO THE STUDY OF THE FRENCH GRAMMAR. Byv Win. P. W;Vilson. 12mo. FRENCH COMPANION, consisting of' familiar conversations on every topic that can be useful: together with models of letters, notes and cards. The whole exhibiting the true pronunciation ot the French Language, the silent letters being printed in Italic throughout the worlr. By Mr. De Rouillon. Second American, from the tenth London edition. By Prof. Mouls. 1 vol. 18Smo. BLACKSTONE'S COMMENTARIES on the Laws of England; with Notes by Christian, Chitty, Lee, Hovendon, and Ryland. Also, a life of the Author, and References to American Cases. By a member of the New York bar. 2 vols. 8vo. DUBLIN PRACTICE OF MID'WIFERY, with Notes and Additions. By Dr. Gilman. 12mo. BLAIR'S LECTURES ON RHETORIC; abridged, with questions for the use of Schools. 18mo. ENGLISH HISTORY; adapted to the use of Schools, and young persons. Illustrated by a map and engravings, by Miss Robbins. Third edition. 1 vol. 12mo. ENGLISH EXERCISES; adapted to Murray's English Grammar, consisting of Exercises in Parsing', instances of False Orthography, violations of the Rules of Syntax, Defects in Punctuation; and violations of the Rules respecting Perspicuous and Accurate ~WTriting. Designed for the benefit of private learners, as well as for the use of Schools. By Lindley Murray. 1Smo. RYAN'S ASTRONOMY on an improved plan, in three Books; systematically arranged and scientifically illustrated with several cLts and engravings, and adapted to the instruction of youth, in Schools and-Academies. 1Smo. MYTHOLOGICAL FABLES; translated by Dryden, Pope, Con-reve, Addison, and others; prepared expressly for the use of Youth. 12mo. YOUTH'S PLUTARCH, or Select Lives of Greeks and Romans By Miss Robbins. 18mo. Dean's Stereotype Edition, BONNYCASTLE'S INTRODUCTION TO ALGEBRA; CONTAINING THE INDETERSHINATE AND DIOPHANTINE ANALY SI S, AND THE APPLICATION OF ALGEBRA TO GEOMETRY. REVISED, CORRECTED AND ENLARGED By JAMES RYAN AUTHOR OF " AN ELEMENTARY TREATISE ON ALGEBRA THEORETICAL AND PRACTICAL," &C. TO WHICH IS ADDED A LARGE COLLECTION OF PROBLEAMS FOR EXERCISE, ORIGINAL A ND SELECTED, BY JOHN F. JENKINS, A.M., PRINCIPAL OF THE MECIHANICS' SOCIETY SCHOOL. NEW YORK: W. B.'DtAN, PRINTER & PUBLISHER, 2 ANN STREET. Entered according to the Act of Congress, in the year 1845, by W. E. D[)AN, ill the Clerk's office of the D)strict Court of the Southern District of Now York. ADVERTISEMENT TO THE SECOND NEtW YORK EDITION. IT would be superfluous to advance any thing in commendation of "Bonnycastle's Introduction to Algebra," as the number of European editions, and the increase of demand for it since its publication in this country, are sufficient proofs of its great utility. But to make it universally useful both to the tutor and scholar, I have given, in this edition, the answers that were omitted by the author in the original. In the course of the work, particularly in Addition, Subtraction, Multiplication, Division, Fractions, Simple Equations, and Quadratics, I have added a great variety of practical examples, as being essentially necessary to exercise young students in the elementary principles. Several new rules are introduced, those of principal note are the following: Case 12.. Surds, containing two rules for finding allny root of a Binomial Surd, the Solution of Cubics by Converging Series, the Solution of Biquadratics by Simpson's and Euler's methods: all these rules are investigated in the plainest manner possible, with notes and remarks, interspersed throughout the work, containing some very useful matter. There is also given all.the Diophantine Analysis, contained in Bonnycastle's Algebra, Vol. I. Svo. 1820, being a methodical abstract of this part of the science, which comprehends most of the methods hitherto known for resolving problems of this, kind, and will be found a ready compendium for suchedl~ may acquire some knowledge of the ANALYTIc A RT. JAMES RYAN. New York, January 1, 1822. PREFACE. TI-IE powers of the mind,'like those of the body, are increased by frequent exertion; application and industry supply the place of genius and invention; and even the creative faculty itself may be strengthened and improved by use and perseverance. Uncultivated nature is uniformly rude and imbecile, it being by imitation alone that we at first acquire knowledge, and the means of extending its bounds. A just and perfect acquaintance with the simple elements of science, is a necessary step towards our future progress and advancement; and this assisted by laborious investigation and habitual inquiry, will constantly lead to eminence and perfection. Books of rudiments, therefore, concisely written, well digested, and methodically arranged, are treasures of inestimable value; and too many attempts cannot be made to render them perfect and complete. When the first principles of any art or science are firmly fixed and rooted in the mind, their application soon becomes easy, pleasant and obvious; the understanding is delighted and enlqrged; we conceive clearly, reason distinctly, and form just and satisfactory conclusions. But, on the contrary, when the mind, instead of reposing on the stability of truth and received principles, is wandering in doubt and uncertainty, our ideas will necessarily be confused and obscure; and every step we take must be attended with fresh difficulties and endless perplexity. That the grounds, or ftudamental parts of every science, are dull and unentertaining, is a complaint universally made, and a truth not to be denied; but then, what is obtained with difficulty is usually remembered with ease; and what is purchased with pain is often possessed with pleasure. The seeds of knowledge are sown in every soil, but it is by proper culture alone that they are cherished and brought to maturity. A few years of early and assiduous application never fails to procure us the reward of our industry; and who is there, who knows the pleasures and advantages which the sciences afford, that would think his time, in this case, misspent, or his labours useless q Riches and honours are the gift of fortune, casually bestowed, or hereditarily received, and are frequently abused by their possessors; but the superiority of wisdom and knowledge is a pre-eminence of merit, which originates with the man, and is the noblest of all distinctions. Nature, bountiful and wise in all things, has prt;vlded us with an infinite variety of scenes, both for our instruction and entertainment; and, like a kind and indulgent parent, admits all her children to an equal participation of her blessings. But as the modes, situations, and circumstances of life are various, so accident, habit, and education, have each their predominating influence, and give to every mind its partic 8 PREFACE. ular bias. Where examples of excellence are wanting, the attempts to attain it are but few; but eminence excites attention, and produces imitation. To raise the curiosity, and to awaken the listless and dormant' powers-of younger minds, we have only to point out to them a valuable acquisition, and the means of obtaining it; the active principles are immediately put into motion, and the certainty of the conquest is ensured from a determination to conquer. But of all the sciences which serve to call forth this spirit of enterprise and inquiry, there are none more eminently useful than Mathematics. By an early attachment to these elegant and sublime studies, we acquire a habit of reasoning, and an elevation of thought, which fixes the mind, and prepares it for every other pursuit. From a few axioms, and evident, principles, we proceed gradually to the most general propositions and remote analogies; deducing one truth from another, in a chain of argument well connected and logically pursued; which brings us at last, in the most satisfactory maimner, to the conclu-. sion, and serves as a general direction in all our inquiries after truth. And'it -is not only in this respect that mathematical learning is so highly valuable; it is, likewise, equally estimable for its practical utility. Almost all the works of art and devices of man, have a dependance upon its principles, and are indebted to it for their origin and perfection. The cultivation of these admirable sciences is, therefore, a thing of the utmost importance, and ought to be considered as a principal part of every liberal and well-regulated plan of education. They are the guide of our youth, the perfection of our reason, and the foundation of every great and noble undertaking. From these considerations, I have been induced to compose an intro-:ductory course of mathematical science; and from the kind encouragement which I have hitherto received, am not without hopes of a continuance of the same candour and approbation. Considerable practice as a teacher, and a. long attention to the difficulties and obstructions which retard the progress of learners in general, have enabled me to accom~modate myself the more easily to their capacities and understandings. And as an earnest desire of promoting and diffusing useful knowledge is the chief motive for this undertaking, so no pains or attention shall be wanting to make it as complete and perfect as possible. The subject of the present performance is ALGEBRA; which is one of the most important and useful branches of those sciences, and may be justly considered as the key to all the rest. Geometry delights us by the simplicity of its principles, and the elegance of its.demonstrations; Arithmetic is confined in its object, and partial in its application; but Algebra, or the Analytic Art, is general and comprehensive, and may be: applied with success in all cases where truth is to be obtained and proper data can be established. To trace this science to its birth, and to point out the various alterations and improvements it has undergone in its progress, would far exceed the limits of a preface.*.It will be sufficient to observe that the invention is of the highest antiquity, and has challenged the praise and adoration of' all ages. Dioph?,anuts, a Greek mathematician, of Alexandria in Egypt, who ~flourished in or about the third century after Christ, appears to have been the first, among the ancients, who applied it to the solution of indeterminate or unlimited.problems; but it is to the moderns that we are principally indebted for the most curious refinements ofthe art, and its great and extensive usefulness in every abstruse and difficult inquiry.'7',,ton, Maclaurin, Sanderson, Simpson, * Those who &iAe desirous of a knowledge of this kind, may consult the Introduction to nlyv Treatise on Algebl'a; where they will find a regular historical detail of the rise and progress of'the science, from.its first - ide beginnings to the present tiles. PREFACE. 9 and Emerson, among our own countrymen, and ClairasZt, Euler, Lagracnge, and Lacroix, on the continent, are those who have particularly excelled in this respect; a.nd it is to their works that I would refer the young student, as tile patterns of elegance and perfection. The following compendium is formed entirely upon the model of those writers, and is intended as a useful and necessary introduction to them. Almost every subject, which belongs to pure Algebra, is concisely and distinctly treated of; and no pains have been spared to make the whole as easy and intelligible as possible. A great number of elementary books have already been written upon this subject; but tLere are none, which I havd yet seen, but what appear to me to be extremely defective. Besides being totally unfit for the purpose of teaching, they are generally calculated to vitiate the taste, and mislead the judgment. A tedious and inelegant method prevails through the whole, so that the beauty of the science is generally destroyed by the clumsy and awkward manner in which it is treated; and the learner, when he is afterwards introduced to some of our best writers, is obliged, in a great measure, to unlearn and forget every thing which he has been at so much pains in acquiring. There is a certain taste and elegance in the sciences, as well as in every branch of polite literature, which is only to be obtained from the best authors, and a judicious use of their instructions. To direct the student in his choice of books, and to prepare him properly for the advantages he may receive from them, is therefore, the business of every writer who engages in the humble but useful task of a preliminary tutor. This information I have been careful to give, in every part of the present performance, where it appeared to bd in the least necessary; and, though the nature and confined limits of my plan admitted not of diffised observations or a formal enumeration of particulars, it is presumed nothing of real use and importance has been omitted. My principal object was to consult the ease, satisfaction, and accommodation of the learner; and the favourable reception the work has met with from the public, has afforded me the gratification of believing that my labours have not been unsuccessfully employed. CONTENTS. PAGE DEFINITIONS,..... 1 1 Addition, 17 Subtraction,. _ e _20 Multiplication,- 21 Division,. 25 Algebraic Fractions, - - 32 Involution, or the Raising of Powers, - - 45 Evolution, or the Extraction of Roots, 48 Of Irrational Quantities, or Surds, -- -53 Of Arithmetical Proportion and Progression, - -. 78 Of Geometrical Proportion and Progression, -. 82 Of Equations, - -'. 88 Of the Resolution of Simple Equations, - - 90 Miscellaneous Questions,. - 105 Of Quadratic Equations, - - - - 112 Questions producing Quadratic Equations, - - - - 122 Of Cubic Equations, - 129 Of the Solution of Cubic Equations, - 131 Of the Solution of Cubic Equations by Converging Seri's - 13 Of the Resolution of Biquadratic Equations, 11.6 To find the Roots of Equations by Approximation, - 15 To find the Roots of Exponential Equations, -.. Of the Binomial Theorem, -. 0 -Of the Indeterminate Analysis, ~ - l65 Of the Diophantine Analysis, - - -174 Of the Summation and Interpolation of Series, 203 Of Logarithms, - 222 Multiplication by.Logarithms, -. 233 Division by Logarithms, - 235 The Rule of Three by Logarithms, 236 Involution, by Logarithms,. 238 Evolution, by Logarithms, - - - -. 240 A Collection of Miscellaneous Questions, -.... - 242 APPENDIX, on the Application of Algebra to Geometry,. - - 247 ALGEBRA. ALGEBRA is the science which treats of a general method of performing calculations, and resolving mathematical problems, by means of the letters of the alphabet. Its leading rules are the same as those of arithmetic; and the operations to be performed are denoted by the following characters: + plus, or more, the sign of addition; signifying that the quantities between which it is placed are to be added together. Thusl a + b shows that the number, or quantity, represented by b, is to be added to that represented by a; and is read a plus b. - minus, or less, the sign of substraction; signifying that the latter of the two quantities between w.hich it is placed is to be taken from the former. Thus a - b shows that the quantity represented by b is to be taken from that represented by a: and is read a minus b. Also, a-.b represents the difference of the two quantities a and b, when it is not known which of them is the greater. X into, the sign of multiplication; signifying that the quantities between which it is placed are to be multiplied together. Thus, a X b shows. that the quantity represented by a is to be multipled by that represented by b; and is read a into b. The multiplication of simple quantities is also frequently denoted by a point, or by joining the letters together in the form of a word. Thus, a X b, a, b, and ab, all signify the product of a and b; also, 3 X a, or 3a, is the product of 3 and a; and is read 3 times a. by, the sign of division; signifying that the former of the two quantities between which it is placed is to be divided by the latter. Thus, a - b, shows that the quantity represented by a is to be divided by that represented by b; and is read a by b. or a divided bv b. 12 DEFINITIONS. Division is also frequently denoted by placing one of the two quantities over the other, in the form of a fraction Thus, b -. a and - both signify the quotient of b divided a-b by a; and + signifies that a - b is to be divided by aac. a +- c. = equal to, the sign of equality; signifying that the quantities between which it is placed are equal to each other. Thus, x = a + b shows that the quantity denoted by x is equal to the sum of the quantities a and b; and is read x equal to a plus b. Any two algebraic expressions are said to be identical, when they are of the same value, for all the values of the letters of which they are composed. * Thus (x + a) X (x - a) = X2-a2, whatever numeral values may be given to the quantities represented by x and a. > greater than, the sign of majority; signifying that the former of the two quantities between which. it is placed is greater than the latter. Thus a > b sh6ws that the quantity represented by a is greater than that represented by b; and is read a greater than b. < less than, the sign of minority; signifying that the former of the two quantities between which it is placed is less than the latter. Thus, a < b shows that the quantity represented by a is less than that represented by b; and is read a less than b.: as, or to, and:: so is, the signs of an equality of ratios; signifying that the quantities between which they are placed are proportional. Thus, a: b:: c: d denotes that a has the same ratio to b that c has to d, or that a, b, c, d, are proportionals; and is read, as a is to b so is c to d, or, a is to b as c is to d. v the radical sign, signifying that the quantity before which is placed is to have some root of it extracted. * Woodhouse, in his Principles of Analytical Calculation, says that X2 - a2 is not generall/ = (x-a). (.2r+a): for instance, the particular case of x - a is to be excluded; the proof essentially demanding this circumstance, to wit, that x - a be a quantity, or that x be greater than a. Euler calls x - I =.x - 1 an identical eqlcation; and shows that x is indeterminate, or that any number whatever may be substituted for it. See Euler's Algebra, page 289, Vol. I.-ED. DEFINITIONS. 13 Thus,,Va is the square root.of a;',/a is the cube root of a; and'/ a is the fourth root of a; &c. Thee roots of quantities are also represented by figures placed at the righthand corner of' them, in the form:of a fraction. k L Thus, a2 is the square root of a; a3 is the cube root of a; and a" is the nth root of a, or a root denoted by any number n. In like manner, a2 is the square of a; a3 is the cube of a; and alM is the mth power of a, or any power denoted by the number m. Go is the sign'of infinity, signifying that the quantity standing before it is of an unlimited value, or greater than any quantity that can be assigned, The coefficient of a quantity is the number or letter which is prefixed to it. Thlus, in the quantities 3b, — -b, 3 and- x are the coefficients of b; and a is the coefficient of x in the quantity ax. A quantity without any coefficient prefixed to it is supposed to have I or unity; and when a quantity has no sign before it, + is always understood. Thus, a is the same as + a, or + la; and - a is the same'as - a. A term is any part or member of a compound quantity, which is separated from the rest by the signs + or - Thus a and b are the terms of a + b; and 3a, - 2b, and + 5cd, are the terms of 3a - 2b + 5cd. In like manner, the terms of a product, fraction, or proportion, are the several parts or quantities of which they are composed. a Thus, a and b are the terms of ab, or of -; and a, b, c, d, are the terms of the proportion a: b:: c: d. A factor is one of the terms, or multipliers which form the product of two or more quantities..Thus, a and b are the factors of ab; also, 2, a, and b2, are the factors of 2ab2; and a- and b. — x are the factors of the product (a, - x) X (b - x). A composite number, or quantity, is that which is produced by the multiplication of two or more terms or factors. Thus, 6 is a composite number, formed of the factors 2 and 3, or 2 X 3; and 3abc is a composite quantity, the factors of which are 3, a, b, c. 2~ 14 DEFINITIONS. Like quantities, are those which consist of the same letters or combinations of letters; as a and 3a, or 5ab and 7ab, or 2a2b and 9a2b. Unlike quantities, are those which consist of different letters, or combinations of letters; as a and b, or 3a, and a2, or 5ab2 and 7a2b. Given quantities, are such as have known values, and are generally represented by some of the first letters of the alphabet; as a, b, c, d, &c. Unknown quantities, are such as have no fixed values, and are usually represented by some of the final letters of the alphabet; as x, y, z. Simple quantities are those which consist of one term only; as 3a, 5ab, - 8a2b, &c. Compound quantities, are those which consist of several terms; as 2a + b, or 3a -2c, or a - 2b - 3c, &c. Positive, or affirmative quantities, are those which are to be added; as a, or + a, or + 3ab, &c. Negative quantities are those which are to be subtracted: as - a, or - 3ab, or — 7ab2, &c. Like signs, are such as are all'positive, or all negative; as +- and +-, or - and - Unlike signs, are when some are positive and others negative; as + and -, or - and +. A monomial, is a quantity consisting of one term only; as a, 2b, - 3a-b, &c. A binomial, is a quantity consisting of two terms, as a + b, or a - b; the latter of which is, also, sometimes called a residual quantity. A:trinomial, is a quantity consisting of three terms, as a + 2b - 3c; a quadrinomial of four, as a - 2b + 3c -d, an-d a polynomial, or multinomial, is that which has many terms. The power of a quantity, is its square, cube, biquadrate, &c.; called also its second, third, fourth power, &c.; as aa, a3, a4, &c. The index, or exponent of a quantity, is the number which denotes its power or root. Thus, - 1 is the index of a-', 2 is the index of a2, and 2 of az or Va. When a quantity appears without any index, or exponent, it is always understood to have unity, or 1. Thus, a is the same as a', and 2x is the same as 2x1; the 1, in such cases, being usually omitted. . DEFINITIONS. 15 A rational quantity, is that which can be expressed in finite terms, or without any radical sign, or fractional index; as a, or -a, or 5a, &c. An irrational Quantity, or Surd, is that of which the value cannot be accurately expressed in numbers, as the square roots of 2, 3, 5. Surds are commonly expressed by means of the radical sign V/; as V2, Va, 4/a3, or a fractional in1 1 3 dex; as 22, a2, an, &c A square or cube number, &c.. is that which has an exact square or cube root, &c. Thus, 4 and' a2 are square numbers; and 64 and 3a3 are cube numbers, &c. A measure of any quantity, is that by which it can be divided without leaving a remainder. Thus, 3 is a measure of 6, 7a is a measure of 35a, and 9ab of 27 a2b2. Commensurable quantities, are such as can be each divided by the same quantity, without leaving a remainder. Thus, 6 and 8, 2,/2 and 3 V/2, 5a2b and 7ab2, are commensurable quantities; the common divisors being 2, /2, and ab. Incommensurable quantities, are such as have no common measure, or divisor, except unity. Thus, 15 and 16, V2 and /3, and a q+ band a2 + b2 are incommensurable quantities. A multiple of any quantity, is that which is some exact ~number of times that quantity. Thus, 12 is a multiple of 4, 15a is a multiple of 3a, and 20a2b2 of 5ab. The reciprocal of any quantity, is that quantity inverted, or unity divided by it. a 1 Thus, the reciprocal of a, or - is -; and the reciprocal of La a b is -. A function of one or more quantities, is an expression into which those quantities enter, in any manner whatever, either combined, or not, with known quantities. * This definition of a Surd, or irrational Quantity, is due to Robert Adrian, LL. D., Professor of Mathematics and Natural Philosophy in the University of Pennsylvania, who had first published it in his edition of Hutton's Course of MAathenatics.-ED. f16 DEFINITIONS. Thus, a - 2x, ax +23x, 2x - a (a2 —2)2, axs", a%, &C., are functions of x; and atxyba 2, ay+x (ax2 -- by2)2 &c., are functions of x and y. A vinculum, is a bar —, or parentheses ( ), made use of to collect several quantities into one. Thus a -- b X c, or (a + b) c, denotes that the compound quantity a + b. is to be multiplied by the simple quantity c; and Vab + c2, or (ab ~ c2)2, is the square root of the compound quantity ab -c2. Practical Examples for computing the numveral Values of various Algebraic Expressions, or Combinations of Letters. Supposing a = 6, b = 5, c = 4, d = 1, and e.= 0. Then 1. a2 + 2ab - c + d = 36 + 60 - 4 - 1 = 93. 2. 2a- 3a2b -+ c3 = 432 -540 + 64- -44. 3. a2 X-(a + b) - 2abc - 36 X 11 - 240 - 156. 4. 2a V (b2 ac) + V (2ac + C2) =. 1 X + 8 = 20. 5. 3a V (2cc +C2), or 3a (2ac + c2)2 = 18 V 64 = 144. 6. V [2a2 - (2ac + 2) 1 (72 - V- 64) = (72 8) = V 64 = 8. 2a - 3c 4bc 12 +- 12 80 24 6d- +4e /(2ac c2) 6 + 0' - (48 + 16) 6 80 + = 14. Required the numeral values of the following quantities supposing a,, c, e, to be 6, 5, 4, 1, and 0, respectively, as above. 1. 2a2 + 3bc - 5d = 127, 2. 5a2b — 10aba2 + 2e = -600 3. 7a2 + b - c X d + e = 253 4. 5 / ab - b2- 2ab - e2= --.613875 a a -b 5. X d - ----- 2a2e = c d 6. 3 V c + 2 V (2a + -b-d) 14 7. a V/(a2 + b2) + 3bc / (a2 - b2) - 245.8589862 8. 3a2b +- I [c2 + v (2ac + c2)] - 542.8844991 2b - c V 5b+ 3 /c - d 9. - 2a 3a c"" 6 2a q - c ADDITION. 17 ADD ITIO N. ADDITION is the connecting of quantities together by means of their proper signs, and incorporating such as are like, or that can be united into one sum; the rule for performing which is commonly divided into the three following cases.* CASE I. When the Quantities are like, and have like stgns. RVLE.-Add all the coefficients of the several quantities together, and to their sum annex the letter or letters belonging to each term, prefixing, when necessary, the common sign. EXAMPLES. 3a - 3ax 2b + 3y 5a - 6ax 5b -- 7y a - ax b + 2y 7a - 2ax 8b + y 12a - 7ax 4b- +4y 28a -19ax 20b - 17y 2ay -2by2 a - 2x2 5ay - 6by2 a - 6x2 4ay -by2 4a- x2 7ay - 8by2 3a - 5x2 16ay -. by2 7a - x2 34ay -1 8by2 16a — 15x' 3ax2 7x - 4y 2a + x2 2ax2 x — 8y 3a + x2 12axz 3x - y a + 2x2 9a:x2 x - 3y 9ad + -3X2 10aX2 4x - 4a-+ x2z 36ax2 6x - 17y 19a + 8$x2 * The term Addition, which is generally used to denote this rule, is too scanty to express the nature of the operations that are to be performed in it; which are sometimes those of addition, and sometimes subtraction, according as the quantities are negative or positive. It should, therefore, be called by some name signifying incorporation, or striking.a balance; in which case, the incongruity here mentioned would be removed. 20 IS ADDITION. CASE IL When the Quantities are like, but have unlike signs. RULE.-Add all the affirmative coefficients into one guni, and those that are negative into another, when there are several of the same kind: then subtract the less of these sums from the greater, and to the difference prefix the sign.of the greater, annexing the common letter or letters as before. EXAMPLES. - 3a 2a — 3x2 6x - 5ay + 7a - 7a 5x2 -3x + 2ay - 8as -3a - x2 x - 6ay - a +a —3x2 2x + ay + la - 7a 6x - 2ay -- 21a2 3ay -.7 3ab +- 7x -33al - ay + 8 - 3ab - 10x -8a2 + 2ay- 9 +- 3ab - 6x +- l0a - 3ay - 11 - ab - 2x + 13aa + 12ay + 13 + 2ab + 7x + 1Oa2 + 13ay -6 + 4ab -4x -2aVx -- 6a2+- 2b 6ax2 + 5x2 +aV x q- 2a2 - 3b - 2ax2a - 6x2 - 3aVt - 5a2 - 8b + 3ax2 -- lOxc +- 7a x _- 4a-2b - 7ax2 + 3x2 4aVcrx -3a2 - 9b +- ax2+ lls - a x - a -- 2b Jr axa + 3x2 CASE III. FWhen the Quantities are unlike; or some like ancd others unlike. RULE.-Collect all the like quantities together, by taking their sums or differences, as in the foregoing cases, and set ADDITION. 19 down those that are'anlike, one after another, with their proper signs. EXAMPLES. 5xy 2xy- 2x2 2ax- 30 4ax 3x2 +- xy 3x2 - 2ax - xay x2 +, xy 5x - 3x2 - 4ax 4X2 - 3xy 3Vx +10 4xy 6x2 + xy 8x2- 20 + ax2 8a2x - 3ax lOb2 - 3a2x - ax 7a x -- 5xy -- b~ 2a2x2 + 3ax2 9x y -5ax 50 +2a2x -ax2 2a2x'2 + xy a2x2 + 120.+i 2ax2 lOa2+2 - 5xy - ax 9b2 + 3a2X2 _ a2x + 170 + 3a2y 2 Vx -17y 2a2- 3a x - 2xya 3 V xy + 10x x2a _ 2a2x2 - 3y2x 2x2y + 25y 3a2 - 13xy - sx2y 12xy — Vxy xy 32a2 + 2xy2 -9y+ 18x'2 20- 65x2 3ay - 3y2x -82y 19 /9 12xzy y 37a2 - 3a / — 12xy;+ 2x-Y+ iV_- 64x- 2a2x2 + _/_ _ ] 20 EXAMPLES FOR PRACTICE. 1. Required the sum of 1 (a + b) and 2 (a -b). Ans. a. 2. Add 5x - 3a + b + 7 and - 4a - 3x + 2b- 9 together. Ans. 2x- 7a+ - 3b — 2. 3. Add 2a + 3b - 4c - 9 and 5a - 3b + 2c - 10 together. Ans. 7a - 2c- 19. 4. Add 3a + 2b - 5, a +- 5b - c, and 6a - 2c + 3 together. Ans. 10a + 7b - 3c -2. 5. Add x3 + ax2+ bx +- 2 and x3 + CX2 - dx.- 1 together. Ans. 2x3 - (a + c) X2- +(b - d) x + 1. 6; Add 6xy - 12x2, - 4x2 + 3xy, 4xS - 2xy, and - 3xy + 4x2 together. Ans. 4xy - 8xa ~~20 ~ SUBTRACTION. I I 7. Add 4ax - 130 +- 3x2, 5x2 +- 3ax -+ 9x2, 7xy —4x2 + 90, and a/ x + 40 - 6x2 together. Ans. 7ax + 8x2 + 7xy. -8. Add 2a2 - 3ab + 2b3-3a2, 3b3 -- 2a + a3- 5c3, 4c2 - 2b3 + 5ab + 100, and 20ab + 16a2 - be - 80 together. Ans. 13a - +22ab- +3b3 + a3 -c3 +- 20 - be. 5a 3c2 7'/ be ab~+x 8a 7(2 9. Add + 9 ~and - ---- b a x db a V/bc ab q- ~x 12 b 6(ab ) together. x d ]3a 4c_ V(bc) ab + x Ans. + --- 5 - 3 10. Add 3a2 + 4bc- e2 + 10, - 5a2 + 6bc + 2e2 - 15, and - 4a2 - 9bc - 1062 + 21 together. Ans. be - Ga2 - 9e2 + 16. SUBTRACTION. SUT'rRACTION is the taking of one quantity from another; or the method of finding the difference between any two quantities of the same kind; which is performed as follows: —* RuLE. —Change all the signs (+ and -) of the lower line, or quantities that are to be subtracted, into the contrary signs, or rather conceive them to be so changed, and then collect the terms together, as in the several cases of addition. EXAMPLES. 5a2 _- 2b xa - 2y + 3 Sxy + 8x- 2 2a2 - 5b 4x2+ 9y- 5 3xy- 8x- 7 3a2 —7b -3x2 _- ly + 8 2xy + 16x + 5 * This rule being the reverse of addition, the method of operation must be so likewise. It depends upon this principle, that to subtract an affirmative quantity from an affirmative, is the same as to add a negative quantity to an affirmative. Thus, according to Laplace, we can write a=a —b-b........(1), a —c = —cfb-b... (); so that if from a we are to subtract + b, or - b; or what amounts to the same thing, if in'a we suppress + b, or - b; the remainder from transformation (1), must be a - b in the first case, and a + b in the second. Also, if from a - c we take away + b, or - b, the remainder, from (2), will be a- c - b, or a — c + b.-ED. MULTIPLICATION. 21 5xy - 18 8y2 - 2y - 5 10 - Sx -3xy -xy + 12 -y2 + 3y- +2 - x + 3 -xy 6xy - 30 9y2 _ -y - 7 7 —7x -2xy -5x2y- 8a 4 v ax -- 22y 52 - / X — 49 + 3xWy-7b 3 ax -5xy2 6x2- 8x - x2 -8x"2y —8a+-7b Vax-2x2y+ 5xy -X2 +- 8x + 2 V/x-49 EXAMPLES FOR PRACTICE. 1, Find the difference of I (a + b) and 1 (a - b). Ans. b. 2. From 3x- 2a - b + 7, take 8 - 3b + a + 4x. Ans. 2b x - -3a - 1. 3. From 3a + b + c - 2d, take b -8c +- 2d- 8. Ans. 3a -+ 9c - 4d + 8. 4. From 13x2 - 2ax + 9b2, take 5x2 - 7ax - b2. Ans. 8x2- 5ax - 10b2. 1J 5. From 20ax - 5v x-t+ 3a, take 4ax+ - 5x2 - a. Ans. 16ax — 10 x + 4a. 6. From 5ab - 2b - c - bc - b, take b2- 2ab + bc. Ans. 7ab +- b2 - c -- 7. From ax3 - bx2 - cx - d, take bx2 + ex - 2d. Ans. ax3 - 2bx - (c- e) x +t- d. 8. From - 6a- 4b - 12c + 13x, take 4x - 9a + 4b - 5c. Ans. 3a + 9x - 8b - 7c, 9. From 6x2y - 3 V (xy) - 6ay, take 3x2y + 3 (xy) - 4ay. Ans. 3x2y - 6 V (xy) - 2ay..! 10. From the sum of 4ax- 150 + 4x2, 5x2 + 3ax ~ 10x2, and 90 - 2ax - 12 V (x); take the sum of 2ax- 80 7x, 7X2 - 8ax - 70, and 30 - 4 \/ (x) - 2x2 + 4a2x2. Ans. 1lax -- 60- - 4a2x. MULTIPLICATION. MULTIPLICATION, or the finding of the product of two or more quantities, is performed in the same manner as in arithmetic; except that it is usual, in this case, to begin the operation at the lefthand, and to prfoceed towards the right, or contrary to the way of multiplying numbers. 22 MULTIPLICATION. The rule is commonly divided into three cases; in each of which it is necessary to observe, that like signs, in multiplying, produce +, and unlike signs,-. It is likewise to be remarked, that powers, or roots of the same quantity, are multiplied together by adding their. indices: thus, I 1 a X a2, or al X a2 = -a3; a2 X a3= -a5; a2 X as =; and am X an - a + n. The multiplication of compound quantities, is also, sometimes, barely denoted by writing them down, with their proper signs, under a vinculum, without performing the whole. operation, as 3ab (a - b). or 2a (a2 + b2). Which method is often preferable to that of executing the entire process, particularly when the product of two or more factors is to be divided by some other quantity, because, in this case, any quantity that is common to both the divisor and dividend, may be more readily suppressed; as will be evident from various instances in the following part of the work.* CASE I. When the factors are both simxple quantities. RULE. —Multiply the coefficients of the two terms together, and to the product annex all the letters, or their powers, belonging to each, after the manner of a word; and the result, with the proper sign prefixed, will be the product required.t * The above rule for the signs may be proved thus: If a, b, be any two quantities, of which 2 is the greater, and B - b is to be multiplied by a, it is plain that the product, in this case, must be less than aB, because t - b is less than B; and, consequently, when each of the terms of the former are multiplied by a, as above, the result will be (B-b) X a = aB - ab. For if it were aB - ab, the product would be greater than aB, which is absurd. Also, if D be greater than b, and A greaterthan a, and it is required to multiply B -b by A - a, the result will be (B -b) X ( - a) = AB - a-bA + ab. For the product of B - b by A is A (a - b), or AB —Ab, and that of -- b by - a, which is to be taken from the former, is - a (B - b) as has been already shown; whence B - b being less than B, it is evident that the part, which is to be taken away must be less than aB; and consequently since the first part of this product is - aB, the second part must be + ab; for if it were - ab, a greater part than aB would be to be taken from A (B — b), which is absurd. t XWhen any number of quantities are to be multiplied together, it is the same thing in whatever order they are placed: thus, if ab is to be MULTIPI,ICATION. 23 EXAMPLES. 12a - 2a -- 5a - 9x2 3b + 4b -6 -- 5bx 36 ab -8ab - 30ax + 45bx3 7ab - 6a2x -2xy2 -7axy - 5ac + 5x -y + 6ay - 35a"bc - 30a2X2 - 2x2y3 -42a2xy2 3a2b 12ac2x -6xyz - a2xy 26a2 - 2xy - ay2z + 2Xy2 6a4b - 24a"23y - 6axy'z2 - 2a2x2y3 CASE II. W7 hen one of the factors is a compoulnd quanttty. RULE. — vlulltiply every term of the compound factor, considered as a multiplicand, separately, by the multiplier, as in the former case; then these products, placed one after another with their proper signs, will be the whole product required. EXAMPLES. 3a -2b 6xy- 8 a2 — 2x - 1 4a 3x 4x'12a 2-8ab 18xZy -24x 4a2x - 8x2 + 4x 12x - ab 35x - 7a 3y y — 2 5a - 2x xy 60ax - 5a2b -.70x2 + 14ax 3xy3 + xy2 — 2xy 13x2 - a2b 25xy +- 3a2 3x2 -xy + 2yi - 2a 13x2 5xr2 - 26ax2 q- 22ab 325x3y + 39a2"x2 15x4 — 53y- 10 x2y2 multiplied by _, the product is either abe, acb, or bca, &c.; though it is usual, in this case, as well as in addition and subtraction, to put them according to their rank in the alphabet. It may here also be observed in conformity to the rule given above for the signs, that (-t-a) X ( -b). or ( -a) X ( — b)+ = ab; and ( -+ ) X (-b), or( - a) X ( + B) _ - 24 MULTIPLICATION. CASE III. When both, thefactors are compound quantities. RULE.-Multiply every term of the multiplicand separately, by each term of the multiplier, setting down the products one after another, with their proper signs; then add the several lines of products together, and their sum will be the whole product required. EXAMPLES. x -- y 5X+- 4y X2+ xy - Y2 x - y 3x - 2y x - Y x2 -- X 15x2 - 12xy x3 + x2y _ y2 + xy + y2 - 10xy _ 8y2 - x2y - xy2 y — y3 2 +2xy+y2 15x2 + 2xy —8y2 x3 * _2xy2 + y3 + y x2 + y x2 + xy +- y2 X - y x2+y x -- y x2 + xy x4 - X2Y x3 +- x2Y + xy2 - xy +- y2 - x2y + _ y2 _ x y3 c2 _ 2 _ x 4+ 2x2y x+ y2 -3 _Y y EXAMPLES FOR PRACTICE. 1. Required the product of x2 - xy + y2 and x + y. Ans. x3 + Y3. 2. Required the product of c3 +- 2y + xy+2 + y3 and x - y. Ans x4.y4. 3. Required the product of x2 + xy ~- y2 and x2 - xy ~ y2. Ans. x4 +- xy2 + y4. 4. Required the product of 3x2 - 2xy +- 5, and xt2+ 2xy - 3. Ans. 3x4 4- 4Xr3y - 42y2 - 4X2 + 16xy- 15. 5. Required the product of 2a2 - 3ax n+~ 4x2 and 5a2 - 6ax - 2 2. Ans. 10a4 - 27a3x + 34ax2 — 18ax3- 8x4. 6. Required the product of 5x3 + 4ax'2 + 3a2Xr + a3, and 2xc2 - 3ax + a2. Ans. 10X5 - 7ax4 - a2xC3 - 3a3c2 +a 5. 47. Required the product of 3x3 + 2x2y2 + 3y3 and 2x3 - 3x2y2 + 5y3. Ans. 6x6 - 5x5y2 - 6xc4y4 + 21x y3 + X2y5 + 15y6. 8. Required the product of x3 - ax2 + bx - c and cx dx + e. Ans.. - ax4-d _ 4 + (b + d + ad + eae) 2 + (ed + b) x — ce. DIVISION. 25 9.' Required the product of the four following factors, viz. I. IM.. IV. (a + b), (a2 +- ab + b2), (a - b), and (a2 - ab + 62). Ans. a6 - b6. 10. Required the product of a3 - 3a2+ + 3ax2 -X3 and a3 - 3a2x q+ 3ax2 -- X. Ans. a6 - 3a4X2 + 3a2x4 -. 11. Required the product of a4 +- a2c2 + c4 and a2 - c2. Ans. a6 - c6. 12. Required the product of a2 + 62 + c2 - ab -ac - bc and a + b +- c. Ans. a3 - 3abe + b3 + c3. D I V I S I 0 N. DIVISION Is the converse of mulplication, and is performed like that of numbers; the rule being usually divided into three cases; in each of which like signs give + in the quotient, and unlike signs -, as in finding their prodllcts.t It is here also to be observed, that powers and roots of the same quantity, are divided by subtracting the index of the divisor from that of the dividend. a3 2r a2. Thus, a3 - a2, or - a a- 3 or — = a6; ae3 a 4 —a3 or --- a2; and a?n -- n, or - am". a an CASE 1'. When the divisor and dividend are both simple quantities. RULE.-Set the dividend over the divisor, in the manner of a fraction, and reduce it to its simplest form, by cancelling the letters and figures that are common to each term. * I would advise the learner to perform the calculation of this example several ways; viz. First, by multiplying the product of the factors I. and II. by the product of the factors III. and IV. Secondly, by multiplying the product of the factors I. and. III. by the product of the factors II. and IV. Thirdly, by multiplying the product of the factors I. and IV. by the product of the factors II. and III. The last method is the most concise. See Euler's Algebra, page 119, Vol. I.-ED. t According to the rule here given for sigs, it follows that +ab -- a.b - ab + nab -tb, -=-b a,,-= ---- = as will readily appear by multiplying the quotient by the divisor; the signs of the product being then the same as would take place in the[: former rule. 3 26 DIVISION. EXAMPLES. 6ab 12ax2 6aB - 2a, or - =3b; and 12ax2 3x, or -— =4ax, 2a 3x a a a —a, or - =1; anda-+ —-a, or --—. a -a -2a Also - 2a - 3a, or — -- -; and 9x -.- 3x4 -- 3x4. 1. Divide 16x2 by 8x, and 12a2x2 by - 8a2x. 3x Ans. 2x, and - - 2 2. Divide -15ay2 by 3ay and - 18ax2y by - 8ax. Ans. - 5y, and 9xy 3. Divide - -a2 by -a2 and ax3 by - -a a. 3 5 5 51 Ans. - 3, and - -a2 X'T2 3 4. Divide 12a2b2 by - 3a2b, and - 15ay by - 3ay2. Ans. - 4b, and 5y6. 5. Divide - 15a2x2 by 5ax2, and 21a2C2X 2 by- 7ac2x4. Ans. - 3a, and - 3ax4. 6. Divide - 172a3c by - 5xXac~2, and 24xy by 8 v (xy). 17x6ac2 - Ans. --- and 3 /(xy) CASE II. When the divisor is a simple quantity, and the dividend a compound one. RULE.-Divide each term of the dividend by the divisor, as in the former case; setting down such as will not divide in the simplest form they will admit of. EXAMPLES. ab + h2 a + b (ab + b2). 2b, or ab - I a + 1 b 2+. 1Oab - 15ax (1Oab - 15ax) -- 5a, or - =2b - 3x. ri r DIVISION. 27 30ax - 48x2 (30ax- 48x2) - 6x, or - 5a- 8x 6x 1. Let 3x3 + 6X2 + 3ax - 15x be divided by 3x. Ans.2 2x + a -5. 2. Let 3abc + 12abx - 9a2b be divided by 3ab. Ans. c + 4x - 3a. 3. Let 40ab'3 + 60a2b -- 17ab be divided by - ab. Ans. - 40a2b2 - 60ab + 17. 4. Let 15a2bc -'12acx2 + 5ad2 be divided by - 5ac. 12X2 da Ans. - 3ab +. --- 5 c 5. Let 20ax + 15ax2 + 10ax + 5a be divided by 5a. Ans. 4x3 + 3x- 3+x + 1. 6. Let 6bcdz + 4bzd2 - 2b22 be divided by 2bz. Ans. 3cd- + 2dZ - bz. 7. Let 14a2 - 7ab - 2lax - 28a be divided by 7a. Ans. 2a-b+3x -4 8. Let - 20ab + 60ab3- 12a2b2 be divided by - 4ab. Ans. 5 - 15b3 3ab. CASE III. When the divisor and dividend are both compound qualities. RuLE. —Set them down in the same manner as in division of numbers, ranging the terms of each of them so, that the higher power of.one of the letters may stand before the lower. Then divide the first term of the dividend by the first term of the divisor, and set the result in the quotient, with its proper sign, or simply by itself, if it be affirmative. This being done, multiply the whole divisor by the term thus found; and, having subtracted the result from the dividend, bring down as many terms to the remainder as are requisite for the next operation, which perform as before; and so on, till the work is finished, as in common arithmetic. EXAMPLES. x + y) x2 + 2xy + y2(x + y x2 +xy xy + y2 xy +y2 2 DIVISION. a + x) a3 + 5a2x+ 5ax2 +x3 (a2 + 4ax +X~ a3 + a2x 4a2 x + 5ax2 4a2x + 4ax2 ax- x3 a X+ a3 -- 3) — 92+ 27x - 27 (x2 - 6x+ 9 a3 - 3X2 -- 6x2+ 27x - 6x + 1 8S 9x- 27 9x- 27 2x- 3ax +- a2) 4x4 - 9az2 + 6a3 - a4 (2x2 + 3ax - 4ax4- 6ax3 + 2a2X2 6ax3 -- 1 la2 2 + 6ax 6ax 3- 9a2 x + 3a'x - 2a2x2 + 3a3x - a' - 2a2x2 + 3ae3x - a4 -NOTE 1. If the divisor be not exactly contained in the dividend, the quantity that remains after the division is finished, must be placed over the divisor, at the end of the, quotient, -in the form of a fraction; thus,* a +- ) a3 - 3 (-ce ax + x 2 - a3 + a2 a. 2. ax2 - a3 a2 + a3 -- 2xI I in the case here given, the operation of division may be considered DIVISION. 29 2yx+y: + y)-X4 + y4(X3 _ X2y + Xy2. y3+ X +y x4 + x3y - v3y + y4 - X3y X2y2 X2y2 -- y4 x.2y2 + Xy3 - -xy3 + Y4 - xy3 —. y4 2y4 2. The division of quantities may also be sometimes carried on, ad infinitum, like a decimal fraction; in which case a few of the leading terms of the quotient will generally be sufficient to indicate the rest, without its being necessary to continue the operation; thus, IT, X X3.4 a f- ) a. (1 a 2 — a +T 4 &t a a2 a a a +- a a a a as terminated, when the highest power of the letter, in the first or leading term of the remainder, by which the. process is regulated, is less than the power of the first term of the divisor; or when the first term of the divisor is not contained in the first term of the remainder; as the succeeding part of the quotient, after this, instead of being integral, as ft ought to be, would necessarily become fractional. * Now, it is easy to perceive that the next or.6th term of the quotient will be - -., and the seventh term x and so on, alternately plus and ta5.6 mintus; this is called the law of contiSuatiion of the series. And the sum of all the terms when infinitely continued is said to be equal to the a 2 fraction -_. Thus we say the vzlgarfraction ~, when reduced to a decimal, is - -222, &c., infinitely continued. The terms in the quotient are found by dividing the remainder by a,, the first term of the divisor; thus, the first remainder -x divided by a, gives -- the second term in the quotient; and the second remainder -- divided by a gives + - the third term, &c. a2 3* 30 DIVISION. a a + CS a2 a3 ad a2 a3 And by -a process similar to thle above, it may be shown that at aX X2 X3 X4 X5 = + —X -+.-+-+ + -- +; +, - &c. a-X a a2 a 3 a4 a5 Where the law, by which either of these series may be continued at pleasure, is obvious.* * In this example, if x be less than a, the series is convergent, or the value of the terms continually diminish; but when a is greater than a, it is said to diverge. To explain this by numbers: suppose a 3, and x =2. a 52 X3 Then, 1+if, x+ -- &c. The corresponding values are, ++4 8 where the fractions or terms of the series grow less and less, and the farther they are extended, the more they converge or approximate to 0, which is supposed to be the last'term or limit. But if a = 2, and a = 3, X. 23 Then 1 + - + 3 + -3, &. The corresponding values are, 3 9 27 &c. n which the terms beeome larger and larger. This is called a diverging series. If x = 1, and a = 1 in the preceding example: a I x2.~, Then, - --- 1 - - + 2 - &e., will be 1 + 1 -- 1-, Ten a a a2 +3 1 &C. Now, because, 1I -- = it has been said that 1 -- 1- 1 -- 1, &c., infinitely continued, is = i; a singular conclusion, when it is perceived from the terms themselves, that their sum must necessarily be either 0 or - 1, to whatever exteht the division is supposed to be continued. The real question, however, results from the fractional parts, which DIVISION. 31 EXAMPLES FOR PRACTICE. I. Let a2 - 2ax - x2 be divided by a- x. Ans. a - x. 2. Let x - 3ax2 +- 3a2x - a3 be divided by x - a. Ans. x2- 2ax + a2 3. Let a3 - 5aax - 5a2 + X be divided by a + a. Ans. a2 q 4ax q- x2 4. Let 2y - 19y2 +- 26y -16 be divided by y- 8. Ans 2y2- 3y + 2. 5. Let 5 +- 1 be divided by x- 1, and. x - 1 by x - 1. Ans. x4 _- 3 + x2 _ x + 1,'and x5 + x4 q- xI q- x2 +- x q- 1. 6. Let 48x3 - 76ax2 - 64ax + 105a3 be divided by 2w - 3a. Ans. 24x2 -2ax -35a 7. Let 4X4 - 9X2 + 6x - 1 be divided by 2x2 + 3x- 1. Ans. 2x. 3x + 1. 8. Let x4 - a2x2 + 2a3x - a4 be divided by x2 - ax + a2. Ans. a2+ ax- a2. 9. Let 6x4 - 96 be divided by 3x- 6, and a5 + x5 by a+- x. Ans. 2X3 + 4x2 + 8x + 16, and a4- ax + a2x2 - ax3 +- x4. 10. Let 32x5+- 243 be divided by 2x- +3, and x6 — a6 by x - a. Ans. 16x4- 24x3 + 36x2 - 54x + 81, and 5-I- ax4 +- a2x3 + a3X2 + a4x +- a5. I1. Let b4 - 3y4 be divided by b - y, and a4 +- 4a2b + 8b4 by a + 2b. Ans. b3 + b2y + by2 +- y3 _ Y and a3-2ab +- 4ab + 16b3 + 24b4 4ab2- 8b _. 8b3 a +2 a 4- 2b 12. Let x2+ px+ -q be divided by +- a, and x3 - px 2+ qx —r by x — a. q- pa + a2 Ans. p a --, and 2 +- (a-p)x - ap a3 - a2p + aq - r +- a q- q - -, X- a 13. Let 1- 5x- +10x2 — 10x3 + 5x4- x5 be divided by 1 -2x + x. Ans. 1 —3x + 3x2 -3. 14. Let a4 + 4b4 be divided by a2 - 2ab + 2b2. Ans. a 2+ 2ab - 2b2. 15. a5 - 5a4x + 10axa2- 10a2x3 + 5ax4 - x5 be divided by a2 - 2lax +-2. Ans. a3 - 3a2x +- 3ax2- X3 16. Let a4 +- b4 be divided by a2 J- ab 2~ 2 + b2. Ans. a2 - ab /2 + b (by the division) is always -+ 2 when the sum of the terms is 0, and -- when the sum is + 1: consequently - is the true quotient in the former case, and 1- - in the other.-ED. 32 ALGEBRAIC FRACTIONS. OF ALGEBRAIC FRACTIONS. ALGEBRAIC fractions have the same names and rules of operation as numeral fractions in common arithmetic; and the methods of reducing them, in either of these branches, to their most convenient forIns, are as follows:CASE I. To find the greatest common measure of the terms of afraction. RULE.-:1-. Arrange the two quantities according to the order of their powers, and divide that which is of the highest dimensions by the other, having first expunged any factor, that may be contained in- all the terms of the divisor, without being common to those of the dividend. 2. Divide this divisor by the remainder, simplified, if necessary, as before; and so on, for each successive remainder and its preceding divisor, till nothing remains, when the divisor last used will be the greatest common measure required; and if such a'divisor cannot be found, the terms of the fraction have no common measure.* NOTE. If any of the divisors, in the course of the operation, become negative, they may have their signs changed, or be taken affirmatively, without altering' the truth of the result; and if the first term of a divisor should not be exactly contained in the first term of the dividend, the several terms of the latter may be multiplied by any number or quantity, that will render the division complete.t * If, by proceeding in this manner, no compound divisor can be found, that is, if the last remainder be only a simple' quantity, we may conclude the case proposed does not admit of any, but isalready in its lowest terms. Thus, for instance, if the fraction proposed were to be a3 + -2acx -3a. zo-.4x3. a + ax+ 2 it is plain by inspection, that it is not reducible by any simple divisor; but to know whether it may not, by a compound one, I proceed as above, and find the last remainder to be the simple quantity 7x9: whence I conclude that the fraction is already in its lowest terms. t In finding the greatest common measure of two quantities, either of them may be multipliedl, or divided, by any quantity, which is not a divisor of the other, or that contains no factor which is common to them both, without in any respect changing the result. It may here also be farther added, that the common measure, or ALGEBRAIC FRACTIONS. 3 EXAMPLES. 1. Required the greatest common measure of the fraction 4 1 t5 + X3' 4 1) x5 + 3 (x x X- orx2+ii 1 X4_1(X2_j X4 + x2 - x2- -X2_1 Whence x2 + 1 is the greatest common measure required. 2. Required the greatest common measure of the fraction X3-_ b2x xz+ 2bx + b' x2 + 2bx + b2) ~X - b2x (x xa + 2bX2 + b2x * - 2bx2 - 2b2x or x +b 2 + 2bx + 2( + b X2 Jr,b bx + 62b bX + b2 Where x + b is the greatest common measure required. 3. Required the greatest common measure of the fraction 3a2 -2a- 1 4a3- 2a3 - 3a -+ 1 divisor, of any number of quantities, may be determined in a similar manner to that given above, by first finding the common measure of two of them, and then of that common measure and a third; ahid so on to the last. * Here, I divide the remainder -- 2z2-2 2b2x by - 2xb, (its greatest simple divisor) and the quotient is x+-b; and then I divide the last divisor by x - b, &c.-En. ALGEBRAIC FRACTIONS. 3a2 - 2 - 1)4a -2a2 - 3a + 1 3 12a3 - 6a2- 9a + 3 (4a 12a'3 -8a - 4a 2a2- 5a 3) 3a2 - 2a - 1 2 6a2- 4a - 2 (3 6a2 — 15a -- 9 Ila- 11 or a - 1 Where, since a — 1 ) 2a2 - 5a + 3 (3a - 3, it follows that the last divisor a - 1 is the common measure required. 4. It is required to find the greatest common measure ot a3 - a3 Ans. x - a. 4 -a4 5. Required the greatest common measure of the fraction a4 _ X4 Ans. a2- x2. a3 -- a2x - ax2+ 36. Required the greatest common measure of- the fraction a4 -+ a2X2 - a4 x4 +aX3 a- a 4 Ans. ax' a 2. 7. Required the greatest common measure of the fraction 7a2 - 23ab +d-b2 5a3- l8a2h+11ab +ll2-6b t Ans. a - 3b. 5cP - 18abl S - l ab2 -6bS' 8. *Required the greatest common measure of the fraction' xa3 - ax2 + bx2 - 2a2x + bax - 2ba2 2 - bx + 2ax - 2ab Ans. x 2a. * This fraction can be reduced by Simpson's rule (page 48) thus:Fractions that have in them more than two different letters, and one of the letters rises only to a single dimension, either in the numerator or denominator, it will be best to divide the said numerator or denominator:(whichever it is) into two parts, so that the said letter may be found in every term of the one part, and be totally excluded out of the other; this being done, let the greatest common divisor of these two parts be found, which will evidently be a divisor to the whole, and by which the division of the quantity is to be tried; as in the following example, where the fraction given is X3 -+ ax2 + bx.2 2a2ox - baz -2 ba2 zx2-b)x Sax- 2ab Ihere the denominator being the least compounded, and b rising therein ALGEBRAIC FRACTIONS. 35 9. Required the greatest common measure of the fraction x4 -. 3ax3 - 8a2x2 + 1 8a3x- 8a4 c -3 2 2 ~' Ans. X2 - 2ax -2a2. x3 -ax _ Sa x + 6aa 10. Required the greatest common measure of the fraction 5a5 - O4ab + 5a3b2 Ans. a -+ b. a3b 2- 2ab q- 2aba q- b4' 11. Required the greatest common measure of the fraction 6i5 + 15a4b - 4a3c2 - 1a2bc2 Ans. 3a" - 2C0 9a3b - 27a2bc - 6abc2 + 18bc' Ans. CASE II. To reduce fractions to their lowest or most simple terms. RULE.-Divide the terms of the fraction by any numbers or quantity, that will divide each of them without leaving a remainder; or find their greatest common measure, as in the last rule, by which divide both the numerator and denomina tor, and it will give the fraction required. EXAMPLES. a2bc x2 1. Reduce and to their lowest terms. a2bc a262 2 A Here — = - Ans. And =. Ans. 5a2b 5b ax + x'2 a + x 2. It is required to reduce 2 + ato its lowest terms. Here cx +- x2 c ac- a2x or c + X 2C + a2x (a2 a2c + a2x Whence c + x is the greatest common measure; cX - X2 X and c + x)-~- = -- the fraction required, a 2C at- 2x ato a single dimension only, I divide the same into the parts x2- + 2ax, and - bx- - 2ab; which, by inspection, appear to be equal to (x-+- 2a)X x, and (x + 2ca) X - b. Therefore z + 2a is a divisor, to both the parts, and likewise to the whole, expressed by (x + 2a) X ( — b); so that one of these two factors, if the fraction given can be reduced to lower terms, must also measure the numerator; but the former will be found to succeed, the quotient coming out z -ax+ -bx - ab, exactly; whence the fraction itself is reduced to x2. — ax + bx - ab -b, which is not reducible farther by x - b, since the division does not terminate without a remainder, as upon. trial will be found. This rule is sometimes of great utility, because it spares great labour, and is very expeditious in reducing several fractions.-ED. 36 ALGEBRAIC FRACTIONS. X3m b2X 3. It is required to reduce + b2 to its lowest terms. x+ 2bx + b2) 3- b2x (x x3 + 2bx2 + b2X - 2bx2- 2b2x 1 orx+ 6b I x2+2bx+b2(x+b x2 + bx bx + b2 bx + b2 Whence x + b is the, greatest common measure; and x + ~) X3 - b2X X2- bx 2 +-2,+ --- the fraction required. x2 + 2bx q- be x + b And the same answer would have been found, if x3 - b2% had been made the divisor instead of x2 + 2bx + b2. x4 - a4 4. It is required to reduce - to its lowest terms. a5 - a22 3,2 + a2 Ans. 3 6ai2 + 7ax - 3x2 5. It is required to reduce 6a2 - lax 32 to its lowest 3a -- x' terms. Ans. - 2x3 - 16x - 6 6. It is required to reduce - 24 - 9 to its lo-west 3 i- 24 - 9 terms. Ans. A. 9X5 + 2x + 4x2 - x -- 1 7. It is required to reduce to'5x4 - 2x3 -- 10x2 — x -- 2 3X3 + X2 + 1 its lowest terms. Ans. + 5X2 + x +2 a~d2C _ C _ _ a%2 + C4 8. It is required to reduce 4a2d 4acd 2ac2+2c to its lowest terms. ad2 + cd2 - ac2 - C3 Ans. 4ad — 2c2 CASE III. To reduce a mixed quantity to an improperfraction. RULE. —Multiply the integral part by the denominator of the fraction, and to the product add the numerator, when it is ALGEBRAIC FRACTrIONS. 37 affirmative, or subtract it when negative; then the result, placed over the denominator, wvill give the improper fraction required. EXAMPLES. 1. Reduce 32 and a - - to improper fractions. C Hjere 3= 3 X 5 +2 15 +2 17 A 5 5 5' b aX c- b ac- b And a - - - Ans. C. a a2~ a2 2. Reduce x -+ - and x - - to improper fractions. a xX Xc + a x2 +a Here c + — --. Ans. x x & a2 — 2 2_,2 _2 2x2 +2 a2 And x * -. Ans. 2x 3. Let 1 - - be reduced to an improper fraction. a a — 2x Ans. 3x — b 4. Let 5a -- be reduced to an improper fraction. 5a2 - 3x + b Ans..a 5. Let x - -- be reduced to an improper fraction. 2a 2ax - a - x2 Ans. 2a 2x — 7 6. Let 5 + -- be reduced to an improper fraction. 3x 17x' -7 Ans. 3x a —a-i 7. Let I - be reduced to an improper fraction. a 2a- x -+ 1 Ans.. * Xv x -2. In adding the numerator a2 —x, the sign -affixed a2 - 2 to the fraction -- denotes that the whole of that fraction is to be subtracted, and consequently that the signs of each term of the numerator must be changed when it is combined with c2; hence the imoro-a2 -+ 2Pxz 22-a2 per fraction is -- o ---- ED. 4 38 ALGEBRAIC FRACTIONS. x-3 8. Let 1 + 2x - --- be reduced to an improper fraction. 5x ~10x2 + 4x + 3 Ans. 5x CASE IV. To reduce an improper fraction to a whole or mixed quantity. RvLE. —Divide the numerator by the denominator, for the integral part, and place the remainder, if any, over the denominator, for thefiractional part; then the two, joined together, with the proper sign between them, will give the mixed quantity required. EXAMPLES. 27 a ax -~ a2 1. Reduce -2 and __+- to mixed quantities. 5 x 27 Here. - 27. 5= 5-. Ans. And aX a = (ax +a2) + x a -—. Ans. X X ax — Y 2. It is required to reduce the fraction - to a whole quantity. Ans. a — x2. ab - 2a2 3. It is required to reduce the fraction s- to a mixed b ~~~~quantity. A~a~ 2 4. It is required to reduce the fraction -- to a mixed a —X quantity. 2X2 Ans. a x + --—. a —x 5. it is required to reduce the fraction - to a whole Y quantity. Ans. if + xy + yY 10x2 - 5x + 3 6. It is required to reduce the fraction - to a ax mixed quantity. 3 Ans. 2x- 1 + CASE V. To reduce fractions to other equivalent ones, that shall have a common denominator. RULE. —Multiply each of the numerators, separately, into ALGEBRAIC FRACTIONS. 39 all the denominators, except its own, for the new numerators, and all the denominators together for a common denominator.* EXAMPLES. a b 1. Reduce and- to fractions that shall have a common b denominator. Here a X c ac H X c- b2 the new numerators. b X c = bc the common denominator. W a b ac b2 Whence and - - and-X the fractions required. b c bc b 2x b 2. Reduce - and - to equivalent fractions having a coma C mon denominator. 2cx ab Ans. _ and ac ac a a+-b 3. Reduce a and to equivalent fractions having a, C common denominator. Ans. ac and ab - b be bc 3x 2b 4. Reduce -2a' 3c' and d, to equivalent fractions having a common denominator. 9cx 4ab 6acd Ans. and 6ac' 6ac' 6ac 3 2x 4x 5. Reduce —, and a -+, to fractions having a comRu 3' g a co5' mon denominator. 45 40x 60a + 48x Ans. - and 60' 60' 60 a 3x a~+ x 6. Reduce - 3, and +, to fractions having a common denominator. 7a2 - 7ax 6ax - 6X2 14a +- 14x Ans. --- an 14a - 14x' 14a -' 14a - 14x' * It may here be remarked, that if the numerator and denominator of a fraction be either both multiplied, or both divided, by the same number or quantity, its value will not be altered; thus 2 2X3 6 3 3 -.3 1 a ac ab a,and = ~~ or C= and —3-3 X3 9' 12 12 — 3 4 b be bbe c which method is often of great use in reducing fractions more readily to a common denominator. 40 ALGEBRAIC FRACTIONS.;CASE VI. To add fractional quantities together. RULE. —Reduce the fractions, if necessary, to a common denominator; then add all the numerators together, and under their sum put the common denominator, and it will give the fractions required.* EXAM-PLES. X 1. It is required to find the sum of and. 2 3 Here x X 2 = 2x the numerators. And 2 X 3 = -6 the common denominator. 3x 2x 5x Whence -- 6-' the sum required. 6 6' a, c e 2. It is required to find the sum of, dandf. - Here a X d Xf= adf c X b xf= cbf the numerators. e X b X d = ebd) And b X d xf-= bdf the common denominator. adf cbf ebd adf+ cf + ebd the sum. Whence - + - + f- bdf bdf bdf bdf 3. It is required to find the sum of a —, and b - c Here, taking only the fractional parts, we shallhave:3 X a t the numerators. 2ax X b- 2abx the numerators. And b X c = be the common denominators. 3cx2 2abx 2abx - 3cx2 Whence a -a —- -+ b - a -- - ---- the sum. be' bc 2x 5x 4. It is required to find the sum of and 39x Ans. 335 5. It is required to find the sum of - and - 2a 5' 15x Jr 2ax __-_Ans 10a * In the adding or subtracting of mixed quantities, it is best to bring the fractional parts only to a common denominator, and then to affix their sum or difference to the sum or difference of the integral parts, interposing the proper sign. ALGEBRAIC FRACTIONS. 41 6. It is required to find the sum of, and. 2131 13x Ans. -> 4x x- 2 7. It is required to find the sum of - and x 7 5 27x - 14 Ans. 35 2x 8x 8. Required the sum of 2a, 3a +-, and a - 9-. 22x 45' 3x a a-x 9. Required the sum of 2a + 3 — - and 5 a a —' a 3a2_- 3ax2~ + 5x2 Ans. 2a 2 + 5a ax 10. Required the sum of 5x + X and 4x - 3 5x 5X2-_ 16x + 9 Ans. 9x + 15x 2a a +- 2x 11. It is required to find the sum of 5x, -, and -—' 4x 8a + 3ax + 6x' Ans. 5x +12x2 CASE VII. To subtract onefractional quantity from another. RuLE.-Reduce the fractions to a common denominator, if necessary, as in addition; then subtract the less numerator from the greater, and under the difference Write the common denominator, and it will give the difference of the fractions required. EXAMPLES. 2x 3cs 1. It is required to find the difference of 2 and 3 3 5 Here 23X X 3 10x the numerators. And 3 X 5 = 15 the common denominator. 4* 42 ALGEBRAIC FRACTIONS. Whence -I 9 = - the difference required. 15 5 15 1 2. It is required to find the difference of x and 2b 2a -4x 3c Hre ( a) X 3c= 3cx -3ac tenumerators. (2a - 4x) X 2b = 4ab -8bx the numerators And 2b X 3c = 6bc the common denominator. 3cx - 3ac 4ab - 8bx 3cx - 3ac-4ab+ 8bx Whence - the 6bc 6bc 6bc difference required. 12x 3x 4x 3. Required the difference of -and Ans. x +7' 35' 4. Required the difference of 15y and 1 -+ 2 11Sy — 1 Ans. 8 5. Required the difference of and a b —c b +c' -2acx Ans. 2 b2 _ C2~ X —-'a x 6. Required the difference of x - andx + - c 2b' 2ba - 2bx - cx Ans. 2bc 7. Required the difference of a + - and a- - 2a2 + 2xa Ans. 2x +7 5x-6 8. Required the difference of ax + - and x -.-. 