UC-NRI 
 

 V 
 
- 
 
 
UNIVERSITY EDITION. -REVISED AND ENLARGED. 
 
 THEORETICAL AND PRACTICAL 
 
 TREATISE ON 
 
 A L G E B R A: 
 
 IN WHICH THE EXCELLENCIES OF THE DEMONSTRATIVE METHODS OF THE 
 
 FRENCH ARE COMBINED "WITH THE MORE PRACTICAL OPERATIONS 
 
 OF THE ENGLISH J AND CONCISE SOLUTIONS POINTED 
 
 OUT AND PARTICULARLY INCULCATED. 
 
 DESIGNED 
 
 SCHOOLS, COLLEGES, AND PRIVATE STUDENTS. 
 
 BY H. N. ROBINSON, A. M. 
 
 FORMERLY PROFESSOR OF MATHEMATICS IN THE UNITED STATES NAVY J AUTHOR 
 
 OF A TREATISE ON ARITHMETIC, ASTRONOMY, GEOMETRY, TRIGONOMETRY, 
 
 SURVEYING, CALCULUS, NATURAL PHILOSOPHY, ctC. AC. 
 
 TWENTY-EIGHTH STANDARD EDITION. 
 
 CINCINNATI: 
 PUBLISHED BY JACOB ERNST 
 
 ANDERSON, GATES & WRIGHT, 112 MAIN STREET. 
 
 D. ANDERSON & CO.: TOLEDO, OHIO. 
 IVISON & PHINNEY, N. YORK. 
 
 1858. 
 
I 
 lift - 
 
 
 Entered, according to Act of Congress, in the year 1847, 
 
 BY HORATIO N.ROBINSON, 
 In the Clerk's Office of the District Court of the State of Ohio. 
 
 according to aa of Congress, in the year 1857, 
 
 BY H. N. ROBINSON, 
 
 in the Clerk's Office of the District Court of the United States far the Northern 
 District of New York. 
 
 EDUCATION OEPT 
 
 STEREOTYPED BY JAMH8 & OU., 
 CINCINNATI. 
 
PREFACE. 
 
 SOME apology may appear requisite for offering a new book to the public on 
 the science of Algebra especially as there are several works of acknowledged 
 merit on that subject already before the public, claiming attention. 
 
 But the intrinsic merits of a book are not alone sufficient to secure its adop- 
 tion, and render it generally useful. In addition to merit, it must be adapted 
 to the general standard of scientific instruction given in our higher schools ; it 
 must conform in a measure to the taste of the nation, and correspond with the 
 general spirit of the age in which it is brought forth. 
 
 The elaborate and diffusive style of the French, as applied to this science, 
 can never be more than theoretically popular among the English; and the se- 
 vere, brief, and practical methods of the English are almost intolerable to the 
 French. Yet both nations can boast of men highly pre-eminent in this science, 
 and the high minded of both nations are ready and willing to acknowledge the 
 merits of the other ; but the style and spirit of their respective productions are 
 necessarily very different. 
 
 In this country, our authors and teachers have generally adopted one or the 
 other of these schools, and thus have brought among us difference of opin- 
 ion, drawn from these different standards of measure for true excellence. 
 
 Very many of the French methods of treating algebraic science are not to b- ; 
 disregarded or set aside. First principles, theories and demonstrations, are tb 3 
 essence of all true science, and the French are very elaborate in these. Yet no 
 effort of individuals, and no influence of a few institutions of learning, can 
 change the taste of the American people, and make them assimilate to the 
 French, any more than they can make the entire people assume French ^ iva- 
 city, and adopt French manners. 
 
 Several works, modified from the French, have had, and now have consid- 
 erable popularity, but they do not naturally suit American pupils. They are 
 not sufficiently practical to be unquestionably popular ; and excellent as they 
 are, they fail to inspire that enthusiastic spirit, which works of a more practi- 
 cal and English character are known to do. 
 
 At the other extreme are several English books, almost wholly practical, with 
 little more than arbitrary rules laid down. Such books may in time make 
 good resolvers of problems, but they certainly fail in most instances to make 
 scientific algebraists. 
 
 The author of this work has had much experience as a teacher of algebra 
 
 961661 
 
iv 
 
 PREFACE 
 
 and has used the different varieties of text books, with a view to test their co n 
 parative excellencies, and decide if possible on the standard most proper to be 
 adopted, and of course he designed this work to be such as nis experience arid 
 judgment would approve. 
 
 One of the designs of this book is to create in the minds of the pupils a love 
 for the study, which must in some way be secured before success can be at^ 
 tained. Small works designed for children, or those purposely adapted to per- 
 sons of low capacity, will not secure this end. Those who give tone to public 
 opinion in schools, will look down upon, rather than up to, works of this kind, 
 and then the day of their usefulness is past. On the other hand, works of a 
 high theoretical character are apt to discourage the pupil before his acquire- 
 ments enable him to appreciate them, and on this account alone such works 
 are not the most proper for elementary class books. 
 
 This work is designed, in the strictest sense, to be both theoretical and prac- 
 tical, and therefore, if the author has accomplished his design, it will be fouru\ 
 about midway between the French and English schools. 
 
 In this treatise will be found condensed and brief modes of operation, no 
 nitherto much known or generally practised, and several expedients are system- 
 atised and taught, by which many otherwise tedious operations are avoided. 
 
 Some applications of the celebrated problems of the couriers, and also of the 
 lights, are introduced into this work, as an index to the pupil of the subsequent 
 utility of algebraic science, which may allure him on to more thorough investi- 
 gations, and more extensive study. 
 
 Such problems would be more in place in text books on natural philosophy 
 and astronomy than in an elementary algebra, but the almost entire absence of 
 them in works of that kinu, is our apology for inserting them here, if 
 apology be necessary. 
 
 Quite young pupils, and such as may not hare an adequate knowledge of 
 physics and the general outlines of astronomy, may omit these articles of ap- 
 plication ; but in all cases the teacher alone can decide what to omit and what 
 to teach. 
 
 Within a few years many new text books on algebra have appeared in differ 
 cut parts of the country, which is a sure index that something is desired 
 something expected, not yet found. The happy medium between the theo- 
 retical and practical mathematics, or, rather, the happy blending of the two, 
 which all seem to desire, is most difficult to attain ; hence, many have failed 
 in their efforts to meet the wants of the public. 
 
 Metaphysical theories, and speculative science, suit the meridians of France 
 and Germany better than those of the United States. But it is almost impos- 
 sible to comment on this subject without being misapprehended ; the author 
 of this book is a great admirer of the pure theories of algebraical science, for 
 it is impossible to be practically skillful without having high theoretical acquire- 
 ments. It is the man of theory who brings forth practical respite 1 but it is not 
 theory alone it is theory long and wel applied. 
 
PREFACE v 
 
 Who will contend that Watt, Fitch, or Fulton, \vcre ignorant or inattentive 
 to every theory concerning the nature and power of steam, yet they are only 
 known as practical men, and it is almost in vain to look for any benefactors of 
 mankind, or any promoters of real science from those known only as theo- 
 rists, or among those who are strenuous contenders for technicalities and forms. 
 
 We are led to these remarks to counteract, in some measure, if possible, that 
 false impression existing in some minds, that a high standard work on aigebra, 
 must necessarily be very formal in manner and abstrusely theoretical in mat- 
 ter ; but in our view these are blemishes rather than excellencies. 
 
 The author of this work is a great advocate for brevity, when not purchased 
 at the expense of perspicuity, and this may account for the book appearing 
 very small, considering what it is claimed to contain. For instance, we have 
 only two formulas in arithmetical progression, and some authors have 20. 
 We contend the two are sufficient, and when well understood cover the whole 
 theory pertaining to the subject, and in practice, whether for absolute use or 
 lasting improvement of the mind, are far better than 20. The great number 
 onlv serves to confuse and distract the mind ; the two essential ones, can be 
 remembered and most clearly and philosophically comprehended. The same 
 remarks apply to geometrical progression. 
 
 In the general theory of equations of the higher degrees this work is not too 
 diffuse ; at the same time it designs to be simple and clear, and as much is 
 given as in the judgment of the author would be acceptable, in a work as ele- 
 mentary and condensed as this ; and if every position is. not rigidly demonstra- 
 ted, nothing is left in obscurity or doubt. 
 
 We have made special effort to present the beautiful theorem of Sturm in 
 such a manner as to bring it direct to the comprehension of the student, and if 
 we have failed in this, we stand not alone. 
 
 The subject itself, though not essentially difficult, is abstruse for a learner, 
 and in our effort to render it clear we have been more circuitous and elaborate 
 than we had hoped to have been, or at first intended. 
 
 We may apply the same remarks to our treatment of Horner's method of 
 solving the higher equations. 
 
 Brevity is a great excellence, but perspicuity is greater, and, as a goncral 
 tiling, the two go hand in hand; and these views have guided us in preparing 
 the whole work ; we have felt bound to be clear and show the rationale of 
 every operation, and the foundation of every principle, at whatever cost. 
 
 The Indeterminate and Diophantine analysis are not essential in a regular 
 course of mathematics, and it has rot been customary to teach them in many 
 institutions ; for these reasons we d > not insert them in our text book. The 
 teacher or the student, however, will find the-m in a concise form in a key to 
 thin work. 
 
PREFACE TO THE 27TH (ENLARGED) EDITION. 
 
 THIS is a progressive age, and teachers, and schools, and books, rnusl 
 progress, proportionally, or be left in the rear. 
 
 To keep up with the spirit of the times, and make this volume more 
 complete and valuable, we have added some twenty-four pages of what we 
 conceive to be very valuable matter. 
 
 In the demonstration of the binomial theorem we originally adopted 
 that method, which, to judge from our own personal experience and obser- 
 vation as a teacher of others, was the one most easily comprehended by 
 learners but unfortunately, that method was not complete; it demonstrated 
 only so far as the actual operation was carried, and the further continua- 
 tion of the law could only be inferred. This being the case, we have 
 added another demonstration in this edition, which may require a little 
 more mental discipline to clearly comprehend, but which is, in itself, per- 
 fect and complete. 
 
 We have enlarged the practical part of logarithms, and inserted two 
 small tables, which will enable one to obtain the logarithm of any number 
 whatever ; and by the examples and illustrations here given, a student can 
 obtain a very good knowledge of logarithms, and their uses, in a smaller 
 compass than can be found in any other book known to us. 
 
 We have also added several new practical artifices in the solution of the 
 higher equations, which we hope will not be overlooked. And to the 
 progressive teachers of the United States the whole is respectfully dedi- 
 cated, by their obedient servant, 
 
 THE AUTHOR 
 
 EI.CRIDGE, N". Y., September, 1857. 
 
 VI 
 
CONTENTS. 
 
 INTRODUCTION 9 
 
 SECTION I. 
 
 Addition .13 
 
 Subtraction I ( 
 
 Multiplication 1 ! 
 
 Division 2(> 
 
 Negative Exponents i27 
 
 ALGEBRAIC FRACTIONS 34 
 
 Greatest Common Divisor '31 
 
 Least Common Multiple 42 
 
 Addition of Fractions 44 
 
 Subtraction of Fractions 46 
 
 Multiplication of Fractions 47 
 
 Division of Fractions 49 
 
 SECTION II. 
 
 Equation of one unknown Quantity 52 
 
 Question producing Simple Equations. 62 
 
 Equations of two unknown Quantities .70 
 
 Equations of three or more unknown Quantities 76 
 
 Problems producing Simple Equations of two or more unknown quanti- 
 ties 87 
 
 Interpretation of negative values in the Solution of Problems 94 
 
 Demonstration of Theorems 97 
 
 Problem of the Couriers 98 
 
 Application of the Problem of the Couriers 102 
 
 SECTION III. 
 
 INVOLUTION , 106 
 
 Some application of the Binomial Theorem 108 
 
 Evolution 113 
 
 Cube Root of Compound Quantities 121 
 
 Cube Root of Numerals 123 
 
 Brief method of Approximation to the Cube Root of Numbers 124 
 
 Exponential Quantities and Surds 128 
 
 PURE EQUATIONS 1 33 
 
 Binomial Surds 141 
 
 Problem producing pure Equations 147 
 
 Problem of the Lights ..151 
 
 Application of the Problem 152 
 
 vii 
 
vin CONTENTS. 
 
 SECTION IV. 
 Quadratic Equations 157 
 
 Particular mode of completing a Square, (Art. 99) 160 
 
 Special Artifices in resolving Quadratics, (Art. 106) 169 
 
 Quadratic Equations containing two or more unknown Quantities 175 
 
 Questions producing Quadratic Equations 183 
 
 SECTION V. 
 
 A rithmetical Progression ] 89 
 
 Geometrical Progression 1 95 
 
 flarmonical Proportion 199 
 
 Problems in Progression and Harmonical Proportion 200 
 
 Geometrical Proportion 205 
 
 SECTION VI. 
 
 Binomial Theorem its Demonstration, &c 218 
 
 ' " its General Application 223 
 
 Infinite Series 227 
 
 Reversion of a Series 234 
 
 Exponential Equations and Logarithms 239 
 
 Use and application of Logarithms 256 
 
 Compound Interest and Annuities 258 
 
 SECTION VII. 
 
 General Theory of Equations 263 
 
 Newton's Method of Division, (Art. 167) 276 
 
 Equal Roots 279 
 
 Transformation of Equations, (Art. 171) 282 
 
 Synthetic Division 290 
 
 Sturm's Theorem 309 
 
 Newton's method of Approximation 317 
 
 Homer's Method 31 9 
 
 Its application to Numerical Roots 333 
 
 Expedients to be used in solving Equations 337 
 
 Recurring and Binomial Equations 340 
 
 Practical examples for solution 343 
 
 General method of Elimination (extra) 345 
 
 APPENDIX. 
 
 Inequality 349 
 
 Differential method of Series , 350 
 
 Specific Gravity 357 
 
 Maxima and Minima 359 
 
ELEMENTS OF ALGEBRA. 
 
 INTRODUCTION. 
 
 DEFINITIONS AND AXIOMS. 
 
 ALGEBRA is a general kind of arithmetic, an universal analysts, 
 or science of computation by symbols. 
 
 Quantity or magnitude is a general term applied to everything 
 which admits of increase, diminution, and measurement. 
 
 The measurement of quantity is accomplished by means of an 
 assumed unit or standard of measure ; and the unit must be the 
 same, in kind, as the quantity measured. In measuring length, we 
 apply length, as an inch, a yard, or a mile, &c. ; measuring area, 
 we apply area, as a square inch, foot, or acre ; in measuring 
 money, a dollar, pound, &c., may be taken for the unit. 
 
 Numbers represent the repetition of things, and when no ap- 
 plication is made, the number is said to be abstract. Thus 5, 13, 
 200, &c., are numbers, but $5, 13 yards, 200 acres, are quanti- 
 ties. 
 
 In algebraical expressions, some quantities may be known, 
 others unknown ; the known quantities are represented by the 
 first or leading letters of the alphabet, a, b, c, d, &c,, and the 
 unknown quantities by the final letters, z, y, x, u, &c. 
 
 THE SIGNS. 
 
 (1) The perpendicular cross, thus +, called plus, denotes ad' 
 dition, or a positive value, state, or condition. 
 
 (2) The horizontal dash, thus , called minus, denotes siib- 
 traction, or a negative value, state, or condition. 
 
 (3) The diamond cross, thus X, or a point between two quan- 
 tities, denotes that they are to be multiplied together. 
 
UO: ; V ! J v INTRODUCTION. 
 
 (4) A horizontal line with a point above and below, thus -~, 
 denotes division. Also, two quantities, one above another, as 
 numerator and denominator, thus ~, indicates that a is divided 
 by b. 
 
 (5) Double horizontal lines, thus =, represent equality. Points 
 between terms, thus a : b :: c : d, represent proportion, and ar 
 read as a is to b so is c to d. 
 
 (6) The following sign represents root J ; alone it signifies 
 square root. With small figures attached, thus *.J 4 ,J 5 N /, &c., 
 indicates the third, fourth, fifth, &-., root. 
 
 Roots may also be represented by fractions written over a 
 
 quantity, as a a 3 ? , &c., which indicate the square root, 
 the third root, and fourth root of .* 
 
 (7) This symbol, of>6, signifies that a is greater than b. 
 This * , a<^b, signifies that a is less than b. 
 
 (8) A vinculum or bar , or parenthesis ( ) is used to con- 
 nect several quantities together. Thus -j-6-{-cX#,or (a-}-b-{-c)x, 
 denotes that a plus b plus c is to be multiplied by x. The bar 
 
 may be placed vertically, thus, which is the same 
 
 as (a d-\-e}y, or the same as ay dy-\-ty without the 
 
 d 
 
 +4 
 
 vinculum. 
 
 (9) Simple quantities consist of a single term, as a, b, ab, 3#, 
 &c. Compound quantities consist of two or more terms coli- 
 nected by their proper signs, as a-\-x, 3b-\-2y, 7ab 3xy-{-c, <fcc 
 A binomial consists of two terms ; a trinomial of three ; and a 
 polynomial of many, or any number of terms above two. 
 
 (10) Quantities multiplied together are called factors of the 
 product, and factors are called coefficient in respect to each other. 
 Thus, in the expressions 3#, abx ; 3 and x are factors of the 
 quantity 3a?, and 3 is the coefficient of x. Also, a, b, and x are 
 factors of abx ; a is the coefficient of bx ; b is the coefficient 
 of ax ; and ab is the coefficient of x. 
 
 (11) The measure of a quantity is some exact fact or of that 
 quantity : thus, 5 is the measure of 20 or 25. 
 
 *The adoption and utility of this last mode of notation, which ought to 
 be exclusively used, will be explained in a subsequent part of this worU. 
 
EXERCISES ON NOTATION. u 
 
 (12) The root of any quantity is some equal factor of that 
 quantity. The square root is one of two equal factors ; the cube 
 root one of three equal factors, and so on. 
 
 (13) The multiple of any quantity is some exact number of 
 limes that quantity. 
 
 Other terms will be defined as we use them. 
 AXIOMS. 
 
 Axioms are self-evident truths, and of course are above demon 
 stration ; no explanation can render them more clear. The fol- 
 lowing are those applicable to algebra, and are the principles on 
 which the truth of all algebraical operations finally rests: 
 
 Axiom 1. If the same quantity or equal quantities be added to 
 equal quantities, their sums will be equal. 
 
 2. If the same quantity or equal quantities be subtracted from 
 equal quantities, the remainders will be equal. 
 
 3. If equal quantities be multiplied into the same, or equal 
 quantities, the products will be equal. 
 
 4. If equal quantities be divided by the same, or by equal 
 quantities, the quotients will be equal. 
 
 5. If the same quantity be both added to and subtracted from 
 another, the value of the latter will not be altered. 
 
 6. If a quantity be both multiplied and divided by another, 
 the value of the former will not be altered. 
 
 7. Quantities which are respectively equal to any other quan- 
 tity are equal to each other. 
 
 8. Like roots of equal quantities are equal. 
 
 9. Like powers of the same or equal quantities are equal. 
 
 EXERCISES ON NOTATION. 
 
 When definite values are given to the letters employed, we 
 can at once determine the value of their combination in any alge- 
 braic expression 
 
12 ELEMENTS OF ALGEBRA. 
 
 Let a=5 6=20 c=4 d=l 
 Then a+6 c=5+20 4 or a+6 c=21 
 
 a&+aH-d=5X 20+5X4+1 = 121 
 
 SECTION I. 
 
 ADDITION. 
 
 (Art. 1.) Before we can make use of literal or algebraical 
 quantities to aid us in any mathematical investigation, we must 
 not only learn the nature of the quantities expressed, but how to 
 add, subtract, multiply, and divide them, and subsequently learn 
 how to raise them to powers, and extract roots. 
 
 The pupil has undoubtedly learned in arithmetic, that quanti- 
 ties representing different things cannot be added together ; for 
 instance, dollars and yards of cloth cannot be put into one sum ; 
 but dollars can be added to dollars, and yards to yards ; units can 
 be added to units, tens to tens, &c. So in algebra, a can be 
 added to a, making 2a ; 3a can be added to 5a, making Sa. As 
 a may represent a dollar, then 3a would be 3 dollars, and 5 
 would be 5 dollars, and the sum would be 8 dollars. Again, a 
 may represent any number of dollars as well as one dollar ; for 
 example, suppose a to represent 6 dollars, then 3a would be 18 
 dollars, and 5a would be 30 dollars, and the whole sum would be 
 48 dollars. Also, 8 is 8 times 6 or 48 dollars ; hence any num- 
 ber of 's may be added to any other number of 's by uniting 
 their coefficients ; but a cannot be added to &, or 4 to 36, or to 
 any other dissimilar quantity, because it would be adding unlike 
 things, but we can write a-\-b and 3+36, indicating the addi- 
 tion by the sign, making a compound quantity, 
 
ADDITION. 13 
 
 Let the pupil observe that a broad generality, a wide latitude 
 must be given to the term addition. In algebra, it rather means 
 uniting-, condensing, or reducing terms, and in some cases, the 
 sum may appear like difference, owing to the difference of signs. 
 Thus, 4a added to a is 3a ; that is, the quantities wilted can 
 make only 3a, because the minus sign indicates that one a must 
 be taken out. Again, 76-f-36 46, when united, can give only 
 66, which is in fact the sum of these quantities, as 4b has the 
 minus sign, which demands that it should be taken out; hence to 
 add similar quantities we have the following 
 
 RULE. Add the. affirmative, coefficients into one sum and the 
 negative ones into another, and take their difference with the 
 sign of the greater, to which affix the common literal quan- 
 tity. 
 
 EXAMPLES FOR PRACTICE. 
 
 5a llx + 5#6 6a-f56 7cd + 8xy 
 
 2a 2x Gab Ga4b 2cd + 3xy 
 
 Sum la IQx ab +6 9cd+llxy 
 
 5fl-f& 
 
 cdy+ax 
 
 4x 6 
 
 3a-l-c 
 
 2cdy '3 ax 
 
 2^+10 
 
 7o 26+c 
 
 4cdy-}-3ax 
 
 -3x+ 7 
 
 3a--36 4c 
 
 Icdy ax 
 
 6rc 12 
 
 Sum 12 46 2c 9x 1 
 
 N. B. Like quantities, of whatever kind, whether of powers or 
 roots, may be added together the same as more simple or rational 
 quantities. 
 
 Thus 3a 2 and 8a* are 11 2 , and 76 3 -f-36 3 =106 3 . No matter 
 what the terms may he, if they are only alike in kind. Let the 
 reader observe that 2(a-\-b~)-\-3(a-{-b) must be together 5(a-\-b), 
 mat is, 2 times any quantity whatever added to 3 times the 
 same quantity must be 5 times that quantity. Therefore, 
 
 {-y=l t Jx J ry 1 for Jx+y, which represents the 
 square root of x-\-y, may be considered a single quantity. 
 
 (Art. 2.) To find the sum of various quantities we have the 
 following 
 
14 ELEMENTS OF ALGEBRA. 
 
 RULE. Collect together all those that are alike, by willing 
 their coefficients, and then write the different sums, one after 
 another, with their proper signs. 
 
 EXAMPLES. 
 
 1. 
 
 2. 
 
 3. 
 
 3xy 
 
 , 9x*y 
 
 6xy 12 x? 
 
 2ax 
 
 7x*y 
 
 4^ +3xy 
 
 5xy 
 
 -\-3axy 
 
 ~{-4x* 2xy 
 
 Gax 
 
 ~4x*y 
 
 - -3^-{-4a? 2 
 
 Sum Sax 2xy 
 
 3axy 2x?y 
 
 4xy 8^ 
 
 4. 
 
 5. 
 
 6. 
 
 1 4ax 2 x z 9-j-lO Jax 5; y 
 
 5ax-{-3xy 2x-}-7 l Jxy-{-5y 
 
 Sy z 4ax 5y-\-3,Jax-\-4y 
 
 ^4-406^ 3a^+26 10 4Jax+4y 
 
 7. Add 2xy2a z , 3a z -{-xy, a z -\-xy, 4a? 3xy, 2xy 2a*. 
 
 Ans. 4a?-{-3xy. 
 
 8. Add 8aV 3xy, Sax 5xy, Qxy 5ax, 2a z x z -\-xy, 
 5ax 3xy. rfns. I0a z x z -+-5ax xy. 
 
 O. Add 2a?* 10y, 3jxy-{-Wx, 2x z y-\-25y, \2xy,Jxy, 
 Sy-f-nV^. 
 
 Ans. 2x z y+12xy-{-Wx-\- 
 
 10. Add 26a>-12, 3x*2bx, 5a^ 3 Jx, 
 a!*+3. dns. 
 
 11. Add 106 2 3bx*, 2b z x z b z , 1026^, b z x z 20, 
 3bx z +b z . Ans. 10^ 2 +36 2 a> 2bx z 10. 
 
 12. Add 2a 2 Zax^+x 2 , 2ax* -\3xy-\-9, Wtfxy i. 
 
 JJns. 12a 2 ax*+x* Uxy+4. 
 
 1. Add 9&c 8 18ac* 156c 3 --ac 9ac 2 246c 3 , 9ac 2 2 
 
 . ac 2. 
 