8 21 86x - 99 Ans. ax - i68 3x- 5 9 Required the difference of 2x + -, and 3x + 11x- 10 32x + 5 15 ~ 105 ALGEBRAIC FRACTIONS. 43 10. Required the difference of a + aX and a(a + x) a(a - x)' 4x Ans. aCASE VIIL To multiply fractional quantities together. RuLE.-Multiply the numerators together for a new nunmerator and the denominators for a new denominator; and the former of these being placed over the latter, will give the product of the fractions, as required.* EXA MPLES. x 2x 1. It is required to find the product of - and. 6 9~ xeX2x 2x2 x2 Here = = the product required. 6 X 9 54 27 x 4x 2. It is required to find the continued -product of - - 2' 5' I Ox and 21' x X 4x X 10x 40x3 4x3 Here X 4 - the product. 2 X 5 X 21 210 21 a-+x 3. It is required to find the product of - and a-. a a- x x X (a + x) + a_ _ Here - the product. aX (a-x) a2'-ax 3x 5X 4. It is required to find the product of 2 and Tf2 3b' 5X3 Ans. b 2x 3xa 5. It is required to find the product of -- and 2-. 3x3 Ans. ~~ —- ~~~~- h5a' * When the numerator of one of the fractions to be multiplied, and the denominator of the other, can be divided by some quantity, which is common to each of them, the quotients may be used instead of the fractions themselves. Also, when a fraction is to be multiplied by an integer, it is the same' thing whether the numerator be multiplied by it, or the denominator divided by it. Or if an integer is to be multiplied by a fraction, or a fraction by an integer, the integer may be considered as having unity for its denominator, and the two be then multiplied together as usual. 44 ALGEBRAIC FRACTIONS. 2x 4x2 6. It is required to find the continued product of 3u, 4 a 8axi3 and Ans. 1 a+x 21a + 21x' 2x 3ab 7. It is required to find the continued product of -,-, a C 5ac and 5ac Ans. 15ax. bx b 8. It is required to find the product of 2a + - and 3a -. a ax 2b b2 Ans. 6a2 + 3bx — - X a2 9. It is required to find the, continued product of 3x, x 4t-1 xv: - 1 3' -- 3x +1 and - Ans. - 2a' a + b' 2~a2 - 2ab' 10. It is required to find the continued product of +.. a- a-' a2 - b2 ax as - a2% and a + - Ans. ax -iX2 a - X' CASE IX. To divide one fractional qzantity by another. RULE.-Multiply the denominator of the divisor by the numerator of the dividend, for the numerator; and the numerator o the divisor by the denominator of the dividend, for the denominator. Or, which is more convenient in practice, multiply the dividend by the reciprocal of the divisor, and the product will be the quotient required.* EXAMPLES. x 2x 1. It is required to divide x by 3 9* Here 2x x 9 9x 3 H.ere X 2. = = — 12 Ans. 3 9 3 2x 6x 2 2a A4c 2. It is required to divide -b by d-. b'd * WVhen a fraction is to be divided by an integer, it is the same thing whether the numerator be divided by it, or the denominator multiplied by it. Also, when the two nunmerators, or the two denominators, can be divided by some common quantity, that quantity may be thrown out of each, and the quotients used instead of the fractions first proposed. INVOLUTION. 45 2a d 2ad ad Here - 4= 4c -2 Ans. x~a x~-b 3. It is required to divide a by x + a 5x +-,a 5x2 +- 6ax Asa2' Here - X - 2 Ans. x - b x q-b " -b 2,2 X 4. It is required to divide 3 + 3 by + a 2xu2 x +a: 22 (x +- a) 2x Here X- - = - a3+X3 x X (a,+ X') _.a 7x 3 7r2 5. It is required to divide -by 3 Ans. 7xu 5 X 15' 4X2 4x 6. It is required to divide -by 5x. Ans. x+I- 1 2x C +- 1 7. It is required to divide - -- by - t Ans. 6 3- 4x x x 5 8. It is required to divide - by Ans. 2ax ~- x2 x 9. It is required to divide 3 X —----- by 2a- x3 c Ans. a e c2 ~- cx +- X2 CX4., of any given quantity. - b4 X - th10. It is required to divide by X'" + b6 Ans. -- IN'VOLUTION. INVOUTIOed is the raising of powers from ansty proposed root; or the method of finding the squarer cub., biq.adrate, &c., of any given quantity. RULE 1. —Multiply the index of the quantity by the index of the power to which it is to be raised, and the result will be the power required. Or multiply the quantity into itself as many times less one as is denoted by the index of:the power, and the last product will be the answer. Note. When the sign of the root is +-, all the powers of it will be +-; and when the sign is -, all the even powers 46 INVOLUTION. will be -, and the odd powers -: as is evident from multiplication.* EXAMPLES. a, the root. a2 the root. a' = square. a4 = square. as = cube. a6 = cube. a4 - 4th power. as - 4th power. a -= 5th power. a10 = 5th power. &c. &c. - 3a the root. - 2ax2 the root. + 9a' = square. - 4a2X4 = square. - 27a3 - cube. - 8a3;6 = cube. + 81a4 = 4th power. + 16a8 = 4th power. &c. &c. X 2ax2 -the oot. the root. a~c 3b a2 4a2X4 - square. + 9b2 = square. x3 80ax6 cube. - _ = cube. ao 27b3 x~ 16a4x8 - 4th power. + 16ai -- 4th power. a 81bT &c. &c. x - a the root.. x + a the root. a- a x + a x2 ax x2 + ax - ax Jr a2 + ax -a2 oc - 2ax + a2 square. x2 + 2ax + a2 square. x - a x + a X3 - 2axa +- a2x x3 3ax2 + a2 - ax2 + 2ax - a- ax2 - 2a2x - a3 -- 3ax2 +- 3a2x - a3 cube. x3 - 3ax2 +- 3a 2x as cube. * Any power of the product of two or more quantities is equal to the same power of each of the factors multiplied together. And any power of a fraction is equal to the same power of the numerator divided by the like power of the denominator. Also, amn raised to the nth power is amn; and - am raised to the nth power is i:: amn, according as n is an even or an odd number. INVOLUTION. 47 EXAMPLES FOR PRACTICE. 1. Required the cube, or third power, of 2a2. Ans. 8a6. 2. Required the biquadrate, or fourth power, of 2a2x. Ans. 16a3x4. 3. Required the cube, or third power, of - - y3. Ans. -- 8 xSy 3a2x 4. Required the biquadrate, or fourth power of 5b2 81a8x4 Ans. -- 625b8 X 5. Required the fourth power of a + x;' and the fifth power of a - y. Ans. a4 + 4a3x +- 6a2 2 + 4ax +-X4, and a5 - 5a4y + 10a3y2 - 10aC2y3 - 5a4 -_ y5. RULE 2.-A binomial or residual quantity may also be readily raised to any power whatever, as follows:1. Find the terms without the coefficients, by observing that the index of the first, or leading quantity, begins with that of the given power, and decreases continually by 1, in every term to the last; and that in the following quantity, the indices of the terms are 1, 2, 3, 4, &c. 2. To find the coefficients, observe that those of the first and last terms are always I; and that the coefficient of the second term is the index of the power of the first: and for the rest, if the coefficient of any term be multiplied by the index of the leading quantity in it, and the product be divided by the number of terms to that place, it will give the coefficient of the term next following. Note. The whole number of terms will be one more than the index of the given power; and when both terms of the root are +, all the terms of the power will be +; but if the second term be -, all the odd terms will be +, and the even terms -; or, which is the same thing, the terms will be + and - alternately.* * The rule here given, which is the same in the cases of integral powers as in the binomial theorem of Newton, may be expressed in general terms, as follows:mn m on-1 i M-1?t7-2 t - I?-2 n -3 (a+b) — a. fma+ b -- m,. 2 a- b2 +' 2 3 a b3, &c. mA -m wz-I m- m-2 n-1 n-2 n-s ( / a ma bm u * a b2 -An 2 * a 3 b3, 48 EVOLUTION. EXAMPLES. 1. Let a + x be involved, or raised to the fifth power. HIere the terms,, without the coefficients, are, a., a4x, a3x2, aaX, 4 ax X5. And the coefficients, according to the rule, will be 5X4 10X3 10X2 5X1 1, 5, 2 3 4' I or 1, 5, 10, 10, 5, 1, Whence the entire fifth power of a +x is a5 + 5a4x + 10a3x +- 10a2x3 + 5ax4 - x5. 3. Let a - x be involved, or raised, to the sixth power. Here the terms, without their coefficients, are, 6 a5 42, &3; a2X4 5 5 26 aa, X, 4X, ax3 a4, aX. And the coefficients, found as before, are, 6X 5 15X420 X 3'15X2 6 X1 2' 3' 4' 5 6 or 1, 6, 15, 20, 15, 6, 1. Whetnce the entire sixth power of a - x is a6 - 6a5x - 15a4x2 - 20a3x3 + 15a2x4 - 6ax5 d- x6. 3.,Required the fourth power of a + x, and the fifth power of a- x. Ans. a4- 4a + 6a22+ 4a3 +4, and a5 - 5a4x + 10a3x2 - 10a2x3 + Sax4 - xA. 5. Required the sixth power of a +- b, and the seventh power of a-y. Ans. a6 + 6a5b + 15a4b2 + 20aSbs3 + 15a2b4 + 6ab5 + be, and a7 - 7a6y + 21 a5y2 - 35a4y3 + 35ay4 - 21 ay5 q+ 7ay6 - y7. 6. Required the fifth power of 2 + x, and the cube of a - bx + c. Ans. 32 + 80x -- 80xc2 +- 40a3 - 10x4 + x5, and a3 + 3a2c -P 3ac" -- c" - 3a2bx - 6acbx - 3c2bx + 3ahb2X2 + 3cb2hx2 b3 3. EVOLUTION. EVOLUTION, or the extraction of roots, is the reverse of involution, or the raising powers; being the method of finding the square root, ~ube root, &c., of any given quantity. CASE I. Tofind any root of a simple quantity. RULE.-Extract the root of the coefficient for the numeral &c. which formulae will also equally hold when m is a fraction, as will be more fully explained hereafter. It may also be farther observed, that the sum of the coefficients in every power, is equal to the number 2 raised to that power. Thus 1 + I - 2, for the first power; 1 - 2 + 1 = 4 = 22, for the square; 1 + 3 - 3 -1 8 23, for the cube, or third power; and so on. EVOLUTION. 49 part, and the root of the quantity subjoined to it for the literal part; then these joined together, will be the root required. And if the quantity proposed be a fraction, its root will be found by taking the root both of its numerator and denominator. Note. The square root, the fourth root, or any other even root, of an affirmative quantity, may be either + or -. Thus a2' + a or - a, and 4 b4 = + b or - b, &c. But the cube root, or any other odd root, of a quantity, will have the same sign as the quantity itself. Thus, V a3 a; / -a'==-a; and V —a5 — a, &c-* It may here, also, be farther remarked, that any even root of of a negative quantity is unassignable. Thus, - a2 canriot be determined, as there is no quantity, either positive or negative, (+ or -), that, when multiplied by itself, will produce - a2. EXAMIPLES. 1. Find the square root of 9x2-; and the cube root of 8x3. Here 92 = 9 X / x2 3 X = 3x. Ans. And Q/83x3= v8 X Vx3= 2 X X =2x. Ans. 2. It is required to find the square root of a antd the 8a'X3 cube root of - - a2xa 2 aa22 ax s8a3x3 2ax Here I/= 4C2 v 4c2 2c 2 7 3c 3. It is required to find the square, root of 4%2x6. Ans. 2ax3. 4. It is required to find thei cube root of - 125a'3x6. Ans. - 5ax2. 5. It is required to find the 4th root of 256a4x8. Ans. 4ax.3 4a4 6. It is required to find the square root of 2a2 Ans. 3xy' ~8a' ~ 2a 7. It is required to find the cube root of 1256. Ans. * The reason why -+ a and - a are each the square root of a2 is obvious, since, by the rule of multiplication, (+ a) X ( — a) and (- a) X ( —a) are both equal to a2. And for the cube root, fifth root, &c., of a negative quantity, it is plain, from the same rule, that (-a) X (-a) X (-a)=-a3; and (-a3) X (- a2) =-a5. And consequently 3V' a3 = -a, and 5v- a =-a. 50 EVOLUTION. 32anx'~ 8. It is required to find the 5th root of - 243 - 2ax2 Ans. 3CASE 11I. To extract the square root of a compound quantity. RULE 1.-Range the terms, of which the quantity is composed, according to the dimensions of some letter in them, beginning with the highest, and set the root of the first term in the quotient. 2. Subtract the square-of the root thus found, from the first term, and bring down the two next terms to the remainder -for a dividend. 3. Divide the dividend, thus found, by double that part of the root already determined, and set the result both in the quotient and divisor. 4. Multiply the divisor, so increased, by the term of the root last placed in the quotient, and subtract the product from the dividend; and so on, as in common arithmetic. EXAMPLES. I. Extract'the square root of x4 - 4x3 +- - 4x + 1. X4- 4x3 + 6x2 - 4x + 1 (x2 - 2x + 1 2x - 2x) - 4a3 - 6xc2 - 4cx + 4x2 2 - 4x +- 1) 2x2 - 4x + 1 2X2 -_ 4x +- 1 Ans. xc2 - 2x + 1, the root required. 2. Extract the square root of 4a4 + 12acx + 13a2cc + axc36,- 4. 4a4 + 12a3x + 13a2cc2 + 6axc +- xc4 (2a2 + 3arx + x2 4a4 4a2 + 3ax) 12a3x + 13a2ac 12ac3x + 9a2xa 4aF + 6ax + cx2) 4a2x2 + 6a=3 + X4 4a2cc + 6axa3 +- x4 *e EVOLUTION. 51 Note. When the quantity to be extracted has no exact root, the operation may be carried on as far as is thought necessary, ortill the regularity of the terms show the law by which the series would be continued. EXAMPLE. i. It is required to extract the square root of 1 - x. x x2 a3 5x4 2 8 16 128' 2+ All ~2 + - 4 X2 a3 X4 -- + _ - 4 Y16 8 46-/ 8 16 64 256 5X4 xi5 x 64 +64 256THere, if the numerators and denominators of the two last terms be each multiplied by 3, which will not alter their values, the root will become x x2 3 x3 3.5x4 3.5.7X5 I w -h + - 2.4 + _3 - X + 5 c.' 2 2.4' 2.4.6 2.4.6.8' 2.4.6.8.10' where the law of the series is manifest. EXAMPLES FOR PRACTICE. 2. It is required to find the square root of a4 + 4a3x + 6ax2 ~ 4ax3 ~ +4. Ans. a2 - 2ax - x2. 3. It is required to find the square' oot of x4 - 2x3 + 3 1 2 -- 1-6' Ans. x2 _- e 4 4. It is required to find the square root of 4x6 - 4x4 + 12x 3- x+2 - 6x - 9. Ans. 2x3-+- x -3. EVOLUTION. 5. Required the square root of x6 + 4x5 + 10x4 + 20x3 + 25x2 + 24x -- 16. Ans. X3 - 2X2 + 3x + 4. 6. It is required to extract the square root of a2 + b. b b2 b3 5b4 Ans. a + r —+ - 128a& 7. It is required to extract the square root of 2, or of 1 + 1. Ans. + A -I +- + —L1 + A-L, &C. CASE HII. To find any root of a compound quantity. Ru-r,. —Find the root of the first term, which place in the quotient; and having subtracted its corresponding power from that term, bring down the second term for a dividend. Divide this by twice the part of the root above determined, for the square root; by three times the square of it, for the cube root, and so on; -and the quotient will be the next term of the root. Involve the whole of the root, thus found, to its proper power, which subtract from the given quantity, and divide the first term of the remainder by the same divisor as before; and proceed in this manner till the whole is finished.* EXAMPLES. 1. Required the square root of a4 - 2a3x + 3a2x2 - 2zx3 + x4. a4- 2a3x + 3a2x2- -2ax3 + x4 (a2 - ax - C2 a4 2a2) - 2a3x a4 - 2aw + a2X2 2a,) 2cax2 a4 - 2a3x + 3a2x2 - 2ax3 + x4 v As this rule, in high powers, is often found to be very laborious, it may be proper to observe, that the roots of various compound quantities may sometimes be easily discovered, as follows:Extract the roots of all the simple terms, and connect them together by the signs -+- or -, as may be judged most suitable for the purpose; then involve the compound root, thus found, to its proper power, and if it be the same with the given quantity, it is the root required. But if it IRRATIONAL QUANTITIES, OR SURDS. 53 2. Required the cube root of xa + 6x5 - 40x3 + 96x - 64. 6 + 6x5 - 40x3 +- 96x - 64 (x2 - 2x - 4 3x4) 6x5 x6 + 6x2 + 12x4 + 8x3 3x4) - 12x4 x6 +- 6x5 - 40x3 +- 96x- 64 3. Required the square root of 4a2 - 12ax - 9x2. Ans. 2a - 3x. 4. Required the square root of a2 + 2ab + 2ac + b2 + 2bec +c2. Ans. a + b+c. 5. Required the cube root of xG - 6x5 + 15x4 - 20x3- + 15x2 - 6x + 1. Ans. 3- 2x- +1. 6. Required the 4th root of 16a4 -96adx + 216a22 _ 216ax3 + 81x4. Ans. 2a - 3x. 7. Required the 5th root of 32x5 - 80x4 + 80x3 - 40x2 + lOx -1. Ans. 2x -1. OF IRRATIONAL QUANTITIES, OR SURDS. IRRATIONAL Quantities or Surds, are those of which the values cannot be accurately expressed in numbers; and are usually expressed by means of the radical sign V/, or by fractional indices; in which latter case, the numerator shows the power the quantity is to be raised to, and the denominator its root. Thus, V/2, or 2, denotes the square root of 2; V aS, or a3 the cube root of the square of a, &c.* be found to differ only in some of the signs, change them from + to-, or from - to +-, till its power agrees with the given one throughout. Thus, in the third example next following, the root is 2a - 3, which is the difference of the roots of the first and last terms; and in the fourth example, the root is a +- b + c, which is the sum of the roots of the first, fourth, and sixth terms. The same may also be observed of the sixth example, where the root is found from the first and last terms. * A quantity of the kind here mentioned, as for instance / 2, is called an irrational number, or a surd, because no number either whole or fractional, can be found, which, when multiplied by itself, will produce 2. But its approximate value may be determined to any degree of exactness, by the common rule for extracting the square root, being 1 and certain non-periodic decimals, which never terminate. 5* 64 IRRATIONAL QUANTITIES, OR SURDS, CASE I. To reduce a rational quantity to the form of a surd. RuLE.-Raise the quantity to a power corresponding with that denoted by the index of the surd; and over this new quantity place the radical sign, or proper index, and it will be of the form required. EXAMPLES. 1. Let 3 be reduced to the form of the square root. Here 3 X 3 = 32- 9; whence V/9. Ans. 2. Reduce 2x2 to the form of the cube root..1 Here (2X2)3 = 8x"; whence V 8x6, or (8 6)3. 3. Let 5 be reduced to the form of the square root. Ans. v (25). 4. Let - 3x be reduced to the form of the cube root. Ans. V - (273). 5. Let - 2a be reduced to the form of the fourth root. Ans. -4V(16a). 6. Let a2 be reduced to the form of the fifth root, and la ~- /b, -/ and a to the form of the square root. 2a bVa Ans. a l0, V (a+ 2 V ab b), (), and 62. Note. Any rational quantity may be reduced by the above rule, to the form of the surd to which it is joined, and their product be then placed under the same index or radical sign. EXAMPLES. Thus 2 V/2 = v/4 X /2 =- V/(4 X 2) = / 8 And 2 3 4 = 8 X 3V 4 -V3 (8 X 4) = 39 32 Also 3 / a = / 9 X / a = / (9 X a) = v 9a And W- 4a = =/1 XV4a= V -( X 4a) =-3Va 1. Let 5 v 6 be reduced to a simple radical form. Ans. V/ (150). 2. Let /5a be reduced to a simple radical form. Ans. (5). 2a 9 3. Let -3 -/ 2 be reduced to a simple radical form. 32a Ans. 2/a CASE II. To reduce quantities of different indices, to others that shall have a given index. R ULE.-Divide the indices of the proposed quantities by IRRATIONAL QUANTITIES, OR SURDS. 55 the given index, and the quotients will be the new indices for those quantities. Then, over the said quantities, with their new indices, place the given index, and they will be the equivalent quantities required. EXAMPLES. 1. 1 1. Reduce 32 and 2 3 to quantities that shall have the index l. Here I. = X 6-= 3, the 1st index; And ---- 2, the 2d index. 3 6 3 1 3 1 I 1 Whence (33)6 and (22)6, or 276, and 46, are the quantities required. 1! 2. Reduce 5 2 and 63 to quantities that shall have the common index 6 Ans. 1256 and 366. 3. Reduce 22 and 44 to quantities that shall have the com1 mon index -. Ans. 168 and 168. 4. Reduce a2 and a2 to quantities that shall have the common index. Ans. (aS)4 and (a2)4. 5. Reduce a2 and b3 to quantities that shall have the common index 1 mon index - Ans. (a3)6 and (b4)6. Note. Surds may also be brought to a common index, by reducing the indices of the quantities to a common denominator, and then involving each of them to the power denoted by its numerator. EXAMPLES.. Reduce 3 2 and 4 3 to quantities having a common index 1. Reduce 3 and 4 3to quantities having a Common index. L a L A Here 32 = 36 = (33)6 = (27)6 And 43 = 46 (42) = (16)6 Whence (27)6 and (16)6. Ans. 2. Reduce 43 and 54 to quantities that shall have a common index. Ans. 25612 and 125T. 56 PIRRATIONAL QUANTITIES, OR SURDS. 3. Reduce a2 and a3 to quantities that shall have a common index. Ans. (aS)6 and (a)6 4. Reduce a3 and bV to quantities that shall have a common index. Ans. (a4)TW and (b3)i. 5. Reduce a' and b", to quantities that shall have a common index. 1 1 Ans, (am)n and (bn) rnn, CASE III. To reduce surds to their most simple forms. RULE,-Resolve the given number, or quantity, into two factors, one of which shall be the greatest power contained in it, and set the root of this power before the remaining part, with the proper radical sign between them.* E XAMPLES. 1. Let V48 be reduced to its most simple form. Here V/ 48 — (16 X 3)-4 3. Ans. 2.. Let 3V 108 be reduced to its most simple form. HIere V 108 -3V (27 X 4) -- 334. Ans. Note 1. When any number, or quantity, is prefixed to the surd, that quantity must be multiplied by the root of the factor above imentioned, and the product be then joined to the other part, as before. EXAMPLES. 1. Let 2 V 32 be reduced to its most simple form. Here 2 V 32 2 V(16 X 2) 8 V 2. Ans. 2. Let 5, /24 be reduced to its most simple form. Here 53V 24 = 5V (8 X 3) = 10V 3. Ans. Note 2. A fractional surd may also be reduced to a more convenient form, by multiplying both the numerator and denominator by such a number, or quantity, as will make the denominator a complete power of the kind required; and then joining its root, with 1 put over it, as a numerator, to the other part of the surd.t * When the given surd contains no factor that is an exact power of the kind required., it is already in its most simple form. Thus, /V 15 cannot be reduced lower, because neither of its factors, 5 nor 3, is a square. t The utility of reducing surds to their most simple forms, in order to have the answer in decimals, will be readily perceived from considering IRRATIONAL QUANTITIES, OR SURDS. 57 EXAMPLES. 1. Let 2/ be reduced to its most simple form. Here v 7 = 49 = / 49 X 14) = 7 14. Ans. 2 2. Let 33 - be reduced to its most simple form. Here 3 = 3V 2-3 125 - X 50) -3 50 Ans 1 / 125 5 EXAMPLES FOR PRACTICE. 3. Let V/ 125 be reduced to its most simple form. A1ns. 5 5. 4. Let V 294 be reduced to its most simple form. Ans. 9V46. 5. Let 3/ 56 be reduced to its most simple form. Ans. 233) A. 6. Let V/ 192 be reduced to its most simple form. Ans. 4 3. 7. Let 7/ 80 be reduced to its most simple form. Ans. 287V5. 8. Let 93V 81 be reduced to its most simple form. Ans. 27 3. 9. Let 3192 be reduced to its most simple form. 12_1 6 Ans. 24 /30. 10. Let 7 3V0 be reduced to its most simple form. Ans. 128 5. 18. Let V 98a2X be reduced to its most simple form. Ans. 27a/ 2x 12. Let / (3 - a22) be reduced to its most simple form. Ans. (x - a2). the first question above given, where it is found that V/ 27 = 7 1/ 14* in 0.hich case it is only necessary to extract the square root of the ohole number 14, (or to find it in some of the tables that have been calculated for this purpose,) and then divide it by 7; whereas, otherwise, we must have first divided the numerator by the denominator, and then have fo11un the root of the uotient, for the surd part; or esimple have determined the root botih of the numerator and denomre inator, and then divided one by the other; which are each of them troublesome processes when performed by the col-mmon rules; and in the next example for the cube root, the labour w-vold be much greater IRRATIONAL QUANTITIES, OR SURDS. CASE IV. To add surd quazntities together. RULE.-When the surds are of the same kind, reduce them to their simplest forms as in the last case; then, if the surd part be the same in them all, annex it to the sum of the rational parts, and it will give the whole sum required. But if the quantities have different indices, or the surd part be not the same in each of them, they can only be added togetheer by the signs + and -. EXAMPLES. 1. It is required to find the sum of v 27 and -1/v 48. Here V27 — /. (9 X 3) = 3 V 3 And V48= V(16X3)=4V3 Whence 7 v 3 the sum. 2. It is required to find the sum of 3 500 and / 108. Here V 500 = V(125 X< 4) =5 V 4 And V 108 = 3(27 X 4) = 3 / 4 Whence 8 V 4 the sum. 3. It is required to find the sum of 4 V 147 and 3 V 75 Here 4 V147= 4AV(49 X 3) =28/V3 And 3V/75 = 3V/(25'X 3)=15V3 Whence 43 vt 3 the sum. 2 1 4. It is required to find the sum of 3 V and 2 V O 2 10 3 Here 3 V -= 3 / —- v10 5 25 5 I 10 2 And 2V-=2V- -- /!00 10 v 100 10 4 Whence - V/ 10 the sum. EXAMPLES FOR PRACTICE. 5. It is required to find the sum of V 72 and V 128. Ans. 14 Va (2). 6. It is required to find the sum of V 180 and V 405. Ans. 15 /(5). 7. It is required to find the sum of 3 V40 and V/ 135. Ans. 93 V (5). IRRATIONAL QUANTITIES, OR SURDS. 59 8. It is required to find the sum. of 4 / 54 and 5 V 128. Ans. 32 V/(2). 9. It is required to find the sum of 9 v 243 and 10 V 363 Ans. 191 V(3). 2 27 10. It is required to find the sum of 3 / - and 7 V - 3 50' Ans. 3 -to V/(6). 1 1 11. It is required to find the sum of 121V and 3 3 -2. Ans. 6-/(2). 12. It is required to find the sum of l- V a2b and 4 4bx4. Ans. (Z a 3 ) CASE V. Tofind the difference of surd quantities. RULE.-When the surds are of the same kind, prepare the quantities as in the last rule; then the difference of the rational parts annexed to the common surd, will give the who'e difference required. But if the quantities have different indices, or the surd part be not the same in each of them, they can only be subtracted by means of the sign -. 1. It is required to find the difference of V 448 and V 112. Here V/ 448 = / (64 X 7) =8 V 7 And 112 = V (16 X 7)= 4 V 7 Whence 4 /.7 the difference. 2. It is required to find the difference of 3 i192 and 3V24. Here V/192 ='V(64 X 3) = 4V3 And V24 = V( 8 X 3) = 2V3 Whence 23,/ 3 the difference. 3. It is required to find the difference: of 5 V 20 and 3 V/ 45. Here 5 /20 5 /(4 X 5)=.10/5 And 3V/45'3/(9X 5)= 925 Whence / 5 the difference. 32 2 1 4 It is required to find the difference of - and - V 4 3' 5 6 60 IRRATIONAL QUANTITIES, OR SURDS. Here 432 = V 6 = 3.V6 = 1A6 4 3 4 9 12 4 2 1 2 6 2 1 And -36 30 / 6 = - 6 Whence - v 6 the di] 60 ference, or answer required. EXAMPLES FOR PRACTICE. 1. It is required to find the difference of 2 V 50 and /V 18. Ans. 7 V (2). 2. It is required to find the difference of 3\/ 320 and V 40. Ans. 2 V(5). 3 5 3. It is required to find the difference of / and V/ 5 27. Ans. 4 (1 5). 4. It is required to find the difference of 2 V~ and V/ 8. Ans. V (2). 5. It is required to find the difference of 3 3 -1 and 3v72. Ans. V (9). r.2 9 6. It is required to find the difference of 3v and V/-' 32 Ans. 7. It: is required to find the difference of V/80a4x and V 20a2x3. Ans. (4a2 - 2ax) V' (5ax) 8. It is required to find the difference of 83a36b and 2, ab. Ans. (8a - 2a2) V (b). Note. The two last answers may be written thus, (2ax - 4a2) V (5x), and (2a2 - a)' (b); or (4a2 2ax) V 5x (8a 2a2) V b. CASE VI. To multiply surd quantities together. RuLE. —IWhen the surds are of the same kind, find the product of the rational parts, and the'product of the surds, and the two joined together, with their common radical sign between them, will give the whole product required; which may be reduced to its most simple form by Case III. But if the surds are of different kinds, they must be reduced to a common index,, and then multiplied together as usual. It is also to be observed, as before mentioned, that the pro IRRATIONAL QUANTITIES, OR SURDS. 61 duct of different powers, or roots of the same quantity, is found by adding their indices. EXAMPLES. 1. It is required to find the product of 3 V 8 and 2 V/ 6. Here 3V / 8 Multiplied 2 V 6 Gives 648 = 6 V16 X 3 = 24V3. Ans. 1 2 3 5 2. It is required to find the product of - V ~ and - V 6 2 3 4 6' Here 12 2 3 3 5 Multiplied - V - 4 6 3 10 3 5 3 15 Gives'VI3. It is required to find the products of 2" and 33 3 1 1 Here 2-.26 = (23) = 8 And 33 36(32)6 = 9 Whence (72)6. Ans. 4. It is required to find the product of 5 V a and 33 a. 1 3 Here 5 Va = 5a 5a6 1 o And 33Va= 3a3 = 3a6 Whence 15a = 15 (a5) or 15 6V aS. Ans. EXAMPLES FOR PRACTICE. 5. It is required to find the product of 5 V 8 and 3 V 5. Ans. 30 V (10). 6. It is required to find the product of V 18 and 5 A 4. Ans. 103 (9) I 2 7. Required the produ6t of - V 6 and5 V 9. 4 15 Ans. V (6). 8. Required the product of 2V 18 and 5 V 20. Ans. 15 V (10). 6 62 IRRATIONAL QUANTITIES, OR SURDS. 9. Required the product of 2 V/ 3 and 13h V 5. Ans. 27 V (15). 10. Required the product of 724a3 and 1201a4. Ans. 8706al'2. 11 Required the product of 4- +2 V 2 and 2 - V2. Ans 4. 12. Required the product of (a + b)'t and (a + b)m. m'n Ans. (a + b) m. CASE VII. To divide one surd quantity by another. RULE..-When the surds are of the same kind, find the quotient of the rational parts, and the quotient of the surds, and the two joined together, with their common radical sign between them, will give the whole quotient required. But if the surds are of different kinds, they must be reduced to a common index, and then be divided as before. It is also to be observed, that the quotient of different powers or roots of the same quantity, is found by subtracting their indices. EXAMPLES. 1. It is required to divide 8 V/ 108 by 2 V/ 6. 8 V 108 Here - = -4V/18 = 4V(9 X2) = 12 /2. Ans. 2. It is required to divide 83V 512 by 4V 2. S /51t2 Here 2 = 2V3 256 = 2V/ (64 X 4)= 83 4. Ans. 4V/2 3. It is required to divide 2/ 5 by /V 10. Ans. 2 3 )V5 3 5 3 10 3 Here 2 = 3/ V= 2V - A/ 10. Ans. \/2 2 2 2 4 4 4. It is required to divide V7 by 37. 1 a V7 772 76 3-1 I-Iere -=7 6 —7 Ans.;V7 73 76 5. It is required to divide 6 V 54 by 3 V 2. Ans. 6 V 3. 6. It is required to divide 4'/ 72 by 2V 18. Ans. 23 4. 3 1 1 7. It is required to divide 5-V/ by /I 4 1352 33 Ans. ~ 3. IRRATIONAL QUANTITIES, OR SURDS. 63 5 f2 f2 3 8. It is required to divide 4 / V by2 -5 2 / Ans. v2. 42 1 2 9. It is required to divide 422/ a by 22 3/ ab. Ans. - -. 16 (b2) 10. It is required to divide 322 a by 134 Va. 5 4 648 i Ans.- aG 275 32 91. 11. It is required to divide 9-an by 4-1am 825 n-; Ans. - a 424 12. It is required to divide v 20 + V 12 by V 5 +- / 3. Ans. V4 or 2. Note. Since the division of surds is performed by subtracting their indices, it is evident that the denominator of any fraction may be taken into the numerator, or the numerator into the denominator, by changing the sign of its index. am Also, since -- 1, or - am- = aO, it follows, that the am expression a0 is a symbol equivalent to unity, and consequently, that it may always b'e replaced by 1 whenever it occurs.+ * To what is above said, we may also farther observe, 1. That 0 added. to or subtracted from any quantity, makes it neither greater nor less; that is, a+O = a, and a-O = a. 2. Also, if nought be multiplied or divided by any quantity, both the product and the quotient will be nought; because any number of times 0, or any part of 0, is 0; that is, 0 Xa, oraX= 0, and _ = 0. 3. From this it likewise follows, that nought divided by nought, is a finite quantity, of some kind or other. For since O X a = 0, or 0 = O X a, it is evident that - =a. 4. Farther, if any finite quantity be divided by 0, the quotient will be infinite. For let - - q, then if b remain the same, it is plain, the less a is, the a greater will be the quotient q; whence, if a be indefinitely small, q will 64 IRRATIONAL QUANTITIES, OR SURDS. EXAMPLES. 1 a- 1 a — 1. Thus, -- or a-; and = -, or a —n. a 1 arh 1 b bGa- a-n 1 bm 2. Also, - or ba-2, and or n 3. Let be expressed with a negative index. Ans. a-2 4 Let a-i be expressed with a positive index. Ans. 2a 5. Let be expressed with a negative index. a+x Ans. (a + x)-1. 6. Let a (aa - x2)-1 be expressed with a positive index. a Ans. - (a -2) 3 CASE VIII. To involve, or raise surd quantities to any power. RULE.-WVhen the surd is a simple quantity, multiply its index by 2 for the square, by 3 for the cube, &c., and it will give the power of the surd part, which, being annexed to the proper power of the rational part, will give the whole power required. And if it be a compound quantity, multiply it by be indefinitely great: and consequently, when a is 0, the quotient q will be infinite; that is, b I or- -=G. 0' Which properties are of frequent occurrence in some of the higher parts of the science, and should be carefully remembered, 1 - Since, therefore, -- is the same as (a +b). Let us suppose, m a+b the general formula, n -1; and we shall have for the coefficients -1 - I / —-1- - &c- 3 ~= — 1f; 2 - 3 1; 1, &c. and for the powers of a we have ant=aa-2; an-2 _ 1 -- 1 1 11b b2 b3 b4 b5 an-3= — &C.; sothat(a+ b) =. -..+ — - a,4' Xab — a a2 a3a a4 as a &c., which is the same series that is found by division. For more on this subject see the Binomial Theorem, (further on) or Euler's Algebra. IRRATIONAL QUANTITIES, OR SURDS. 65 itself the proper number of times according to the usual rule.* EXAMPLES. 1. It is required to find the square of -a3' 3 2_'_ 3 H-ere (-a3)2=3 =4a2 a2. Ans. 2 2. It is required to find the cube of - v 3. Here X- 32 = —- 27= 7(9 X 3)= = /3. Ans. 27 27 27 3.' It is required to find the square of 3 / 3. Ans. 9 3,/ 9. 4. It is required to find the cube of 17 /21. Ans. 103173 /(21). 5. It is required to find the 4th power of V /6. Ans. -3 6 6. It is required to find the square of 3 d- 2 / 5. Ans. 29- +12 V5. 7. It is required to find the cube of / x + 3 V/y. Ans. x x +- 27y V x + 9x V y +- 27y V y. 8. It is required to find the 4th power of V 3 - V 2. Ans. 49 - 20V 6. CASE IX. To find the roots of surd quantities. RULE.-When the surd is a simple quantity, multiply its index by -1 for the square root, by ~- for the cube root, &c., and it will give the root of the surd part; which being annexed to the root of the rational part, will give the whole root required. And if it be a compound quantity, find its root by the usual rule.t * When any quantity that is affected by the sign of the square root is to be raised to the second power, or squared, it is done by suppressing the sign. Thus, (0/a)2, or v/a X/a=a; and (V/ab)2, or V/(a+b)XV(a+b)=a+b. t The nth root of the mth power of any number a, or the rnth power of the nth root of a, is an. Also, the %th root of the mth root of any number a, or the mth root of the nth root of a, is amln. From which last expression, it appears, that the square root of the square root of a is the 4th root of a; and that the cube root of the square root of a, or the square root of the cube root of a, is the 6th root of a; and so on for the fourth, fifth, or any other numerical root of this kind. 6* 66 IRRATIONAL QUANTITIES, OR SURDS. EXAMPLES. 1. It is required to find the square root of 9 V 3. _ 1 1. 1 1 Here (9V3)2 92 X33 2X =-92 X 36 =3 3. Ans. 2. It is required to find the cube root of 8 V 2. "-~~(~yZ)' ( 2~ x. Here (V2) ( X (2 X) (26) 1V 2. Ans. 3. It is required to find the square root of 103. Ans. 10V (10) 8 4 4. It is required to iind the cube root of -a4 27 Ans. Va 3/a. 16a -t 5. It is required to find the 4th root of -1 Ans. ~-a6' 81 6. It is required to find the cube root of a a/ o 3' Ans. V/, or S V(3a). 7. It is required to find the square root of x2 - 4x V a + 4a. Ans, x-2V/a. 8. It is required to find the square root of a + 2 v ab + b. Ans. V/a +-Vb. CASE X. To transform a binomial, or a residual surd, into a general surd. RULE.-Involve the given binomial, or residual, to a power corresponding with that denoted by the surd; then set the radical sign of the same root over it, and it will be the general surd required. EXAMPLES. 1. It is required to reduce 2 + v/ 3 to a general surd. Here (2 + V/3)2 4 + 3 + 4 V3= 7 + 4 3; therefore 2 + V/ 3 = v (7 + 4 / 3), the answer. 2. It is required to reduce V 2 + V 3 to a general surd. Here ( 2 +- /3)2=2 + 3+2 6 =5 +2 V6; therefore V 2 + / 3 = V (5 + 2 / 6), the answer. 3. It is required to reduce' 2 +- V4 to a general surd. Here ( V 2 +- 3/4)3=6 + 6 2 + 6 V3 4; therefore A/ 2 + V/4 = 3A 6 (1 + ~ 2 + 3/ 4), the answer. 4. It is required to reduce 3 - V 5 to a general surd. Ans. V (14 - 6 v 5). IRRATIONAL QUANTITIES, OR SURDS. 67 5. It is required to reduce V 2 -2 / 6 to a general surd. Ans. /V(26 —8 V3). 6. It is required to reduce 4 - / 7 to a general surd. Ans. V (23- 8 V 7). 7. It is required to reduce 2 V 3 - 3 V 9 to a general surd. Ans.' (162 V/9 - 108 3 - 219). CASE XI. To extract the square root of a binomial, or residual surd. PvuLrx.* —Substitute the nulmbers, or parts, of which the given surd is composed, in the place of the letters, in one of * Prop. 1. The square root of a quantity cannot be partly rational an-d par'tly a quadratic surd. If possible, let -/ = a + - in; then, by squarljng both sides, nt = a2- + 2a / m q-m, and by transposition 2a / m nl -- a 9= — a2 - m; therefore V/ m a rational quantity, which 2a iLs contrary to the supposition. A quantity of the form 1/V a, is called a quadratic surd. Prop. 2. - In any equation x- / y = a +- V b, consisting of rational quLa,;ltities and quadratic'squrds, the rational parts on each side are equal, aid also the irr'ational parts. If' x be not equal to a, let x = a+mm; then a + m- /y = a-+ xb, or n - + v y = - b; that / b is partly rational and partly a quadratic surd, which is impossible, (Pi'op. 1.);. x= a, and /y -= v b. In like manner, ifx — /y -=- v- b; then.x = a, and — y = —b. Prop. 3. -If twuo quadratic surds, 1/ x and /Vy, cannot be reduced to others which have t/he same irrational part, their product is irrational. If possible, let / xy r='z. where r is a whole number or a fraction: Then Xy = r2x2s, and y = r'x;.-. Vy = r V x; that is, V y and -Vx may be -so reduced as to have the same rational part, which is contrary to the supposition. Prop. 4. One quadratzc surd, / x, cannot be made up of two others, ~/ m c and / n, wAich have not the same irrational part. If possible, let V/ x = V m + / n; then by squaring both sides, x = m -- 2- / inn + n, and x-m — n = 2V i nn, a rational quantity equal to an irrational, which is absurd. Prop. 5. T'he square root of a binomial, one of whose terms is a quadratic surd, and the other rational, may sometimes be expressed by a binomial, one or both of whose te'rms are quadratic surds.. Let a + V b be the given binomial, and suppose / (a — +/ b) = x - y; where.x and y are one or both quadratic surds; then, (see Ryan's Elementary Treatise on Algebra, Art. 367,) /V(a —/b)=rx-y;.-. by multiplication, /(a2 —b) -2 -y2. Also, by squaring both sides of the first equation, a+ /b = x;2+2,xy q y2, and (Prop. 2.). a -- -2y+2. Hence by addition, a - (a2 - b) = 2.2, and by subtraction, a - / (as2-b) = 2y2;.. theroot x+y =c/ i -a- a s/ (a2-b) t+ s oja-oV(a —b) From this conclusion it appears, that the square root of a + / b can 68 IRRATIONAL QUANTITIES, OR SURDS. the two following formula, according as it is a binomial or a residual, and it will give the root required. v (a + 1/ b) = v [a + 1 v (a2 - b)] + v [a - A (a2- b) 2, (a - b) = v [a+ i V(a - b)] - v [- Ia - v 4a- b)] And if the second part of the binomial, or residual, in this case, be an imaginary surd, the same theorems will still hold, by only changing - b into + b, as below. tv( + v- b) - v [2la - v(a2 + b)] + 1/ I[a - v(a2 + b)] ( - v-b) = v + - v + )] - /[ -a- (a + b)] Where it is to be observed, that the only cases that are useful in this extraction, are, when a is rational, and a2 - b in the first of these formulah, or a2 + b in the latter, is a complete square. EXAMPLES. 1. It is required to find the square root of 11 + v' 72, or v(11 -+ 6 2). Here, I/ia~ +V (a2- b)= -V -+ -11/ (121 - 72) -1, + 7 =3; and av -V'ab Ad 121 - 72 = -7- -a ='2. v2tsC a- 2 2 f1/1I1 -V 2 21-T Whence v(ll1 -+ 6 v/ 2) = 3 +- v 2, the answer required. 2. It is required to find the square root of 3 - 2 v 2 Here,, + -k (a - b)=~,- + I A,' (9-8)- V' - = 1/2; and 8a, (a _ b) - - (9 -8) ( - 85 -1; Whence v/(3 - 2 v 2) = V2 - 1, the answer required. 3. It is required to find the square root of 6 ~i 2 v 5. Ans. V/ 5 ~: 1. 4. It is required to find the square root of 23:t 8 v 7.Ans. 4 4- / 7. 5. It is required to find the square root of 36 -i 10 1/ 11. Ans. 5 -: / (11). 6 It is required to find the square root of 33 ~- 12 v 6. Ans. 2 v 6::: 3. 7. It is required to find the square root of 1 + 4 v - 3, or 1 + lv-48. Ans. 2 +,-3. be expressed by a binomial of the form x + y, one or both of which are quadratic surds, when a2-b is a perfect square. By a similar process it might be shown that the square root of a —l/b, or (subject to the sb-am 1imitation-b- a- -b subject to the same limitation.-ED. IRRATIONAL QUANTITIES, OR SURDS. 69 8. It is required to find the square root of 3 ~: 4 v - 1, or 3 i:;t- 16. o Ans. 2=t -- 1. 9. It is required to find the square root of - 1 + / -- 8. Ans. 1 +- - 2. 10. It is required to find the square root of a2 + 2x A, (a2 -. X2). Ans. x + v (a2 _-T). 11. It is required to find the square root of 6 + 2 -v 2 - V (12) - (24) Ans. 1 + v 2 - / 3. FOR TRINOMIAL, QUADRINOMIAL SURDS, &C. RULE.-Divide half the product of any two radicals by a third, gives the square of one radical part of the root; this repeated with different quantities, will give the squares of all the parts of the root, to be connected by + and -. But if any quantity occur oftener than once, it must be taken but once. For if x + y + z be any trinomial surd, its square will be a2 + y2 + z2 + 2ay + 2xz + 2yz; then if half the product of any two rectangles as 2xy X 2xz (or 2x2yz) be divided by 2x2yz some third 2yx, the quotient 2/z = x2, must needs bel*he 2yz square of one of the parts; and the like for the rest. EXAMPLE 1. To extract the square root of 10 +- (24) +- / (40) + v- (60). v (24) X V (40) d/ (24) X v (60) 2 V(60) an 2(40) 9 = 3, and (4 2) X(214) -) /(25) = 5. And the root 2 (24) is ~2 +- 3 -+ v5. EXAMPLE 2. It is required to find the square root of 14 + v (32) -,V(48) + V (80) - V (24) + v (40) - V (60). V(32 X 48) _ V(24) Here (, this produces nothing. 2 v (80) v 5 Again, (32x ) (16) 4. And - v( 2 V(24) =(2' (24) /(32 X 40) (48 X 24) =- (25) —5;nd 4-42; and2(3 2 -v (60) 2 A(32) = / -3; and V(32 X 80) V 9 = 3; and 2V (4 = / (16)= 4, &c., therefore 2 A/ (0) - 70 IRRATIONAL QUANTITIES, OR SURDS. the parts of the root are v 4, v 5, v 3, v 2, V 4, &c., and the root is 2 +- v 2 - v 3 + v 5; for, being; squared, it produces the surd quantity given. CASE XII. To extract any root (c) of a binomial surd. RULE 1.* —Let the quantity be A - B, whereof A is the greater part and c the exponent of the root required. Seek * Let the sum or difference of two quantities, x and y, be raised to a power whose exponent is c, and let the 1st, 3d, 5th, 7tA, 4tc., terms of that power, collected into one sum, be called A, and the ~rest of the te'rms, in the e:ven places, called B; the difference of the squares of A and B shall be equal to the difference of the squafres of x and y raised to the same power c. For the terms in the power c of x +- y, writing for their coefficients respectively, 1, c, d, e, &c., are.xc-i cxc C-y dx -2y- + exC-3y3 + &C = A + B; and the same power of x —y (changing the signs ll the even places) is xC - cxe-ly + dxC'2y2 - exC-3y-3 +, &c. - A-B. And therefore, (x + y)C (x - y)e or (x2 - y2)C = (A + B) (A - ) B) AS-4- B2. Let one, or both of the quantities, x, iy, be a quadratic surd, that is, let c- + y, the c root of the proposed binomial A + B belong to one of these forms, p + l/ /q, k /v p- q, or k -Vp +- l V q. And it follows that, 1. If x + - = p + - v/ q, c being any whole number, A, the sum of the odd terms, will be a rational number; and B, the sum of ths terms in the even places, each of which involves an odd power of y, will be a rational number multiplied into the quadratic surd V/ q. 2. Let c, the exponent of thle root sought, be an odd number, as we may always suppose it, because if it is even, it may be halved by the extraction of the square root, till it becomes odd: and let x- +y= k v/p +q. Then A will involve the surd V'p, and B will be rational. 3. But if both members of the root are irrational, x- y= (kVp + lVq) A and B are both irrational, the one involving -Vp, and the other the surd v q. And in all these cases, it is easily seen that when x is greater than y, A will be greater than B. From this composition of the binomial A + B, we are led to its resolution, as in the above rule, by these steps.'W~hen A is rational, and A2- B2 is a perfect power. 1. By the theorem just demonstrated, A2- B2 = (x2 -y 2) accurately; and therefore extracting the c root of A2 - B2, it will be x2 - y2; call this root n. 2. Extract in the nearest integer, the c root of A+ B, it will be (nearly) x +y; which put = r. 3. Divide xs —y (= n) by x+- (= -r) the quotient is (nearly) x.-y; and the sum of the divisor and quotient is (more neasly) 2x; that n rae _ is, if an integer value of x is to be found, it will be the nearest to - IRRATIONAL QUANTITIES, OR SURDS. 71 the least number n whose power nc is divisible by A2 - B2, the quotient being Q compute /' [(A + B) X V/ Q] in the nearest integer number, which suppose to be r. Divide 4. x2 —(x2-y2) =y2; or, (i-r) 2-= —y2: whence Y. -n Y /(? )- 2- nf; and therefore, putting t _ r the rootsought xz y=t +-t-V(t2 - n); the same expression as in the rule, when Q= 1, s = 1; that is, when A2 —B2 is a perfect c power, and the greater number, A, is rational. II. raWhen A is irrational, and Q = 1. By the same process, $_;- r (= T) and y = v (T2-n). But seeing A is supposed irrational, and c an odd number, x will be irrational likewise: and they will both involve the same irreducible surd /-p, or s, which is found by dividing A by its greatest rational divisor. Write, therefore, for x or T, its value t X s, and xf- y = ts- V/(t2s2 —!). III. If the c root of A2 - B2 cannot be taken, multiply A2 —B2 by a number CQ, such that the product may be the (least) perfect c power nba ( A2 Q-_ B2 0.). And now (instead of A 4- B) extract the c root of (A + B) X v Q, which found as above, will be t s + V (t22s - i); and consequently the c root of A + B will be t s + V (t 2 s2-en),'divided by the c root of -Q/'; that is, ts+ -/ (t2 2-n) In the operation, it is required to find a number 0Q, such, that (A2 - B2) X 0Q may be a perfect c power; this will be the case if Q be taken equal to (A2 - B)'-1; but to find a less number which will answer this condition, let A2 - B2 be divisible by a, a,.....(m); b, b..... (na); d, d,.....(); &c. in succession, that is, let A2 - B2 = abn dr, &c.; also, let Q.=a by dZ, &c.; (A2-B2)X.=-ami+ Xbf+Zf Xd r+ &c. which is a perfect cth power, if x, y, z, &c., be so assumed that en + x, n 4- y, r + z, are respectively equal to c, or some multiple of c. Thus, to find a number which multiplied by 180 will produce a perfect cube, divide 180 as often as possible by2, 3, 5, &c., and it appears that 2.2.3. 3.5 = 180; if, therefore, it be multiplied by 2.3. 5.5, it becomes 2.3 3. 53, or (2.3. 5)3, a perfect cube. If A and B be divided by their greatest common measure, either integer or quadratic surd, in all cases where the cth root car. be obtained by this method, Q. will either be unity, or some power of 2 less than 2c. Ift' the residual A- B be given, it is evident from its genesis by involution, that the same rule gives its root x py. See Universal'Arithmetic, p. 139, Dr. Waring's Med. Alg. p. 287, or Maclaurin's Alg. p. 124. 72 IRRATIONAL QUANTITIES, OR SURDS. A v Q by its greatest divisor, arid let the quotient be s, and let n r2 t, the nearest integer. Then the root 2s ts - / (t2s2 - ), if the c root of A ~ — B can be extracted. It is proper to observe that this rule, which was first given by NEWTON in the Universal Arithmetic, fails, when t = exactly; in which case, instead of taking t the nearest inn r/r teger value of 2 —-it must be taken equal to I. See Ryan's Key to the second New York edition of Bonnycastle's Algebra. EXAMPLE. What is the cube root of v 968 - 25. We have A2 —B2= 343 =7 X 7 X 7. Q X7'3- n3, whence n = 7, and Q=_ 1. Then 3/ [(A + B) X vQ] = 3V/ 56 + = r= 4. A V Q — v 968 — 22 v 2, and thb n radical part v 2 1 s, and 4 t -2 2, inthe near 2s 2 v 2 est integer. And ts = 2 v 2, (t2s - n) = v (S -7) = 1. 2 A/2 q- I 6'(Q = 1). And the root is 2 v2 - 1, whose cube, upon trial, I find to be v 968 + 25. RULE 2.* —Let the surd, that is to have its root extracted, * Thus, let nv/(a + / b) = x + V y; and we shall have by involution, a.+ b = (z +- /y) An equation, which, by expanding the righthand member, and comparing the rational and irrational parts, gives n n(v —-1) n-2 __(n —1) (n-2) (n-3) n-4 a x J 2 X 2.3.4 y gab _ on-i _ (,~-1) (~-2)2z.3s v/y q..... V/b = zn- 1 /y q. q,, &e. Or, which is the same thing under a different form, a = 1a ( Vs/ )n +(-vy)n Vb =1 (x+ Vy)n -(x-vy)n a Whence, by squaring each of these equations, and subtracting the latter from the former, we shall have a —b- t (. +' Vy)2n + 2 A/ (x2 —)n +(x2 —V' y)m -,.,,_ ( +,-. 22- n.L~(.,-.2.n IRRATIONAL QUANTITIES, OR SURDS. 73 be of the form 3 (a +- / b), or V/ (a -./ b). Thlen if a2 - b be a perfect integral cube, and some whole number can be found, that, when substituted for n, will make n3 - 3 iV (a2- b)] n = 2a, the roots of the two expressions, in this case, will be 3 (a + / b) -= n + 1 V [n2 - 4 (a -_ b)] 3 (a - V b) = n - i [n - 4 V (2 - b)] And if the second part of the binomial, or residual, be an imaginary surd, and a2 + b be a perfect integral cube, the extraction may be effected by finding the integral value of n in the following equation as before: 3 - 3 [ / (a + b)] n = 2a. In which last case, the roots of the two. expressions will be, v (a+V - b) =-' n+ v[n2- 43V(a2 + h)l VJ (a -V -- b) =n - V [n2 _- 4 3/ (a2 + b)] each of which formulae may be obtained by barely changing the, sign of b in the former. EXAMPLE. It is required to find the cube root of 10 i: 6 V 3, or 10 ~,/ (108). Here a-10, and b-108; whence. V (a2 b) = 3 (100 -108) = -2, and n3 — 3 C3 (2- b)] n = 20, or n3 + 6n a 20, where it readily appears from inspection, that n 2. Whence 3V (10- V+ 108) (4-4 X -2) 1 - V (12) = 1 + V 3, and 3V (10- V 108) _V(4 —4X- 2)= 1 - V12 = 1 -V'3. EXAMPLES FOR PRACTICE. 1. Required the cube root of 68 - / 4374. 4 —V6 Ans. Or, by rejecting the terms that destroy each other, and then multiplying by -,! a2- b = (X2 I-y)n, or x2-y= (a2 -b)n' Where, supposing a2 — b to be a complete power of the xth degree, let (a2- h)n be put = c. Then, since X2 - y = c, and consequently y 2 x2 - c, if this value be suhqb(? - 1) n n(n-1) (n —2) (n-3) stituted for y, in the equation xn - ). -'n-2y -' -.3.4 n2 -4y2, &c., -=a, we shall obtain an an equation, in which the value of z, as before mentioned, is irrational, when the extraction required is possible. See Wood, or Ryan's Algebra.-ED. 7 74 IRRATIONAL QUANTITIES, OR SURDS. 2. Required the cube root of 11 + 5 - 7 Ans. 1+~ V2 3 Required the cube root of 2 v 7 + 3 v 3.,/6 + v3 Ans. 4. Required the fifth root of 29 v 6 + 41 v 3. V/6 + -/ 3 Ans. 9 5. Required the cube root of 45: 29 ~v 2. Ans. 3 + v'2, and 3 - 2. 6. Required the cube root of 9 i 4 v 5, or 9 ~ A/ 80. Ans. x + 21 5, and -- ~ 5. 7. Required the cube root of 20 i 68 1/ - 7. Ans. 5 + / - 7, and 5 - v - 7. 8. It is required to find the cube root of 35 ~i1 69 v - 6. Ans. 5 + / 6, and 5- v- 6. 9. It is required to find the cube root of 81 = v - 2700.? Ans. -' 3+ 2v-3, and -3-2 v 3. CASE XIII. To find such a multiplier, or multipliers, as will make aly binomial surd rational. RULE.t-1. When one or both of the terms are any even roots, multiply the given binomial or residual, by the -same ~ Whenever it can be done, the operation, in cases of this kind, ought to be abridged, by dividing the given binomial by the greatest cube that it contains, and then finding the root of the quotient; which, being multiplied by the root of the cube by which the binomial was divided, will give the root required. Thus, in the example above given, 81-V5/ —2700 — 27 X (3 - v-1 -4 ), where the root of 3 + v — I — being now more easily found to be - 1 -+ 2 V/-, or -+ - - 3, we shall have, by multiplying by 3, (which is the cube root of 27), - 3+2 V - 3, as above. Also this is useful, in Cardan's rule for cubic equations; thus, 3V/ (81 -V (-2700))- /V(81 —v(-2700))=-3 X 2=-6, or= -- X 2= -3, or a X 2 = 9, the imaginary parts vanishing, by the contrariety of their signs. See De Moivre's Appendix to Sanderson's Algebra, Universal Arithmetic, or Maclaurin's Algebra. -t If a multiplier be required, that shall render any binomial surd, -whether it consist of evern or odd roots, rational, it may be found by substituting the given numbers, or letters, of which it is composed, in the places of their equals, in the following general formula:Binomial nVa ~:: n/ b. Multiplier V an-'1 =F d an-"-2 b +-:Van- 3 b tF V an —4 b3 +, &c.; where the upper sign of the multiplier must be taken with the upper IRRATIONAL QUANTITIES, OR SURDS. 75 expression, with the sign of one of its terms changed; and repeat the operation in the same way, as long as there are surds, when the last result will be rational. 2. When the terms of the binomial surd are odd roots, the rule becomes more complicated; but for the sum or difference of two cube roots, which is one of the most useful cases, the multiplier will be a trinomial surd, consisting of the squares of the two given terms and their product, with its sign changed. sign of the binomial, and the lower with the lower; and the series continued to X terms. This multiplier may be derived from observing the quotient which arises from the actual division of the numerator by the denominator of the following fractions: thus,. ____= Xa -X = n-x-2y + 3n-x 2~, +3 &C.,,., +- yn-1 to u terms, whether n be even or odd. II.-tI = zri-i- a -2ytJ- xa'-3 —, &c.,..... — yn-1 to n z+y terms, where n is an even number. 7III l Yn=yn-1ixn-2qyJ +x —3y2-, &C.. ~, Vyn-1 to n xt-y terms, when n is an odd number. Now, let rn = a, yn- b; then z = Ax a, y = n/ b, and these fractions a-b a-b a+-b severally become -, and -: Ata-Vb a ~ nv a + r b Va~ Vb aJ a -- 1 -- And, since x-'= n a,,- = -2 an-2, &c., also y2 e/b2, y3 ='V b3, &c., therefore, a- b a/ —Vb = A a2-1 + an-2b + v,dan-3b2+, &c.... + bto a terms; where a may be any whole number whatever. And, a+-b - %n an — - an-2b I- v an-3b2 -, &c.,...:: V ba-l to n terms;. where the terms b and / bn-1l have the sign —, when a is an odd number; and the sign -, when n is an even number. NPow, since the divisor multiplied by the quotient gives the dividend, it appears from the foregoing operations, that, if a binomial surd of the for?, V a - n7V be qlitiplziedby IV an-lb V an-2b -+, &c.,...+ ~Vbn-1, (n being any whole number whatever), the prodtct quill be a - b, a rational quantity; and if a binomial surd of the form / a-f r b be mldtiplied by 7n/ an-' -'V an-2b +-'/ al-bb2 -, &c.,... ~X /-ban-1, the product will be ad-b, or a -b; according as the index a is an odd or an even number. See my Elemen.tary Treatise on Algebra, Theoretical and Practical.-ED. 76 IRRATIONAL QUANTITIES, OR SURDS. EXAMPLES. 1. To find a multiplier that shlall render 5 + -v 3 rational. Given surd 5 +- / 3 Multiplier 5 - V 3 Product 25 - 3 = 22, as required. 2. To find a multiplier that shall mnake V 5 + V 3 rational Given surd V 5 + V 3 Multiplier / 5 - V 3 Product 5 - 3 = 2, as required. 3. To find multipliers that shall make V 5 - 4V 3 rational Given surd 4 5 +- 4 3 1st multiplier t 5-4 V 3 1st product v5 - V 3 2d multiplier V 5 + V 3 2d product 5 - 3 = 2, as required. 4. To find a multiplier that shall make 3'V7 + V3 rational. Given surd V 7 + V 3 Multiplier V 72 - V/(7 X 3)- +V 32 7 + 3V(3 X 72) -V (3 X 72)- _V(7 X 32) +- 3V (7 X 32) + 3 Product 7 + 3 10, as was required. 5. To find a multiplier that shall make V/5 - / x rational. Ans. V5 +- v,/ x 6. To find a multiplier that shall malke v a + V b rational. Ans. V a - 6/b. 7. To find a multiplier that shall make a +- V b rational. Ans. a -V /b. 8. Ib is required to find a multiplier that shall make 1 - 3, 2a rational. Ans. 1 + / 2a -+ 3V 4a2. 9. It is required to find a multiplier that shall make 33- 1 V 2 rational. Ans. 3V 9 + V 3' 6 + - V 4. 10. It is required to find a multiplier that shall mrake. 3 4V(a?) + 4v (b3), or a + 0b rational. Ans 4V a9 - (6b3) + V (a36) - V V, IRRATIONAL QUANTITIES, OR SURDS. 77 CASE XIV. To reduce a fraction, whose denominator is either a simple or a compound surd, to another that shall have a rational denominator. RvLE. —1.-When any simple fraction is of the form multiply each of its terms by,va, and the resulting fraction b / a will be —a b Or, when it is of the form p 7 multiply them by V a2, and b /a2 the result will be a b And for the general form -/. multiply by t/ an-1, and the be/ anresult will be — a 2. If it be a compound surd, find sucn a multiplier, by the last rule, as will make the denominator rational; and multiply both the numerator and denominator by it, and the result will be the fraction required. EXAMPLES 1. Reduce the fractions 2A and to others that shall V-3 V -5' have rational denominators. 2 2 / 3 2 / 3 3 3 4V/53 Here X; and 5-. 5X 349 53 6 4V 53 6 3,X 53 453 125 the answer required. 2. Reduce to a fraction whose denominator shall be rational. 3 V5 + V2 3V 5 + 3 2 3V5+3V2 Here — X -- /5 — V2 v5 + V/2 5 - 2 3 V 5 + V 2 - = l/ — S/ ~ 5 + V/ 2 the answer required. 3. Reduce x/2 to a fraction, whose denominator shall be rational. 7* 78 ARITHMETICAL PROPOinTI);|ON AND PR"OGRESSION. V2 /2 X (3 +-'2) 3V/2+2 24-3/V2 Here 3-A 2 (3 - V2) X (3 + V2) 9- 2 7 2 3 -= v 2 the answer required. 4. Reduce V6 to a fraction that shall have a raV 7 + V 3 V (42) - V/(18) tional denominator. Ans. 5. Reduce - to a fraction that shall have a rational 3+Vx 8/ 3- x-/ x denominator. Ans. - -X xe 9-x ca —/b 6. Reduce to a fraction, the denoi*nator of which a+v l /b' b -2a\/ b shall be rational. Ans. 10 7. Reduce a, to a fraction that shall have a raV7 — V5 tional denominator. Ans. 5 X [3V (49) ~ u/ (35) - 3V (25)1 ~ 3 8. Reduce ~ to a fraction that shall have a ra*+ 9 + 10 tional denominator. 3 V 9 - 33a/ (10) ~V/(300), Ans. As.19 4 9. Reduce -+ V 5to a fraction that shall have a rational denominator. Ans. 4 - 10 - 2 V2 - (2 + V5) XV5 OF ARITHMETICAL PROPORTION AND PROGRESSION. ARITHMETICAL PRORORTION,: is the relation which two quantities of the same kind, have to two others, when the difference of the first pair is equal to that of tihe second. Hence, three quantities are said to be in arithmetical proportion, when the difference of the first and second is equal to the difference of the second and' third. Thus, 2, 4, 6, and a, a + b, a + 2b, are quantities in arithmetical proportion. And four quantities are said to be in arithmetical proportion, when'the difference of the first and second is equal to the difference of the third and fourth. ARITHMETICAL PROPORTION AND PROGRESSION. 79 Thus, 3, 7, 12, 16, and a, a+- b, c, c +-, are quantities in arithmetical proportion. ARITHMETICAL PROGRESSION, is when a series of quantities increase or decrease by the same common difference. Thus, 1, 3, 5, 7, 9, &c., and a, a +- d, a +- 2d, a -- 3d, &c., are increasing series in arithmetical progression, the common differences of which are 2 and d. And 15, 12, 9, 6, &c., and a, a - d, a - 2d, a - 3d, &c., are decreasing series in arithmetical progression, the cornmon differences of which are 3 and d. The most useful properties of arithmetical proportion and progression are contained in the following theorems: — 1. If four quantities are in arithmetical proportion, the sum of the two: extrenles will be equal to the sum of the two means. Thus, if the proportionals be 2, 5, 7, 10, or a, b, c, d, then will 2 +2 —1:0 = 5 + 7, and a + d = b - c. 2. And if three quantities be in arithmetical proportion, the sum of the two extremes will be double the mean. Thus, if the proportionals be 3,. 6, 9, or a, b, c, then will 3 + 9 = 2 X 6 = 12, and a + c = 2b. 3. Hence an arithmetical mean between any two quantities is equal to half the sum of those quantities. Thus-: an arithmeticall mean between 2 and 4 is = _3; and between5 and6itis= 5 56. And an arithmetical mean between a and b is.* 2 4, In any continued arithmetical progression, the sum of the two extremes is equal to the sum of any two terms that are equally distant from them, or to double the middle term, when the number of terms is odd. * If two, or more arithmetical means between any two quantities be required, they may be expressed. as below. 2a +b: a+b. _ Thus, = +b and -- = two arithmetical means between a and.b, 3 a being the less extreme and b the greater. Adaf+b (n-1) a+ 2r b (an 2) a + 3b a+ nb And na ( — l (~ -- --—')a — &c. to =anynumi-t +' n-1 - -- b-a ber. (n) of arithmetical means between a and b; where -y- is the common difference; which,, being added to a, gives the first of these means; and then again to this last, gives the second; and so on. 80 ARITHMETICAL PROPORTION AND PROGRESSION. Thus, if the series be 2, 4, 6, 8, 10, then will 2 + 104 + 8-= 2 X 6 = 12. And, if the series be a, a + d, a + 2d, a + 3d, a + 4d, then will a + (a + 4d) = (a + d) + (a + 3d) —- 2 X (a + 2d). 5. The last term of.any increasing arithmetical series is equal to the first term plus the product of the common difference by the number of terms less one; and if the series be decreasing, it will be equal to the first term minus that product. Thus, the nth term of the series a, a + d, a + 2d, a + 3d, a - 4d, &c., is a + (n- l)d. And the nth term of the series a, a - d, a - 2d, a - 3d, a- 4d, &c., is a - (n - 1)d. 6. The sum of any series of quantities in arithmetical progression is equal to the sum of the two extremes multiplied by half the number of terms. Thus, the sum of 2, 4, 6, 8, 10, 12, is = (2 + 12) X = 14 X 3 = 42. And if the series be a + (a + d) - (a +- 2d) + (a + 3d) + (a + 4d), &c... + 1, and its sum be denoted by S, we shall have S = (a + 1) X 2, where I is the last term, and n the number of terms. Or, the sum' of any increasing arithmetical series may be found, without considering the last term, by adding the product of the common difference by the number of terms less one to twice the first term, and then multiplying the result by half the number of terms. And, if the series be decreasing, the sum will be found by subtracting the above product from twice the first term, and then multiplying the result by half the number of terms, as before. Thus, if the series be a + (a +- d) - (a +- 2d) -- (a - 3d) + (a + 4d), &c., continued to n terms, we shall have S - 2a +(n-1l)d' X 2. And if the series be a - (a - d) + (a- 2d) +- (a - 3d) + (a - 4d), &c., to n terms, we shall have (*) The sum of any number of terms (n) of the series of natural numbers 1,'2, 3, 4, 5, 6, 7, &c., is = - (n —. ARITME'TICAL PROPO-RTION AND PROGRESSION. 81 EX AMPLES. 1. The first term of an increasing arithmetical series is 3, the common difference 2, and the nLumber of terms 20; required the suym of the series. First, 3 + 2 (20 -1) =3 + 2 X 19 3 + — 38 = 41, the last term. 20 20 And (3 + 41) X =2 4 44 X 10-440, the sum 2 2 required. Or,!2X3+(20-1) X2 X- 2=(6+19X2)X10= (6 + 38) X 10 = 44 X 10 = 440, as before. 2. The first term of a decreasing arithmetical series is 100, the common difference 3, and the number of terms 34; required the sum of the series. First, 100- 3(34 - 1) -= 100 - 3 X 33 = 100 - 99 = 1, the last term. 34. 34 And (100 + 1) X - 101 X- = 101 X 17= 1717, the 2 2 sum required. 34=(2oo-33x3) Or, 42 X100-(34 —1)X3 X =(200-33X3)X 17 = (200 - 99) X 17 = 101 X 17 = 1717, as before. 3. Required'the suni of the natural numbers 1, 2, 3, 4, 5, 6, &c., continued to 1.000 terms.'Ans. 500500. 4. Required the slim of the odd numbers 1, 3, 5, 7, 9, &c., continued to 101 terms. Ans. 10201. 5. How many strokes do the clocks of Venice, which go on to 24 o'clock, strike in a day? Ans. 300. 6. Required the 365th term of the series of even numbers 2, 4, 6, 8, 10, 12, &c Ans. 730. 7. The first term of a decreasing arithmetical series is 10, 100X101 Thus, 1 + 2. 3 +- 4~- 5, &c., continued to 100 terms, is = -- - = 50 X 101 — 5050. Also the. sum of any number of terms (it) of the series of odd numbers 1, 3, 5, 7, 9, 11, &c., is = n2. Thus, 1 j -3 + 5 + 7 + 9, &c., continued to 50 terms, is = 502=2500. And if any three of the quantities, a, d, n, S, be given, the fourth may be found from the equation S = i2a - (,l) d X, or (a+l )XWhere the upper sign -[- is to be used when the series is increasing, and the'lower sign - when it is decreasing; also the last term I = a i: (it - 1)d, as above. 82 GEOMETRICAL PROPORTION AND PROGRESSION. the conlmon difference 1, and the number of terms 21; required the sum of the series. Ans. 140. 8. One hundred stones being placed on the ground, in a straight line, at the distance of a yard from each other, how far will a person travel, who shall bring them one by one, to a basket, placed at the distance of a yard from the first stone? Ans. 5 miles and 1300 yards. OF GEOMETRICAL PROPORTION AND PROGRESSION. * GEOMETRICAL PROPORTION, is the relation which two quantities of the same kind have to two others, when the * If there be taken any four proportionals, a, b, c, d, which it has been usual to express by means of points: thus, a:b::c:d, this relation will be denoted by the equation / = i where the equal ratios are represented by fractions, the numerators of which are the antecedents, and the denominators the consequents. Hence, if each of the two members of this equation be multiplied by bd, there will arise ad = be. From which it appears, as in the, common rule, that the product of the two extremes of any four proportionals is equal to that of the means. And if the third c, in this case, be the same as the second, or c = b, the proportion is said to be continued, and we have ad - b2, or b= ad; where it is evident, that the product of ths extremes of three proportionals is equal to the square of the mean; or, that the mean is equal to the square root of the product of the two extremes. Also, if each member of the equation ad =be be successively divided by bd, dc, ae, &c., the results will give a c Ia:b::c:d a b O'r the --- a: c::b d b d i proportions a a = - b a d c a c j j:a &c. &c. So that, by following this method, we can easily obtain all the transfobrmations of the terms of the proportion, that can be made to agree with the equations ad = be. In like manner, from the same equality b - 2, there will result, by ma no ma mc multiplication, the following equivalent forms: - n —d; b - d Which, being converted into proportions, become ma: mb::'e:'nd and ma: ~ib:: me: nd. And, by taking any like powers, or roots, of the different sides of the same equation, we have; —- =-. Or, putting the bn d GEOMETRICAL PROPORTION AND PROGRESSION. 83 antecedents, or leading terms of each pair, are the same parts of their consequents, or the consequents of the antecedents. terms in the form of a proportion, a': b.' cmn: dn. In which cases m and n, may be any whole or fractional numbers whatever. Again, if there be taken the several equations, d |-= which correspond a, b c: cd _ h > with )e:f::g:' -_ j Ithe proportions i k 1: i &c. &c. we shall have, by multiplying their like terms, aXeXi,&c. -cXgX, &c e bXfX,&c. dXhXm,&c. Or, by putting the expression in the form of a proportion, aei, &c., C C:bfk, &c.,:' cgl, &c.,: dlm, &c. Also, taking -- d, asbefore,weshall have, by multiplication, b —; and by augmenting or diminishing ima me ma J nxb each side of the equation by 1; -=: 1 t - 1; or nb nd nb emc ~- ssd; which, being expressed in the form of a proportion, gives ma - nb:nb:: mc- nid: nd; or vma::i -wb:ma s d nb:: xd. And if the abovementioned equation - -= be put, by a similar multiplication of its terms, under the form - -p, and then augmented or qb qd' diminished by 1, as in the last case, there will arise pa + qb: pc -- qd:: qb: qd. Whence, dividing each of the antecedents of these two anama 4- nl'ib b logies by their consequents, the result will give -~ - = and me -end nd d; pa i- qb qb b p2c:i -d- q-d d'. And, consequently, as the two righthand members b ma i4- b pa i 4qb of these expressions are each -, we shall have rlm' mend pc - qd Or, by converting the corresponding terms of this equation into a proportion ma 4- nb: mc ~- xd:: pa -4-qb: qpc ~ — qd. Also, because the a c a b common equation - = d gives =, if the latter be put under the equivama rb pa — pb lent forms -d and = we shall obtain, by a similar process, mia 4- nc: pa 4- qc:: mb 4- d: pb 4- qd; which two analogies may be considered as general formula for changing the term sof the proportion a: b:: c: d, without altering its nature.' Thus, by supposing m, si, p, q, to be each = 1, and taking the antecedents with the superior signs, and 84 GEOMETRICAL PROPORTION AND PROGRESSIO1N. And if two quantities only are to be compared together, the part or parts, which the antecedent is of its consequejlt, or the consequent of the antecedent, is called the ratio; observing, in both cases, always to follow the same method. Hence, three quantities are said to be in geometrical proportion, when the first is to the same part, or multiple, of the second, as the second is. of the third. Thus, 3, 6, 12, and a, ar, ar2, are quantities in geometrical proportion. And four quantities are said to be in geometrical proportion, when the first is the same part, or multiple, of the second, as the third is of the'fourth. Thus, 2, 8, 3, 12, and a, ar, b, br, are geometrical proportionals. Direct proportion, is when the same relation subsists between the first of four terms and the second, as between the third and fourth. Thus, 3, 6, 5, 10, and a, ar, b, br, are in direct proportion. Inverse, or reciprocal proportion, is when the first and second of four quantities are directly proportional to the reciprocals of the third and fourth. the consequents with the inferior, we have a + b: a — b: c -- d c -d, and a + c: a - c:: b d: b - d; which forms, together with several of those already given, are the usual transformations of the common analogy pointed out above. In like manner, by taking sr, i, and p each = 1, and q - 0, there will arise a 6b: a:: c i d-: c, and a t c': a:: b i d:b; each of which proportions may be verified by making the product of the extremes equal to that of the means, and observing that ad = bc. Lastly, taking any number of equations of the form before used, for a c e g expressing proportions, to, &c.; which, according to the b d J hi common method, are called a series of equal ratios, and are usually denoted by a: b: c: d:: e f:: g: h::,&c., we shall necessarily - c have, from the fractions being all equal to each other, 6q — =q, e g -= q, - q, &c. And, by multiplyinf q by each of the denominators, a = bq, c = dq, e =fq, g = hq, &c. Whence, equating the sum of all the terms on the lefthand side of these equations, with those on the right, we have a-+c+e+g —, &c. =(b+d — f+lr+, &c.,)q. And consequently, by division, and the properties of proportionals before shown, a-+c+e-g+, &c. a -c ac-c+e b+ d f+ h+, &c. b b + d - +df=', which results show thlat, in a series of equal ratios, the sum of any number of the antecedents is to that of their consequents, as one, or more of the antecedents, is to one, or the same number of consequents. GEOMETRICAL PROPORTION AND PROGRESSION. 85 Thus, 2, 6, 9, 3, and a, ar, br, b, are inversely proportional; because 2, 6, -3' and a ar,' are directly pro9.3''r 12 portional. GEOMETRICAL PROGRESSION, is when a series of quantities have the same constant ratio; or which increase, or decrease, by a common multiplier, or divisor. Thus, 2, 4, 8, 16, 32, 64, &c., and a, ar, ar2, ar3, ar4, &c., are series in geometrical progression. The most useful properties of geometrical proportion and progression are contained in the follpwing theorems:1. If three quantities be in geometrical proportion, the product of the two extremes will be equal to the square of the mean. Thus, if the proportionals be 2, 4, 8, or a, b, c, then will 2 X 8 - 42, and a X c = b. 2. Hence, a geometrical mean proportional, between any two quantities, is equal to the square root of their product. Thus, a geometric mean between 4 and 9 is - 36 = 6. And a geometric mean between a and b is - / ab.* 3. If four quantities be in geometrical proportion, the product of the two extremes will be equal to that of the means. Thus; if the proportionals be 2, 4, 6, 12, or a, b, c, d; then will 2 X 12 = 4 X 6, and a X d = b X c. 4. Hence, the product of the. means of four proportional quantities, divided by either of the extremes, will give the other extreme; and the product of the extremes, divided by either of the means, will give the other mean. Thus, if the proportionals be 3, 9, 5, 15, or a, b, c, d; then will 9X 15, and 3 X5 = 9: also, X = d, and 3 5 a aXd= b. — =6. C * If two or more geometrical means between any two quantities be required, they may be expressed as below: V3 a2b and V3/ab2 =two geometrical means between a and b. 4Va3b, 4V/anb2, and Vab3 - three geometrical means between a and b. And generally, 1 1 1 -- n-i- n-2 (anb)n+l, (a b2)n +1, (a b3) n+l = any number (n) of geometrical means between a and b. Where n is the ratio; so that if a be multiplied by this, it will give the firs of these means; and this last being again multiplied by the same, will give the second; and so on. S 8G GEOMETRTCAL PROPORTION AND PROGRESSION. 5. Also, if any two products be equal to each other, either of the terms of one of them, will be to either of the terms of the other, as the remaining term of the last is to the remaining term of the first. Thus, if ad- bc, or 2 X 15 = 6 X 5, then will any of the following forms of these quantities be proportional:Directly, a: b:: c: d, or 2: 6:: 5:15. Invertedly, b: a:: d: c, or 6: 2:: 15: 5. Alternately, a: c:: b: d, or 2: 5:: 6: 15. Conjunctly, a: a - b:: c: c + d, or 2: 8:: 5: 20. Distinctly, a: b - a:: c: d - c, or 2: 4:: 5:-10. Mixedly, b + ca: b- a': d + a: dd- b, or 8':4:: 20: 10. In all of which cases, the product of the two extremes is equal to that of the two means. 6. In any continued geometrical series, the product of the two extremes is equal to the product of any two means that are equally distant from them; or to the square of the mean, when the number of terms is odd. Thus, if the series be 2, 4, 8, 16, 32; then will 2 X 32 - 4 X 16 = 82. 7. In any geometrical series, the last term is equal to the product arising from multiplying the first term by such a power of the ratio as is denoted by the, number of terms less one. Thus, in the series 2, 6, 18, 54, 162, we shall have 2 X34 = 2 X 81: 162. And in the series, a, ar, ar2, ar., ar4, &c., continued to n terms, the last term will be I= -ar 8. The sum of any series of' quantities in geometrical progression, either increasing or decreasing, is found by multiplying the last term by the ratio, and then dividing the difference of this product and the first term by the difference between the ratio and unity. Thus, in the series 2, 4, 8, 16, 32, 64, 128, 256, 512, we shall have 512 X 2 -2 _ 1024- 2 - 1022, the sum of the 2 -1 terms. Or the same rule, without considering the last term, may be expressed thus:- Find such a power of the ratio as is denoted by the number of terms of the series; then divide the difference between this power and unity, by the difference between the ratio and unity, and the result, multiplied by the first term, will be the sum of the series. GEOMETRICAL PROPORTION AND PROGRESSION. 87 Thus, in the series a + ar / ar2 + arl + ar4, &c., to ar we shall have S = _-ZY r - I Where it is to be observed, that if the ratio, or common multiplier, r, in this last series, be a proper fraction, and consequently the series a decreasing one, we shall have, in that case, a - ar + ar2 + ar3 + ar4, &c., ad infinitum - 9. Three quantities are said to be in harmonical proportion, when the first is to the third, as the difference between the first and second is to the difference between the second and third. Thus, a, b, c, are harmonically proportional, when a: c a - b: b - c, or a: c:: b - a: c -b. And c is a third harmonical proportion to a and b, when ab 2a —b' 10. Four quantities are in harmonical proportion, when the first is to the fourth, as the difference between the first and second is to the difference between the third and fourth. Thus, a, b, c, d, are in harmonical proportion, when a:d::a-b:c- d, or a:d::b —a:d-c. And d isa ac fourth harmonical proportional to a, b, c, when d- b in each of which cases it is obvious, that twice the first term must be greater than the second, or otherwise the proportionality will not subsist. 1l. Any number of quantities, a, b, c, d, e, &c., are in harmonical progression, if a: c:: a -b: b - c; b: d:: b -c:c —d; c: e:: c-d: d-e; &c. 12. The reciprocals of quantities in harmonical progression, are, in arithmetical progression. Thus, if a, b, c, d, e, &c., are in harmonical progression, a' b' c' d'e',&c., will be in arithmetical progression. 13. An harmonical mean between any two quantities, is equal to twice their product divided by their sum. 2ab Thus -= an harmlonical mean between a andeb.* a +b e In addition to what is here said, it may be observed, that the ratio of two squares is frequently called duplicate ratio; of two square roots, sub-duplicate ratio; of two cubes, triplicate ratio; and of two cube roots, sub-triplicate ratio; &c. 88 OF EQUATIONS. EXAMPLES. 1. The first term of a geometrical series is 1, the ratio 2, and the number of terms 10; what is the sum of the series? Here 1 X 29 = 1 X 512 = 512, the last term. 512 X 2-1 1024- 1 And 2 - = 1023, the sum required. 2 —1 1 2. The first term of a geometrical series is - the ratio - 2' 3' and the number of terms 5; required the sum of the series. 1, 1\ 4 1 1 1 Here - X(j =X the last term. 162{ XX. _ ~ 121 3 121 And = -- 243 X 1 the sum. 2 3 243 3. Required the sum of 1, 2, 4, 8, 16, 32, &c., continued to 20 terms. Ans. 1048575. 1111 1 i 4. Required the sum of 1,,, 1- &c., continued 21 4' 87 16.32' to 8 terms. Ans. 1 - 128' 1 1 1 1 5. Required the sum of 1, 3, 9- 27' 81 &c., continued to 3' 9' 27' 81' 9841 10 terms. Ans. 1 19683' 6. A person being asked to dispose of a fine horse, said he would sell him on condition of having a farthing for the first nail in his shoes, a halfpenny for the second, a penny for the third, twopence for the fourth, and so on, doubling the price of every nail. to 32, the number of nails in his four shoes; what would the horse be sold for at that rate? Ans. 44739241. 5s. 3-d. OF EQUATIONS. THE DOCTRINE OF EQUATIONS, is that branch of algebra which treats of the methods of determining the values of unknown quantities by means of their relations to others which are known. This is done by making certain algebraic expressions equal to each other (which formula, in that case, is called an equation), and then working by the rules of the art, till the quantity sought, is found equal to some given quantity, and consequently becomes known. The terms of an equation are the quantities of which it is OF EQUATIONS. 89 composed; and the parts that stand on the right and left of the sign =, are called the two members, or sides, of the equation. Thus, if x-= a + b, the terms are x, a, and b; and the meaning of the expression is, that some quantity x, standing on the lefthand side of the equation, is equal to the sum of the quantities a and b on the righthand side. A simple equation is that which contains only the first power of the unknown quantity: as, x + a - 3b., or ax = be, or 2x + 3a2 = 5b2; Where x denotes the unknown quantity, and the other letters, or numbers, the known quantities. A compound equation is that which contains two or more different powers of the unknown quantity; as, x2 + ax = b, or 3 - 4x2 +- 3x = 25. Equations are also divided into different orders, or receive particular names, according to the highest power of the unknown quantity contained in any one of their terms: as quadratic equations, cubic equations, biquadratic equations, &c. Thus, a quadratic equation is that in which the unknown quantity is of two dimensions, or which rises to the second power: as, x2= 20; x2 2+- ax= b, or 3x2 +- 10x = 100. A cubic equation is that in which the unknown quantity is of three dimensions, or which rises to the third power: as, x3= 27; 23- 3x —35; or 3-a- a + bx — c. A biquadratic equation is that in which the unknown quantity is of four dimensions, or which rises to the fourth power: as 4 = 25; 54 - 4x- 6; or x4 - a3- + - c = d. And so on for equations of the 5th, 6th, and other higher orders, which are all denominated according to the highest power of the unknown quantity contained in any one of their terms. The root of an equation is such a number or quantity, as, being substituted for the unknown quantity, will make both sides of the equation vanish, or become equal to each other. A simple equation can have only one root; but every compound equation has as many roots as it contains dimensions, or as is denoted by the index of the highest power of the unknown quantity, in that equation. Thus, in the quadratic equation x2 + 2x = 15, the root, or value of x, is either ~ 3 or - 5; and, in the cubic equation X3 - 9x2 + 26x = 24, the roots are 2, 3, and 4, as will be found by. substituting each of these numbers for x. In an equation of an odd number of dimensions, one of its 8* 90 RESOLUTION OF SIMPLE EQUATIONS. roots will always be real: whereas, in an equation of an evel, number of dimensions, all its roots may be imaginary; as roots of this kind always enter into an equation by pairs. Such are the equations x2 - 6x + 14 = 0, and x4 - 2x3 - 9x2 2 10x + 50 = 0.* OF THE RESOLUTION OF SIMPLE EQUATIONS, Containing only one unknown Quantity: THE resolution of simple, as well as of other equations, is the disengaging the unknown quantity, in all such expressions, from the other quantities with which it is connected, and making it stand alone, on one side of the equation, so as to be equal to such as are known on the other side; for the performing of which, several axioms and.processes are required, the most useful and necessary of which are the following: —t CASE I. Any quantity may be transposed from one side of an equation to the other, by changing its sign; and the two members, or sides, will still be equal. Thus, if x + 3 = 7; then will x = 7 - 3, or x = 4. * To the properties of equations abovementioned, we may here farther add:1. That the sum of all the roots of any equation is equal to the coefficient of the second term of that equation, with the sign changed. 2. The sum of the products of every two of the roots, is equal to the coefficient of the third term, without any change in its sign. 3. The sum of the products of every three terms of the roots, is equal to the coefficient of the fourth term, with its sign changed. 4. And so on, to the last, or absolute term, which is equal to the product of all the roots, with the sign changed or not, according as the equation is of an odd or an even number of dimensions. See, for a more particular account of the general theory of equations, Vol. II. ot Bonnycastle's Treatise on Algebra, 8vo., 1820; or Ryan's Elementary Treatise on Algebra, 12mo., 1824.-ED. t The operations required for the purpose here mentioned, are chiefly such as are derived from the following simple and evident principles:1. If the same quantity be added to, or subtracted from, each of two equal quantities, the results will still be equal; which is the same, in effect, as taking any quantity -from one side of an equation, and placing it on the other side, with a contrary sign. 2. If all the terms of any two equal quantities, be multiplied or divided, by the same quantity, the products, or quotients thence arising, will be equal. 3. If two quantities, either simple or compound, be equal to each other, any like powers, or roots, of them will also be equal. All of which axioms will be found sufficiently illustrated by the processes arising out of the several examples annexed to the six different cases given in the text. RESOLUTION OF SIMPLE EQUATIONS. 91 And, if - 4 + 6 — 8; then will x 8 + 4 - 6 = 6. Also, if x - a + b = c - d; then willx = — a- b + c - d. And, if 4x - 8 = 3x + 20; then 4x - 3x = 20 -+ 8, and consequently x = 28. From this rule it also follows, that if a quantity be found on each side of an equation; with the same sign, it may be left out of both of them; and that the signs of all of the terms of any equation may be changed from + to -, or from - to +, without altering its value. Thus, if x + 5 - 7 + 5; then, by cancelling, x = 7. And if a - x = b - b-c; then, by changing the signs, x-a= c- b, or x= a +c-b. EXAMPLES FOR PRACTICE. 1. Given 2x + 3 — = x +- 17 to find x. Ans. x —14. 2. Given 5x - 9 - 4x -+ 7 to find x. Ans. x =16. 3. Given x + 9 - 2 = 4 to find x. Ans. x=- 3. 4. Given 9x -8 — 8 x - 5 to find x. Ans. x -3. 5. Given 7x — + 8 - 3 - 6x + 4 to find x. Ans. x =-1. CASE II. If the unknown quantity, in any equation, be multiplied by any number, or quantity, the multiplier may be taken away, by dividing all the rest of the terms by it; and if it be divided by any number, the divisor may be taken away, by multiplying all the other terms by it. Thus, if ax = 3ab - c; then will x = 3b -- a And, if 2x + 4 - 16; then will x + 2 = 8, or x = 8 - 2 = 6. Also, if -- = 5 + 3; then will - = 10 + 6 = 16. f2x And, if -7 - 2 = 4; then 2x - 6 = 12, or by division, x -3 = 6, or x = 9. EXAMPLES FOR PRACTICO,. 1. Given 16x - 2 - 34 to find x. Ans. x = 2. 2. Given 4 - 8 = - 3x 13 to find x. Ans. x = 3. 3. Given 10x - 19 = 7x + 17 to find x. Ans. x = 12. 4.- Given 8x-3 + 9 — 7x 9 + 27 to find x. Ans. x = 2 4d 5. Given 3ax - 3ab - 12d. Ans. x = b +ab6 92 RESOLUTION OF SIMPLE EQUATIONS. CASE III. Any equation may be cleared of fractions, by multiplying each of its terms, successively, by the denominators of those fractions, or by multiplying both sides by the product of all the denomiiinators, or by any quantity that is a multiple of them. X X 3x Thus, if +- 5, then, multiplying by 3, we have x + 3 4 4 = 15; and this, multiplied by 4, gives 4x + 3x - 60; whence, 60 4 by addition, 7x =-60, or x = --- 8 7 7' x X And, if -+ - = 10; then, multiplying by 12, (which is a multiple of 4 and 6,) 3x + 2x = 120, or 5x- 120, or x= 120 5 - 24. It also appears, from this rule, that if the same number, or quantity, be found in each of the terms of an equation, either as a multiplier or divisor, it may be expunged from all of them, without altering the result. Thus, if ax = ab + ac; then by cancelling, x = b + e. cb And -if - + -; then, x + b=c, or x = c-b. a a a EXAMPLES FOR PRACTICE. 3x x 1. Given -- + 24 to find x. Ans. x=195. 2 4 2. Given - + - 62 to find x. Ans. x - 60. x —3 x x- 19 3. Given 2 q- 20 to find x. Ans. x =9. 2 3 2 Given x+1 x2 x ~-3 4. Given + -- 16- to find x. 2 3 4 Ans. x =13..ex - a x 2; a 4- h 5. Given + -= —+ — to find x. b c a d a=cb acb2.-a2cd acnd'+ abd - 2cbd CASE IV If the unknown quantity, in any equation, be in the form of a surd, transpose the terms so that this may stand alone, on one side of the equation, and the remaining terms on the other RESOLUTION OF SIMPLE EQUATIONS. 93 (by Case I.); then involve each of the sides to such a power as corresponds with the index of the surd, and the equation will be rendered free from any irrational expression. Thus, if V/ x-2=-3; then will x/ -3 + 2 = 5, or, by squaring, x 52 = 25. And if / (3x + 4) = 5; then will 3x + 4 = 25, or 3x = 25 21 4- =21, or x = -- =7. 3 Also, if V/(2x +- 3) + 4 = 8; then will 3V (2x + 3) = 8-4 = 4, or 2x + 3 = 43= 64, and consequently 2x 64 - 3 =61, 61 1 or x=-6 —_- 302 z EXAMPLES FOR PRACTICE. 1. Given 2 V/ x + 3 9 to find x. Ans. x - 9. 2. Given v (x - 1) -2 = 3 to find x. Ans. x -24. 3. Given 3 (3x + 4) + 3 = 6 to find x. Ans. x = 723 4. Given V(4 + x)= 4- v/ to find x. Ans. x= 2 —. 5. Given V/(4a2 + x2) = 4V (4b64 + X4) to find x. b4-4aa4s Ans. x- = / (- 4a4) CASE V. If that side of the equation which contains the unknown quantity, be a complete power, the equation may be reduced to a lower dimension, by extracting the root of the said power on both sides of the equation. Thus, if x2 - 81; then x = v 81 = 9; and if x3 = 27, then x = q 27 - 3. 33 Also, if 3x2 - 9 = 24; then 3x2 24 + 9 = 33, or x3- 3 3 = 11, and consequently x== v11. And, if x2 + 6x + 9 = 27; then, since the lefthand side of the equation is a complete square, we shall have, by extracting the roots, x-{+ 3 v/27 =/(9X3) =3 /3, or x = 3 /3 -3. EXAMPLES FOR PRACTICE. 1. Given 9X2 - 6 = 30 to find x. Ans. x =- 2. 2. Given x3 + 9 = 36 to find x. Ans. x=-3 81 3. Given x2 - x 3- + - to find x. Ans. x = 4 4 4 2 a 4. Given X2 X ax + - -4 to find x. Ans. x- b -2. 5. Given x2 +- 14x + 49 121 to find x. Ans. x 4 94 RESOLUTION OF SIMPLE EQUATIONS. CASE VI. Any analogy, or proportion, may be converted into an equation, by making the product of the two extreme terms equal to that of the two means. Thus, if 3x:16:: 5: 6; then 3 X 6- = 16 X 5, or 18x 80 40 4 = 0, or x =- _ 4 - 22x 2cx And if-: a:: b: c; then will- = ab, or 2cx - 3ab; or, 3 ~~~3 3ab by division, x- 2 2c x 4x Also, if 12-x: -4: 1; then 12 - x = 2x, or2x 2 2 12 + x = 12, and consequently x = = 4. EXAMPLES FOR PRACTICE. 3 20ab 1. Given x: a:: 5bc: cd to find x. Ans. 3d 2. Given 10 -: x::3: 1 to find x. Ans. x = 3L 3 3' 3. Given 8 + 8x: 4x:: 8: 2 to find x. Ans. x- 1. 4. Given x: 6 - x:: 2: 4 to find x. Ans.x- 2. a2 5. Given 4x:a:: 9 x: 9 to findx. Ans. X- - 1!6' MISCELLANEOUS EXAMPLES. 1. Given 5x - 15 = 2x + 6 to find the value of x. Here 5x - 2x = 6 + 15, or 3x = 6 + 15 = 21; and there21 fore x = T= 7. 2. Given 40 - 6x - 16 = 120 - 14x, to find the value of x. Here 14x- 6x 120 - 40 + 16; or 8x- 136 - 40 = 96; 96 and therefore x- 8 = 12. 3. Given 3x2 _ 1Ox = 8x + - 2, to find the value of x. Here 3x - 10 = 8 + x, by dividing by x; or 3x - x = 8 + 10 = 18, by transposition. 18 And consequently 2x = 18, or x 9. 4. Given 6ax3 - 12abx2 = 2axs + 6ax2:, to find the value of x. Here 2x - 4b = x -- 2, by dividing by 3ax2:; or 2x - x — = 2 + 4b; and therefore x =- 4b + 2. RESOLUTION OF SIMPLE EQUATIONS. 95 5. Given x 2 + 2x — 1 = 16, to find the value of x. Here x + 1 = 4, by extracting the square root of each side. And therefore, by transposition, x = 4 - 1 = 3. 6. Given 5ax - 3b = 2dx + c, to find the value of x. Here 5ax - 2dx = c ~- 3b; or (5a - 2d)x = c +- 3b; and c +- 3b therefore, by division x 5a - 2d a x a 7. Given - - + 10, to find the value of x. 2x 2x 6x Here x - + - - 20; and 3x - 2x+- 60; or 12x - 8x + 6x = 240; whence 1Ox - 240, or x =- 24. x —3 x x-19 8. Given -- + — 20- --- to find the valueof x 2 3 2 2x Herex-3+ — =40-x+19; or 3x -9+2x=120 - 3a + 57; whence 3x + 2x + 3x -120+ 57 + 9; that is, 8x -- 186, or x == 23 2x 9. Given V -- + 5 = 7, to fihd the value of x. 2x. 2x Here V/ -- = 7 - 5 = 2; whence, by squaring, - 22 = 4, and 2x = 12, or x =6. 2a2 10. Given x - (a2 - X2) = (a2, to find the value of X. Here x V (a2 + X2) +- a2 + x2 = 2a2; or x (a2 +a2) = a2 -2, and S2.(a+ - 2)= _ a4-2a2x' + X4; whence a2x2 -- X4 = a4 - 2a2X2 + X4, and a2x2 = a4 - 2a'x2; therefore 4 a2 a2 3a2Xa2 = a, or X2 - - and consequently x = V 3a2 3 3 1 3 a = a - = a -/:-='/ 3, the answer required. 9: 3 EXAMPLES FOR PRACTICE. 1. Given 3x - 2 + 24 = 31, to find the value of x. Ans. -- 3. 2. Given 4 - 9y= 14- 1 ly, to find the value of y. Ans. y = 5. 3 Given x +- 18=3x- 5, to find the value of x. Ans. x — 11~. 4. Given + - + + = 11, to determine the value of x. Ans. x = 6. 96 RESOLUTION OF SIMPLE EQUATIONS 5. Given 2x -- + 1 = 5x - 2, to find the value of x. ~2 ~~~~6 Ans. x=77' x x x 7 6. Given - + ----, to determine the value of x. 2 3 4 10' 1 Ans. x-=15' 7. Given + + -= 4- -4, to find the value of x. Ans. x=3 — 13' 8. Given 2 + - 3% = V/ (4 + 5x), to find the value of x. Ans. x= 12. 9. Given x + a +, to find the value of x. a Ans. x - - 2a 10. Given V + v (a + x) = V (a +' to find the value of x- Ans. x= - ~3' ax- b a bx bx- a 11. Given -+ a _ - to find the value ofx. 3b Ans. x=3a- 2b' 12. Given V/ (a + - 2) - (b4 +- 4), to find the value of x. b4 - a4 Ans. — V 2a2 13. Given v (a + x) +- (a - x) = / ax, to find the value of x. 4a2 Ans. x- 4a2 a a 14. Given + -- b to determine the value of x. 1 +ix 1-x Ans. x b - 2a b' 15. Given a +- =,/ [a + - x V (b2 +- x2)], to find the value of x. 62 Ans. x= - -a 4a 16. Given /(x2 + 3a2) _ ~ i(2 - 3a-2) = x v/a, to find the value of x. 9a3 Ans. aC =4, —, 4 - 4a' RESOLUTION OF SIMPLE EQUATIONS. 97 17. Given V/ (a + x) + V (a - x) - b, to find the value of x Ans. xa= V(4a - 2) 18. Given 3 (a - x) 3V (a - x)= b, to find the value of x Ans. x \/a2 -(3 ) V// 3b 19. Given V a + / x = V ax, to find the value of x. Ans. x - (Va- 1)2' 20. Given aV/( + )~V-l )+- 1 = a, to determine the value of x. a Ans. - x (a a) — v/ (C - 4) 21. Given V(a2+ ax) = a -V (a2 - ax), to find the value of x. Ans. x= V- 3. 22. Given V (a2 -- X2) (- ) = a v (1 -?x), to find the value of x. Z _ 1\ Ans. a= / 2 — 3 23 Given V (x- +a) = c - /(x - b), to find the value of x. Ans. x = —2c b c 4be 24. Given V/ -- — + V --— 2 to find the value a- ~ a-x X2' of x. /D - c\ Ans. x= -a b~J,.C Of the resolution of simple equcations, containing two unknown quagntities. When there are two unknown quantities, and two independent simple equations involving them, they may be reduced to one, by any of the three following rules:RULE 1.-Observe which of the unknown quantities is the least involved, and find its value in each of the equations, by the methods already explained; then let the two values, thus found, be put equal to each other, and there will arise a new equation with only one unknown quantity in it, the value of which may be found as before. * Thes rule depends upon the well-knoen axiom, that thine ts which are equal to the same thins are equal to each other, and the two following methods are founded on principles which are equally simple and obvious. 9 98 RESOLUTION OF SIMPLE EQUATIONS. EXAMPLES.. Given 2 + 3y 23 to find the values of x and y. 23 - 3y Here, from the first equation, x = 2 10 + 2y And from the second, x- 1 -; 23 - 3y 10 ~ 2y Whence we have 2 - 10 + y 2 5 Or 115 - 15y = 20 + 4y, or 19y = 115 - 20 95. 95 23 - 15 That is, y = — 5, and x= =4. 19 2 2. Given x +- y - a to find the values of x and y Here, from the first equation, x = a - y, And from the second x = b + y, Whence a - y b + y, or 2y = a - b, a-b And therefore y - 2 -, and x = a -- y. a -b a-b Or by substitution, x = a - - 2 2 2 3. Given a Y 8 to find the values of x and v 2y Here, from the first equation, x = 143 And from the second, x = 24 - o; 2y 3g Therefore, by equality, 14 - 3y 24 - - And consequently, 42 - 2y = 72 - 2, Or, by multiplication, 84 - 4y = 144 - 9y; And, therefore, also 5y = 144 - 84 = 60, 60 24 Or, by division,y =_ = 12, and x = 14 -- = 6. Y 5 3 EXAMPLES FOR PRACTICE. 1. Given 4x + y = 34, and 4y + x = 16, to find the values of x and y. Ans. x = 8, y= 2. 2. Given 2x +3y =6 16, and 3x- 2y = 11, to find the values of x and y. Ans. x = 5, y = 2. RESOLUTION OF SIMPLE EQUATIONS. 99 3G e 3y 9 3x 2y 61 3 Gi — en - 2-0' and - 1 2- to find the 5 4 20 4 5 120' values of x and y. Ans. x =, y - 4. 2 G-2y — } 4. Given 2 to find x and y. Ans. x = a + b, and y = a — b 6. Given 2+3 to find x and y. 3- 2: -=Ans. x -- 12, and y= 6. 6. G8iven y-22 a- y - 9 to find x and y. ~. x ~ y ~ ~ 4 ~.3 Ans. x = 12, and y - 9. 72~x 3y 7. Given x + y = 80, and-2 -= —,to find x and y. Ans, x = 42 JL, and y = 37 a_7. 8 Given y - 6, and x= y +- 6,to find x andy. Ans. x = 24, and y 18. RULE 2.-Find the values of either of the unknown quantities in that equation in which it is the least involved; then substitute this value in the place of its equal in the other equation, and there will arise a new equation with only one unknown quantity in it; the value of which may be found as before. EXAMPLES. 1. Given 3 + 2y = 17 to find the vralues of x and y. From the first equation, x = 17 - 2y2; which value, being substituted for x, in the second, gives 3 (17 - 2y) - y =2, Or 51 — 6y - y 2, or7y=51 —2=-49. 49 Whence y — = 7, and x = 17- 2y - 3. 7 2. Given x + y — 3 1 to find the values of x and y. From the first equation, x- =13 - y; which value, being substituted for x, in the second, Gives 13 - y - y = 3, or 2y- 13 - 3-= 10. 10 Whence y- 2 — 5, and x = 13 - y 8. 3. Given I: y:: a: b to find the values of x and y. ~x2 qY 2? =c 100 RESOLUTION OF SIM'IPLE EQUATIONS. Here the analogy in the first, turned into an equation, ay gives bx = ay, or x = b, And this valuei substituted for x in the second, gives (b)2S +yc or -+ y =C bc. Whence we have a2y2 + b2y2 = b2c, or y2 =2 - b"' C C And, consequently, y = b, arnd = a a, EXANMPLES FOR PRACTICE. I. Given — + 7y - 99, and + -7x 51, to find the values of x andy. Ans. x 7, andy= 14. x a 3~ Y X 2y — x 2. Given — 12-= - 8, and XY- 8 — Y - 2 5 3 4 27, to find the values of x and y. Ans. x = 60, y = 40. 3. Given x + y = s, and 2 - y2= d, to find the values of x and y. s +- d S — d Ans. x= -,2 Yy 2s 2s 2s 4.'Given 5x- 3y= 150, and 10Ox - 15y-= 825, to find X and y. Ans. x = 45, and y = 25. 5. Givenx + y 16, and x: y:: 3: 1, to find x and y. Ans. x = 12, and y = 4. 6. Given + -12, and + x = 9, to find x and y. 2 2 Ans. x = 10, and y = 4. 7. Givenx:y::3:2,and x2_y2 = 20, to find x and y. Ans. x = 6, and y — 4. x Y +Y X 2x — y 8. Given 12 -Y- 413an dXY+ -16 27, to find x and y. Ans. x = 60, and y = 20. RULE 3.-Let one or both of the given equations be multiplied, or divided, by such numbers, or. quantities, as will make the term that contains one of the unknown quantities the same in each of them; then, by adding, or,ubtracting, the two equations thus obtained, as the case may require, there will arise a new equation, with only one unknown quantity in it, which may be resolved as before.* * The values of the unknown quantities in the two literal equations a+ — by = c, add a'-{-+b'y =_ c', may be found in general terms, by multiplying the first by a', and( the second by a, and then working according RESOLUTION OF SIMPLE EQUATIONS. 101 EXAMPLES. 1. Givern 3 - 2y - 140 to find the values of x and y. First, multiply the second equation by 3, and it will give 3x- +6y = 42. Then, subtract the first equation from this and it will give 6y- 5y = 42- 40, or y = 2. Whence, also, x =z 14 - 2y = 14 - 4 = 10. 2. Given. 2x - 5y -96 to find the values of x and y. Multiply the first equation by 2, and the second by 5; then 10x - 6y = 18, and 10x +- 25y -80. And if the former of these be subtracted from the latter, 62 there will arise 31y = 62, or y - == 2. 9 -3Y 15 Whence, by the first equation, x = - ---- - = 3. 5 5 EXAMPLES FOR PRACTICE. 1. Given -- + 6y = 21, andY + = 23 - 5x, to find x and y., Ans. x = 4, and y = 3. 2. Given 3x4- 7y = 79, and 2y = + 2, to find x and y. Ans. x = 10, and y = 7. 3. Given 30x + 40y = 270, and 50Ox + 30y = 340, to find x and y. Ans. x - 5, and y = 3. 4. Given 3x-3y-=2x+2y, and x+y:xy::3:5, to find x and y. Ans. x = 10, and y = 2. 5. Given x2y +- xy2 = 30, and x3 y3 = 35, to find x and y. Ans. x = 3, and y =2. c/ - ca, to the last rule, when the results, so determined, will be y = a' — a" ab' -ba" and x =b b; which solution may be applied t6 any particular case ab'- ba" of this kind, by substituting the numeral values of a, b, a', b', in the place of the letters, and observing, when either of them is negative, to change the signs accordingly. Where the numerator is the difference of the products of the opposite coefficients in the order in which! is not found, and the denominator is the difference of the products of the opposite coefficients taken from the orders that involve the two unknowxn quantities. Coefficients are of the same order which either affect no unknown quantity, as c and c'; or the saime unknown quantity in the different equations, as a and a'. Coefficients are opposite when they affect the different unknown quantitles in the different equations, as a and b', a' and b.-ED. 9* 102 RESOLUTION OF S0IMPLE EQUATIONS. 3x2- 5y 2x ~.y x -2y x, / 6. Given 2x 3, and - -r 2 5 4 2 3 to find x and y. Ans. x = —12, and y = 6. 7. Given x + y: a:: x - y: b, and x2 y- y C, tofind the values of x and y. a-b c a-b c Ans. x= C6V- 6 2 b v2 ab' 8: Given ax + by = c, and dx + ey -=f, to find the values of x and y. ce- bf af- de Ans. x- y _ — Ans e - d' ae - bd' 9. Given x y-a, and - y2 = b, to find the values of x and y. a2 +- b aM —b Ans. x- -- = — _' 10. Given x2 + xy = a, and y2 + xy b=, to find the values of x andy. a b Ans.- -l-/v(a + b) (a - b)' Of the resolution of simple eqluations, containing three or more unknoqwn quantities. When there are three unknown quantities, and three independent simple equations containing then, they may be reduced to one by the following method.* RULE.-Find the values of one of the unknown quantities in each of the three given equations, as if all the rest were known; then put the first of these values equal to the second, and either the first or second equal to the third, and there will arise two new equations with only two unknown quantities in them, the values of which may be found as in the former case; and thence the value of the third Or, multiply each of the equations by such numbers, or quantities, as will make one of their terms the same in them all; then having subtracted any two of these resulting equations from the third, or added them together, as the case may ~ The necessity for observing that the given equations in this and other similar cases are so proposed as to be-independent of each other, will be obvious from the following example:x —Jy+z -5; 2x-+Jy-z:7; x;-3 r -2-z2; where, if it were required to determine the value of x, y, and z, it will be found, by eliminating x from each of them, and then equating'the results, that 5v -3 z 3, and 5y - 3z =-3; which equations, being. identical, or both the same, furnish no determi-. nate answer. And, in effect, if the three equations be properly examined, it will be found, that the third is merely the difference of the first a-nd second, and consequently involves no condition but what is contained in the other two. RESOLUTION OF SIMPLE EQUATIONS. 103 require, there will remain only two equations, which may be resolved by the former rules. And in nearly the same way may four, five, &c., unknown quantities be exterminated from the same number of independent simple equations; but, in cases of this kind, there are frequently shorter and more commodious methods of operation, which can only be learnt from practice.* EXA MPLES. +Y +Z =29 1. Given x 2y + 3z62 to find x, y,and.. +.ly + 4lz 10 Here, from the first equation, x = 29 - y - z. From the second, x = 62 - 2y - 3z. 2 1 And from the third, x 20 - y- -z Whence 29 - y - = 62 -2y - 3z, And, also, 29-y -z = 20 y- 1, From the first of which, y = 33 - 2z, 3 And from the second, y = 27 -a, Therefore 33 -2 - 2 27- 2, or z 12, Whence,' also y = 33 - 2z = 9, And x = 29 -y - = 8. 2x + 4y - 3z = 22 ) 2. Given 4x- 2y -+ 5z-= 18 to find x, y, and z. 6x+7y —z =63 * The values of the unknown quantities in the three literal equations, ax-. by —cz = d; a'. - b'y +i-c'e — a=d'; a"x-[-b"yJ-+c"z =d"; may be exhibited in general terms, like those before mentioned, as follows:db'c" - dcb " +- cd'b" - bd'c"' - bc'd"i - cb'dd" ab c" -ac'b" + ca'b" - ba'c"' bc'a" -cioCa"' ad'c" - ac'd" -t- ca'd" - da'c" +- dc'a" - cd'a" = ab'c" -ac"b" q-ca'b " —ba'c" + bc'a" -cb'a"' ab'd"- ad'b" 4- da'b" — ba'd " + bd'a"-db'a" ab'c" - ac'b" - ca'b" - ba'c" 4- bc'a" - cb'/a",which formul, by substitution, may be employed for the resolution of any numeral case of this kind, as in the instance of two equations before given. The numerator of any of these equations, such as z, consists of all the different products, which can be made of three opposite coefiicients taken from the orders in which z is not found; and the denominator consists of all the products that can be made of the three opposite coefficients taken from the orders which involve the three unknown quantities. 415 RESOLUTTrION OF SIMPLE EQUATIONS. Here multiplying the first equation by 6, the second by 3, and the third by 2, we shall have 12x + 24y - 18z- = 132, 12x- Gy + 15z = 54, 12x + 14y- 2z= 126. And, subtracting the second of these equations successively from the first and third, there will arise 30y - 33z -- 78,.20y - 17z- 72. Or, by dividing the first of these two equations by 3, and then multiplying the result by 2, 20y- 22z = 52, 20y - 17z = 72, Whence, by subtracting the former of these from the latter, we have 5z = 20, or z - 4. And, consequently, by substitution and reduction, y 7, and x = 3. 3. Given x +tz- - z= 53, x - 2y - 3z - 105, and x+ 3y~ 4z = 134, to find the values of x, y, and z. Ans. x =24, y-6, andz= 23. 4. Given + +2Y+ -= 32, x + 4y= -z = 15 and 1 1 1 + 5y +- z = 12, to find the values of x, y, and z. Ans. x = 12, y = 20, z = 30. 5. Given 7x +- 5y + 2z = 79, 8x- + 7y + 9z = 122, and x + 4y + 5z = 55, to find the values of x, y, and z. Ans. x- 4, y=9, z = 3. 6. Given x - y = a, x - z = b, and y q-z = c, to find the values of x, y, and z. a - b - c a + b - c b —-a - c Ans. y - -- =- andz --- 2 2' 2'2 7. Givenx- ++z- _62,x-+ + — 47, and- x-~ - 2 3 4 3 4 5 4 5 6 -38, to find x, y, and z. Ans. x 24, y = 60, a.nd z = 120. 8. Given z - y -- x + 100, y - 2x - 2z - 100, and z - 100 == 3x + 3y, to find x, y, and z. Ans. x -- 9a, y = 45- T, and z -- 63 7T 9. Given x+ y +z= 7, 2x-3 =-y+ 3, and 5x+5z = 3y + 19, to find x, y, and z. Ans. x - 4, y = 2 and a- I. 10. Given 3x +- 5y 4- =- 25, 5x - 2y r 3z = 46, and 3y + 5z - x = 62, to find x, y, and z. Ans. x = 7, y =8, andz = 9 RESOLUTION OF SIMPLE EQUATIONS. 105 11. *Given y - z 13, x +y- u 17, x +z+ -. 18, and y' + z + u = 2 1, to find x, y, z, and u. Ans. x — 2, y = 5, z = 6, and u= 10. MISCELLANEOUS QUESTIONS, PRODUCING SIMPLE EQUATIONS. THE usual method of resolving algebraical questions, is first to denote the quantities, that are to be found, by x, y, or some of the other final letters of the alphabet; then, having properly examined the state of the question, perform with these letters, and the known quantities, by means of the common signs, the same operations and reasonings, that it would be necessary to make if the quantities were known, and it was required to verify them, and the conclusion will give the result sought. Or, it is generality best, when it can be done, to denote only one of the unknown quantities by x or y, and then to determine the expression for the others from the nature of the question; after which the same method of reasoning may be followed, as above. And, in some cases, the substituting for the sums and differences of quantities, or availing ourselves of any other mode, than a proper consideration of the question may suggest, will greatly facilitate the solution. 1. What numnber is that whose third part exceeds its fourth part by 16? Let z = the number required. will be 1 1 Then its part will be -x, and its - part 4X. 1 1 And therefore x - x== 16, by the question, 3 4'That is, x- - x = 48, or 4x- 3x 192, Hence x = 192, the number required. 2. It is required to find two numbers such, that their sum shall be 40 and their difference 16. Let x denote the least of the two numbers required, Then will x + 16 - the greater number, And x + x + 16 = 40, by the question, * This can be resolved by proceeding after the same manner as equations involving three unknown quantities; but the resolution of' it mav be greatly facilitated; by introducing into the calculation, beside the principal unknown quantities, a new unknown quantity arbitrarily assumed, such as, for example, the sum of all the rest; and when a little practised in such calculations, they become easy. 106 RESOLUTION OF SIMPLE EQUATIONS. 24 That is, 2x -40 - 16, or x = - 12 = leastfnumber. And x + 16 = 12 -+ 16 - 28 = the greater number required. 3. Divide 10001. between A, B, and c, so that A shall have 721. more than B, and c 1001. more than A. -Let x = B's share of the given sum, Then will x + 72 = A's share, And x + 172 = c's share, Hence their sum is x + x + 72 + x -+ 172, Or 3x + 244 = 1000, by the question, That is, 3x- 1000- 244 - 756, 756 Or x- 2521, = B.'s share, Hence x + 72 = 3241. = A's share, And x + 172 = 4241. = c's share, Also, as above, 2521. = B's share. Sum of all = 10001. the proof. 4. It is required to divide 10001. between two persons, so that their shares of it shall be in the proportion of 7 to 9 Let x = the first person's share, Then will 1000 - x = second person's share, And x: 1000 - x:: 7: 9, by the question, That is, 9x = (1000 - x) X 7 = 7000 - 7x, 7000 Or 9x +-'7x= 7000, or x ='00 - 4371. 10s. = 1st share, and 1000 - x = 1000 - 4371. 10s. - 5621. 10s. = 2d share. 5. The paving of a square court with stones, at 2s. a yard, will cost as much as the enclosing it with pallisades, at 5s. a yard; required the side of the square. Let x = length of the side of the square sought, Then 4x = number of yards of enclosure, And x2 = number of yards of pavement, Hence 4x X 5 = 20x = price of enclosing it, And,2 X 2 = 2v2 = the price of the paving, Therefore 2x2 - 20x, by the question, Or, 2x = 20, and x == 10, the length of the side required. 6. Out of a cask of wine, which had leaked away a third part, 21 gallons were afterwards drawn, and the cask being then gauged, appeared to be half full; how much did it hold? Let x = the number of gallons the cask is supposed to have held. Then it would have leaked away ~x gallons. 3 RESOLUTION OF SIMPLE EQUATIONS. 107 Whence there had been taken out of it, altooether, 21 -- Ix gallons, 1 1 And therefore 21 + x- Ix by the question, That is, 63 - x = 2x, or 126 4- 2x = 3x. Consequently, 3x - 2x = 126, or'x = 126, the number of gallons required. 7. What fraction is that, to the numerator of which if 1 be added. its value will be 3' but if 1 be added to the denominator, its value will be 4' Let the fraction required be represented - x +1 1 x 1 Then —:- 3- and = -, by the question. y 3' y y-l 1 Hence 3x + 3 = Y, and 4x y -, or x = 4Therefore 3 (Y4 ) +3 -y, or 3y +3 + 12= 4y. y +l1 15 + I 16 That is, y = 15, and x - - 41 4 4 4 4 Whence the fraction that was to-be found is. 8. A market-woman bought in a certain number of eggs at 2 a penny, and as many others at 3 a penny, and having sold them out again, altogether, at the rate 5 for 2d., found she had lost 4d.; how many eggs had she? Let x - the number of eggs of each sort, Then will x-= the price of the first sort, 1 And ~ x = the price of the second sort, But 5: 2:: 2x (the whole number of eggs): 5, Whence 5 = the price of both sorts, when mixed together at the rate of 5 for 2d 108 RESOLUTION OF SI.MPLE EQUATIONS. 1 1 4x And consequently nx + -x - - = 4, by the question. That is, 15x + 10x- 24x=- 120, or x = 120, the number of eggs of each sort, as required. 9. If A can perform a piece of work in 10 days, and B in 13; in what time will they finish it, if they are both set about it together? Let the time sought be denoted by x. Then 1- - the part done by A in one day, And-3 = the part done by B in one day. a X Consequently -- ~- 1 (the whole work), That is, 13x +- 10x = 130, or 23x = 130. 130 15 Whence 23 5 - days, the time required. 23 23 10. If one agent A, alone, can produce an effect e, in the time a, and another agent B, alone in the time b; in what time will both of them together produce the same effect? Let the time sought be denoted by x, ex Then a: e x: - = part of the effect produced by A. a en And b: e:: xn: = part of the effect produced by B. ex en Hence - +-b = e, (the whole effect) by the question. g X Or - + b = 1 by dividing each side by e. aa Therefore, x -- - a, or bx + ax - ab, ab Consequently, x- = - time required. 11. How much rye, at 4s. 6d. a bushel, must be mixed with 50 bushels of wheat, at 6s. a bushel, so that the mixture may be worth 5s. a bushel? Let x = the number of bushels required. Then 9x is the price of the rye in sixpences, And 600 the prices of the wheat in ditto, Also (50 + x). X 10 the price of the wheat in ditto, Whence 9x + 600 = 500 + 1 Ox, by the question. RESOLUTION OF SIMPLE EQUATIONS. 109 Or, by transposition, 10x - 9x = 600 - 500. Consequently x- = 100, the number of bushels required. 12. A labourer engaged to serve for 40 days, on condition that for every day he worked he should receive 20d., but for every day he was absent he should forfeit 8d.: now, at the end of the time he had to receive 11. 11s. 8d.: how many days did he work, and how many was he idle? Let the number of days that he worked be denoted by x. Then will 40 - x be the number of days he was idle, Also 20x the sum earned, and (40 - x) X 8, Or 320 - 8x the sum forfeited, Whence 20x - (320 - 8x) -- 380d. ( = 11. lis. 8d.), by the question, That is 20x - 320 - 8x = 380, Or 28x = 380 + 320 -- 700, Consequently, x = 2-8 - 25, the numnber of days he work28 ed, and 40 - x = 40 - 25 = 15, the number of days he was idle. QUESTIONS FOR PRACTICE. 1. It is required to divide a line, of 15 inches in length, into two such parts, that one may be three fourths of the other. Ans. 8# and 61. 2. My purse and money together are worth 20s., and the money is worth 7 times as much as the purse, how much is there in it? Ans. 17s. 6d. 3. A shepherd being asked how many sheep he had in his flock, said, if I had as many more, half as many more, and 7 sheep and a half, I should have just 500; how many had he? Ans. 197. 4. A post is one fourth of its length in the mud, one third in the water, and 10 feet above the water, what is its whole length? Ans. 24 feet. 1 1 5. After paying away ~ of my money, and then I of the 5 remainder, I had 72 guineas left; what had I at first? Ans. 120 guineas. 6. It is required to divide 3001. between A, B, and c, so that A may have twice as much as B, and c as much as A and n together. Ans. A 1001., B 501., c 1501. 7. A person, at the time he was married, was 3 times as old as his wife: but after they had lived together 15 years, he was only twice as old; what were their ages on the weddingday? Ans. Bride's age 15, bridegroom, 45. 10 110 RESOLUTION OF SIMPLE EQUATIONS. 8. What number is that from which, if 5 be subtracted, two thirds of the remainder will be 40? Ans. 65. 9. At a certain election, 1296 persons voted, and the successful candidate had a majority of 120; how many voted.for each? Ans. 708 for one, and 588 for the other. 10. A's age is double of B's, and B's is triple of c's, and the sum of all their ages is 140; what is the age of. each? Ans. A's 84, B's 42, and c's 14 11. Two persons, A and B, lay out equal sums of money in trade; A gains 1261., and B loses 871., and A's money is now double of B's; what did each lay out? Ans. 3001. 12. A person bought, a; chaise, horse, and harness for 601.; the horse came to twice the price of the harness, and the chaise to twice the price of the horse and harness; what did he give for each? Ans. 131. 6s. 8d. for the horse, 61. 13s. 4d. for the harness, and 401. for the chaise. 13. A person was desirous of giving 3d. a piece to some beggars, but found he, had not money enough in his pocket by 8d.. he therefore gave them each 2d., and had then 3d. remaining; required the number of beggars? Ans. 11. 14. A servant agreed to live with his master for 81. a year, and a livery, but was turned away at the end of seven months, and received only 21. 13s. 4d. and his livery; what was its value? Ans. 41. 16s. 15. A person left 5601., between his son and daughter, in such a manner, that for every half crown the son should have, the daughter was to have a shilling; what were their respective shares? Ans. Son 4001., daughter 1601. 16. There is a certain number, consisting of two places of figures, which is equal to four times the sum of its digits; and if 18 be added to it, the digits will be inverted; what is the number? Ans. 24. 17. Two persons, A and B, have both the.same income:; A saves a fifth of his yearly, but B, by spending 501. per annum more than A, at the end of four years finds himself 1001. in debt; what was their income? Ans. 1251. 18. When a company at a tavern came to pay their reckoning, they found, that if there had been three persons more, nthey would have had a shilling apiece less to pay, and if there had been two less, they would have had a shilling apiece more to pay; required the number of persons, and the quota of each. Ans. 12 persons, quota of each 5s. 19. A person at a tavern borrowed as much money as, he had about him, and out of the whole spent is.; he then went to a second tavern, where. he also borrowed as much as he RESOLUTION OF SIMPLE EQUATIONS. 111 had now about him, and out of the whole spent Is.; and going on, in this manner, to a third and fourth tavern, he found, after spending his shilling at the latter, that he had nothing left; how much money had he at first? Ans. 11led. 20. It is required to divide the number 75 into two such parts, that three times the greater shall exceed seven times the less by 15. Ans. 54 and 21. 21. In a mixture of British spirits and water, 2 of the whole plus 25 gallons was spirits, and -- part minus 5 gallons was water; how -many gallons were there in each? A.ns. 85 of wine, 35 of water, 22. A bill of 1201. was paid in guineas and moidores, and the number of pieces of both sorts that were used were just 100; how many were there of each, reckoning the guineas at 21s., and the moidores at 27s.? Ans. 50. 23. Two travellers set out at the same time from London and York, whose distance is 197 miles: one of them goes 14 miles a day, and the other 16: in what time will they meet? Ans. 6 days 13~- hours. 24. There is a fish whose tail weighs 91b., his head weighs as much as his tail and half his body, and his body weighs as much as his head and'his tail; what is the whole weight of the fish? Ans. 721b. 25. It is required to divide the number 10 into three such parts, that if the first be multiplied by 2, the second by 3, and the third by 4, the three products shall be all equal. Ans. 4-83,'3 1 23. 26. It is required to divide thie number 36 into three such parts, that l the first, I of the second, and i of the third, shall be cll equal to each other. Ans. The parts are 8, 12, and 16. 27. A person has two horses, and a saddle, which of itself is worth 501.; now, if the saddle be put on the back of the first horse, it will make his value double that of the second, and if it be put on the back of the second, it will make his value triple that of the first; what is the value of each horse? Ans. One 301. and the other 401. 28. If A give B 5s. of his nloney, B will have twice as much as the other has left; and if B give A 5s. of his money, A will have three times as much as the other has left: how much has each? Ans. A 13s. and B. Ils. 29. What two numbers are those whose difference, sum,:and product, are to each other, as the numbers 2, 3, and 5 respectively? Ans. 10 and 2. 30. A person in play lost a fourth of his money, and then 112 QUADRATIC EQUATIONS. won back 3s., after which he lost a third of what he now had, and then won back 2s.; lastly, he lost a seventh of what he then had, and after this found he had but 12s. remaining; what had he at first? Ans, 20s. 31. A hare is 50 leaps before a greyhound, and takes 4 leaps to. the greyhound's 3, but 2 of the greyhound's leaps are as much as 3 of the hare's; how many leaps must the greyhound take to catch the hare? Ans. 300. 32. It is required to divide the number 90 into four such parts, that if the first part 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 be all equal? Ans. The parts are 18, 22, 10, and 40. 33. There are three numbers whose differences are equal, (that is, the second exceeds the first as much as the third exceeds the second), and the first is to the third as 5 to 7; also the sum of the three numbers is 324; what are those numbers? Ans. 90, 108, and 126. 34. A man and his wife usually drank out a cask of beer in 12 days, but when the man was from home it lasted the woman 30 days; how many days would the man alone be in drinking it? Ans. 20 days. 35. A general, ranging his army in the form of a solid square, finds he has 284 men to spare, but increasing the side by one man, he wants 25 to fill up the square.; how many soldiers had he? Ans. 24000. 36. If A and B together can perform a piece of work in 8 days, A and c together in 9 days, and B and c in 10 days, how many days will it take each person to perform the same work alone. Ans. A 14-A days, B 174I{, and c 23'7. QUADRATIC EQUATIONS. A QUADRATTC EQUATION, as before observed, is that in which the unknown quantity is of two dimensions, or which rises to the second power, and is generally divided into simple or pure, and cognpoulnd or adfected. A simple or pure quadratic equation, is that which contains only the square, or second power, of the unknown quantity, as, 2 b b ax2 = b, or x2 = -; where x = r-. a a A compound or adfected quadratic equation, is that which contains both the first and second power of the unknown quantity, as, b e axa + bx- =, or x2 -— X- -. a a QUADRATIC EQUATIONS. 113 In which case it is to be observed, that every equation of this kind, having any real positive root, will fall under one or other of the three following forms:a 2a \ 1. x2 ax = b... where x - - a J b). a a2 2. x —ax==b... wherex-+ q-, /Q-+b). 3. x2 - a-X,-b. where x -- b ) Or, if the second and last terms be taken either positively or negatively, as they may happen to be, the general equation 2 b ax2: bx = c, or xa - x = i -, a a which comprehends all the three cases above-mentioned, may be resolved by means of the following rule:RULE.-Transpose all the terms that involve the unknown quantity to one side of the equation, and the known terms to * It may be observed, with respect to these forms, that In the case x2 - ax —b=o, where x =_ -a + / (a2- + b), or -_ a /- /(las2 + b), the first value of x must be positive, because V (la2 + b) is greater than 1/'a2, or its equal -na; and its second value will evidentlybe negative, because each of the terms of which it is composed is negative. 2. In the case r2 -ax-b = o, where x =.a +- V/ (4a2 + b) or ad- / (a.2+b), the first value of x is manifestly positive, being the sum of two positive terms: and the second value will be negative, because V/ (a2 - -b) is greater than A/4,la, or its equal Ia. 3. In the case zs-2 ax2 b = o, where x. a -v + / (a2s-b), or a a-V (a2 - b), both the values of x will be positive, when 1a2 is greater than b; for its first value is then evidently positive, being composed of two positive terms; and its second value will also be positive; because V (Ia 2-b ) is less than 1/,a2, or its equal La. But if 4a2, in this case, be less than b, the solution of the proposed equation is impossible; because the quantity la2 - b. under the radical, is then negative; and consequently V/ (Ia2 - b) will be imaginary, or of no assignable value. 