ADDITION. 15 
 
 14. Add 3m 2 1, 6am 2w 2 -f-4, 7 8a?/i-}-2w 2 , and 
 6ro 2 -f2am-H. ^ns 9ro 2 +ll. 
 
 15. Add 12a 13a6-i-16a;r, 8 4m+2y, 6a-f-7a6 2 4- 
 \%y 24, and lab lQax-{-4m. 
 
 rfns. 6a 6a6+ 1 4y -f-7o6 2 1 6. 
 
 16. Add 72ax 4 8a# 3 , 38z 4 3ay 4 +7ay*, S+l2ay\ 
 -607/4-12 34az 4 +5i/ 3 9oy 4 . ^ns. 2ai/ 8 +20. 
 
 Add a-(-6 and 3 56 together. 
 
 Add Go; 56-f a+8 to 5a4x-\-4b3. 
 
 Add a+26 3c 10 to 36 4a+5c-}-10 and 56 c. 
 
 Add 3a-f-6 10 to c da and 4c-J-2 36 7. 
 
 Add 3a 2 -l-26 2 c to 206 3 2 -{-6c 6. 
 
 (Art. 3.) When similar quantities have literal coefficients, we 
 may add them by putting their coefficients in a vinculum, and 
 writing the term on the outside as a factor. 
 
 Thus the sum of ax and bx is 
 
 Add 
 
 EXAMPLES. 
 1. 
 
 ax+bif 
 
 2. 
 
 ay+cx 
 3ay-{-2cx 
 4y -{-Qx 
 
 Sum (-}-- 
 Add 
 
 lc+4d)x-\-(b-\-3a-{-7)y 2 ( 
 
 3. 
 
 3x-\-2xy 
 bx-^cxy 
 (a -\-b}x-{-2cdxy 
 
 4. 
 
 ax-}-7y 
 7ax 3y 
 2x -\-4y 
 
 Sum 
 
 5. Add 8oH-2(;r+a)-r-36, 9ax+6(x-{-a) 96, and 
 66 7ax 8(a?+). 
 
 6. Add (a-\-b),Jx and (c+2a 6)7a? together. 
 
 7. Add 28o 3 (aM-57/)+21, 18 13a 3 (#-f-5i/), 
 
 . 18a+13. 
 
16 ELEMENTS OF ALGEBRA. 
 
 8. Add 17a(x-h3t/)-J-12aW, 8 18ay 
 
 SUBTRACTION. 
 
 (Art. 4.) We do not approve of the use of the term subtraction, 
 as applied to algebra, for in many cases subtraction appears like 
 addition, and addition like subtraction. We prefer to use the 
 expression^/u/mg 1 the difference. 
 
 What is the difference between 12 and 20 degrees of north 
 latitude ? This is subtraction. But when we demand the differ- 
 ence of latitude between 6 degrees north and 3 degrees south, the 
 result appears like addition, for the difference is really 9 de- 
 grees, the sum of 6 and 3. This example serves to explain the 
 true nature of the sign minus. It is merely an opposition to 
 the sign plus ; it is counting in another direction; and if we 
 call the degrees north of the equator plus, we must call those 
 south of it minus, taking the equator as the zero line. 
 
 So it is on the thermometer scale ; the divisions above zero 
 are called p!us t those below minus. Momsy due to us may be 
 called plus ; money that we owe should then be called minus, 
 the one circumstance is directly opposite, in effect, to the other. 
 Indeed, we can conceive of no quantity less than nothing, as 
 we sometimes express ourselves* It is quantity in opposite cir- 
 cumstances or counted in an opposite direction; hence the differ' 
 ence or space between a positive and a negative quantity is 
 their apparent sum. 
 
 As a further illustration of finding differences, let us take the 
 following examples, which all can understand : 
 
 From 16 16 16 16 16 16 
 
 Take 12 8 2 2 4 
 
 Differ, 4 8 14 16 18 ~ 20 
 
 Here the reader should strictly observe that the smaller the 
 number we take away, the greater the remainder, and when the 
 subtrahend becomes minus, it must be added. 
 
SUBTRACTION. 17 
 
 From I2a 120 12a 120 120 12a 
 
 Take 20 16 12a 9 60 a 
 
 Diff. _8 4a 30 60 lla 
 
 When a greater is taken from a less, we cannot have & posi- 
 tive or plus difference, it must be minus. 
 
 From 200 100 50 50 100 
 Take lla lla lla lla b b 50 
 Diff. 90 a 60 lla +b b5a 50 
 
 Here it will be perceived, that any quantity subtracted from 
 zero will be the same quantity with its sign changed. 
 
 (Art. 5.) Unlike quantities cannot be written in one sum, (Art 
 1,) but must be taken one after another with their proper signs : 
 therefore, the difference of unlike quantities can only be ex- 
 pressed by signs. Thus the difference between a and b is a ft, 
 a positive quantity if a is greater than 6, otherwise it is negative 
 From a take b c, (observe that they are unlike quantities). 
 
 OPERATION. 
 From a-fO-f-0 
 
 Take Q+b c 
 
 Remainder, or difference, a 6-f-c 
 
 This formal manner of operation may be dispensed with ; the 
 ciphers need not be written, and the signs of the subtrahend need 
 only be changed. 
 
 From the preceding observation, we draw the following 
 
 GENERAL RULE FOR SUBTRACTION, OR ALGEBRAIC DIF- 
 
 FERENCES. 
 
 Change the signs of the subtrahend, or conceive them to be 
 clanged; then proceed as in addition. 
 
 EXAMPLES. 
 
 1. 2. 3. 
 
 From 4a+2x 3c 3ax-{-2y a-\-b 
 
 Take a-\-4x 6c xyly a & 
 
 Remainder, 3a 2a?+3c Sax xy-\-^y 26 
 
18 ELEMENTS OF ALGEBRA. 
 
 4. 5. 6. 
 
 From 2# 2 3x-^-y 2 7a+2 5c < 
 
 Take x* 4aH-a a-J-2-1- c %x 
 
 Rem. 3r>-l-#-- 2 a 8a * 6c 
 
 7. 8. 
 
 Fmm 8x 2 3xy-}-2y z -}- c ax-\-bx-\-cx 
 
 Take a,' 2 Qxy-\-3y z 2c 
 
 Diff. 
 
 9. 
 
 From ax-\-by-{-cz 
 
 Take mx ny pz 
 
 Diff. 
 
 (Art. 6.) From a take b. The result is 6. The minus 
 sign here shows that the operation has been performed ; b was 
 positive before the subtraction ; changing the sign performed 
 the subtraction; so changing the sign of any other quantity 
 would subtract it. 
 
 11. From 3 take (ab-\-x c ^/), considering the terms in 
 the vinculum as one term, the difference must be 3a (ab-\-x 
 c y), but if we subtract this quantity not as a whole, but term 
 by term, the remainder must be 3a ab aj-j-c-f-y. 
 
 That is, when the vinculum is taken away, all the signs 
 within the vinculum must be changed. 
 
 12. From 3Qxy take (40#y 26 2 +3c 4d). 
 
 Rem. 2b 2 IQxy 3c-{-4c/. 
 
 13. From Jx-\-y+3ax 12 take 
 
 Rem. 5ax 3jx-}-v 12 b. 
 
 14. Find the difference between 6?/ 2 2i/ 5 and 8^/ 2 
 4-12. 
 
 15. From 3a 6 2o;+7 take 
 
MULTIPLICATION. 19 
 
 10. From 3p-\-q-\-r 25 take q8r-}-2s 8. 
 
 dns. 3j-f-9r 4s+8 
 
 17. From 13a 2 2ax+9x* take 5a z lax x 2 . 
 
 An*. Sa*+5ar-\-lOx* 
 \_ 
 
 18. From 20xy S^a-f 3i/ take to/+5a 2 y 
 
 19. From the sum of Gorfy Ilax*, and 8;r 2 i/+3a;r 3 , take 
 42^ 4 ax ^ ttf Diff. 
 
 20. From the sum of 15a 2 6-r-8ctfo 3 and 
 
 take the sum of 12a 2 & 3cdx 8 and 16+cdx 4a z b. 
 
 Diff. 
 
 21. From the difference between Sab I2cy and 3ab-\- 
 y take tlie sum of Sab Icy and ab-\-cy 
 
 Diff. 5ab Wcy. 
 From 2+26 take ab. 
 From ax-}-bx take .T bx. 
 From a-{-c+6 take a-{-c b. 
 From 3a?+2y+2 take 5x-\-3y+b. 
 From 6-f 2j?+c take 5a+6a? 3c. 
 From 4a 2x 2 take 6a 2a? 2. 
 From 12^2^7/4-3 take 7+Gy+Wx. 
 
 MULTIPLICATION. 
 
 (A.rt. 7.) The nature of multiplication is the same in arithmetic 
 and algebra. It is repeating one quantity as many times as there 
 are units in another ; the two quantities may be called factors, 
 and in abstract quantities, either may be called the multiplicand ; 
 the other will of course be the multiplier. 
 
 Thus 4X5. It is indifferent whether we consider 4 repeated 
 5 times or 5 repeated 4 times ; that is, it is indifferent which we 
 call the multiplier. Let a represent 4, and b represent 5, then 
 the product is aXb ; or with letters we may omit the sign and 
 the product will be simply ab. 
 
20 ELEMENTS OF ALGEBRA. 
 
 The product of any number of letters, as ab c d, is abed. 
 
 The product of x y z is xyz. 
 
 Tn the product it is no matter in what order the letters are 
 placed ; xy and yx is the same product. 
 
 The product of axXby is axby or abxy. Now suppose 
 =6 and 6=8, then ab=4S, and the product of axXby would 
 be the same as the product of GxXSy or 4Sxy. From this we 
 draw the following rule for multiplying simple quantities : 
 
 Multiply the coefficients together and annex the letters, one 
 after another, to the product. 
 
 EXAMPLES. 
 
 1. Multiply 3x by 7 a. Prod. 21 ax. 
 
 2. Multiply 4y by Sab Prod. IZaby. 
 
 3. Multiply 36 by 5c, and that product by Wx. 
 
 Prod. ISQbcx 
 
 4. Multiply Qax by I2by by 7ad. Prod. 5Q4aaxydb. 
 
 5. Multiply Sac by 116 by xy. 
 
 6. Multiply af by pq by 4. 
 
 In the above examples no signs were expressed, and of course 
 plus was understood ; and it is as clear as an axiom that plus 
 multiplied by plus must produce plus, or a positive product. 
 
 (Art. 8.) As algebraic quantities are liable to be affected by 
 negative signs, we must investigate the products arising from 
 them. Let it be required to multiply 4 by 3, that is, repeat 
 the negative quantity 3 times ; the whole must be negative, as 
 the sum of any number of negative quantities is negative. 
 Hence minus multiplied by plus gives minus, aXb gives 
 ab ; also a multiplied by b must give ab, as we may 
 Conceive the minus b repeated a times. 
 
 (Art. 9.) Now let it be required to multiply 4 by 3, that 
 is, minus 4 must be subtracted 3 times ; but to subtract minus 
 4 is the same as to add 4, (Art. 5,) giving a positive or plus 
 quantity; and to subtract it 3 times, as the 3 indicates, will 
 give a product of +12. 
 
 That is, minus multiplied by minus gives phis. 
 
MULTIPLICATION. 21 
 
 This principle is so important that we give another mode of 
 illustrating it: 
 
 Required the product of a b by a c. 
 
 Here a b must be repeated a c times. 
 
 If we take a b, a times, we shall have too large a product, 
 as the multiplier a is to be diminished by c. 
 
 That is a b 
 
 Multiplied by a 
 
 Gives ua ab 1 which is too great by a b repeated 
 
 c times, or by ac cb, which must be subtracted from the former 
 product; but to subtract we change signs, (Art. 5,) therefore the 
 true product must be aa ab ac-}-cb. 
 
 That is, the product of minus b by minus c gives plus be, 
 and, in general, minus multiplied by minus gives plus. 
 
 But plus quantities multiplied by plus give plus, and minus by 
 plus, or plus by minus, give minus ; therefore we may say, in 
 short, 
 
 77m/ quantities affected by like signs, when multiplied toge- 
 ther, give plus, and when affected by unlike signs give minus. 
 
 (Art. 10.) The product of a into b can only be expressed by 
 ab or ba. The product of abed, &c., is abed; but if b c and d 
 are each equal to a, the product would be aaaa. 
 
 The product of aa into aaa is aaaaa ; but for the sake of 
 brevity and convenience, in place of writing aaa, we write a 3 . 
 The figure on the right of the letter shows how many times the 
 letter is taken as a factor, and is called an exponent. The pro- 
 duct of a 3 into a 4 is a repeated 3 times as a factor, and 4 times 
 as a factor, in all 7 times ; that is, write the letter and add the 
 exponents. 
 
 EXAMPLES. 
 
 What is the product of a 9 by a 5 ? Jlns. a*. 
 
 What is the product of x* by # 8 ? Am. x l \ 
 
 What is the product of i/ 2 by y* by t/ 5 ? JJns. y w . 
 
 What is the product of a n by a m ? Ans. a" + "'. 
 
 >Vhat is the product of 6V by bx. Ans. b*x*. 
 
 What i s the product of ac by ac 2 by aV ? Ans. ^ 5 c 6 . 
 
22 ELEMENTS OF ALGEBRA. 
 
 If adding numeral exponents is a true operation, it must be 
 equally true when the exponents are literal. 
 
 N. B. When the exponent is not expressed, one is understood, 
 for a is certainly the same as a 1 , or once taken. 
 
 (Art. 11.) Every factor must appear or be contained in a pro- 
 duct. Thus ax z multiplied by bx 3 must be abx 5 . Now if a 6 
 and b ==10 the product would be 60# 5 . 
 
 Multiply 3a 2 by 7 3 . Product 21a 5 . 
 
 From this we draw the following rule for the multiplication 
 of exponential quantities. 
 
 Multiply the, coefficients and add the exponents of the same 
 latter. Jill the letters must appear in the product. 
 
 EXAMPLES. 
 
 Multiply 4a 4 by 3. Prod. 12 5 . 
 
 Multiply 3x z by 2x*. Prod. O.r 5 . 
 
 Multiply 3x by 7x* by 3a?y. Prod. 63aVy. 
 
 What is the product of 2ax z , 4axy, 7abx ? 
 
 Prod. 5Ga?x*by. 
 What is the product of *2& n , 3a m x, and ax 1 
 
 Prod. Ga n + m +\x*. 
 
 Multiply 9a?x by 4x. Prod. 36a 2 .r 2 . 
 
 Multiply ITtfW by 7ac. Prodf. 119 4 6 2 c 4 . 
 
 Multiply Ilfl 5 i 2 c by 10 5 i 8 c 9 , Prod 110a 10 6 10 c 10 . 
 
 Multiply 121i 2 c 3 ^ by 5a A bxy 2 . Prod. 605a 4 i 3 c 3 ^y. 
 
 Multiply 77a 3 cz 4 by 61 2 ^. Prod. 4Q97a 5 bcx*. 
 
 Multiply 117a6 2 c 3 a: by 2 3 i 2 c. Prod. 234aWc*x. 
 
 Multiply 9a z x by Gx. 
 Multiply 9 ax 2 by 7 ax. 
 Multiply 7 ax by 4. 
 Multiply Sac by 2cx by 4c. Prod 24c s .z 
 
 (Art. 12.) When one compound quantity is to be multiplied 
 or repeated as many times as there are units in another, it is 
 evident that the multiplicand must be repeated by every term of 
 the multiplier. 
 
MULTIPLICATION. 
 
 Thus the product of a-\-b-}-c by x-\-y+z. 
 It is evident that a-\-b-{-c must be repeated x times, then y 
 times, then z times ; and the operation may stand thus : 
 
 Product by x ax-\-bx-\-cx 
 Product by y ay-{-by-\-cy 
 
 Product by z az-}-bz-}-cz 
 
 Entire Product ax-{-bx-}-cx-{-ay-}-by-\-cy-}-az-{-bz-{-cz. 
 
 From the foregoing articles we draw the following general 
 rule for the multiplication of compound quantities. 
 
 Multiply all the terms of the multiplicand by each terra of 
 the multiplier, observing that like signs, in both factors, give 
 plus, and unlike, minus. 
 
 Write each term of the product distinctly by itself, with its 
 proper sign, and afterwards condense or connect the terms as 
 much as possible, as in addition. 
 
 EXAMPLES. 
 
 1. 2. 
 
 Multiply 2ax 3x 3x-\- 2y 
 
 By 2x -\-4y 4x 5y 
 
 Partial product 4ax 2 - 6x* l2x*-\- Sxy 
 
 2d part. prod. Saxy I2xy 15xy ICty 2 
 
 Whole prod. 4ax-{-Saxy6x z I2xy 12^ 7xyWy* 
 
 3. Multiply 2x z -{-xy 2?/ 2 
 
 By 3x 3^ 
 
 Partial product Qx*-\-3x~y Gxy 2 
 
 2d partial product 6x*y 3xy 2 -}-6y 3 
 
 Whole product Gx 3 3x 2 y 9xy*+6y* 
 
 4. Multiply 3a 2 2a5 b z by 2 4b. 
 
 'Prod 6a 3 16a 2 6-f-6& 2 -f 4#>. 
 
 5. Multiply x*xy-\-if by x+y. Prod, 
 
24 ELEMENTS OF ALGEBRA. 
 
 6. Multiply a 2 3ac-f-c 2 by a c. 
 
 Prod, a 3 4a 2 c+4c 2 c 8 . 
 
 7. Multiply a-f-6 by a-\-b. Prod. a 2 +2a&-f-& 2 - 
 
 8. Multiply x+y by #+# /Vod. rc 2 +2a < i/+3/ 2 . 
 
 9. Multiply a b by a b. Prod. a 2 2ab-\-b': 
 
 10. Multiply x y by x y. Prod, x* 2xy-\-y\ 
 
 (Art. 13.) By inspecting all the problems, from the 7th to the 
 10th, we shall perceive that they are all binomial quantities, and 
 the multiplicand and multiplier the same. 
 
 But when a number is to be multiplied into itself the product 
 is called a square. Now by inspecting the products, we find 
 that the square of any binomial quantity is equal to plus ; the 
 squares of the two parts and twice the product of the two parts. 
 
 N. B. The product of the two parts will be plus or minus, 
 according to the sign between the terms of the binomial. 
 
 Let us now examine the product of o-{-b into a b. 
 a -}-b 2m +2n 
 
 a b 2m 2n 
 
 a?-\-ab 4m 2 -\-4mn 
 
 ab b 2 4mn 4?i 2 
 
 Product a 2 b* 4m* in 2 
 
 Multiply 20-J-3& by 2a 3b. Prod. 4a*Qb 2 . 
 
 Multiply 3y-\-c by 3yc. Prod. 9j/ 2 c 2 . 
 
 Thus, by inspection, we find the product of the sum and 
 difference of two quantities is tqual to the difference of their 
 squares. 
 
 The propositions included in this article are proved also in 
 geometry. 
 
 (Art. 14.) We can sometimes make use of binomial quantities 
 greatly to our advantage, as a few of the following examples will 
 show : 
 
 1. Multiply a-r-6-f-c, by a+6 f c. 
 
 Suppose a-\-b represented by s, then it will be s-J-c. 
 
MULTIPLICATION. 25 
 
 The square of this is s 2 -f-2se-fc- ; restoring the value of s, and 
 we have (a-}-b)~+2(a+b)c-\-c 2 . 
 
 2. Square x-\-y z. Let x-}-y=s. 
 
 Then ( z) 2 =s 2 2sz+z*=(x-\-y} 2 2(x+y)z-{-z*. 
 
 &. Multiply x-\-y-\-z by x+y z. Prod. (x-\-y} 2 z*. 
 
 4. Multiplj 2a; 2 3a:+2 by a? 8. 
 
 I 2X 3 19^+260; 10. 
 
 5. Multiply ax-\-by by ax+cy. 
 
 Prod. d*x 2 +(ab-\-ac)xy+cby*. 
 
 6. Multiply %x+y by 
 
 7. Multiply a 3 -f 2a 2 6+2a6 2 +6 
 By a 3 
 
 Prod. 
 
 8. Multiply ^ ^ a? +f 
 By \x + 2 
 
 Product, la 8 -}- V a^Js-K 
 9. What is the product of a m +6 m by o n +6 n ? 
 
 10. What is the product of z 2 fa? by a; 2 ia? ? 
 
 ^fns. as* f a?+ 1 a; 2 
 
 11. What is the product of 4z 3 -|-8;r 2 -r-16aM-32 by 3x 0? 
 
 4 192. 
 
 12. What is the product of a*-}-a 2 b-{-ab 2 -{-b 2 by a b ? 
 
 w?ns. a 4 ** 
 3 
 
26 ELEMENTS OF ALGEBRA. 
 
 DIVISION. 
 
 (Art. 15.) Division is the converse of multiplication, the pro* 
 duct being called a dividend, and one of the factors a divisor. If 
 a multiplied by b give the product &, then ab divided by a must 
 give b for a quotient, and if divided by b, give a. In short, if 
 one simple quantity is to be divided by another simple quantity, 
 the quotient must be found by inspection, as in division of num- 
 bers. 
 
 EXAMPLES. 
 
 1. Divide IGab by 4a. *An&. 4b 
 
 2. Divide 2 lacd by 7c. rfns. 3a(l 
 
 3. Divide ab*c by ac. Am. 6 2 . 
 
 4. Divide Zaxy by 2bc. Ans. ?^ 
 
 In this last example, and in many others, the absolute division 
 cannot be effected. In some cases it can be partially effected, 
 and the quotients must be fractional. 
 
 5. Divide 3acx* by ac?/. Ans. - . 
 
 * J y 
 
 Qb 
 
 6. Divide 72te by 8abx. Ans. . 
 
 7. Divide 27aby by llabx. Ans. 
 
 (Art. 16.) It will be observed that the product of the divisoi 
 and quotient must make the dividend, and the signs must con- 
 form to the principles laid down in multiplication. The follow- 
 ing examples will illustrate : 
 
 8. Divide 9y by 3?/. jlns. 3. 
 
 9. Divide 9i/ by 3y. Ans. +3. 
 3O. Divide +9y by 3y. Ans. 3. 
 
 * The terra quotient would be more exact and technical here ; but, in re- 
 sults hereafter, we shall invariably use the term Ans., as more brief and ele- 
 gant, and it is equally well understood. 
 
DIVISION. 27 
 
 (Art. 17.) The product of a 3 into 2 ; s a 5 , (Art. 10,) that is, 
 in multiplication we add the exponents ; and as division is the 
 converse of multiplication, to divi le powers of tKe same letter, 
 we must subtract the exponent of the divisor from that of the 
 dividend. 
 
 Divide 2a by a 4 . Ans. 2a*, 
 
 Divide a 7 by a 6 . Jlns. a. 
 
 Divide IGr 5 by 4x. tfns. 4x*. 
 
 Divide ISaxy* by Say. Jlns. 5xy z . 
 
 Divide 63a m by 7a n . Aw. 9"- n . 
 
 Divide I2ax n by Sax. JLns. 
 
 Divide 7rf6 by 
 
 Divide 5a 2 ^ by -7fl 4 a; 2 . ^ns. -^, 
 
 Divide 117aW by 78 5 6c 4 . . 
 
 o 
 
 Divide 96a6c by \Za*bc*d. Ans. 
 
 Divide a^c 2 by 
 
 n 
 
 Divide 27a 3 6 4 cd 2 by 21 abed. Ans. -d>b*d. 
 
 Divide 14a6 2 cc? by Ga^c 2 . ns. . 
 
 ottC 
 
 (Art. 18.) The object of this article is to explain the nature of 
 negative exponents. 
 
 Divide a 4 successively by a, and we shall have the following 
 quotients : 
 
 1 1 1 
 a 3 , a 2 , a, 1, ~, ^, ^, &c. 
 
 Divide fl 4 again, rigidly adhering to the principle that to 
 divide any power of a by a, the exponent becomes one less, and 
 we have 
 
28 ELEMENTS OF ALGEUKA. 
 
 a , a 2 , a 1 , a, or 1 , or 2 , er 3 , &c. 
 
 Now these quotients must be equal, that is, a 3 in one series 
 equals a 3 in the other, and 
 
 rt 2 =a 2 , a=a l , l=a, -=<*- 1 =a~* =a~* 
 a a 2 a 3 
 
 Another illustration. We divide exponential quantities by 
 subtracting the exponent of the divisor from the exponent of the 
 dividend. Thus a 5 divided by a 2 gives a quotient of 5 ~ 2 =o 3 . 
 a 5 divided by a 7 =a 5 ~ 7 =a~ 2 . AVe can also divide by taking the 
 dividend for a numerator and the divisor for a denominator, thus 
 
 i 5 I 1 
 
 =, therefore 2 =a~ 2 (Axiom 7.) 
 
 a 1 a 2 a 2 
 
 From this we learn, that exponential factors maybe changed 
 from a numerator to a denominator^ and the reverse, by chang- 
 ing the signs of the exponents. 
 
 Thu< -ax~* ~ -=a-- 
 
 US ' ^~ 3*r'~3a 3 x n 
 
 Divide d>bc by a*b 2 c~ l . Am. ar l b~ l c*. 
 
 Observe, that to divide is to subtract the exponents. 
 