4. It may be also further observed, that there is a fourth case of the formX2 TX- ax- +b = o, where x - _ -- Va (~a2s-b), or x =-a-/V (4a2 - b), the two values of x will be both negative, or both imaginary, according as 1as is greater or' less than b; the imaginary roots, when they occur, being here of the forms - (a' + c' V/ — 1) and - (a'-Cv' — 1). From which it follows, that if all the terms of a quadratic equation, when brought to the lefthand side, be positive, its two roots will be both negative, or both imaginary; and conversely, if each of the roots be negative, or each imaginary, the signs of all the ternus will be positive. So that, of all quadratic equations, which can have any real positive root, that of the third form, zs - ax.- b = o, is the only one, where the solution for certain numeral values of a and b will become impossible 10* 114 QUADRATIC EQUATIONS. the other; observing to arrange them so that the term which contains the square of the unknown quantity may be positive, and stand first in the equation. Then, if this square has any coefficient prefixed to it, let all the rest of the terms be divided by it, and the equation will be brought to one of the three forms above-mentioned. In which case, the value, of the unknown quantity x is always equal to half the cocficient, or multiplier of x, in the second term of the equation, taken with a contrary sign, together wvith ~- the square root of the square of this number and the known quantity that forms the absolute or third term of the equation.* Note.-All equations, which have the index of the unknown quantity, in one of their terms, just double that of the other, are resolved like quadratics, by first finding the value of the square root.:of the first term, according to the method used in the above ri~e, and then taking such a root, or power of the result, as is\denoted by the reduced index of the unknown quantity. Thus, if there be taken any general equation of this kind, as, x2mn ~" axm - b, we shall have, by taking the square root of x2?1, and observing the latter part of the rule, * This rule, which is more commodious in its practical application than that usually given, is founded upon the same principle; being derived from the well-known property, that in any quadratic, x2 i ax - b, if the square of half the coefficient a of the second term of the equation be added to each of its sides, so as to render it of the form z2 k a.x +la2 = I a2 -I- b, that side which contains the unknown quantity will then be a complete square; and, consequently, by extracting the root of each side, we shall have x::i: ~ _ = -4- (l a2~ -- b), or x — TF I a V- /(Va (2 ~ b), which is the same as the rule, taking a and b in +- or - as they may happen to be. It may here also be observed, that the ambiguous sign:J:, which denotes both + and -, is prefixed to the radical part of the value of x in every expression of this kind, because the square root of any positive quantity, as a2, is either +-a or -a; for ( —a) X (+a), or (" X ( —a) are each - -Ia2: but the square root of a negative quantiy, as - a2 is imaginary, or unassignable, there being no quantity, either positive or negative, that, when multiplied by itself, will give a negative product. To this we may also further add, that from the constant occurrence of the double sign before the radical part of the above expression, it necessarily follows, that every quadratic equation must have two roots; whichLre either both real, or both imaginary, according to the nature of the question. QUADRATIC EXQUATIONS. 1t15 >2 2 x-= -- - (a +-),b andx- -> I ~/ -) b And if the equation, which is to be resolved, be of the following form, rn Xm - aX2 b we shall necessarily have, according to the same principle, X2~ a _ (-+ +b), and x= a,~Q-Cb) -. EXAMPLES. 1. Given x2 + 4x = 140, to find the value of x. Here x2 + 4x= 140. by the question, Whence x = - 2 i- V, (4 + 140), by the rule, Or which is the same thing, x = - 2 t- V/ 144. Wherefore x - 2 +- 12 = 10, or - 2 - 12 =-14, Where one of the values of x is positive and the other negative. 2. Given x2 - 12x + 30 = 3, to find the value of x. Htere a - 12x -- 3 - 30 = - 27, by transposition, Whence x = 6 ~ V/ (36 - 27), by the rule, Or, which is the same thing, x = 6 -i v/ 9, Therefore x = 6 + 3 = 9, or = 6 - 3 = 3. WhLere it appears that x has two positive values. 3. Given 2x2 + 8x - 20 = 70, to find the value of x. Here 2x2 + 8x = 70 + 20 = 90, by transposition. And x2 + 4x — 45, by dividing by 2, Whence x = - 2 ~- v (4 + 45), by the rule, Or, which is the same thing, x = - 2 ~ — V 49. Therefore, x = - 2 +- 7 = 5, or = - 2 - 7 -- 9. Where one of the values of x is positive and the other negative. 4. Given 3X2 - 3x + 6 = 5 L, to find the value of x. Here 3x - 3x -L - 6 =- by transposition. 3 2 And 2 _ X -- - by dividing by 3, Whence x -- =,/ - -), by the rule. 2 1 1 1 Or, by subtracting from., x - - ~' V 9 4 2 36 1 1 2 1 I 1 Therefore x -- + - = 2, or = 2-6 - 3' In 3which2 6case has two positive values In which case x has two positive values 116 QUADRATIC EQUATIONS. 5. Given x2. - - + 20} -42-, to find the value of x. 1 1 Here - 2 - x - 422- - 20, =_ 22X, by transposition, And x 2 = 444, by dividing by 2, or multiplying by 2, Whence we have x = + 44 ), bthe rule, 1 1 400 Or, by adding I and 44~ together, x = i -, 3 9, 1 1 Therefore x - + 6= - 7, or _ 6- -- 6-1 Where one value of x is positive and the other negative. 6. Given ax2 + bx = c, to find the value of x. b c Here x2 + -X - -, by dividing each side by a. a a b b2 C\ Whence, by the rule, x -- 2a - 4a2 a' b b2 + 4ac Or, multiplying c and a by 4a, X = - a- V 4a2 b1 1 Therefore x= — - - - (b' +4ac). 2a 2aV 4ac). 7. Given ax2 - bx -+ c:= d, to find the value of x. Here ax - bx = d -- c, by transposition, b d-c And x2 - -x —,by dividing by a. a a Whence x= 2- a +- V - 42 ) by the rule, b1 1 Or, multg d —c and a by 4a, x2:=a:a V 4a(d-c)~b2 }. 8. Given x4 + ax2 = -- b, to find the value of x. Here x4 + aX2 = b, by the question, Or, X2 = - a a +b ---- - 2 —V(a2+- 4b), by 2 4 2 2 the rule, Whence x=*V A/ -- i 1 (4b + a)4 by extraction of roots. 1 G1 1 9. Given,8 -- X3 --- to find the value of x. 2 4 32 QUADRATIC EQUATIONS. 117 1 1 1 Here - = -- 3, by the question, 2 4 32 1 1 And x6 -- - =- by multiplying by'2. 2 16 Whence x3 -=: v/ -,- by the rule, And consequently x = 3/_-= V ~ = - 3 2. 4 8 2 10. Given 23 + 3x3 2, to find the value of x.. I Here 2x 3 + 3x3 = 2, by the question, And x3 + -2 1"3 by dividing by 2, 3n x 3 9 3 5 1, Whence x3 - o + 1)- 4 - - - or —2. 4 1 -6 4 4 2-_ Therefore, x = (2)3 or (- 2) =-8. 11. Given x4 - 12xa3 + 44x2 - 48x = 9009 (a), to find the value of x. This equation may be expressed as follows:2(x2 - 6x)2 + 8 (x2 - 6x) = a, * The biquadratic equation X4 - 12:r3 - 44x2 - 48x = a can be easily exhibited under the form (X2 — 6X)2 + 8 (x2 - 6x) = a by the following method:z4 - 12x3 - 44x2- 48x (z2 - 6x X4,X2 - 6x) - 12x.r + 44x2- 48 -12:3a + 36x2 x2- 6x) 8z2 -48x (8 8x.2- 48x * * Consequently, (x2 - 6x)2 + 8 (x2 - 6x) = -4-12x3+ 44x2 -48 x = a; for since, in extracting the square root of any quantity, the square of the root thus found plus the remainder, is always equal to the proposed quantity. In a similar manner, the biquadratic equation x4- 2ax3- 5a2.x +- 4a3 x = d, may be exhibited under the form (.2;2 + a.z)2 + 4a2 (.z2 + az) = d which can be resolved by the rule, page 126, for resolving quadratic equations. HIence it follows, that if the remainder, after having found the first two terms of the square root, according to the rule page 50, can be resolved into two such factors, so that the factor containing the unknown -118 ~ QUADRATIC EQUATIONS.'Whence x2 - 6x = - 4 i -/ (16 + a), by the common rule, And, by a second operation, x-=3 i- v) 9 -4 4 / (16 + a), Therefore, by restoring the value of a, we have x = 3 i: ~V (5 i V/9025), Or, by extraction of roots, x = 13, the Ans. rEXAMPLES FOR -PRACTICE.* 1. Given x2 - 8 +- 10 = 19, to find the value of x. Ans. x-= 9. 2. Given xa - x - 40 = 170, to find the value of x. Ans. x-= 15. 3. Given 3x2 -,2x - 9 = 76, to find the value of x. Ans. x =5. I 2_ 1 4. Given X2 3 x + 7A = 8, to find the value of x. Ans. x=- 1-. 1 1 5. Given - x -- V/x = 22x, to find the value of x. 2 ~ Ans. x — 49. 6. t.Given x + v (5x + 10) = 8, to find the value of x. Ans. x = 3. quantity shall be equal to the terms of the root fhus found; the proposed biquadratic may be always reduced to a quadratic form, as above. See Ryan's Algebra, page 396.-ED. * The unknown quantity in each of the following examples, as well as in those given above, has always two values, as appears from the common rule; but the negative and imaginary roots being, in general, but seldom used in practical questions of this kind, as here suppressed, t In some quadratic equations involving radical quantities of the form V (ax+ b), both the values of x, found by the ordinary process, will not answer the proposed equation, except we take the radical quantity with the double sign -i. In resolving the above example, two values of x, that is 18 and 3 are found: but it appears, that 18 does not answer the condition of the equation, except we take the radical quantity a/ (5x q- 10) with the sign -. Now, since these two values of x are found from the resolution of the equation x2- 21 x =- 54; it necessarily follows that each of them, when substituted for x, must satisfy that equation; which may be verified thus; in the first place, by substituting 18 for x, in the equation X2- 21x =-54. wehave (8)2 - 11 X 18 -54, or 324-378 — 54, that is, -54 — 54, or, by transposition, 0 = -0. Again, substituting 3 for x, we have (3)2 —21 X3 — 54, or 9 — 63 — 54,.'. 54- 54- 0, or 0 = 0. And as the equation xi - 21x - - 54, may be deduced from the equation + v (5x -+ 1.)) = 8 -, or - (5x + 10) = 8 —x; it is evident that the radical quantity 1 (5xn- 10) must be taken with the double sign:4:, in the primitive equation, in order that it would be satisfied by the values, 18 and 3, of x, found above; that is,- 18 answers to the sign -, and 3 to the:sign +-. See Ryan's Elementary Treatise on Algebra, Theoretical and Practical, where this subject is clearly illustrated.-ED. QUADRATIC EQUATIONS. 119 7. Given V (10 + x) — V (10 + x) = 2, to find the value of x. Ans. x — 6 8. Given: 2x4 - X2 + 96 969, to find the. value of x. Ans. x=2V6. 9. Given 6 + 20x3 - 10 = 59, to find the value of x. Ans. x —3 /3 10. Given 3x2' - 2xn + 3 = 11, to find the value of x. Ans. x ==/ 2. 11. Given 54Vx - 3 Vx = 1-1, to find the value of x. 13 1 Ans. 3 —, or - 2 1 2 12. Given x v (3 + 2x2) - + -2 to findvalue of x. Ans. 1= V/ ( —3 + 3 Z 2). 13. Given X v ( -- = -, to find the value of x, Ans. x = -'(1+'2). 14. Given - (1 - xa) = x2, to find the value of x. Ans. X=(4 5 _ 3) 15. Given / (- 1) =V'(2- b2), to find the value of x. 1Ans. x= — -a +q'V (8b2 + a2).16. Given /(1 -.- X2) 2 (1 - -- _ ):-= 5, to find the value of x. A 1 Ans. x= + — /41 17. Given'(- )+ X(1- )=, to find the value of x. Ans. x --- -- I 5. 18. Given 4n- 2x3n - Xn -- 6, to find the value of x. Ans. x-=V( 1+L,/13). 19. Given ax - 2 x3 + X - a, to. find the value of x. Ans. = 1 v 3 4 V(a+a +) 120 QUADRATIC EQUATIONS. When there are more equations and unknown quantities than one, a single equation, involving only one of the unknown quantities, may sometimes be obtained by the rules before laid down for the solution of simple equations; and, in this case, one of the unknown quantities being determined, the others may be found bry substituting its value in the remaining equations. EXAMPLES. 1. Given } Z+ ~y - 28 to find the values of x and y. 28 Here, from the second equation, we have y = -; and 784 by substituting this in the first, xa+ -2 = 65, or X4 - 65x2 784. Whence, by the common rule before given, we have 2 ~v(4 _784) 4 Or, by reducing the parts under the last radical, and ex-,65 33) tracting the root, x = i: -/ i - =7, or - 4, and con28 28 sequently, y= 7, or - = 4, or - 7. Or the solution, in cases of this kind, may often be more readily obtained by some of the artifices frequently made use of upon these occasions; which can only be learned from experience: thus, taking, as before, (1.) x2 + y2 - 65, (2.) xy = 28, we shall have, as in the former method, by multiplying by 2, 2,xy = 56, and, by adding this equation to the first, and subtracting it from the same, x2 + 2xy + y2 _ 121, and x2 - 2xy + y2 - 9. Whence, by extracting the square roots of each of these last equations, there will arise x + y = - 11, and x - y -= 3, and consequently, by adding and subtracting these, we shall have 2x - t- 14, or - 7, or - 7, and y = 4, or - 4. It will also sometimes facilitate the operations, by substituting for one of the unknown quantities the product of the other, and a third unknown quantity; which method may be applied with advantage whenever the sum of the dimensions of the unknown quantities is the same in every term of the equation. 2. Given; cmy - 2y2-6 6 to find the values of x and y. QUADRATIC EQUATIONS. 121 Here, agreeably to the above observation, let x = vy, then V2y2 + vy2 = 56, and vy2 + 2y2 -- 60, whence, from the first of these equations, y2 = and from the second, y2 60 Therefore, by equating the righthand member 60 56 of these two expressions, we shall have ~2 = or v v-t-2 +v or 60v2 + 60v - 56v + 112. And, by transposing 56v, and 1 28 dividing the result by 60, v2 + v =. Hence, by the, 15- 15 common rule for quadratics, we have v - =30( 90() 28\ 1 41 4 + -) = — - - And, consequently, by the former 15 30 30 3' 60 60 part of the process, y2 - 18, or y / v~ ~2 1o -+ 2 (18)=3V/2, andx=-vy=3- X3V/2-=4V/2. The work may also be sometimes shortened, by substituting for the unknown quantities, the sun and difference of two other quantities; which method may be used when the unknown quantivies, in each equation, are similarly involved. 2 2 -x — - 12: 3. Given 718Y X to ~fnd the values of x and y. Here, according to the above observation, let there be assumed x z + v, and y = - v. Then, by adding these two equations together, we shall have x + y 2z 12, or z = 6, also, since x = 6 + v, y = 6 - v, and by the first equation 3 +- i -- 18xy, we shall obtain, by substitution, (6 + v)3 + (6 - v)3 - 18 (6 + v) (6 - v), or, by involving the two parts of the first member, and multiplying those of the second, 432 + 36u = -648 - 1Sv2, whence, by transposition, 216 54v2 - 216; and by division, v' 2-= 4; or v i=:: 2. 54 And therefore, by the first assumption, and the first part of the process, we have x = z v =6 2 = 8, or 4, and y z-v- = 6 2 -- 4, or 8. 1.1 122 QUADRATIC EQUATIONS. QUESTIONS PRODUCING QUADRATIC EQUATIONS. The methods of expressing the conditions of questions of this kind, and the consequent reduction of them, till they are brought to a quadratic equation, involving only one unknown quantity and its square, are the same as those already given for simple equations. 1. To find two numbers such that their difference shall be 8, and their product 240. Let x equal the least number. Then will x + 8=- the greater, And x (x + 8) = x2 + 8x = 240, by the question, Whence x =-4 +- (16 + 240) =-4 + 256 by the common rule, before given, Therefore x = 16 - 4 = 12, the less number, and x +- 8 = 12 + 8 = 20, the greater. 2. It is required to divide the number 60 into two such parts, that their product shall be 864. Let x = the greater part, Then will 60 - x = the less, And x (60 -x) = 60x — 2 = 864, by the question, Or, by changing the signs on both sides of the equation, X2 - 60x = - 864, Whence x = 30 =~ V/(900 - 864)= 30i v 36 = 30 6 by the rule, And consequently, x - 30 + 6.- 36, or 30 - 6 = 24, the two parts sought. 3. It is required to find two numbers such, that their sum shall be 10 (a), and the sum of their squares 58 (b). Let x = the greater of the two numbers, Then will a - x the less, And x2 + (a - x)2 = 2x2 - 2ax + a2 = b, by the question, Or 2x2 - 2ax = b - a2, by transposition, b - aAnd xa- ax = 2, by division. _a 2 b - + a: 1 (2b-aa). Whence x -=2 j- )=-~ — V (2b a2) 2 4 \~d~ 2 2 by the rule, And if 10 be put for a, and 58 for b, we shall have 10 1 x =2-'2 + / (116 - 100)-= 7, the greater number, 10 1 And 10 -x = -- — V (116- 100) = 3, the less. 2 2 QUADRATIC EQUATIONS. 123 4 Having sold a piece of cloth for 341., I gained as much per cent. as it cost me; what was the price of the cloth? Let c = pounds the cloth cost, Then will 24 - x = the whole gain, But 100: *: x': 24 - x, by the question, Or, x2 = 100 (24 - x) - 2400 - 100x, That is x2 + 100x = 2400, Whence x = -50 + V/ (2500 + 2400) = -50 + 70 - 20, by the rule, And consequently, 201. = price of the cloth. 5. A person bought a number of sheep for 801., and if he had bought four more for the same money, he would have paid 11. less for each; how many did he buy? Let x represent the number of sheep, 80 Then will - be the price of each, 80 And + 0= price of each, if x + 4 cost 801. x+ 4 80 80 But -_ + 1, by the question, x x+4 80x Or, 80 4 + x, by multiplication. And 80x + 320 = 80x + x'2 + 4x, by the same, Or, by leaving out 80x on each side, x2 + 4x - 320, Whence x -- 2 +- V (4 + 320) - -2 + 18, by the rule, And consequently, x -- 16 the number of sheep. 6. It is required to find two numbers, such that their sum, product, and difference of their squares, shall.be all equal to each other. Let x = the greater number and y = the less. rhen x y = x- y2 by the question. x - y Hence 1 = — -- -x-y, orx- y -- 1, by 2d equation. And (y + 1) + y- = y (y + 1), by 1st equation, That is, 2y+ 1 -y2 y; y2 y=l. Whence, y=1 + /( +1 + 5, by the rule, 1 1 Therefore, y = + + v5 - i.6180 3 1 And x = y + 1 = - 2- / 5 = 2.6180... Where... denotes that the decimal does not end. 124 QUADRATIC EQUATIONS. 7. It is required to fnd four numbers in arithmetical progression, such that the product of the two extremes shall be 45, and the product of the means 77. Ijet x = least extreme, and y = common difference, Then x, x + y, x + 2y, and x + 3y, will be the four nmnbers, Hence: ~(x + y)( + 2y) = 2+ 3xy + 2y = 77 by the question, And' 2y -= 77 - 45 = 32, by subtraction, 32 Or, y2 - -= 16 by division, and y = V 16 = 4. Therefore, x2 + 3xy = + 12x = 45, by the Ist equation, And consequently, x - 6 + V (36 + 45) -6 + 9 = 3, by the rule. Whence the numbers are 3, 7, 1], and 15. 8. It is required to find three numbers in geometrical progression, such that their sum shall be 14, and the sum of their squares 84. Let x, y, and z, be the three numbers, Then, cz - y2, by the nature of proportion, And x2"+ y + z2 - 84 by the question, Hence, x + z - 14 - y, by the second equation, And Ax + 2zx + z2 = 196 -- 28y + y2, by squaring both sides, Or x2 -- 2 + 2y2 = 196 - 28y + y2 by putting 2y2 for its equal 2.rz, That is, X2 ~- y2 + Z2 = 196 - 28y, by subtraction, Or, 196 - 28y = 84, by equality, 196 - 84 Hence y = 28 4, by transposition and division. 16 Again, xz = y 2 16, or x -, by the 1st equation, 16 And x + y + z - - 4 + z = 14, by the 2d equation, Or, 16 + 4z +- Z2= 14z, or z2 -l 0 — 16, Whence z = 5 ~+' V/(25 - 16) = 5 i3 = 8, or 2 by the rule9 Therefore, the three numbers are 2, 4, and 8. 9. It is required to find two numbers, such that their sum shall be 13 (a), and the sumi of their fourth powers 4721 (b). Let x = the difference of the two numbers sought, 1 1 a -1 Then will a + x, or - the greater numberk QUADRATIC EQUATIONS. 125 a aor the less, And 1 a - x, or the less) But (a -6)4- 6 = b, by the question, Or (a + r)4 q+ (a - X)4 - 16b, by multiplication, Or 2a4 + 12a2X2'4 24 = 16b, by involution and addition, And x4 + a 622 - 8b - a4, by transposition and division, Whence 2 — 3a2 -+ v (9a4 +- 8b - a4) - 3a + v 8 (a4 + b), by the rule, And ax- V/ [- 3a2 + 2 V 2 (a4 + 6)], by extracting the root, Where, by substituting, 13 for a, and 4721 for b, we shall have x = 3, 13+ 1x t6 Therefore - -+ - 2 8, the greater number, 2 2 13 — x 10 And - 5, the less number. 2 The sum of which is 13, and 84 + 54 = 4721. 10. Given the sum of two numbers equal s, and their product =p, to find the sum of their squares, cubes, biquadrates, &c. Let x and y denote the two numbers; then (.) x + = s, (2.) y =p. From the square of the first of these equations take twice the second, and we shall have (3.) x? + y2 _ S2 _ 2p- = sum of the squares. Multiply this by the 1st equation, and the product Wvill be x3 + xy2 _+ ocy + y3 = S3 _ 2sp. From which subtract the product of the first and second equations, and there will remain (4., ax3 - y S3 - 3sp = sum of the cubes. Multiply this likewise by the 1st, and the product will be x4 + Xy3 + axy + y4 = S4- 3s2p; from which subtract the product of the second and third equations, and there will remain (5.) X4 + y4 - s - 4s2p + 2p2 = sum of the biquadrates. And, again multiplying this by the 1st equation and subtracting from the result the product of the second and fourth, we shall have (6.) a5 + y5 -_ 5-3 - 5sp + 5sp2 = sum of the fifth powers. And so on; the expression for the sum of any powers in general being xca' + yn = s- - msm-2p + (m-3) m-4 m(m -4)(m-5) + (m - 5)(m - 6)(m - 7) 4 2'3 2-3'4'l'l q 126 QUADRATIC EQUATIONS. &c Where it is evident that the series will terminate when the index of s becomes = 0. EXAMPLES FOR PRACTICE. 1. It is required to divide the number 40 into two such parts, that the sum of their squares shall be 818. Ans. 23 and 17. 2. To find a number such, that if you subtract it from 10, and then multiply the remainder by the number itself, the product shall be 21. Ans. 7 or 3. 3. It is required to divide the number 24 into two such parts, that their product shall be equal to 35 times their difference. Ans. 10 and 14. 4. It is required to divide a line, of 20 inches in length, into two such parts that the rectanole of the whole and one of the parts shall be equal to the square of the other. Ans. 105 - 1Q, and 30-10 /V5. 5. It is required to divide the number 60 into two such parts, that their product shall be to the sum of their squares in the ratio of 2 to 5. Ans. 20 and 40. 6. It is required to divide the numbqir 146 into two such parts, that the difference of their square roots shall be 6. Ans. 25 and 121. 7. What two numbers are those wvhose sum is 20 and their product 36? Ans. 2 and 18. 8. The sum of two numbers is~ 1 and the sum of their reciprocal 3L-; required the numbers. Ans. - and-. 9. The difference of two numbers is 15, and half their product is equal to the cube of the less number; required the numbers. Ans. 3 and 18. 10. The difference of two numnbers is 5, and the difference of their cubes 168.5; required the numbers. Ans. 8 and 13. 11. A person bought cloth for 331. 15s., which he sold again at 21. 8s. per piece, and gained by the -bargain as much as one piece cost him; required the number of pieces.. Ans. 15. 12. What two numbers are those, whose sum, multiplied by the.,greater, is'equal to 77, and whose difference, multiplied by the, less, is equal to 12? Ans. 4 and 7. 13. A grazier bought as many sheep as cost him 601., and after reserving 15 out of the number, sold the remainder for 541., and gained 2s. a head by them: how many sheep did he buy? Ans. 75. 14. It is required to find two numbers, such that their QUADRATIC EQUATIONS. 127 product shall be equal to the difference of their squares, and the sum of their squares equal to the difference of their cubes. _/Ans. 1A/5 and J(5 +- /5). 15. The difference of two numbers is 8, and the difference of their fourth powers is 14560; required the numbers.* Ans. 3 and 11. 16. A company at a tavern had 81. 15s. to pay for their reckoning; but before the bill was settled, two of them went away; in consequence of which those who remained had 10s. apiece more to pay than before; how many were there in company? Ans. 7. 17. A person ordered 71. 4s. to be distributed among some poor people; but before the money was divided, there came in, unexpectedly, two claimants more, by which means the former received a shilling apiece less than they would otherwise have done; what was their number at first? Ans. 16 persons. 18. It is required to find four numbers in geometrical progression such, that their sutn shall be 15, and the sum of their squares 85. Ans. 1, 2, 4, and 8. 19. The sum of two numbers is 11, and the sum of their fifth powers is 17831; required the numbers. Ans. 4 and 7. 20. It is required to find four numbers in arithmetical pro* In solving this question, the reduced equation, found by the usual methods of operation, will be of the form 23 + aZ - b; which is a cubic equation, and therefore cannot be resolved by the ordinary rules of quadratics; but such equations may sometimes be reduced to the form of a quadratic, and then resolved according to the rules already given. Whenever, in a cubic equation of the form X 3+ ax = b, b can be divided into two factors m and n, so that mi2+ a = s, then the cubic equation can be resolved as a quadratic: thus, in the cubic equation x3 + 6. = 20, 20 = 2 X 10, and 22 6 = 10. Now, multiplying both the sides of the equation by x, we have x4 -- 6x2 = 10 X 2x, adding (2x)2 to both sides, X4 + 10s X = (29z)2 + 10 (22x);. completing the square, X +- 10.2 -+- 25 = (2xz)2 +- 10 (2x) — 25, and extracting the root, x2 — 5 = 2x + 5;.'. by transposition, xu2 = 2x, and x.= 2, or - 0. This method, as well as some other similar artifices, is of no utility when the divisor has not integral roots, and even then it can be resolved more readily by Newton's Ml]ethod of Divisors. It is proper to observe, that cubic equations of the, form x3 + azx2 + bx - c, may be also exhibited under the form of a quadratic, from which, by completing the square, the value of the unknown quantity will be determined. For instance, the cubic equation -3 + 2ax2 + 5a2 + 4a3 = 0, may be reduced to the form (x2 + ax)2 + 4a22 (- + ax) = 0; thus, multiply the given equation by x, we have x4 + 2ax3 + 5a2x2 + 4a3 x= 0; which may be readily exhibited under the above form. See Ryan's Elementary Treatise on Algebra, Practical and Theoretical. (Art, 423.).-ED. l28 WQUADRATIC EQIVATIONS. gression, such, that their common difference shall be 4, and their continued product 176985. Ans. 15, 19, 23, and 27. 21, Two detachments of foot being ordered-to a station at the distance of 39 miles from their present quarters, begin their. march at the same time; but one party, by travelling - of a mile an hour faster than the other, arrive there an hour sooner; required their rates of marching. Ans. 31 and 3 miles per hour. 22. It is required to find two numbers, such, that the square of the first plus their product shall be 140, and the square of the second minus their product 78. Ans. 7 and 13. 23. It is required to find two numbers, such that their difference shall be 13-, —, and the difference of their cube roots 1~. Ans. 155, and 21o 24. It is required to find three numbers in arithmetical progression, such that the sum of their squares shall be 93; and if the first be multiplied by 3, the second by 4, and the third by 5, the sum of the products shall be 66. Ans. 2, 5, and 8. 25. The sum of three numbers in harmonical proportion is 191, and the product of the first and third is 4032; required their numbers. Ans. 72, 63, and 56. 26. It is required to find four numbers in arithmetical progression, such that if they are increased by 2, 4, 8, and 15 respectively, the sums shall be in geometrical progression. Ans. 6, 8, 10, and 12. 27. It is required to find two numbers, such, that if their difference be' multiplied into their sum, the product will be 5; but if the difference of their squares be multiplied into the sum of their squares, the product will be 65. Ans. 3 and 2. 28. It is required to divide the number 10 into two such parts, that if the square root of the greater part be taken from the greater part, the remainder shall be equal to the square root of the less part added to the less part. Ans. 5 + V 19 and 5 -:/V19. 29. It is required to find two numbers, such that if their product be added to their sum, it shall make 61; and if their sum be taken from the sum of their squares, it shall leave 88. Ans. 7- V+ 2 and 7 -'2. 30. It is required to find two numbers, such that their difference multiplied by the difference of their squares, shall be 576; and their sum multiplied by the sum of their squares, shall be 2336. Ans. 5 and 11. 31. It is required to find three numbers in continual proportion, Whose sum shall be 20, and the sum of their squares 140. Ans. 6- + V 3,-, 61, and 63 - V 3-5. .CUBIC EQUATIONS. 129 32. It is required to find two numbers whose product shall be 320, and the difference of their cubes to the cube of their difference, as 61 is to unity. Ans. 20 and 16. 33. The sum of 700 dollars was divided among four persons, A, B, C, and D, whose shares were in geometrical progression; and the difference between the greatest and least, was to the difference between the two means, as 37 to 12 What were all the several shares? Ans. 108, 144, 192, and 256 dollars. OF CUBIC EQUATIONS. A cubic equation is that in which the unknown quantity rises to three dimensions; and, like quadratics, or those of the higher orders, is either simple or compound. A simple or pure cubic equation is of the form b b ax3 = b, or X3 where x -3V-, a a A compound cubic equation is of the form 3 ax b, X - ax2 - b, or x3 + ax2 + bx - c, in each of whlich the known quantities a, b, c, may be either + or-. Or either of the two latter of these equations may be reduced to the same form as the first, by taking away its second term; which is done as follows:RULE.-Take some new unknown quantity, and subjoin to it a third part of the coefficient of the second term of the equation with its sign changed; then, if this sum, or difference, as it may happen to be, be substituted for the original unknown quantity and its powers in the proposed equation, there will arise an equation wanting its second term. Note. —The second term of any of the higher orders of equations may also be exterminated in a similar manner, by substituting for the unknown quantity some other unknown quantity, and the 4th, 5th, &c., part of the coefficient of its second term, with the sign changed, according as the equation is of the 4th, 5th, &c. power.* ~ Equations may be transformed into a variety of other new.equations, the principal of which are as follows:1. The equation.z4 - 4x3 - 19x2 + 106 - 120 - 0, the roots of which are 2, 3, 4, and -5, by chaliging the signs of the second and fourth terms, becomes x4 +- 4.3 - 19x2 — 106 -- 120 = 0, the roots of which are 5, — 2, -3, and -4. 2. The equation x3 + x2 - 10. +- 8 = 0, is transformed, by assuming x ='y-4 into b3 — ll1y2 + 30y = O, or y2 — lyj + 30 = 0; the roots of which are greater than those of the former by 4. 3. The equation x3- 6x2 - 9.- 1 =_ 0, may be transformed into one 1.30 CUBIC EQUATIONS. EXAMPLES. 1. It is required to exterminate the second term of the equation Xa3 +'Sax2 = b, or X3 3ax2 -- b = 0. 3a Here x = z - -~- = z - a, x3r 3 3 - 3az2 + 3a2z - a Then - 3a z2.- 6a2z + 3a3 b -b Whence z3 - 3a2z + 2a3 - b = O, Or z3 - 3a2 = b - 2a3, -in which equation the second power (Z2), of the unknown quantity, is wanting. 2. Let the equation 3-_ 12x2 4- 3x - 16, be transformed into another that shall want the second term. Here x z- +4. (z~+4)3=z3+ 12,at+48z +64 Then - 12(z + 4)2= _ 122 - 96z - 192_ +3(z+4)= 4) +3z+12 Whence z3 - 45z - 116 = - 16' Or, z3 - 45z = 100 which is an equation where _2, or the second term, is wanting, as before. 3. Let the equtation - 6x2 = 10, be transformted into another that shall want the second term. Ans. z3 - 12z - 26. 4. Let y3 - 15y2 + 8ly = 243, be transformed into an equation that shall want the second term. Ans. 3 - 6x = 88. which shall want the third term, by assuming x = y + e, and in the resulting equation, let 3e2 - 12e - 9, or c2 —4e 3 — 0, in which the values of e are 1 and 3; then assume x = y+- 3, or y + 1, and the resulting equation will be y3 - 3y2- I = 0, an equation wanting the third term. 4. The equation 6x2 - 61xz2 6-=-1 0,j by assuming x = -, may be transformed into.3 - 6Y + lly — 6 = 0; the roots of which are to be reciprocals of the former. 5. The equation 3x3 - 13x2 + 14xl + 16 = 0, by assuming x = 3, may be transformed into,/3 3- 13y2 + 42y + 144 = 0, the roots of which are three times those of the former.-ED. CUBIC EQUATIONS. 131 3 x2 7 9 5. Let the equation x3 - - 2+ - 0, be transform4 8 16 ed into another that shall want the second term. 11 3 Ans. y3 +y-t Y = 46. Let the equation x4-+ 8 3-5e-+ 10x -4 0, be transformed into another, that shall want the second term. Ans. y4 _ 29y2 + 94y;- 92. 0, 7. Let the equation x — 3x + 3x 2 - 5x - 2:- O, be transformed into another, that shall want the third term. Ans. y4- +y3 - 4y - 6 0. 8. Let the equation 3x3 - 2x + 1 0, be transformed into another, whose roots are the reciprocals of the former. Ans. y3 - 2y2 _+ 3 = 0. 9. Let the equation 2-~4 -1~ x -- _ + 1_- O, be transformed into another, in which the coefficient of the highest term shall be unity, and the remaining terms integers. Ans. y4 - 3y3 + 12y2 - 162 y + 72 0. OF THIE SOLUTION OF CUBIC EQUATIONS. RULE. —Take away the second term of the equation when necessary, as directed in the preceding rule. Then, if the numeral coefficients of the given equation, or of that arising from the reduction above-mentioned, be substituted for a and b in either of the following formulae, the result will give one of the roots, as required.* e If, instead of the regular method of reducing a cubic equation of the general form X3+a z2 +bx+ = 0, to another, wanting the second term, as pointed out in the preceding article, there be put x = I (y — a), we shall have, by substitution and reduction, y3 +-(9b - 3a2) 9ab - 27c - 2a3; where, since the value of y can be determined by either of the formula given in this rule, the value of x will also be known, being x- 3 (y-a).: And if b = 0, or the original equation be of the following form, x3 -ax2 -+c - 0, the reduced equation will be y3 - 3a2y= - 2a3 - 27c, where the value of y, being found as above, we shall have, as before, x - 3 - a), which formulra, it may be observed, are more convenient, in some cases, than those restlting~ from the preceding article; as the coefficients, thus obtained, are always integers.; whereas, by the former method, they are frequently fractions. 132 CUBIC EQUATIONS. +3 + ax = 6. X or 2 2 4 f27) 3 b b24 2 a Where, it is to be observed, that when the coefficient a, of a3 the second term of the above equation, is negative, 27 as also 3, in the formula, will be negative; and if the absolute term b be negative, 2 in the formula will also be negative;'b2 but will be positive.* It may likewise be remarked, that when the equation is of the fornm X3 - ax = b, T* his method of solving cubic equations is usually ascribed to Cardan, a celebrated Italian analyst of the 16th century; but the authors of it were Scipio Ferrcus and Nicholas Tartalea, who discovered it about the same time, independently of each other, as is proved by Montucla, in his Histoire des Mliatltematiqwes, Vol. I. p. 568, and more at large in Hutton's Mathematical Dictionary, Art. Algebra. The rule above given, which is similar to that of Cardan, may be demonstrated as follows:Let the equation, whose root is required, be.3 - a.v = b, And assume y + z = x, and 3yx =-a. Then, by substituting these values in the given equation, we shall have 3+3y/2z + 3yz2+z3 +a X (zy3 - z) = y3 z- 3+ 3yz X (y + z) a X (y —Z) y3- -a X ( -- z) - X (y z) = b, or y3 _. And if, fr'om the square of this last equation, there be taken 4 times the cube of the equation yz = — a, we shall hlave y6 —2y3z3 -+ z = b2 +- 24 3, or y —3 = 3 (b2 + 24 a3). But the sum of this equation and y3 + z3 = b, is 2y3 = b + -/ (b2 + 4 4 24 a3), and their difference is 2z3 = b - V (b3+ 7 4a3); whence y 3 P[b - a/ (b3 + — 7 a3)], and z = 3V [hLb - V (b2 + -i- a3)]. From which it appears, that y +-z, or its equal x, is 3 [b3- + V (4b2~ + -' aa)] — 3' [' - (~b2- + -. a3)], which is the theorem; Or, since z is -- it will be y a- =y- or x3y' 33 tb-f, V(4b2-+ va3)] - -a, the same as the rule. CUBIC EQUATIONS. 133 b and 27is greater than or 43 greater than 2'7b2 the solution of it cannot be obtained by the above rule; as the question, in this,instance, falls under what is usually called the Irreducible Case of cubic equations.* EXAMPLES. 1. Given 2X' - 12x2 + 36x =-= 44, to find the value of x. Here x3 - 6X2 + 18- = — 22, by dividing by 2. And, in order to exterminate the second term, Put X = z +- =z +2, (z + 2)3 = z3 + 622 q+ 12z + 8 Then - 6 (z -- 2) = 6z2-24z -24 = 22, 18(z+-2) = 18az+36 Whence a3 + 6z + 20 = -22, or za + 6z = 2, And consequently, by substituting 6 for a and 2 for b, in the first formula, we shall have 2 4 216~ 2 4 216 v2 FV4 27 ) /T 2 ~4 -27 * It may here be farther observed as a remarkable circumstance in the history of this science, that the solution of the Irred'ucible Case above-mentioned, except by means of a table of sines, or by infinite series, has hitherto baffled the united efforts of the most celebrated Mathematicians in Europe; although it iswell known that all the three ioots of the equation are, in this case, real; whereas in those that are resolvable by the above formula, only one of the roots is real; so that, in fact, the rule is only applicable to such cubics as have two equal, or two impossible roots. The reason why the assumptions, made in the note to the former part of this article with respect to the solution of the equation x3 - ax - b, are found to fail in the case in question (and it does not appear that any other can be adopted) is, that the two auxiliary equations, 3yz = - a and y3 - -3 = b, which in this case become 3yz = a, and y3 + z3-=-, a3 ~ or 33 = —, and y3 — 3 =, cannot take place together; being in27 consistent with each otber. For the greatest product that can be formed of the two quantities V3 +- z3 is -when they are all equal to each other; or since y3 + Z3 b, when each of these = lb; in which case their product is = b2. a3 a3 But, as above shown, Y3Z3 = - by the question; therefore, when -7> 27 27 the two conditions are incompatible with each other; and conequently 4, the solution of the problem, upon that supposition, can only be obtained by imaginary quantities. 12 i134 CUBIC EQUATIONS. v[1 +(1 +8)] + [1 -(1 +8)]= V(1 +V9)+ V (1 —V 9)= (1 + 3) + 3 (1 - 3) = 32V 4-V 2, Therefore.x = z -+ 2 =- / 4 - 3 2 + 2 = 2 + 1.587401 - 1.259921 = 2.32748, the answer. 2. Given x3 - 6x = 12, to find the value of x. Here a being equal to - 6, and b equal to 12, we shall have, by the formula, — 2 x-=-V [6-3+-V(36-8)] — 6 + 6V:(36 - 8) = V (6 + V28) + 3(6 - -- 28)V 3 (6- +5.2915) + V(6 + v28), 2 2 (6 + 2 V (11.2915) + - = 2.2435 + 2 2-.2435 - 2.2435 +.8957 = 3.1392, Therefore'x = 3.1392, the answer. 3. Given 3 - 2x -- 4, to find the value of x. Here a being - -2, and b = - 4, we shall have, by the formula, x=V- -2+ V(4- + 2 -2-V((44-, or 10 10 by reduction, V (- 2+ - /3) — (2 + - 3)= "V ( —2 + 1.9245)- -V(2 +1.9245)= - ('- 0755V/ 3.9245 -.41226 — 1.5773 - 1.9999, or - 2; Therefore x - 2, the answer.* Note.-When one of the roots of a cubic equation has been found, by the common formula as above, or ini any other way, the other two roots may be determined as follows:-'Let the known root be denoted by r, and put all the terms of the equation, when brought to the lefthand side. = 0; then, if the equation, so formed, be divided by ax:i: r according as r is positive or negative, there will arise a quadratic equation, * When the root of the given equation is a whole number, this method only determines it by an approximation of 9s, in the: decimal part, which sufficiently indicates the entire integer; but in most instances of this kind, its Value may be readily found, by a few trials, from the equation itself..Or if, as in the above example, the roots, or numeral values of 3%/ 10 10 (-2:+'-. 3), and-3-V(2 + 9V 3), be determined -according to the:rule laid down in Surds, Case 12, the result will be found equal to — 2 as it ought. CUBIC EQUATIONS. 135 the roots of which will be the other'two roots of the given cubic equation. 4. Given x3 - 15x = 4, to find the three roots or values of x. Here x is readily found, by a few trials, to be equal to 4, and therefore x- 4) x3- 15wx- 4(x2 + 4x + 1 X3 - 4xw2 4xs - 15x 4x2 -- 16x x-4 x — 4 Whence, according to the note already given, x2 + 4x + 1= 0, or x + 4x = - 1; the two roots of which quadratic are - 2 + V 3 and - 2V 3; and consequently 4, -2 V/ 3, and — 2 —V/3, are the three roots of the proposed equation. EXAMPLES FOR PRACTICE. 1. Given x3 + 3x2 - 6x = 8, to find the root of the equa. tion, or the value of x. Ans. x -- 2. 2. Given x3 + x2 = 500, to find the root of the equation, or the value of x. Ans. x — 7.616789. 3. Given 3 _- 3x2 = 5, to find the root of the equation, or the value of x. Ans. x = 3.425101. 4. Given 3 - 6x - 6, to find the root of the equation, or the value of x. Ans. V4 -+ 3V2. 5. Given X3 + 9x = 6, to find the root of the equation, or the value of x. Ans. V 9 - 3V 3. 6. Given X3 + 2X2 - 23x = 70, to find the root of thee quation, or the value of x. Ans. x = 5.134599. 7. Given a3 - 17x2 + 54x = 350, to find the root of 1he equation, or the value of x. Ans. x = 14.954068. 8. Given a - 6x 4, to find the three roots of the equation, or the three values of x. Ans. -2, 1+v3, andl —V3. 9. Given x3 — 5x +- 2x-= - 12, to find the three roots of the equation, or the three values of x. Ans. - 3, 1 + V/ 5, and 1 - V 5. 136 CUBIC EQUATIONS. OF THE SOLUTION OF CUBIC EQUATIONS, BY CONVERGING SERIES. THIS method, which, in some cases, will be found more convenient in practice than the former, consists in substituting the numeral parts of the given equation, in the place of the literal, in one of the following general formulae, according to which it may be found to belong, and then collecting as many terms of the series as are sufficient for determining the value of the unknown quantity, to the degree of exactness required.' 1. + ax =:,b.t 2b 1 2.5 2762 2.5.8.11 V [2(27b2 + 4a3)] 6.9 27b 4 - 6.9.12.15 * The method laid down in this article, of solving cubic equations by means of series, was first given by NiCoLE, in the Memoirs of the Academy of Sciences, an. 1738, p. 99; and afterwards, at greater length, by CLAIRAUT in his Elemens d'Algebre. t With respect to the determination of'the roots of cubic equations by means of series, let there be given, as above, the equation x3-+ax = b, where the root by transposing the terms of each of the two branches of the common formula, is Xz= 3 / V(ib2 +-27 a3) +n-Vb X (iS + -!3)- b }; - orby putting, for the sake of greater simplicity, V (4b2- - 4- a3) = s, and reducing the expression, x = sa Hence, extracting the roots of the righthand member of this equation by the binomial theorem, there will arise 3/ 1 + 2) = 1 (bf2.5 (b \3 2.5.8 (1)\4 1+ b 2 ( \2 2.5 ( 3 2.5.8 J'(1+ \2s31 3.6 3,.6.9 \2s) 3.6 9.12 2s \ o, b 4 2s) -, &C. And consequently, if the latter of these two series be taken from the former, the result, by making the first term of the remainder a multiplier, will give 1 2b6s 1 1-6.9 2s + &c., where, since s = V b2- a3, we shall have ( ) 2s =2 ab, 2 b-4a3' b 4 27b 2 2bsa 2b 2b \2s = \27- 4a3. And 6s [2(27b2 + 4 aS)]' Whence, also, by substitution, we have the above formula CUBIC. EQUATIONS. 137 -27b' ~2 2.5.8.11.14.17 ['27b2 27 4aJ c 6.9.12.15.18.21 27b + 4a3 Or 2b 2__5 1 27b62 8.11 x / [2(27b2 + 4a3)] 6.9 27b2 + 4a3' 12.15 27b2 14.17 / 27b 2 20.23 27 71B 7 (2 + 4C 1+ 2 27b + ] 4a3- 12.21 27b2 + 4a/ 24.27 ( 27b2 272 { 4a3/D, &c. In which case, as well as in all the following ones, A, 13, C &c., denote the terms immediately preceding those in which they are first found. 2. x3 - ax = -b, where lb2 is supposed to be greater than La, or 272 > 4a3. b =2 27b 2-4a3 2.5.8 27b2 - 4a3 x -(2 v/ I 2 2 1-3.6( 272 )7 - 3.6.9.12 27b2 2.5.8.11.14 27b - 4a)3 & *, 3'.6.9.12.15.18 27b2 &c. Z~i ~b C 2 27b2 -4a3 5.8 27b2- 4a3 i 2 3 1 -..6 ( 27b2 ) 9.12 27b2 11.14 27b - 4a 17.20 27b2- 4oa3.-~( 7 C ~ —.4 ) D -, &c. 15.18 272h 21.24 27b2 In which case the upper sign must be taken when b is The root, as found by the common formula, when properly reduced, is i+ b- %/(1b2-_La2)j b 2 (Vb2 - 2 a3)]t Or, putting, as in the last case, - /(46b2 -- a3), or its equal 27b3 )= s, we shall have x =:: 32 +V ) 1-s Whence, extracting the roots of the righthand member of this equation, there will arise /(l1+-s) = 1 2+ Is' - -'3 6 - -.5.8 s, 2 2.5 2.5.8 &tC. wV(l — ~s) -— Is- -"s-.- -. —. 2.. 4 S -, &C. &.>1g3.6 3.6.9 3.6.9.126 And, consequently, by adding the two series together, and taking the 1st term of the result as a multiplier, we shall have x = i: 2 1- 3 - 21 3.6 -.568 2-5 &811.14 b &. Or, & Orby substituting a 3.6.9.12s 3,6.9eu2.15.18'treybusittn for its equal s, we get the above expression. 12* .138 CUBIC EQUATIONS. positive, and the under sign when it is negative; and the same for the first root in the two following cases:3. x3 - ax-=:~ b, where lb2 is supposed to be less than 27 aI, or 27b2 < 4a3. b 2 4a3 - 27b2. 2.5.8 4a3 - 27b2) =2 3.6 27b2 3.6.9.12 27b2 2.5.8.11.14 4a3 — 27b) -_ &.. Or, 3.6.9.12.15.18 272 * This expression is obtained from the last series, by barely changing the signs of the numerator and denominator in each of its terms; which does not alter their value. Hence, in order to determine the other two roots of the equation, let that above found, or its equivalent expression, -V Ib + -V(4b- -1- a3) + v b —/ (b2 —7 - ) =_ i r. - Then, according to the formula that has been before given for these roots in the former part of the present article, we shall have z —F:[ i V[- 3 /[(b + v( 2b2 - a3)]-. /[bb-V (~b2- L X3)] }1 Or, 4 22 2 putting b v (b2- 1, a3) = s, and reducing the expression, x- F -:: 2 3(1 + s) - V/(1- s). Whence, extracting the cube roots of the righthand member of this equation, there will arise 2 2.5 2.5.8 3V (1 —S) = 1 —- S2 3 - S-, &c. 3.6 3.6.9 3.6.9.12 And, consequently, by taking the latter of these series from the former, and making the first term of the remainder a multiplier, we shall have = =F r (sib) V-3 2512.5 2.5.8.11 25111417 1:i(~b~[v -t.-: sj..l_ e si4-, &C. 2 3 6.9 6.9.12.15 6.9.12.15.18.21 ~ 2 _f27b2 -4a3 / 272-43 But since s= v (~b' — 2 a3) 2 s7 = 27b2 27 2 b 27b2 4a3-2712 T 4a30-M2 &c, and also V!bXV 3 Mb2' -b- and also 3 b X/ —3.27b — 4a3 v X V 4a3 —27b2 A/(4a3 —27b 2) -b, if these values be substituted for their equals, in the last series, the result will give the above expressions for the two remaining roots of the equation. CUBIC EQUATIONS. 139 b 2 43 2 5.8 4a' — 272) 36 27b ) 9.12 27b2 11.14 4a3 - 276 17.20 4a3 - 27b2 B + 15.18k 27b 22 21.24 27b &. which series answers to the irreducible case, and must be used when 2a3 is less than 27b/. And if the root thus found be put = r, the other two roots nmay be expressed as follows: — r V/(4a3 - 27) 1 2.5 4a3 - 27b2\ 2.5.8.11 2 9V 2/)2 6.9k 2t72 6.9.12.15 4 a: - 27b. 72 _2.5.8.11.14.17 4a3 -27b2" 3. 27b2 3.6.9.12.15.18.21 27b2 Or, z —:r V/ (4a3 -27b2) 2.5 4a - - 27b2 8.1 ] 2 9 3 2b2 6.9 27b2/ 12.15 4a3 — 27b2 14.17 4a3- 27b2 20.23 4a - 27b2 24B - C2 L 27b2 18.21 27b2 c+ 24.27k 27b2 D -, &C. Where - -r, or + -r, must be taken according as b is positive or negative; and the double signs::~ must be considered as + in one case, and - in the other, as usual..4. x3- ax —:] b, where lb2 is still supposed to be less than -1a7, or 27ba < 4a3. 4.x 2b 6 2.5 27b2 3,/ [2(4a - 27b2)] 6.9 4a3 —27b2 2.5.8.11 27b2 2.5.8.11.14.17 27b2 6.9 12.15 4a'3 —272 - 6.9.12.15.18.21 4a3-27);b2 &c.. *Or, * By transposing the terms of the common formula, as in the first case, we shall havez x= V (tb2 —l, a3) -ibj -v' I (/b2 - 1 a3) l-b }. Or, by putting, for the sake of simplicity as before, V (4b2 aa) s, and reducing the equation x= s (12 )- P Whence, extracting the roots of the righthand number, as in the fornler instances, 3l.:_ b 2 (b 2- 2.5 )...8 b._. /\ 2s T \2S/ 3.6\2Ss 3.6.9 2.9. 12S, 140 CUBIC EQUATIONS. 2b 2.5 27b2 8.11 X =:= ~/[2(4- 27b)] 1 6.9 4a- 27i2) 12.15 27b2 14.17 27'b2 20.23 27b2 4c3 - 27b2B 18.21 (4a32 b27) + 24.27 4 - 27b2) b / 2 b 2.5 /b 2.5.8 /b\ I- 3 11- iI —- 1 12 -- 1-132s) 3 (2s) 3. 6 (2S2 3.6.9 \2s 3.6.9.12 3s-' &e. And consequently, by taking the latter of these series from the former, and making the first term of the result a multiplier, we shall have s 2-.5sb + 2.6.811 b{ ]+ 2.5.8.11.14.17 b + =6s 6.9 2s ] 6.9.1-2.15 2sj 6.9.12.15.18.o21 k2s )P-, l b 2,7br2 &c. 5. But since s = V (ib2 — a3), we shall have ) 2 = 27b2 /b \ / 27b2 b 2bs3 2b -- 4a3- 27\)4 4a3-27b2' &c., and 6s - 2b 2b 63V (42-217 a3) v 2(4a3 - 27b2) Whence, if these values be substituted for their equals in the last series, there will arise the above expression for the first root of the equation. And if we put the root thus found, or its equiValent expression: V(b2 24 7 a3)+1nb }-3 v/ (b2 1- a3)- ib }-4 r we shall have, accordingto the formula before given for the other two roots, 4 i'Ot3 j x=T'F 2~ r23/[V b- 2 I - ( 1 a3) + [] + V [ (Ib2-1 a3) -'bj] I. Or, taking, as before, V (4b2- x1 ba)=s, and simplifying the result, x-=FF 2 s 31 + + ( Whence, by extracting the roots of the righthand side of this equation, there will arise lb b\2/b \25 b 2.5.8 V\ 2+ 2s 625 3.62s)/ 3.6.9 2s) 3.6.9.12() &c. 8y (i b)=1 - -: ( bs ) —3.6\- 2- 3.6,9 3.6.9.12( s) - &c. And, consequently, if the latter of these series be added to the former, we shall have, by making the first term of the result a multiplier, r6 2 2'.5 5 6 2.5.8.11.14 (= )F-+,&; b since: s- - ).4 ) s f~2 (36 2s'3:6.s3.6.9.12 ('b 6 &-, &c, But since s=V/(-142 — -L ( 2biT2i=) e, r shall 2s. 4.27 CUBIC EQUATIONS. 141 D -, &c., which series also answers to the irreducible case, and must be used when 2a3 is greater than 27b2. And if the root thus found, be put = r, as before, the other two 4a-3 - 27b2 {I roots may be expressed thus: x - = F 2 i 1+ 2 27b2 2.5.8 27b2 2.5.8.11.14 3.6 4d'- 27b. -3.6.9.12 74a3- 27b2 3.6.9.12.15.18 2762 (;4a_ 2.7b2), &c. Or, r 4-4a3 27b2 2 27b2 5.8 X:= =F - 6V -1 I- (4 a -- 2 4 + -3 x4a- 27b) 9.12 27b2 11.14 27b2 17.20 27b2 ( --— ) + - B ( — )C- 2( ) 4-a3 27b 125.18 4a3 - 2b2 21.24 4a3 - 27b2 D +, &c. Where the signs are to be taken as in the latter part of the preceding case. EXAMPLES. 1 Given x3 + 6x = 2, to find the value of x. Here a-= 6, and b = 2, whence 27ba2 27 X 4 1 1 -7a -!- - 4a -and 27b2+74a3 27X 44 +4 216 1 +8 9' 2b 4 L/[2(27ba +4a')] V3, [2(4 X 27 + 4 X 216)] 4 4 2 3/81 V648= +8 )-3 - =-h -.-~- Consequently, 2 V(27 + 8 X 27) 6 /9 27 27 by formula 1, we shall have 1 1.0000000 (A) 2.5 1 -.9x- A.0205761 (n) 6.9 9 8.11 1 — X- ~B.0011177 (c) 12.15X 9 14.17 1 141X-c.0000782 (D) also have ( )2- 27- 27b-a &c., and consequent. 2s 27b2- 4a3 43 - 272 {' ~y- 74as 27b,2 ly, s — 3=V -/( - ) Hence, if these values be substituted for their equals in the above series, the result will give the above expressions for the two remaining roots of the equation. 142 CUBIC EQUATIONS. 20:23 1 — X- D.0000062 (E) 24.27 9 26.29 1 3.} E..0000005 (F) 30.33 9 1.0217787 Log. 1.0217787 0.0093570 Log.,/648 0.9371916 Colog. 27 8.5686362 No. 3274801 - 1.5151848. Therefore x --.3274801 2. Given x3 -- 9x - 12, to find the real value of x,. Here a = 9 and b = 12; 12 27ba - 4ac3 27X 144-4X273 whence 23 —- 2 6 and 2 -V27b2 27 X 144 144 - 108 36 1 144 144 4 Consequently, by formula 2, we shall have 1 1.0000000 (A) 2 1 - X-GX (A) -.0277778 (B) 5.8 1 5.8X1(-) -.0025720 (c) 9.12 11.14 1 x —X (c) -.0003667 (D) 17.20 1 1 2 yX- (D) -.0000619 (E) 23.26 1 2-26 X I (E) -.0000114 (F) 29.32 1 -- — X- (F) —.0000022 (G) Sum.0307920 Comp..9692080 Log. 969208 -- 1.9864137 CUBIC EEQUATIONS. 143 Log. 2 V36, or Log. V 48 0,5604137 No. 3.522334.5478274 therefore x = 3.522334. 3. Given X3 - 12x = 16, to find the three values of x. aIere c-= 12 and b = 15; b 15 4a3 -:27:b2 Whence 2 3/ - = 2 3V- =-3V 60 and 2 2 27b 2 4.123 - 27.152 256 - 225 _ 31 27.152 225 225 Consequently, by formula 3, we shall have 1 +1.0000000 (A) 2 31 +- -X- A +0.0153086 (B) 3.6 225 5. 8 31 -- X B — 0.000781-2 (c) 9.12 225 11.14:31 - x.4 X31 c +0.0000614 (D) 15.18 225 17.20 31 21720 X22 D -0.0000057 -(r,) 21.24X 2- ~ D2a 23.26 31 -— X -25 E +0.0000006 (F) -27.30 225 Sum of + Terms +1.0153706 Sum of - Terms -.0007869 Difference 1.0145837 Log. 1.014837.0062880 Log. V 60.5927171'No. 3.971962.5990051 Therefore the affirmative value of x or first root, r= 3.971962. V(4a3 - 27b2) V'837 _ (9 X 93) _ 93 Agaia, 9, gb~ =9V450 4a- 9'32b2 9V/450 39V/450 433 - 27b2 31 and 27b2 225 Hence, +1 1.0000000 (A) 2.5 31 i~.-X- X.. 1 -,0255144 (B) 6.9 225 i44 CUPIfC EQUATIONS. 8.11 31 + 8.1- 31 B +0017186 (c) 12.15 225 14.17 31 -— X c -.0001490 (D) 18.21 X225 20.23 31 3+'-4 2 X22 D t+.0000145 (r,) 24.27 225 26.29 31 -X E —.0000014 (F) 30.33 225 Sum.9760683 Log. 9760683 - 1.9894802 Log. V 93 0.9842415 Colog. 3V450 9.1155958 Colog. 3 9.5228787 No..4099445 - 1.6121964 r Also - -1.9859810 Last No:k:0.4099445 Result - 1.5760365 Or — 2.3959255 Whence the three roots or valucs of x are 3.971962 - 1.5760365, and — 2.3959255. 4. Given x3 - 6x = 2, to find the three values of x. -2b - 4 /Here [2(4a3 - 272)] V [2(4.63 -- 27.4)] -4 4 -2 2V49 276b 4]- -3 =- and 33V [2(4.8 - 4)] 6 7 7 321 4a3-27b 4.27 1 1 4.63 - 27.4 - 8-1 7Hence, by the formula 4, we shall have 1 1.0000000 (A) 2.5 1 6-.X-A -- 0264550 (B) 6.9 7 8.11 1 + 12.15 X- +.0018476 (c) 14..17 1 ~X- ~C ~-.0001662(D) 18.21 7 CUBIC EQUATIONS. 145 20.23 1 2+ 4 XD +.0000168 (E) 24.27 7 26.29 l - X- -.0000018 (F) 30.33 7 Sum +.9752414 Log..9752414 -.9891120 Log. 2 0.3010300 L. 3/49 0.5633987 Colog. 21 8.6777807 No. 329870 -1.5313214 Therefore one of the negative roots or values of x, is.339870 - - r. 4a3 - 27b2 4.63 - 27.4 Again, V =V, = V(63. 2-7) 6 4 4 27b2 1 -, 89, and 4a3 - 27b2 7 Hence, I1 1.0000000 (A) 2 1 + X —A +0.0158730 (B) 3.6 7 5. 8 1 -"5 - 8X- lB -.0008398 (c) 9.12 7 11.14 1 + — X1 -C +.0000684 (D) 15.18 7 17.20 1 — X- D -.0000066 (E) 21.24 7 23.26 1 -+2-f3.26 ~- E1 F +.0000002 (F) 27.30 7 Sum 1.0150952 Log. 1.0150952.0065070 Log. V 189.3794103 No. 2.431741.3859173 Therefore - +.169935 2 13 146 BIQUADRATIC EQUATIONS. Last number i2.431741 Result +2.601676 Or -2.261806 And consequently, +2.601676, -2.261806, and -.339870, are the three roots required. EXAMPLES FOR PRACTICE. i. Given x3 + 9x = 30, to find the root of tile equation, or the value of Ax Ans. x = 2.180849. 2. Given x3 - 2x 5, to find the root of the equation, or the value of x. Ans. x - 2.0945515 3. Given X3 - 3x = 3, to find the root of the equation, or the value of x. Ans. 2.103803 4. Given x3 - 27x =36, to find the three roots or values of a. Ans 5.765722, - 4.320684, and - 1.445038. 5. Given x3 - 48x2 = - 200, to find the root of the equation, or the value of x. Ans. 47.9128. 6. Given x3 - 22x = 24, to find the root of the equation, or the value of x. Ans. 5.1622'~7 OF BIQUADRATIC EQUATIONS. A biquadratic equation, as before observed, is one that rises to the fourth power, or which is of the general form - X4 -- ax3 q- bx2 q cx -- d -O. The root of which may be determined by means of the following formula; substituting the numbers of the given equation, with their proper signs, in the places of the literal coefficients, a, b, c, d. RULE 1.*-Find the value of z in the cubic equation z3 + (1ac- 1 b-d z 1 1 + + * This 108 8 24 * This method is that given by Simpson, p. 120 of his Algebra, which consists in supposing the given biquadtatic to be formed by taking the difference of two complete squares, being the same in principle as that of Feqrrari. Thus, let the proposed equation be ofthe form.4- + cz3 -+ b.X G.- + — d -=0 (1), having all its terms complete; and assume (x2 + 2- +-p)2(Ox + r ) ='l + 4. 3 ~a b.Z2 + cX + d. Then, if,2 lax +- p and q: -- r be actually involved, we shall have Yt ++.4 + |p t-,V |p. +p~ = - 4+p. X +p2+~. - 2I /. -- -4 ax3+bx2+ex+d. q2 - 2qr And consequently, by equating the homologous terms, there will arise BIQUADR-eAT1C EQUATIONS. 147 by one of the former rules; and let the root, thus determined, be denoted by r. Then find the two values of x in each of the following quadratic equations:2 +-(a +vl 2 (r- b) } ) _-(r +b) + l(r f2 4 36 + b)2 - d IX2 ~ (-a- a ~ 2 (2r - 6))=- (r+ b) - /(r and they will be the four roots of the biquadratic required. 1. 2p —-a2q -- b 2p -a -2-b = -- q2 2. ap-2qr -- c or ap-c = 2qr 3. 2-r2 =d p2-d = 1.2 where, since the product of the first and last of the absolute terms of these equations is evidently equal to 4 of the square of the. second, we shall have 2p3 + (4 a2 - b)p2 - 2dp — d (Ia2 - b) - ~ (as2p2 — 2acp + c2). Or, by bringing the unknown quantities to the lefthand side, and the known to the right, and then dividing by 2, p3 —p+2 p-i (ac -4d)p = (c2 +a2d) — bd. (2.) From which last equation p can be determined by the rules before given for cubics. And since, from the preceding equations, it appears that q = -/(2p+ a2-b) and:. = or / (p2-d), 3t is evident that the several values of z can be obtained from the quantities thus found. For, because x4 + a.x3 + bas: — cx- + d, or its equal (x2+ - ax-p)2 — (qx: -')2 - 0, it is plain that (X2 - x + p)2 -(qx + r) 2. And, therefore, by extracting the roots of each side of this equation, there will arise z-2 + az-{-p q -' r; or x2 + (L a - q). x = -p. VWhence, by substituting the above values of p, q, and r, for their equals, and transposing the terms, we shall have.2 + 2ta -r V (2p +ias-b) z +p T — (p2-d) = 0, for the case where ap - c is positive; and x2 +- tj a / 2p -+ Ia2- b) p i A(p2-d)/ 0, for the case where ap - c is negative: which two quadratics give the four roots of the proposed equation. b And by putting p = z +6- in the reducing equation (2), in order to destroy its second term, the several steps of the investigation may be made to agree with the expressions given in the above rule. 148 BIQUADRATIC EQUATIONS. EXAMPLES. 1. Given the equation x4- 10' + 35x2 - 50x + 24 = 0, to find its roots. Here a =- 10, b = 35, c — 50, and d —24. Whence, by substituting these numbers in the cubic equation, I b-~-d I 1 b Z3 +r (-4 ac 2- - d)Z - P8 +b3 (C + da- 2 (ac+ 1 12 108 +8 2)a + 8d), we shall have the following reduced equation, 13 35 23 _ - 7 12 108 which, being resolved according to the rule before laid down for that purpose, gives =,(35 — ISV 3)+ V( 35-ls - 3). But, by the rule for binomial surds, given in the former part of the work, 7 1 3V(35 + 18V - 3) —- -+ V- 39 and V(35- 18V — 3) 7 1- v 7 — 37 Wheretrfore x - + 3- / - 3 -+ -3 And if this number be substituted for r, - 10 for a, 35 for b, and 24 for d, in the two quadratic equations, 2+ (2 -+ ae +2 (r- b) ) =-(r + b) + (r + 1 b)2,d -.(5 a - / + 2 (. - b) ) x = - (r+ 6). + )2 - d, they will become, after reducing them to their most simple terms, Xi2 3x = -- 2 and xa - 7x=- 12: from the first of which 4x = - / = 2, or 1, and 2 4 2 2 7 1 7 1. from the second, x = - v 4 = 2 k 2 4 or 3; Whence the four roots of the given equation are 1. 2, 3, and 4. BIQUADRATIC EQUATIONS. 149 Or, when its second term is taken away, it will be of the formn x4 + 6b2 + cx + d 0, to which it can always be reduced; and in that case, its solution may be obtained by the following, rule: RULE 2.-Find the value of z in the cubic equation 1 1 1 1'- ( b62 _+d)Z= j-y3+ C2 - bd, and let the root thus determined be denoted by r, Then find the two values of x in each of the following quadratic equations:I i ~ + [/12 } - b) ]2 = - (r + - b) + J v(r +r b)2 - } 1 1 1 IL, 2 _ [( r. f)}]x=- ~+ b)V(r + b)2-dj, and they will be the four roots of the biquadratic equation required.* e The method of solving biquadratic equations was first discovered by Louis Ferrari, a disciple of the celebrated Cardan before mentioned; but the above rule is derived from that given by Descartes, in his Geomletry, published in 1637, the truth of which may be shown as follows:Let the given or proposed equation be X4 - +z2 + b. + c = 0, and conceive it to be produced by the multiplication of the two quadratics x 2pz — p q = 0; and z2 +j-'xs + s -0. Then, since these equations, as well as the given one, are each = 0 there will arise, by taking their product,:c 4 +(p + r) x3 -[- (s - q -[-r):2' + (ps -- qr).z+ qs = -X +aX2- bz +c. And consequently, by equating the homologous terms of this last equation, we shall have the four following equations, p+r==O; s-+qpr=a; ps+ —qr=b; qs=c; Or,' —p; s+q=- a+p2; s-q=-; qs=c. P Whence, subtracting the square of the third of these from that of the second, and then changing the sides of the equation, we shall have b2 a2 + 2ap2 + p4 — = 4qs, or 4c; or p6 + 2ap4 +- (a2-4c) p~2 = b2. p2 Where the value of p may be found by the rule before given for cubic equations. b Hence, also, since s + q - a-v pS, and s - q - -, there will arise, by p addition and subtraction, 1 1 b 1 1 b =-a+- p2+-; q= a+ -p2 — 2 2 2p 2 2 where p being known, s and q are likewise known. And, consequently, by extracting the roots of the two assumed quad. 13* 150 BIQUADRATIC EQUATIONS. Or the fo.ur roots of the given equation, in this last case, will be as follows:Z = _ A21 r b 2 3 -6 6~jf~ 2 3 6kr-,-j a = + riSV2 (?-3 b) }Jr b- -2/ [(r + 6- b)! -3}j x = 12 -2 (r - b,) } v 2 2 3 V[(r + 6) ) 2. Givept x4 + 12x - 17 - 0, to find th'e four roots of the equation. Here a-O b =, c-=12, andd=- 17; Whence, by substituting these nlumbers in the cubic equation, i3 1 12. 