 Divide Uab*cd by 6a 2 bc*. dns. = 
 
 (Art. 19.) A compound quantity divided by asimple quantity, is 
 effected by dividing each term of the compound quantity by the 
 simple divisor. 
 
 EXAMPLES. 
 
 1. Divide Sax 15x by 3x. Ana. a 5. 
 
 . Divide 8a^+12^ by 4x*. Jlns. 2x+3. 
 
 3. Divide 3bcd+l2bcx 96 2 c by 3bc. Jlns. d-\-4x 3b. 
 
 4. Divide lax-\-3ay 7bd by lad. 
 
 x 3 y , b 
 
 tins. - T r^H 
 
 d Id a 
 
DIVISION. 29 
 
 ft. Divide 15 2 6c \5acx*-\-5ad* by 5ac. 
 
 a 
 fins. 
 
 c 
 6. Divide Wx*I5x 2 25a? by 5x. Jlns. 2x* 3x 5. 
 
 V Divide -Wab+GQab* by Qab. 
 
 . I Oft 9 . 
 
 o 
 
 8. Divide 36 2 6 2 -f 60 2 & 6a& by 1206. 
 
 O. Divide lOrx cry-{-2crx by cr. 
 
 10. Divide 10uy-r-16d by 2d. 
 
 11. Divide 6?/ 18aceH-24a by Ca. 
 
 12. Divide mx amx-\-m by m. 
 
 (Art. 20.) We now come to the last and most important ope- 
 ration in division, the division of one compound quantity by an- 
 other compound quantity. 
 
 The dividend may be considered a product of the divisor into 
 the yet unknown factor, the quotient ; and the highest power of 
 any letter in the product, or the now called dividend, must ba 
 conceived to have been formed by the highest power of the same 
 letter in the divisor into the highest power of that letter in the 
 quotient. Therefore, both the divisor and the dividend must 
 be arranged according to the regular powers of some letter. 
 
 After this, the truth of the following rule will become obvious 
 by its great similarity to division in numbers. 
 
 RULE. Divide thejirst term of the dividend by the first term 
 of the divisor, and set the result in the quotient.'* 
 
 Multiply the whole divisor by the quotient thus found, and 
 subtract the product from the dividend. 
 
 The remainder will form a new dividend, with which pro- 
 ceed as before, till the first term of the divisor is no longer 
 contained in the first term of the remainder. 
 
 The divisor and remainder, if there be a remainder, are then 
 
 * Divide the first term of the dividend and of the remainders by the first 
 term of the divisor ; be nut troubled about other terms. 
 
30 ELEMENTS OF ALGEBRA. 
 
 to be written in the form of a fraction, as in division of num- 
 bers. 
 
 EXAMPLES. 
 
 Divide a*+2ab+& by a+b. 
 
 Here, a is the leading letter, and as it stands first in both the 
 dividend and divisor, no change of place is necessary. 
 
 OPERATION. 
 
 ab 
 
 ab+b* 
 ab+b 2 
 
 Agreeably to the rule, we consider that a will be contained in 
 a 2 , a times ; then the product of a into the divisor is a*-\-ab, and 
 the first term of the remainder is ab, in which a is contained 6 
 times. We then multiply the divisor by 6, and there being no 
 remainder, a-}-b is the whole quotient. 
 
 Divide a 3 +3 *#-{- Sao? +# 3 by x+a. 
 
 As the highest power of a stands in the first term of the divi- 
 dend, and the powers of a decrease in regular gradation from 
 term to term, therefore we must change the terms of the divisor 
 to make a stand first. 
 
 OPERATION. 
 
 a 3 -!- a z x 
 
 2a z x-\-2ax z 
 
DIVISION. 8] 
 
 a. a c) 3 4a 2 c-|-4ac 2 C 3 (fl 2 3ac4-c 
 
 a 3 a z c 
 
 3 . a 2 40-HK 6a 2 -f- 1 2a 8(02 
 
 2a 2 -f- 8a -8 
 
 4. Divide 6a: 4 96 by 6z 12. .#n$. a*-f 23*4-43+8. 
 
 5. Divide a 2 ft 2 by a 6. ./#ns. a+6. 
 
 6. Divide 25z 6 ^ 2x? 8x 2 by 5r 4;r*. 
 
 (Art. 21.) We may cast out equal factors from the dividend 
 and divisor, without changing the value of the quotient; for 
 amxy divided by am gives xy for a quotient ; cast out either of 
 the common factors a or m from both dividend and divisor, and 
 we shall still have xy for a quotient. This, in many instances, 
 will greatly facilitate the operation. Thus, in the 4th example, 
 the factor 6 may be cast out, as it is contained in all the terms ; 
 and in the 6th example the factor a? 2 may be cast out ; the quo- 
 tients will of course be the same. 
 
 T. Divide a 9 -+4ax+4x*+y* by a+2x. 
 
 8. Divide 6a 4 -j-9a 2 15a by So 55 3a. 
 
 (Observe Art. 21.) rfns. 2 8 +2+5. 
 
 9. Divide a,- 6 y 6 by x*+2x*y+2xy z +y*. 
 
 Ans. x*2x*y -\-2xy* y*. 
 
32 ELEMENTS OF ALGEBRA. 
 
 10. Divide * ( 2 -f l>)tf+b z by ax b. 
 
 tfns. x ax b 
 
 11. Divide 1 by 1 a. J2ns. 1-f a+a 2 -f a 8 , &c., &c 
 
 12. Divide aM-+?+l by +*. 
 
 N. B. We may multiply both dividend and divisor by the 
 same number as well as divide them. 
 
 13. Divide 1 5*/-H<ty 2 Wy*+5y 4 y 5 by 1 27/-f-7/ 2 . 
 
 Am. 1 3y4-3y 2 2/ 3 
 
 14. Divide 4 -f 4/; 4 by a z 2ab+2b 2 . 
 
 Am. a z -\-2ab-\-2b z . 
 
 15. Divide a? 6 a? 4 -|-a- 3 ^ 2 +2^~ 1 by a^+a? 1. 
 
 ^/w. a: 4 
 
 16. Divide a 5 or 5 by a #. 
 
 Am. 
 
 JT. Divide & 5 -f?/ 5 by 
 
 18. Divide 3 +5a 2 ?/-j-5?/ 2 -{-i/ 3 by a-fy. 
 
 Ans. 
 
 If more examples are desired for practice, the examples in 
 multiplication may be taken. The product or answer may be 
 taken for a dividend, and either one of the factors for a divisor ; 
 the other will be the quotient. 
 
 Also, the examples in division may be changed to examples in 
 multiplication ; and these changes may serve to impress on the 
 mind of the pupil the close connection between these two opera- 
 tions. 
 
 (Art. 22.) In the following examples the dividends and divi- 
 sors are given in the form of fractions, and the quotients are the 
 terms after the sign of equality. Let the pupil actually divide, 
 and observe the quotients attentively. 
 
 a; 2 -a 2 
 
 1. - - = 
 
 x a 
 
DIVISION, 33 
 
 2. ^Zl^ 
 x a 
 
 3. 
 
 4. 
 
 a? a 
 
 Hence we may conclude that in general x m m is divisible by 
 x a, m being any entire positive number. 
 
 That is, ~=x m - l -\-ax m -* J r - - - a rn ^x-\-a m ~ l 
 
 The quotient commencing with a power of x, one less than 
 m, and ending with a power of a, one less than m. 
 
 These divisions show, that the difference of two equal powers 
 of different quantities is always divisible by the difference of 
 their roots. 
 
 (Art. 23.) By trial, that is, actual division, we shall find that 
 
 .r 2 a z 
 
 -; =x a. 
 x-\-a 
 
 x-}-a 
 
 = x 5 ax*+ aV 
 
 x-\-a 
 
 &c. &c. &c. &c. 
 
 From which we learn that the difference of any two equal 
 powers of different quantities, is also divisible by the sum of their 
 roots when the exponent of the power is an even number. 
 
 (Art. 24.) By actual division we find that 
 
 --. 
 x-\-a 
 
 x-\-a 
 
 And in general, we may conclude that the sum of any two 
 equal powers of different quantities, is divisible by the sum of 
 their roots when the exponent of the power is an odd number. 
 
34 ELEMENTS OF ALGEBRA. 
 
 Li Art. 22, if we make a=l the formulas become 
 x 2 I 
 
 =aH-l. 
 
 , &c. 
 
 8 1 
 
 a? 1 
 If we make a?=l, what will the formulas become ? 
 
 Make the same substitutions in articles 23 and 24, and exa- 
 mine the results. 
 
 By inspecting articles 22, 23, and 24, we find that 
 
 =a? 3 a s , &c.. 
 
 for the product of the divisor and quotient must always produce the 
 dividend. These principles point out an expedient of condensing 
 a multitude of terms by multiplying them by the roots of the terms 
 involved. Thus, a^iaa^+aVzta^+a 4 ? can be condensed to 
 two terms by multiplying them by # a, the root of the first and 
 last term, with the minus sign where the signs are plus in" the 
 multiplicand, and with the plus sign where the signs are alter- 
 nately plus and minus. See examples in Art. 22 and 24. 
 
 ALGEBRAIC FRACTIONS. 
 
 (Art. 25.) We shall be very brief on the subject of algebraic 
 fractions, because the names and rules of operations are the same 
 as numeral fractions in common arithmetic ; and for illustration, 
 shall, in some cases, place them side by side. 
 
 CASE i. To reduce a mixed quantity to an improper frac- 
 tion, multiply the integer by the denominator of the fraction^ 
 and to the product add the numerator, or connect it with its 
 proper sign, -{- or ; then the denominator being set under 
 this sum, will give the improper Jraction required. 
 
ALGEBRAIC FRACTIONS. 35 
 
 EXAMPLES. 
 
 1. Reduce 2f and G-TT to improper fractions. 
 
 Jhu. y and *+* 
 
 These two operations, and the principle that governs them, 
 aie exactly alike. 
 
 2. Reduce 5J and +y to improper fractions. 
 
 An,. y .ad ?*+*" 
 
 3. Reduce 4 | and a -- to improper fractions. 
 
 x 
 
 , ax b 
 
 MS. y and -- 
 x 
 
 , T 4 1 3# 
 
 . Keduce 5 -- - and 20 -- to improper fractions. 
 
 5. Reduce 5-|- y to an improper fraction. 
 
 6. Reduce 12-J -- -r to an improper fraction. 
 
 7. Reduce 4-f-2a?H to an improper fraction. 
 
 c 
 
 8. Reduce 5a; - - to an improper fraction. 
 
 3 
 
 9. Reduce 3 9 -- T-TT to an improper fraction. 
 
 - 
 
 * 
 
 a+3 
 
 CASE 2. TVte converse of Case 1. Tb reduce improper 
 fractions to mixed quantities, divide the numerator by the de- 
 nominator, as far as possible, and set the remainder, (if any,} 
 
36 ELEMENTS OF ALGEBRA. 
 
 over the denominator for the fractional part ; the two joined 
 together with their proper sign, will be the mixed quantity 
 sought. 
 
 EXAMPLES. 
 
 1 . Reduce 4 7 and ^ to mixed quantities. 
 
 Jins. 5J and 0-f-y 
 
 2. Reduce */ and - to mixed quantities. 
 
 dns. 2f and 0-f-- 
 
 3. Reduce to a mixed quantity. 
 
 ab+y 
 
 50- 
 
 y 
 
 4. Reduce - - to a whole or mixed quantity. 
 
 Jlns. 
 
 5. Reduce to a whole number. 
 
 xy 
 
 Am. 2(s?+xy+y 2 ) by (Art. 22.) 
 
 6. Reduce - to a mixed quantity. 
 
 c 
 
 1002 4a-j-6 
 
 7. Reduce - - -- to a mixed quantity. 
 
 5a 
 
 8. Reduce - to a mixed quantity. 
 
 3 
 
 3# 2 I2ax-}-y 9x 
 
 9. Reduce - 2 - to a mixed quantity. 
 
 (Art. 26.) It is very desirable to obtain algebraic quantities in 
 their most condensed form. Therefore, it is often necessary to 
 reduce fractions to their lowest terms ; and this can be done as 
 in arithmetic, by dividing both numerator and denominator by 
 
ALGEBRAIC FRACTIONS. 37 
 
 their obvious common factors, or for their final reduction, by 
 their greatest common measure. If the terms have no common 
 measure, the fraction is already to its lowest terms. 
 
 The principle on which these reductions rest is that of divi- 
 sion, explained in (Art. 21). 
 
 CASE 3. To find the greatest common measure of the terms 
 of a fraction, divide the greater term by the less, and the last 
 divisor by the remainder, and so on till nothing remains ; then 
 the divisor last used will be the common measure required. 
 
 But note, that it is proper to arrange the quantities according 
 to the powers of some letter, as is shown in division. 
 
 N. B. During the operation we may cast out, or throw in a 
 factor to either one of the terms which is not a factor in the 
 other, as such a factor would make no part of the common 
 measure, and the value of quantities is not under consideration. 
 
 Thus, the fraction - has a4-b for its greatest common 
 a 2 b* 
 
 measure; and this quantity is not affected by casting out the 
 factor b from the numerator, and seeking the common measure 
 
 a+b 
 
 of the fraction . 
 a 2 b 2 
 
 (Art. 27.) To demonstrate the truth of the rule for finding the 
 greatest common measure, let us suppose D to represent a divi- 
 dend, and d a divisor, q the first quotient and r the first re- 
 mainder. 
 
 Tn short, let us represent successive divisions as follows : 
 
 d)D(q 
 dq 
 
 
 
 Now, in division, the dividend is always equal to the product 
 of the divisor and quotient, plus the remainder, if any. 
 
38 ELEMENTS OF ALGEBRA. 
 
 Therefore, r=r'q" 
 
 and d=rq'-}-r' 
 
 and D=dq -f-r. 
 
 As r^=r'q", the last divisor r' is a factor in r (there being no 
 remainder) ; that is, r' measures r. 
 
 Now as r' measures r, it measures any number of times r, 
 or r<7'-f-r', orrf; therefore r' measures d. 
 
 Again, as r' measures d and r, it measures any number of 
 times d -\-r ; that is, it measures dq-\-r or D. 
 
 Hence r', the last divisor, is a common measure to both D 
 
 and d, or of the terms of the fraction __ 
 
 d 
 
 We have now to show that r' is not only the common mea- 
 sure of D and d, but the greatest common measure. 
 
 In division, if we subtract the product of the divisor and quo- 
 tient from the dividend, we shall have the remainder. 
 
 That is, D dq=r, and drq'r'. 
 
 Now, every common measure of D and d is also a measure 
 of r, because D dq=r ; for the same reason every common 
 measure of r and d is also a measure of r'. But the greatest 
 measure of r' is itself. This final remainder is, therefore, the 
 greatest common measure of D and d. 
 
 EXAMPLES. 
 1. Find the greatest common measure of the two terms of 
 
 the fraction - - and with it, reduce the fraction to its lowest 
 
 terms. 
 
 CONSIDERATION AND OPERATION. 
 
 The denominator has a 3 as a factor to all its terms, which is 
 not a factor in the numerator ; hence this can form no part of 
 the common measure, or the common measure will still be there 
 if this factor is taken away. 
 
 We then seek the common measure of a 4 1 and a?-\-l. 
 
ALGEBRAIC FRACTIONS. 
 
 a 2 ! 
 
 Hence a 2 ~\-l is the common measure, which, used as a divisor 
 to both numerator and denominator, reduces the fraction to 
 a 2 ! 
 
 " a 3 ' 
 
 2. Find the greatest common measure, and reduce the frac- 
 tion 
 
 Divide this rem. by y 2 zy 2 ?/ 3 
 
 ay 2/ 2 
 *y af 
 
 3. Find the greatest common measure and reduce the frac- 
 uon 
 
 Common measure # ^y. 
 Fraction reduced 
 
 Ans. Greatest common measure 2 +a^y. Reduced frac- 
 
 .ion *& 
 5axy. 
 
40 ELEMENTS OF ALGEBRA. 
 
 Find the greatest common measure of 
 and a 2 c+2abc+b 2 c. 
 
 Reject the common factor c in one of the quantities, 
 
 a'-f 2a z b+ ab* 
 
 4. Find the greatest common divisor, and reduce the fraction 
 to its lowest terms. 
 
 Here we find that neither term is divisible by the other ; but 
 if these quantities have a common divisor, such divisor will still 
 exist if we multiply one of the terms by any number whatever, 
 to render division possible. 
 
 Therefore take 4a 3 2a 2 3a-\- I 
 Multiply it by 3 
 
 2__2a l)12a 3 6a 2 9a+ 3(4a 
 12a 3 8a 2 4a 
 
 2a 2 5a+ 3 
 Multiply by 3 
 
 9(2 
 4a 2 
 
 Divide by 11 lla-j-11 
 
 a 1 
 a 1 
 
 Jins Greatest common measure a 1. Reduced fraction 
 
 j*<zi:L_ 
 
 4^+201* 
 
ALGEBRAIC FRACTIONS. 41 
 
 5. Find the greatest common measure and reduce the frac- 
 
 to its lowest terms ' 
 
 Jlns. Common divisor a 2 x 2 . Reduced fraction . 
 
 a x 
 
 <* Find the greatest common measure, and reduce the frac- 
 tion to its lowest terms. 
 
 Jlns. Common divisor a? y 2 . Red. frac. 
 
 . Reduce 
 
 g. Reduce 
 
 
 , 
 a 2 x z 
 
 to its lowest terms. 
 tins. 
 
 a+x 
 
 to its lowest terms. ^ & 
 
 3a x 
 
 Ans. 
 
 9. Reduce 
 
 10. Reduce 
 
 - to its lowest terms. 
 So 3 
 
 Am. 
 
 bx 
 
 to its lowest terms. 
 
 
 11. What is the value of -- 
 
 Ans. 
 
 12. Find the greatest common divisor of 12a 4 24a 3 6+ 12a 2 6 2 , 
 and 8o 3 6 2 - 
 
 (A rt. 28.) We may often reduce a fraction by separating both 
 numerator and denominator into obvious factors, without the 
 formality of finding the greatest common divisor. The follow 
 ing are some examples of the kind : 
 4 
 
42 ELEMENTS OP ALGEBRA. 
 
 1. Reduce -- - to its lowest terms. 
 
 a(a 2 b 2 ) _a(ab}(a+b} 
 
 2. Reduce - to its lowest terms. Jlns. 
 
 - T IV AfcO lUWtOl 1C HHO. VJ.IIO. - - 
 
 ^2 J ,p | 
 
 3. Reduce ; to its lowest terms. Jlns. . 
 
 zy+y y 
 
 CX-}- CX 2 C-}-CX 
 
 4. Reduce to its lowest terms. Jins. . 
 
 acx-{-abx ac-\-ab 
 
 2# 3 .I6x 6 
 
 5. Reduce to its lowest terms. Jlns. I. 
 
 3^3 24x 9 
 
 (Art. 29.) To find the least common multiple of two or 
 more quantities. 
 
 The least common multiple of several quantities is the least 
 quantity in which each of them is contained without a remainder. 
 
 Thus, the least common multiple of the prime factors, a, b, c, 
 x, is obviously their product abcx. Now observe that the same 
 product is the least common multiple also, when either one of 
 these letters appears in more than one of the terms. Take a, 
 for example, and let it appear with b, c, or x f or with all of 
 them, as a, ab, c, ax, or a, 6, c, ax, the product abcx is still 
 divisible by each quantity. Therefore, when the same factor 
 appears in any number of the terms, it is only necessary that it 
 should appear once in the product; that is, once in the least 
 common multiple. If it should be used more than once, the 
 product so formed would not be the least common multiple. 
 
 From this examination, the following rule for finding the taast 
 common multiple will be obvious : 
 
 RULE, Write the given quantities, one after another, and 
 draw a line beneath them. Then divide by any prime factor 
 that will divide two or more of them without a remainder, 
 bringing down the quotients and the quantities not divisible, 
 to a line below. Divide this second line as the first, forming 
 
ALGEBRAIC FRACTIONS. 43 
 
 a third, fyc., until nothing but prime quantities are left. Then 
 multiply all the divisors and the remainders that are not divis- 
 ible, and their product will be the least common multiple. 
 N. B. This rule is also in common arithmetic. 
 
 EXAMPLES. I 
 
 1. Required the least common multiple of Sac, 4a 2 , 
 8c, and ex. 
 
 2a)Sac 4a 2 I2ab Sac ex 
 
 2c)4c 
 
 2a 
 
 Qb 
 
 4c 
 
 ex 
 
 2)2 
 
 a 
 
 3b 
 
 2 
 
 X 
 
 1 
 
 a 
 
 36 
 
 1 
 
 X 
 
 Therefore 2a X 2c X 2 X a X 36 X x=24a 2 cbx. 
 
 Here the divisor 2c will not divide 2a, but the coefficient of 
 c will divide the coefficient of a, and we let them divide, for it 
 is the same as first dividing by 2, and afterwards by c. From 
 the same consideration we permit 2c to divide ex, or let the let- 
 ter c in the divisor strike out c before x. 
 
 By the rule we should divide by 2 and by c separately ; but 
 this is a practical abbreviation of the rule. 
 
 2. Required the least common multiple of 27 a, 156, 9a6, and 
 3a 2 . Ans. 135a 2 6. 
 
 3. Find the least common multiple of (a 2 x 2 }, 4(a x), 
 and (a+x). tins. 4(a 2 #*). 
 
 Find the least common multiple of ax 2 , bx, acx, and 
 a 2 x 2 . Am. (a?a?)acbx?. 
 
 5. Find the least common multiple of a-f-6, a b, a 2 -{-ab-{-b 2 , 
 and a 2 ab+b 2 . Ans. a 6 b 6 . 
 
 The least common multiple is useful many times in reducing 
 fractions to their least common denominator. 
 
 CASE 4. To reduce fractions to a common denominator. 
 
 (Art. 30.) The rule for this operation, and the principle on 
 which it is founded, is just the same as in common arithmetic, 
 merely the multiplication of numerator and denominator by the 
 
44 ELEMENTS OF ALGEBKA. 
 
 same quantity. The object of reducing fractions to a common 
 denominator is to add them, or to take their difference, as diffe- 
 rent denominations cannot be put into one sum. 
 
 RULE. Multiply each numerator by all the denominators, 
 except its own, for a new numerator, and all the denominators 
 for a common denominator. 
 
 Or, find the, least common multiple of the given denomina- 
 tors for a common denominator; then multiply each denomi- 
 nator by such a quantity as will give the common denomina- 
 tor, and multiply each numerator by the same quantity by 
 which its denominator was multiplied. 
 
 EXAMPLES. 
 
 1. Reduce and to a common denominator. 
 x 2c 
 
 4ac 3bx 
 
 rfns. - and - . 
 
 2cx 2cx 
 
 2a 3a-f-26 
 
 8. Reduce and - to a common denominator. 
 b 2c 
 
 4ac , Bab+2 I? 
 Jlns. and --- 
 2bc 2bc 
 
 3. Reduce and , and 4d to a common denominator. 
 3x 2c 
 
 , 9bx , 24cdx 
 
 Jins. - and and - . 
 Gcx Gcx Gcx 
 
 4. Reduce 7-, - , ~ , to fractions having a common de- 
 c x-\-a 
 
 Jlns. acx-{-a 2 c (bx-\-b)(x-}-a) bey 
 
 bcx -\-abc b ex -\-abc ' bcx-\-abc' 
 
 (Art. 31.) CASE 5 Addition or finding the sum of fractions. 
 
 RULE. Reduce the fractions to a common denominator ; 
 and the sum of the numerators, written over the common deno- 
 minator, will be the sum of the fractions. 
 
ALGEBRAIC FRACTIONS. 45 
 
 EXAMPLES. 
 
 3x 2x X 
 
 1 Add , , and - together 
 57 * 
 
 I28.r 
 
 // * j o -. ___ _. _ _ .^ -_-,-._ > _ 
 
 105 105 
 
 , a . a-\-b 
 2. Add T and ~. Ans. 
 
 T ~. . - - 
 
 b c be 
 
 3. Add -, - and - . 
 
 f and . 
 
 a 6 a 4- 6 
 
 a-f-3 2a 5 , 
 
 5. Add 2a+- - and 4aH --- . .^/?5. 6H 077- 
 
 8x* %ax %abx 8cz* 
 
 6. Add a r- and b-{ Jins. a-\-o-\ . 
 
 b c uc 
 
 x 2 2;r 3 
 
 7. Add 5.r+ -- and 4x . 
 
 o ox 
 
 Ans. 
 
 3.r b , 6 a?. 
 
 8. What is the sum of 26-{ -, 7 - and - ? 
 
 5 6 x b 
 
 2 5bx 
 
 Mt ___ f) Q,j O 
 
 9 What is the sum of 5y-l-^ - and 4y -- C 1 ? 
 
 d o 
 
4C ELEMENTS OF ALGEBRA. 
 
 1.0. What is the sum of 50, , and ^~ 
 
 Ans. 
 
 (Art. 32.) CASE G. Subtraction or finding difference. 
 
 RULE. Reduce the fractions to a common denominator, and 
 subtract the numerator of that fraction which is to be sub- 
 tracted from the numerator of the other, placing the difference 
 over the common denominator. 
 
 EXAMPLES. 
 