3 + 1. - (-b2 + d)z b- ~ _- bd, 12 108 8 3 we shall have, after simplifying the results, z2 + 17z = 18, Where it is evident, by inspection, that z = 1. And if this number be substituted for r, 0 for b, and - 17 for d in the two quadratic equations in the above rule, their solution will give x= - /2 + V/(- +- i, 18) = - v2 -+ V( - -+ 3 /V2) x= - 2 2-/ (- + v 18)= - x/2 - /(-+ ~ 3 /2) -+ 2 2 + v (_2- 18) - + V 2 + V( 3 2 —- ) x — + 2 -- 2 (- - V — 18) = + 1/V-2 - V/( 3 V 2) Which are the four roots of the proposed equation; the first two being real, and the last two imaginary. RULE' 3. —The roots of any biquadratic equation of the form x4 + Ca.X2 + bx -+ c= 0, may also be determined by the ratics, -.2+ pxE +q = 0, and x-2+ rx- s = 0, or its equal xz —px+s = we shall have= _ — p ~ (-p2- / q); x _ = p ~i V/ (4rp2 —s);.which expression, when taken in + and -, give the four roots of the proposed biquadratic as was required. * This method, which differs considerably from either of the former, consists in supposing the root of the given equation, x4. ax2 — bz+ c = 0(1), to be of the following trinomial surd form z = i/v + / q -/ r; where p, q, ~, denote the roots of the cubic equation yv —fy2-gy l_ (2), of which the coefficients f, g, and the absolute term h, are the unknown quantities that are to. be determined. BIQUADRATIC EQUATIONS. 151 iollowing general formulee first given by EvLER; which are remarkable for their elegance and simplicity. Find the three roots of the cubic equation z3 + 2az2 + (a2 - 4c) z = b2, by one of the former rules before given for this purpose; and let them be denoted by r', r", and r"' Then, agreeably to the theory of equations before given, we shall have p+ - q-=- f; pq- pr+-qr g; pqr =h. And by squaring each side of the formula expressing the value of x, x2 =p+'-t- 2 v q- +2 /r 2 qr. Or, by substituting f for its equal - (p - q-+ r), aind bringing the term, so obtained, to the other side of the equation x.2 +f = 2 / Pq- 2 Vp r-2 V/ qr. Also, by again squaring each side of this last expression, we shall have x4 +Jr 2f,2 -+-f2 = 4pq - 4pr + 4qA- + 8 A/p2qr r 8 V q2p r + 8 V r2pq. Or, substituting 4g for its equal 4pq-+-4pr +-4qr, and bringing the term to the other side as before,.X4 + 2f.z2 -[2 - 4g -= 8 / pq (V p - V q - r). But since, from what has been above laid down, we have / p- q/ Q /V r = x, and V/pq' -= / Ah, if these be put for their equals in the last equation, it will become, by this substitution, z4 + zf-2 - 82X +f2 4g = 0. Whence, comparing these coefficients with those of the given equation, there will arise 2f-=a; -8-/A- =b; f2-4g = c; or, a b2 a2 c 2f=; 64; g 16 4 And-consequently, by substituting these values in the assumed cubic equation (2), we shall have 1 b2 y-3 I- ay2q-.- (a2 -4c)y -= (3), 16 64 the three roots of which last equation, when substituted for v, q, and r, in the formula x = V/p ~ —V' a- a/ q, will give, by taking each term of the expression both in + and -, all the four values of x. Or, in order to render this result more commodious in practice, by freeing it from fractions, let / = g. Then, by substitution and reduction, we shall have the corresponding equation z3 + 2aZ2 - (a2-4c) z = b2, (4), the three roots of which are each, evidently, four times those of the former. Hence, using this instead of equation (3), and denoting its roots by r', /", g"', the last mentioned formula, taking each of its terms in +- and -, as before, will give the values of x, as' in the above expressions. Note.-If we were to take all the possible changes of the signs, in this case, which the terms of the assumed formula admit of, it would appear that x should have eight different values; but it is to be observed, that, according to the first part of the above investigation, the product V/p X 1/ q X / r — / h, or 8b; and consequently, that when b is positive, either all the three radicals must be taken in A-, or two in -, and one in +-; and when b is negative, they must either be all -, or two +- and one - which considerations reduce the number of roots to four. 1f52 BIQUADRATIC EQUIATIONS. Then, we shall have, Whern b is positive, When b is negative, -v r'- Vr"-V /r"' q- +/r' +- / r" + Vr"''X 2 *. - 2 -v /r' r- r" + Vr7' ~ vr' - Vr" - Vr" 2 2 + V r'- + a/ \ "' - V?- - Vr- +rV X- - _ - X — 2 2 2 I 2 ~Xi!: —-:, f. 2 Note.-If the three roots r', r", r"', of the auxiliary cueic equation be aJl real and positive, the four roots of the proposed equation will also be real; and if one of these roots be positive and the other two imaginary, or both of them negative and equal to each other, two of the roots of the given equation will be real, and tWo imaginary; which are the only cases that produce real results. 3. Given x4 - 25x2 + 60x - 36 = 0, to find the four roots of the equation. Here a -25, b = 60, and c — 36; Whence, by substituting these values for their equals in the cubic equation above given, we shall have Z3 - 2 X 25 z2 _- (25 + -4 X 36) z- 602, or z3 - 502 + 769 z -- 3600: the three roots of which last equation, as found by trial, or by one of the former rules, are 9, 16, and 25, respectively; whence x = -(- / 9 - / J 6'- /25) = (- 4 - 5)= - x — (-V- 9 +- 16 + v 25) (-3 + 4 + 5)- +.3 x =2 (+ V 9 -v 16 + v 25) —- (+ 3 -4 + 5) - + 2 x =2 (+ A/9 + V, 16 -,/25) 2- } (+ 3 + 4 - 5) = I 1 And consequently the four roots of the proposed equation are l, 2, 3, and - 6. EXAMPLES FOR PRACTICE. 1. Given x4 -55x2 -_ 3Qx 504=-0, to find'the four roots, or values of. x. Ans. 3, 7, - 4, and - 6. 2. Given x4 - 2X3 -7x2- _ 8 _ —-12, to find the four roots, or values of x. Ans. 1, 2, - 3, and - 2. 3. Given x4 - 8X3 + 14x2 + 4x = 8, to find the four roots, or values of x. Ans. 3 + V, 3 1 3 5 4. Given X4 -- 17X2 - 20x - 6 - 0, to find the four roots, or values of x.. 2 v -2- 2. RESOLUTION OF EQUATIONS BY APPROXIMATION. 153 5. Given x4 - 3x2 - 4x 3, to find the four roots, or values of x. Ans. 13 -+- V —3, — ~ V-3. 6. Given x4- 19x3+ 132x2- 302x - 200 -= 0, to find the four roots or values of x. Ans. 4.27768,.80955 roots, or values of x. A +s d-6.956377, + V'(- 9.3686).'7. Given x4 - 27x3 + 162x2 + 356x - 1200 = 0, to find the four roots, or values of x. Ans. 2.05608, - 3.00000 13.15306, 14.79086 8. Given x4 — 12xZ + 12x - 3 — 0, to find the four roots, -or values of i. Ans.!.606018, - 3.907378 or Xalues of x. Ans 2.858083,.443277. 9. Given x4 - 36X2 + 72x - 36 = 0, to find the four roots, or values of x. 0.8729836, 1.2679494 4.7320506, - 6.8729836. 10. Given x4-12x3 +- 47x2 - 72x +- 36 = 0, to find the roots, or values of x. Ans. 1, 2, 3, and 6. 11. Given x4 +- 24x3 - 114x2-24x + 1 - 0, to find the roots, or values of x. Ans. ( 197 - 14, 2 + V/5 -V 197 - 14, 2 - V5. 12. Given x4 - 6x3 - 58x2 - 1 ] 4x - 11 -- 0, to find the roots, or values of x. Ans.: f / 3 + - i /(17 f+i -21 73). OF THE RESOLUTION OF EQUATIONS BY APPROXIMATION. EQ UATIONS of the fifth power, and those of higher dimensions, cannot be resolved by any rule or algebraic formula that has yet been discovered; except in some particular cases where certain relations subsist between the coefficients of their several terms, or when the roots are rational; and, for that reason, can be easily found by means of a few trials. In these cases, therefore, recourse must be had to some of the usual methods of approximation; among which, that commonly employed is the following, which is universally applicable to all kinds of numeral equations, whatever may be the number of their dimensions, and, though not strictly accurate, will give the value of the root sought to any required degree of exactness. RuLE.-Find by trials, a number nearly equal to the root sought, which call r; and let z be made to denote the difference between this assumed root, and thb true root x. Then, instead of x, in the given equation, substitute its 154 RESOLUTION OF EQUATIONS BY APPROXIMIATION. equal r +t z, and there will arise a new equation involving only z and known quantities. Reject all the terms of this equation in which z is of two or more dimensions; and the approximate value of z may then be determined by means of a simple equation. And if the value, thus found, be added to, or subtracted from, that of r, according as r was assumed too little or too great, it will give a near value of the root required. But as this approximation will seldom be sufficiently exact, the operation must be repeated, by substituting the number thus found for r in the abridged equation exhibiting the value of z; when a second correction of z will be obtained, which, being added to, or subtracted from, r, will give a nearer value of the root than the former. And by again substituting this last number for r, in the abovementioned equation, and repeating the same process as often as may be thought necessary, a value of x may be found to any degree of accuracy required. Note.-The decimal part of the root, as found both by this and the next rule, will, in general, about double itself at each operation; and therefore it would be useless, as well as troublesome, to use a much greater number of figures than those in the several substitutions for the values of r.* EXAMPLES. 1. Given X3 + x2 + x =- 90, to find the value of x by approximation. Here the root, as found by a few trials, is nearly equal to 4. Let, therefore, 4 = r, and r + z = x. T3 = r3 + 32z + 3rz2 + Z3 Then 2 = r2 + 2sz + Z2 - 90. x — r -z And by rejecting the terms z3, 3rz2, and Z2, as small in comparison with a, we shall have r' - r" + r + 3r2z + 2rz +- z - 90; * It may here be observed, that if any of the roots of an equation be whole numbers, they may be determined by substituting 1, 2, 3, 4, &c., successively, both in plus and in minus, for the unknown quantity, till a result is obtained equal to that in the question; when those that are found to succeed, will be the roots required. Or, since the last term of any equation is always equal to the continued product of all its roots, the number of these trials may be generally diminished, by finding all the divisors of that term, and then siustituting them both in plus and minus, as before, for the unknown quantity, when those that give the groper result will be the rational roots sought; but if none of them are found to succeed, it mnay be concluded that the equation cannot be resolved by this method; the roots, in that case. being either irrational or imaginary. RESOLUTION OF EQUATIONS BY APPROXIMATION. 155 90-r3- r2 — r 90-64-16-4 6 Whence z -- - —.10. 3r2- +2r + 1 48 + 8 + 1 57And consequently, x = 4.1, nearly. Again, if 4.1 be substituted in the place of r, in the last equation, we shall have 90 - r'r2 - r 90-68.921-16.81 —4.1 z _-= --- -.00283; 3r2+ 2r +- 1 50.43 - 8.2+1 And consequently, x = 4.1 d+.00283 - 4.10283 for a second approximation. And if the first four figures, 4,102, of this number be again substituted for r, in the same equation, a still nearer value of the root will be obtained; and so on, as far as may be thought necessary. 2. Given X2 + 20x 100, to find the value of x by approximation. Ans. x 4.1421356. 3. Given X3 - 9X2 + 4x = 80, to find the value of x by approximation. Ans. x = 2.4721359. 4. Given x4 - 38x3 + 210x2 + 538x + 289 = 0, to find the value of x by approximation. Ans. x = 30.53565375. 5. Given xa' + 6X 4 — 103 - 12a s- 207x +- 110 - 0, to find the value of x by approximation. Ans. 4.46410161. The roots of equations, of all orders, can also be determined, to any degree of exactness, by means of the following easy rule of double position; which, though it has not been generally employed for this purpose, will be found in some respects superior to the former, as it can be applied, at once to any unreduced equation consisting of surds, or compound quantities, as readily as if it had been brought to its usual form RULE 2.-Find, by trial, two numbers as near the true root as possible, and substitute them in the given equation instead of the unknown quantity, noting the results that are obtained from each. Then, as the difference of these results is to the difference of the two assumed numbers; so is the difference between the true result, given by the question, and either of the former, to the correction of the number belonging to the result used; which correction being added to that number when it is too little, or subtracted from it when it is too great, will give the root required nearly. And if the number thus determined, and the nearest of the two former, or any other that appears to be more accurate, be now taken as the assumed roots, and the operation be repeated as before, a new value of the unknown quantity will be ob 156 RESOLUTION OF EQUATIONS BY APPROXIMATION. tained still more correct than thle first; and so on, proceeding in this manner as far as may be judged necessary.' EXAMPLES. 1. Given x3 + x2+ =- 100, to find an approximate value of Here it is soon found by a few trials, that the value of x lies between 4 and 5. Hence, by taking these as the two assumed numbers, the operation will stand as follows:First Sup. Second Sup. 4.. x.. 5 16.. 2.. 25 64 1. 125 84 Results 155 155.. 5.. 100 Therefore 84.. 4.. 84 71:1:: 16:.225 And consequently x = 4 +.225 =.4.225, nearly. * The above rule for Double Position, which is much more simple and commodious than the one commonly employed for this purpose, is the same as that which was first given at p. 311 of the octavo edition of my Arithmetic, published in 1810. To this we may further add, that when one of the roots of an equation has been found, either by this method or the former, the rest may be determined as follows:Bring all the terms to the lefthand side of the equation, and divide the whole expression, so formed, by the difference between the unknown quantity (x) and the root first found; and the resulting equation will then be depressed by a degree lower than the given one. Find a root of this equation, by approximation, as in the first instance, and the number so obtained will be a second root of the original equation. Then by means of this root, and the unknown quantity, depress the second equation a degree lower, and thence find a third root; and. so on, till the equation is reduced to a quadratic; when the two roots of this, together with the former, will be the roots of the equation required. Thus in the equation x3 - 15z2-+ 63.x = 50, the first root is found by approximation to be 1.02804. Hence, x — 1.02804) X3 - 15x2 + 63 -- 50 (x2 — 13.97195.x- + 48.63627 = 0. And the two roots of the quadratic equation, x2 - 13,97195x — 48.63627, found in the usual way, are 6.57653 and 7.39543. So that the three roots of the given cubic equation x3 — 15z2 + 63x - 50, are 1.02804, 6.57653, and 7.39543, their sum being. = 15, the coefficient of the second term of the equation, as it ought to be when they are right. RESOLUTION OF EQUATIONS BY APPROXIMATION. 157 Again, if 4.2 and 4.3 be taken as the two assumed numbers the operation will stand thus:First Sup. Second Sup. 4.2.... 4.3 17.64.. 18.49 74.088. x3. 79.507 95.928 Results 102.297 102.297.. 4.3.. 102.297 Therefore 95.928.. 4.2.. 100 6.369:.1 2.297.036. And consequently, x = 4.3 -..036 = 4.264, nearly. Again, let 4.264 and 4.265 be the two assumed numbers; then First Sup. Second Sup. 4.264 x.. 4.265 18.181696.. x..18.190225 77.526752.. x3.. 77.581310 99.972448 Results 100.036535 Therefore, 100.036535 4.265 100 99.972448 4.264 99.972448.064087:.001::.027552: 0004299 And consequently, x -= 4.264 +.0004299 = 4.2644299, very nearly. 2. Given (}x2 - 15)2 + x v = 90, to find an approximate value of x. Here, by a few trials, it will be soon found, that the value of Xq lies between 10 and 1 1; which let, therefore, be the two assumed numbers, agreeably to the directions given in the rule. Then, Fzrst Sup. Second Sup. 25.. (1 — 15)2. 84.64 31.622. xaix.. 36.482 56.622 Results 121.122 121.122. 11.. 121.122 56.622.. 10.. 90 Hence 64.5 1:: 31.122:.482. And consequently, x = 11 -.482 = 10.518. 14 1538 EXPONENTIAL EQUATIONS. Again, let 10.5 and 10.6 be the two assumed numbers Then, First Sup. Second Slup. 49.7025.. (-Lxs- 15)2.. 55,830784 34.0239.. x.. 34.511099 83.7264 Results 90.341883 Hence, 90.341883.. 10.6.. 90.341883 83.7264.. 10.5.. 90 6.615483. -.1::.341883:.0051679 And consequently, x = 10.6 -.0051679 = 10.5948321, very nearly. EXAMPLES FOR PRACTICE. 1. Given X3 + - 102 + 5x = 2600, to find a near approximate value of x. Ans. x = 11.00673. 2. Given 2x4 -163 - 402 -_ 30x + 1 -- 0, to find a near value of' x. Ans. x =- 1.284724. 3. Given X5 + 2x4 + 3x, + 4x2 + 5x == 54321, to find the value of x. Ans. 8.414455. 4. Given V (7x3 + 4x2) + V (20x2 — 10x)-= 28, to find the value of x. Ans. 4.510661. 5. Given /[144x"- (X2 + 20)2] + v [196x2 - (x2 + 24)2] = 114, to find the value of x. Ans. 7.123883. OF EXPONENTIAL EQUATIONS. An exponential quantity is that which is to be raised to some -unknown power, or which has a variable quantity for its index; as, 1 1 ax, a,, 2 eor X, &c. And an exponential equation is that which is formed between any expression of this kind and some other quantity whose value is known; as, a=-b,- = a, &c. Where it is to be observed, that the first of these equations, when converted into logarithms, is the same as X log. a- b. or x =; and the second equation xZ = a log. ab is the same as x log. x = log. a. In the latter of which cases, the value of the unknown EXPONENTIAL EQUATIONS. 59 quantity x may be determined, to any degree of exactness, by the method of double position, as follows:RULE.-Find, by trial, as in the rule before laid down, two numbers as near the number sought as possible, and substitute them in the given equation x log. x = log. a, instead of the unknown quantity, noting the results obtained from each. Then, as the difference of these results is to the difference of the two assumed numbers, so is the difference between the true result, given in the question, and either of the former, to the correction of the numnberbelonging to the result used; which correction, being added to that number when it is too little, or subtracted from it when it is too great, will give the root required, nearly. And if the number thus determined, and the nearest of the two former, or any other that appears to be nearer, be taken as the assumed roots, and the operation be repeated as before, a new value of the unknown quantity will be obtained still more correct than the first; and so on, proceeding in. this manner as far as may be thought necessary. EXAMPLES. 1. Given xS = 100 to find an approximate value of x. Here, by the above formula, we have x log. x = log. 100 = 2. And since x is readily found, by a few trials, to be nearly in the middle between 3 and 4, but rather nearer the latter than the former, let 3.5 and 3.6 be taken for the two assumed numbers. Then log. 3.5 =.5440680, which, being multiplied by 3.5, gives 1.904238 = first result; And log. 3.6 =.5563025, which, being multiplied by 3.6, gives 2.002689'for the second result. Whence, 2.002689.. 3.6. 2.002689 1.904238. 3.5.. 2..098451.l.002689:.00273 for the first correction; which, taken from 3.6, leaves x=- 3.597Z7, nearly. And as this value is found, by trial, to be rather too small, let 3.59727 and 3.59728 be taken as the two assumed numbers Then log. 3.59728 = 0.555974243134677 to 15 places. The log. 3.59727 = 0.555973035847267 to 15 places, 160 BINOMIIAL -THEOREM. which logaiithms, multiplied by their respective numbers, give the following products:1.999995025343512 both t rue to the last figure. 1.999985122662298 both true to the last figure. Therefore, the errors are 4974656488 and 14877337702 and the difference of errors 9902681214 Now, since only 6 additional figures are to be obtained, we may omit the three last figures in these errors; and state thus: as difference of errors 9902681: difference of sup. 1:: error 4974656; the correction 502354, which, united to 3,59728, gives us the true value of x = 3.59728502354.* 2. Given xx= 2000, to find an approximate value of x. Ans. x -- 4.82782263. 3. Given (6x)- = 96, to find the approximate value of x. Ans. x- = 1.8826432. 4. Given x- = 123456789i to find the value of x. Ans. 8.6400268. 5. Given xZ - x-= (2x - xC), to find the value of x. Ans. x = 1.747933. OF THE BINOMIAL THEOREM. The binomial theorem is a general algebraical expression, or formula, by which any power, or root of a given quantity, consisting of two terms, is expanded into a series; the form of which, as it was first proposed by.Newton, being as follows:m n - n (P + P) pn[lZ —2-)Q+ (Pn ( n )' W ] /m- 2n\ Or, m m - n r m —n m - 2n (P + PQ)n = p + - A - 57- Q J n- 3n -nDQ, &c. 4n When P is the first term of the binomial, Q the second * The correct answer to this question has been first given by Doctor Adrain, in his edition of Hutton's Mathematics, who plainly proves that Hutton's answer, which is the same as Bonnycastle's, is incorrect. DSee Hutton's Mathematics, Vol. I. p. 263,. Y. Edition. —ED. BINOMIAL THEOREM. 16I1 term divided by the first, - the index of the power, or root, and A, B, C, &C., the terms immediately preceding those in which they are first found, including their signs + or -. Which theorem may be readily applied to any particular case, by substituting the numbers, or letters, in the given example, for P, Q, m, and n, in either of the above formulte, and then finding the result according to the rule.* EXAMPLES. 1. It is required to convert (a + )2- into an infinite series. x m 1 Here P - a, Q -- - or m = 1, and n = 2; a n 2 2 when'ce on m B = (a2) = (a2) = a = A, m i a X X -- AQ= X. X-=- =B, n 2 1 a2 2a * This celebrated theorem, which is of the most extensive use in algebra and various other branches of analysis, may be otherwise expressed as follows:- +xn n 97b\.- I en Int n m - 2n \I &c.] m Or, (a+x) -= On a+ [1+-t x-~ n m +n(2 + _ 7m+ m+2n m n' [1 —x +n 2n \a-x 2n 3n \a+x' &c.J Or, (a-Fvx)n = 2a [1+ m a 2;)+mm+ (a m)2_ m+ nn* nf( X3 n a or x 2n nax t 2) n 3n a+x &c. It may here also be observed, that if na be made to represent any whole, or fractional number, whether positive or negative, the first of these expressions may be exhibited in a more simple form, ( )m m m-+ m(m-1) rn-a m.(m —1) (-2) mn-3 (a+x) =a $ ma x + a X2+ a X3. e(m — 1) (m —2)..... [m —(n —1)] am-'xm *...1.2.3.4. Where the last term is called the general term of the series, because if 1, 2, 3, 4, &c., be substituted successively for n, it will give all the rest. 14' 162 BINOMIAL THEOREI. m-n 1 -2 x x x 2n 4 X 2a a2 2.4a m - 2n 1-4 xZ x 3x3 -- -co X X - -- Ds 3n 6 2.4a3 a2 2.4.6a5 m - 3n 1 - 6 3x3 x 3.5x4 4n 8 2.4.6a X a2 2.4.6.8a7 m - 4n 1- 8 3.5x x 3.5.7x5 5n — }-10 2.4.6.8a17 a 2.4.6.8.10a9 &3c. &c. &c. Therefore (a2 + x)i = x x2 3x3 3.5x4 3.5.7x5 x -- ~, a + 2a 2.4a3 2.4.6a5 2.4.6.8a7 + 2.4.6.8.10 a?'Where the law of formation of the several terms of the series is sufficiently evident. 2. It is required to convert ( 6)2' or its equal (a + b)-2, into an infinite series. b m Here P - a, Q =., and - = - 2, or m — 2, and n = 1 a n whence mn m = (a)= _. = = A, P _ (a- — A) m 2 1 6 2b -AQ"- X X - B, n 1 a' a aa m - n -2 - 1 2b b 3b2 BQ_ X 3 X -=- =C 9 2n 2 a- a4 m - 2n -2-2 3b6 b 4b3 -Q = -,- x- X --- a=,, 3n 3 a4 a a m- 3n - 2 -3 4b3 b 5b4 -- DQ= -& - X -- =Eq 4n 4 a a a &c. &c. &c. 1 1 2b 3b2 4b3 5b4 Consequently, (a b)2= - a4 a & a2 3. It is required to convert - - or its equal a2 (a - ) 2 -1(aa - x) a, into an infinite series. Here x m 1 2 a 2 n - P~, = ——,an-' EBINOMIAL THEOREM. 163 whence m m -~ 1 pn = (a')n = (a) = —, a' m 1 1 a — AQ= — X X — = —B, n 2 a a2 a3 m-n -1-2 x x 3x2 BQ — X X — = 5-'C, 2n 4 2aa a" 2.4a5 m —2n -1 -4 3x2 x 3.5x3 3n 6 2.4a5 aX - 2.4. 6a mn — 3n -1i - 6 3.5x 3.7x4 DQ= X X -- -E, 4n X 2.4.66.8a9 &c. &c. ac. Therefore, 1 1 1 {,;rA 3 rz2 3.5 ax)- a 2 6 a3 2.4.6 2.4.6.8 ( +, & And @2 -a +.) 4 3 +2 2.4.6 -~ + 2.-.6.8' -- + + (a2a - )F () +, &c. 4. It is required to convert V 9, or its equal (8 + 1)3, into an infinite series. 1 rn 1 Here P =-8, Q and -3, or m =1 and n =3;' n 31 Whence, in r 1 pr = (8)3n= 8 = 2 A, m 1 2 1 1 -AQ- X- X- B, n 3 1 2 - 3.22 m —n 1-3 1 1 1 2n -6 X 322 3.6.2 m —2n 1 —6 1 5 3n Q = 9 X 3.6. 24X 23 3.6.9.27 rn-3n 1-9 5 1'5. 8 4n 12 3.6.9.27 X2a 32 6 9 1Z 21= m-4n 1-12 5.8 1 5.8.11 -- EQ - 5n 15 x 3.6.9.1:2.210 X 23 = 3.6.9.12.15.213 &c. &c. &c. 164 BINOMIAL THEORElM. Therefore, V/ 9 - 1 1 5 5.8 5.8.11 2+ _; *.52+ + 2+ 3.22 3.6.24 3.6.9.2' 3.6.9.12.210 3.6.9.12.15.2'3' &c. 5. It is required to convert V2, or its equal /(1 + 1), into an infinite series. 1 1 3 3.5 3.5.7 Ans.1+- -+ - -+& 2 2.4.6 2.4.6.8 2.4.6.8.10' 6. It is required to convert 5v/7, or its equal (8- 1)31 into an infinite series. 1 1 5 5.8 Ans. 2 — 4 5 &C. 3.22 3.262 3.6.9.2 3.6.9.12.210 7. It is required to convert 5V 240, or its equal (243 - 3)5, into an infinite series. 1 4 4.9 4.9.14 5.33 5.10.37 5.10.15.3" 5.10.15.20.3'51 8. It is required to convert (a iX) 2 into an infinite series. Asa4(X x2 3X3 3'5x4: c 2a 2.4a2 2.4.6a. 2.4.6.8a4 9. It is required to convert (a + b)3 into an infinite series. I ltb 2b2 2.5b3 2.5.8b4 Ans. a' I l:4 4. C. 1 3a 3.6a2 3.6.9a3 3.6.9.12a4'10. It is required to convert (a - b)4 into an infinite series. 6S b 3b62 37b 3.7.116b4 Ans. a4. &c. 4a 4.8a2 4.8.12a? 4.8.12.16a4 11. It is required to convert (a + x)3 into an infinite series. A a 1 x ~' 4x3 4.7x4 4.47.10 5 3a — 9a2 92a3 92.12a4 92. 12.15a5 12. It is required to convert (1 - x)5 into an infinite series. 2x 2.3x2 2.3.8x3 2.3.8.13x4 Ans. It -&C.1 5 5.10 5.10.15 5.10.15.20 1 13. It is required to convert:, or its equal (a -t x)2 (a:i x) into an infinite series. 1An. 3x2 3.5x3 3.5.7X4 &. Ant s.11 F 2aF + - --- + =F C INDETERMINATE ANALYSIS. 165 a 14. It is required to convert -, or its equal (a:$=)3 (a i:: x), into an infinite, series. e( x 4X2 4.7xS3 4.7.10x4 & 3a + 3.6a2 3.6.9 + 3.6.9.12a4 = 15. It is required to convert —', or its equal (1+ X)5 -_1. (1 + x) ~, into an infinite series. x 6x2 6.11 x3 6.11.16x4 Ans. 1 — 5 - 16. it is required to convert Or its equal (a + x) (a2 - a2), into an infinite series. x X2 x3 3x4 3x5 5x6 5x1 Ans. I +- - a q - - -- -' &c. ~Ans.1++-2aa 2a3 8a4 85 + 6a6 16a7' OF THE INDETERMINATE ANALYSIS. IN the common rules of Algebra, such questions are usually proposed as require some certain or definite answer; in which case it is necessary that there should be as many independent equations, expressing their conditions, as there are unknown quantities to be determined; or otherwise the problem would not. be limited. But in other branches of the science, questions frequently arise that involve a greater number of unknown quantities than there are equations to express them; in which instances they are called indeterminate or unlimited problems, being such as usually admit of an indefinite number of solutions; although, when the question is proposed in integers, and the answers are required only in whole positive numbers, they are, in some cases, confined within certain limits, and in others the problem may become impossible. PROBLEM 1.-To find the integral values of the unknown quantities x and y in the equation ax - by = ~- c, or ax + by = c. Where a and b are supposed to be given whole numbers, which admit of no common divisor, except when it is also a divisor of c. 166 INDETERMINATE ANALYSIS. RULE I.-Let wh denote a whole, or integral number; and reduce the equation to the form'by::I c c - by x=- wh, orx - -wh. a a 2. Throw all whole numbers out of that of these two expressions, to which'the question belongs, so that the numbers d and e, in the remaining parts, may be each less than a; then dy e e - dy wh, or w — = W. a a 3. Take such a multiple of one of these last formulae, corresponding with that abovementioned, as will make the coefficient of y nearly equal to a, and throw the whole numbers out of it as before. Or, find the sum or difference of —, and, the expression ay above used, or any multiple of it that comes near -, and the result, in either of these cases, will still be = wh, a whole number. 4. Proceed in the same manner with this last result; and so on, till the coefficient of y becomes = 1, and the remainder some number r; then y-+r = wh, = p, and y = ap F r. Where p may be 0, or any integral number whatever that makes y positive; and, as the value of y is now known, that of x may be found from the given equation, when the equation is possible.* NoTE.- Any indeterminate equation of the form ax - by =:a: c, in which a and b are prime to each other, is always possible, and will admit of an infinite number of answers in -whole numbers. But if the proposed equation be of the form ax + by = c, the number of answers will always be limited; and, in some cases, the question is impossible; both of which circum* This rule is founded on the obvious principle, that the sum, difference, or product of any two whole numbers, is a whole number; and that if a number divides the whole of any other number and a part of it, it will also divide the remaining part. INDETERMINATE ANALYSIS. 167 stances may be readily discovered from the mode of solution above given.* EXAMPLES. 1. Given 19x - 14y = 11, to find x and y in whole numbers. 14y +-11 19y Here x = = wh., and also = wh. 19 19 lsy 14y + 1 1 5y - 11 Whence, by subtraction,!9 4 __ _- 1 = wh. 19 19 19 5y - 11 20y - 44 y- 6 Also, 1- X 4 -- - = y -2 +- wh. 19 19 19 And, by rejecting y - 2, which is a whole number, y-6 Wh. =-p. 19 Whence we have y = 19p + 6. 14y +- l 14(19p + 6) +- ]1 266p+- 95 19 19 19 14p +- 5. Whence, ifp be taken = 0, we shall have x = 5 and y = 6, for their least values; the number of solutions being obviously indefinite. 2. Given 2x + 3y = 25, to determine x and y in whole positive numbers. 25 - 3 1 — Here x — = =12 - y 2 2 Hence, since x must be a whole number, it follows that - must also be a whole number. 2 * That the coefficients a and b, when these two formulae are possible, should have no common divisor, which is not at the same time a divisor of c, is evident; for if a = mnd, and b = me, we shall have ax i: by mdx -4- mey = c; and consequently, dxzr + ey = _. But d, e, x, y, being supposed to be whole numbers, - must also be a whole number, which it cannot be except when m is a divisor of c. Hence, if it were required to pay 1001. in guineas and moilores only, the question would he impossible; since, in the equation 21:-r+ 27y = 2000, which represents the conditions of the problem, the coefficients, 21 and 27, are each divisible by 3, whilst the absolute term 2000 is not divisible by it. See my Treatise on Alcgbra, for the method of resolving questions of this kind by means of Continuned Fractiones. 168 INDETERMINATE ANALYSIS. 1 —y Let, therefore, - wh.= p; 2 Then 1 - y = 2p, or y = 1 - 2p. and since x= 12 - y + 1-y 12- (1 - 2p) +p= 12 + 3p-l, We shall have x = 11 + 3p, and y = 1 - 2p; Where p may be any whole numnber whatever, that will render the values of x and y in these two equations positive. But it is evident, from the value of y, that p must be either O or negative; and consequently, from.that of x, that it must be 0, — 1, - 2, or - 3. Whence, if p = p -1, p -2, p 3, Then xj= 11 =, 8, = 5, x -2, y-1, y —3, y=5, y —7. Which are all the answers in whole positive numbers that the question admits of. 3. Given 3x -- 8y - 16, to find the values of x and y in whole numbers. 8y -- 16 2y 1- 2y-_ - Here x=- -2y- 5 -- wh.; or - wh. 3 3 3 Also, 2 - 2 =4y - 2 Y —= wh. 3 -2 3 3 Or, by rejecting y, which is a whole number, there will remain y-h- w= w.=p. Therefore, y = 3p + 2, 8y-16 8(3p +2) -16 _ 24p _p. Andx —3 _ —- p. 3 3 3 Where, if p be put = 1, we shall have x = 8 and y = 5 for their least values; the number of answers being, as in the first question, indefinite. 4. Given 21 x +- 17y - 2000, to find all the possible values of x and y in whole numbers. 2000 - 17y9 5 — 17y Here x = 95~ - wh. 21 21 Or, omitting the 95, wh.; 21 21 5 - 17y 4y 5-wh; Consequently, by addition, + -F-2 -- 4 = 21 21 21 4y -+- X 20y -+ 25. 4 + 20y h Also, -X5 X= 21 21 h; 21. 21 21 INIYETERMINATE ANALYSIS. 169 Or, by rejecting the whole number 1, 4 wh.; 21 And, by subtraction, 2 y- 2y y -4 Wi. - p; 21 21 21 Whence y = 21 p + 4, Andx 2000 - 17~ 2000 - 17(2lp + 4) = 92 17p. 21 21 Where, ~if p be put = 0, we shall have the least value of y = 4, and the corresponding or greatest value of x = 92. And the rest of the answers will be found by adding 21 continually to the least value of y, and. subtracting 17 from the greatest value of x; which being done, we shall obtain the six following results:x z92 75 58 41124 7 y -'4 25 46 67 88 109 These being all the solutions, in whole numbers, that the question admits of. Note 1. —When there are three or more unknown quantities, and orfly one equation by which they can be determined, as ax + by + cz = d, it will be proper first to find the limit of the quantity that has the greatest -coefficient, and then to ascertain the different values of the former, from 1 up to that extent, as in the following question:5. Given 3x + 5y + 7z 100, to find all the different values of x, y, and z in whole numbers.' Here each of-the least integer values of x and y are 1, by the question; whence it follows, that 100 - 5 — 3 100 - 8 92 7 7 7 Consequently, z cannot be greater than 13, which is also the limit of the number of.answers; though they may be considerably less. * If any indeterminate equation, of the. kind above given, has one or more of its coefficients, as c, negative, the equation may be put under the form ax - 6by = d+- cz, in which case it is evident that an indefinite number of values may be given to the second side of the equation by means of the indefinite quantity z; and consequently, also, to x and y in the first. And if the coefficients a, b, c, in any such equation, have a common divisor, while d has not, the question, as in the first case, becomes Impossible. 15 170 INDETERMINATE ANALYSIS. By proceeding, therefore, as in the former rule, we shall have 1 00 - 5y - 7z 1 — 2y-z 3 =- 33 - y - 2z+ — - wA. 3 3 And, by rejecting 33 - 2z, 1 -2y -z 3y 1 -2y -z y- 1 -z — 3 — -= wh.; or - + - wh.; ~3 3 3 3 Whence Y - 1- =P And y = 3p + z-1. And consequently, putting p 0, we shall have the least value of y = z - 1; where z may be any number, ffoin 1 up to 13, that will answer the conditions of the question. When, therefore, z - 2, we have y 1, 100 - 19 And x - - - 27. 3 Hence, by taking z = 2, 3, 4, 5, &c., the corresponding values of x and y, together with those of z, will be found to be as below: —z= 2 3 4 5 6 7- 8 y- 1 2 3 4 5 6 7 x=27 23 19 15 11 7 3 Which are all the integral values of x, y, and z, that can be obtained from the given equation. Note 2.-If there be three unknown quantities, and only two equations for determining them, as ax +- by + cz = d, and ex J fy + gz = A, exterminate one of these quantities in the usual way, and find the values of the other two from the resulting equation, as before. Then, if the values, thus found, be separately substituted in either of the given equations, the corresponding values of the remaining quantities will likewise be determined: thus:6. Let there be given x - 2y - z = 5, and 2x + y - z = 7, to find the values of x, y, and z. Here, by multiplying the first of these equations by 2, and subtracting from the second the product, we shall have 3 +- 5y 2y 3z - 5y = 3, or = -- 1 + y = h; And consequently 2Yor/ 3y- y wh. = p. — And consequently 3 3 Whence y = 3p. INDETERMINATE ANALYSIS. 171 And, by taking p = 1, 2, 3, 4, &c., we shall have y -- 3, 6, 9, 12, 15, &c., and z = 6, 11, 16, 21, 26, &c. But from the first of the two given equations x 5 + 2y - z; whence, by substituting the above values for y and z, the results will give x =.5, 6, 7, 8, 9,.&c. And therefore the first six values of a-, y, and z, are as below:51 6 7 8 9 10 y=3 6 9 12 15 18 z= 6 11 16 21 26 31 Where the law by'which they can be continued is sufficiently obvious. EXAMPLES. POR PRACTICE. 1. Given 3x = 8y - 16, to find the least values of x and y in whole numbers. Ans. x = 8, y = 5. 2. Given 14x = 5y + 7, to find the least values of x and y in whole numbers. Ans. x = 3, y = 7. 3. Given 27x - 1600- 16y, to find the least values of x and y in whole numbers. Ans. x — = 48, y = 19. 4. It is required to divide 100 into two such parts, that one of them may be divisible by 7, and the other by 11. Ans. The only parts are 56 and 44. 5. Given 9x + 13y -- 2000, to find the greatest value of x, and the least value of y in whole numbers. Ans. x -- 215, y =-5. 6. Given I lx + 5y = 254, to find all the possible values of x and y in whole numbers. Ans. x = 19, 14, 9, 4; y.-= 9, 20, 31, 42. 7. Given 17x + 19y + 21z 4()00, to find all the answers in whole numbers which the question admits of. Ans. 10 different answers. 8. Given 5x + 7y + 11z = 224, to find all the possible values x, y, anid z, in whole positive numbers. Ans. The number of answers is 59. 9. It is required to find in how many different ways it is possible to pay 201, in h/if-guineas and half-crowns, without using any other sort of coin. Ans. 7 different ways. 10. I owe my friend a shilling, and have nothing about me but guineas, and he has nothing but louis-d'ors; how must I contrive to acquit myself of the debt, the louis being valued at 17s. apiece, and the guineas at 21s.? Ans. I must give him 13 guineas, and he must give me 16 louis. 172 INDETERMINATE ANALYSIS. 11. How many gallons of British spirits, at 12s., 15s., and 18s. a gallon, must a rectifier of compounds take to make a mixture of 1'000 gallons that shall be worth 17s. a gallon? Ans. 1111 at 12s.; lll at 15s.; and 777 - at 18s. PROPLEm 2. —To find such a whole number as, being divided by other given numbers, shall leave given remainders. RULE 1.-Call the number that is to be determined x, the numbers by which it is to be divided a, b, c, &c., and the given remaindersf, g, h, &c. 2. Subtract each of the remainders from x, and divide the differences by a, b, c, &c., and there will arise x-f X — g - -h -f -— g,,, &c., = whole numbers. a C 3. Put the first of these fractions —' p, and substitute the value of X, as found in terms of p, from this equation, in the place of x, in the second fraction. 4. Find the least value of p in this second fraction, by the last problem, which put.- r, and substitute the value of x, as found in terms of a, in the place of x in the third fraction. Find, in like manner, the least -value of r, in this third fraction, which put = s. and substitute the value of x, as found in the terms of s, in the fourth fraction as before. Proceed in the same way with the next following fraction, and so on to the last; when the value of x, thus determined, w-ill give the whole number required. EXAMPLES. 1.' It is required to find the least whole number, which, being divided by.17, shall leave a remainder of 7, and when divided by 26, shall leave a remainder of 13. Let x — the number required. x-7 x -13 Then -- and -- whole numbers. 17 26 x —7 And, putting -- p, we shall have x == 17p + 7. Which value of x, being sub-stitdted in the second fraction, 17p + 7- 13 17p-6 gives -- - = wh. 26 -26 But it is obvious that - is also = wh. And consequently, 26 17p — 6 9p+6 26 26 26 INDETERMINATE ANALYSIS. 173 Or X9p 3= =p ~ 26 26 26 Where, by rejecting p, there remains P26. = r. Therefore, p = 26r - 18; Hence, if r be taken, - 1, we shall have p 8. And consequently, x = 17p + 7= 17 X 8 + 7 143, the number sought. 2. It is required to find the least whole number, which, being divided by 11, 19, and 29 shall leave the remainders 3, 5, 10 respectively. Let x = the number required. x-3 x- 5 x-10 Then -- and = -whole numbers. 11' 19' 29 -3 And, putting p —, we shall have x = l1p + 3. Which value of x, being substituted in the second fraction, gives 11 - 2 gives 9 - = w. Or 1p — 2 P.w 19 19 19 3p -4 And, by rejecting p, there will remain 19 - wo 19 3p - 4 18p - 24 18p - 5 Also, by multiplication, -X 6 = 19 19 19 - l —wh.;.18p- 5 Or, by rejecting the 1, 19-= wh. 19p But I P is likewise = wh. 19 19p l57 — p+S-5 Whence 19P -- = ISP — 5 P+ wh., which put = r. Then we shall have p = 19r - 5, and x = 11 (19r - 5) + 3 = 209r - 52. And if this value be substituted for x in the third fraction, there will arise 209r - 62 6r -4 7r - 2 + --- = wh.; 29 29 Or, by neglecting 7r - 2, we shall have the remaining part 6r - 4 of the expression -- = wh.; 29 15 15' .174 DIOPHANTINE ANALYSIS. But, by multiplication, 6r - 4 30r- 20 r - 20 -29 X5- 29 29 r - 20 Or, by rejecting r, there will remain 29 = wh., which put = s. Then r = 29s + 20; where, by taking s = 0, we shall.have r = 20. And consequently, 2 = 209r- 52 - 209 X 20 - 52 = 4128, the number required. 3. To find a number, which, being divided by 6, shall leave the remainder 2, and when divided by 13, shall leave the remaindeir 3. Ans. 68. 4. It is required to find a number, which, being divided by 7, shall leave 5 for a remainder, and if divided by 9, the remainder shall be 2. Ans. 47, 110, &c. 5. It is required to find the least whole number, which, being divided by 39, shall leave the remainder 16, and when divided by 56, the remainder shall be 27. Ans. 1147. 6. It is required to find the least whole number, which, being divided by 7, 8, and 9, respectively, shall leave the remainders 5, 7, and 8. Ans. 215. 7. It is required to find the least whole number, which, being divided by each of the nine digits, 1, 2, 3, 4, 5, 6, 7, 8, 9, shall leave no remainders. Ans. 2520. 8. A person receiving a box of oranges observed, that when he told them out by 2, 3, 4, 5, and 6. at a time, he had none remaining; but wlhen he told them out by 7 at a time, there remained 5; how many oranges were there in the box? Ans. 180. OF THE DIOPHANTINE ANALYSIS. TiIs branch of Algebra, which is so called from its inventor, Diophantus, a Greek mathematician of Alexandria in Egypt, who flourished in or about the third century after Christ, relates chiefly to the finding of square and cube numnbers, or to the rendering'certain compound expressions free from surds: the method of doing which is by making such substitutions for the unknown quantity, as will reduce the resulting equation to a simple one, and then finding the value of that quantity in terms of the rest. It is to be observed, however, that questions of this kind do DIOPHANTINE ANALYSIS. 175 not always admit of answers in rational numbers, and that when they are resolvable in this way, no rule can be given, that will apply in all the cases that may occur; but as far as respects a particular class of these problems relating to squares, they may generally be determined by means of some of the rules derived from the following formula:PROBLEM 1.-To find such values of x as will make V (ax2 + bx + c) rational, or ax2 + bx + c = a square.* RULE 1. — When the first term of the formula is wanting, or a- 0,- put the side of the square sought = n; then bx + c n2. And consequently, by transposing c, and dividing by the 2 _ n -c coefficient b, we shall have x =;where n may be any number taken at pleasure. 2. When the last term is wanting, or c = 0, put the side of the square sought = nx, or, for the sake of greater genemx rality -; then, in this case, we shall have ax2 + bx — 5.2X 2 — 2 And consequently, by multiplying by n2, and dividing by b4n x, there will arise an2x + bn2 = m2x, and x =,,where m2 and n, both in this and the following cases, may be any whole numbers whatever, that will giVe positive answers.t 3. When the coefficient a, of the first term, is a square number, put it = d2, and assume the side of the square sought n 2Gdm m2 dx + -; then,d2X2 +bx d++ 2 2dm _xn n n+ And consequently, by cancelling d2x2, and multiplying by * The coefficients a, b, of the unknown Quantities, as well as the absolute term c, are here supposed to be all integers; for if they were fractions, they could be readily reduced to a common square denominator; which, being afterwards rejected, will not alter the nature of the question; since any square number, when multiplied or divided by a square number, is still a square. t The unknown quantity x, in this case, can always be found in integers when b is positive; and, in Case 4, next following, its integral value can always be determined, whether b be positive or negative. See Vol. II. of Bonnycastle's Treatise on Algebra, Art. (H.) 1|7dG6 ~ T6ADIOPHANTINE ANALYSI$S. n2, we shall have bn2+ cn2 - 2dmnx + m2, and x = m2 Cn2 -n2 _ 2dm7n 4. When the last term c is a square number, put it = e2, mX and assume the side of the square sought - - + e; then na n ax + bx + eII 2 q+- x+ e2. And consequently, by cancelling e2, and dividing by a, we shall have ax + b = mn2x 2enr bn2 - 2emn -- andx m2 76 n: an2 5. When the given formula, or general expression, axG2 + bx + c can be divided into two factors of the form fx + g and hx + k, which it always can when b2 - 4ac is *a square, let there be taken (fx + g) X (hx + k) (fx + g)2; then, by regn2 - kn2 duction, we shall have x — were it may be hne2 - fma observed, that if the square root of b2 - 4ac, when rational, be'put =, the two factors abovementioned, will be b —J' b+s' ax +, and x -+ * 2 2a And, consequently, by substituting them in the place of the former, ve shall have, am2 (b -') - n2 (b- d+) 2a (n2 - am2) 6, When the formula, last mentioned, can be- separated into two parts, one of which is a square, and the other the product of two factors, its solution may be obtained by putting the sum of the square and the product so formed, equal. to the square of the sum of its roots, and - times one of the n * These factors are found by putting the given formula ax2+x -c -+ 0, and then determining its roots; which, by the rule for quadratics, are x= — 2- -' (b2 — 4ac), and x- - 2 /(b2- 4ac). 2a 2a 2a 1a Whence, if b2 - 4ac be a square, of which the root is a, we shall have a, - 2a and + a- -t -, for the divisors of axz2- bx -c, or ax + 2-5 and x + +-, for its two factors, as in the above rile. DIOPHANTINE ANALYSIS. 177 factors, and then finding the values of x as in the former instances. 7. These being all the cases of the general formula that are resolvable by any direct rule, it only remains to observe, that, either in these, or other instances of a different kind, if we can find, by trials, any one simple value of the unknown quantity which satisfies the condition of the question, an expression may be derived from this that will furnish as many other values of it as we please. Thus, let p, in the given formula ax2 + bx + c, be a value of x so found, and make ap2 + bp + c = q2. Then, by putting x = y + p, we shall have ax2 + bx + ca(y + p)2+ b(y +p)+c = ay2+ (2ap + b)y + ap+ b+c or ax2 + bx +- c = ay2+- (2ap + b)y + q2. From which latter expression, the values of y, and consequently those of x, may be found as in Case 4. Or, because c = q 2 - bp - ap2, if this value be substituted for c, in the original formula ax2 + bx + c, it will become a (2 _ p2) + b ( -_ p) + q2, or q2 ( + (-p) X (x + a b) =a square; which last expression can be resolved by Case 6. It may here, also, be farther observed, that by putting the 2 ax —b given formula, ax2 +- bx + c =, and takin-g x =-2- we shall have, by substituting this value for x in the former of these expressions, and then multiplying by 4a, and transposing the terms ay2.- (b2x -4ac) 2; or putting, for the sake of greater simplicity, b2 - 4ac - b', this last expression may then be exhibited under the form ay2 + b' =.z2, where it is obvious, that if ay2 + (b2 - 4ac), or its equal ay2 + b', it can be made a square, ax2 + bx + c will also be a square. And as the proposed formula can always be -reduced to one of this kind, which consists only of two terms, the possibility or impossibility of resolving the question, in this state of it, can be more easily perceived.* * It may here be observed, that an infinite number of expressions, of the kind ay2 +- (b2 - ac), or ay2 + b' z2,,here mentioned, are wholly irresolvable; among which we may reckon'2y2 _: 3, 57y2 ~:: 6, 7y2 ~- 5, &c. none of which can ever become squares, whatever number, either whole or fractional, be substituted for y; although there are a variety of instances in which the value of y may be found, even in integers, so as to render the formula ay2 + b = z2. For a further detail of which circumstances, as well as for other particulars relating to this part of the subject, see the second volume of Eler's Algebra, or the second volume of Bonxycastle's Algebra. 178 DIOPHANTINE ANALYSIS. EXAMPLES. 1. It is required to find a number, such that if it be multi. plied by 5 and then added to 19, the result shall be a square. Let x = the required number: then, as in Case 1, 5x + a - 19 19 - 2, or x 5 -; where It is evident that n may be any number whatever greater than V 19. Whence, if n be taken = 5, 6, 7, respectively, we shall 25- 19 36 — 19 49 - 19 have ---— 1 or -= 3, or -— 6; 5 5 5 the latter of which is the least value of x, in whole numbers, that will answer the conditions of the question; and consequently 5x + 19 = 5 X 6 19 = 30 + 19 = 49, a square number as was required. 3. It is required to find an integral number, such that it shall be both a triangular number and a square. It is here to be observed, that all triangular numbers are,2 1 of the form; and therefore the question is reduced to 2 x2 $- X 2X2 -. 2+2 the making -, or its equal, a square. — 2~~~~ 4 Where, since the divisor 4 is a square number, it is the same as if it were required to make 2x2a + 2x a square. Let, therefore, 2x2 + 2x = (_)2- - agreeably to the method laid down in Case 2. Then,'by dividing by x, and multiplying the result by n2, the equation will become 2n2x + 2n2 m2x, or (m2 - 2n2) 2 2a e xc= 2n; and consequently2; where, if n be m? - 22' x - x 64 -8 taken = 2, and m = 3, we shall have x = 8, and -=-272 - 2 36, which is the least integral triangular number that is at the same time a square. 3. It is required to find the least integral number, such that if 4 times its square be added to 29, the result shall be a square. Here it is evident, that this is the same as to make 4x2 - 29 a square. And, as the first term in the expression is a square, let DiOPHANTINE ANALYSIS. 179 m 4m m2 4UP + 29 = (2x -- m)2 4X2 x + -; agreeably to Case 3. 4m m2 4m m" Then, -- x +-= 29, or -X= 29 - and consen f n- n 29n2 - m2 quently, x = 4mn i where, if m and n be each taken 29 —1 =1, we shall have x = ---- 7, and 42 + 29 = 4 X 49 + 29 = 225 = (15)2, which is a square number, as was required. 4. It is required to find such a value of x as will make 7x2 - 5x + I a square. Here the last term 1 being a square, let there be taken, according to Case 4, ~m M 92 2m 72 _5x 1 + I = - 1)2-_ X2 -_-w +1 Then, by rejecting the 1 on each side of the equation, mr2 2m and dividing by x, we shall have 7x - 5 -= x- -, and nm n 2ran- 5n"2 consequently, x =; where, if m and n be each 2-5 3 taken -= 1, the result will give x = 1- - =6-, or by tak48- 45 ing n = 3, and m = 8, we shall have x 64 3, which makes 7 X 32 - 5 X 3 + 1 = 49 72, as required. 5. It is required to find such a value of x as will make 8w2 + 14x + 6 a square. Here, by comparing this expression with the general formula, ax 4- bx +- c, we shall have a = 8, b = 14, and c = 6. And, as neither a nor c, in the present instance, are squares, but b2 - 4ac = 196 - 192 = 4 is a square, the given expression can be resolved, by Case 5, into the two following factors, 8x + 6 and x + 1. Let, therefore, 8x2 - 14x +-6 = (8x + 6) (x + 1) = 2- (w +- 1)2, agreeably to the rule there laid down. Then there will arise, by dividing each side by x + 1, m= 8x + 6 = (x i+ 1). And consequently, by multiplication and reduction, we shall 180 DIOPHANTINE ANALYSIS. m2 _ 6n2 have, in this case, x = 82n _ m2; where it appears that, in M2 order to obtain a rational answer, -2 must be less than 8 and greater than 6. Whence, by taking m = 5, and n = 2, we shall have x = 25- 24 1 8 14 400 20\ 32 25 = 7' which makes + -- 6 = 49 - )2,as required. 6. It is required to find such a value of x as will make 2x2 - 2 a square. Here, by comparing this with the general formula ax2 + bx + c, as before, we shall have a = 2, b = 0, and c — 2. And, as neither a nor c are squares, but b62 - 4ac = - 4ac - - 4 (2 X -2) = 16 is a square, the root of which is 4, the given expression can be resolved, by Case 5, into the two factors 2x - 2, and x + 1, or 2 (x - 1), and (x ~- 1), which is evident, indeed, in this case, from inspection. -Let, therefore, 2 2- = 2.(x - 1) X(x + 1) =2 (x +1)2, agreeably to the rule; and there will arise, by division, m/2 2x - 2 - (x + 1). And consequently, by multiplication, 2n2~- m22 and reducing the result, we shall have x = 2n" — m'; where, by taking n = 1, and m - 1, we shallhave x =-3, and 2x2 - 2=18 - 2 = 16 =(4)2; or, taking n —- 2 and m 3, the result will give x =- 17. But as x enters the problem only in its second power, + 17 may be taken instead of - 17; since either of them give 2x2 - 2 = 576 = (24)2. 7. It is required to find such a value of x as will make 5x2 + 36x + 7 a square. Here, by comparing the expression with the general formula, we shall have a = 5, b = 36, and c = 7.'And as neither a nor c are squares, but b2 - 4ac -=1296 - 140 = 1156 = (34)2 is a square, it can be resolved, as in the last example, into the two factors, 5x + 1, and x + 7. Whence, putting 5x"2 + 36x + 7 = (5x + 1) X (x + 7) = 2 (x + 7)', there will arise, by dividing, by x + 7, 5x + 1 = - (x + 7). DIOPHANTINE ANALYSIS. 181 And consequently, by multiplication, and reducing the 7m2 n2 resulting expression, we shall have x = - -; where, 5n2 m2 taking m= 2, and n = 1, the substitution will give = 7 X 4-1 5 X I 4= 27, which makes 5 X (27) -- 36 X 27 + 7 = 4624 = (68)2, as required. 8. It is required to find such a value of x as will make 6X2 - 13x + 10 a square. Here, by comparing the given expression with the general formula ax2 + bx - c, we have a' 6, b = 13, and c= 10. And as neither a, c, nor 62 - 4ac, are squares, the question, if possible, can only be resolved by the method pointed out in Case 6. In order, therefore, to try it in this way, let the first simple square 4, be subtracted from it, and there will remain, in that case, 6x2 ~ 13x + 6. Then since (13)2 - 4 (6 X 6)= 169 - 144 = 25 is now a square, this part of the formula can be resolved, by Case 5, into the two factors 3x +.2, and 2x + 3. Whence, by assuming,, according to the rule, 6x'2 + 13x +~10=4+(3x+2)X(2x+3)=-2 +-(3x+2) =24+ 4r m2 (3x + 2)+ -2 (3x - 2)2, we shall have, by cancelling the 4 on each side, and dividing by 3x + 2, 2x + 3 = 4m?z + -2 (3x + 2). And consequently, by multiplying by n2, and transposing the terms, we have 2n2x - 3m2nx 4mrn + 2m2 - 3n2, or 4mn + 2m2 - 3n2 2n2 - 3mn2 Where, putting in 2, and n = 3, the result will give x 24-8 - 27 5 - 18 - 12 — 6; or, if m be taken = 13, and n = 17, we shall 4 X 17 X 13+ 2 X (13)2 — 3 X (17)2 355 have x-= — - = 5, 2(17)2 - 3(13)2 71 Which makes 6 X (5)2 + 13 X 5 + 10 = 225 = (15)2, as required. 9. It is required to find such a value of x as will make 13a -- 15x +7 a square. 182 DIOPHANTINE ANALYSIS. Here, by comparing this with the general formula, as before, we have a = 13, b =- 15, and c = 7. And as neither a, b, nor b2 - 4ac, are squares, the answer to the question, if it be resolvable, can only be obtained by Case 6. In order, therefore, to try it in that way, let (1 - x)2 or 1 - 2x + xa be subtracted from the given expression, and there will remain 12x2 + 17x + 6. And as (17)2 - 4 (6 X 12), which is'= 1, is now a square, this part of the formula can be resolved by Case 5, into the two factors 4x.- 3 and x +- 2.'Whence, assuming 13 2+15+7=(1 -x)2+(4 +3) X (3x+2)=i(1 -)+ n n m 2)2, we shall have, by cancelling (1 - x)2, and dividing by 2m nm2 3x + 2, 4x + 3 -(1 - x) + - (3x + 2); and consequently, by multiplying by n2, and transposing the terms, there will arise 4n2X + 2mnx - 3mn2- = 2mn + 2m2 - 3l2, or x = 27ln + 2n2 - 3-n2 422 +- 2mn - 3mz2' Where, putting m and n each -- 1, we shall have x= 2+2-3 13 15 13 45 63 4-2+2 3 =,which makes - + - + 7 = -- ---- 2 1 1 )2, as required. 10. It is required to find such a value of x as will make 7x2 + 2 a square. Here it is easy to perceive that neither of the former rules will apply. But as the expression evidently becomes a square when x = 1, let, therefore, x = 1 + y, according to Case 7, and we shall have 7x - 2 = 9 +- 14y + 7y2; Or, putting 9 + 14y + 7y2 = (3 + _ y)2, according to the rule, and squaring the righthand side, 9 + 14y + 7y2 =9 + 6m m2 wY+ -y2 n yq-1y2. Hence, rejecting the 9s and dividing the remaining terms by y, we have 7n2y + 14n2 = 6mn + m2y; and consequently, DIOPHANTINE ANALYSIS. 183 6mn - 14n2 6imn - 14n2 y — 7n2_ m,and x = 7n2_2; where it is evident that m and n may be any positive or negative numbers whatever. If, for instance, m and n be each taken - 1, we shall have 4 1 y = — and x -3. Or, since the second power of x only enters the formula, we may take, as in a former instance, x = -1, which value makes 7x2 + 2 = -7- + 2 -7 + 18 -_ -A a square. Or, if m = 3 and n -1, we'shall have x = 17, and 7xa2 + 2 - 7 X (17)2 + 2 = 2025 = (45)2, a square as before. And by proceeding in this manner, we may obtain as many other values of x as we p ease. PROBLEM 2.-To find such values of x as will make'/ (ax' + bx2 + cx + d) rational, or ax' +- bx2 + x d c = a square. This problem is much more limited and difficult to be resolved than the former; as there are but a few cases of it'that admit of answers in rational numbers, and in these the;rules for obtaining them are of a very confined nature, being mostly such as are subject to certain limitations, or that admit only of a few simple answers, which, in the instances here mentioned, may be found as follows;RULE 1.-WVhen the third and fourth terms of the formula are wanting, or c and d are each = 0, put the side of the square sought = nx, than ax3 + bx2 = n2'Z. And consequently, by dividing each side of the equation by n2 b x2, we shall have ax + b = n2, or x = -, where n may be any integral or fractional number whatever. 2. When the last term d is a square, put it = e2, and assume the side of the required square = e +- x, and the proposed 2e ~2 formula is e2 + cx + 6x' + ax3 = e' + cx + ec2..Whence, by expunging the terms e2 + cx, which are common, and dividing by x2, we shall have 4ae2x +. 4be2 = C2; C2 _ 4beg and consequently, = - 4ae2 c 4be2 _ c2 Or if, the same,case, there be put e +- c + 8e3 2 for the side of the required square, we shall have, by squaring, 184 DIOPHANTINE ANALYSIS. e2 + cx +- bx' + ax3= e2 + cx + bxt + e (4e - c2) 3 8e0 {4be22- ca)4 (x4 4. And, as the first three terms (e2 + cx + bx2) 64e6 are now common, there will arise, by expunging them, and then multiplying them by 64e6, 64ae6- = 8ce2 (4be -- c2) x3 + (46e2 - C2)24. Whence, by dividing each side of this last equation by x3, and reducing the result, we shall have 64ae6 - 8ce2 (4be2 - c2) IX = (4be -_ c2)2 which last method gives a new value of x, different from that before obtained. It must be observed, however, that each of these forms fail when the second and third terms of the given formula are wanting, or b and c each = 0.* 3. When neither of the above rules can be applied to the question, the formula can be resolved, by first finding, by trial, as in the former problem, some value of the unknown quantity that makes the given expression a square: in which case other values of it may be determined from this, when they are possible, as follows:Thus, let p be a value of x so found, and make ap3 + bp2 j- cp + d = q'; Then, by putting x = y + p, we shall have ap3 + bp2 + cp + d = a (y +p)3 + b (y + p)2 + C (y + p) + (z = ay3 + (3ap + b) y2 + (3ap2 + 2ap + c) y + apa + bp2 + cp + d, or ax' + bx2 + cx J+ d = ay3 + (3ap + b) y2 + (3ap2 + 2ap + c) y + q2. From which latter form, the value of y, and consequently that of x, may be found by either of the methods given in Case 2. It may also be further remarked, that if the given formula, * In the first of these methods, the assumed root, e +- x, is determined by first taking it in the form e + nzr, and then equating the second term of it, when squared, with the corresponding term of the original formula; when it will be found that Xn --. In like manner the assumed root e + xe+ x2, in the second method, is determined by first taking it in the form e -x +n-mxz, and then equating the second and third terms of it, when squared, with the corresponding terms of the given formula, when it will be found that c 4be2-c2 na= and m = 2e ~~~8e3 DIOPHANTINE ANALYSIS. 185 in any case of this kind, can be resolved into factors, such that one of them shall be a square, it will be sufficient to make the remaining factor a square, in order to render the whole expression so; since a square, multiplied or divided by a square, is still a square.* EXAMPLES. 1. It is required to find such a value of x as will make llx3 + 3x2 a square. Let the given expression 11x3 + 3x2 - n2a2: agreeably to (Case 1. Then, by dividing by x2, we shall have 11 x + 3 - n; n' - 3 and consequently, x; where n may be any number, positive or negative, that is greater than / 3. Taking, therefore, n = 2, 3, 4, 5, &c., respectively, we 1 6 13 shall have, in this case, x or 2, the last of which is the least integral answer that the question admits of. 2. It is required to find such values of x as will make X3 - 2x'2 + 2x + 1 a square. Here the last term 1, being a square, let 1 + 2x - 2xa + X3- (1 + x)2 = 1 + 2x + -2, agreeably to the first part of Case 2. Then, since the first two terms, on both sides of the equation, destroy each other, we shall have x3 - 2x2 =- t2, or -3 3x2, and consequently x = 3; which, by substitution, makes 1 + 2x -2x2 + 3 = 1 + 6-18 + 27 = 16 a square, as required. Again, by putting x -= y + 3, according to Case 3, we shall * The method of determining the factors of which any formula is composed, when it can be done, is to put the given expression = 0, and then find the roots r, r', &c., of the equation so formed; each of which will give a factor - r, x —', and these are generally easily discovered, as we here seek only the rational roots, which are always divisors of the absolute term, or of that which does not contain z. Thus the formula z3 — X2 -x + 1 is resolvable into the factors (1 - x) X (1-x) X (1- x), or (1 -x)2 X (1 x); and by putting 1 +x-. = %2, we have x = n2 - 1; where, if X? be taken equal to any number whatever, 3 -X2 -. + 1 will be a square; though, by any other mode of solution, it would be difficult to find even tawo or three values of x. It may here also be observed, there are but few questions in this problem that can be determined in whole numbers. Several of them, likewise, admit only of one answer, and others are totally irresolvable, either in integers or fractions. Thus, if it were required to make X3 + 1 a square, the only positive value of x that renders this possible, is 2; and the making of 3x2 - 1 a square, is impossible. 16* 186 DIOPHAINTINE ANALYSIS. have 1 + 2x -2x2 X3 = 1 + 2 (y + 3) —2 (y + 3)2 + (y- +3)3 = 16 + 17y-+ 7y2 + y3. And consequently, by making 16 + 17y + 7y2 + y3 = 17 289 (4 + 8 y)2_ 16 + 17y + - y2, agreeably to the first part of Case 2, by cancelling 16 + 17y, there will arise 7y2 + y3 = 289 2 289 y, or y + 7 =289 289 - 448 159 Whence y= -- = —andx = 3 - 64 64 64 159 192 - 159 33 4 = 64 -- -6' for another value of X. 64 64 64' Which number, being substituted in the original formula, 429025 /655\ 2 makes 1 +I2-2x2+x 3262144 5x- 2 a square, as before. 3. It is required to find such values of x as will make 3x3 - 5x2 + 6x + 4 a square. Here, 4 being a square, let 4 + 6x - 5x2 + 3x3 = (2 + ax)2 4 + 6x + -x ~2, as in the first part of Case 2. Then, since the first two terms on each side of the equation destroy each other, we shall have 33 - 5x2 — x2, or 5+-4 29 3x - 5 —, and consequently, in this case, x 4 3 12' 29 29 45 Whence (2 +- X -)2-= (2 + 2)2 (-)2 a square, as 12 - 8 8 was required. Or, by the second method of the same Case, let 4 + 6 x 29 87 - 5x2 + 3x- (2 + -X- + —)2 -4 + 6x- 5x - 6 3 + 841 X 4; then, as the first three terms on each side of this 256 841 87 equation destroy each other,.we shall have 2 —-4 6 x3 256 16 841 87 =3x3, or -- - ---- 3, or 841lx- 1392 768; and con256 16 1392 + 768 2160 sequently, x 841 84 which is another value of x, that, being substituted in the original formula, will make it a square. DIOPHANTINE ANALYSIS. 187 4. It is required to find such values of x as will make X3 + 3 a square. Hdre, it is evident that the expression is a square when x -- 1. Let, therefore, x 1 + y, and we shall have 3 + x3 = 4 + 3y + 3y2 + y3. And, as the first part of this is a square, make, according to the first part of Case 2, 4 + 3y + 3y2 +- y =(2 +-y)2 4 + 3y+ - fy2. Then, because the first two terms on each side of the equation destroy each other, we shall have 3 + 3y2 = Ay92, or y + 3 o -- 9 9 -48 39 39 Whence y = -- 3 16 -- = —-, and x =1 - -= 1he6cey 16= 16161 16- 39 23; which is a second value of x. 16 16 3 39 y)2 Again, let 4 - 3y + 3y2 y3 (2-+ y + y2)2+ 4 64 17 1 521 3y + 3y2 + - Y3 + 401521 y4 according to the second part of 128 4096 Case 2. Then, as the first three terms on each side of the equation 1521 117 destroy each other, we shall have 4- y4 + __ y3, or 1521 117 4096 128 352 352 1873 Whence, also, y- 152' and 1 + - - 52which 1521 1521 521 is a third value of x. And by proceeding in the same way with either of these new values of x as with the first, other values of it may be obtained; but the resulting fraction will become continually more complicated in each operation. PROBLEM 3.-To find such values of x as will make V (ax4 + bx3 + cx2 + dx + e) rational, or ax4 + bx3 + cx2 + dx + e = a square. The resolution of expressions of this kind, in which the indeterminate, or unknown quantity, rises to the fourth power, is the utmost ~limit of the researches that have hitherto been made on the formula affected by the sign of the square root; and in this problem, as well as in that last given, there are only a few particular cases that admit of answers in rational numbers; the rest being either impossible, or such as afford 188 DIOPHANTINE ANALYSIS. one or two simple solutions; which may generally be found as follows:' — RULE 1.-When the last term e, of the given formula, is a square, put it =f2, and make f2 + dx + Cx2 + bx3 + a X4 d 4cf2 - d222 2 d(4cf2 - d2) (f + - X + - 2)2 _f2 + dL + Cx - + 1X3 + 2f 8f f (4cf - d2) _ 64f6 Then, by expunging the first three terms, which are common to each side of the equation, there will remain bx3 + 4 (4._ - 2) 3 + (4f" - d 2)24 ax4 c X3 - 44. And consequently, by Sf4 64ff dividing by x3, and reducing the result, we shall have x - 64bf6 - Sdf2 (4cf2 - d2) (4Cf2 - d2)2 -- 64af6 which form fails when the coefficients c and d, or b and d, are each - 0. 2. When the coefficient a, of the first term of the formula, is a square, put it -- g2, and make g2x4 + bx3 + CX2 + dx + e 6 4c2- b2 b(4c-2 -2b)2 (g-X2 +_ X + 3 )2 = 44 rbx3+CX2+ 0 4 -AX (4cg2 -. b2)2 b(4cg _ b2) (4ce - b2)2 + g6 64g' Then, dx + e - )(c-62 (4cr 62)2 and conse(4cg2 - b2)2 - 64eg6 qtuently, x 64dg _ 8b (4cg2 _ b); which form likewise fails under similar circumstances with the former. 3. When the first and last terms of the formula are both squares, put a — g2, and e — f2, and make f2 + dx + cx2 + d d bx3 + g2x4 (f + X + gx2)2 f2 + dx + (2fg + )2 + 2f i4f2 dg d.2 dg. dg 3- g2x4. Then, cx2 +6 bx3 (2fg - a)2 ~- f 4f2 f And consequently, x.2 (2fg- c) - d2. f(bf- dg) Or, because g enters the given formula only in its second * As an instance of what is above said,.it may be observed, that the only value of x that renders the formula 2x4 - 3x2 + 2 a square, is 1; and the formula x4 —x-2+F1 can never be a square, except when x=+, or -1. DIOPHANTINE ANALYSIS. 189 power, it may be taken either negatively or.positively; and consequently, we shall also have - 4 ( + f(bf + dg) So that this mode of solution furnishes two different answers. Also, if there be taken -for another supposition f2 + dx b bf + CX2 4+ bX3 +24 g = (f+ - X +g2)2-f2 +- X + (2fg b2 -d -- 2 2 + bx3 +d g24, hence, by cancelling, dx + cx2 = bfx 62 g (bg — hf) ~r (2fg - ) x2; and consequently, x-Ih2 ( gf (= f — ) And because f enters the given formula only in the second power, it may be taken either negatively or positively; and (bg + bf) consequently, we shall also have x = b g2(2fg c) So that this solution likewise furnishes two values for x, which are each differetnt from the former. But these forms all fail under similar circumstances with those of the second Case. 4. When neither the first nor the last terms are squares, the formula cannot be resolved in any other way than by first endeavouring to discover, by trials, some simple value of the unknown quantity that will answer the conditions of the question, and then finding other values of it according to the methods pointed out in the last two problems. Thus, let p be a value of x so found, and make ap4 + bp3' — cp2 + dp + e = q2. Then, by putting x = y + p, we shall have ap4 - bp3 + cp' + dp + e = a (y + p)4 + b (y + p)3 + C (y + p)2 + d (y + p) -+ e ay4 + (ap + b) y3+ (6ap2 + 3bp +c) y2 + (4ap3, + 3bp2 + 2cp + d)y + ap4 +.bp3 + cp + C dp + e, or ax4 + bx3 + c~X + do - e = ay4 + (ap + b)y3 + (6ap2 - 3bp + c) y +- (4ap3 +- 3bp2 ~ 2cp +- d) y + q2. From which last formula, the value of y, and consequently that of x, may be found by Case 1. EXAMPLES. 1. It is required to find such a value of x as will make 1 -- 2x- +3x2 - 4x3 54 a square. Here,'the first term 1, being asquare, let'1 - 2x 4- 3x24x + 5x4 =(1 - x+ x2)2 =1 — 2- + 3x2 - 2x3 + x4, agree ably'to the method in Case 1. Then we shall have 5x4 4x3 - x4 - 2x3. And consequently, 5x — 4 = x -'2; whence x = 4 190 DIOPHANTINE ANALYSIS. And consequently, 1 - 2x + 3x2 - 4x3 + 5x4 = 1 - 1 - 5 9 -- + 1which is a square number, as was required. 2. It is required to find such a value of x as will make 4x4 _ 2x3 - x2 + 3x - 2 a square. Here, the first term being a square, let 4x4 - 2x3 - x2 5 5 25 3x -2 =(2x2 - x - — )2 = 4x4 _ 2x3_ 2x2 - X -- 16 12 + 6 + 256' according to the method in Case 2. 5 f25 5 Then we shall have 3x -2 -- x + 2- or 3x - — x=2 16 256 16 25.4 —. TWhence 768x - 80x = 512 + 25; and conse256 512 + 25 537 quently, x 768-80 - 688 Or, if we put x = -, the formula in that case will become 4 2 1 3 y y y2 y And, therefore, multiplying this by y4, which is a square, it will be 4 -2y - y + 3y - 2y4. Where the first term being now a square, if the expression, so transformed, be 688 1 resolved by Case 1, we shall have y = -7, and x 537 688, as before. 3. It is required to find such values of x as will make 1 + 3x + 7x2 - 2x3 + 4x4 a square. Here, both the first and last terms being squares, let 1 + 3.x + 7X2 2x3 + 4X4 = (1 +3x + 2)2= 1 3x. 25 6x + 4X4, according to the method in Case 3. 25 Then, we shall have 6ax + -- x2 = 7x2 - 2x3; or 6x + 2x 25 3 -7 - —; and, by reduction, a= 32. And if we put the same formula, 1 + 3x + 7x2 - 2x3 + 4x4 - (1+ -- x- 2x2)2 = 1 + 3x -7x -- 623 + 4X4, we shall ha've, by cancelling, 7xr2 - 2x'3 - - -x2 - 6x3; whence 6x - 35 35 2x= - --- 7= -; or = 4 a416 DIOPHANTINE ANALYSIS. 191 And, in a similar manner, other values of x may be found, by employing the method of substitution pointed out in the latter part of Case 3. 4. It is required to find such values of x as will make 2x4 - 1 a square. Here, 1 being an obvious value of x, let, according to Case 4, x = 1 + y. Then 2x4- 1 =2(1 -y)4 -1 =2(1 +4y+-6y2+4y3 + y4) -1. 1 = + 8y + 12y2 +- 8y3 + 2y4. And since the first term of this last expression is now a square, we shall have, by Case 1, 1 + 8y + 12y2 + 8y3 + 2y4 = (1 + 4y - 2y= )2 -1 + 8y + 12y2 - 16y3 - 4y4. Whence, as the three first terms of the two members of this equation destroy each other, there will remain 4y4 - 16y.3 = 2y4 + 8yS; or y 12; and consequently x = 1 + y - J 3; which value being substituted for x, makes 2x4 - 1 = 57121 = (239)2, as required. And if 13 be now taken as the known value of x, and the operation be repeated as before, we shall obtain, for another value of x, the complicated i10607469769 fraction 2447192159 PROBLEiM 4.-To find such values of x as will make aV (a3x' bx2 + cx + d) rational, or ax3 + bx2 + cx + d = a cube. This formula, like the two latter of those relating to squares, cannot be resolved by any direct method, except in the cases where the first or last terms of the expression are cubes; it being necessary, in all the rest, that some simple number answering the conditions of the question, should be first found by trial before we can hope to obtain others, but when this can be done, the problem, in each of the cases here mentioned, may be resolved as follows:RULE 1. —When the last term d of the given formula is a cube, put it = e3, and mnake e3 + cx + bxa + ax3 - C C2 C3 (e + T 2-)3= + ~ + ~ 3 + 27- 6 XThen, by expunging the two first terms on each side of the equation, which are common, there will remain aXs3 + C3 C2 bx2 -= - 3 + 3 x2; whence, by division and reduction, we shall have 27ae6x - 27be8 -- cax + 9cse3, and consequently x-9e(3b -- c2); which form fails when the coefficients b C3 - 27aea6 and c, or a and c, are each equal 0. 192 DIOPHANTINE ANALYS~TI.. 2. When the coefficients a of the first term is a cube, put it = f3, and make f3x3 + bx2 + cx + d — (fx + b )3 = -f33 3f2 bo.. b3 + bx2 + a —z + bx. Then, by expunging the two first terms on each side of the equation, as before, there will remain cx +-'id 3- X + 2-f; whence, by multiplying by 27f6 we shall have 27f6cx + 27df6 -- 9h2f3x + b3, and consequently, x -- b3- 27df46 9f3 (3cf3 - b2); which form likewise fails when b and c, or b and d, are each = 0. 3. When the first and last terms are both cubes, put a = f and d - e3 and make es + cx + bX2c +fx3 = (e — C)3 = e3 + 3-fe2x + 3+f2ef2.fx. Then cx + bex = 3fex -- 3f ex'; Whence we shall have bx - 3fe = 3f2 - c, and conse3fe2 - c quently, x-b 3f; which formula may also be resolved by either of the two first cases. 4. When neither the first nor the last terms are cubes, let p be a value of x, found by inspection or by trials, and make ap3 + bl2 +- p + d = q3. Then, by putting x:= y + p, we shall have ap3 + bp2 + cp + d= a (y + p)3 + h (y + p)2 + C (y + p) + d = ay3 + (3ap + b) y= + (3ap2 J 2bp + c) y + cap~ r bp2 ~ cp + d, or ax' + bx2 +- cx - d = ay3 - (3ap- +b) y2- +(3ap2 + 2bp + c)y + q3. From which latter form, the value of y, and consequently that:of x, may be found as in Case 1. EXAMPLES. 1. It is required to find such a value of x as will make x2- x +- 1 acube. Here, the last: term being a cube, let the root of the cube sought = 1 ~- -x according to Case 1. Then, by cubing, we shall have I +x + xa = 1 + x + ~xP + 3 And since the two first terms on each side of this equation destroy each other, there will remain x2 = -x2 + x i3. Whence, dividing by x2, we shall have -L x + - = 1, or x +9 = 27; and consequently, x = 27 - 9 - 18; which num DIOPHANTINE ANALYSIS. 193 ber, by substitution, makes 1 + - A- x2 = 1 + 18 + 324 = 343 73 a cube number, as was required. And if we now take this value of x, and proceed according to the method employed in Case 4, we shall obtain x - 137826 1538653; which last number will also lead, in like manner, to other new values. 2. It is required to find such a value of x as will make X3 + 32- + 133 a cube. Here, the first term being a cube, let its root = 1 + x, according to Case 2.'Then, by cubing, we shall have 133 + 3x2 x3 = ( 1 + x)3 1 + 3x + 3x2 -+ x. And since the two last terms of this equation destroy each other, there will remain 1 + 3x = 133, or 3x = 133 - 1 13_ 132; whence x 3 = 44, and x+3x2 133 -- 91125 = (45)3, a cube number, as was required. And if 45 be now taken as a known value of x, other values of it may be found, as in the last example. 3. It is required to find such a value of x as will make 8 + 28x + 89x2 - 125x3 a cube. Here, let the root sought = 2 - 5x, according to Case 3. Then, by cubing, we shall have 8 A- 28x + 89x2 - 125x3 (2 - 5 x) 8 - 60x -1- 1502 -- 125x3. knd since the'first and last terms of this equation destroy ea h other, there will remain 28x +- 89x2-= - 60x + 15Ox2. Whence, by dividing by x, and transposing the terms, we Fsall have 150x - 89x = 28 + 60, or 61x -= 88; and conse88 quently x =. 61 And as this formula can also be resolved either by the first or second Case, other values of x may be obtained, that will equally answer the conditions of the question. 4. It is required to find such a value of x as will make 2x3- 3x - 7 a cube. Here, - I being a value of x that is readily found, by inspection, let x = y - 1, agreeably to Case 4. Then, by substitution, we shall have 2x3 - 3x +7 2(y - 1)' - 3 (y - 1) + 7 = 2y - 62 4+'3y +- 8. And as the last term of this expression is a cube, let 21 3 1 8 + 3y - 6y + 2y = (2 + y)3 = 8 + 3y +f + 17 194 DIOPHANTINE ANALYSIS. according to Case 1. Then, by expunging the equal terms on 1 each side, there will remain 2y3 - 6y2 = ] y2 + 64 y. Whence, dividing by y2, and reducing the terms, we shall have 128y —384 =y + 24, or 127y -408; and conse408 408 281 quently, y-1-, and x-127 1 12='127 127 127 Which number, by substitution, makes 2x3 - 3x + 7 = 2 X (281)3 281 45118016 356 reuire - -3 X 19-7 -]7 = — ==- \12-7 as required. (127)3 127 2048383 -127 And, by talking this last as a new value of x, others may be determined by the same method. PROBLEM 5.- Of the resolutions of double and triple equalities When a single formula, containing, one or more unknown quantities, is to be transformed into a perfect power, such as a square or a cube, this is called, in the Diophantine Analysis, a simple equality; and when two formula, containing the same unknown quantity or quantities, are to be each transformed to some perfect power, it is then called a double equality, and so on; the methods of resolving which, in such cases as admit of any direct rule, are as follows:RULE 1. —In the case where the unknown quantity does not exceed the first degree, as in the double equality, ax - b —, and cx - d = l, let the first of these formulae ax b z2, and the second cx +- d = w2. Then, by equating the two values of x, as found from these equations, we shall -have cz2 + ad - be = aw2, or acza + a (ad - bc) = a2w2. And since the quantity on the righthand side of this equation is now a square, it only remains to find such a value of z as will make, when the question is resolvable, acz2 + a (ad - be) = E; which being done, according to the method -2 b pointed out in Problem 1, we shall have x =a 2. When the unknown quantity does not exceed the second degree, and is found in each of the terms of the two formulae; as in the double equality ax2 bx = 0, and cx2 + dx = ~. Let x= -; then, by substitution, and multiplying each of the resulting expressions by y2, we shall have a + by, and c - dy +-10, 'DIOPHANTINE ANALYSIS. 195 from which last formulae, the value of y, when the question is possible, and consequently that of x, may be determined as in Case 1. But if it were required to make the two general expressions ax2 + bx + c = O, and dx2 + ex +f= L, the solution could only be obtained in a few particular cases, as the resulting equality would rise to the fourth power. In the case of a triple equality, where it is required to make ax + by = l, cx +- dy - =O, and ex +fy = -, let the first of them ax + by = u2, the second cx + dy = v2, and the third ex q-fy -: w2. Then, by first eliminating x in each of these equations, and afterwards y in the -two resulting equations, we shall have (af - be)v2 - (cf - de) t2 = (ad - bc) w2; or, putting v =uz, and reducing the terms, the result will give af —be cf_ — d_ _ o the simple equality af2-be z2 where the ad - bc ad e - bc righthand member being a square, it only remains to find a value of z that will make the lefthand member a square; which, when possible, may be done by Problem 1. Hence, having z, we have, as above, v - ut; and the first d - b2 az 2-c 2 two equations will give x -- t — u and y - - ad -bc ad - bc where u may be any whole or fractional number whatever. But if the three formulae, here proposed, contained only one variable quantity, the simple equality, to which it would be necessary to reduce them, would rise, as in the last case, to the fourth power, and be equally limited with respect to its solution. 4. In other cases of this kind, all that can be done is to find successively by the former rules, several answers, when one is known; and if neither this nor any of the abovementioned modes of solution are found to succeed, the problem under consideration can only be determined by adopting some artifice of substitution that will fulfil one or more of the required conditions, and then resolving the remaining formulte, when they are possible, by the methods already delivered for that purpose; but as no general precepts can be given, for obtaining the solution in this way, the proper mode of proceeding, in such cases, must chiefly depend upon the skill and sagacity of the learner. EXAMPLES. 1. It is required to find a number x, such that x + 128 and Xi + 192 shall be both squares. 196 DIOPHANTINE ANALYSIS. Here, according to Case 1, let x + 128 -w2, and x + 192 2 2 Then, by eliminating x, and equating the result, we shall have w2 - 128 - z- 192, or w2 + 64 z2. And as the quantity on the righthand side of the equation is now a square, it only remains to make wt -+ 64 a square. For which purpose, put its root - to + n; then w2 + 64 = W2 + 2nw +- n2, or 2nw -- n2 = 64; and consequently, w = 64 -n2 2 —'; where, taking n, which is arbitrary, = 2, we shall 2n 64 —4 60 have wo-:= 15; and consequently x -w2 - 128 = 152 - 128:225 - 128 97, the answer. 2. It is required to find a number x, such that x2 + x and x"-. shall be both squares. Here, according to Case 2, of the last Problem, let x 1 1 I 1 1; then we shall have to make -2+ -, and --- squares; y Y Y2 Y or, by reduction, 2+ y) =, and 2(1 - Or, since a square number, when divided by a square number, is still a square, it is the same as to make 1 + y= —, and 1 — y-=; For this purpose, therefore, let 1 +-y - z2, or y = z2 1; then I - y = 2 - z2; which is also. to be made a square. But as neither the first- nor the last terms of this formula are squares, we must, in order to succeed, find some simple number that will answer the condition required; which, it is evident from inspection, will be the case when z- 1. Let, therefore, z = 1 - wu, agreeably to Problem 1, Case 7, and. we shall have 1- y=-2-z 2 - (1-w)21 + 2w - w2;. or y w2 - 2w; ()r, putting 1 - nw for the root of the former of these exporessions, there will arise, by squaring, 1 + 2w w2= 1 - 2nw + n2w2. Whence, expunging the 1 on each side,.and dividing by w, we shall have 2 - w = — 2n + n2w; and consequently 2n2 + 2 1 1 _ (n-+ 1)2 nw2 -I r Iy = ws - 2w 4n - 4n where, in order to render the value of x-positive, n may be taken equal to any proper fraction whatever. DIOPHANTINE ANALYSIS. 197 Or if, for the sake of greater generality, - be substituted for n, we shall have (7i2 +1- n2)2 4mn (n2 - m2)' where m and n may now be taken equal to any integral numbers whatever, provided n be made greater than m. 25 If, for instance, n = 2 and m = 1, we shall have x = A-; 169 and if n _ 3 and m = 2, x 1 9-; and so on, for any other number. 3. It is required to find three whole numbers in arithmetical progression, such, that the sum of every two of them should be a square. Let x, x ~ y, and x + 2y, be the three numbers sought; and put 2x + y = U2, 2x + 2y = v2, and 2x + 3y wa, agreeably to Case 3. Then, by eliminating x and y from each of these equations, we shall have va -- u w2 v2, or 2vu2 - us _ v2 And if we now put v = uz, there will arise 2u2z2 _- u2 w2; or, by dividing by it2 2z2 - 1 -; where the righthand member being a square, it only remains to make 2za - 1 a square, which it evidently is when z = 1. But as this value would be found not to answer the conditions of the question, let z= 1 - p; then 2Z2 - 1 = 2 (1 - p)2 - 1 = 1 - 4p -'2p2. And consequently, if this last expression be put = (1 - up)2, we shall have, by squaring, 1 - 4p + 2p2 -- 1 - 2np + np2, or - 4 +- 2p= -2n + np; whence 2n - 4 2n - 4 n2 -2n- 2 2p and zl- -- a' - 2 n- 2 =-2 - 2 Or if, for the sake of greater generality, - be substituted for n in the last expression, we shall have m2 - 2ran -+ 2n2 12 2n2 And since, by the two first equations, y = v -- 2 -- u22 - Ua2 -2 2 -1)u2, and xZ1(2U2 - - (2 -_ Z) U2, it iS evident that z must be some number greater than 1,. and less than /2 1:7t 198 DIOPHANTINE ANALYSIS. If, therefore, m = 9 and n- 5, we shall have;81 - 90 + 50 41 241 u2 720 Z ] _-5() = =31 X,and y =-X. Z 81 - 50 31 2n YX Or, taking u - 2 X 31, x = 482, and y - 2880, we haie x = 482, x + y = 3362, and x + 2y = 6242, which are the numbers required. 4. It is required to divide a given square number into two such parts that each of them shall be a square.' Let a2 = given square number, and x2 and a2 - a2 its two parts. Then, since X2 is a square, it only remains to make a2 _- 2 a square. For which purpose let its root = nx - a, and we shall have a2 - X2 =i n2x2 - 2anx + a2, or -x - n2x2 - 2anx; 2a~n whence, by reduction, _2 the root of the -first part, wn2 + 1' 2an2 an2 -- a and nx - a - - a + p the root of the second. n2-{- 1 n2 - 1 Therefor' an an2 - a\ Therefore n2 q _ 1} and 2 —— _1)2 are the parts required; \ / n2 +J / where a and n may be any numbers taken at pleasure, provided n be greater than 1. 5. It is required to divide a given number, consisting of two known square numbers, into two other square numbers. Let a2+ b2 be the given numbers, and X2, y2, the two required numbers, whose sum, _2 + y2, is to be equal to a2 + b2. Then it is evident, that if x be either greater or less than a, y will be accordingly less or greater than b. Let, therefore, x = a + mz, and y = b - nz, and we shall have a2 - 2amz + m2z2 +- b2 - 2bnz + n2z2 = a,2 + b2. Or, by transposition and rejecting the terms which are common to each side of the equation, n2z 2 + n2Z2 = 2bnz - * To this we may add the following useful property:If s and r be any two unequal numbers, of which r is the greater, it can then be readily shown, frinm the nature of the problem, that 2rs, S2 —r2, and s2 +- r2. will be the perpendicular, base, and hypothenuse of a right-angled triangle. From which expressions, two square numbers:may be found, whose sum or difference shall be square numbers; for (2-rs)2 +-(s2 —r2)i-= (s2 4- r2)2, and (s2 + r2)'- (2rs)2 = (s2-r2), or (s2 +- 2)2 -(s2 -r2)2 (2rs)2; where s and r may be any numbers whatever, provided r be greater than s. DIOPHIANTINE ANALYSIS. 199 2amz, or m2z -+ n2z = 2bn - 2am; whence 2bn - 2am 2bmn ~ a(n' - rM2) 2amnn+b(m2. —n2) z= -- - na,x -, 1/ - m2 q- -n. mJ7 + n2 r n2 + m2 + n2 where mn and n may be any numbers, taken at pleasure, provided their assumed values be such as will render the values of x, y, and z, in the above expressions, all positive. 6. It is required to find two square numbers, such that their difference shall be equal to a given number. Let d- the given difference; which resolve into two factors a, b, making a the greater and b the less. Then, putting = the side of the less square, and x + b = side of the greater, we shall have (x + b))2 - x2 = z2 + 2b +J b2 - x2 = d =(ab), or 2bx + b2 = d = (ab). Whence, dividing each side of this equation, by b, we shall a-b have x == - = the side of the less square sought, and a-b a+ b. +~ b = + b = the side of the greater. ~2 2 If, for instance, d = 60, take a X b = 30 X 2, and we shall 30 —2. 30+2 have x= -- 14, and x 2 16, or 162- 142 2 2 256 - 196 = 60, the given difference. 7. As an instance of the great use of resolving formula of this kind into factors, let it be proposed, in addition to what has been before said, to find two numbers, x and y, such that the difference of their squares, x2 - -y2, shall be an integral square. Here the factors of x2 - y2, being x + y and a - y, we shall have (x + y) X (x - y) = x - y2. And since this pro-'duct is to be a square, it will evidently become so, by making each of its factors a square, or the same multiple of a square. Let there be taken, therefore, for this purpose, X + y = mr2, x. y - ims2. Then, by the question, we shall have (x + y) X (a - y) or its:equal 2 - yo- mr2s2; which is evidently a square, whatever may be the value of rn, r. s. But by addition and subtraction, the. above equations give, when properly reduced, - m (r' + s2) _m (_ Z__ S2);2' 2:where, as above said, m, r, and s, may be assumed at pleasure. Thus, if we take in = 2, we shall have x = r + s2a, and y = r" = s2, which expressions will obviously give integral 200 DIOPHANTINE ANALYSIS. values of x and y, if r and s be taken = any integral numbers. 8. It is required to find two numbers, such that., if either of them be added to the square of the other, the sums shall be squares. Let x and ybe the numbers sought; and consequently fa2 + y and y2 + x the expressions that are to be transformed into squares. Then, if r — x be assumed for the side of the first square, we shall have a2 + y = r2 - 2rx + x2, or y = r - T2 y 2rx; and consequently, x= 2r And if s + y be taken for the side of the second square, we shall have y2 r - Y = S2 + 2sy + y2; or, by reducing the 2r. equation,.r2- y = 4rsy + 2rs2, and consequently, by rer22 —r2 2r2s + S2 duction, y 1 and x 4rs 1; where r and s may be any numbers, taken at pleasure, provided r be greater than 2s2. 9. It is required to find two numbers, such that their sum and difference shall be both squares. Let x and x2 _- be the two numbers sought; then, since their sum is evidently a square, it only remains to make their difference, x2 - 2x, a square. For this purpose, therefore, put the root = x - r, and we shall have X2 - 2x = x2- 2rx +- r2; Or, by transposition, and cancelling x2 on each side of the equation, 2rx - 2x = r2: whence and x 2x = 2= - 2r- 2' r - 1 where r may be any number, taken at pleasure, provided it be greater than 2. 10. It is required to find three numbers, such that not only the sum of all three of them, but also the sum of every two, shall be a square number. Let 4x, xc2 - 4x, and 2x + 1, be the three numbers sought; then, 4x + (X2 - 4x) xa2, (x2 - 4x) + (2x +- ]) x2 -2x - 1, and 4x + (x2- 4x) + (2x + 1)=- X2 2 + 1, being all squares, it only remains to make 4x + (2x + 1), or its equal, 6x + 1, a square. For which purpose, let 6x + 1 n2, and n2-1 we shall have, by transposition and division, = 6; DIOPHANTINE ANALYSIS. 201 4n2 - 4 (n2 _ 1)2 4n2-4 2n2 - 2 whence' -- - and - 1, or 6 36 6 6 2nP2- 2 n4 - 26n2 + 25 n2- 2 their equals, and -, are the 3 36 3 numbers required. Where n may be any number, taken at pleasure, provided it be greater than 5. QUESTIONS FOR PRACTICE. 1. It is required to find a number x, such that x + 1 and i - I shall be both squares. Ans. x = — 4. 2. It is required to find a number x, such that x +- 4 and x + 7 shall be both squares. Ans. -- -.e 3. It is required to find a number x, such that 10 — t, and 10 - x shall be both squares. Ans. x - 6. 4. It is required to find a number x, such that x2 + 1 and ~+l 1 shall be both squares. Ans. 95. It is required to find three integral square numbers, such that the sum of every two of them shall be squares. Ans. (528)2, (5796)2, and (6325)2. 6. It is required to find two numbers, x and y, such that x2 +- y and y +~ x shall be both squares. Ans. x- =, and y =- -. 7. It is required to find three integral square numbers that shall be in harmonical proportion. Ans. 25, 49, and 1225. 8. It is required to find three integral cube numbers, x3, y3, and z3, whose sum may be equal to a cube. Ans. x - 3,-y = 4, z = 5. 9. It is required to divide a given square number (100) into two such parts, that each' of them may be a square number. Ans. 64, and 36. 10. It is required to find two numbers, such that their difference may be equal to the difference of their squares, and that the sum of their squares shall be a square number. Ans. 4 and a. 11. To find two numbers, such that if each of them be added to their product, the same shall be both squares. Ans. - and -. 12. To find three square numbers in arithmetical progression. Ans. 1, 25, and 49. 13. To find three numbers in arithmetical progression, suCh that the sum of every two of them shall be a square number. Ans. 120~, 8401, and 15601. 14. To find three numbers, such that if to the square of 202 -DIOPHANTINE ANALYSIS. each, the sum of the other two be added, the three sums shall be all squares. Ans. 1, -3, and J36 15. To find twb numbers in proportion as 8 is to 15, and such that the sum of their squares shall be a square number. Ans. 576 and 1080. 16. To find two numbers, such that if the square of each be added to their product, the sums shall be both squares. Ans. 9 and 16. 17. To find two whole numbers such, that the sum or difference of their squares, when diminished by unity, shall be a square. Ans. 8 and 9. 18. It is required to resolve 4225, which is the square of 65, into two other integral squares. Ans. 2704 and 1521. 19. To find three numbers in geometrical proportion, such that each of them, when increased by a given number (19), shall be square numbers. Ans. 81, -, and 2, -. 20. To find two numbers, such that if their product be added to the sum of their squares, the result shall be a square number. Ans. 5 and 3, 8 and 7, 16 and 5, &c. 21. To find three whole numbers such, that if to the square of each the product of the other two be added, the three sums shall be all squares. Ans. 9, 73, and 328. 22. To find three square numbers, such that their sum, when added to each of their three sides, shall be all square numbers. Ans. -J4 ~ 19.5_ o and _-LLL - roots required. 62 92 09 6292 0 2920 -- 23. To find three numbers in geometrical progression, such, that if the mean be added to each of the extremes, the sums, in both cases, shall be squares. Ans. 5, 20, and 80. 24. To find two numbers such, that not only each of them, but also their sum and their difference, when increased by unity, shall be all square numbers. Ans. 3024 and 5624. 25. To find three numbers such, that whether their sum be added to, or subtracted from, the square of each of them, the numbers thence arising shall be all squares. Ans. 59~6 e 9 6- and 2. 26. To find three square numbers such, that the sum of their squares shall also be a square number. Ans. 9, 16, and -2 ~' 27. To find three square numbers such, that the difference of every two of them shall be a square number. Ans. 485809, 34225, and 23409. 28. To divide any given cube number (8), into three other cube numbers. Ans. 1, 674 and -'}2 29. To find three square numbers such, that the difference SUMMATION OF INFINITE SERIES. 203 between every two of them and the third shall be a square number. Ans. 1492, 2412, and 2692. 30. To find three cube numbers such, that if from each of them a given number (1) be subtracted, the sum of the remainders shall be a square number. Ans. 3a-9a - 3Al- o and 8 OF THE SUMMATION AND INTERPOLATION OF INFINITE SERIES. THE doctrine of Infinite Series is a subject which has engaged the attention of the greatest mathematicians, both of ancient and modern times;. and when taken in its whole extent, is, perhaps, one of the most abstruse and difficult branches of abstract mathematics. To find the sum of a series, the number of the terms of which is inexhaustible, or infinite, has been regarded by some as a paradox, or a thing impossible to be done; but this difficulty will be easily removed, by considering that every finite magnitude whatever is divisible in infinitum, or consists of an indefinite number of parts, the aggregate, or sunm of which, is equal to the quantity first proposed. A number actually infinite, is, indeed, a plain contradiction to all our ideas; -for any number that we can possibly conceive, or of which we have any notion, must always be determinate and finite; so that a greater may still be assigned, and a greater after this; and so on, without a possibility of ever coming to an end of the increase or addition. This inexhaustibility, therefore, in the nature of numbers, is all that we can distinctly comprehend by their infinity: for though we can easily conceive that a finite quantity may become greater and greater without end, yet we are not, by that means, enabled to form any notion of the ultimatum., or last magnItude, which is incapable of farther augmentation. Hence we cannot apply to an infinite series the common notion of a sum, or of a collection of several particular numbers, which are joined and added together, one after another; as this supposes that each of the numbers composing that sum, is known and determined. But as every series generally observes some regular law, and continually approaches towards a term, or limit, we can easily conceive it to be a whole of its own kind, and that it must have a certain real value, whether that value be determinable or not. Thus, in many series, a number is assignable, beyond which no number of its terms can ever reach, or, indeed, be ever 204. SUMMATION OF INFINITE SERIES. perfectly equal to it; but yet may approach towards it in such a manner as to differ from it by less than any quantity that can be named. So that we may justly call this the value or sum of the series; not as being a number found by the common method of addition, but such a limitation of the value of the series, taken in all its infinite capacity, that, if it were possible to add all the terms together, one after another, the sum would be equal to that number. In other series, on the contrary, the aggregate, or value of the several terms, taken collectively, has no limitation; which state of it may be expressed by saying, that the sum of the series is infinitely great; or, that it has no determinate or assignable value, but may be carried on to such a length, that its sum shall exceed any given number whatever. Thus, as an illustration of the first of these cases, it may be observed, that if r be the ratio, g the greatest term, and I the least, of any decreasing geometric series, the sum, according to the common rule, will be (rg - I) (r - 1): and if we suppose the less extreme I to be diminished till it becomes - 0, the sum of the whole series will be rg - (r - 1): for it is demonstrable that the sum of no assignable number of terms of the series can ever be equal to that quotient; and yet no number less than it will-ever be equal to the value of the series. Whatever consequences, therefore, follow from the, supposition of rg - (r - 1) being the true and adequate value of the series taken in all its infinite capacity, as if all the parts were actually determined, and added together, no assignable error can possibly arise from them, in any operation or demonstration, where the sum is used in that sense; because, if it should be said that the series exceeds that value, it can be proved, that this excess must be less than any assignable difference; which is, in effect, no difference at all; whence the supposed error cannot exist, and consequently rg' — (r- 1 ) may be looked upon as expressing the true value of the series. continued to infinity. We are, also, farther satisfied of the reasonableness of this doctrine, by finding, in fact, that a finite quantity is frequently convertible into an infinite series, as appears ia; the case of circulating decimals. Thus, two thirds expressed decimallv is -9 -.66666, &c., - - 1 o-o t - 6 q+ Tw-6oL +, &c., continued ad infinituni. But this is a geometric series, the first term of which is %-, and the ratio Jl; and therefore the sum of all its terms, continued to infinity, will evidently be equal to —, or the number from which SUMMATION OF INFINITE SERIES. 205 it was originally derived. And the same may be shown of many other series, and of all circulating decimals in general. With respect to the processes by which the summation of various kinds of infinite series are usually obtained, one of the principal is by the method of differences pointed out and illustrated in Prob. 4, next following. Another method is that first employed by James and John Bernoulli, which consists in resolving the given series into several others of which the summation is known; or by subtracting from an assumed series, when put = s, the same series, deprived of some of its first terms; in which case a new series will arise, the sum of which will be known. A third method, which is that of Demoivre, consists in putting the sum of the series - s, and multiplying each side of the equation by some binomial or trinomial expression, which involves the powers of the unknown quantity x, and certain known coefficients; then taking x, after the process is performed, of such a value that the assumed binomial, &c., shall become = 0, and transposing some of the first terms, a series will arise, the sum of which will be known as-before. Each of which methods, modified so as to render it more commodious in practice, together with several other artifices for the same purpose, will be found sufficiently elucidated in the miscellaneous questions succeeding the following problems:PROBLEM 1. —Any series being given to find its several orders of differences. RULE 1.-Take the first term from the second, the second from the third, the third from the fourth, &c., and the remainders will form a new series, called the first order of differences. 2. Take the first term of this last series from the second, the second from the third, the third from the fourth, &c., and the remainders will form another new series, called the second order of differences. 3. Proceed, in the same manner, for the third, fourth, fifth, &c., order of differences; and so on till they terminate, or are carried as far as may be thought necessary.* EXAMPLES. 1. Required the several orders of differences of the series 1, 22, 32, 42, 52, 62, &c. * When the several terms of the series continually increase, the differences will all he positive; but when they decrease, the differences will be negative and positive alternately.. 1s 206 SUMMATION OF INFINITE SERIES. 1, 4, 9, 16, 25, 36, &c. 3, 5, 7, 9, 11, &c., 1st diff. 2, 2, 2, 2, &c., 2d diff. 0, 0, 0, &c., 3d diff. 2. Required the different orders of differences of the series 1, 2s, 33, 43, 53, 6i, &c. 1, 8, 27, 64, 125, 216, &c. 7, 19, 37, 61, 91, &c., 1st diff. 12, 18, 24, 30, &c., 2d diff. 6, 6, 6, &c., 3d diff. 0, O,, c., 4th diff. 3. Required the several orders of differences of. the series 1, 3, 6, 10, 15, 21, &c. Ans. 1st, 2, 3, 4, 5, &c.; 2dj 1, 1, 1, &c. 4. Required the several orders of differences of the series 1, 6, 20, 50, 105, 196, &c. Ans. 1st, 5, 14, 30, 55, 91, &c.; 2d, 9, 16, 25, 36,' &c.; 3d, 7, 9, 11, &c.; 4th, 2, 2, &c. 5. Required the several orders of differences of the series 1 1 1 1 1 &C. 2' 4' 8' 16' 32' 1 1 1 1"1 1 Ans. 1st, &c.; 2d, 16 3 &c.; 4 8' 6 8 165 321 1. 1 1 3d, — —'&c.; 4th,-, &c. 3d 16' 32 32 PROBLEM 2. —Any series a, b, c, d, e, &c., being given to find the first term of the nth order of differences. RULE. —Let d stand for the first term of the nth differences. n — I1 n- n-2 Then will a-nb - n. 2c —n. d + n 2 2 3 n —I n —2 n-3.. 4 e &c., to n + 1 terms =-, when n is an 2 8 4 even number. n-1 n-i n —2 n-1 And:-a a- nb -- n. c + n. n. 2 2 3 2 n-2 n-3 3. — e, &Jc., to n + 1 terms =;, whenn is an odd number.* * When the terms of the several orders of differences' happen to be very great, it will be more convenient to take the logarithms of the quantity concerned whose differences will be smaller: and when the operation is finished, the quantity answering to the last logarithm may be easily found. SUMIMIATION OF INFINITE SERIES. 207 EXAMPLES. 1. Required the first term of the third order of differences of the series 1, 5, 15, 35, 70, &c. here a, b, c, d, e, &c., - 1, 5, 15, 35, 70, &c., and n= 3. n —1 n-1 n-2 Whence-a +nb - n. c - n. 3 =2 3 a + 3b- 3c + d - I + 15 - 45- +35 4 the first term required. 2. Required the first term of the fourth order of differences of the series 1, 8, 27, 64, 125, &c. Here a, b, c, d, e, &c., = 1, 8, 27, 64, 125, &c., and n _ 4. n-l n-1 n-2 n —1 Whence a —nb n2-c2 3- d+n. 2 n-2 n-3 — e=a-4b-'6c -4d+e= 1 - 32 +-1623 4 256 + 125 — 0; so that the first term of the fourth order is 0. 3. Required the first term of the eighth order of differences of the series 1, 3, 9, 27, 81, &c.*- Ans. 256. 4. Required the first term of the fifth order of differences of the series I, 1 1 &C. Ans. -- 2', 4' S' 16' 32' 64' &c. PROBLEM 3. —To find the nth term of the series a, b, c, d, e, &c., when the differences of any order become at last equal to each other. RULE.-Let d', d". d"', div, &c., be the first of the several orders of differences, found as in the last problem. n-1 n-i n-2 n — n-2 n-3 Then will a + - d' +- 2- d" 2 3 n- n n-2 n-3 n-4 d"' —. div, &c., = nth term required. EXAMPLES. 1. It is required to find the twelfth term of the series, 2, 6, 12, 20, 30, &c. 2, 6, 12, 20, 30, &c. 4, 6, 8, 10, &c. 2, 2, 2, &c. o, 0, &c. ~ The labour in questions of this kind may be often abridged, by putting ciphers for some of the terms at the begining of the series; by which means several of the differences will be equal to 0, and the answer, on that account, obtained in fewer terms. 208 SUMMATION OF INFINITE SERIES. Here 4 and 2 are the first terms of their differences. Let, therefore, 4 - d', 2 = d", and n = 12. n —1 n-1 n-2 Thena+ d dd'+- 1' +55d"=2 + 44 + 110 = 156 = 12th term, or the answer required. 2. Required the twentieth term of the series 1, 3, 6, 10, 15, 21, &c. 1, 3, 6, 10, 15, 21, &c. 2, 3, 4, 5, 6, &c. 1, 1, 1, 1 &c. 0, 0, 0, &c. iHere 2 and 1 are the first terms of the differences. Let, therefore, 2 - d', 1 = d", and n = 20. n — n —1 n-2 Then a d'- d'- =1 q+ 19d' + 171d".- 1 + II 1 2 38 + 171 =_ 210 20th term required. 3. Required the fifteenth term of the series 1, 4, 9, 16, 25, 36, &c. Ans. 225. 4. Required the twentieth term of the series 1, 8, 27, 64, 125, &c. Ans. 8000. 1 1 1 5. Required the thirtieth term of the series 1, 10 3' 6' 10' 15' 1 1 & &c. Ans. -. PROBLEM 4.* —To find the sum of n terms of the series a, b, c, d, e, &c., when the differences of any order become at last equal to each other. RULE. —Let d', d", d'", div, &c., be the first of the several orders of differences. n- n- 1 n-2 n-1 Then will na Jr- d' + n. 2 d" -+ n. 2 2 3 2 n-2 n-3 n-i1 n-2 n-3 n-4 -- d'" +- n.. *div, &c., 3'4 2 3 4 5 to the sum of n terms of the series. e When the differences in this or the former rule are finally = 0, any term, or the sum of any number of the terms, may be accurately determined; but if the difference do not vanish, the result is only an approximation; which, however, may be often very usefully applied in resolving various questions that may occur in this branch of the subject, and which will become continually nearer the truth as the differences diminish. SUMMAlION OF INFINITE SERIES. 209 EXAMPLES. 1. Required the stm of n terms of the series, 1, 2, 3, 4, 5, 6, &c. Here 1,2, 3, 4, 5, 6, &c. 1,I, 1, 1, 1, &;c. 0, 0, 0, 0, &c. Where I and 0 are the first terms of the differences Let, therefore, n = 1, d' - 1, and d" = 0. n-1 n2_ -n n2+n Then will na+ n. -d' 2 snn 2 2 2 of n terms, as required. 2. Required the sum of n terms of the series 12, 22, 32, 42. 52, &c., or 1, 4, 9, 16, 25, &c. Here 1, 4, 9, 16, 25, &c. 3, 5, 7, 9, &c. 2, 2, 2, &c. 0, 0, &c. Where. 3 and 2 are the first terms of the differences.. Let, therefore, a = 1, d 3, and d" 2. n-1 n-1 n —2 Then will na -n.. - d' +-n d" n +- 3n. 2 +fl~23 2 n- 1 n-1 n -2 3n2a- 3n n - 3n2 + 2n + —2n. n ~ + 2- - ~ 2 3 2 3 n X (n + 1) (2n + 1) -=6 -- sum of n terms as required. 3. Required the sum of n terms of the series 13, 23, 33, 43, 53, &C., or 1, 8, 27, 64, 125, &c. Here 1, 8, 27, 64, 125, &c. 7, 19, 37, 61, &c. 12, 18, 24, &c. 6, 6, &c. 0, &c. Where the first terms of the differences are 7, 12, ald 6. Let, therefore, a = 1, d' = 7, d" 12, and d"' - 6. n- - n-l n-2 n- I Then will na + n.- - d' + n.' 2 - 3 d+ n.-2- n-2 n —3 n - I n- I n-2 n-2 n3 + d=n7n 1 12n. n — +6n. 3 4 2 3 3 n —1 n-2 n-3 7n' - 7n 2 3 4 + 2 n4.-6+ I -6n.+ 4n 14n7 — i4n 8n3 -24n q+ 16n 4 +4 71 4 84 18* 210. SUMMATION OF INFINITE SERIES. n4 - 6n73 + 11n2 - 6n n4 - 2n3 + n2 - sum of n terms as 4 4 required. 4. Required the sum of n terms of the series 2, 6, 12, 20, 30, &c. Ans. X (n +1) X (n+ 2 3 5. Required the sum of n terms of the series 1, 3, 6, 10, n n-+1 n'+-2 15, &c. Ans. Y' 2 * 3 6. Required the sum of n terms of the series 1, 4, 10, 20, n+1 nq+2 n+3 35, &c. Ans.. 4 7. Required the sum of n terms of the series 14, 24, 34, 44, n5 n4 n3 n &c., or 1, 16, 81, 256, &c. Ans. + - - 3-T0 8. Required the sum of n terms of the series 15, 25, 35, 45, - 6 n5 55n4 712 55, &c. Ans.6 - +5 -12 6 2 12l PROBLEM 5.-The series a, b, c, d, e, &c., being given, whose terms are a unit's distance from each other, to find any intermediate term by interpolation. RULE.-Let x be the distance of any term y, that is to be interpolated from the first term, and d', d", d"',&c., the first terms of the differences. x —! x —1 — 2 x —1 Then will a - xd'+ - x - d"+- x. --— d"' - 2 2 2 x-2 x-3 div, &c., = y. 3'4 EXAMPLES. 1. Given the logarithmic sines of 10 0', 10 1', 10 2', and 10 3', to find the log. sine of 10 1' 40". Here 1 0' 1~ 1 1~ 1 2' 1~ 3' Sines 8.2418553 8.2490332 8.2560943 8.2630424 71779 70611 69481 -1168 -1130 38 Whence the first terms of the differences are 71779, -- 1168, and 38. Let, therefore, x = 1~ 1' 40"- 1I 0' = 1' 40" = 1- -= dis tance of y, the term to be interpolated; and d' = 71779, d" = — 1168, and d"' = 38. SUMMATION OF INFINITE SERIES. 211 x —1 x 1 x-2 Then will y = a +- xd' + x. — d" + x.- d"=a 22 3 5 d 5 + - d d"'=f 8.2418553 +.0119631 ~0000694 - 3 9 81.0000002 = 8.2538232 s= ine of 10 1' 40", as was required. 2. Given the series 5&c, to find the term 5 51' 52' 53' 54& which stands in the middle between the two terms -2 and 1 1 Ans. 3. Given the natural tangents of 880 54', 880 55', 880 56', 880 57', 88~ 58', 88Q 59', to find the tangent of 88~ 58'18". Ans. 55.711144. PROBLEM 6.-Having given a series of equidistant terms, a, b, c, d, e, &c., whose first differences are small, to.find any intermediate term by interpolation. RLE. —Find the values of the unknown quantity in the equation which stands against the given number of terms, in the following table, and it will give the term required * — 1. a - b - 0. 2. a - 2b ~- c = 0. 3. a — 3b + 3c - d = O. 4. a -4b+6c-4d +e- 0. 5. a - 5b - 10c - 10d + 5e -f= 0. 6. a-6b - 15c -20d + 15e-6f +g=0. n-1 n-1 n —2 a — nb n. n. I c - n.. — n d Jr 2 2 3 n — 1 n-2 n-3 *2. -- e, &;c., = 0. 2 3 4 EXAMPLES. 1. Given the logarithms of 101, 102, 104, and 105, to find the logarithms of 103. Here the number of terms is 4. And against 4, in the table, we have a - 4b + 6c - 4d + 4X (b + d) - (a - e) e = 0; or c - value of the unknown quantity, or term to be found.. The more terms are given, in any series of this kind, the more accurately will the equation that is to be used approximate towards the true result, or answer required. 212 SUaIMATION OF INFINITE SERIES. a = 2.0043214 Where, taking the logs. of b = 2.0086002 01, 102, 104, and 105. d = 2.0170333 e — 2.0211893 And consequently, 4 X (b + d) = 16.102.5340 a + e = 4.0255107 6)12.0770233 2.0128372 =- log. of 103, as required. 2. Given the cube roots of 45, 46, 47, 48, and 49, to find the cube root of 50. Ans. 3.684031. 3. Givenr the logarithms of 50, 51, 52, 54, 55, and 56, to find the logarithm of 53. Ans. 1.7242758695. PROmiSCUOUS EXAMPLES RELATING TO SERIES. 1. To find the sum (s) of n terms of the series 1, 2, 3, 4, 5, &c. First, 1+2+3+4+-5,&c.,... ns. Ad n +. (n - i) + (n - 2) + ( - 3) + (n - 4), &c. T.......'y = — 1 -s; Therefore, by addition, (n + 1) + (n + 1) + (n + 1) + (n + 1) + (n + 1), &c.... *+ (* + 1)= 2s. it2 2 n And consequently, n (n +- 1) = 2s; or s 2-= sum required. 2. To find the sum (s) of n terms of the series 1, 3, 5, 7, 9, 11, &c. First, 1 + 3 + 5 + 7 9, &c... (2n - 1) s. Ad (2n - l)+(2n - 3)+(2n-5)+.. + 1 =s. Therefore, by addition, 2n +- 2n +- 2n +2n + 2n +, &c.,..... 2n — 2s. And consequently, 2n X n = 2s; 2n; Or s 7 -- = n sum required. 3. Required the sum (s) of n terms of the series a - (a + d) + ( + 2d) + (a + 3d) + (a + 4d), &c. First, a -+ (a +- d) +- (a- +2d) + (a + 3d), &c., *... —,a + (n- l) d = s. And a + (nd - d) + a + (nd - 2d) + a + (.d - 3d) + a + (nd -4d), &c., a = s. SUMMATION OF INFINITE SERIES. 213 Therefore, by addition, a - (nd - d) - 2a +- (nd - d) + 2a + (nd,- d), &c., + 2a + (nd- d) = 2s. And consequently, (2a -1- nd - d) X n = 2s; n Or s = (2a + nd - d) x - = sum required. Or the same may be done in a different manner, as follows:a +- (a - d) ~ (a +- 2d) -- (a + 3d) +- (a +- 4d), &c.' — ~~~~~~~~~~~= S. (+~-t-l — 1 +l+1,&c.)X [a (-+-O- 1 +- 2+ 3 +4, &c.) X a But n terms of 1 +- 1 -f' 1 +- 1 +- 1, &c., — = n. And n terms of 0 +'I - 2 +- 3 + 4, &c.,=n.n-1 ~nn X (nX — ) d 1) And nterms of 0+q-1~q-2~43~-4, &c.,= Whence s =- na + 2 a d(n - ) which is the same answer as before. 4. To find the sum (s) of n terms of the series 1, x, x2, x3, x4, &C. First, 1 + x + X2 + x3 + X4, &c.,.....-l=s And x q- X2 q- X3 q X44 Xa &c.,.... = SX. Whence, by subtraction, xn - 1 = sx - s, Xn - I Or s = - = sum required. x —-1 And when x is a proper fraction, the sum of the series, continued ad infinitum, may be found in the same manner. Thus, putting 1 +- x +- x2 +- x +- x4 + X5, &c., = s. We shall have x +- x2 +- x3 +- x4 +- x5, &c., = sx, And consequently, - 1 = - s; or s - sx- = 1, 1 Whence s =1 sum of an infinite number of terms, as was to be found. 5. Required the sum (s) of the circulating decimal.999999, &c., continued ad infinitum. 9 9 9 9 First,.999999, &c., = -0 t- 10 + 10 00 &c., = 10 10 00 200 000 1 1 1 1 9(+ 1+ - 100+ &c.,) ==s 1 1 1 1 s or, i-6 $ —6 + r -6-6 oo0+ ao. = 1 1 1 10s Therefore, 1 + - + — ~ + -4' &C., -—. 10 100 1000 10s s 9s And consequently, 10 - -s - =; 9 99 214 $SUMlMATION OF INFINITE SERIES. Whence s = 1 _ sum of the series. 6. Required the sum (s) of the series a2 + (a + d)2 + (a +- 2d)j2 + (a + 3d)2 + (a + 4d)2, &c., continued to n terms. Here, First, a2 = a2 (a- d)2 a2 + 2 X lad+ ld2 (a +2d)2= a2 +- 2 X 2ad + 4d2 (a +- 3d)2= a2 +- 2 X 3ad + 9d2 (a + 4d)2 = a +- 2 X 4ad +- 16d2, &c. &c. Whence Sum Qf n terms of(l 1 1 1 + 1+ 1 +, &c.) a2, +... ditto of (O + 1 + 2 3 + 4+-, &c.) 2ad2, +-... ditto of(O + 1 +4 9 +16, &c.)d. But n terms of 1 + 1 + 1 +- t +-, &c., = n. And of0- 1 +2 + 3 +4, ac., _ 1.2 Also of O + 1 + 4 9, &c., = ( )(2n- 1) 1.2.3 Therefore s -- na2 + n (n - 1) ad +- n(n- 1)(2n- d2 _ 1.2.3 the whole sum of the series to n terms. 7. Required the sum (s) of the series a3 + (a + d)3 + (a + 2d)3 + (a + 3d)3 + (a + 4d)3, &c., continued to n terms. First, a3 = Ca3, (a + d)3 =a3 +3 X la2d + 3 X lad2 + 13, (a+2d)'3 a3+ 3 X 2a2d + 3 X 4ad 2 8d3, (a + 3d)3 = a3 + 3 X 3a2d- +3 X 9ad' + 27d'J, (a + 4d)3 = aa + 3 X 4a2?d + 3 X 16ad2+ 64d3, (x+ 5d)3 = a3 t 3 X 5a2d + 3 X 25ad2 + 125d3, &c. &c. Whence Sum of n terms of (1 + 1 + 1 + 1, &c.) a3, +... ditto of (O + 1 + 2 + 3 + 4, &c.) 3a2d, s = +-... ditto of(0 + 1 + 4 + 9 + 16, &c.) 3ad2, -+... ditto of (O+ - 1 8 + 27 + 64, &c.)d3, But n terms of 11 + - 1 +I 1 + 1, &C.,- n. n (nU - 1) Ditto... of 0 + 1 +2 + 3 +- 4, &,. 1). n(n — 1)(2n- 1) Ditto... of O +- 1 -+ 4 +- 9 +- 16, &c., = 1.2.3 n4 - 23 f+ na Ditto... of + + 8 +27 64, &c., = - -. 4 SUMMATION OF INFINITE SERIES. 215 n(n- 1) 3a2d n(n- 1) (2n -!)3ad2-2 Therefore, s na + 2 + 31) ) —3a+ 2 6 )? 4 2n' + n2) d (m14 -=2n3 + 8z2) X' =sum of n terms, as was to be found. 4 8. Required the sum (s) of n terms of the series 1 + 3 + 7 + 15 + 31, &c. The terms of this series are evidently equal to I, (1 + 2), (1 + 2 + 4), (1 + 2 + 4 + 8), &c., or to the successive sums of the geometrical series 1, 2, 4, 8, 16, &c. Let, therefore, a = 1 and r = 2, and we shall have a + a- ar + a+ a3 + r4, &c., = 1 + 2 + 4 + 8 + 16, &c. But the successive sums of 1, 2, 3, 4, &c., terms of the series are, ar - a. (-= ( 1) x --- 2.- -(r-1)X — r-1 r —1 ar? -a a ar3- a a 3. = (e- 4)x ar4 -- a 4 4. =(r-1 ) X — r -1 r_ — &c. &c. a n terms of r+ r+ 2 2 4, &C. Therefore, s ~ -rI - n ternls of + 1 + 1 + 1, &c. But 1 + 1 - + 1 - 1 + 1 + 1 - 1, &c., =-. r And r + ra + r3 + r+4 +, &c., - (rn- 1) X —. r (r"-I) a a WhencesX -- X - nX -2(2n l) —n= r- r-1 r —1 whole sum required. 9. It is required.to find the sum of n terms of the: Series 1 3 7 15 31+ 63 -+-+ —+-+-+5-' &c' 1 2 4 8 16 32' Here the terms of this series are the successive sims of the 1 i 1 1 1 geometrical progression - + - &c. 1 2 4 8 16 Let, therefore, a = 1 and r = 2; then will +1 1 + 1 1 a a a a a a f+ -++- &c.,=a+-++ + + + +, 1 2 8r Ar-,jr 216 SUMMATION OF. INFINITE SERIES. But the successive sums of 1, 2, 3, 4, &c., terms of the series are, (r — 1) X a a (r1._-1) X - 1 (r- 1) X (r r - 3. ( 1) X X (r- 1) Xr r-) x —. (r4 —) X a X a &c. &c. Therefore, i n terms of r +- r +-{- r -+- r, &ce. r —1 1 - 1 These being the two series'deriv.ed from the above expressions; But r -+ r r + r + r + r, &c., = nr. 1 1 1 1 +1 r r2 r+3 -(r n- -- Whence a rn -- 1 (n- 1) 2n + 1. X (nr - (r= sum r — 1 (r- - 1)' 2n-1 required. 10. Required the sum (s) of the infinite series of the recipro1 1 1 1 cals of the triangular numbers + -- ~ -6 + -, &c. 1 3 6 10 15 L 1 1 1 Let 3 + 3 + 1+ 0, &c., ad infinitum = s. O I 1 1 Or, -1-+. 3 &c............. S.'1.1 1.3 2.3 2.5' 1.2 2.3 + 3.4 + 4.5 2 That is, I. -{ - + -~' 4-' &C....'(1 2)(2 3 3 4) (4 5 20 1 1 1 1 1 i 1 &, c. Or, 3 4 5 6 &c. s 1 Whence or s = 2 = sum required. t2 10 SUMMATION OF INFINITE SERIES. 217 1P. And if it be required to find the sum of n terms of the 1 1 1 t 1 same series, - -+ ~&+ — &c. 1 3 6 10 15' 1 I 1 1 1 1 Let z q - - q- -{-, c&., to -. 1 1 1 1 1 1 Then z- -+ 3- + +, &c., to-. 1 2 3 4 5 n 1 1 1 1 1 1 1 And z- ++ - +3 - +~,&co, to 1 n+1 2 3 4 5 1. 1 1 1 1 1 1 n Therefore 1, -- 2&c. to 1 n+1 2 6 12 20 n n+1' n 1n u 1 1 1 Or q- -+ &c., to n+l 2 6 12 20 n(n -1)' 2n 1 1 1 1 2 W1hence -- +- - + &c., to n+1 13 6 10 n (u ) 1 1 1 1 1 2 2n Or+-q- - q- q- - ~ &c.+ to -- = sumof n terms of the series, as was required. 1 1 12. Required the sum of the infinite series 1.23 + 2-3.4 + + &c. 3.4.5 4.5.6 Let z +-3 -- J+ 5- &c., ad infinitum. I._1I 1 i I Then z - 2- ~ + + -+ 5-, &c., by transposition. 1 1 1 1 And 1 2-. - +- - 4- +-, &c., by subtraction. 1.2 2.3 3.4 4.5 1 1 1 1 1 &C Or 1 -2.3 4 5- 6 &c., by transposition. 2 2.3 3.4 4.5 5.6' 1 4 6 8 Whence 1.4.3. - 3- &c., by subtraction. 2 1.4.3 2.9.4 3.16.5 O1 2 2 2 &. Or.2 2.-.3 +.3r + 3.2.4 + 4.2-3-5.6' 1 1 1 1 And - 2 = -+ -- +, &c. 2 1.2.3 2.3.4 3.4.5 1 1 1 1 1 But - - 2 =;therefore - -+ - 2 4 1.2.3 2.3.4 3.4.5 4.5.6' 19 218 SUMMATION OF INFINITE SERIES. ad infinitum, = I, which is the sum required. 13. And if it were required to find the sum of n terms of 1 1 1 1 the same series 1 — + - 5 &C. 1.2.3 2.3.4 3.4.5 4.5.6 1 1 1 1 Let z +~ — - +, &C., to 1.2 2.3 3.4 n(n +) 1 1 1 1' 1 Then z 2 - + -, &c., to 2 2.3 3.4 4.5 n(n+) A 1nd z-1 1 1 1 1 2 (n + 1)(n + 2) 2.3 3.4 4.5 5.6 -7 +- &c., continued to ( - 1) ( - terls. 1 1 2 2 2 Therefore - -+ - - + + &c.2 2 (n -+- 1)(n+ - n) 1.2.3 2.3.4 3.4.5' to n terms, by subtraction. Whence1 1 I 1 1 4 2 (n + 1) (n + 2)- 1.2.3 2.3.4 3.4.5' &c., to n terms, by division. i 1 1 And consequently, -2Jr - + -3 &c., continued to 1 1 n terms - - = sum required. -4 2 (n+1)(n+2) 14. Required the sum (s) of the series - + 1- + 2 4 8 16 -' &c., continue d ad infnitum. Let x- and. s = — The1 =(1 - x _- x + 4, &.) And z = (1 + x) X (x - x, + -3- x4 + x', &c.) Whence, by multiplication, x - x2' x4+x5, &c. 1 +x X - X2 + 13 - X4 + XI, &C. +- xa - X3 + 4 - X5, &c. Whose sum is - x + - 0 -+- 0 O-0. &c. SUMMATION OF INFINITE SERIE$. 219 Therefore, z = x, and x - x2 + 3 - x -- x5, &c = 1 +x' Or - + 6 +' &c., = -. — = _ sum required. 2 4 8 16 32 1+i 3 15. Required the sum of the series 12 4+ 3 + 4 42 8 16 &c., continued ad infinitum. 1 z Let.x = 2 and s = - ) 2 Then, - x 2x2+ 3x3+ 4X4 +'5X, &e. And z = (1 - )2 X (x + 2x2+ 3x3 + 4x4 + 50, &c.) Whence, by multiplication, x + 2x2 + 3x + 4x4, &c. 1 — 2x +xa x + 2x2 + 3X3 -- 4x4, &c. 2x2 - 4x3 - 6x4, &Co + X3 + 2x4, &c. Whose -sum is =x - 0 O+O+O, &c. Therefore z= x, And x + 2x.+ 3x + 4x4+ 5x, &c., -( 1 2 3 4 5 6 Or2 + ~+ - + +- -+,:&c., =(1 2 2=surn 2 4 8 16 3+2 64(11)2 of the infinite series required. 16. It is required to find the sum (s) of the series + - 4 9 16 25. 9- - 2q - 81 - 2 -, &c., continued adiinfnitum. 1 z Let x = - and ( — =s.) Then - - x Z- 4x+ -" 9x+ -- +16x4+'26xs, &e. (1 -X)3 And z = (1 - x)3 X (x + 4x2 + 9x - + 16x4, &c.) = x +- x, as will be found by actual multiplication. Therefore x + x2 z, And x + 4a + 93 16x4, &c., + x (1 + x) 220 SUMMATION OF INFINITE SERIES. Or, 1 4 9 16 -(1 ( ) 3 -+=+-~- I -I = sum required, 3 + 9 17 819 (1 ( )3) 2 17. Required the sum (s) of the series _. -+ + 7n mr a - 2d a 3d. — +2d +'3 — d-, &c.,. continued ad infinitum. 1 z Let a _ -, and s - r mn(1 - X)2' ZT a a " d a + 2d a - 3d Then, -, &C m(lI3c)2 m mrX m Mr M &C 0z a -- +d a+.d a+- 3d Or, + —-— + -- +, &c. (1 — x~a r ra r That is (I - x]2a + (a + d) -x (a - 2d) X2 + (a +- 3d) x' +( -( 4d)x4, &c. And z==(1 -)2X a + (a + d)x +(a + 2d) x2+(a+3d)x3, &C., 1= (1 - x) a+ dx, as will appear by actually multiplying by (1 — ).. Therefore, z = (1 - x) a -- dz; and consequently, -- + a -d- c+ — a(r- sum of the mzr m2 7c. = m (r - 1)2 infinite series required. EXAMPLES. 1. Required the sum of 100 terms of the series 2, 5, 8, 11, 14, &c. Ans. 15050. 2. Required the sum of 50 terms of the series 1 + 22 + 3a +. 42 - 52', &c. Ans. 42925. 3. It is required to find the sum of the series I -- 3x + 6X2 + 10X3 +-15 4, continued ad ijfinitum, &c., when x is less than 1. 1 Ans. (1.- x)" 4. It is required to find the sum of the series 1 + 4x + 10'z + 20x3 + 35X4, &c., continued ad infinitum, when x is less than 1. 1.Ans. — s ~ (1 - x)4 5. It is required to find the sum of the infinite series, 1 1 1 5. -- - ~5 - &cI Ans. aor 1.3 3.5 5.T7 7,9' 1-0 2' SUMMATION OF INFINITE SERIES. 221 6. Required the sum of 40 terms of the series (1 X 2) -+-(3 X 4) + (5 X 6) + (7 X 8), &c. Ans. 86884. 2x - 1 7. Required the sum of n terms:of the series 2 +.2x 2x - 3 2x - 5 2x-7'2S - n -+ —+ -- -&+C. Ans. n 2 —x) 8. Required the sum of the infinite series 1 + 2 1.2.3.4 2.3A4.5 1 1 1 +.- +-, &c. Ans. — 3.4.5.6 4.5.6.7' 18 9. Required the sum of the series 1 + a + + - + 10 20 35' 3 1 &c., continued ad infnitum. Ans., or I 2 10. It is required, to find the sum of n terms of.the series 1 + 8x + 27x2 + 64x3 + 125x4, &c., continued ad infinitum. I + 4x + x2 Ans. (+ X4 (1 - x)4 12.2 11. Required the sum of n terms of the series + 4 + 3 4 5 6 1 1 r +r - 1 -+ ++; c+'&C. Ans. ~') -' i r r r_ ( (r - n -)2 12. Required the sum of the series 26+ 8 + + 1 1 - &.,.. + 18.12 2n(1 +2n)' Ans 3 5n + 3n Ans..