 7x 2x 1 2lxAx-\-2 l7x-\-2 
 
 1. From -take- . An*. - - - =__. 
 
 take . Eq. fractions - 2 
 
 Difference or *ftns. 
 
 nt O/y rp 
 
 3. From 5 take ~. Diff. ^ 
 
 o 7 1 
 
 4. From ^ take . Ans. 
 
 7 y 63 
 
 2a b Ba 4b 
 
 9 From ; subtract -7 . 
 4c 3o 
 
 126c 
 
 . llfl 10 , 3a 5. 
 
 6. From 3a-\ subtract 2a-\ . 
 
 15 7 
 
 tins, a- 
 
 . From x-\ ~~* subtract ^. fins, x 
 
 x 2 xy 
 
 ab . 2b4a 5ad 5bd 4bc-\-Sac 
 
 8. From - ~ take . Jim. -- - 
 2c bd IQcd 
 
ALGEBRAIC FRACTIONS. 47 
 
 . x x a . cx-\-bx~ ab 
 
 9. From 3x-\--r take x . fins. 2x-{ -, 
 
 be be 
 
 a-\-b a b 
 
 10. Find the difference between - r and J-T-. 
 
 a b a+b 
 
 4ab 
 tins. 
 
 a 2 ft 2 
 
 11. prom take . fa. 4. 
 
 xy xy 
 
 CASE 7. Multiplication of fractions. 
 
 (Art. 33.) The multiplication of algebraic fractions is just the 
 fame in principle and in fact, as in numeral fractions, hence the 
 rule must be the same. 
 
 It is perfectly obvious, that f multiplied by 2 must be -|, and 
 multiplied by 3 must be f ; and the result would be equally ob- 
 vious with any other simple fraction ; hence, to multiply a frac- 
 tion by a whole number, we must multiply its numerator. 
 
 It is manifest that doubling a denominator without changing its 
 numerator halves a fraction, thus ; double the 2, and we have 
 ?, the half of the first fraction. 
 
 Also f , double the 5 gives T \, the half of |. In the same 
 manner, to divide a fraction by 3 we would multiply its denomi- 
 nator by 3, &c. In general, to divide a fraction by any num- 
 ber, we must multiply the denominator by that number. 
 
 Now let us take the literal fraction -r, and multiply it by c, the 
 
 ac 
 product must be -j- . 
 
 Again, let it be required to multiply - by -. Here the mul- 
 
 tiplication is the same as before, except the multiplier c is divided 
 by d ; therefore if we multiply by c we must divide by d. But 
 
 the product of - by c is ; this must be divided by d 1 and we 
 ac 
 
 shall have - ; for the true product of ,- by -=. 
 bd b J d 
 
48 ELEMENTS OF ALMEBliA. 
 
 From the preceding investigation we draw the following rule 
 to multiply fractions : 
 
 RULE. Multiply the numerators together for a new nume- 
 rator, and the denominators together for a new denominator. 
 
 N. B. When equal factors, whether numeral or literal, 
 appear in numerators and denominators, they may be canceled, 
 or left out, which will save subsequent reductions. 
 
 EXAMPLES. 
 
 1. Multiply by * and . Al . a ^L. 
 
 7 b J x c ex 
 
 In this example, b in the denominator of one fraction cancels 
 in the numerator of another. 
 
 iviuiiipiy 
 
 < 
 
 3. Multiply : 
 4U Multiply 
 5. Multiply* 
 
 fi TVTiiltiT-wlir 
 
 f\f\ U ' J f\ f 1 \ 
 
 ou o{a~\~x\ 
 
 **+% bv 2 
 
 18 
 
 Ans. -. 
 5x 
 
 (a-x)a 
 
 2 x 2 , 2a 
 
 Vtr 
 
 2i/ a-j-a? 
 ar 2 i/ 2 3? a 
 
 2 onrl 
 
 y 
 
 Ans. a. 
 
 , ana 
 a; 07+y a; y 
 
 a?+l , ^ 1 
 
 N. B. Reduce mixed quantities to improper fractions. 
 
 7. What is the continued product of , - : : and 
 
 a-\-b ax-\-x* 
 
 ax 2 (rt b) 
 
 ax x 
 
 4v* 15|/ 30 
 
 8. Multiply ^-^ by -* . AM. 
 
 * Separate into factors when separation is obvious. 
 
ALGEBRAIC FRACTIONS. 49 
 
 9. Multiply - by -775. #* 
 
 a-\-b ab b* b 
 
 10. Multiply -^P-, by jgg^gj* ^ n5 ' 3 (+ :F )- 
 11 Required the continued product of -= r-, -^-T j and 
 
 w ~~~O flr~~it 
 
 .5n*. (+*). 
 
 19. Multiply + by -. .*.. 
 
 _ . 
 14. Mult,ply - by 
 
 4 2 166 2 , 56 
 
 15. MulUply --- by 
 
 8a . +32afe+32ft . 
 
 CASE 8 Division of Fractions. 
 
 (Art. 34.) To acquire a clear understanding of division in 
 fractions, let us return to division in whole numbers. 
 
 The first principle to which we wish to call the attention of 
 the reader, is, that if we multiply or divide both dividend and 
 divisor of any sum in division, by any number whatever, we do 
 not affect or change the quotient. (Art. 21.) 
 
 Thus, 2)6(3 4)12(3 8)24(3 &c 
 
 The second principle to which we would call observation is, 
 that if we multiply any fraction by its denominator, we have the 
 numerator for a product. 
 
 Thus, 5 multiplied by 3 gives 1, the numerator, and | by 5 
 
 gives 2, and -r multiplied by b gives a, <fec. 
 5 
 
50 ELEMENTS OF ALGEBRA. 
 
 Now let it be required to divide j- by -. 
 
 The quotient will be the same if we multiply both dividend 
 and divisor by the same quantity. Let us multiply both terms 
 
 by ?, the denominator of the divisor, and we have -=- to be di 
 
 vided by the whole number c. But to divide a fraction by a 
 whole number we must multiply the denominator by that num- 
 
 ber. (Art. 33.) Hence j~ is the true quotient required. 
 
 We can mechanically arrive at the same result by inverting 
 the terms of the divisor, and then multiplying the upper terms 
 together for a numerator, and the lower terms for a denominator; 
 therefore to divide one fraction by another we have the following 
 
 RULE. Invert the terms of the divisor, and proceed as in 
 multiplication. 
 
 EXAMPLES. 
 
 a+b . c 
 
 1. Divide by p-r. rfns. 
 
 c a-\-b 
 
 2. Divide 5X by -. 
 
 a c 
 
 c 5x 5 ex 
 
 Operation: divisor inverted -rX = r- 
 
 b a ab 
 
 10c* 
 
 3. Divide - by -- ,. 
 
 a x a 2 & 
 
 I5ab (a-\-x}(a x) 3b(a+x) 
 
 Operation, - X^ - - -- '-. ^ns. ^-- -. 
 a x lOac 2c 
 
 2* 2 7 as 
 
 4. Divide - by --s. Ans. 
 
 x-\-a J x* -\-1ax-\- a* 
 
 Operation, 
 
 Divide into factors, in all such cases, and cancel. 
 
ALGEBRAIC FRACTIONS. 
 
 51 
 
 
 a 2 -\-ax-}-x 2 ' 
 70# 15 
 
 5 7 25 * 
 
 /?? ~~ 
 
 5 5 
 
 x 
 
 18x 21 
 Jlns. ^ r . 
 
 ' x+l J 3 ' 
 1O. Divide hv 
 
 3a 2 
 
 , l . Dmde ____ + _ by 
 
 Am. 
 
 12. Divide by 
 
 5 5 
 
 ,_.... na nx . ma mx 
 
 13. Divide TT - by i ~ 7 . 
 
 - 
 
 Ans. 
 m 
 
 14. Divide 12 by 
 
 15. Divide 
 
 , a 
 by -. 
 x J x 
 
 ,., T,. ., x b 3cx 
 
 lo. Divide 5 by r . 
 
 J 
 
 Jlns. 
 
 I2x 
 
 a*-fax-\-x* 
 
 Jlns. b+- 
 a 
 
 Divide a by the product of : into . 
 
 X+y xy 
 
 Jlns. 
 
 18. Divide ^-i-rr b y the product of ^-^ ; into ~. 
 ofx._i_A\ J 2 a 4-6 
 
52 ELEMENTS OF ALGEBRA. 
 
 SECTION II. 
 CHAPTER I. 
 
 Preparatory to the solution of problems, and to extended in- 
 vestigations of scientific truth, we commenced by explaining the 
 reason and the manner of adding, subtracting, multiplying, and 
 dividing algebraic quantities, both whole and fractional, that the 
 mind of the pupil need not be called away to the art of perform- 
 ing these operations, when all his attention may be required on 
 the nature and philosophy of the problem itself. 
 
 For this reason we did not commence with problems. 
 
 Analytical investigations are mostly carried on by means 
 
 OF EQUATIONS. 
 
 (Art. 35.) An equation is an algebraical expression, meaning 
 that certain quantities are equal to certain other quantities. Thus, 
 3-f-4=7 ; a-{-6=c ; #4-4=10, are equations, and express that 
 3 added to 4 is equal to 7, and in the second equation that a 
 added to b is equal to e, <fcc. The signs are only abbreviations 
 for words. 
 
 The quantities on each side of the sign of equality are called 
 members. Those on the left of the sign form the first member, 
 those on the right the second. 
 
 In the solution of problems every equation is supposed to con- 
 tain at least one unknown quantity, and the solution of an equa- 
 tion is the art of changing and operating on the terms by means 
 of addition, subtraction, multiplication, or division, or by all these 
 combined, so that the unknown term may stand alone as one 
 member of the equation, equal to known terms in the other 
 member, by which it then becomes known. 
 
 Equations are of the first, second, third, or fourth degree, 
 according as the unknown quantity which they contain is of the 
 first, second, third, or fourth power. 
 
 ax-\-b=3az is an equation of the first degree or simple 
 equation* 
 
EQUATIONS. 53 
 
 ax z -\-bx=3ab is an equation of the second degree or quadratic 
 equation. 
 
 ax s -\-bx 2 -\-cx=2a 4 b is an equation of the third degree. 
 ax*-{-bx 3J rcx 2 -\-dx=2ab 5 is an equation of the fourth degree. 
 
 We shall at present confine ourselves to simple equations. 
 
 (Art. 36.) The unknown quantity of an equation may be 
 united to known quantities, in four different ways : by addition, 
 by subtraction, by multiplication, and by division, and further 
 by various combinations of these four ways as shown by the 
 following equations, both numeral and literal : 
 
 NUMERAL. LITERAL. 
 
 1st. By addition, aH-6=10 x-\-a=b 
 
 2d. By subtraction, x 8 = 12 x c=d 
 
 3d. By multiplication, 20#=80 axe 
 
 4th. By division, 7~ 16 ~d = ~^~ a ' 
 
 5th. ar+6 8+4=10+2 3, x+a b+c=d+c, &c., 
 are equations in which the unknown is connected with known 
 quantities by both addition and subtraction. 
 
 y* or 
 
 2aH-r= 21 > ax-\~Cj are equations in which the unknown 
 
 o 
 
 is connected with known quantities by both multiplication and 
 division. 
 
 Equations often occur, in solving problems, in which all of 
 these operations are combined. 
 
 (Art. 37.) Let us now examine how the unknown quantity can 
 be separated from others, and be made to stand by itself. 
 Take the 1st equation, or other similar ones. 
 
 x -f6=10 x-\-a=b 
 
 Take equal quantities 6= 6 a=a from both 
 
 members, and #=10 6 x=b a the 
 
 remainders must be equal. (Ax. 2.) Now we find the term 
 sdded to x, whatever it may be, appears on the other side with 
 a contrary sign, and the unknown term x being equal to known 
 terms is now known. 
 
64 ELEMENTS OF ALGEBRA. 
 
 Take the equations 
 Add equals to both memb. 
 Sums are equal #=124- 8 x=d-{-c (Ax. 1.) 
 
 Here again the quantity united to x appears on the opposite 
 side with a contrary sign. 
 
 From this we may draw the following principle cr rule 3f 
 operation : 
 
 Any term may be transposed from one member of an equa- 
 tion to the other, by changing its sign. 
 
 Now 20#=80. axe. If we divide both members by the 
 coefficient of the unknown term, the quotients will be equal. 
 
 (Ax. 4.) Hence #=f J=4. x=-. 
 
 That is, the unknown quantity is disengaged from known 
 quantities, in this case, by division. 
 
 7* "7* 
 
 Again, take the equations -=16; ~=g-\-a. 
 
 Multiply both members by the divisor of the unknown term, 
 and we have o?=16X4. x=gd-\-ad. Equations which must 
 be true by (Ax. 3.), and here it will be observed that x is libe- 
 rated by multiplication. 
 
 From these observations we deduce this general principle : 
 
 That to separate the unknown quantity from additional 
 terms we must use subtraction; from subtracted terms we 
 must use addition ; from multiplied terms we must use divi- 
 sion ; from divisors we must use multiplication. 
 
 In all cases take the opposite operation. 
 
 EXAMPLES. 
 
 1. Given 3x 4=7# 16 to find the value of #. Ans. #=3. 
 
 2. Given 3#-}-9 1 5x=0 to find the value of x. 
 
 Am. x. 
 
 3. Given 4y-}-7y-\-2l 3-\-y to find y. Ans. y=5h- 
 
 4. Given Sax c=b 3ax to find the value of x. 
 
 Jlns. *=*-- C 
 8a 
 
EQUATIONS. 55 
 
 5. Given ax z -{-bx=9x z -{-cx to find the value of x in terms 
 
 of a, b, and c. jlns. x= -. 
 
 a 9 
 
 N. B. In this last example we observe that every term of the 
 equation contains at least one factor of x ; we therefore divide 
 every term by x, to suppress this factor. 
 
 (Art. 38.) In many problems, the unknown quantity is 
 often combined with known quantities, not merely in a simple 
 manner, but under various fractional and compound forms. 
 Hence, rules can only embody general principles, and skill and 
 tact must be acquired by close attention and practical application : 
 but from the foregoing principles we draw the following 
 
 GENERAL RULE. Connect and unite as much as possible all 
 the terms of a similar kind on both sides of the equation. Then, 
 to clear of fractions, multiply both sides by the denominators, 
 one after another, in succession. Or, multiply by their con- 
 tinued product, or by their least common multiple, (when such 
 a number is obvious,) and the equation will be free of fractions. 
 
 Then, transpose the unknown terms to the first member of 
 the equation, and the known terms to the other. Then unite 
 the similar terms, and divide by the coefficient of the unknown 
 term, and the equation is solved. 
 
 EXAMPLES. 
 
 1. Given x-K#-{-3 7=61, to find the value of x. 
 Uniting the known terms, after transposition, agreeably to the 
 rule of addition, we find x-}-zX=9. Multiply every term by 
 2, and we have 2x-\-x=lS. Therefore x=6. 
 
 2. Given 2x-{-$x-{-ix 3a=46-f 30, to find x. 
 
 N. B. We may clear of fractions, in the first place, before we 
 condense and unite terms, if more convenient, and among literal 
 quantities this is generally preferable. 
 
 In the present case let us multiply every term of the equation 
 by 12, the product of 3X4, and we shall have 
 
 24;r-f9a:-|-4tf 360=486+ 36a. 
 Transpose and unite, and 37#=48&-f-72a 
 
56 ELEMENTS OF ALGEBRA. 
 
 Divide by 37, Md *= 
 
 3. Given 5r-f-#-f-*#=39, to find the value of x. 
 
 Here are no scattering terms to collect, and clearing of frac- 
 tions is the first operation. 
 
 By an examination of the denominators, 12 is obviously their 
 least common multiple, therefore multiply by 12. Say 12 
 halves are 6 whole ones, 12 thirds are 4, 12 fourths are 3, &c. 
 
 Hence, 6x+4x+ 3#=39 X 12 
 
 Collect the terras, 13a?=39X 12 
 
 Divide by 13, and a?= 3X12=36, Am. 
 
 N. B. In other books we find the numerals actually multiplied 
 by 12. Here it is only indicated, which is all that is necessary. 
 For when we come to divide by the coefficient of x, we shall 
 find factors that will cancel, unless that coefficient is prime to all 
 the other numbers used, which, in practice, is very rarely the 
 case. 
 
 4. Given a?-hia?-H#=0, to find x. 
 
 This example ia essentially the same as the last. It is identi- 
 cal if we suppose a=39. 
 
 Solution, 
 
 Or, 13#=12a 
 
 12a 
 
 Divide and X= -T^ 
 
 L*i 
 
 Now if a be any multiple of 13, the problem is easy and 
 brief in numerals. 
 
 , 3x 11 5x 5 , 97 7# 
 5. Given 2H -- = - + - to find the value of x. 
 
 16 o Z 
 
 Here 16 is obviously the least common multiple of the deno- 
 minators, and the rule would require us to multiply by it, and 
 such an operation would be correct ; but in this case it is more 
 easy to multiply by the least denominator 2, and then condense 
 like terms. Thus, 
 
, EQUATIONS. 57 
 
 Multiply by 2, and we have 
 
 Recollect that we can multiply a fraction by dividing its deno- 
 minator. Also observe that we can mentally take away 42 from 
 both sides of the equation, and the remainders will be equal. 
 
 (Ax. 2.) 
 
 Then 7= 
 
 8 4 
 
 Multiply by 8, and 
 
 3 X .11 =10a> 10+440 56x ; 
 Transposing and uniting terms, we have 
 
 49#=441 ; 
 By division, x=9, 
 
 6. Given ar+2^ + ll=fa?-f 17, to find x. 
 
 If we commence by clearing of fractions, we should make 
 comparatively a long and tedious operation. Let us first reduce 
 it by striking out equals from both sides of the equation. We 
 can take 1 1 from both sides without any formality of transposing 
 or changing signs ; say drop equals from both sides, (Ax. 2.) and 
 reduce the fraction f#=|a?. 
 
 All this can be done as quick as thought, and we shall have 
 
 Multiply by 4, then 
 
 +10=3+24, or =a 
 5 o 
 
 Hence, 7#=70, or xlQ Am. 
 
 7. Given %x 5-KaM-8-Kx 10=100 6 7 to f.nd the 
 value of x. 
 
 Collecting and uniting the numeral quantities, we have 
 
 -^=; 
 Multiply every term by 60, and we have 
 
 Collecting terms, 47a?=94.60 
 
 Divide both sides by 47, and x= 2.60=120 Ans. 
 
58 ELEMENTS OF ALGEBRA. 
 
 (Art. 39.) When equations contain compound fractions and 
 simple ones, clear them of the simple fractions first, and unite, 
 as far as possible, all the simple terms. 
 
 EXAMPLES. 
 
 6#-f-7 , 7x 13 
 
 8. Given ~ ~-+ a , Q = ~- to find the value of a\ 
 y tx-po o 
 
 Multiply all the terms by the smallest denominator, 3. That 
 is, divide all the denominators by 3, and 
 
 Multiply by 3 again, and 6a?+7+ " 
 
 Drop 6#-f-7, and - - =5. 
 2>x~\~ 1 
 
 Clear of fractions, 21 a: 39=10a?-f 5. 
 
 Drop IQx and add 39, and we have 11#=44, or x=4. 
 
 9. Given -- - -- =- to find the value of x. 
 21 4x 11 3 
 
 Observe that ~ may be expressed in two parts, thus, 
 
 7x 16 7x x 
 
 -j-ry Observe also, that -=-. Hence these terms may 
 2121 2 1 o 
 
 be dropped, the remainders must be equal. Transpose the 
 
 16 x-\-S 
 
 minus term, then = -- . 
 21 4x 11 
 
 Clear of fractions, and 64^11X16=21^4-21X8. 
 
 Drop 21a; and observe that 11 X 16 is the same as 22 X 8. 
 
 Then 432- 22X8=21X8. Let =8, 
 
 Then 43z 22a=21. 
 
 Transpose 22a and 43#=43a. 
 
 Hence x=a. But a=8. Therefore x=8. 
 
 N. B. We operate thus, to call attention to the relation of 
 quantities, and to form a habit of quick comparison, which will, 
 in many instances, save much labor and introduce the pupil into 
 the true spirit of the science 
 
EQUATIONS. 59 
 
 1O. Given = r~\"7 to find tne value of or. 
 
 36 ox 4 4 
 
 9a? 
 
 By a slight examination we perceive that is equal to %x. 
 
 oD 
 
 Hence these terms may be left out, as they balance each other. 
 
 Therefore 
 
 Clear of fractions, and 25a? 20 =36# 108. 
 Transpose 25x 36#=20 108. 
 
 Unite and change signs, and 11#=88 or a?=8, Jim 
 
 . 36 . 5#-}-20 4x . 86 t 
 11. Gl ven _ + _+ _ is = T +_ to find *. 
 
 By taking equals from both sides, we have 
 
 5#t-20 
 
 - - rz%' By reduction a? =4. 
 Qx 16 
 
 12. Given - --- ^= 6 ^ -- r to find * 
 
 42 4 
 
 Multiply by 4, to clear of fractions, and 
 
 3tf 2#-f-2=24;r 20# 13. Reduced x=5. 
 
 (Art. 40.) When a minus sign stands before a compound 
 quantity, it indicates that the whole is to be subtracted ; but we 
 subtract by changing signs, (Art. 5). The minus sign before 
 
 y. __ I 
 
 - in the last example, does not indicate that the x is minus, 
 
 but that this term must be subtracted. When the term is multi- 
 plied by 4, the numerator becomes 2x 2, and subtracting it 
 we have 2#-|-2. 
 
 Having thus far explained, we give the following unwrought 
 equations, for practice : 
 
 O/v n\ 
 
 13. Given =-+24 to find the value of x. Jns. 19],. 
 
 14. Given zx-\-lx=lQ to find the value of x. rfns. 24. 
 
60 ELEMENTS OF ALGEBRA. 
 
 15. Given ^-+|.=20^i_ to find z. Ana 9 
 
 16. Given ^i+^?=16-? to find *. Am. 13 
 
 . Given 2 *-+,5= to find *. 
 
 3 5 
 
 , a ~. 2#+l a?+3 
 
 18. Given x --- _=__ to find x. tins. x=l3 
 
 19. Given 5#+5#+*tf+ya?=77 to find * ^s. #=60. 
 
 20. Given 5a?+|a>Ha?=130 to find a?. rfns. x=120 
 
 21. Given ^af+^+T^a^QO to find x. Jlns. x=l20 
 
 22. Given ^+^+^=82 to find y. ^ns. y=84 
 
 23. Given 5a?-FIa;+5a?=34 to find x. tins. 
 
 24. Given 11^4-+++=315 to find x. 
 
 25. Given 3/++++=146 to find y. Ans. y=56 
 
 26. Given _29=_-30 to find a?, ^n*. a?=2 
 
 ar+2 a? 2 
 
 There is a peculiar circumstance attending this 26th example, 
 and the 4th example of Art. 42, which will cause us to refer to 
 them in a subsequent part of this work. 
 
 N. B. In solving equations 19, 20, 21, 22, and 23, use no 
 larger numbers than those given, indicating and not performing 
 numeral multiplications. 
 
 (Art. 41.) Every proportion may be converted into an equa- 
 tion. Proportion is nothing more than an assumption that the 
 same relation or the same ratio exists between two quantities, 
 as exists between two other quantities. 
 
 That is, Ji is to B as C is to D. There is some relation be- 
 tween A and B. Let r express that relation, then /?=r*#. But 
 
EQUATIONS. 61 
 
 the relation between C and D is the same (by hypothesis) as 
 between Ji and B. Hence D=rC. 
 
 Then in place of A : B : : C:D 
 we have Ji \rA\\ C:rC 
 
 Multiply the extreme terms, and we have 
 
 Multiply the mean terms, and we have 
 
 Obviously the same product, whatever quantities may be re- 
 presented by either .#, or r, or C. 
 
 Hence, to convert a proportion into an equation, we have the 
 following 
 
 RULE. Place, the product of the extremes equal to the pro- 
 duct of the means. 
 
 (Art. 42.) The relation between two quantities is not changed 
 by multiplying or dividing both of them by the same quantity. 
 Thus, a : b :: 2a : 2b, or more generally, a : b :: na : nb, for the 
 product of the extremes is obviously equal to the product of the 
 means. 
 
 That is, a is to b as any number of times a is to the same 
 number of times b. 
 
 We shall take up proportion again, but Articles 41 and 42 are 
 sufficient for our present purpose. 
 
 EXAMPLES. 
 
 1. Given 3.r 1 : 2;r-f-l :: 3x : x to find x. 
 (By Art. 41.) 3x 2 x=6x*+3x. 
 Transpose and unite, and we have Q=3x*-{-4x. 
 Divide by x, and 3x4-4=0 or x f , Jlns. 
 
 3 3x 
 
 2. Given '- : : : 6 : 5x 4 to find x. 
 
 The first two terms have the same relation as 5 : Jar, or oa 
 2 : x. Hence 2 : x :: 6 : 5x 4. 
 
 Product of extremes and means, lOa: S=Gx or x=Z. 
 
 X. 
 
 Am. #=:3. 
 