~ i- S =- - 16" 32 +- 48n — 16ne 1 1 1 13. Required the sum of the series + + + 3.s8 6.12 9.1 1 1 1 n 12.20 WP * * * 3n (4 + 4-) *Ans. a = sos1 o — t fowng 1-2 12 +i2ni * The symbol A, made use of in these, and some of the following series, denotes the sum of an infinite number of terms, and S the sum of n terms. 19* 222 LOGARITHIMS. 14. Required the sum of the series +- 6 2.7 7.12 12.17 6 6 17.22 (5n- 3).(5nn + 2) 3 3n Arns. C -, s _ 5 2-+ 5n 15. Required the sum of the series k_- k — 3.6 6.8 9.10 12.12- &c...: (: 12.12 3n (4+ 2n) 1 7n n Ans. 24 2 (3 +- 6n) 4 (6 + 6n)~ 2 3 4 16. Required the sum of the series + 3.5 5.7 7.9 5 1 ~ - n 1 1 1 Ans. = - 12 12 4 (3 + 4n)' 5 6 17. Required the sum of the series + 2 347 8.4+ n 3.4.5 4.5.6-'.&c,. + (1 +.(+ 2 ) 3 3 2 1 Ans. 7-, s - - + 2 2 2 1+ 2+mn OF LOGARITHMS. LOGoARITHMS are a set of numbers that have been computed and formed into tables, for the purpose of facilitating many difficult arithmetical calculations; being so contrived, that the addition and subtraction of them ansswers to the multiplication, The series here treated of are such as are usually called algebrascal, which, of course, embrace only a small part of the whole doctrine. Those, therefore, who may wish for fiarher information on this abstruse but highly curious subject, are referred to the Miscellanea Asnallytica of of Demoivre, Sterling's Metlod Differ., James Bernouilli's de Seri-. lnfiei., Simpson's Mlifbh. Disse't., TWaring's Medii Aralj/t., Clark's translation of Lor-nac's Series, the various works of Euler, and Lacroix Traite d6e Calcl. Diff. et I1t., where they will find nearly all the materials that have been hitherto collected respecting this branch of analysis. LOGARITHMS. 223 and division of natural numbers with which they are made to correspond.* Or, when taken in a similar but more general sense, logarithms may be considered as the exponents of the powers to which a given or invariable number must be raised, in order to produce all the common,. or natural numbers. Thus, if a -y, a t _= y,, a"a = yUt, &c. then will the indices x, a', x", &c., of the several powers of a, be the logarithms of the numbers y, y', y", &c., in the scale, or system, of which a is the base. So that from either of these formulae it appears, that the logarithm of any number, taken separately, is the index of that power of some other number, which, when involved in the usual way, is equal to the given number. And since the base a, in the above expressions, can be assumed of any value, greater or less than 1, it is plain that there may be an endless variety of systems of logarithms, answering to the same natural number. It is likewise farther evident, from the first of these equations, that when y - 1, x will be = 0, whatever may be the value of a; and consequently, the logarithm of 1 is always 0, in every system of logarithms. And if x = 1, it is manifest from the same equation, that the base'a will be = y; which base is therefore the number whose proper logarithm, in the system to which it belongs, is 1. * This mode of computation, which is one of the happiest and most useful discoveries of modern times, is due to Lord Napier, Baron of Merchiston, in Scotland, who first published a table of these numbers in the year 1614, under the title of Cano;n Mirificmn Logarith7moru;n; which performance was eagerly received by the learned throughout Europe, whose efforts were immediately directed to the improvement and extensions of the new calculus that had so unexpectedly presented itself. Mr. Henry Briggs, in particular, who was, at that time, professor of geometry in Gresham College, greatly contributed to the advancement of this doctrine, not only by the very advantageous alteration which he first introduced into the system of these numbers' by making 1 the logarithm of 10, instead of 2.3025852, has had been done by Napier; but also by the publication, in 1624 and 1633, of his two great works, the Arithrmetica Lopgreith.mica and the Trigonomeltrica Britaxica, both of which were formed upon the principle abovementioned; as are, likewise, all our common logarithmic tables at present in use. See, for farther details on this part of the subject, the Introduction to my Treatise of Plane and Spherical Trigonomet'ry, 8vo. 2d edit., 1813; and for the construction and use of the tables, consult those of Sherwin, Hutton, Taylor, Callet, and Borda, where every necessary information of this kind may be readily obtained. t214 LOGARITHMS. Also, because al- = y, and ax' - y' it follows from the multiplication of powers, that ax X ax', or ax +- "' = yy'; and consequently, by the definition of logarithins, given above, x — x' = log. yy', or, log. yy' = log. y + log. y'. And, for a like reason, if any number of the equations aX = y, x' — = y',` ax" = yw", &c., be multiplied together, we shall have ax + -' +'", &Cc., - yy'y", &c.; and consequently, x + x' + x", &c., = log. yy'y," &c.; or, log. yy'y", &c.; = log. y "- log. y' + log. y", &c. From which it is evident, that the logarithm of the product of any number of factors is equal to the sum of the logarithms of those factors. Hence, if all the factors of a given number, in afiy case of this kind, be supposed equal to each other, and the sum of them be denoted by m, the preceding property will then become log. y"m = m log. y. From which it appears, that the logarithm of the mth power of any number is equal to m times the logarithm of that number. in like manner, if the equation ax = y be divided by a.' = y-, we shall have, from the nature of powers, as before, or av-2;' -=; and, by the definition of logarithms laid down in the first part of this article, x - x' = log. or log. = log. y- log. y'. Hence, the logarithm of a fraction, or'of the quotient arising from dividing one number by another, is equal to the logarithm of the numerator minus the logarithm of the denominator. And if each member of the common equation ax = y be raised to the fractional power denoted by we shall have, n m m in that case, n = y; And'consequently, by taking the logarithms, as before, m mn - - = log. yn, or log. yn log. y. n n Where it appears that the logarithm of a mixed root, or power, of any number, is found by multiplying the logarithm LOGARITHMIS. 225 of the given number by the numerator of the index of that power, and dividing the result by the denominator. And if the numerator m, of the fractional index, be in this case taken equal to 1, the above formula will then become [ 1 log. ys - 1 log, y. n From which it follows, that the logarithm of the nth root of any number is equal to the nth part of the logarithm of that number. Hence, besides the use of logarithms, in abridging'the operations of multiplication and division, they are equally applicable to the raising of powers and extracting of roots; which are performed by simply multiplying the given logarithm by the index of the power, or dividing it by the number denoting the root. But although the properties here mentioned are common to every system of logarithms, it was necessary, for practical purposes, to select some one of them from the rest, and to adapt the logarithms of all the natural numbers to that particular scale.. And, as 10 is the base of our present system of arithmetic, the same numniber has accordingly been chosen for the base of the logarithmic system, now generally used. So that, according to this scale, which is that of the common logarithmic tables, the numbers 10 4 10 -, 10-2, 10-', 100, 101, 102, 103, 104, &c. Or, 1, 10,100, 1000, 10000, &c., ~1006' 1000' 100' 10' have for their logarithms - 4 3 - 3, - 2,-, 0, 1, 2, 3, 4, &c. WVhich are evidently a set of numbers in arithmetical progression, answering to another set in geometrical progression; as is the case in every system of logarithms. And therefore, since the common or tabular logarithm of any number (n) is the index of that power of 10, which, when involved, is equal to the given number,.it is plain, from the following equation, 10x = n, or 10-x - n that the logarithms of all the intermediate numbers, in the above series, may be assigned by approximation, and made to occupy their proper places in the general scale. It is also evident, that the logarithms of 1, 10, 100, 1000, 226 LOGARITHMS. &c., being 0, 1, 2, 3, &c., respectively, the logarithm of any number, falling between 0 and 1, will be 0 and some decimal parts; that of a number between 10 and 100, 1 and some decimal parts; of a number between 100 and 1000, 2 and some decimal parts; and so on,'for other numbers of this kind. 11 1 And, for a similar reason, the logarithms of 1 000' 10' 100' 1000' &c., or of their equals.1,.01,.001, &c., in the descending part of the scale, being - 1, - 2, - 3, &c., the logarithm of any number, falling between 0 and 1, will be - 1, and some positive decimal parts; that of a number between.1 and.01, - 2, and some positive decimal parts; of a number between.01 and.001, - 3, and some positive decimal parts; &c. Hence, likewise, as the multiplying or dividing of any n-number by 10, 100, 1000, &c., is performed by barely increasing or diminishing the integral part of its logarithm by 1, 2, 3, &c., it is obvious that all numbers which consist of the same figures, whether they be integral, fractional, or mixed, will have, for the decimal part of their logarithms, the same positive quantity. So that, in this system, the integral part, of any logarithm, which is usually called its index or characteristic, is always less by 1 than the number of integers which the natural number consists of; and for decimals, it is the number which denotes the distance of the first significant figure from the place of units. Thus, according to the logarithmic tables in common use, we have Numbers. Logarithms. 1.36820' 0.1361496 20.0500 1.3021144 335.260 2.-5253817.46521 1.6676490.06154 2.7891575 &c. &c. WThere the sign - is put over the index, instead of before it, then that part of the logarithm is negative, in order to distinguish it from the decimal part, which is always to be considered as -+-, or affirmative. Also, agreeably to what has been before observed, the logarithm of 38540 being 4.5859117, the logarithms of any LOGARITHMS. 227 other numbers, consisting of the same figures, will be as follows - Numbers. Logarithms. 3854 3.5859117 385.4 2.5859117 38.54 1.5859117 3.854 0.5859117.3854 1.5859117.03854 2.5859117.003854 3.5859117 Which logarithms, in this case, as well as in all others of a similar kind, whether the number contains ciphers or not, differ only in their indices, the decimal, or positive part, being the same in them all. And, as the indices, or integral parts, of the logarithms of any numbers whatever, in this system, cah always be thus readily found from the simple consideration of the rule abovementioned, they are generally omitted in the tables, being left to be supplied by the operator, as occasion requires. It may here, also, be farther added, that when the logarithm of a given number in any particular system, is known, it will be easy to find the logarithm of the same number in any other system, by means of the following equations:a -= n, and es' = n, or log. n = x, and 1. n = a'. Where log. denotes the logarithm of n, in the system of which a is the base, and l. its logarithm in the system of which e is the base. For, since a = ex, or a,' = e, and ex - a, we shall have for the base a, -, = log. e, or x = a' log. e; and for the base e, --. a, or a' = x 1. a. ~ The great advantages attending the common, or Briggean system of logarithms, above all others, arise chiefly from the readiness with which we canr always find the characteristic or integral part of any logarithm from the bare inspection of the natural number to which it belongs; and the circumstance, that multiplying or dividing any number by 10, 100, 1000, &c., only influences the characteristic of its logarithm, without affecting the decimal part. Thus, for instance, if i be made to denote the index or integral part of the logarithm of any number N, and d its decimal part, we shall have log. N = i + d; log. 10l: X N = (i+ Am) + d; log. 1- (i —m) + d; where it is plain that the decimal part of the logarithm, in each of these cases, remains the same. 28W LOGARITHMS. Whence, by substitution from the former equations log. n = 1. n X log. e; or log. n=. n X I.a Where the multiplier, log. e, or its equal 1, expresses the I.a constant relation which the logarithms of n have to each other in the systems to which they belong. But the only system of these numbers deserving of notice, except that above described, is the one that furnishes what have been usually called hyperbolic or Naperian logarithms, the base e of which is 2.718281828459 Hence, in comparing these with the common or tabular logarithms, we shall have, by putting a in the latter of the above formiulat - 10, the expression log. n = 1. n X /-T' or 1. n = log. n X 1. 10. Where log. in this case denotes the common tabular logarithm of the number n, and 1 its hyperbolic logarithm; the con1 1 stant factor, or multiplies, I — which is --- or its i.10 2,3025850929' equal.4342944819, being what is usually called the mnodulus of the common system of logarithms.* PROBLEM 1. —To compute the logarithm of any of the natural numbers 1, 2, 3, 4, 5, &c. RULE L.-1. Take the geometrical series 1, 10, 100, 1000, 10000, &c., and apply to it the arithmetical series, 0, 1, 2, 3, 4, &c., as logarithms. 2. Find a geometric mean between 1 and 10, 10 and 100, or any other two adjacent terms of the series, betwixt which the number proposed lies. 3. Also, between the mean, thus found, and the nearest extreme, find another geometrical mean in the same manner; * It may here be remarked, that although the common logarithms have superseded the use of hyperbolic or Naperian logarithms, in all the ordinary operations to which these numbers are generally applied, yet the latter are not without some advantages peculiar to themselves; being of frequent occurrence in the application of the Fluxionary Calculus, to many analytical and physical problems, where they are required for the finding of certain fluents, which could not be so readily determined without their assistance; on which account great pains have been taken to calculate tables of hyperbolic logarithms, to a considerable extent, chiefly for this purpose. Mr. Barlow, in a Collection of Mathematical Tables lately published, has given them for the first 10000 numbers. LOGARITHMS. 229 and'1so on, till you are arrived within the proposed limit of the 1ml:-.oer whose logarithm is sought. 4. Find, likewise, as many arithmetical means between the corresponding terms of the other series 0, 1, 2, 3, 4, &c., in the same order as you found the geometrical ones, and ]he last of these will be the logarithm answering to the number required. EXAMPLES. Let it be required to find the logarithm of 9. Here the proposed number lies between 1 and 10. First, then, the log. of 10 is 1, and the log. of I is 0. There V (10 X 1) a/10 = 3.1622777 is the geometrical mean; And - (1 + 0) --. 5 is the arithmetical mean; Hence the log. of 3.1622777 is.5. Secondly, the log. of 10 is 1, and the log. of 3.1622777 is.5. Therefore / (10 X 3.1622777) = 5.6234132 is the geometrical mean; And - (1 +. 5) -. 75 is the arithmetical mean; Hence the log. of 5.6234132 is.75. Thirdly, the log. of 10 is 1, and the log. of 5.6234132 is 75; Therefore V/(10 X 5.6234132) = 7.4989422 is the geometrical mean; And 1 (1 +.75) —. 875 is the arithmetical mean; Hence the log. of 7.4989422 is.875. Fourthly, the log. of 10 is 1, and the log. of 7.498422 is 875; Therefore VP(10 X 7.4989422)= 8.6596431 is the geometrical mean; And -1 (1 +.875).9375 is the arithmetical mean; Hence the log. of 8.6596431 is.9375. Fifthly, the log. of 10 is 1, and the log. of 8.6596431 is.9375. Therefore V (10 X 8.6596431) = 9.3057204 is the geometrical mean. And l (1 +.9375) =.96875 is the arithmetical mean;. Hence the log. of 9.3057204 is.96875. Sixthly, the log. of 8.6596431, is.9375, and the log. of 9.3057204 is.96875; Therefore V(8.6596431 X 9.3057204) = 8.9768713 is the geometrical mean. "And ~ (.9375 +.96875) =_.953125 is the arithmetical mean; 20 230 LOGARITHMS. Hence the log. of 8.9768713 is.953125. And, by proceeding in this manner, it will be found, after 25 extractions, that the logarithm of 8.9999998 is.9542425, which may be taken for the logarithm of 9, as it differs from it so little that it may be considered as sufficiently exact for all practical purposes. And in this manner were the logarithms of all the prime numbers at first computed. RULES 2. — When the logarithm of any number (n) is known, the logarithm of the next greater number may be readily found from the following series, by calculating a sufficient number of its terms, and then adding the given logarithm to their sum. Log. (n - 1)l n -log. M + - 1 2n+1 3(2n + 1 )3 5(2n + 1)5 1 1 + 7(2n +- )' 9(2n +1)9 11(2n ~+ 1)11 Or, SM A 3B Log. (n + 1) = log. n + I + )2 e2n+1 3(2n+1)2 5(2n+1)2 5e 7D.9E;~ - + _&c. 7(2n + )2 9(2n - 1)2 11 (2n- +1)2' Where a, B, c, C&c., represent the terms immediately preceding those in which they are first used, and a' - twice the modulus -.8685889638.... * EXAMPLES. 1. Let it be required to find the common logarithm of the number 2. Here, because n + 1 -= 2, and consequently n = 1 and 2n + 1 = 3, we shall have M': 8685889638 =2nD+ 1 3 -.289529654 (A) A.289529654 =+.010723321 (n) 3 (2n + 1)2 — 3.32 * It may here be remarked:, that the difference between the logarithms of any two consecutive numbers is so much the less as the numbers are greater; and consequently, the series which comprises the latter part of the above expression will in that case converge so much the faster. Thus, log. n and log. (-f- 1), or its equal, log. n+-log. (1- -), will obviously differ but little from each other when n is a larger number. LOGARITHMS. 231 3n 3 X.010723321 __ _ — _- =.000714888 (c) 5 (2n 1 )- 5.32 5c 5 >(.000714888 5c 5 X. 000714888 _=.000056737 (D) 7(2n+1)2: 7.32 7D 7 X.000056737 __________,9(2n+1)2 O O.000004903 (E) 9 (2n + 1)2 9.32 9E 9 X.000004903 -1 (2n 1) X 132 _=.000000446 (F) 11 (2n + 1 )- 11.32 lI 11 X.000000446 1 _-a 1F 113X.0000 _=.000000042 (G) 13 (2n+ 1)2 13.32, 13G 13 X.000000042 — (_n+1 -= -.000000004 (H) 15 (2n -+ -1)2 15.32 Sum of 8 terms...301029995 Add log. of 1..000000000 Log. of 2...301029995 Which l)ogarithm is true to the last figure inclusively. 2. Let it be required to compute the logarithm of the number 3. Here, since n+ 1 = 3, and consequently, n = 2, and 2n + 1 = 5, we shall have is'.868588964 M.8G8588-.. =.173717793 (A) 2n +- 1 5 A.173717793 32 =) 3529 - -.002316237 (B) 3(2n + 1)2 3.52 3B 3 X.002316237 =-__ ____- =.000055590 (c) 5(2n +1)2 5.52 5c 5 X.000055590;~ (~,~I-. =.000001588 (D) 7(2n+ 1)2 7.52 7. 7 X.000001588.. ---- =.000000050 (E) 9 (2n + 1)2 9.52 9E 9 X.000000050 =__ =.000000002 (F) 11 (2n +- t)2 11.52 Sum of 6 terms..176091260 Log. of 2.. 301029995 Log. of 3...477121255 232 LOGARITI MS. Which logarithm is also correct to the nearest unit in the last figure. And in same way we may proceed to find the logarithm of any prime number. Also, because the sum of the logarithms of any two numbers gives the logarithm of their product, and the difference of the logarithms the logarithm of their quotient, &c.; we may readily compute, from the above two logarithms, and the logarithm of 10, which is 1, a great number of other logarithms, as in the following examples:3. Because 2 X 2 = 4, therefore }.301029995 log. 2 Mult. by 2 2 gives log. 4.602059990 4. Because 2 X 3 6, therefore }.301029995 to log. 2 add log. 3.477121255 gives log. 6.778151250 5. Because 2- 8, therefore log. 2.301029995 Mult. by 3 3 gives log. 8.903089985 6. Because 32 — 9, therefore log. 3.477121255 Mult. by 2 2 gives log. 9.954242510 7. Because -_-a = 5, therefore 1.000000000 2 from log. 10 1.000000000 take log. 2.301029995 gives log. 5.698970005 8. Because 3 X 4 12, there-.477121255 fore to log. 3.477121255 add log. 4.602059991 gives log. 12 1.079181246 And thus, by computing, according to the general formula, MULTIPLICATION BY LOGARITHMS. 233 the logarithms of the next succeeding prime numbers 7, 11, 13, 17, 19, 23, &c., we can find, by means of the simple rules before laid down for multiplication, division, and the raising of powers, as many other logarithms as we please, or may speedily examine any logarithm in the table. MULTPILICATION BY LOGARITHMS. TAKE out the logarithms of the factors from the table, and add them together; then the natural number, answering to the sum, will be the product required. Observing, in the addition, that what is to be carried from the decimal part of the logarithms is also affirmative, and must, therefore, be added to the indices, or integral parts, after the manner of positive and negative quantities in algebra. Which method will be found much more convenient, to those who possess a slight knowledge of this science, than that of using the arithmetical complements. EXAMPLES. 1. Multiply 37.153 by 4.086, by logarithms. Nos. Log s. 37.153.... 1.5699939 4.086.... 0.6112984'Prod. i51.8071. 2.1812923 2. Multiply 112.246 by 13.958, by logarithms. Nos. Logs. 112.246.... 2.0491709 13.958.... 1,1448232 Prod. 1563.128. 3.1939941 3. Multiply 46.7512 by.3275, by logarithms. Nos. Logos. 46.7512... 1.6697928.3275.... 1.5152113 Prod. 15.31102. 1.1850041 Here, the + 1, that is to be carried from the decimals, cancel the - 1, and consequently there remains 1 in the upper line to be set down. 20* 234 MULTIPLICATION BY LOGARITHIMIS.; 4. Multiply,.37816 by.04782, by logarithms. Nos. Logs..37816. 1.5776756.04782.. 2.6796096 Prod..0180836. 2.2572852 Here the + 1, that is to be carried from the decimals, destroys the - 1, in the upper line, as before, and there remains the - 2 to be set down. 5. Multiply 3.768, 2.053, and.007693, together. Nos.- Logs. 7.768.... 0.5761109 s,2.053... 0.3123889.007693... 3.8860997 Prod..059511. 2.7745995 Here the + 1, that is to be carried from the decimals, when added to' 3, makes - 2 to be set down. 6 Multiply 3.586, 2.1046, 8372, and 0294, together. Nos. Logs. 3.586...0.554610 2.1046... 0.323170.8372.. 1.922829.0294.... 2.468347 Prod..1857618. 1.268956 Here the + 2, that is to be carried, cancels the - 2, and there remains the - 1 to be set down. 7. Multiply 23.14 by 5.062 by logarithms. Ans. 117.1347; 8. Multiply 4.0763 by 9.8432, by logarithms. Ans. 40.12383. 9. Multiply 498.256 by 41.2467, by logarithms. Ans. 20551.41. 10. MIultiply 4.026747 by.012345, by logarithms. Ans. 0497102. 11.-VMultiply 3.12567,.02868, and.12379, together, by logarithms. Ans..01109705. 12. Multiply 2876.9,.10674.098762, and.0031598, by logarithms. Ans..0958299. DIVISION BY LOGARITHMS. 2 DIVISION BY LOGARITHMS. FROM the logarithm of the dividend, as found in the tables, subtract the logarithm of -the divisor, and the natural number answering to the remainder, will be the quotient req.ired. Observing, if the subtraction cannot be made in the usual way, to add, as in the former rule, the 1 that is to be carried from the decimal part, when it occurs, to the index of the logarithm, of the divisor, and then this result, with its sign changed, to the remaining index, for the index of the logarithm of the quotient. EXAMPLES. 1. Divide 4768.2 by 36.954, by logarithms. Nos. Logs. 4768.2.... 3.6783545 36.954 c... 1.5676615 Quot. 129.032. 2.1106930 2. Divide 21.754 by 2.4678, by logarithms. NATos. Logs. 21.754.... 1.3375391 2.4678,... 0.3923100 Quot. 8.1518.. 0.9452291 3. Divide 4.6257 by.17608, by logarithms. Nos. Logs. 4.6257... 0.6651725.17608... 1.2457100 Quot. 26.2741.. 1.4194625 Here - 1, in the lower index, is changed into + 1, which is then taken for the index of the result. 4. Divide.27684 by 5.1576, by logarithms. Nos. Logs..27684.... 1.4422288 5.1576... 0.7124477 Quot..0536761 1.7297811 Here the 1 that is to be carried from the decimals, is taken 236 RULE OF THREE BY LOGARITHMS. as - 1, and then added to - 1, in the upper index, which gives - 2 for the index of the result. 5. Divide 6.9875 by.075789, by logarithms. Nos. Logs. 6.9875... 0.8443218.075789.... 2.8796062 Quot. 92.1967. 1.9647156 Here the 1, that is to be carried from the decimals, is added to - 2, which makes - 1, and this put down, with its sign changed, is + 1. 6. Divide.19876 by.0012345;, by logarithms. Nos. Logs..19786... 1.2983290.001.2345... 3.0914911 Quot. 161.0051. 2.2068379 Here - 3, in the lower index, is changed into + 3, and this added to - 1, the other index, gives + 3 - 1 or 2. 7. Divide 125 by 1728, by logarithms. Ans. 0723379. 8. Divide 1728.95 by 1.10678, by logarithms. Ans. 1562.144. 9. Divide 10.23674 by 4.96523, by logarithms. Ans.. 2.061685. 10. Divide 19956.7 by.048235, by logarithms - Ans..413739. 11. Divide.067859 by 1234.59, by logarithms. Ans..0000549648. THE RULE' OF THREE, OR PROPORTION, BY LOGARITHMS. FoR any single proportion, add the logarithms of the second and third terms together, and subtract the logarithm of the first from their sum, according to the foregoing rules; then the natural number answering to the result will be the fourth term required. Bue if the proportion be compound, add together the logarithms of all the terms that are to be multiplied, and from the result take the sum. of the logarithms of the other terms, and the remainder will be the logarithm of the term sought. RULE OF THREE BY LOGARITHMS. 237 Or, the same may be performed most conveniently thus:Find the complement of the logarithm of the first term of the proportion, or what it wants of 10, by beginning at the lefthand, and taking each of its figures from 9, except the last significant figure on the right, which must be taken from 10; then add this result and the logarithms of the other two terms together, and the sum, abating 10 in the index, will be the logarithm of the fourth term, as before. And if two or more logarithms are to be subtracted, as in the latter part of the above rule, add their complements and the logarithms of the terms to be multiplied together, and the result, abating as many lOs in the index as there are logarithms to be subtracted, w;ill be the logarithm of the term required; observing when the index of the logarithm, whose complement is to be taken, is negative, to add it, as if it were affirmative, to 9; and then take the rest of the figures from 9, as before. EXAMPLES. 1. Find a fourth proportion to 37.125, 14.768, and 135,279, by logarithms. Log. of 37.125... 1..5696665 Complement.... 8.4303335 Log. of 14.768. 1.1693217 Log. of 135.279... 2.1312304 Ans. 53.81099... 1.7308856 2. Find a fourth proportional to.05764,.7186, and.34721, by logarithms. Log. of.05764... 2.7607240 Complement.-. 11.2392760 Log. of.7186... 1.8564872 Log. of.34721.... 1.5405992 Ans. 4.328681... 0.6363554 3. Find a third proportional to 12.796, and 3.24718, by logarithms. Log. of 12.796.... 1.1070742 Complement..... 8.8929258 Log, of 3.24718.. 0.5115064 238 INVOLUTION BY LOGARITHMS. Log. of 3.24718.... 0.5115064 Ans..8240216... 1.9159386 4. Find the interest of 2791. 5s. for 274 days, at 4- per cent. per annum, by logarithms. Comp. log. of 100 8.0000000 Comp. log. of 365. 7.4377071 Log. of 279.25.... 2.4459932 Log. of 274... 2.4377506 Log. of 4.5..... 0.6532125 Ans. 9.433296... 0.9746634 5. Find a fourth proportional to 12.678, 14.065, and 100.979, by logarithms. Ans. 112.0263. 6. Find a fourth proportional to 1.9864,.4678, and 50.4567, by logarithms. Ans. 11.88262. 7. Find a fourth proportional to.09658,.24958, and.005967, by logarithms. Ans..02317234. 8. Find a third proportional to.498621, and 2.9587, and a third proportional to 12.796, and 3.24718, by logarithms. Ans. 17.55623, and.8240216. INVOLUTION, OR THE RAISING OF POWERS BY LOGARITHMS. TAKE out the logarithm of the given number from the tables, and multiply it by the index of the proposed power; then the natural number answering to the result, will be the power required. Observing, if the index of the logarithm be negative, that this part of the product'will be negative; but as what is to be carried from the decimal part will be affirmative, the index of the result must be taken accordingly. EXAMPLE3. 1. Find the square of 2.7568, by logarithms. Log. of 2.7568... 0.4402477 Square 7.599946... 0.8804954 2. Find the cube of 7.0851, by logarithms. INVOLUTION BY LOGARITHMS. 239 Log. of 7.0851,.. 0.8503399 3 Cube 355.6475.... 2.5510197 3 Find the fifth power of.87451, by logarithms. Log. of.87451... 1.9417648 5 Fifth power.5114695. 1.7088240 Where 5 times the negative index - 1, being - 5, and - 4 to carry, the index of the power is 1. 4. Find the 365th power of 1.0045, by logarithms. Log. 1.0045*.... 0.0019499 365 97495 116994 58497 Power 5.148888.. 07117135 5. Required the square of 6.05987, by logarithms. Ans. 36.72603. 6. Required the cube of.176546, by logarithms. Ans. 005502674. 7. Required the 4th power of.076543, by logarithms. Ans. 0000343259. 8. Required the 5th power of 2.97643, by logarithms. Ans. 233.6031. 9. Required the 6th power of 21.0576, by logarithms. Ans. 87187340. L0. Required the 7th power of 1.09684, by logarithms. Ans. 1.909864. * The answer, 5.148888, though found strictly according to the general rule, is not correct in the last four figures, 8888; nor can the answers to such questions, relating to very high powers, be generally found true to 6 places of figures by the tables of Log. commonly used; if any power above the hundred thousandth were required, not one figure of the answer here given could be depended on. The Log. of 1.0045 is 00194994108 true to eleven places, which, multiplied by 365, gives.7117285 true to 7 places, and the corresponding number true to 7 places is 5.149067. See Doctor Adrain's edition of H-Iut. Math. Vol. 1. p. 169. 240 EVOLUTION OF LOGARITHMS. EVOLUTION, OR THE EXTRACTION OF ROOTS, BY LOGARITHMS. TAKE out the logarithm of the given number from the table, and divide it by 2 for the square root, 3 for the cube root, &c., and the natural number, answering to the result, will be the root required. But if it be a compound root, or one that consists of both a root and a power, multiply the logarithm of the given number by the numerator of the index, and divide the product by the denominator, for the logarithm of the root sought. Observing, in either case, when the index of the logarithm is negative, and cannot be divided without a remainder, to increase it by such a number as will reader it exactly divisible; and then carry the units borrowed, as so many tens, to the first figure of the decimal part, and divide the whole accordingly. EXAMPLES. 1. Find the square root of 27.465, by logarithms. Log. of 27.465... 2)1.4387796 Root 5.2407.7193898 2. Find the cube root of 35.6415, by logarithms. Log. of 35,6415... 3)1.5519560 Root 3.29093..5173186 3. Find the 5th root of 7.0825, by logarithms. Log. of 7.0825.. 5)0.8501866 Root 1.479235..1700373 4. Find the 365th root of 1.045, by logarithms. Log. of 1.045... 365)0.00'19116 Root 1.000121...0.0000524 5. Find the value of (.001234)3, by logarithms. QUESTIONS IN LOGARITHMS. 241 Log. of.001234... 3.0913152 2 3)6.1826304 Ans..00115047... 2.0608768. Here, the divisor 3 being contained exactly twice in the negative index - 6, the index of the quotient, to be put down, will be - 2. 6. Find the value 6f (.024554)3, by logarithms. Log. of.024554.... 2.3901223 2)5.1703669 Ans..00384754... 3.5851834. Here 2 not being contained exactly in - 5, 1 is added to it, which gives - 3 for the quotient; and the 1 that is borrowed, being carried to the next figure, makes 11, which, divided by 2, gives.58, &c. 7. Required the square root of 365.5674, by logarithms. Ans. 19.11981. 8. Required the cube root of 2.987635, by logarithms. Ans. 1.440265. 9. Required the 4th root of.967845, by logarithms. Ans..9918624. 10. Required the 7th root of.098674, by logarithms. Ans..7183146. 11. Required the value of ( by logarithms. Ans..146895. 12. Required the value of (12 )5, by logarithms. Ans..1937115. MISCELLANEOUS EXAMPLES IN LOGARITHMS. 1. Required the square root of - by logarithms. Ans..1275153. 1 2. Required the cube root of 31419, by logarithms. Ans..6827842. 21 242 MISCELLANEOUS QUESTIONS. 3. Required the.07 power of.00563, by logarithms. Ans..6958821. of 4 4. Required the value of (by logarithms. Ans..042'79825. 1 5 7 5. Required the value of a X.012 V1-, by logarithms. Ans..001165713. X.03,J 15 6. Required the value of 9 71 0 5, by log.a713Z12 -L X.1 9 4,/17 rithms. Ans..0009158638. 7. Required the value of 127({V 19 ~+ lV35 4 14ff - I V28 bylogarithms. Ans. 49.38712. MISCELLANEOUS QUESTIONS. 1. A person being asked what o'clock it was, replied that it was between eight and nine, and that the hour and minute hands were exactly together; what was the time? Ans. 8h. 43 min. 382Y sec. 2. A certain number, consisting of two places of figures, is equal to the difference of the squares of its digits, and if 36 be added to it, the digits will be inverted; what is the number? Ans. 48. 3. What two numbers are those, whose difference, sium, and product, are to each other as the numbers 2, 3, and 5, respectively? Ans. 2 and 10. 4. A person, in a party at cards, bet three shillings to two upon every deal, and after twenty deals found he had gained five shillings; how many deals did he win? Ans. 13. 5. A person wishing to enclose a piece of ground with palisades, found, if he set them a foot asunder, that he should have too few by 150, but if he set them a yard asunder he should have too many by 70; how many had he? Ans. 180. 6. A cistern will be filled by two cocks, A and n, running together, in twelve hours, and by the cock A alone in twenty hours; in what time will it be filled by the cock B alone? Ans. 30 hours. 7. If three agents, A, B, C, can produce the effects, a, b, c, in the times e, f, g, respectively; in what time would they jointly produce the effect d. Auns. d f g MISCELLANEOUS QUESTIONS. 243 8. What number is that, which being severally added to 3; 19, and 51, shall make the results in geometrical progression? Ans. 13. 9. It is required to find two geometrical mean proportionals between 3 and 24, and four geometrical means between 3 and 96. Ans. 6 and 12; and 6, 12, 24, and 48. 10. It is required to find six numbers in geometrical progression such, that their sum shall be 315, and the sum of the two extremes 165. Ans. 5, 10, 20, 40, 80, and 160. 11. The sum of two numbers is a, and the sum of their reciprocals is b, required the numbers. Ans 2 VL / (ab - 4). 12. After a certain number of men had been employed on a piece of work for 24 days, and had half finished it, 16 men more were set on, by which the remaining half was completed in 16 days; how many men were employed at first; and what was the whole expense, at Is. 6d. a day per man? Ans. 32 the number of men; and the whole expense 1151. 4s. 13. It is required to find two numbers such, that if the squares of the first be added to the second, the sum shall be 62, and if the square of the second be added to the first, it shall be 176. Ans. 7 and 13. 14. The fore wheel of a carriage makes six revolutions more than the hind wheel, in going 120 yards; but if the circumference of each wheel was increased by three feet, it would make only four revolutions more than the hind wheel in the same space; what is the circumference of each wheel? Ans 12 and 15 feet. 15. It is required to divide a given number a into two such parts, x and y, that the sum of mx and ny shall be equal to some other given numnber b. b -a a7n - b Ans. x —' andy m - n -n - 16. Out of a pipe of wine, containing 84 gallons, 10 gallons were drawn off, and the vessel replenished with 10 gallons of water- after which 10 gallons of the mixture were again drawn off, and then 10 gallons more of water poured in; and so on for a third and fourth time; which being done, it is required to find hovvw much pure wine remained in the vessel, supposing the two fluids to have been thoroughly mixed each time? Ans. 48`- gallons. 17. A sam of money is to be divided equally among a 244 MISCELLANEOUS QUESTIONS. certain number of persons; now, if there had been 3 claimants less, each would have had 1501. more, and if there had been 6 more, each would have had 1201. less; required the number of persons and the sum divided. Ans. 9 persons; sum 27001. 18. From each of 16 pieces of gold, a person filed the worth of half a crown, and thlen offered them in payment for their original value, but the fraud being detected, and the pieces weighed, they were found to be worth in the whole, no more than eight guineas; what was the original value of each piece? Ans. 13s. 19. A composition of tin and copper, containing 100 cubic inches, was found to weigh 505 ounces; how many ounces of each did it contain, supposing the weight of a cubic inch of copper to be 5~- ounces, and that of a cubic inch of tin 4Lounces? Ans. 420 oz. of copper, and 85 oz. of tin., 20. A privateer, running at the rate of 10 miles an hour, discovers a vessel 18 miles ahead of her, making way at the rate of 8 miles an hour; how many miles will the latter run before she is overtaken? Ans. 72 miles. 21. In how many different ways is it possible to pay 1001. with seven shilling pieces, and dollars of 4:s. 6d. each? Ans. 31 different ways. 22. Given the sum of two numbers -- 2, and the sum of their ninth powers = 32, to find the numbers by a quadratic equation. Ans. 1: V(6 v 34 - 33). 23. Given ys - xy = 666, and x-' + xy 406, to find x and y. Ans. z = 7, and y = 9. 24. The arithmetical mean of two numbers exceeds the geometrical mean by 13, and the geometrical mean exceeds the harmonical mean by 12; what are the numbers? Ans. 234 and 104. 25. Given nxy - yy - = 3, and x6 y2 -- y2 y6 =7, to find the values of x and y. Ans. x = 1( V5 + 1), y -- (,/5 - 1). 26. Given x - y z — = 23, xy -+- z -- yz = 167, and xyz 385, to find x, y, and z. Ans. x =- 5, y = 7, z = 1 27. To find four numbers, x, y, z, and w, having the product of every three of them given; viz. ryz = 231, xyw 420, yzw = 1540, and xzw = 660. Ans. x = 3, y =- 17, z = 11, and to = 20. 28. Given + yz — 384, y + xz-237, and z r xy = 192, to find the values of x, y, and z. Ans. x- 10, y 7, and z= 22. 29. Given x2 xy 108, y2 + yz -- 69, and x2 + xz = 580, to find the values of x, y, and z. Ans. x = 9, y = 3, and z = 20 MISCELLANEOUS QUESTIONS. 245 30. Given - +y y2 = 5, and X4 + X2y2 + y4 = 11 to find the values of x and y by a quadratic. Aus. 2 y V 2 15. Ans. x= ~vtl0 q- i/5, y= ~v 10 — V5. 31. Given the equation a4 -- 6+ 132 - 12x = 5, to find 3 1 the value of x by a quadratic. Ans. - i 2 -13. 2 2 32. It is required to find by what part of the population a people must increase annually, so that they may be double at the end of every century. Ans. By 144th part nearly. 33. Required the least number of weights, and the weight of each, that will weigh any nrumber of pounds from one pound to a hundred weight..Ans. 1, 3, 9, 27, 81. 34. It is required to find four whole numbers such, that the squares of the greatest may be equal to the sum of the squares of the other three. Ans. 3, 4, 12, and 13. 35. It is required to find the least number, which, being divided by 6, 5, 4, 3, and 2, shall leave the remainders, 5, 4, 3, 2, and 1, respectively. Ans. 59. 36. Given the cycle of the sun 18, the golden number 8, and the Roman indiction 10, to find the year. Ans. 1717. 37. Given 256x - 87y: 1, to find the least possible values of x and y in whole numbers. Ans. x 52, and y -= 153. 38. It is required to find two different isosceles triangles such, that their perimeters and areas shall be both expressed by the same numbers. Ans. Sides of the one 29, 29, 40; and of the other 37, 37, 24. 39. It is required to find the sides of three right-angled triangles, in whole numbers, such, that their areas shall be all equal to each other. Ans. 58, 40, 42; 74, 24, 70; 113, 15, 112. 40. Given x - 1.2655, to find a near approximate value of x. Ans. 3.82013. 41. Given xy = 5000, and y = 3000, to find the values of x and y. Ans. = 4.691445, and y = 5.510132. 42, Given x -+ yY - 285, and yO - xy = 14, to find the values of x and y. Ans. x = 4.016698, and y = 2.825716. 43. To find two whole numbers such, that if unity be added io each of them, and also to their halves, the sums, in both cases, shall be squares. Ans. 48 and 1680. 44. Required the two least nonquadrate numbers x and y such, that s2 + y2 and X3 + y3 shall be both squares. Ans. x = 364, andy = 273. 45. It is required to find two whole numbers such, that 21* -I1SCELLANEOTS QUESTIONS. their sum shall be a cube, and their product and quotient squares. Ans. 25 and 100, or 100 and 900, &c. 46. It is required to find three biquadratic numbers such, that their sum shall be a square. Ans. 124, 154, and 20'. 47. it is required to find three numbers in continued geometrical progression such., that their three differences shall be all squares. Ans. 567, 1008, and 1792. 48. It is required to find three whole numbers such, that the sum of difference of any two of them shall be square numbers. Ans. 434657, 420968, and. 150568. 49. It is required to find two whole numbers such, that their sum shall be a square, and the sum of their squares a biquadrate. I Ans. 4565486027761, and 1061652293520. 50. It is required to find four whole numbers such, that the difference of every two of them shall be a square number. Ans. 1873432, 2288168, 2399057, and 6560657. 2 51. It is required to find the sum of the series -- 3 9 3 4 3 -- + - &c., continued to infinity. Ans. - 27 81 4' 3 52. It is required to find the sum of the infinite series 9 27 81 243 3 &C. -—, &c. Ans. - 16 64 256!024' 7e 53. Reqluired the sum of the series 5 + 6 + 7 + 8 + 9 4-, &c., continued to n terms Ans. 2 (i+ + 9). 54. It is required to find how many figures it would take to express the 25th term of the series 2 +2- 2 4 2' -1- 28 4- 20", &c. Ans. 5050446 figures. 55. It is required to find the sumn of 100 terms of the series (1 X2) + (3X 4)-(5 X 6)-+(7 X 8) +(9X 10), &c. Ans. 1343300. 56. Required the sum of 12 + 2" -+ 32 + 42 + 52, &c.,.... + 50, which gives the number of shot in a square pile, the side of which is 50. Ans. 42925. 57. Required the sum of 25 terms of the series 35 + 36 X 2 + 37 X 3 38 X 4 + -39 X 5, &c., which gives the number of shot in a complete oblong pile, consisting of 25 tiers, the number of shot in the uppermost row being 35. Ans. 16575. APPENDIX. OF THE APPLICATION OF ALGEBRA TO GEOMETRY. IN the preceding part of the present performance, I have considered Algebra as an independent science, and confined myself chiefly to the treating on such of its most useful rules and operations as could be brought within a moderate compass; but as the numerous applications, of which it is susceptible, ought not to be wholly overlooked, I shall here show, in compliance with the wishes of many respectable teachers, its use in the resolution of geometrical problems, referring the reader to my larger work on this subject for what relates more immediately to the general doctrine of curves.* For this purpose it may be observed, that when any proposition of the kind here mentioned is required to be resolved algebraically, it will be necessary, in the first place, to draw a figure that shall represent the several parts, or conditions, of the problem under consideration, and regard it as the true one. Then, having properly considered the nature of the question, the figure so formed, must, if necessary, be still farther prepared for solution, by producing, or drawing, such lines in it as may appear, by their connexion or relations to each other, to be most conducive to the end proposed. This being done, let the unknown line, or lines, which it is judged will be the easiest to find, together with those that are known, be denoted by the common algebraical symbols, or letters; then, by means of the proper geometrical theorems, make out as many independent equations as there are unknown * The learner, before he can obtain a competent knowledge of the method of application abovementioned, must first make himself master of the principal propositions of Euclid, or of those contained in my Elements of Geometr'y; in which work he will find all the essential principles of the science comprised within a much shorter compass than in the former. And in such cases where it may be requisite to extend this mode of application to trigonometry, mechanics, or any other branch of mathematics, a previous knowledge of the nature and principles of these subjects will be equally necessary. 38 APPLICATION OF ALGEBRA TO GEOMETRY. quantities employed; and the resolution of these, in the usual m-anner, will give the solution of the problem. But as no general rules can be laid down for drawing the lines here mentioned, and selecting the propere.st quantities to substitute for, so as to bring out the most simple conclusions, the best means of obtaining experience in these matters will be to try the solution of the same problem in different ways; and then to apply that which succeeds the best to other cases of the same kind, when they afterwards occur. The following directions, however, which are extracted, weith some alterations, from Neewton's Universal Arithmetic and Simpson's Algebra and Select E.xercises, will often be found of considerable use to the learner,_ by showing him how to proceed in many cases of this kind, where he would otherwise be left to his own judgment: 1st. In preparing the figure in the manner abovementioned, by producing or drawing certain lines, let them be either parallel or perpendicular to some other lines in it, or be so drawn as to form similar triangles; and, if an angle be given, let the perpendicular be drawn opposite to it, and so as to fall, if possible, from one end of a given line. 2d. In selecting the proper quantities to substitute for, let those be chosen, whether required or not, that are nearest to the known or given parts of the figure, and by means of which the next adjacent parts may be obtained by addition or subtraction only, without using surds. 3d. When in any problem there are two lines, or quantities, alike related to other parts of the figure, or problem, the best way is not to make use of either of them separately, but to substitute for their sum, difference, or rectangle, or the sum of their alternate quotients; or for some other line or line in the figure, to which they have both the same relation. 4th. When the area, or the perimeter, of a figure is given, or such parts of it as have only a remote relation to the parts that-are to be found, it will sometimes be of use to assume another figure similar to the proposed one, that shall have one of its sides equal to unity, or to some other known quantity; as the other parts of the figure, in such cases, may then be determined by the known proportions of their like sides, or parts, and thence the resulting equation required. These being the most general. observations that have hitherto been collected upon this subject, I shall now proceed to eludicate them by proper examples; leaving such farther remarks as may arise out of the mode of proceeding here used, to be applied by the learner, as occasion requires, to the APPLICATION OF ALGEBRA TO GEOMETRY. 249 solutions of the miscellaneous problems given at the end of the present article. PROBLEM l. —The base, and the sum of the hypothenuse and perpendicular of a right-angled triangle being given, it is required to determine the triangle. A Let ABC, right-angled at c, be the proposed triangle; and put BC = b, and AC = x. Then, if the sum of AB and Ac be represented by s, the hypothenuse AB will be expressed by s -x. But, by the well-known property of right-angled triangles (Euc.,, 47,) AC2 BC2- AB2, or x2 - 2 = S -- 2sx + x2. Whence, omitting x2, which is common to both sides of the equation, and transposing the other terms, we shall have 2sx = s2 - b, or s2_ b * = s.. 2s which is the value of the perpendicular Ac; where s and b may be any numbers whatever, provided s be greater than b. In like manner, if the base and the difference between the hypothenuse and perpendicular be given, we shall have,; by putting x for the perpendicular and d + x for the hypothenuse, X2 + 2dx +- d2 - b2 - + x or b2_ d2 2d Where the base (b) and the given difference (d) may be any numbers as before, provided b be greater than d. PROBLEM 2.-The difference between the diagonal of a square and one of its sides being given, to determine the square. * The edition of Edclid referred to in this and all the following problems, is that of Dr. Simson, London, 1801; which may also be used in the geometrical construction of these problems, should the student be inclined to exercise his talents upon this elegant but more difficult branch of the subject. The New York edition of Brewster's Legendre, or Playfair's Euclid, will answer the same purpose. 250 APPLICATION OF ALGEBRA TO GEOMETRY..A D B C Let AC be the proposed square, and put the side BC, or cD, X. Then, if the difference of BD and BC be put = d, the hypothenuse BD will be - x -- d. But since, as in the former problem. B-o + CD2, or 2BC2 = BD2, we shall have 2X2 x2 +' 2dx + d2, or X2 - 2dx = — d2; Which equation, being resolved according to the rule laid down for quadratics in the preceding part of the work, gives - d + d V 2, Which is the value of the side Bc, as was required. PROBLEM 3.-The diagonal of a rectangle ABCD, and the perimeter, or sum of all its four sides, being given, to find the sides. ~ 9D Let the diagonal AC = d, half the perimeter AB + BC = a and the base BC = x; then will the altitude AB = a- X. And since, as in the former problem, AB2 + BC2 = AC2, we shall have a2 - 2ax + x2 + x2 -= d, or -2 a2,2 _ ax = 2 2 Which last equation, being resolved, as in the former instance, gives x = —a ~ = V( 2da _ a2). Where a must be taken greater than d and less than d' x2. PROBLEM 4.-The base and perpendicular of any plane triangle ABC being given to find the side of its inscribed square. APPLICATION OF ALGEBRA TO GEOMETRY. 251 B F D C C Let EG be the inscribed square; and put Be = b, AD =p, and the side of the square EH or EF _ x. Then, because the triangles ABC, AEH, are similar, we shall have AD: BC:: AI: EH, or p: b:: (p - x): x. Whence, taking the products of the means and extremes, there will arise px = bp - hx, Which, by transposition and division, gives bp Where b and p may be any numbers whatever, either whole or fractional. PROBLEM 5.-Having the lengths of three perpendiculars, EF, EG, EH, drawn from a certain point E, within an equilateral triangle ABc, to its three sides, to determine the sides. A n D F C Draw the perpendicular AD, and having joined EA, EB, and *EC, put EF = a, EG = b, EI-I c, and BD (which is j Be) = x. Then, since AB, BC, or CA, are each = 2x, we shall have, by Euc. I, 47, AD = V (AB2 - BD2) = (4x2 - x2) = V 3-C = 4 v' 3. And because the area of any plane triangle is equal to half the rectangle of its base and perpendicular, it follows that A ABC =- BC X AD X= X A/ 3' — X 3, A BEC = 3C X EF X X a = ax, A AEC = 1 A X EGX X b bX, a AAEn = 1 AB X EH = X C = ex. But the last three triangles BEC, AEC, AEB, are together equal to the whole triangle ABC; whence x2 V 3 ax + bx + cx, 21Z a9APPLICATION OF ALGEBRA TO GEOMETRY. And consequently. if each side of this equation be divided by x, we shall have xV 3 = a-+ b 6+ c, or = -- /3 Which is, therefore, half the length of either of the three equal sides of the triangle. CoR.-Since, from what is above shown, AD is = x V 3, it follows, that the sum of all the perpendiculars, drawn from any point in an equilateral triangle to each of its sides, is equal to the whole perpendicular of the triangle. PROBLEM 6.-Through a given point p, in a given circle aCBD, to draw a chord cD, of a given length. Draw the diameter APB; and put cD = C, AP - b, PB = C, and cP = x; then will PD = a -X. But, by the property of the circle (Euc. III. 35,) CP X PD = AP X PB; whence X(a - x) = bc, or 2 - ax - - be. Which equation, being resolved in the usual way, gives x-1 a i,/ ( a2 - bc); Where x has two values, both of which are positive. PROBLEM 7.-Through a given point P, without a giver circle ABDc, to draw a right line so that the part CD, intercepted by the circumference, shall be of a given length. Draw PAB through the centre o; and put CD = a, P.A = b, PB = c, and Pc = a; then will PD = X + a. But, by the property of the circle, (Euc. iII, 36, cor.,) PC X PD - PA X PB; whence X (X + a) = be, or x2 + ax = be. APPLICATION OF ALGEBRA TO GEOIETRY. 253 Which equation being resolved, as in the former problem, gives x=- - = 1:,/( 2 + c); Where one value of x is positive and the other negative.* PROBLEM 8.-The base of Bc, of any plane triangle ABC. the sum of the sides AB, AC, and the line AD, drawn from the vertex to the middle of the base, being given, to determine the triangle. A B c Put BD or DC = a, AD = b, AB + AC = s, and AB = x; then will AC = s- x. But, by my Geometry, B. Ir, Prop. 19, ABa + AC2 = 2BDS + 2ADa; whence OZ + (s _-)2 = 2aS + 2b2, or 2 - SX= a2 + b2 — S2. Which last equation, being resolved as in the former instances, gives x- = s sV (a2 + b2 _ { 2), for the values of the two sides &B and AC of the triangle; taking the sign + for one of them, and - for the other, and observing that a2 + b2 must be greater than 41s2. PROBLEM 9.-The two sides AB, sc, and the line AD, bisecting the vertical angle of any plane triangle ABC, being given, to find the base Bc. A B C * The two last problems, with a few slight alterations, may be readily employed for finding the roots of quadratic equations by construction; but this, as well as the methods frequently given for constructing cubic and some of the higher order of equations, is a matter of little importance in the present state of mathematical science; analysis, in these cases, being generally thought a more commodiois instrument than geometry. 22 254~ APPLICATION OF ALGEBRA TO GEOMETRY. Put AB a, c = b, c, AD and Be -cx; then, by Euc. vi., 3, we shall have AB (a): AC (B): D: DC. Arid consequently, by the compositions of ratios, (Euc. v, 18,) ax a — b: a::X;: BD = _ and a-l- b: b:: a: Dc= - a+b But, by Euc. VI., 13, 13D X DC + AD2 = AB X AC; wherefore, also, abx2 2 (a + C2 -ab, or abx -= (a + b)2 x (ab - c2). From which last equation, we have ab _ C2 x = (a+ b) a- b Which is the value of the base Bc, as required. PROBLEM 10.-Having given the lengths of two lines AD, BE, drawn from the acute angles of a right-angled triangle ABc,- to the middle of the opposite sides, it is required to determine the triangle. A JU D c Put AD = a, BE = b, CD or icB-=, and cE or -cA; then, since (Euc. I, 47) cD2 + CA?, = AD2, and cE2 - cB2 - BE2, we shall have x 4y2 = 2, y2 q 4X2 = 2. Whence, taking the second of these equations from four times the first, there will arise 15y2= 4a2 - b2, or 4a2 _ b2 y= v 15 And, in like manner, taking the first of the same equations from four times the second, there will arise 15x2 = 4b2'- a2, or 4b2 a x V-5 15 APPLICATION OF ALGEBRA TO GEOMETRY. 255 Which values of x and y are half the lengths of the base and perpendicular of the triangle; observing that b must be less than 2a and greater than 2 a. PROBLEM l1.-Having given the ratio of the two sides of a plane triangle ABC, and the segments of the base, made by a perpendicular falling from the vertical angle, to determine the triangle. A 13 i Put BD = a, DC = b, AB X, AC = y, and the ratio of the sides as m to n. Then, since by the question, AB: AC:: m: n,l and by B. II, Prop. 16, of my Elements of Geometry, AB2 - AC2 = BD2 DC2, we shall have a y:: m: n, and 2 _ y2 a2 _ b2. But, by the first of these expressions, no = my, or y -- whence, if this be substituted for y in the second, there will arise;-2 - Xs= a2 _ b2, or (m2 _,l2) 2 = mn2 (a.- b2). And consequently, by division and extracting the square root, we shall have a2 b2 X = m -/V,- and m2_ n2' a2 b2 y = n v m- n —; which are the values of the two sides AB, AC, of the triangle, as was required. PROBLEM 12.-Given the hypothenuse of a right-angled triangle ABC, and the sides of its inscribed square DC, to find the other two sides of the triangle. 256 APPLICATION OF ALGEBRA TO GEOMIETRY. D E Put AB — h, DE or DF = S, AC = X, and CB = y; then, by similar triangles, we shall have AC (X): CB (y):: AF ( - S): FD (S). And consequently, by multiplying the means and extremes, xy - sy =- s, or xy = s(x + y), (1). But since, by Euc I, 47, Ac + CB2 = AB2, we shall likewise have x + y2 = h2.... (2). Whence, if twice equation (1) be added to equation (2), there will arise 2+2X yy2h2+2s(x +y), or (x + y) - 2s (x + y) = h2; Which equation, being resolved after the manner of a quadratic, gives x + y = s ~ V (7 1~+ S2), or y =s -x i v (hA2 + S2). Hence, if this value be substituted for y in equation(1), there will arise! - x,/ (h2 + 3'2) = s s ~, / (h2 + s2)1, or x2 _s (A2 + xlS~x= S- (+ 2 + S2)1. And consequently, by resolving this last equation, we shall have _x l s~ V (h2 + 2)~ [ t h2 _ S L. 2 V (h2+ S2); and 3yf= V-4t V (h2 + s'2) =F V I/h2-J _ s s / T Sv(.2 + S2i; Which are the values of the perpendicular Ac and base Be, as was required. PROBLEM 13.-Having given the perimeter of a rightangled triangle ABC, and the perpendicular CD, falling from the right angle on the hypothenuse, to determine the triangle. U; — A APPLICATION OF ALGEBRA TO GEOMETRY. 257 Put p = perimeter, CD = a, AC = x, and BC = y; then AB =P - (x + y). But, by right-angled triangles (Euc. I. 4'7,) AC2 + BC2 = AB2; whence xa2 y2 = 2 _ 2p (x +y) + X2 + 2xy + y2. Or, by transposing the terms and dividing by 2, p (x + y) - p2 = xy..... )And since, by similar triangles, AB: BC:: AC: CD, we shall also have, by multiplying the means and extremes, AB X CD = BC X AC, or ap - a(x + y) = xy....... (2). Whence, by comparing equation (1) with equation (2), there will arise (a + p) X (x + y) = ap -+ _p2. Whence, ( s a+ 2p) X + y --, or p(a+ a-p) And if these values be now substituted for x + y and y in equation (2), the result, when simplified and reduced, will give (a -- p) x p (a + — p) X = ap2. From which last equation, and the value of y above found, we shall have or (a+ ( p) P Or AC 2 ( ) 2 (a +p) / (a - p) 2a; and orc 2p(a + p) xI (a_ p) 2a (ap) And if the sum of these two sides be taken from p, the result will give 2a AB = p - (X + y) - ) Which expressions are, therefore, respectively equal to the values of the three sides of the triangle.. PROBLEM 14.-Given the perpendicular, base, and sum of the sides of an obtuse angled plane triangle ABC, to determine the two sides of the triangle. A 2 D 22* 252. APPLICATION OF ALGEBRA TO GEOMETRY. Let the perpendicular AD = p, the base BC =-b, the sum of AB and AC = s, and their difference = x. Then, since half the difference of any two quantities added to half their sum, gives the greater, and, when subtracted, the less, we shall have AB (s + ), and AC = ( - ). But, by Euc. I, 47, CD - AC2 - AD2a, or CD = j/ ~ (S - ) -— PI; and, by B. II, 12, AB2 = BC2 + AC2 + 2BC X CD; wvhence +2=' _- )2+ 2V (S- X)2 v.2_, or s - 2Y = 2b l (Sr - And if each of the sides of this last equation be squared. there will arise, by transposition, and simplifying the result, (s2 - 2) X2 = b2 (S2 _ b2) - 4b2p5 or =V b(1 - _ b2). Whence, by addition- and subtraction, we shall have s - b6 AB 4 - and 2 2 s2- b 2' Which are the sides of the triangle, as was required. PROBLEM 15.-It is required to draw a right line BFE from one of the angles B of a given square BD, so that the part FE, intercepted by DE and Dc, shall be of a given length. Av D 1D B C Bisect FE in o, and put AB or BC = a, FG or GE — b, and BG == X; then will BE = - + b and BF = -b. But since, by right-angled triangles, AE2 = BE2 A- B2, we shall have AE = V/(X+ b)2- a2j. And because the triangles BCF, EAB, are similar, BF: BC BE: AE:, or a (x = b) (X - bs) / s Jr b)2 - a,%. APPLICATION OF ALGEBRA TO GEOMETRY. 259 Whence, by squaring each side of this equation, and arranging the terms in order, there will arise X4 - 2 (a + b2) 2 = b2(2a2 _ — 2). Which equation, being resolved after the manner of a quadratic, will give = I a2 b2+:a V (a2 + 4b2). And consequently, by adding b to, or subtracting it from, this last expression, we shall have BE -v a 2 +. 2 (a2+ 4b2) +'b, or BF -— 7 / iab2 (a 2 q- 4b2)- b. Which values, by determining the point E, or F, will satisfy the problem. Where it may be observed, -that the point G lies in the circumference of a circle, described from the centre D, with the radius FG, or half the given line. PROBLEM 16. —The perimeter of a right-angled triangle ABC, and the radius of its inscribed circle being given, to determine the triangle. E3 D C Let the perimeter of the triangle = p, the radius or, or OE, of the inscribed circle - r, AE - X, and BD -- y. Then, since in-the right-angled triangles AEO, AFO, OE, is equal to OF, and OA is common, AF will also be equal AE, or x. And in like manner it may be shown, that BF is equal to BD, or y. But by the question, and Euc. I, 47, we have (x + r) -1- (y J r) ~- (x J+ y) = p, and (x + r)2 + (y + r)2 = (x + y)2. Or, by adding the terms of the first, and squaring those of the second, x~ +- y-=2-p- r, and r(x + y) = xy - i. Hence, since, in the first of these equations, y - (+p - r) - a, if this value be substituted for y in the second, there will arise x2- ( p - r)X= -- - pr, Which equation, being resolved in the usual manner, gives 260 APPLICATION OF ALGEBRA TO GEOMETRY. x- ( p- r) 1V (1 p- r)2 - pr, and y = ( (p - r) = i (1p-r)2 -pr. And consequently, if r be added to each of these last expressions, we shall have AC-=(1 p + r) ~: / -(Lp- r)2 - J2r and 41 7' BC (Lp + r ) a /) (1p -( r)2 _p r for the values of the perpendicular and base of the triangle, as was required. PROBLEM 17.