02 ELEMENTS OF ALGEBRA. 
 
 y K 
 
 4. Given - : (#5) : : f : to find x. 
 
 fins. -=5, 
 
 5. Given a?-{-2 : a : : b : c to find the value of x. 
 
 dns. j?= 2. 
 
 c 
 
 6. Given 2a? 3 : x 1 :: 2x : #-f-l to find the value of x. 
 
 Jlns. x=3 
 
 7. Given z-J-6 : 38 x : : 9 : 2 to find x. fins. x=30. 
 
 8. Given #+4 : x 11 : : 100 : 40 to find x. Jlns. x=2l. 
 
 QUESTIONS PRODUCING SIMPLE EQUATIONS. 
 
 (Art. 43.) We now suppose the pupil can readily reduce a 
 simple equation containing but one unknown quantity, and he is, 
 therefore, prepared to solve the following questions. The only 
 difficulty he can experience is the want of tact to reason briefly 
 and powerfully with algebraic symbols ; but this tact can only 
 be acquired by practice and strict attention to the solution of 
 questions. We can only give the following general direction : 
 
 Represent the unknown quantify by some symbol or letter, 
 and really consider it as definite and known, and go over the 
 same operations as to verify the answer when known. 
 
 EXAMPLES. 
 
 1. What number is that whose third part added to its fourth 
 part makes 21 1 Jlns. 36. 
 
 The number may be represented by x. 
 
 Then $x-}-Zx=2l. Therefore x=3G. 
 
 2. Two men having found a bag of money, disputed about 
 the division of it. One said that the half, the third, and the 
 fourth parts of it made $130, and if the other could tell how 
 much money the bag contained, he might have it all. How 
 much money did the bag contain 1 Jlns. $120. 
 
 (See equation 20, Art. 40.) 
 
EQUATIONS. 63 
 
 3. A man has a lease for 20 years, one-third of the time pas. 
 is equal to one-half of the time to come. How much of the time 
 nas passed ? 
 
 Let x= the time past. 
 
 Then 20 x= the time to come. 
 
 x 203? 
 
 By the question --= . Therefore x=l2 Am. 
 o Z 
 
 4. What number is that, from which 6 being subtracted, and 
 the remainder multiplied by 11, the product will be 121? 
 
 Let 0:= the number. 
 
 Then (a: 6)11=121, or x 6=11 by division. 
 
 Hence x=l7. 
 
 5. It is required to find two numbers, whose difference is 6, 
 and if | of the less be added to of the greater, the sum will 
 be equal to ^ of the greater diminished by of the less. 
 
 Let x the less. Then #-f-6= the greater. 
 
 By the question %x-\ = x. 
 
 5 o 
 
 Drop |a? from both sides and add \x to both sides, and we 
 
 have ~- =2, or x=2, the less number. 
 
 
 We may clear of fractions in full, and then transpose and unite 
 terms, but the operation would be much longer. 
 
 6. After paying | and ^ of my money, I had $66 left ; how 
 much had I at first? JLns. $120. 
 
 7. After paying away 5 of my money, and | of what remained, 
 and losing ^ of what was left, I found that I had still $24. How 
 much had I at first ? rfns. 60. 
 
 8. What number is that from which if 5 be subtracted, of 
 the remainder will be 40 ? Jlns. 65, 
 
 9. A man sold a horse and a chaise for $200 ; 5 of the price 
 of the horse was equal to | of the price of the chaise. What 
 was the price of each? Jlns. Chaise $120. Horse $80. 
 
(M ELEMENTS OF ALGEBRA. 
 
 1O Divide 48 into two such parts, that if the less be divided 
 by 4, and the greater by 6, the sum of the quotients will be 9. 
 
 rfns. 12 and 36. 
 
 11. An estate is to be divided among 4 children, in the fol- 
 lowing manner : 
 
 The first is to have $200 more than | of the whole. 
 The second is to have $340 more than ^ of the whole. 
 The third is to have $300 more than i- of the whole. 
 And the fourth is to have $400 more than j of the whole. 
 What is the value of the estate ? Ans. $4800 
 
 12. Find two numbers in the proportion of 3 to 4, whose 
 sum shall be to the sum of their squares as 7 to 50. 
 
 Jlns. 6 and 8. 
 
 N. B. When proportional numbers are required, it is generally 
 most convenient to represent them by one unknown term, with 
 coefficients of the given relation. Thus, numbers in proportion 
 of 3 to 4, may be expressed by 3x and 4x, and the proportion 
 of a to b may be expressed by ax and bx. 
 
 13. The sum of $2000 was bequeathed to two persons, so 
 that the share of A should be to that of B as 7 to 9. What was 
 the share of each? Jins. #'s share $875, B's share $1125. 
 
 14. A certain sum of money was put at simple interest, and 
 in 8 months it amounted to $1488, and in 15 months it amounted 
 to $1530. What was the sum 1 Ans. $1440. 
 
 Let a?= the sum. The sum or principal subtracted from the 
 amount will give interest: therefore 1488 x represents the 
 interest for 8 months, and 1530 x is the interest for 15 months. 
 
 Now whatever be the rate per cent, double time will give 
 double interest, &c. Hence 8 : 15 :: 1488 x : 1530 x. 
 
 N. B. To acquire true delicacy in algebraical operations, it is 
 often expedient not to use large numerals, but let them be repre- 
 sented by letters. In the present example let =1488 Then 
 a-f-42=1530, and the proportion becomes 8: 15:: a z:a-\- 
 42 x. 
 
EQUATIONS. 05 
 
 Multiply extremes, &c., 8-f-8-42 Sx=l5aI5x. 
 
 Drop Sa and Sx. We then have 8'42=7a 7#. 
 
 Dividing by 7 and transposing xa 48=1440, Am. 
 
 15. A merchant allows $1000 per annum for the expenses 
 of his family, and annually increases that part of his capital 
 which is not so expended by a third of it ; at the end of three 
 years his original stock will double. What had he at first / 
 
 rfns. $14,800. 
 
 Let x= the original stock, and a=1000. 
 
 To increase any quantity by its part is equivalent to multi- 
 
 \ 3C \Ct 
 
 plying it by . Hence - is his 2d year's stock. 
 
 16. A man has a lease for 99 years, and being asked how 
 much of it had already expired, answered that f of the time past 
 was equal to of the time to come. Required the time past and 
 the time to come. 
 
 Assume #=99. Am. Time past, 54 years. 
 
 17. In the composition of a quantity of gunpowder 
 The nitre was lOlbs. more than f of the whole, 
 
 
 ~r . 
 The sulphur 4 Ibs. less than \ of the whole, 
 
 The charcoal 2 Ibs. less than \ of the nitre. 
 
 What was the amount of gunpowder? Jlns. 69 Ibs. 
 
 18. Divide $183 between two men, so that y of what the first 
 receives shall be equal to T \ of what the second receives. What 
 will be the share of each ? Ans. 1st, $63 ; 2d, $120. 
 
 19. Divide the number 68 into two such parts that the differ- 
 ence between the greater and 84 shall be equal to 3 times the 
 difference between the less and 40. Ans. Greater, 42 ; Less, 26. 
 
 20. Four places are situated in the order of the letters A, B, 
 C, D The distance from A to D is 34 miles. The distance 
 from A to B is to the distance from C to D as 2 to 3. And 4 
 of the distance from A to J?, added to half the distance from C 
 to D, is three times the distance from B to C. What are the 
 respective distances ? 
 
 Ans. From A to 7?=12 ; from B to C=4 ; frorp C to//=18, 
 6 
 
66 ELEMENTS OF ALGEBRA. 
 
 21. A man driving a flock of sheep to market, was met by a 
 party of soldiers, who plundered him of 5 of his flock and 
 more. Afterwards he was met by another company, who took 
 5 what he then had and 10 more: after that he had but 2 left. 
 How many had he at first? fins. 45. 
 
 22. A laborer engaged to serve for 60 days on these comli 
 tions : That for every day he worked he should have 75 cents 
 and his board, and for every day he was idle he should forfeit 25 
 cents for damage and board. At the end of the time a settlement 
 was made and he received $25. How many days did he work, 
 and how many days was he idle ? 
 
 The common way of solving such questions is to let x= the 
 days he worked ; then 60 x represents the days he was idle. 
 Then sum up the account and put it equal to $25. 
 
 Another method is to consider that if he worked the whole 60 
 days, at 75 cents per day, he must receive $45. But for every 
 day he was idle, he not only lost his w^es, 75 cents, but 25 
 cents in addition. That is, he lost $1 every day he was idle. 
 
 Now let x= the days he was idle. Then x= the dollars 
 he lost. And 45 #=25 or #=20 the days he was idle. 
 
 23. A boy engaged to carry 100 glass vessels to a certain 
 place, and to receive 3 cents for every one he delivered, and to 
 forfeit 9 cents for every one he broke. On settlement, he re- 
 ceived 2 dollars and 40 cents. How many did he break ? 
 
 rfns. 5. 
 
 24. A person engaged to work a days on these conditions : 
 For each day he worked he was to receive b cents, for each day 
 he was idle he was to forfeit c cents. At the end of a days he 
 received d cents. How many days was he idle ? 
 
 ab d . 
 
 Jlns. -T-, days, 
 o-f-c 
 
 25. It is required to divide the number 204 into two such 
 parts, that f of the less being taken from the greater, the remain- 
 der will be equal to f of the greater subtracted from 4 times the 
 less. Arts. The numbers are 154 and 50.* 
 
EQUATIONS. 67 
 
 (Art. 44.) We introduce this, and a few following problems, 
 to teach one important expedient, not to say principle, which is, 
 not always to commence a problem by putting the unknown 
 quantity equal to a single letter. We may take 2;r, 3;r, or nx 
 to represent the unknown quantity, as well as #, and we may 
 resort to this expedient when fractional parts of the quantity are 
 called in question, and take such a number of ar's as may be 
 divided without fractions. 
 
 In the present example we do not put x= to the less part, as 
 we must have f of the less part. It will be more convenient to 
 put 5x= the less part. Then f of it will be 2x. Put a=204. 
 
 26. A man bought a horse and chaise for 341 () dollars. 
 Now if J of the price of the horse be subtracted from twice the 
 price of the chaise, the remainder will be the same as if | of the 
 price of the chaise be subtracted from 3 times the price of the 
 horse. Required the price of each. 
 
 Ans. Horse $152. Chaise $189. 
 
 N. B. Let Sx= the price of the horse. 
 Or let 7x= the price of the chaise. 
 Solve this question by both of these notations. 
 
 27. From two casks of equal size are drawn quantities, which 
 are in the proportion of 6 to 7; and it appears that if 16 gallons less 
 had been drawn from that which now contains the less, only one 
 half as much would have been drawn from it as from the other. 
 How many gallons were drawn from each 1 Jlns. 24 and 28. 
 
 N. B. Let Qx and 7x equal the quantities drawn out. 
 
 28. Divide $315 among four persons, ./?, B, C, and D, giving 
 B as much and more than Jl ; C 5 more than Jl and B toge- 
 ther ; and D | more than A, B and C. What is the share of 
 eadi ? Ans. ./? $24. B $36. C $80, and D $175. 
 
 If we take x to represent #'s share, we shall have a very 
 complex and troublesome problem.* But it will be more simple 
 by making 6x=J2's share. 
 
 * Taking x for yTs share, and reducing their sum, gives Equation 24, 
 Art. 40. 
 
68 ELEMENTS OF ALGEBRA. 
 
 Thus, let 6x=tfs share. 
 
 Then 9xJB % s share. 
 
 And l5z-\-5x=C's share. 
 
 Also 35#-j =Z)'s share. 
 
 Sum 70*4^=315 
 
 280oH-35#=315X4 
 315^=315X4 
 
 x=4 Hence 6a?=24, *#'s sh. 
 
 549. A gamester at play staked ^ of his money, which he lost, 
 but afterwards won 4 shillings ; he then lost ? of what he had, 
 and afterwards won 3 shillings ; after this he lost ^ of what he 
 had, and finding that he had but 20 shillings remaining, he left 
 off playing. How much had he at first ? rfns. 30 shillings. 
 
 30. A gentleman spends of his yearly income for the sup- 
 port of his family, and J of the remainder for improving his 
 house and grounds, and lays by $70 a year. What is his in- 
 come? Jlns. 9X70 dollars, or more generally, 9 times the 
 sum he saves. 
 
 31. Divide the number 60 (a) into two such parts that their 
 product may be equal to three times the square of the less num- 
 ber ? Ans. 15 and 45, or |a= the less part. 
 
 32. After paying away ? and \ of my money, I had 34 (a) 
 dollars left. What had I at first ? 
 
 rfns. 56 dollars. General answer y^X28. 
 
 33. My horse and saddle are together worth 90 (a) dollars, 
 and my horse is worth 8 times my saddle. What is the value 
 of each ? rfns. Saddle $10. Horse $80. 
 
 34. My horse and saddle are together worth a dollars, and 
 my horse is worth n times my saddle. What is the value of 
 
 each ? Jlns. Saddle r Horse 
 
EQUATIONS. 69 
 
 35. The rent of an estate is 8 per cent greater this year than 
 last. This year it is 1890 dollars. What was it last year? 
 
 Ans. $1750. 
 
 36 The rent of an estate is n per cent, greater this year than 
 last. This year it is a dollars. What was it last year 1 
 
 100 
 
 dollare ' 
 
 37. A and B have the same income. A contracts an annual 
 debt amounting to 1 of it ; B lives upon | of it ; at the end of 
 two years B lends to A enough to pay off his debts, and has 32 
 (a) dollars to spare. What is the income of each ? 
 
 Ans. $280 or (35a). 
 
 38. What number is that of which |, | and f added together 
 
 make 73 (a) ? . a 84a 
 
 v ' Am. 84. General Ans. . 
 
 73 
 
 39. A person after spending 100 dollars more than I of his 
 income, had remaining 35 dollars more than 5 of it. Required 
 his income. Ans. $450. 
 
 40. A person after spending (a) dollars more than \ of his 
 income, had remaining (6) dollars more than of it. Required 
 his income. 
 
 41. There are two numbers in proportion of 2 to 3, and if 4 
 be added to each of them, the sums will be in proportion of 5 
 to 7? Ans. 16 and 24. 
 
 42. It is required to find a number such, that if it be increased 
 by 7, the square root of the sum shall be equal to the square 
 root of the number itself, and 1 more. Ans. 9. 
 
 43. A sets out from a certain place, and travels at the rate of 
 7 miles in 5 hours ; and 8 hours afterward B sets out from the 
 same place in pursuit, at the rate of 5 miles in 3 hours. How 
 long and how far must B travel before he overtakes A ? 
 
 Ans. 42 hours, and 70 raijes 
 
70 ELEMENTS OF AUJEURA. 
 
 SIMPLE EQUATIONS. 
 CHAPTER II. 
 
 (Art. 45.) We have given a sufficient number of examplos t 
 and introduced the reader sufficiently far into the science pre- 
 vious to giving instructions for the solution of questions contain- 
 ing two or more unknown quantities. 
 
 There are many simple problems which one may meet with 
 in algebra which cannot be solved by the use of a single un- 
 known quantity, and there are also some which may be sohid 
 by a single unknown letter, that may become much more simple 
 by using two or more unknown quantities. 
 
 When two unknown quantities are used, two independent 
 equations must exist, in which the value of the unknown letters 
 must be the same in each. When three unknown quantities are 
 used, there must exist three independent equations, in which the 
 value of any one of the unknown letters is the same in each. 
 
 In short, there must be as many independent equations as 
 unknown quantities used in the question. 
 
 For more definite illustration let us suppose the following 
 question : 
 
 ji merchant sends me a bill of 16 dollars for 3 pair of shoes 
 and 2 pair of boots ; afterwards he sends another bill of 23 
 dollars for 4 pair of shoes and 3 pair of boots, charging at the 
 same rate. Wliat was his price for a pair of shoes, and what 
 for a pair of boots ? 
 
 This can be resolved by one unknown quantity, but it is far 
 more simple to use two. , 
 
 Let x= the price of a pair of shoes, 
 
 And y= the price of a pair of boots. 
 
 Then by the question 3a?-|-2y=16 
 
 And 4*-}-3r/=23. 
 
 These two equations are independent ; that is, one cannot be 
 converted into the other by multiplication or division, notwith- 
 standing the value of a: and of y is the same in both equations. 
 
 Having intimated that this problem can be resolved with one 
 
EQUATIONS. 7! 
 
 unknown quantity, we will explain in what manner, before we 
 proceed to a general solution of equations containing two un- 
 known quantities. 
 
 Let #= the price of a pair of shoes. 
 Then 3x= the price of three pair of shoes. 
 And 16 3x the price of two pair of boots. 
 
 I / O/y 
 
 Consequently - = the price of one pair of boots. 
 Now 4 pair of shoes which cost 4#, and 3 pair of boots which 
 cost - being added together, must equal 23 dollars. 
 
 That is, 4x4-24^=23. 
 
 Or, 1 #=0. Therefore a? 2 dollars, the price 
 
 of a pair of shoes. Substitute the value of x in the expression 
 
 1 A Q'v* 
 
 - and we find 5 dollars for the price of a pair of boots. 
 2 
 
 Now let us resume the equations, 
 
 =23 (B) 
 
 FIRST METHOD OF ELIMINATION. 
 
 (Art. 46.) Transpose the terms containing y to the right hand 
 sides of the equations, and divide by the coefficients of x, and 
 
 From equation (rf) we have x= - - (C) 
 
 o 
 
 And from (B) we have x=- ~^-^- (B) 
 
 4 
 
 Put the two expressions for x equal to each other. (Ax. 7.) 
 And == - . 
 
 An equation which readily gives y=5, which, taken as the 
 value of y, in either equation (C) or (Z>) will give #=2. 
 
 This method of elimination, just explained, is called th > 
 method by comparison. 
 
72 ELEMENTS OF ALGEBRA. 
 
 SECOND METHOD OF ELIMINATION. 
 (Art. 47.) To explain another method of solution, let us again 
 resume the equations : 
 
 3x-\-2y=lG (A) 
 4#-f3i/=23 (if) 
 
 The value of x from equation (#) is x=i(lQ 2y). 
 
 Substitute this value for x in equation (2?), and we have 
 4X (16 2y)+3y^=23, an equation containing only y. 
 
 Reducing it, we find y=5 the same as before. 
 
 This method of elimination is called the method by substitu- 
 tion, and consists in finding the value of one unknown quantity 
 from one equation to put that value in the other which will cause 
 one unknown quantity to disappear. 
 
 THIRD METHOD OF ELIMINATION. 
 
 (Art. 48.) Resume again Sx-\-2y=l6 (.#) 
 4x+3y=23 () 
 
 When the coefficients of either x or y are the same in both 
 equations, and the signs alike, that term will disappear by sub- 
 traction. 
 
 When the signs are unlike, and the coefficients equal, the term 
 will disappear by addition. 
 
 To make the coefficients of x equal, multiply each equation 
 by the coefficient of x in the other. 
 
 To make the coefficients of y equal, multiply each equation 
 ly the coefficient of y in the other. 
 
 Multiply equation (A) by 4 and l2x-\-8y=Q4 
 Multiply equation (E) by 3 and 12#-|-9i/=69 
 
 Difference y = 5 as before. 
 
 To continue this investigation, let us take the equations 
 2x+3y=23 (A) 
 5x 2y=iO (JB) 
 
 Multiply equation (A) by 2, and equation (B) by 3, and we 
 have 4x-{-6y=46 
 
EQUATIONS, 73 
 
 Equations in which the coefficients of y are equal, and the 
 signs unlike. In this case add, and the y's will destroy each 
 other, giving 19#-=76 
 
 Or a?=4. 
 
 This method of elimination is called the method by addition 
 and subtraction. 
 
 FOURTH METHOD OF ELIMINATION. 
 
 (Art. 48.) Take the equations 2x-\-3y=23. (.#) 
 And 5x 2t/=10. () 
 
 Multiply one of the equations, for example (,/#), by some inde- 
 terminate quantity, say m. 
 
 Then 2mx-\-3my=23m 
 
 Subtract () 5x 2y=lO 
 
 Remainder, (C) (2m 5)a?+ (3m+2)y=23m 10 
 
 As m is an indeterminate quantity, we can assume it of any 
 value to suit our pleasure, and whatever the assumption may be, 
 the equation is still true. 
 
 Let us assume it of such a value as shall make the coefficient 
 of y, (3m-l-2)=0. 
 
 The whole term will then be times ?/, which is 0, and equa- 
 tion (C) becomes 
 
 (2m -5)#=23m 10 
 
 23m 10 , nx 
 
 f *=-2^f W 
 
 But 3w-f2=0. Therefore i= |. 
 Which substitute for m in equation (/)), and we have 
 
 __23 X f 10 _ 23 X 2^30_ 76^ i 
 x ~ 2Xf -5 = ~ 2X2 15~ -19~~ 
 
 This is a French method, introduced by Bezout, but it is too 
 indirect and metaphysical to be much practised, or in fact much 
 known. 
 
 Of the other three methods, sometimes one is preferable and 
 sometimes another, according to the relation of the coefficients 
 and the positions in which they stand. 
 7 
 
74 ELEMENTS OF ALGEBRA. 
 
 No one should be prejudiced against either method, and in 
 practice we use either one, or modifications of them, as the case 
 may require. The forms may be disregarded when the princi- 
 ples are kept in view. 
 
 (Art. 49.) To present these different forms in the most general 
 manner, let us take the following general equations, as all par- 
 ticular equations can be reduced to these forms. 
 
 ax-\-by =c (J3) 
 
 Observe that a and a', may represent very different quantities, 
 BO b and b' may be different, also c and c' may be different. In 
 special problems, however, a may be equal to a', or be some 
 multiple of it ; and the same remark may apply to the other 
 letters. In such cases the solution of the equations is much 
 easier than by the definite forms. Hence, in solving definite 
 problems great attention should be paid to the relative values of 
 the coefficients. 
 
 First method. 
 
 Transpose the terms containing y and divide by the coeffi- 
 cients of a?, and 
 
 also x= c ^y (C) 
 a' 
 
 Therefore -=- (Axiom 7.) 
 
 a a 
 
 Clearing of fractions, give a'c a'by=ac f ab'y. 
 Transpose, and (ab' a'b)y=ac' a'c. 
 
 T j- ac ' a ' c 
 
 By division yr, 77- 
 
 y ab' ab 
 
 When y is determined, its value put in either equation marked 
 (C) will give x. 
 
 Second method. 
 From equation (.#) ' ' 
 
EQUATIONS. 75 
 
 Which value of x substitute in equation (B) and 
 a'c a'by 
 
 Clearing of fractions and transposing a'c, we have 
 
 ab 'y a'by=ac' a'c 
 ~ ac' a'c 
 
 Ot y=^=b 
 
 The same value of y as before found. 
 
 Third method. 
 
 Multiply equation (rf) by ', and equation (B) by a. 
 And a'ax-\-a'by=a'c. 
 
 Also a'ax-}-ab'y=ac' 
 
 Difference (ab' ~a'b}y=ac' a'c 
 
 _ ac' a'c . 
 
 Or y=r, -- TL same va ^ ue as by the 
 
 ab' ab 
 
 Dther methods. 
 
 Fourth method. 
 
 Multiply equation (JT) by an indefinite number m, 
 And amx-}-bmy=mc 
 
 Subtract (B] a'x-\- b'y=c' 
 
 And (am a']x-}-(bm b']y=mc c'. 
 
 Now the value of m may be so assumed as to render the 
 coefficient of #=0, or am a'=0. 
 
 Then (bm b')y=mcc' 
 
 But am a'=0, or m= . 
 
 a 
 
76 ELEMENTS OF ALGEBRA. 
 
 Put this resultant value of m, in equation ((7), and 
 
 cX 
 
 ca' ac' 
 
 bX--b> 
 a 
 
 by multiplying both numerator and denominator by a. 
 
 (Art. 50.) The principles just explained for elimination be- 
 tween two quantities may be extended to any number, where the 
 number of independent equations given are equal to the number 
 of unknown quantities. For instance, suppose we have the 
 three independent equations : 
 
 ax+by+cz=d (A) 
 
 a'x+b'y+c'z=d' (B) 
 a"x+b"y-}-c"z=d" (C) 
 
 We can eliminate either a?, or y, or z, (whichever may be 
 most convenient in any definite problem) between equations (A] 
 and (J5,) and we shall have a new equation containing only two 
 unknown quantities. We can then eliminate the same letter be- 
 tween equations (B} and ((7,) or (A] and (C*,) and have another 
 equation containing the same two unknown quantities. 
 
 Then we shall have two independent equations, containing two 
 unknown quantities, which can be resolved by either of the four 
 methods already explained. 
 
 (Art. 51.) Another theoretical method, from the French, we 
 present to the reader, more for curiosity than for any thing else. 
 
 Multiply the first equation (./?,) by an indefinite assumed num- 
 ber m. Multiply the second equation (B] by another indefinite 
 number n, and add their products together. Their sum will be 
 
 (am-}-a'n)x-\-(bm-\-b'n)y-i-(cm-{-c'n}z=dm-i-nd' 
 Subtract eq. (C) a"x +b"y c"z=d". 
 
 And (am-\-a'n a"}x-\-(bm-\-b'n &")7/-j-(cra4-c'n c"}z 
 
 =dm-\-nd' d" 
 
EQUATIONS. 77 
 
 As m and n are independent and arbitrary numbers, they can 
 be so assumed that 
 
 am-\-a'n "=0 and bm+b'n &"=0. 
 Then am-\-a'n=a" (1) and bm-{-b'n=b" (2) 
 
 And z= - : -- - (3) 
 
 cm+c'n c" 
 
 From equations (1) and (2) we can find the values of m and 
 , which values may be substituted in equation (3,) and then z 
 will be fully determined. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Given \ , > to find the values of x and y. 
 
 ? 12#+7=1005 
 
 We can resolve this problem by either one of the four methods 
 just explained. But we would not restrict the pupil to the very 
 letter of the rule, for that in many cases might lead to operations 
 unnecessarily lengthy. 
 
 If we take the third method of elimination, we should multi- 
 ply the first equation by 12, the second by 8 ; but as the coeffi- 
 cients of x contain the common factor 4, we can multiply by 3 
 and 2, in place of 12 and 8. That is, multiply by the fourth 
 part of 12 and 8. 
 
 In practice even this form need not be observed, we may de- 
 cide on our multipliers by inspection only. 
 
 Three times the 1st gives 24#-f-15?/=204 
 
 Twice the 2d gives 24a?-J-14i/=200 
 
 Difference gives y4 
 
 Substituting this value of y in first equation, and 
 8#-f-20=68 or x=Q. 
 
 In solving this, we have used modifications of the 3d and 2d 
 formal methods. 
 
78 ELEMENTS OF ALGEBRA. 
 
 For exercise, let us use the 4th method. 
 
 8 mx -|- 5my = 68m 
 Take l2x-\- 7y = 100 
 
 7)y=68m -100 
 Assume 8m 12=0. 
 68m 100 
 
 Then =-- But m== - 
 
 ,, P 68X1100 204200 
 
 Therefore =-__ == 
 
 2. Given \ ^J= J 9 J to find * and y. 
 
 If we multiply the first of these equations by 3, the coefficients 
 of y will be equal, and the equations become 
 
 15a?-r-6y=57, 
 And 7a? 6?/=9. 
 
 To eliminate y, we add these equations (the signs of the terms 
 containing y being unlike), and there results 
 
 This value of x put in the 1st equation gives 
 
 And 2/= 2 - 
 
 3. Given ?-t^-J-6?/=21 and 01^4-5^=23 to find x and y 
 
 Clear of fractions and reduce. We then have 
 
 #-{-24^=76 
 And 15ff-f y =63. 
 
 In this case there are no abbreviations of the rules, as the 
 coefficients of the unknown terms are prime to each other. 
 Continuing the operation, we find a?=4, y=3. 
 
EQUATIONS. 79 
 
 2^? 3t/ 
 4. Given #-{-?/= 17 and ~~T to & n & x an( ^ y 
 
 Owing to the peculiarity of form in the 2d equation, it is most 
 expedient to resolve this by the 2d method. 
 
 From the 2d, x=^. Then ^-f y=17. 
 
 O o 
 
 Clearing of fractions, 9*/+8y = 17 X 8. 
 
 Or, 17y=17X8, or y=8. 
 
 Hence, a?=9. 
 
 C ^a:+8i/=1947 
 5. Given < J isi S to values of a? and y. 
 
 Here we observe that both x and y are divided by 8, x in one 
 equation, and y in the other ; also, x and y are both multiplied 
 by 8. 
 
 (Art. 51.) All such circumstances enable us to resort to many 
 pleasant expedients which go far to teach the true spirit of al- 
 gebra. 
 
 Add these two equations, and ^-f-8(#-}-y)=325. 
 
 8 
 
 Assume x-\-ys. 
 
 Or let s represent the sum of x-\-y, then js-f-8s=325. 
 Clear of fractions, and s-{- 64s =325X8. 
 Unite and divide by 65 and 5=5X8. 
 
 Or x-{-y=5a. (A.) By returning to the value of s, and put- 
 ting a=8. 
 
 Multiply the 1st equation by 8, and 
 
 Subtract (rf) x + i/=5a 
 
 Rem. 
 
 Divide by 63 and y3a=24. Whence #=2a=16. 
 Let the pupil take any one of the formal rules for the solution 
 of the preceding equations, and mark the difference. 
 
 6. Given $x-{-3y=2} and ^y-J-8.r=29 to find x and y. 
 
 Jins. o:=9. =6 
 
80 ELEMENTS OF ALGEBRA. 
 
 7. Given 4#-f-t/=34 and 4y-\-x=l6, to find ;raiid^. 
 
 Ans. x=S. y=2 
 
 8. Given 5#-{-5?/=14 and 5a;+5y=ll, to find x and y. 
 
 ?. #=24. v=6 
 
 9. Giver. #-Hi/=8 and zX-}-y=7 to find a 1 and y. 
 
 Ans. z=Q. 2/=4. 
 
 10. Given \x+7y=99 and iy-}-7,r=51 to find x and y. 
 
 flns. x=7. =14 
 
 11. Given 
 
 4 
 
 fins. x=6Q. i/=40 
 
 12. Given -+-=6 and | =10, to find x and y. 
 
 x y x y 
 
 Multiply the first equation by c, the second by a, and we shall 
 have 
 
 etc , be 
 ] --- =6c 
 x y 
 
 ac . ad 
 
 -- --- =10tf. 
 
 By subtraction (be ad)-Qc 10a 
 
 f be ad 
 
 Therefore =v. 
 
 6c 10a y 
 
 13. Given =28 (Jft and + = (B) to find 
 x y x y 3 ^ ' 
 
 tlie values of x and y. 
 
 21 21 
 
 Divide equation (#) by 7, and =4. 
 
 x y 
 
 Divide this result by 21, and =-- (C\ 
 
 x y 21 ^ J 
 
EQUATIONS. 81 
 
 Multiply (C) by 17, gives ^ H =5? (/>) 
 
 M>Q 210 
 
 Subtract () from () and we have = -. 
 
 1 3 1 
 
 Divide by 73, and -= =- or y=7. 
 
 Putting this value in equation (C) and reducing we find x=3. 
 
 14. Given -+-=- 1 and -+-=-+- to find the values 
 x^y y x^y 0^2 
 
 of * and y Jns. *=4 and y=2. 
 
 ?. #=300. 7/=350. 
 
 $3? 24v 2 -H 
 . Given 3x+Gy+l= ^ ^ , 
 
 to find a: and y. 
 
 151 1 6.r , 9^/110 
 And 3o?= -\ H 
 
 4y 1 3y 4 
 
 .tfns. a?=9. y=2. 
 
 In the first equation actually divide the numerator by the de- 
 nominator, then drop equals from both sides. 
 
 17. Given ^ ' ~j!^"~L-oo C t0 find the values of x and ^* 
 
 ^ns. a:=2. i/=5 
 
 18. Given < ^ . I to find x and y. 
 
 19. Given x-}-y =8 and a? 2 y 2 =16 to find a: and t/. 
 
 20. Given 4(#-f-2/)=9(a; y) and a? 2 y 2 =36 to find * and y. 
 
82 ELEMENTS OF ALGEBRA. 
 
 21. Given x ly :: 4 : 3 and a? ^2/ 3 =37, to find # and y. 
 
 Ans. #=4. y=3. 
 
 22. Given #-fr/=a and x* y z =ab to find x and y. 
 
 a-\-b a /> 
 
 Ans. a?= . 2/ = -2~ 
 
 23. Given -- = and - -j =| to find a; and y. 
 
 Jlns. #=4. lo 
 
 24. Given $(a:+2)+ 8y=31 and 4(y+5)-}-10^=192 to 
 find the values of a; and y. .tfns. x=19. y=3 
 
 25. Given 3a?+7y=79 and 2t/-j-5iC=19 to find the values 
 of x and y. Jlns. #=10. y=7. 
 
 26. Given |(#-f t/)+25=a? and %(x-}-y) 5=y to find 
 the values of x and y. Jlns. #=85. y=35. 
 
 27. Given x 4=y+l and 5#-^=^ ^+37 to find the 
 
 o 4 6 
 
 values of x and y. Ans. x=S. y3 
 
 28. Given 4 ^?=T/ 17| and ^=f+2 to find the 
 
 D O O 
 
 values of a? and y. jJns. x=W. y=2Q. 
 
 7x 21 . x . 3x 19 2a?+v 
 
 29. Given - }-y_=4-{ ^ and -f- 
 
 9# 7 3y4-9 4#+5y 
 -- =-*-T --- rtr 3 to find a? and y. Ans. #=9. y=4. 
 
 
 3O. Given 
 
 23 # 2 
 
 73 3y 
 
 --- 2 
 
 to find x and 
 
 3 
 Jlns. #=21. 
 
EQUATIONS. 83 
 
 CHAPTER HI. 
 
 Solution of Equations involving three or more unknown 
 quantities. 
 
 (Art. 52.) No additional principles are requisite to those given 
 in articles 49 and 50. 
 
 EXAMPLES. 
 
 f X + y+ z = 9 ] 
 
 1. Given \ #+2?/+3z 16 [ to find a?, y, and z. 
 [ a?+3y+4*=21 J 
 
 By the 1st method, transpose the terms containing y and z in 
 each equation, and 
 
 x= 9 T z, 
 
 Then putting the 1st and 2d values equal, and the 2d and 3d 
 values equal, gives 
 
 9 y z=16 2y 3z, 
 1 6 2y 3z =2 1 3y 40. 
 
 Transposing and condensing terms, and 
 
 y=72z, 
 
 Also, 2/ = 5 z, 
 
 Hence, 5 z=7 2z, or z=2 
 
 Having z=2, we have 2/=5 ^z=3, and having the values of 
 both z and y, by the first equation we find x=4. 
 
 f 2x-\-4y 3z=22 1 
 
 a. Given < 4x2y-}-5z=l8 [to find values of a?, y and z 
 [ Gx-}-7y z=63 J 
 
 Multiplying the first equation by 2, 4x-}-Sy 6z~44 
 And subtracting the second, 4x 2y-\- 5z=18 
 
 The result is, (rf l( llz=26 
 
54 ELEMENTS OF ALGEBRA. 
 
 Then multiply the first equation by 3, Gx-{-l2y 9z=GG 
 And subtract the third, 6#-|- 7y z =63 
 
 The result is, (B) 5y Sz= 3 
 
 Multiply the new equation (B) by 2, Wy 16z= 6 
 
 And subtract this from equation (.#) IQy llz=26 
 
 The result is, 5z=20 
 
 Therefore z= 4 
 
 Substituting the value of z in equation (E] and we find y=7. 
 Substituting these values in the first equation, and we find x=3. 
 
 ( 3x+9y-\-8z=4l 1 
 
 3. Given \ 5x-\-4y 2z=2Q \ to find a?, y and z. 
 [ llx+ty 6^=37 J 
 
 To illustrate by a practical example we shall resolve this by 
 the principles explained in (Art. 51.) 
 
 3mx-\-9my -\-Smz = 41m 
 5nx -\-4ny 2nz =2Qn 
 
 Sum (3ra+5n)#+(9m-f4n)?/-f-(8w 2n)z=41m+20w 
 Take llx +7 6z=37 
 
 Rem. (3m4-5n 11) a; (7 9m 4n)y+(8m 27i+6)z= 
 
 Assume 3m-}-5n=ll (1) 
 
 And 9m-f4n=7 (2) 
 
 From equations (1) and (2) we find wi= T 3 T and w=f i 
 These values substituted in equation (3) we have 
 
 _ 41 x T 3 T 4-2QX f f -37 
 ~ ~8X T 3 T 2Xff+ 6 
 
 Multiply both numerator and denominator by 11, and we shall 
 
 123+520407 10 
 have z=- 
 
 24 52-r- 66 10 
 
EQUATIONS. 85 
 
 Putting this value of z in the 1st and 2d equations, we shall 
 have only two equations involving x and y, from which the 
 values of these letters may be determined. 
 
 These equations can be resolved with much more facility by 
 multiplying the 2d equation by 4, then adding it to the 1st to 
 destroy the terms containing z. 
 
 Afterwards multiplying the 2d equation by 3, and subtracting 
 the 3d equation, and there will arise two equations containing x 
 and ?/, which may be resolved by one of the methods already 
 explained. 
 
 (Art. 53.) When three, four, or more unknown quantities with 
 as many equations are given, and their coefficients are all prime 
 to each other, the operation is necessarily long. But when sev- 
 eral of the coefficients are multiples, or measures of each other, 
 or unity, several expedients may be resorted to for the purpose 
 of facilitating calculation. 
 
 No specific rules can be given for mere expedients. Exam- 
 ples alone can illustrate, but even examples will be fruitless to 
 one who neglects general principles and definite theories. Some 
 few expedients will be illustrated by the following 
 
 EXAMPLES. 
 
 1. Given x+y z=25 to find x, y, and z. 
 [ x yz= 9 J 
 
 Subtract the 2d from the 1st, and 2z=6. 
 Subtract the 3d from the 2d, and 2?/=16. 
 Add the 1st and 3d, and 2#=40. 
 
 ( 
 
 2. Given I x y 4 to find x, y and z. 
 I*** = 6 j 
 
 Add all three, and 3#=3 6 or #=12. 
 
 { xy-z= 6 I 
 3. Given < 3y x z=12 to find a?, y and z 
 
86 ELEMENTS OF ALGEBRA. 
 
 Assume x-\-y-\-z=s. Add this equation to each of the given 
 equations, and we then have 
 
 2#= 6+5, (A) 
 
 4</=12+5, (B) 
 
 8z=24+5. (C) 
 Multiply (.#) by 4, and (B) by 2, and we have 
 
 8#=24+4s, 
 
 82=24+ s. 
 
 By addition, 85=3 X 24-}- -7s. Or 5=72. 
 Put this value of s in equation (.#) and we shall have 
 22=6+72. Or #=3+36=39, &c. 
 
 4. Given #+T/=100, 3/+|z=100, z+4#=100 to find 
 ar, t/, and z. 
 
 Put =100. Ans. #=64, y=72, and z=84. 
 
 r = 10 
 
 5 Given 
 
 u-\-v-\-y+z=l2 
 
 to find the value of each. 
 
 v+#+t/+2'=14 
 Here are Jive letters and five equations. Each letter has the 
 same coefficient, one understood. Each equation has 4 letters, 
 z is wanting in the 1st equation, y in the 2d, <fec. 
 Now assume w+v+#+?/+;z=5. 
 Then 5 z=10 (\ 
 
 s #=12 
 s v=13 
 s w=14 
 
 Add, and 5s -5=60 Or 5=15. 
 Put this value of s in equation (.#), and z=5, &c. 
 
 6. Given #+y=a, #+z=&, y+z=c. 
 Add the 1st and 2d, and from the sum subtract the 3d. 
 Jlns. #= 
 
EQUATIONS 
 
 8? 
 
 Given 
 
 8. Given 
 
 O. Given 
 
 1O. Given - 
 
 2ati+yf* I 
 
 3y=u-}-x-\-z I to find the value of w, x, y, 
 4z=u+x+y [ and z. 
 u=x 14 J 
 
 j. w=26, #=40, y=30, z=24. 
 
 2 f #=24. 
 7 rfns. J y=60. 
 
 3 z=120- 
 
 f ar=i l T- 
 
 \ 2/ = T 5 T 
 I ^-T 7 i 
 
 x=20 
 
 x-\-a= 
 
 y 2z=40 
 4y x-\-3z=35 
 t=l3 
 
 u 49 
 
 (Art. 54.) Problems producing simple equations involving two 
 or more unknown quantities. 
 
 1. Find three numbers such, that the product of the 1st and 
 2d, shall be 600 ; the product of the 1st and 3d, 300 ; and the 
 product of the 2d and 3d, 200. 
 
 rfns. The numbers are, 30, 20, and 10. 
 
 2. Find three numbers, such that the^rstf with 5 the sum of 
 the second and third shall be 120 ; the second with | the differ- 
 ence of the third and first shall be 70 ; and the sum of the three 
 numbers shall be 190. Ans. 50 ; 65 ; 75. 
 
 3. A certain sum of money was to be divided among three 
 persons, .#, B, and C, so that #'s share exceeded j of the shares 
 of B and C by $120 ; also the share of B exceeded f of the 
 shares of A and C by $120; and the share of C, likewise, ex- 
 ceeded f of the shares of *# and B by $120. What was each 
 person's share? 
 
 tins. JT* share, $600 ; 's, 480 ; and C"s 360. 
 
88 ELEMENTS OF ALGEBRA. 
 
 4. Jl and B, working at a job, can earn $40 in 6 days ; A and 
 C together can earn $54 in 9 days; and B and C$80 in 15 
 days. What can each person alone earn in a day? 
 
 Let A earn x, B earn y, and C earn z dollars per day, then, 
 By the question, 6x-J- 6y=40 
 9.2=54 
 
 Dividing the equations by the coefficients of the unknown 
 quantities, we have, , _ AO 
 
 - 
 
 See Problem 6. (Art. 53.) 
 
 A man has 4 sons. The sum of the ages of the first, second 
 and third is 18 years ; the sum of the ages of the first, second 
 and fourth is 16 years ; the sum of the ages of the first, third 
 and fourth is 14 years ; the sum of the ages of the second, third 
 and fourth is 12 years. What are their ages? See Problem 5. 
 (Art. 53.) rfns. Their ages are, 8, 6, 4, 2. 
 
 5. .#, B and C sat down to play, each one with a certain num- 
 ber of shillings ; Jl loses to B and C as many shillings as each 
 of them has. Next B loses to Jl and C as many as each of 
 them now has. Lastly, C loses to Jl and B as many as each of 
 them now has. After all, each one of them has 16 shillings. 
 How much did each one gain or lose ? 
 
 Let x= the number of shillings Jl had at first 
 y B's shillings, and 
 z C's shillings. 
 
 Then, by resolving the problem, we shall find #=26, y=14 
 and z=S. Therefore, Jl lost 10 shillings, B gained 2, and (78. 
 
 N. B. When the equations are found, divide the 1st by 4, the 
 2d by 2, and then compare them with Ex. 3. (Art. 53.) 
 
 6. A gentleman left a sum of money to be divided among four 
 servants, so that the share of the first was the sum of the 
 shares of the other three ; the share of the second, | of the sum 
 of the other three ; and the share of the third, k the sum of the 
 
EQUATIONS. 89 
 
 other three ; and it was found that the share of the last was 14 
 dollars less than that of the first. What was the amount of 
 money divided, and the shares of each respectively 1 
 
 Ans. The sum was $120 ; the shares 40, 30, 24 and 26. 
 
 Observe Prob. 7. (Art. 53,) in connection with this problem. 
 
 7. A jockey has two horses, and two saddles which are worth 
 
 15 and 10 dollars, respectively. Now if the better saddle be put 
 on the better horse, the value of the better horse and saddle 
 would be worth of the other horse and saddle. But if the 
 better saddle be put on the poorer horse, and the poorer saddle 
 on the better horse, the value of the better horse and saddle is 
 worth once and f^ the value of the other. Required the worth 
 of each horse ? Ans. 65 and 50 dollars. 
 
 8. A merchant finds that if he mixes sherry and brandy in 
 quantities which are in proportion of 2 to 1, he can sell the mix- 
 ture at 78 shillings per dozen; but if the proportion be 7 to 2 he 
 can sell it at 79 shillings per dozen. Required the price per 
 dozen of the sherry and of the brandy ? 
 
 Ans. Sherry, 81s. Brandy, 725. 
 
 In the solution of this question, put a=78. Then a-f-l=79. 
 
 9. Two persons, A and J9, can perform a piece of work in 
 
 16 days. They work together for four days, when A being 
 called off, B is left to finish it, which he does in 36 days. In 
 what time would each do it separately 7 
 
 Ans. A in 24 days ; B in 48 days. 
 
 1O What fraction is that, whose numerator being doubled, 
 and denominator increased by 7, the value becomes f ; but the 
 denominator being doubled, and the numerator increased by 2, 
 the value becomes |? Ans. . 
 
 11. Two men wishing to purchase a house together, valued 
 at 240 (a) dollars ; says A to J3, if you will lend me of your 
 money I can purchase the house alone ; but says B to A, if you 
 lend me J of yours, I can purchase the house. How much 
 money had each of them ? Ans. A had $160. B $120. 
 
 8 
 
90 ELEMENTS OF ALGEBRA. 
 
 12. It is required to divide the number 24 into two such parts, 
 that the quotient of the greater part divided by the less, may be 
 to the quotient of the less part divided by the greater, as 4 to 1 
 
 rfns. 16 and 8. 
 
 13. A certain company at a tavern, when they came to settle 
 their reckoning, found that had there been 4 more in company, 
 they might have paid a shilling a-piece less than they did; but 
 that if there had been 3 fewer in company, they must -have paid 
 a shilling a-piece more than they did. What then was the num- 
 ber of persons in company, and what did each pay ? 
 
 rfns. 24 persons, each paid 7s. 
 
 14. There is a certain number consisting of two places, a unit 
 and a ten, which is four times the sum of its digits, and if 27 be 
 added to it, the digits will be inverted. What is the number? 
 
 Ans. 36. 
 
 NOTE. Undoubtedly the reader has learned in arithmetic that 
 numerals have a specific and a local value, and every remove 
 from the unit multiplies by 10. Hence, if x represents a digit 
 in the place of tens, and y in the place of units, the number must 
 be expressed by lQx-\-y. A number consisting of three places, 
 with x, y and z to represent the digits, must be expresset^ by 
 
 15. A number is expressed by three figures ; the sum of these 
 figures is 1 1 ; the figure in the place of units is double that in 
 the place of hundreds, and when 297 is added to this number, 
 the sum obtained is expressed by the figures of this number re- 
 versed. What is the number ? Jlns. 326. 
 
 10. Divide the number 90 into three parts, so that twice 
 the first part increased by 40, three times the second part in- 
 creased by 20, and four times the third part increased by 10, may 
 be all equal to one another. 
 
 . *ftns. First part 35, second 30, and third 25. 
 
 17. A person who possessed $100,000 (,) placed the greater 
 part of it out at 5 per cent, interest, and the other part at 4 per 
 
EQUATIONS. 91 
 
 cent. The interest which lie received for the whole amounted 
 to 4640 (b~) dollars. Required the two parts. 
 
 Am. 64,000 and 36,000 dollars. 
 
 General Ansiver. (1006 4) for the greater part, and (5 
 1006) for the less. 
 
 18. A person put out a certain sum of money at interest at a 
 certain rate. Another person put out $10,000 more, at a rate 1 
 per cent, higher, and received an income of $800 more. A third 
 person put out $15,000 more than the first, at a rate 2 per cent, 
 higher, and received an income greater hy $1,500. Required 
 the several sums, and their respective rates of interest. 
 
 Jlns. Rates 4, 5 and 6 per cent. Capitals $30,000, $40,000 
 and $45,000. 
 
 19. A widow possessed 13,000 dollars, which she divided 
 into two parts and placed them at interest, in such a mannei, that 
 the incomes from them were equal. If she had put out the first 
 portion at the same rate as the second, she would have drawn 
 for this part 360 dollars interest, and if she had placed the se- 
 cond out at the same rate as the first, she would have drawn for it 
 490 dollars interest. What were the two rates of interest ? 
 
 Ans. 7 and 6 per cent.* 
 
 20. There are three persons, .#, B and C, whose ages are as 
 follows : if J5's age be subtracted from ./2's, the difference will 
 be C"s age ; if five times J5's age and twice C"s age, be added 
 together, and from their sum JFs age be subtracted, the remain- 
 der will be 147. The sum of all their ages is 96. What are 
 their ages ? Jlns. 's 48, J5's 33, <7's 15. 
 
 21. Find what each of three persons, ./?, B and (7, is worth, 
 from knowing, 1st, that what Ji is worth added to 3 times what 
 B and C are worth, make 4700 dollars ; 2d, that what B is worth 
 added to four times what A and C are worth make 5800 dollars; 
 3d, that what C is worth added to five times what Ji and B are 
 worth make 6300 dollars. 
 
 tins. A is worth 500, B 600, C 800 dollars. 
 
 tt*;e brief solution to these two problems, 18 and 19, in Key to Algebra. 
 
92 
 
 ELEMENTS OF ALGEBRA. 
 
 Put 5= the sum that v?, B and C are worth, to make an aux- 
 iliary equation. 
 
 22. Find what each of three persons, /?, B, C, is worth, 
 knowing, 1st, that what Jl is worth added to / times whauE? and 
 C are worth is equal to p; 2d, that what B is worth added to m 
 times what Ji and C are worth is equal to q; 3d, that what C is 
 worth added to n times what Ji and B are worth is equal to r. 
 
 Let #=r#'s capital, y=B <> 8 1 and 2 = C"s. 
 Then x-^ly-^-lzp^ is the first equation. 
 
 Assume x-{- y+ *=* Multiply this equation by / and 
 subtract the former, and (/ \}x=ls p. 
 
 Is p 
 
 r> -i * ms *7 
 
 By a similar operation, y= J 
 
 And, 
 
 m 1 
 
 ns r 
 
 (A) 
 
 By addiuon, 
 
 ns r 
 
 This equation may take the following form : 
 
 =(- 1 
 
 +-^+- 
 
 Now as the terms in parenthesis are fully determined, of 
 known value, we may represent the first by a, the second by b t 
 and this last form becomes 
 
 s=5 b 
 
 By transposition, &c. (a l)s=6 
 
 b 
 
 Therefore 
 
 *= 
 
 a 1 
 
 This known value of * put in each of the equations marked 
 (A), and the values of x, y and z will be theoretically deter- 
 mined. 
 