-From one of the extremities A, of the diameter of a given semicircle ADB, to draw a right line AE, so that the part DE, intercepted by the circumference and a perpendicular drawn from the other extremity, shall be of a given length. h ~ ~ 3 ~IC Let the diameter AB = d, DE = a, and AE S x; and join BD. Then, because the angle ADB is a right angle, (Euc. II, 31,) the triangles ABE, ABD, are similar. And consequently, by comparing their like sides, we shall have AE: AB: AB: AD, or x d:: d: x - a. Whence, multiplying the means and extremes of these proportionals, there will arise xI - ax - d2, Which equation, being resolved after the usual manner, gives x= a + V/( a2 + d2). PROBLEM 18.-To describe the circle through two given points A, B, that shall touch a right line CD given in position. A P C 1 i D APPLICATION OF ALGEBRA TO GEOMETRY. 261 Join AB; and through o, the assumed centre of the required circle, draw FE perpendicular to AB.; which will bisect it in E, (Euc. III, 3.) Also, join OB; and draw EH, OG, perpendicular to CD; the latter of which will fall on the point of contact G, (Euc. iII. 18.) Hence, since A, E, B, H, F, are given points, put EB = a, EF =-a, EH = C, and lo = x; which will give OF = b - x. Then, because the triangle OEB is right-angled at a,, we shall have OB2 = EO2 + EB2, or B = (X2 + a2). But, by similar triangles, FE: EH::FO: O or oB; or b: c:: b - x: oB; whence, also, OB = - (-). And consequently, if these two values of OB be put equal to each other, there will arise, 2( + 2a) = Cb (- 0). Or, by squaring each side of this equation, and simplifying the result, (b2 _ C2) X2 + 2b6C, = 62 (c2 - a2). Which last equation, when resolved in the usual manner, gives bC2 C4 C2 - aa2 b2 _ 2 (b _ C2)2 + b2 _ c, for the distance of the centre o from the chord AB; where b must evidently be greater than c, and c greater than a. PROBLEM 19.-The three lines Ao, BO, co, drawn from the angular points of a plane triangle ABC, to the centre of its inscribed circle, being given, to find the radius of the circle and the sides of the triangle. Let o be the centre of the circle, and, on AO produced, let fall the perpendiculars CD; and draw oE, OF, OG, to the points of contact, E, F, G. Then, because the three angles- of the triangle ABC are, together, equal to two right angles, (Euc. r, 32,) the sum of 262 APPLICATION OF ALGEBRA TO GEOMETRY. their halves oAc + onr OBE will be equal to one right angle. But the sum of the two former of these, OAC + OCA, is equal to the external angle Doc; whence the sum of DOc + OBE, as also of Doc + oc1, is equal to a right angle: and consequently, OBE -- OCD. Let, therefore, AO - a, BO = b, co -- c, and the radius oE, OF, or OG - X. Then, since the triangles BOE, COD, are similar, Bo: OE:: co:oD, or b::: c: OD; which gives Cx C2X2 C CD - and CD = (c2 - b2 ), or b v (b2 - X2). Also, because the triangle _oc is obtuse angled at o, we shall have (Euc. II, 12,) ACG2 - AO02 + C02 + 2Ao X OD; or 2acx b (a' + C2)+ 2acx AC = 2/(a2 + ct + -, or V/( - b' But the triangles ACD, AOF, being likewise similar, AC: CD: AO: OF, or b (a2 + 2) + 2cx v( ):' (b2 -_ X2))' a: x. Whence, multiplying the means and extremes, and squaring the result, there will arise b~2 b (b + C2) + 2), 1-= c% (b2 -;). Or, by collecting the terms together, and dividing by the coefficient of the highest power of x, ab ac bc X abc + 2c + 2-b + 2a 2 From which last equation x may be determined, and thence the side of the triangle.* PROBLEM 20.-Given the three sides AB, BC, CD, of a trapezium ABRCD, inscribed in a semicircle, to find the diameter or remaining side AD. A DD e This, and the following problem, cannot be constructed geometrically, or by means only of right lines and a circle, being what the ancients usually denominated solid problems,-from the circumstance of their involving an equation of more than two dimensions; in which cases they generally employed the conic sections, or some of the higher orders of curves. MISCELLANEOUS PROBLEMS. 263 Let AB s=a, Bc = b, cD = C, and AD= x; then, by Euc. v, Prop. D, AC X BD = AD X BC +AB X CD = bx + ac. But ABD, ACD, being right angles, (Euc. InI, 31, we shall have AC = / (AD2 - DC2), or v (2 _ C2), and BD = (AD2 - AB2), or V (x2 _ a2). Whence, by substituting these two values in the former expressions, there will arise V/ (.2 c C2) X / (,2 _ 2) = bx + ac. Or, by squaring each side, and reducing the result, 3 - (a2 + b2 - c2) x 2abc. From which last equation the value of a may be found, as in the last problem.* MISCELLANEOUS PROBLEMS. PROBLEM I. To find the side of a square, inscribed in a given semicircle, whose diameter is d. Ans Ans. 1 d /5. PROBLEM II. Having given the hypothenuse (13) of a right-angled triangle, and the difference between the other two sides (7), to find these sides.t Ans. 5 and 12. PROBLEM III. To find the side of an equilateral triangle, inscribed in a circle whose diameter is d; and that of another circumscribed about the same'circle. Ans. Ad V/ 3, and d V/ 3. PROBLEM IV. To find the side of a regular pentagon, inscribed in a circle, whose diameter is d. Ans. -l d/ /(10 - 2 / 5). PROBLEM V. To find the sides of a rectangle, the perimeter of which shall be equal to that of a square, whose side is a, and its area half that of a square. Ans. a + - a V 2 and a - a V 2. e Newton, in his Universal Arithmetic, English edition, 1728, has resolved this problem in a variety of different ways, in order to show that some methods of proceeding, in cases of this kind, frequently lead to more elegant solutions than others; and that a ready knowledge of these can only be obtained by practice. t Such of these equations as are proposed in numbers, should first be resolved generally, by means of the usual symbols, and then reduced to the answers above given, by substituting the numeral values of the letters in the results thus obtained. 264 MISCELLANEOUS PROBLEMiS. PROBLEM VI. Having given the side (10) of an equilateral triangle, to find the radii of its inscribed and circumscribing circles. Ans. 2.8868 and 5.7736. PROBLEM VII. Having given the perimeter (12) of a rhombus, and the sum (8) of its two diagonals, to find the diagonals. Ans. 4- + 2 and 4 -/ 2. PROBLEM VIII. Required the area of a right-angled triangle, whose hypothenuse is o$, and the base and perpendicular x22 and Ax. Ans. 1.029085. PROBLEM IX. Having given two contiguous sides (a, b) of a parallelogram, and one of its diagonals (d), to find the other diagonal. Ans. V /2a2 + 2b2 - d2). PROBLEM X. Having given the perpendicular (300) of a plane triangle, the sum of the two sides (1150), and the difference of the segments of he base (495), to find the base and the sides. Ans. 945, 375, and 780. PROBLEM XI. The length of three lines drawn from the three angles of a plane triangle, to the middle of the opposite sides, being 18, 24, and 30, respectively: it is required to find the sides. Ans. 20, 28.844, and 34.176. PROBLEM XII. In a plane triangle, there is given the base (50), the area (796) and the difference of the sides (10), to find the sides end the perpendicular. Ans. 36, 46, and 33.261. PROBLEM XIII. Given the base (194) of a plane triangle, the line that bisects the vertical angle (66), and the diameter (200) of the circumscribing circle, to find the other two sides. Ans. 81.36587 and 157.43865. PROBLEM XIV. The lengths of two lines that bisect the acute angles of a right-angled plane triangle being 40 and 50 respectively, it is required to determine the three sides of the triangle. Ans. 35.80737, 47.40728, and 59.41143 MISCELLANEOUS PROBLEMS. 265 PROBLEM XV. Given the altitude (4), the base (8), and the sum of the sides (12), of a plane triangle to find the sides. 4 4 Ans. 6 + - A/ 5 and 6- a/ 5. 5 5 PROBLEM XVI. Having given the base of a plane triangle (15), its area (45), and'the ratio of its other two sides as 2 tp 3, it is required to determine the lengths of these sides. Ans. 7.7915 and 11.6872. PROBLEM XVI[o Given the perpendicular (24), the line bisecting the base (40), and the line bisecting the vertical angle (25), to determine the triangle Ans. The base 250 Acbns. T-he base 5 7. 7 From which the other two sides may be readily found. PROBLEM XVIII. Given the hypothenuse (10) of a right-angled triangle, and the difference of two lines drawn from its extremities to the centre of the inscribed circle (2), to determine the base and perpendicular. Ans. -8.08004 and 5.87447. PROBLEM XIX. HIaving given the lengths (%t, b) of two chords, cutting each other at right angles, in a circle, and the distance (c) of their point of intersection from the centre, to determine the diameter of a circle. Ans. V/ -(ac2 + bh2) + 2c21. PROBLEM XX. Two trees, standing on a horizontal plane, are 120 feet asunder; the height of the highest of which is 100 feet, and that of the shortest 80; whereabouts itl the plane must a person place himself so that his distance from the top of each tree, and the distance of the tops themselves, shall be all equal to each other. Ans. 20 V 21 feet from the bottom of the shortest, and 40 / 3 feet from the bottom of the other. PROBLEM XXI. Having given the sides of a trapezium, inscribed in a circle, equal to 6, 4, 5, and 3, respectively, to determine the diameter of the circle. 1 Ans. - /(130 X 153) or 7.051595. 23 266 MISCELLANEOUS PROBLEMS. PROBLEM XXII. Supposing the town A to be 30 miles from B, B 25 miles from a, and c 20 miles from A; whereabouts must a house be erected that it shall be at an equal distance from each of them? Ans. 15.118556 miles from each. PROBLEM XXIII. Given the area (100) of an equilateral triangle ABC, whose base BC falls on the diameter, and vertex A in the middle of the arc of a semicircle; required the diameter of the semicircle. Ans. 20 V 3. PROBLEM XXIV. In a plane triangle, having given the perpendicular(p), and the radii (r, R) of its inscribed and circumscribing circles, to determine the triangle. 2r /V (2pR. — 4rR - r2) Ans. The base - p - 2r PROBLEM XXV. Having given the base of a plane triangle equal to 2a, the perpendicular equal to a, and the sum of the cubes of'its other two sides equal to three times the cube of the base; to determine the sides. Ans. a(2 +! V/6) and a (2 — V 6). 3 3 267 ADDITIONAL PROBLEMS FOR PRACTICE. I. Simple Equations involving only one unknown quantzty 1. Find that number to which 75 being added, the sum shall be quadruple the required number. Ans. 25. 2. A person has 1341 pieces of money, among which are 31 times as many of silver as of copper. How many has he of each? Ans. 298 copper, and 1043 silver. 3. A man pays his four labourers $40, giving the second twice, the third 3 times, and the fourth 4 times as much as the first. What did each receive? Ans. 4, 8, 12, and 16. 4. A legacy of 1080 dollars is to be divided between two persons, so that one is'to receive $7 as often as the other $1. What are the shares? Ans. 135 and 945. 5. Divide 40 marbles between two boys in the proportion of 5 to 3. Ans. 25 and 15. 6. Two travellers, being 360 miles apart, set out at the same time towards each other, one travelling 8 miles an hour, the other 10. How far does each travel before they meet? Ans. 160 and 200 miles. 7. What two numbers are those whose difference is 36, and which are to each other as 7 to 4? Ans. 84 and 48. 8. Four persons gain $2040, of which B takes twice as much as A, C twice as much as A and B, and D as much as B and C. What are their shares? Ans. 120, 240, 720, and 960. 9. A man sells oxen, cows, and sheep, of each an equal number, for $700. He received for each ox $50, for each cow $17, and for each sheep $3. How many were there of each sort? Ans. 10. 10. A person buys 2 oranges, 4 lemons, and 10 apples, for 64 cents. He gives thrice as much for a lemon as for an apple, and as much for an orange as for a lemon and two apples..Whaat. did each cost? Ans. An apple 2, a lemon 6, and an orange 10 cts. 11. Find that number whose 3 part exceeds its } part by 12. Ans. 144. 12. What sum of money is that whose third, fourth, and fifth parts amount to $94? Ans. 120. 268 SIMPLE EQUATIONS INVOLVING 13. What number is that which being divided by 15, the dividend, divisor, and quotient shall amount to 79? Ans. 60. 14. A man lost 4 of his estate, and had $4500 left. What had he at first? Ans. 8100. 15. A person bestowed on two beggars Is. 8d., giving the first 1 and the second -ll of what he had. How much had he? Ans. 8 shillings. 16. A and B talking of their ages, A says to B, your age is thrice and a third of tmy age, and the sum of both is 52. What are their ages? Ans. A's 12 and B's 40 years. 17. A mail driving his geese to market, was met by another, who said, Good-morrow, master, with your hundred geese; said he, I have not a hundred, but if I had as many more, and half as many miore, and two geese and a h'alf, I should have a. hundred. How mIany had he? Ans. 39. 18. What number is that whose third part, less 4, is equal to its fourth, plus 25? Ans. 348. 19. A fathcer, at his death, left ~1000 to be divided between his son and daughter, in such a manner that 5 part of his share should exceed 4 part of hers by ~10. How imust the money be divided? Ans. The son, ~5771. The daughter, ~422-. 20. A gentleman would procure $10 in Iquarters and dimes, and have just 64 pieces. How many of each kind must he get? Ans. 24 qrs. and 40 dimes. 21. The sum of two numbers is 40; if the greater be multiplied by 2, and the less by 3, the difference of the products will be 15. Required the numbers. Ans. 27 and 13. 22. A cask' which held i46 gallons, was filled with a mixture of brandy, wine, and water. In it there were 15 gallons of wine more than there were of brandy, and as imuch water as both wine and brandy. What quantity was there of each? Ans. 29, 44, and 73 gallons respectively. 23. A man hired a labourer on condition that for every day he wrought he should have 12d., and for every day he was idle lie should forfeit 8d. After 390 days, neither was in debt. I-ow many working days were there? Ans. 156. 24. A son asking his father how old he was, received the following reply: My age, says the father, 7 years ago was 4 times as great as yours at that time; but 7 years hence,' if we live, my age will be only double of yours: it is required to find the age of each. Ans. 14 and 35. 25. A mercer having cut 19 yards from each of three equal pieces of silk, and 17 from another of the same length, found that the remnants together were 142 yards. What was the length of each piece e Ans. 54 yards, ONLY ONE UNKNOWN QUANTITY. 269 26. A cistern can be emptied by three apertures; by the first in an hour, by the second in an hour and a half, and by the third in two hours. In what time will all three running together empty it? Ans. In 27-~9 minutes. 27. A grocer has a quantity of tea, by which at 80 cents a pound, he will gain $2,50, but at 60 cents a pound, he will lose $7,50. What is the quantity and prime cost? Ans. 501b. at 75 cts. 28. How old is he, to the number of whose months, if you add its half and its eighth, and subtract a unit, there will be left the square of 21? Ans. 22 yrs. 8 mo. 29. Said A to B, double my money for me, and I will give you 6d out of the stock; with the remainder he applied to C, to double it and gave him also sixpence. He repeated the offer to D and then 6d was all he had left. What had he to begin with? Ans. 6d. 30. Find a number, which being added to itself, and the sum.multiplied by the same, and from the product the same subtracted, and lastly, the remainder divided by the same, the quotient shall be 13. Ans. 7. 31. A draper bought three pieces of cloth, which together measured 159 yards. The second piece was 15 yards longer than the first, and the third 24 yards longer than the second. What was the length of each. Ans. 35, 50 and 74 yards respectively. 32. A said he was 4 years older than C; B told them that his age exceeded the sum of theirs by 9 years; upon which D observed that his age was 45 and equal to the sum of the three. Required the several ages. Ans. A, 11; B, 27; C, 7; D, 45. 33. A boy with a basket of peaches met three companions, and gave half his peaches to the first, who returned him 10; to the second he gave half of what remained, and received back 4. He then halved the remainder with the third, who returned one, and eighteen only remained in his basket. How many were there at first? Ans. 100. 34. One being asked his age, answered that if — I of the years he had lived were multiplied by A of that number it would give his age. How old was he? Ans. 96. 35. A General having lost a batttle, found that he had only half his army plus 3600 men left, fit for action; one-eighth of his men plus 600 being wounded, and the rest, which were one-fifth of the whole army, either slain, taken prisoners, or missing. Of how many men did his army consist? Ans. 24000. 36. If George lives till the quadruple of T of his years, added 23* 270 SIMPLE EQUATIONS INVOLVIN'G to half of them +50 amounts to as much above 100, as his years will actually be below 100, what will be his age? Ans. 36. 37. A anrid B being at play, severally cut packs of cards, so as to take off more than they left. Now it happened that A cut off twice as many as B left, and B cut off seven times as many as A left. How were the cards cut by each? Ans. A cut off 48, and B cut off 28 cards. 38. A courier, who travels 60 miles a day, had been dispatched five days, when a second is sent to overtake him, in order to do which he must travel 75 miles a day. In what time will he overtake the former? Ans. 20 days. 39. WMhat are those two numbers whose sum is 150 and their difference 38? Ans. 56 and 94. 40. A person being asked the time of day answered that the time past from noon was equal to 4 of the time to midnight. What was the hour? Ans. 20 mini. past 5. 41. Required two numbers in the proportion of 3 to 4 whose stln is J of their product. Ans. 21 and 28. 42. What number is that to which if I add 12 and from JL of the sum subtract 12, the remainder shall be 12? Ans. 276. 43. In a certain school there are 306 scholars of which l study Algebra, and the numbers in the Latin, French, SWriting and Reading classes are as 1, 3, 6, and 7. How many students in each? Ans. i5 in Latin, 45 in French, 90 Read., and 105 Write. 44. There are three casks of wine whose contents are in proportion of 2, 3 and 5; from each of which if 10 gallons be drawn the whole quantity will be diminished in the proportion of 4 to 3. Required their contents severally. Ans. 24, 36 and 60. 45. Four places are situated in the order of the four letters A, B, C, D. The distance from A to D is 34 miles, the distance fromi A to B d: istance frolm C to D: 2:3, and onefourth of the distance from A to B added to half the distance from C to D, is three times the distance from AB to C. What are the respective distances? Ans. AB —12, BC: -I4) and CD =18 miles. 46. A trader maintained himself for 3 years at the expense of ~50 a year and in each of those years augmented that part of his stock which was not so expended by 13 thereof. At the end of the third year his original stock was doubled. What was that stock? Ans. ~740. 47. It is required to divide the number 104 into four such ONLY ONE UNKNOWN QUANTITY. 271 parts, that the first being increased by 6, the second diminished by 5, the third multiplied by 4, and the fourth divided by 3, may all be equal. Ans. 14, 25, 5 and 60. 48. Find a number which being increased by 4, and the sum multiplied by 3, produces the same result as half the said number multiplied by 8, and the product diminished by 8. Ans. 20. 49. Two persons received equal sums of money; the first paid away ~25, and the second ~60, and then it appeared that the former had twice as much as the latter. ~ What money did each receive? Ans. ~95. 50. One being asked the time said it was between 10 and 11 and the hour and minute hands were together. Required the time. Ans. 54 6 minutes past 10. 51. A courier sets out from a certain place and travels at the rate of 15 miles in 4 hours; and 12 hours after, another sets out from the same place, and travels the same road at the rate of 9 miles in 2 hours. I demand how long and how far the first must travel before he is overtaken by the second? Ans. 72 hours and 270 miles. 52. It is required to divide the number 99 into five such parts, that the first may exceed the second by 3; be less than the third by 10; greater than the fourth by 9; and less than the fifth by 16. Ans. 17, 14, 27, 8, and 33. 53. A and B began to trade with equal sums of money. In the first year A gained 40 dollars, and B lost 40; but iln the second A lost one-third of what he then had, and B gained a sum less by 40 dollars, than twice the sum that A had lost; when it appeared that B had twice as much money as A. What money did each begin with? Ans. 320 dollars. 54. A boy being caught stealing apples, was told by the owner that he should escape punishment if he would take a certain number of apples, and lay down at the first gate, half he had and half an apple over, and repeat this process with the remainder at the second gate, and also at the third, without dividing an apple at either, and then have one left. If he accomplished the task, how many apples did he take? Aris. 15. 55. Find two numbers in the proportion of 3 to 4, so that if 9 be added to each, the sums will be as 6 to 7. Ans. 9 and 12. 56. Upon measuring the corn produced in a field, being 105 bushels, it appears that it had yielded only one fourth part more than was sown. How much was sown? Ans. 84 bushels. 57. Given the sum of two numbers 12, and the difference of their squares 72, to find the numbers. Ans. 9 and 3. 58. If A can build 40 feet of wall in two days, B 13 feet in 2-72 SIMPL EQUATIONS INVOLVING one day, and C 50 feet in 3 days, how many days will all three be employed in building 596 feet? Ans. 12 days. 59. What number is that whose fourth part squared, is equal to its eighth part cubed? Ans. 32. 60. Three years ago A's age was half of B's, and 9 years hence it will be 5 of it. Required the age of each. Ans. 21, and 39. 61. A woman can get 25 apples for a penny, and twenty-five pears for three pence, if she lays out Is. 8d. for 200 apples and pears together. How many of each will she have? Ans. 50 apples, and 150 pears. 62. Find three numbers, such that the sum of the first and second shall be 15, of the first and third 16, and of the second and third 17. Ans. 7, 8, and 9. 63. A farmer would mix wheat at 4 shillings a bushel with rye at 2s. 6d. the bushel, so that the mixture may consist of 100 bushels at 3s. 2d. a bushel. How many bushels of each sort must be taken? Ans. 444- and 55-. 64. A steamboat whose rate in still water is 10 miles per hour, descends a river whose velocity is 4 miles an hour, and returns in 10 hours. How far did she proceed? Ans. 42. 65. A gardener and his servant each dug a square piece of ground, whose side is as many feet long, as the labourer is years old. The gardener dug four times as much as his servant, and the sum of their ages is 45. What is the age of each? Ans. 30 and 15 years. 66. A and B make a joint stock of ~500, by which they gain ~160, whereof A had for his share ~32 more than B. What was each person's stock? Ans. A's 300, and B's 200. 67. Find two numbers in the proportion of 9 to 7, so that the square of their sum and the cube of their difference may be equal. Ans. 288, and 224. 68. Divide 100 into two parts, so that the difference of their squares may be 1000. Ans. 55, and 45. 69. A certain sum of money being put out at simple interest, amounts to ~297 12s. in 8 months; and in 15 months to ~306. What is the sum, and what the rate of interest? Ans. ~288 at 5 per cent. 70. A waterman can row down stream, on a certain river, keeping in the middle, 5 miles in 3 quarters of' an hour; but it takes him a full hour and a half to return, though he keeps near shore, where the current is but half as strong as in the middle. What is the velocity of the river per hour? Ans. 21 miles. 71. In a right angled triangle, given one leg 30, and the dif ONLY ONE UNKNOWN QUANTITY. 273 ference between the other leg and the hypothenuse 10; to find the two latter. Ans. The hypothenuise 50, the other leg 40. 72. Given the base 12 and perpendicular 8, of any triangle, to find the side of the inscribed square. Ans. 445. 73. It is proposed to divide the number 10 into two parts, such that the greater multiplied by the less, may be to the greater, as 3 to 1. Ans. 7 and 3. 74. Find two numbers, such that if half the less be taken from the greater, 14 may remain; and if 1 be added to four times the less, the sum will be double the greater. Ans. 9, and 18~. 75. A person has two sorts of wine, one worth 20 pence a quart, and the other 12 pence; from which he would mix a quart to be worth 14 pence. How much of each must he take? Ans. - of the first, and of the second. 76. A labourer reaps 40 acres of barley and oats, and receives 4 shillings an acre for the oats. If the price per acre for the barley had been one shilling more, it would have been to the price per acre of the oats, as 3 to 2; and he receives in all ~9 5s. How many acres were there of each sort? Ans. 15 of oats, and 25 of barley. 77. Before noon, a clock which is too fast and points to afternoon time, is put back five hours and forty minutes; and it is observed that the time before shown, is to the true time, as 29 to 105. Required the true time. Ans. 15 min. before 9 A. M. 78. A and B began to pay their debts. A's money was at first two-thirds of B's; but after A had paid ~1 less than two-thirds of his money, and B ~1 more than seven-eighths of his, it was found that B had only half as much as A had left. What sum had each at first? Ans. A had ~72, and B ~108. 79. A regimentof 594 menis to be raised from three towns, A, B, and C. The contingents of A and B are in the proportion of three to five; and of B and C, in the proportion of eight to seven. Required the number raised by each. Ans. 144 by A, 240 by B, and 210 by C. 80. A besieged garrison had such a quantity of bread, as would, if distributed to each man-at 10 ounces a day, last 6 weeks; but having lost 1200 men in a sally, the governor was enabled to increase the allowance to 12 ounces per day, for 8 weeks. Required the number of men at first in the garrison? Ans. 3200. 81. A man gives the first beggar he meets i1 of the pence he had and 4d more; to the second, 6 of the remainder and 274 SIMPLE EQUATIONS INVOLYING 8d. more; and so on, increasing 4d. every time, till he had no. thingleft; and then all the beggars had equal shares. Required the number of beggars, and of pence at first. Ans. 120d. and 5 beggars. 82. A gentleman gave in charity ~46; a part in equal portions to 5 poor men, and the rest in equal portions to 7 poor women. Now a'man and a woman had between thenn ~8. What was given to the men, and what to the women? Ans. The men received ~25, the women ~21. II. Simple Equations involving two or more unknown quantities. 1. Find two numbers, such that if the first be multiplied by 4 and the second by 8, the sum of the products will be 184; but if the first; be multiplied by 8 and the second by 4, the difference of the products will be 8. Ans. 10 and 18. *2. What two numbers are those of which if the first be increased by 6 it will be three times the greater, and if the seccond be diminished by 4, it will be ~ of the first? Ans. 15 and 7. *3. A and B together own 1320 acres of land, A sells 3 of his share, and B X of his, and the same number of acres remain to each. What were their shares? Ans. 720 and 600. 4. What fraction is that which, by subtracting 4 from both its terms, becomes ~, and by adding 9 to both, becomes? Ans. 15 5. A party were to divide their expenses equally. Had there been 3 persons more and each paid 5 cents more, the bill would have been $3.75 more; but if there had been 10 less, and each had paid 7 cents less, it would have been $9.56 less. How many persons were there, and how much did each pay? Ans. 1.8 persons, paying 90 cts. each. *6. Coffee and sugar were cheaper before the war than at present; coffee by 2 cts. and sugar by 4 cts. a pound, and yet the price of the former was double that of the latter. Now the prices are as 3 to 2. At what rate are they sold? Ans. Coffee 18 and sugar 12 cts. *7. What are those numbers, whose sum is 20, and the quotient obtained by dividing the greater by the less is A? Ans. 12 and 8. 8. What fraction is that, whose numerator being doubled, and denominator increased by 7, the value becomes -; but the denominator being doubled, and the numerator increased by 2, the value becomes -3-? Ans.. TWO OR MIORE UNKNOWN QUANTITIES. 275 9. There are two numbers, such, that ~ the greater added to 1 the less is 13; and if the less be taken from l1 the 2 3 greater, the remainder is nothing. What are the numbers? Ans. 18 and 12. 10. There is a number expressed by two digits, which is double the sum of its digits, and 9 added to four times the number will invert the digits. What is the number? Ans. 18. *11. A number consists of two places of figures, and their sum is 4 times the units figure; moreover, if 8 be added to f of the number, the digits will be inverted. Required the number? Ans. 93. 12. There is a cistern into which water is admitted by three cocks, two of which are exactly of the same dimensions. When they are all open, five-twelfths of the cistern is filled in 4 hours; and if one of the equal cocks be stopped, sevenninths of the cistern is filled in 10 hours and 40 minutes. In how many hours would each cock fill the cistern? Ans. Each of the equal ones in 32 hours, and the other in 24. *13. A and B speculate with different sums; A gains 1501, B loses 501, and now A's stock is to B's as 3 to 2. But had A lost 501. and B gained 1001, then A's stock would have been to B's as 5 to 9. What was the stock of each? Ans. A's was 3001. and B's 3501. *14. Two persons can perform a piece of work together in 21 days. After working in company 9 days, one leaves the other to complete the work alone, which he does in 20 days more. How many days would it take each to perform the whole alone? Ans. 52-l and 35. *15. Six persons comparing their gains found that A's and B's amounted together to $30; C's -and D's to $140; E's and F's to 75; also that A had half as much as C, D four times as much as E, and F five times as much as B. What was the gain of each? Ans. A $20, B 10, C 40, D 100, E 25, and F 50. 16. Two persons, A and B, had a mind to purchase a house rated at 1200 dollars;.says A to B, if you.give me 2 of your money, I can purchase the house alone; but says B to A, if you will give me 3ths of yours, I shall be able to purchase the house. How much money had each of them? Ans. A had 800 and B 600. 17. A Vintner bought 6 dozen of port wine and 3 dozen of white, for 121. 12 shillings; but the price of each afterwards falling a shilling per bottle, he had 20 bottles of port, and 3 dozen and 8 bottles of white more, for the same sum What was the price of each at first? Ans. the price of port was 2s. and of white 3s. per bottle, 276 SIMPLE EQUATIONS INVOLVING 18. It is required to find two numbers, such that 2 of one added to 2 of the other shall make 10: and if the first be multiplied by 3, and the second by 4, 1l1 of the sum shall be 6. Ans. 10 and 9. *19. A may-pole was broken by the wind in such a manner that 4 times the upper or broken part, added to 6 times the remaining part, was equal to 5 times the whole and 28 over: and the proportion of the former to the latter part was 9 to 16. Required the height at first. Ans. 100 ft. 20. The contents of two casks differ by 8 gallons. Each is worth as many shillings per gallon, as there are gallons in the other: and both together are worth ~128. How many gallons in each? Ans. 40 and 32. 21. A lot of land was sold at the rate of $2.25 per foot, and came to $5400. Had it been 10 feet shorter, its value would have been $4725. What was its length and breadth? Ans. 80 by 30 ft. 22. Find a number consisting of three digits, such that the middle digit shall be double the first, or left-hand digit, the sum of the digits 12, and if 792 be added to the number, the digits will be reversed. Ans. 129. 23. Find three numbers, such that the first with half the others, the second with one third of the others, and the third with one fourth of the others, shall always make 34. Ans. 10, 22, and 26. *24. A person having mixed a certain quantity of brandy and wrater, found that if he had mixed 6 gallons more of each, there would have been 7 gallons of brandy for every 6 of water; but if he had mixed 6 less of each there would have been 6 gallons of brandy for every 5 of water. How many gallons of each were mixed? Ans. 78 of brandy and 66 of water. 25. A person had two casks, the larger of which he filled with ale, and the smaller with cider. Ale being half a crown and cider 1 s. per gallon, he paid ~8 6s.; but had he filled the larger with cider, and the smaller with ale, he would have paid;~11 5s. 6d. How many gallons did each hold? Ans. The larger contained 18, and the smaller 11 gallons. 26. D and E each cut a separate pack of cards, in such a manner that the cards cut off by D, added to those left by E made 47; also the number. left by both exceeded the number cut by both, by 50. How many cards did each cut? Ans. 11 and 16. 27. What are the dimensions of a rectangular court, which if lengthened 7 feet, and made 4 feet broader, would contain 363 feet more; but if made 4 feet shorter and 3 feet narrower, would be diminished 208 feet? Ans. 40 long and 25 wide. TWO OR MORE UNKNOWN QUANTITIES. 277 28. A Merchant finds that if he mixes sherry and brandy in quantities which are in the proportion of 2 to 1, he can sell the mixture at 78s. per dozen; but if the proportion be as 7 to 2, he must sell it at 79 shillings a dozen. Required the price of each liquor. Ans. The price of sherry was 81s., and of brandy 72s. per dozen. *29. Two persons discoursing of their money-said A to B, give me 25 dollars, and I shall have as much as you; said B *to A, give me 22 dollars, and I shall have as much again as you. What had each? Ans. A 116, B 166. *30. Two pieces of muslin were bought, of equal fineness, but of unequal lengths; one for ~4 10s., the other for 5 guineas. Now if each had been 3 yards longer, they would have been to one another as 7 to 8. What was the length of each, and the price per yard? Ans. 18 and 21, at 5s. *31. A farm, consisting of two kinds of land, is let at an annual rent of ~390, the pasture being valued at 30s. per acre, and the arable at ~3. Now the number of acres of arable is to half the excess of the arable above the pasture, as 5 to 1. Required the quantity of each. Ans. 60 acres pasture, and 100 acres arable. 32. A sets out express from C towards D, and three hours afterwards B sets out from D towards C, travelling 2 miles an hour more than A. When they meet, it appears that the distances they have travelled are in the proportion of 13 to 15; but had A travelled five hours less, and B had gone 2 miles an hour more, they would have been in the proportion of 2: 5. How many miles did each go per hour, and how many hours did they travel before they met? Ans. A went 4, and B 6 miles an hour, and they travelled 10 hours after B set out. 33. A grocer bought a cask of wine and another of gin for $210, the wine at $1.50 a gallon, and the gin at $0.50 a gallon. He afterwards sold 2 of his wine and 37 of his gin for $150, by which he gained $15. How many gallons were there of each? Ans. 126 of wine, and 42 of gin. *34. A vintner has two casks of wine, from each of which he draws 6 gallons, and finds the remainders in the proportion of 4 to 7. He then puts into the less 3 gallons, and into the greater 4 gallons, and the proportion is that of 7 to 12. How many gallons were there in each at first? Ans. 38 and 62. 35. When wheat was 8 shillings a bushel, and rye 5 shillings, a man wished to fill his sack with a mixture of the two for the money he had with him. He found that if he bought 24 278 SIMPLE EQUATIONS INVOLVING 10 bushels of wheat, and laid out the rest of his money in rye, he would not fill his sack by two bushels; and if he took 10 bushels of rye, and then filled his sack with wheat, he would have 5 shillings remaining. How many bushels would the sack hold, and how much money had he? Ans. 15 bushels, and 95 shillings. *36. The weight of the first of two loaded wagons was to that of the second as 3 to 5. A portion of the load of the weightier wagon, however, having been taken out, and as much again being removed from the other, the weight of the first was to that of' the second as 1 to 2, and both together weighed 21 tons. How much did each weigh at first, and how much was taken from each? Ans. One 9 tons, the other 15, and 1 ton taken from the latter, and 2 from the former. 37. A and B play at backgammon, and A bets three shillings to two on every game. After a certain number of games, it appears that A has won 3 shillings; but had he bet 5s. to 2, and lost one game more out of the same number, he would have lost 30 shillings. How many games did they play? Ans. 34. 38. Some smugglers discovered a cave which would exactly hold their cargo, which consisted of 13 bales of cotton and 33 casks of wine. While they were unloading, a revenue cutter hove in sight, and they sailed away with 9 casks and 5 bales, leaving the cave two thirds full. How many bales or casks would it hold? Ans. 24 bales, or 72 casks. 39. Find two numbers, the greater of which shall be to the less as their sum to 42, and as their difference to 6. Ans. 32 and 24. 40. In one of the corners of a rectangular garden there is a fish-pond, whose area is one ninth of the whole garden; the periphery of the garden exceeding that of the pond by 200 yards. Also, if the greater side be increased by 3 yards, and the other by 5 yards, the garden will be enlarged by 645 square yards. The pond is a rectangle, about the same diameter as the garden. What is the length of each side? Ans. 90 and 60 yards. 41. A and B engage together in play; in the first game A wins as much as he had and four shillings more, and finds he has twice as much as B. In the second game B wins half as much as he had at first and one shilling more, and it appears that he has then three times as much as A. What sum had each at first? Ans. A had 6s. and B 8s. 42. A farmer parting with his stock, sells to one person 9 horses and 7 cows for 300 dollars: and to another, at the TWO OR MORE UNKNOWN QUANTITIES. 279 same prices, 6 horses and 13 cows for the same sum. What was the price of each? Ans. The price of a cow was 12 dollars, and of a horse 24 dollars. *43. A bill of ~26 5s. was paid with half guineas and crowns, and twice the number of half guineas exceed 3 times the number of crowns by 17. How many were there of each? Ans. 40 half guineas and 21 crowns. 44. Two labourers, A_ and B, received ~5 17s. for their wages, A having been employed 15, and B 14 days; and A received for working four days, 11 shillings more than B did for three days. What were their daily wages? Ans. A had 5s. and B 3s. a dav. "45. A boat has two sets of sails, the second worth 4ths of the boat, and both sets together valued at ~45. Now the worth of the boat with the second set, is to its worth with the first, as 11 to 12. Required the respective values. Ans. The boat ~35: the 1st set ~25, and the 2d set ~20. *46. A garden, of which the annual rent is ~5O0,is occasionally let with one of two houses. Now the rent of the garden and first house is to the rent of the second house, as 5 to 3, and the rent of the second house is to that of the first, as 3 to 4. At what yearly rent are the houses let? Ans. The first at ~200, and the other at ~150. 47. There are 3 vessels which together hold 127 gallons; now 3 times the first with 4 times the second will fill the third; and four times the first with twice the second make 70 gallons. Required the content of each. Ans. 8, 19, and 100 gallons respectively. 48. Find three numbers, such that ~ of the first, J of the second, and J. of the third shall make 62; 3 of the first, I of the second, and 1 of the third, 47; and 4 of the first, 1- of the second, with 6 of the third, 38. Ans. 24, 60, and 120. 49. Required those three numbers, whereof any one, added to half the other two, will make 51. Ans. Impossible with unequal numbers. Each must be 251. 50. A, B, and C can together finish a piece of work in 9 days; A, B, and D, in 10 days; A, C, and D, in 11 days; and B, C, and D, in 12 days. In how long time can all four together finish it? Ans. 7 1797 days. The problems which are thus marked (*), may be neatly solved with only one unknown quantity; and it will be a useful exercise for the learner to attempt their solution by both methods. 280 QUESTIONS PRODUCING III. Questions producing Pure Equations of the second and higher degrees. 1. What two numbers multiplied together give the product 245; and if the greater be divided by the less, the quotient is 5? Ans. 35 and 7. 2. Find two numbers whose sum is to the less, as 10 to 3, and the difference of whose squares is 640? Ans. 28 and 12. 3. There is a rectangular field containing 192 square rods, and the sum of the length and breadth: is seven times their difference. Required its dimensions. Ans. 16 long, and 12 wide. 4. What are the contents and value of that field whose sides are as 10 to 3;-which is worth as many dollars per acre as there are rods in the length, and will amount, at that rate, to 40 times as many dollars as there are rods in the width? Ans. 12 acres. $960 value. 5. A person has a quantity of sugar for sale. If he. sells it for three times as many cents as there are pounds, he will get as much above $7, as he will get less than $7 by selling it for half as many cents as there are pounds. How much has he? Ans. 20 lbs. 6. Take a certain number and multiply it by 7, add 15 to the product, and multiply this sum by twice the number. Then divide by 90, and subtract from the quotient ~ of the number, and 140 will remain. Required the number. Ans. 30. 7. What two numbers are as 2 to 3, the difference of whose cubes is 152? Ans. 4 and 6. 8. Two pieces of cloth which together amount to 20 yards were sold, each for as many dollars per yard, as it contained yards; and the sums received for the two were to each other as 9 to 4. How many yards were there in each piece? Ans. 12 and 8 yds. 9. There is a wall containing 5400 cubic feet. The height is 5 times the thickness, and the length 8 times~its height. What are the dimensions? Ans. 3ft. thick, 15ft. high, and 120ft. long. 10. It is required to find that number to which 20 being added, and from which 10 being subtracted, the square of the sum added to twice the square of the remainder shall be 17475. Ans. 75. 11. Divide 100 into two parts, so that their product may be 2100. Ans. 70 and 30. Note. Let x= the excess of the greater part above 50. PURE EQUATIONS. 281 12. Find two numbers, one of which is triple the other, and the sum of their fourth powers is 1312. Ans. 2 and 6. 13. A parallelogram contains 1296 square yards; and if the length be divided by the breadth the quotient will be nine. Required the dimensions. Ans. 108 and 12. 14. A farmer sells a meadow at such a rate that the price of an acre was to the number of acres as 2 to 3. If he had received 270 dollars more for it, the price per acre would have been to the number of acres as 3 to 2. Required the number and price. Ans. 18 acres at $12. 15. Find those two numbers whose difference multiplied by the greater gives 160, but multiplied by the less, gives only 96. Ans. 20 and 12. 16. A number of boys set out to rob an orchard, each carrying as many bags as there were boys, and each bag capable of holding 6 times as many apples as there were bags. After filling their bags, they found the whole number of apples was 1536. How many boys were there? Ans. 4. 17. A departs from London toward Lincoln at the same time that B leaves Lincoln for London. When they met, A had travelled 20 miles more than 13, having gone, as far in 62 days as B had in all; and it appeared that B would not reach London under 15 days. What is the distance between those cities, and how far had each travelled? Ans. 100 miles distance: A having gone 60, and B 40. Note. Let x = A's miles per day, y= the number of days for each, and z= B's miles per day. 18. It is required to find three numbers, such that the sum of the first and second multiplied by the third may be 63; the sum of the second and third multiplied by the first may be 28; and the sum of the first and third multiplied by the second may be 55. Ans. 2, 5, and 9. 19. Required those two numbers whose sum is 18, and the sum of their squares 170. Ans. 7 and 11. Note. A neat solution may be had by substituting m+n for x, and m-n- for y. 20. Find a number, such that twice its cube may exceed 106 as much as one fifth of the square root of its sixth power falls short of 70. Ans. 2 3/ 10. 21. It is required to find two numbers, whose sum, product, and the sum of whose squares shall be equal. Let x+y and x-y be the two numbers, Ans. It is impossible. 22. It is required to divide the number 18 into two such parts that their squares may be as 25 to 16. Ans. 10 and 8. 24* 282 PROBLEMS PRODUCING 23. Two workman, 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 those days, received 75 shillings; but B, who had played 7 days, received only 48 shillings. Now had B played only 4 days, and A had played 7, they would have received exactly alike. For how many days were they engaged, and how many did each work? Ans. Enlgaged for 19 days. A worked 15, and B 12, A at 5 shillings, and B at 4, per day. 24. What two numbers are those whose difference multiplied by the less produces 42, and by their sum 133? Ans. 13 and 6. 25. In a mixture of rum and brandy, the difference between their quantities, is to the brandy, as 100 is to the number of gallons of rum; and the same difference is to the rum, as 4 to the number of gallons of brandy. H-low many gallons were there of each? Ans. 25 of rum, and 5 of brandy. 26. A Farmer has 2 cubical stacks of hay; the side' of one is 3 yards longer than the side of the other, and the difference of their contents is 117 solid yards.. Required the side of each. Ans. 5 and 2 yards. 27. The captain of a privateer descrying a trading vessel 7 miles ahead, sailed 20 miles in direct pursuit of her, and then observing that the trader steered at right angles to her former course, changed his own. course so as to overtake her without making another tack; the privateer running 10 knots, and the trader 8 knots per hour. How long did the chase continue? Ans. 2~ hours. 28. Given the product of two numbers, 12, and the sum of their cubes, 224, to find the numbers. Ans. 6 and 2: 29. Find two numbers, so that the fifth power of one shall be to the cube of the other as 972 ta 125. Ans. 6 and 10. 30. What three numbers are as 4, 4, and -, the sum of whose squares is 549. Ans 18, 12, and 9. 2 y2 31. Given -— }-18 31. Given t find the values of x and y. and a — y =12 Note. Put z-+v —x, and z-v=y, x=8, or 4 andy=4 or 8. 32. Required those three numbers whose product divided by the first and second gives 200; by the first and third, 150, and by the second and third, 120. Ans. 10, 20, and 30. 33. What two numbers are those whose product is 320, and the difference of their: cubes is to the cube of their difference as 61 to unity? Ans. 20 and 16. 34. Required four numbers in arithmetical progression, PURE EQUATIONS. 283 which being increased by 2, 4, 8 and 15 respectively, the sums shall be in geometrical progression. Ans. 6, 8, 10 and 12. 35. In a right angled triangle. given the hypothenuse 60, and the radius of the inscribed circle 12; to find the other sides. Ans. 36 and 48. 36. Of four numbers in geometrical progression there are given the sum of the two least =20, and the sum of the two greatest -45, to find the numbers. Ans. 8, 12, 18 and 27. 37. The sum of ~700 was divided among four persons, whose shares were in geometrical progression; and the difference between the greatest and least, was to the difference between the two means, as 37 to 12. What were the several shares? Ans. 108, 144, 192, and 256 pounds. 38. Find a number consisting of two digits, such that if it be multiplied by the right hand digit, the product will be 159, the sum of the squares of the digits being 34; and if the sum of the digits be multiplied by the right hand digit, the product will be 24. Ans. 53. 39. Required two, numbers whose sum is five-sixths of their produzt, and if the second be multiplied by the square of the first, and the first by the square of the second, the sum of those products shall be 30. Ans. 3 and 2. Note.-The negative roots will also answer the conditions. iV. Affected Quadratic Equations involving one Unknown Quantity. 1. The expenses of a party amount to $10; and if each pays 30 cents more than there are persons, the bill will be settled. How many are there? Ans. 20. 2. A persoB took $6 with him to distribute equally among the paupers of his parish; but as 5 of them were absent, the remainder received 10 cents apiece more than they otherwise would. How many paupers were there? Ans. 20. 3. Divide 36 into three such parts, that the second may exceed the first by 4, and the sum of all their squares may be 464. Ans. 8, 12 and 16. 4. Divide 100 into two parts, so that the sum of their square roots shall be 14. Ans. 64 and 36. 5. Find two numbers whose difference is 10, and if 600 be divided by each, the difference of the quotients shall, also be 10. Ans. 20 and 30. 6. Two droves of oxen differing 6 in number were each purchased for ~672, those in one drove being valued at ~2 a head. more than those in the other. Of how many did 284 AFFECTED QUADRATICS INVOLVING either drove consist, and at what rate per ox were they pur. chased? Ans. The less drove of 42 at 161., the other of 48 at 141. 7. In a fleet of transports, the square root of half the number of ships expressed the number bound for the West Indies; 8 of the fleet were for the Mediterranean, and the remaining 8, for the coast of Africa. Required the whole number. Ans. 128. 8. Two regiments, whose aggregate strength was 1305 men, stood respectively in solid square, the front of one presenting 3 men more than the front of the other. Of how many did each consist? Ans. 729 and 576. 9. A gentleman pays $180 more for his chaise than for his horse; and the price of the chaise is to that of the horse, as the latter is to 15. What is the price of each? Ans. The chaise 240, and the horse 60. 10. If the square of a certain number be subtracted from 52, and the square root of the remainder be increased by 6, and that sum multiplied by 3, and the product divided by the number itself, the quotient will be 5. Required the number. Ans. 6. 11. One lays out a certain sum of money in goods, which he sold again for 56 dollars, and gained as much per cent. as the goods cost him. What was the first cost? Ans. 401. 12. What two numbers are those whose difference is 15, and half their product equals the cube of the less? Ans. 3 and 18. 13. There are three numbers in continued geometrical progression; the sum of the first and second is 10, and the difference of the second and third is 24. What are the numbers? Ans. 2, 8, 32. Note. —This gives two positive roots, onlyone of which answecs the conditions. 14. In a rectangle, given the difference between the diagonal and the length, 2, and between the diagonal and the breadth, 9. Required the sides. Ans. 8 and 15. 15. The base of a right-angled triangle is 20, and is a geometrical mean between the hypothenuse and the perpendicular. What is the hypothenuse? Ans. 25,4+ 16. There is a certain number consisting of two digits; if they be multiplied together, the square root of the product will equal the right hand digit, and if the square of their difference be divided by the left hand digit, the quotient will be one third of their sum. Required the number. Ans. 93. 17. A man playing at hazard won at the first throw as much money as he had in his pocket; at the second he won ONE UNKNOWN QUANTITY. 285 the square root of what he then had, and 5 shillings more; at the third throw he won the square of all he then had, when he found he had in all, 1121. 16s. With how much did he b]egin to play? Ans. 18 shillings. 18. Find a number from the cube of which if you subtract 19, and multiply the remainder by that cube, the product shall be 216. Ans. 3. 19. Two partners, A and B gained 181. by trade. A's money was in trade 12 months, and he received for his principal and gain 261. Also B's money, which was 301., was in, trade 16 months. What was A's stock? Ans. 201. 20. The plate of a looking-glass is 18 inches by 12, and is to be framed with a frame of equal width, whose area is to be equal to that of the glass. Required the width of the frame. Ans. 3 inches. 2.1. A set out from C towards D, and travelled 7 miles a day. After he had gone 32 miles, B set out from D towards C, and went every day - th of the whole journey, and after travelling as many days as he went miles in one day, he met A. Required the distance from C to D. Ans. 152 miles. Note.-The other value will also answer the conditions. 22. 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. 30s. per week. 23. What number is that to which if 24 be added, the square root of the sum shall be 18 less than the original number? Ans. 25. 24. A wall was built round a rectangular court to a certain height. Now the length of one side. of the court was 2 yards less than 8 times the height of the wall, and the length of the adjacent side was 5 yards less than 6 times the height of the wall; and the number of square yards in the. court, was greater than the number in the wall by 178. Required the dimensions of the court, and the height of the wall. Ans. The sides were 30, and 19, and the height 4 yards. 25. The sum of ~190 was divided among three persons in geometrical- progression, and the greatest share exceeded the least by ~50. Required the shares. Ans. 40, 60, and 90 pounds 26. Given the: sum of two numbers, 20, and the sum of their cubes, 2240, to find the numbers. Ans. 12 and 8. 27. A traveller, bound to a place 14 miles distant, goes 26 miles the first: day, 24 the second, and so on decreasing in 286 AFFECTED QUADRATICS INVOLVING arithmetical progression. In how many days will he arrive at his journey's end? Ans. 7 days. 28. Given 9x+ V/(16x-2+-36X3)-15x2-4, to find the value of x. Ans. xa-. 29. What are those two numbers whose product is 100, and the difference of their square roots 3? Ans. 25, and 4. 40 30. Given /xs ---- =3x; to find the value of x. Ans. x= —4. Affected Quadratic Equations involving more than one unknown quantity. 1. What two numbers are those whose product is 120, and if the greater be increased by 8, and the less by 5, the product of the two numbers thence arising shall be 300? Ans. 10 and 12. 2. A merchant bought a piece of cloth for ~8, and after cutting off 4 yards, sold the remainder for what the whole cost him, by which he gained 2 shillings a yard upon what was sold. How many yards were there, and what did they cost? Ans. 20 yds. at 8s. 3. Find two numbers, whose sum multiplied by the greater produces 130, and whose difference multiplied by the less gives —. Ans. 10 and 3. 4. What number is that, which being divided by the product of its two digits, the quotient is 2; and if 27 be added to it the digits will be inverted? Ans. 36. 5. There are three numbers, the difference of whose differences is 8; their sum is 41, and the sum of their squares 699. What are the numbers? Ans. 7, 11, and 23. 6. There are three numbers, the difference of whose differences is 5; their sum is 44, and their continual product 1950. Required the numbers. Ans. 6, 13, and 25. 7. A farmer sold 80 bushels of barley and 100 bushels of oats for ~65; but he sold 60 bushels more of oats for ~20, than of barley for ~10. What was the price of each? Ans. Barley 10s and oats 5. 8. The fore wheel of a carriage makes 5 revolutions more than the hind wheel in going 60 yards; but if the periphery of each wheel be increased one yard, it will make only one revolution more than the hind wheel in the same space. Required the circumference of each. Ans. The less 3 yds. the other 4. 9. A person bought two cubical stacks of hay for ~41, each MORE THAN ONE UNKNOWN QUANTITIY. 287 of which cost as many shillings per solid yard, as there were yards in the 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 ~ 1 6. 10. A farmer received ~7 4s. for a certain quantity of wheat, and an equal sum, at a price less by is. 6d._per bushel, for a quantity of barley, which exceeded the quantity of wheat by 16 bushels. How many bushels were there of each?Ans. 32 of wheat, and 48 of barley. 11. A student being questioned concerning his age and height, proposed the following equations; x2+ xy=1155, and y2-xy=2914, from which it was required to find his age in years and his height in inches. Ans. His age 15 yrs. his height 62 in. 12. It is required to find two such numbers that their sum being subtracted from the sum of their squares may leave 14; and if their product be added to their sum, it may make 14. Ans. 4 and 2. 13. The hypothenuse of a right-angled triangle is 50 feet, and the side of the inscribed square is 171 feet. Required the base and perpendicular. Ans. 30 and 40 ft. 14. There are three towns, A, B and C; the road from B to A forming a right. angle with that from B to C. Now a person travels a certain distance from B towards A, and then crosses by the nearest way, to the road leading from C to A, and finds himself 3 miles from A and 7 from C. Arriving at A, he finds he has gone farther by one-fourth of the distance from B to C, than he would have done, had he not left the direct road. Required the distance of B from A and C. Ans. From C, 8 or 6 miles, and from A, 6 or 8. 15. A took cucumbers to market, and B took thrice as many eggs, and they found that if B gave all his eggs for the cucumbers, A would lose 10 pence, as they were then selling. A therefore reserves two-fifths of his cucumbers, by which B would lose six pence, at the same rate. But B selling the cucumbers at 6 pence apiece, gains upon the whole the price of six eggs. Required the number of eggs and cucumbers, and their price. Ans. 30 eggs at a penny, and 10 cucumbers at 4 pence apiece. 16. Find two numbers whose difference is 279, and the difference of their cube roots 3. Ans. 343 and 64. 17. The product of five numbers in arithmetical progression is 945, and their sum is 25. Required the numbers. Ans. 9, 7, 5, 3, and 1. 18. A gentleman divided ~X10 among three servants, in 288 AFFECTED QUADRATICS, &c. geometrical progression; the first had ~90 more than the last How muCh:had each? Ans. 120, 60, and 30 pounds. 19. There are three numbers in.geometri:oal progression, the greatest of which exceeds the least by 15. Also the difference of the squares- ofthe greatest:and least, is to the sum of the squares of all three, as 5 to 7. Required the numbers. Ans. 5, 10, and 20. 20. There is a number consisting of three digits, the first of which is to the second as the second to the third; the number itself is to the sum of its digits, as 124 to 7; and if 594 be added to it, the digits will be inverted. What is the number? Ans..248. 21. There are three numbers in arithmetical progression, and the square of the first added to -the product of the other two is 16; the square of the second added to the product of the other two is 14. What are the numbers? Ans. 1, 3, and 5. 22. The sum of three numbers in geometrical progression is 3, and the product of the mean and the sum of the extremes is 30. Required the numbers. Ans. 1, 3, 9. 23. Given x2+xy-12 ) and xy-2y2- 1 to find the values of x and y Ans. x-3, and y=l1. Note. Put vy —x. The other root produces a surd. 24. Find two numbers whose product is 300; and if 10 be added to the less, and 8 subtracted from the greater, the product of the sum and remainder shall still be 300. Ans. 15 and 20. 25. Of four numbers in arithmetical progression, the sum ot the squares of the means is -400; and the sum of the squares of the extremes, 464. Required the numbers. Ans. 8, 12, 16, and 20. 26. The sum of two numbers added to the square root of their sum, equals 12, and the sum of their cubes is 189. What are the numbers? Ans. 5 and 4. 27. Given -vx~-Vy=5 to find the values and x-2 Vxy+y — /x+ Vy=O of x and y. Ans. x=9, and y=4. 28. Given x-y-4 to findand y. and x4-y-y=2320 to find and y. Ans. x=7 and, y-3. Note. Put m.I-n for x, and m-n for y. a2. Givend~ 7 s5 }to find x and y, the former being the less. Ans. x-=2, and y=5. Note. Substitute as before.