 EQUATIONS. 93 
 
 23. Three brothers made a purchase of $2000 (a;) the first 
 wanted in addition to his own money the money of the second, 
 the second wanted in addition to his own 3 of the money of the 
 third, and the third required in addition to his own of tho 
 money of the first. How much money had each ? 
 
 Jlns. 1st had, $1280; 2d, $1440; and the 3d, $1680. 
 Gen. Jlns. 1st had if a; 2d, if a; and the 3d \ \a. 
 See Prob. 6. (Art. 53.) 
 
 24. Some hours after a courier had been sent from A to B, 
 which are 147 miles distant, a second was sent, who wished to 
 overtake him just as he entered B, and to accomplish this he 
 must perform the journey in 28 hours less time than the first did 
 Now the time that the first travels 17 miles added to the time the 
 second travels 56 miles is 13f hours. How many miles does 
 each go per hour ? Jlns. 1st 3, the 2d, 7 miles per hour. 
 
 25. There are two numbers, such that the greater added to 
 I the lesser, is 13 ; and if the lesser is taken from 5 the greater, 
 the remainder is nothing. Required the numbers. 
 
 rfns. 18 and 12. 
 
 26. Find three numbers of such magnitude, that the 1st with 
 the 5 sum of the other two, the second with of the other two, 
 and the third with * of the other two, may be the same, and 
 amount to 51 in each case. J%ns. 15, 33, and 39. 
 
 27. Jl said to B and C t " Give me, each of you, 4 of your 
 sheep, and I shall have 4 more than you will have left." B said 
 to A and C, " If each of you will give me 4 of your sheep, I 
 shall have twice as many as you will have left." C then said 
 to A and B, " Each of you give me 4 of your sheep, and I shall 
 have three times as many as you will have left." How many 
 had each ? rfns. Ji 6, B 8, and C 10. 
 
 28. What fraction is that, to the numerator of wnich if 1 be 
 added, the fraction will be j : but if 1 be added to the denomina- 
 tor, the fraction will be ? 1 dns. T 4 T . 
 
 29. What fraction is that, to the numerator of which if 2 be 
 
U4 ELEMENTS OF ALGEBRA. 
 
 added, the fraction will be f ; but if 2 be added to the denomina- 
 tor, the fraction will be } ? Ans. f . 
 
 SO. \Vhat fraction is that whose numerator being doubled, and 
 its denominator increased by 7, the value becomes f ; but the do- 
 nominator being doubled, and the numerator increased by 2, the 
 value becomes | ? Ans. J. 
 
 31. If A give B $5 of his money, B will have twice as much 
 money as A has left; and if B give A $5, A will have thrice as 
 much as B has left. How much had each ? 
 
 Ans. .tf $13, and .#$11. 
 
 32. A corn factor mixes wheat flour, which cost him 10 shil- 
 lings per bushel, with barley flour, which cost 4 shillings per 
 bushel, in such proportion as to gain 43f per cent, by selling the 
 mixture at 11 shillings per bushel. Required the proportion. 
 
 Am. The proportion is 14 bushels of wheat flour to 9 of 
 barley. 
 
 33. There is a number consisting of two digits, which num- 
 ber divided by 5 gives a certain quotient and a remainder of one, 
 and the same number divided by 8 gives another quotient and a 
 remainder of one.. Now the quotient obtained by dividing by 5 
 is double of the value of the digit in the ten's place, and the quo- 
 tient obtained by dividing by 8 is equal to 5 times the unit digit. 
 What is the number ? Ans. 41. 
 
 Interpretation of negative values resulting from the solution 
 
 of equations. 
 
 (Art. 55.) The resolution of proper equations drawn from 
 problems not only reveal, the numeral result, but improper enun- 
 ciation by the change of signs. Or the signs being true algebraic 
 language, they will point out errors in relation to terms in com- 
 mon language, as the following examples will illustrate : 
 
 1. The sum of two numbers is 120, and their difference is 
 100 ; what are the numbers ? 
 
 Let x be the greater and y the less. Then 
 a?+y=120 (1) 
 a^=160 (2) 
 The solution gives #=140, and ?/ 20. 
 
EQUATIONS. 95 
 
 Here it appears that one of the numbers is greater than the 
 sum given in the enunciation, yet the sum of x and y, in the al- 
 gebraic sense, is 120. 
 
 There is no such abstract number as 20, and when minus 
 appears it is only relative or opposite in direction or condition 
 to plus, and the problem is susceptible of interpretation in an al- 
 gebraic sense, but not in a definite arithmetical sense. 
 
 Indeed we might have determined this at once by a considera- 
 tion of the problem, for the difference of the two numbers is 
 given, greater than their sum. But we can form a problem, an 
 algebraic (not an abstract) problem that will exactly correspond 
 with these conditions, thus : 
 
 The joint property of two men amounts to 120 dollars, and 
 one of them is worth 160 dollars more than the other. What 
 amount of property does each possess ? 
 
 The answer must be -f-140 and 20 dollars ; but there is no 
 such thing as minus $20 in the abstract ; it must be interpreted 
 debt, an opposite term to positive money in hand. 
 
 2. Two men, A and B, commenced trade at the same time ; 
 Ji had 3 times as much money as B, and continuing in trade, Jl 
 gains 400 dollars, and J9 150 dollars; now A has twice as much 
 money as B. How much did each have at first? 
 
 Without any special consideration of the question, it implies 
 that both had money, and asks how much. But on resolving the 
 question with x to represent ^?'s money, and y B's, we find 
 
 #=300 
 And y= 100 dollars. 
 
 That is, they had no money, and the minus sign in this case 
 indicates debt; and the solution not only reveals the numerical 
 values, but the true conditions of the problem, and points out the 
 necessary corrections of language to correspond to an arithmeti- 
 cal sense, thus : 
 
 A is three times as much in debt as B ; but A gains 400 dol- 
 lars, and B 150 ; now A has twice as much money as B. How 
 much were each in debt ? 
 
 As the enunciation of this problem corresponds with the real 
 
96 ELEMENTS OF ALGEBRA. 
 
 circumstance of the case, we can resolve the problem without a 
 minus sign in the result. Thus : 
 
 Let z= B'B debt, then 3x= J?s debt 
 
 150 x J?'s money, 400 3x= .#'s money 
 Per question, 400 3#-=300 2x. Or a?=100. 
 
 3. What number is that whose fourth part exceeds its third 
 part by 12 ? J2ns. 144. 
 
 But there is no such abstract number as 144, and we cannot 
 interpret this as debt. It points out error or impossibility, and 
 by returning to the question we perceive that a fourth part of any 
 number whatever cannot exceed its third part; it must be, its third 
 part exceeds its fourth part by 12, and this enunciation gives the 
 positive number, 144. Thus do equations rectify subordinate 
 errors, and point out special conditions. 
 
 4L A man when he was married was 30 years old, and his 
 wife 15. How mapy years must elapse before his age will be 
 three times the age of his wife? 
 
 Jlns. The question is incorrectly enunciated ; 7 years before 
 the marriage, not after, their ages bore the specified relation. 
 
 5. A man worked 7 days, and had his son with him 3 days , 
 and received for wages 22 shillings. He afterwards worked 5 
 days, and had his son with him one day, and received for wages 
 18 shillings. What were his daily wages, and the daily wages- 
 of his son ? 
 
 Jfns. The father received 4 shillings per day, and paid 2 shil- 
 lings for his son's board. 
 
 6. A man worked for a person ten days, having his wife with 
 him 8 days, and his son 6 days, and he received $10.30 as com- 
 pensation for all three; at another time he wrought 12 days, his 
 wife 10 days, and son 4 days, and he received $-13.20; at an- 
 other time he wrought 15 days, his wife 10 days, and his son 12 
 days, at the same rates as before, and he received $13.85. What 
 were the daily wages of each ? 
 
 rfns. The husband 75 cts., wife 50 cts. The son 20 cts. ex- 
 pense per day. 
 
EQUATIONS. 97 
 
 7. A man wrought 10 days for his neighbor, his wife 4 days, 
 and son 3 days, and received $11.50 ; at another time he served 
 9 days, his wife 8 days, and his son 6 days, at the same rates as 
 before, and received $12.00 ; a third time he served 7 days, his 
 wife 6 days, and his son 4 days, at the same rates as before, and 
 he received $9.00. What were the daily wages of each ? 
 
 *fins. Husband's wages,$ 1.00; wife ; son 50 cts. 
 
 8. What fraction is that which becomes f when one is added 
 to its numerator, and becomes | when 1 is added to its denomi- 
 nator ? 
 
 Jlns. In an arithmetical sense, there is no such fraction. The 
 algebraic expression, ~y|, will give the required results. 
 
 (Art. 58.) By the aid of algebraical equations, we are enabled 
 not only to resolve problems and point out defects or errors in 
 their enunciation, as in the last article, but we are also enabled 
 to demonstrate theorems, and elucidate many philosophical truths. 
 The following are examples : 
 
 Theorem 1. It is required to demonstrate, that the half sum 
 plus half the difference of two quantities give the greater of the 
 two, and the half sum minus the half difference give the less. 
 
 Let x= the greater number, y= the less, 
 
 s= their sum, d their difference. 
 
 Then x+y= s (A] 
 
 And xy=d (B) 
 
 By addition, 2a?= s-f- d 
 
 Or x=*s-}-zd 
 
 Subtract (S) from (.#) and divide by 2, and we have 
 y=%s |c? 
 
 These last two equations, which are manifestly true, demon- 
 strate the theorem. 
 
 Theorem 2. Four times the product of any two numbers, is 
 equal to the square of their sum, diminished by the square of 
 their difference. 
 
 9 
 
98 ELEMENTS OF ALGEBRA 
 
 Let x= the greater number, and y= the less, as in the lasi 
 theorem. 2x=s -\-d 
 
 And 2=s d 
 
 By multiplication 4xy=s 2 d 2 a demonstration of the 
 theorem. 
 
 Many other theorems are demonstrable by algebra, but we de- 
 fer them for the present, as some of them involve quadratic equa- 
 tions, which have not yet been investigated ; and we close the 
 subject of simple equations by the following quite general prob- 
 lem in relation to space, time and motion. 
 
 To present it at first, in the most simple and practical manner, 
 let us suppose 
 
 Two couriers, A and B, 100 miles asunder on the same road 
 set out to meet each other, A going 6 miles per hour and B 4. 
 How many hours must elapse before they meet, and how far 
 will each travel? 
 
 Let x= Jl's distance, y j#'s, and t the time. 
 Then ^-}-?/=100 ' (1) 
 
 As the miles per hour multiplied by the hours must give the 
 distance each traveled, therefore, 
 
 x=6t and y=U (2) 
 
 Substitute these values in equation (1) and 
 (6+4)f= 100 
 
 1 rjn 
 
 Therefore, *=r ( 3 ) 
 
 100X6 100X4 
 
 And *=8l=- 2/ =4/= 
 
 From equation (3,) we learn that the time elapsed before the 
 couriers met was the whole distance divided by their joint mo- 
 tion per hour, a result in perfect accordance with reason. From 
 equations (4,) we perceive that the distance each must travel is 
 .he whole distance asunder multiplied by their respective mo- 
 tions and divided by the sum of their hourly motions. 
 
 Now let us suppose the couriers start as before, but travel in 
 the same direction, the, one in pursuit of the other. B having 
 
EQUATIONS. 99 
 
 IQO miles the start, traveling four miles per hour, pursued by 
 
 A, traveling 6 miles per hour. How many hours must elapse 
 before they come together, and what distance must each travel? 
 
 Take the same notation as before. 
 
 Then x y=100 (1.) As A must travel 100 miles more than 
 
 B. But equations (2,) that is, x=Qt and y=t, are true under 
 all circumstances. 
 
 Then (64)*= 100 
 
 And i 
 
 The result in this case is as obvious as an axiom. A has 100 
 miles to gain, and he gains 2 miles per hour, it will therefore re- 
 quire 50 hours. 
 
 But it is the precise form that we wish to observe. It is the 
 fact that the given distance divided by the difference of their ma 
 tions gives the time, and their respective distances must be this 
 time multiplied by their respective rates of motion. 
 
 Now the smaller the difference between their motions, the 
 longer the time before one overtakes the other ; when the differ 
 ence is very small, the time will be very great ; when the differ 
 ence is nothing, the time will be infinitely great ; and this is in 
 perfect accordance with reason ; for when they travel equally 
 fast one cannot gain on the other, and they can never come to- 
 gether. 
 
 If the foremost courier travels faster than the other, they must 
 all the while become more and more asunder ; and if they have 
 ever been together it was preceding their departure from the 
 points designated, and in an opposite direction from the one they 
 are traveling, and would be pointed out by a negative result. 
 
 (Art. 59.) Let us now make the problem general. 
 
 Two couriers, A and B, d miles asunder on the same road, 
 set out to meet each other; A going a miles per hour, B going 
 b miles per hour. How many hours must elapse before they 
 meet, and how far will each travel? 
 
100 ELEMENTS OF ALGEBRA. 
 
 Taking the same notation as in the particular case, 
 Let x *#'s distance, y= .5's, and /= the time. 
 
 Then x-\-y=d (I) x=at ybt (2) 
 
 Therefore (a-\-b)t=d Or t==i ~TJ ) ( 3 ) 
 
 ad bd 
 
 If a=b, then x=kd and y=%.d. A result perfectly obvious, 
 the rates being equal. Each courier must pass over one half 
 the distance before meeting. 
 
 If =0 #=== , . =0 and y=-=d. That is, one 
 0+6 
 
 will be at rest, and the other will pass over the whole distance. 
 
 (Art. 60.) Now let us consider the other case, in which one 
 courier pursues the other, starting at the same time from dif- 
 ferent points. 
 
 Let the line CD represent the space the couriers are asunder 
 when the pursuit commences, and the point E where they come 
 together. C D E 
 
 j j j 
 
 The direction from C towards D we call plus, the other direc- 
 lion will therefore be minus. 
 
 Now as in the 2d example, (Art. 58.) 
 
 Put x = CE =^'s distance 
 y=DE=JS 1 s distance 
 Then xy=CD=d (1) 
 
 As before, let t= the time. Then 
 
 Therefore at bt=d 
 
 And t=^ (3) 
 
 For the distances we have 
 
 ad , . t/tt . N 
 
 x= (4) and y= , (o) 
 
 06 v ' 9 " h 
 
EQUATIONS. 101 
 
 By an examination of these equations, it will be perceived fazA 
 x and y will be equal when a is equal to b, yet d still exists as 
 a difference between them. This is in consequence of x and y 
 in that case being so very great that d is lost in comparison. So 
 all values are great or small only in comparison with others or 
 with our scale of measure. 
 
 To make this clear, let us suppose two numbers differ by one 
 and if the numbers are small, the difference may be regarded as 
 considerable; if large, more inconsiderable; if still larger, still 
 more inconsiderable, &c. If the numbers or quantities be infin- 
 itely great, the comparative small quantity may be rejected. 
 Thus : 
 
 5 and 6 differ by 1, and their relation is as 1 to 1.2. 
 
 Also, 50 and 51 differ by 1, and their relation is as 1 to 1.02. 
 500 to 501 are as 1 to 1.002, <fcc. The relation becoming nearer 
 and nearer equality as the numbers become larger, and when the 
 numbers become infinitely great the difference is comparatively 
 nothing. 
 
 When a=b a ft and x ~~ft a symbol of infinity. 
 
 If we suppose b greater than a, a b will become negative, and 
 as x and y refer to the same point, that point must then be in the 
 backward direction from that we suppose the couriers are mo- 
 ving, and will show how far they have traveled since that event. 
 
 If in the equations (3), (4) and (5), c?=0, and at the same 
 time a=bj then we shall have /=-, #=- and 2/=~ ' which 
 
 shows that - is a symbol of indetermination, it being equal to 
 
 several quantities at the same time. If e?=0 the two couriers 
 were together at commencement ; and if they travel in the same 
 direction, and equally fast, they will be together all the while, 
 and the distances represented by x and y will be equal, and of 
 
 all possible values. Hence - may be taken of any value what- 
 
102 CLEMENTS OF ALGEBRA 
 
 evsr^ amV-may Ve'-vntidMo take a particular value, to correspond 
 to any other circumstance or condition.* 
 
 APPLICATION. 
 
 (Art. 61.) We have hitherto considered CD a right line; but 
 the equations would be equally true, if we consider CD to be 
 curved, and indeed we can conceive the line CD to wind about 
 a perfect circle just forming its circumference, and the point E 
 upon the circle, CE being a little more than one circumference. 
 
 This being understood, Equation (3,) (Art. 60,) gives us a so- 
 lution to the following problems. 
 
 1. The hour and minute hands of a clock are together at 12 
 o'clock. When are they next together ? 
 
 * The 26th equation (Art. 40), if resolved in the briefest manner, will show 
 
 
 the influence of the factor rr. In the equation referred to, add 30 to both 
 
 members and divide the numerator of the second member by its denominator, 
 
 5^4-5 
 and we have - -- [-1=6. Drop 1, and divide both members by 5, we then 
 
 x4-\ 
 have =1, or x-\-l=x-}-2. Hence 1=2, a manifest absurdity 
 
 But all our operations, yea, and all our reasonings have been correct, but 
 we did not pay sufficient attention to dividing the numerator by the denomi- 
 
 nator, which was -7 - . Taking 6 for the quotient, which it would be in 
 
 (x t) 
 
 every case except when x 2=0, leads to the absurdity ; which absurdity, 
 in turn, shows that x 2=0, or #=2. 
 
 As another illustration of the influence of this symbol, take the identical 
 equation 100=100, or any other similar one. 
 This is the same as 96-f-4=96-f-4 
 
 Transposing, . 4 4=96 96 
 
 Resolving into factors, 1(4 4) =24 (4 4) 
 
 Dividing by the common factor, and 1=24 ; 
 
 1 
 
 But, to restore equality, - in this case must equal , or 24. 
 
 
 Hence we perceive that - is indeterminate, in the abstract, but may be ren- 
 
 dered definite in particular cases. 
 
EQUATIONS. 103 
 
 By the equation, t j- this problem and all others like it 
 
 are already resolved. All we have to do is to determine the 
 values of </, , and b. 
 
 There are 12 spaces (hours) round the dial plate of a clock ; 
 hence d may represent 12. a and b are the relative motions ol 
 the hands, a moves 12 spaces or entirely round the dial plate 
 while b moves one space. Hence a=12, 6=1, and a 6=11. 
 
 Consequently, t=\\lh. 5m. 27 T \s. 
 
 Again. We may demand what time the hour and minute 
 hands of a clock are together between 3 and 4. 
 
 From 12 o'clock to past 3 o'clock there are 3 revolutions to 
 
 q v/ i o 
 
 pass over in place of one, and the solution is therefore t 
 
 and so on for any other hour. 
 
 2. WJiat time between two and three o'clock will the hour 
 and minute hands of a clock make, right angles with each other? 
 
 Here the space that the one courier must gain on the other is 
 two revolutions and a quarter, or 21 d. 
 
 Hence t=\\Wk=\\ X J=H=2h. 27 T 3 T m. 
 
 3. What time between 5 and 6 will the two hands of a clock 
 make a right line ? 
 
 Here one courier must gain 5 revolutions, or d in the equa- 
 tion must be multiplied by 5 = y . 
 
 Hence, t=\\ X V =6h. 
 
 That is, the hands make a right line at 6 o'clock, a result man- 
 ifestly tri'3. 
 
 This simple equation enables us to determine the exact time 
 when the two hands of a clock shall be in any given position. 
 
 We may apply this equation to a large circle, as well as to a 
 small one; it may apply to the apparent circular course of the 
 heavens, as well as to a dial plate of a clock ; and the application 
 is equally simple. 
 
 The circle of the heavens, like all other circles, is divided into 
 
104 ELEMENTS OF ALGEBRA. 
 
 360 degrees ; and the sun and moon apparently follow each other 
 like two couriers round the circle. 
 
 In one day the moon moves on an average 13. 1764, (divisions 
 of the circle,) and the sun apparently 0. 98565, or not quite one 
 division of the circle. The moon's motion being most rapid. 
 corresponds to a in the equation, and the sun's apparent motion 
 to b. Then &=13.1764 0.98565=12.19075 ; and the 
 time required for one courier to gain on the other the required 
 
 space, in this case a revolution of 360 degrees, or / .-= 
 
 360 a ~" b 
 
 which gives 29.5305887 days, or 29 days, 12 hours, 
 \Z. 19075 
 
 44 minutes, 3 seconds, which is the mean time from one change 
 of the moon to another, called a synodic revolution. 
 
 These relative apparent motions of the sun and moon round 
 the circular arc of the heavens, are very frequently compared to 
 the motions of the hour and minute hands of a clock round the 
 dial plate ; and from the preceding application of the same equa- 
 tion we see how truly. 
 
 We may not only apply this equation to the mean motions of 
 the sun and moon, but it is equally applicable to the mean mo- 
 tions of any two planets as seen from the sun. To appearance, 
 the two planets would be nothing more than two couriers mo- 
 ving in a circle, the one in pursuit of the other, and the time be- 
 tween two intervals of coming together, (or coming in conjunc- 
 tion, as it is commonly expressed,) will be invariably represen- 
 ted by the equation 
 
 To apply this to the motion of two planets, we propose the 
 following problem : 
 
 The planet Venus, as seen from the sun, describes an arc of 
 1 36'/)er day, and the earth, as seen from the same point, de- 
 scribes an arc of 59'. Jit what interval of time will these two 
 bodies come in a line with the sun on the same side ? 
 
 Here a=l36' 6=59' rf=360 
 
 Therefore, a 6=37'; and as the denominator is 
 
EQUATIONS. 105 
 
 minutes, the numerator must be reduced to minutes also ; hence 
 the equation becomes 
 
 360X60 , 
 
 =583.8 days, nearly. 
 
 We have not been very minute, as the motions of the planets 
 are not perfectly uniform, and the actual interval between succes 
 sive conjunctions is slightly variable. Hence we were not par- 
 ticular to take the values of a and b to the utmost fraction. A 
 more rigid result would have been 583.92 days. Half of this 
 time is the interval that Venus remains a morning and an even- 
 ing star. 
 
 (Art. 63.) This equation, as simple as it may Ep' er.r, is one 
 practical illustration of the true spirit and utility oi analysis by 
 algebra. 
 
 The principles and relations of time and motion are fixed and 
 
 d 
 invariable, and the equation t=^^ stands conspicuously as 
 
 one of these relations. 
 
 If t can be determined by observation, as it may be in respect 
 to the earth and the superior planets, the mean daily motions of 
 the planets can be determined ; as f?=360, a=59' 08" the mean 
 motion of the earth, and suppose b the motion of Mars, for ex- 
 ample, to be unknown. 
 
 When unknown, represent it by x. 
 
 Then t= - or at txd. 
 a x 
 
 at d 
 
 Therefore x= -. . 
 t 
 
106 ELEMENTS OF ALGEBRA. 
 
 SECTION III. 
 
 INVOLUTION. 
 CHAPTER I. 
 
 (Art. 64.) Equations, and the resolution of problems producing 
 equations, do not always result in the first powers of the un- 
 known terms, but different powers are frequently involved, and 
 therefore it is necessary to investigate methods of resolving equa- 
 tions containing higher powers than the first ; and preparatory 
 to this we must learn involution and evolution of algebraic 
 quantities. 
 
 (Art. 65.) Involution is the method of raising any quantity to 
 a given p v =h. Evolution is the reverse of involution, and is 
 the method of Determining what quantity raised to a proposed 
 power will produce a given quantity. 
 
 As in arithmetic, involution is performed by multiplication, and 
 evolution by the extraction of roots. 
 
 The first power is the root or quantity itself; 
 
 The second power, commonly called the square, is the quan- 
 tity multiplied by itself ; 
 
 The third power is the product of the second power by the 
 quantity ; 
 
 The fourth power is the third power multiplied into the quan- 
 tity, &c. 
 
 The second power of a is a X# or a z 
 
 The third power is a z Xa or a 3 
 
 The fourth power is a 3 Xa or a 4 
 
 The second power of a 1 is a 4 X 4 or a 8 
 
 The third power of a 4 is a 8 Xa 4 or a 12 
 
 The nth power of a 4 has the exponent 4 repeated n times, 
 or a 4 ". Therefore, to raise a simple literal quantity to any 
 power, multiply its exponent by the index of the required 
 power. 
 
 Raise x to the 5th power. The exponent is 1 understood, 
 and this 1 multiplied by 5 gives x 5 for the 5th power. 
 
 
INVOLUTION. 107 
 
 Raise x 3 to the 4th power Am. a? 12 . 
 
 Raise y 1 to the third power. Ans. y 21 . 
 
 Raise x n to the 6th power. Am. x* n . 
 
 Raise x n to the mth power. Am. a? 17 *. 
 
 Raise ay? to the 3d power. A*is. a*x*. 
 
 Raise ab*x 4 to the 2d power. .tfns. 2 6V. 
 
 R*.ise c 2 7/ 4 to the 5th power. .tfns. c 10 */ 20 . 
 
 (Art. 66.) By the definition of powers the second power is 
 any quantity multiplied by itself; hence the second power of 
 ax is a 2 ^ 2 , the second power of the coefficient a, as well as the 
 other quantity x ; but a may be a numeral, as 6#, and its second 
 power is 36x*. Hence, to raise any simple quantity to any 
 power, raise the numeral coefficient, as in arithmetic, to the 
 required power, and annex the powers of the given literal 
 quantities 
 
 EXAMPLES. 
 
 1. Required the 3d power of 3a;r 2 . Am. 27V. 
 
 2. Required the 4th power of |?/ 2 . Jins. Ify 8 . 
 
 3. Required the 3d power of 2x. Ans. Sx 3 . 
 4U Required the 4th power of 3x. Jlns. 8 la; 4 . 
 
 Observe, that by the rules laid down for multiplication, the 
 even powers of minus quantities must be plus, and the odd powers 
 minus. 
 
 5. Required the 2d power of - . Ans. - ^-. 
 
 6. Required the 6th power of . Ans. 
 
 7. Required the 6th power of -j . Ans ^ . 
 
 ^a? x 
 
 (Art. 67.) The powers of compound quantities are raised by 
 the mere application of the rule for compound multiplication. 
 (Art. 12.) 
 
108 ELEMENTS OF ALGEBRA. 
 
 Let a-\-b be raised to the 2d, 3d, 4th, &c. powers. 
 a +b 
 
 tf+ab 
 ab+b* 
 
 2d power or square, a z -}-2ab 
 a+b 
 
 
 3d power or cube, 
 
 The 4th power, a 4 +4a 3 6-f Ga*b 2 -{-4ab s +b< 
 a+b 
 
 The 5th power, 
 
 &c. 
 
 By inspecting the result of each product, we may arrive at 
 general principles, according to which any power of a binomial 
 may be expressed, without the labor of actual multiplication. 
 This theorem for abbreviating powers, and its general application 
 to both powers and roots, first shown by Sir Isaac Newton, has 
 given it the name of Newton's binomial, or the binomial theorem. 
 
 OBSERVATIONS. 
 
 Observe the 5th power : a, being the first, is called the leading 
 term ; and 6, the second, is called the following term. The sum 
 of the exponents of the two letters in each and all of the terms 
 amount to the index of the power. In the 5th power, the sura 
 
INVOLUTION. 109 
 
 of the exponents ol a and b is 5 ; in the 4th power it is 4 ; in 
 the 10th power it would be 10, &c. In the 2d power there are 
 three terms; in the 3d power there are 4 terms; in the 4th 
 power there are 5 terms ; always one more term than the index 
 of the power denotes. 
 
 The 2d letter does not appear in the first term ; the 1st letter 
 does not appear in the last term. 
 
 The highest power of the leading term is the index of the 
 given power, and the powers of that letter decrease by one from 
 term to term. The second letter appears in the 2d term, and its 
 exponent increases by one from term to term as the exponent of 
 the other letter decreases. 
 
 The 8th power of (+&) is indicated thus: (a-f-6) 8 . When 
 expanded, its literal part, (according to the preceding observa- 
 tions) must commence with a 8 , and the sum of the exponents of 
 every term amount to 8, and they will stand thus : a 8 , a 7 6, 6 6 2 
 5 6 3 , a 4 b 4 , 3 6 5 , a 2 6 6 , ab\ b 9 . 
 
 The coefficients are not so obvious. However, we observe 
 that the coefficients of the first and last terms must be unity. 
 The coefficients of the terms next to the first and last are equal, 
 and the same as the index of the power. The coefficients in- 
 crease to the middle of the series, and then decrease in the 
 same manner, and it is manifest that there must be some law of 
 connection between the exponents and the coefficients. 
 
 By inspecting the 5th power of a-f-# we find that the 2d co- 
 efficient is 5, and the 3d is 10. 
 
 The third coefficient is the 2d, multiplied by the exponent of 
 ihe leading letter, and divided by the exponent of the second 
 Letter increased by unity. 
 
 In the same manner, the fourth coefficient is the third multi- 
 plied by the exponent of the leading letter, and divided by the 
 exponent of the second letter increased by unity, and so on from 
 coefficient to coefficient. 
 
HO 
 
 The 4th coefficient is 
 
 The 5th is 
 The last is 
 
 Now let us expand 
 For the 1st term write 
 For the 2d term write 
 
 For the 3d, 
 
 ELEMENTS OF ALGEURA. 
 
 1 ft S/ Q 
 
 =iO 
 
 = 1 understood. 
 
 I 
 
 a 8 
 Sa'b 
 
 28 e & 2 
 56a 5 6 3 
 
 For the 4th, 
 
 For the 5th, 
 
 Now as the exponents of a and 6 are equal, we have arrived 
 at the middle of the power, and of course to the highest coeffi- 
 cient. The coefficients now decrease in the reverse order which 
 they increased. 
 
 Hence the expanded power is 
 
 Let the reader observe, that the exponent of 6, increased by 
 unity is always equal to the number of terms from the beginning, 
 or from the left of the power. Thus, b z is in the 3d term, &c. 
 Therefore in finding the coefficients we may divide by the num- 
 ber of terms already written, in place of the exponents of the 
 second term increased by unity. 
 
 If the binomial (a-\-b) becomes (a+1,) that is, when b be- 
 comes unity, the 8th power becomes, 
 
 Any power of 1 is 1, and 1 as a factor never appears. 
 If a becomes 1, then the expanded power becomes, 
 
INVOLUTION. Ill 
 
 The manner of arriving at these results is to represent the unit 
 by a letter, and expand the simple literal terms, and afterwards 
 substitute their values in the result. 
 
 (Art. 68.) If we expand (a 6) in place of (a-\-b^ the expo- 
 nents and coefficients will be precisely the same, but the princi- 
 ples of multiplication of quantities affected by different signs 
 will give the minus sign to the second and to every alternate term. 
 
 Thus the 6th power of (a b) is 
 
 (Art. 69.) This method of readily expanding the powers of a 
 binomial quantity is one application of the " binomial theorem" 
 and it was thus by induction and by observations on the result 
 of particular cases that the theorem was established. Its rigid 
 demonstration is somewhat difficult, but its application is simple 
 and useful. 
 
 Its most general form may arise from expanding (-{->)". 
 
 When n=3, we can readily expand it; 
 
 When =4, we can expand it ; 
 
 When n= any whole positive number, we can expand it. 
 
 Now let us operate with n just as we would with a known 
 number, and we shall have 
 
 a n - z b 2 , &c. 
 
 We know not where the series would terminate until we 
 know the value of n. We are convinced of the truth of the 
 result when n represents any positive whole number; but let n 
 be negative or fractional, and we are not so sure of the result. 
 To extend it to such cases requires deeper investigation and 
 rigid demonstration, which it would not be proper to go into at 
 this time. We shall, therefore, content ourselves with some of 
 its more simple applications. 
 
 EXAMPLES. 
 
 1. Required the third power of 3o?+2y. 
 We cannot well expand this by the binomial theorem, because 
 the terms are not simple literal quantities. But we can assume 
 3a?=0 and 2y=b. Then 
 o-f-& and 
 
112 ELEMENTS OF ALGEBRA. 
 
 Now to return to the values of a and &, we have, 
 
 X 4i/ 2 =36;r/ 2 . 
 
 Hence ( 
 
 2. Required the 4th power of 2a 2 3. 
 
 Let #=2a 2 2/=3. Then expand (a? y} 4 , and return tho 
 values of x and y, and we shall find the result, 
 16a 8 96a 6 4-216a 4 216 2 +81. 
 
 3. Required the cube of (a-{-b -j-c+rf). 
 
 As we can operate in this summary manner only on binomial 
 quantities, we represent a-\-b by x, or assume x=a-\-b, and 
 
 Then 
 
 Returning the values of x and y, we have 
 
 Now we can expand by the binomial, these quantities con 
 tained in parenthesis. 
 
 4. Required the 4th power of 2a-{-3x. 
 
 Ans. 
 
 5. Expand (x z +3y*) s . 
 
 6. Expand (2a z -\-ax)* 
 
 7. Expand (# I) 6 . 
 
 8. Expand (3* -5) 3 rfns. 27^135^+225^125. 
 
 9. Expand (+2) 4 . Ans. a 4 +8a 3 -f-24a 2 +32a+16 
 
 10. Expand (1 a) 4 . tins. 1 2a-r-| -- g+Y6 
 
 11. Expand (a-r-6+c) 2 . 
 
 12. Expand (a2&) 3 . 
 
 13. Expand (1 2x) 5 . 
 
EVOLUTION. U3 
 
 EVOLUTION. 
 CHAPTER II. 
 
 (Art. 70). Evolution is the converse of involution, or the ex- 
 ti action of roots, and the main principle is to observe how powers 
 are formed, to be able to trace the operations back. Thus, to 
 square , we double its exponent, (Art. 65), and make it a 2 . 
 Square this and we have a 4 . Cube a 2 and we have a 6 . Take 
 the 4th power of x and we have x 4 . The nth power of x 3 is 
 x s> \ 
 
 Now, if multiplying exponents raises simple literal quantities 
 to powers, dividing exponents must extract roots. Thus, the 
 
 square root of a 4 is a 2 . The cube root of a 2 must be a*. The 
 cube root of a must have its exponent, (1 understood,) divided by 
 
 3, which will make a*' 
 
 Therefore roots are properly expressed by fractional expo- 
 nents, i 
 
 The square root of a is a 2 , and the exponents, 5, I, ^, &c. in- 
 dicate the third, fourth, and fifth roots. The 6th root of y? is 
 
 s_ 
 
 a? 6 ; hence we perceive that the numerators of the exponent in- 
 dicate the power of the quantity, and the denominator the root 
 of that power. 
 
 (Art. 71.) The square of ax is aV 2 . We square both 
 factors, and so, for any other power, we raise all the factors to 
 the required power. Conversely, then, we extract roots by 
 taking the required root of all the factors. Thus the cube root 
 of 8x* is 2x. 
 
 A root that can not be exactly expressed is called a surd, or 
 irrational root. The square root of 2 can not be exactly ex- 
 pressed, hence, it is called a surd. A root merely expressed by 
 die radical sign (J ), or by a fractional exponent, is called a 
 radical quantity. The cube root of 9 may be symbolically ex- 
 pressed thus, ^/9, or (9)3, and this is a radical quantity; but 
 numerically the root can not be exactly expressed ; it is, there- 
 fore, a surd. Surd roots may be found approximately. 
 10 
 
114 ELEMENTS OP ALGEBRA. 
 
 The square root of 64a 6 is obviously 8a 3 , and from this and 
 the preceding examples we draw the following 
 
 RULE. For the extraction of the roots of monomials. Ex- 
 tract the root of the numeral coefficients and divide the exponent 
 of each letter by the index of the root. 
 
 EXAMPLES. 
 
 1. What is the square root of 49a 2 # 4 ? Ans. 7ax 2 
 
 2. What is the square root of 25c 10 6 2 1 Jins. 5c 5 b. 
 
 3. What is the square root of 2Qax ? fins. 
 
 In 20, the square factor 4 can be taken out ; the other factor 
 is 5. The square root of 4 is 2, which is all the root we can 
 take ; the root of the other factors can only be indicated as in the 
 answer. 
 
 4. What is the square root of 12a 2 1 ns. 
 
 5. What is the square root of 144a 2 cV/? Jins. I2ac*xy. 
 
 6. What is the square root of 36.r 4 ? fins. db6a?. 
 
 (Art. 72.) The square root of algebraic quantities may be 
 taken with the double sign, as indicating either plus or minus, 
 for either quantity will give the same square, and we may not 
 know which of them produced the power. For example, the 
 square root of 16 may be either -j-4 or 4, for either of them, 
 when multiplied by itself, will produce 16. 
 
 The cube root of a plus quantity is always plus, and the cube 
 root of a minus quantity is always minus. For -\-2a cubed 
 gives -|-8a 3 , and 2a cubed gives 8a 3 , and a may represent 
 
 any quantity whatever. 
 
 i 
 
 EXAMPLES. . 
 
 1. What is the cube root of 125a 3 ? tins. 5a. 
 
 2. What is the cube root of 64# 6 ? tfns. 4x* 
 
 3. What is the cube root of 216y ? rfns. 6y 3 
 
 4. What is the cube root of 729 A 12 ? dns. 9a*x* 
 
 5. What is the cube root of 320 5 ? Jlns. ~ 
 
EVOLUTION. 115 
 
 6. What is the 4th root of 256V ? ns. 
 
 7. What is the 4th root of 16? 
 
 8. What is the 4th root of 64x*y 2 ? 
 
 N. B. The 1th root is the square root of the square root. 
 
 9. What is the 4th root of 20a# ? Ans. 
 
 10. What is the square root of 75 ? ns. 
 
 75=25X3. 
 
 4a 2 # 4 
 
 11. Required the square root of -5-2" ^ n 
 
 . 32 3 # 5 
 13. Required the square root of - . <flns. 
 
 N. B. Reduce the fraction as much as possible, and then ex- 
 tract the root. 
 
 13. Required the square root of , ^ . *fl.ns. r* 
 
 128a 4 
 
 14. Required the nth root of -sr .. rfns. -r-. 
 
 nn 
 
 a 3 2. ~1. -1 
 
 15. Required the nth root of =-. Am. a n b" c n 
 
 Observing that j =b~ } c~ l 
 
 CHAPTER m. 
 To extract roots of compound quantities. 
 
 (Art. 72.) We shall commence this investigation by confining 
 our attention to square root, and the only principle to guide us is 
 the law of formation of squares. The square of a-{-b is 2 -}- 
 2ab-\-b z . Now on the supposition that we do not know the root 
 'of this quantity to be a-}-b, we are to find it or extract it out of 
 the square 
 
 We know that a 2 , the first term, must have been formed by the 
 multiplication of a into itself, and the next te-m is 2X#. That 
 
110 ELEMENTS OF ALGEBRA. 
 
 is twice the root of the first term into the second term of the root, 
 Hence if we divide the second term of the square by twice the 
 root of the first term, we shall obtain b, the second term of the 
 root, and as b must be multiplied into itself to form a square, we 
 add b to 2, and have 2-f-/;, which we call a divisor. 
 
 OPERATION. 
 
 a 2 
 
 2a-\-b}2ab+b* 
 2ab+b 2 
 
 We take a for the first term of the root, and subtract its square 
 ( 2 ) from the whole square. We then double a and divide 2ab 
 by 2a and we find 6, which we place in both the divisor and quo- 
 tient. Then we multiply 2a-\-b by b and we have 2ab-{-b z , to 
 subtract from the two remaining terms of the square, and in this 
 case nothing remains. 
 
 Again, let us take rt+6-f-c, and square it. We shall find its 
 square to be 
 
 a 2 
 
 2ab+b 2 
 2ab+b 2 
 
 2aci-2bc+c z 
 
 By operating as before, we find the first two terms of the root 
 to be a-}-b, and a remainder of 2ac-\-2bc-{-c z . Double the root 
 already found, and we have 2a-}-2b for a partial divisor. Divide 
 the first term of the remainder 2ac by 2a, and we have c for the 
 third term of the root, which must be added to 2a-\-2b to com- 
 plete the divisor. Multiply the divisor by the last term of the 
 root and set the product under the three terms last brought down, 
 and we have no remainder. 
 
EVOLUTION. 117 
 
 Again, let us take a-\-b-\-c to square; but before we square it 
 let the single letter s=a-}-b. 
 
 Then we shall have s-f-c to square, which produces 
 s 2 -f-2sc+c 2 . To take the square root of this we repeat the first 
 operation, and thus the root of any quantity can be brought into 
 a binomial and the rule for a binomial root will answer for a root 
 containing any number of terms by considering the root already 
 found, however great, as one term. 
 
 Hence the following rule to extract the square root of a com- 
 pound quantity. 
 
 Arrange the terms according to the powers of some letter, 
 beginning with the highest, and set the square root of the first 
 term in the quotient. 
 
 Subtract the square of the root thus found from the first 
 term and bring down the next two terms for a dividend. 
 
 Divide the first term of the dividend by double of the root 
 already found, and set the result both in the root and in t/ie 
 divisor. 
 
 Multiply the divisor, thus completed, by the term of the root 
 last found, and subtract the product from the dividend, and so on. 
 
 EXAMPLES. 
 1. What is the square root of 
 
 2a 2 -f26 
 
 2a 2 -f-46 2 40 2 86+4 
 
 2. What is the square root of 1 4b-\-4b 2 -\-2y iby-{-y*1 
 
 Ans. 1 2b -f-y. 
 
 8. What is the sq lare root of 4z 4 4or 3 -{-13r J 6#-J-9 ? 
 
 Ans. 2x? a?-f3. 
 
 4. What is the sa lore root of 4z 4 lGx 3 +24x 2 16#-f4 ? 
 
118 ELEMENTS OF ALGEBRA. 
 
 5. What is the square root of I6x* +24x?+S9x*+6Qx-{' 100? 
 
 Am. 
 
 What is the square root of 4x* 16# 3 +8a: 2 -f-16#-{-4 ? 
 
 Ans. 
 7. What is the square root of 
 
 3?+2xy-\-y 2 -\-6xz -\-6yz-\- 9z*l dm. 
 
 8. What is the square root of 
 
 Ans. 
 
 a 2 b" 
 
 0. What is the square root of ^ 2-\- -? ? 
 
 a b b a 
 
 Jlns. T - or - T 
 
 b a a b 
 
 1O. What is the square root of a.* fcB* 
 
 ^-T/ or y-x 
 ^j^L- d.TM.cM^.. fW&d, J0 & : 
 
 (Art. 73.) Every square root will be equally a root if we change 
 the sign of all the terms. In the first example, for instance, the 
 root may be taken a 2 26+2, as well as a z -\-2b 2, for either 
 one of these quantities, by squaring, will produce the given 
 square. Also, observe that every square consisting of three 
 terms only, has a binomial root. 
 
 (Art. 74.) Algebraic squares may be taken for formulas, cor- 
 responding to numeral squares, and their roots may be extracted 
 in the same way, and by the same rule. 
 
 For example, a-f b squared is a z -}-2ab-{-b 2 , and to apply this 
 to numerals, suppose a =40 and 6=7. 
 
 Then the square of 40 is 2 =1600 
 
 2ab= 560 
 6 2 = 49 
 
 Therefore, (47) 2 =2209 
 
 Now the necessary divisions of this square number, 2209, are 
 not visible, and the chief difficulty in discovering the root is to 
 make these separations. 
 
EVOLUTION. 119 
 
 The first observation to make is that the square of 10 is 100, 
 of 100 is 10000, and so on. Hence, the square root of any 
 square number less than 100 consists of one figure, and of any 
 square number over 100 and less than 10000 of two figures, and 
 so on. Every two places in a power demanding one place in 
 its root. 
 
 Hence, to find the number of places or figures in a root, we 
 must separate the power into periods of two figures, beginning 
 at the unit's place. For example, let us require the square root 
 of 22-09. Here are two periods indicating two places in the 
 root, corresponding to tens and units. The greatest square in 22 
 is 16, its root is 4, or 4 tens =40. Hence a=40. 
 
 22 09(40+7^47 
 a?= 16 00 
 
 2a-K>=80-{-7=87 )609 
 609 
 
 Then 2a=80, which we use as a divisor for 609, and find it 
 is contained 7 times. The 7 is taken as the value of b, and 
 2+6, the complete divisor, is 87, which multiplied by 7 gives 
 the two last terms of the binomial square. 2a6-|-6 2 =560+49 
 609, and the entire root 40+7=47 is found. 
 
 Arithmetically, a may be taken as 4 in place of 40, and 1600 
 as 16, the place occupied by the 16 makes it 16 hundred, and 
 the ciphers are superfluous. Also, 2 may be considered 8 in 
 place of 80, and 8 in 60 (not in 609) is contained 7 times, <fec. 
 
 If the square consists of more than two periods, treat it as two, 
 and obtain the two superior figures of the root, and when obtain- 
 ed bring down another period to the remainder, and consider the 
 root already obtained as one quantity, or one figure. 
 
 For another example, let the square root of 399424 be ex- 
 tracted. 
 
 39-94-24 || 632 
 36 
 
 123 
 
 394 
 369 
 
 1262 
 
 2524 
 2524 
 
120 ELEMENTS OF ALGEBRA. 
 
 In this example, if we disregard the local value of the figures, 
 we have a=6, 2=12, and 12 in 39, 3 times, which gives 6=3 
 Afterwards we suppose =63, and 2a=126, 126 in 252, 2 times, 
 or the second value of b =-2. In the same manner, we would 
 repeat the formula of a binomial square as many times as we 
 have periods. 
 
 EXERCISES FOR PRACTICE. 
 
 1. What is the square root of 8836? Jlns. 94. 
 
 2. What is the square root of 106929? Jlns. 327. 
 
 3. What is the square root of 4782969? Jlns. 2187. 
 
 4. What is the square root of 43046721? Jlns. 6561. 
 
 5. What is the square root of 387420489? Jlns. 19683. 
 
 When there are whole numbers and decimals, point off periods 
 both ways from the decimal point, and make the decimal places 
 even, by annexing ciphers when necessary, extending the deci- 
 mal as far as desired. When there are decimals only, commence 
 pointing off from the decimal point. 
 
 EXAMPLES. 
 
 1. What is the square root of 10-4976? Jlns. 3-24. 
 
 2. What is the square root of 3271-4207? Jlns. 57-19+. 
 
 3. What is the square root of 4795-25731 ? JZns. 69-247+. 
 
 4. What is the square root of -0036 ? Ans. -06 
 
 5. What is the square root of -00032754? rfns. -01809+. 
 
 6. What is the square root of -00103041 ? rfns. -0321. 
 (Art. 75.) As the square of any quantity is the quantity mul- 
 tiplied by itself, and the product of T by j (Art. 64.) is p ; 
 
 hence to take the square root of a fraction we must extract the 
 square root of both numerator and denominator. 
 
 A fraction may be equal to a square, and the terms, as given, 
 not square numbers ; such may be reduced to square numbers. 
 
EVOLUTION. Itl 
 
 EXAMPLES. 
 
 What is the square root of yYj 
 
 Observe T y f = gf . Hence the square root is f . 
 
 1. What is the square root of T 9 / ? rfns. . 
 
 3. What is the square root of J r || ? Ans. J. 
 
 3. What is the square root of \\\ ? Ans. f . 
 
 4. What is the square root of fUJ? ^w*. 
 
 When the given fractions cannot be reduced to square terms, 
 reduce the value to a decimal, and extract the root, as in the last 
 article. 
 
 CHAPTER IV. 
 
 To extract the cube root of compound quantities. 
 
 (Art. 76.) We may extract the cube root in a similar manner 
 as the square root, by dissecting or retracing the combination of 
 terms in the formation of a binomial cube. 
 
 The cube of a+b is a s -\-3a?b-\-3ab 2 -\-b* (Art. 67). Now 
 to extract the root, it is evident we must take the root of the* 
 first term (a 3 ), and the next term is 3a 2 6. Three times the square 
 of the first Utter or terra of the root multiplied by the 2d term 
 of the root. 
 
 Therefore to find this second term of the root we must divide 
 the second term of the power (3a 2 6) by three times the square 
 of the root already found (a). 
 
 When we can decide the value of 6, we may obtain the com- 
 plete divisor for the remainder after the cube of the first term is 
 subtracted, thus : 
 
 The remainder is 3a z b-}-3ab 2 -\-b 3 
 
 Take out the factor 6, and 3a 2 -f-3a6-j-6 2 is the complete 
 
 divisor for the remainder. But this divisor contains 6, the very 
 
 term we wish to find by means of the divisor ; hence it must be 
 
 found before the divisor can be completed. In distinct algebraic 
 
 11 
 
122 ELEMENTS OF ALGEBRA. 
 
 quantities there can be no difficulty, as the terms stand separate, 
 and we find b by dividing simply 3a?b by 3a 2 ; but in numbers 
 the terms are mingled together, and b can only be found by trial. 
 
 Again, the terms 3a 2 -^-3ab-\-b 2 explain the common arithme- 
 tical rule, as 3a 2 stands in the place of hundreds, it corresponds 
 with the words : " Multiply the square of the quotient by 300 y " 
 "and the quotient by 30," (3a,) &c. 
 
 By inspecting the various powers of a-\-b, (Art. 67,) we draw 
 the following general rule for the extraction of roots : 
 
 Arrange the terms according to the powers of some letter ; 
 take the required root of the first term and place it in the quo- 
 tient : subtract its corresponding power from the first term, and 
 bring down the second term for a dividend. 
 
 Divide this term by twice the root already found for the SQUARE 
 root, three times the square of it for the CUBE root, four times 
 the third power for the fourth root, &c., 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: proceed in this manner till the whole root i9 
 determined. 
 
 EXAMPLES. 
 1. What is the cube root of x-}-Qx 5 4Qx?-\-QQx 64 ? 
 
 a* 
 Divisor 3# 4 ) 6r>= 1st remainder. 
 
 Divisor 3# 4 ) I2# 4 =2d remainder. 
 
 2. What is the cube root of 27a 3 +108 2 +144a+64 ? 
 
 Jlns. 3-f-4. 
 
 a- What is the cube root of a 3 Qa z x+I2ax 9 ' 8x* ? 
 
 Jlns. a 2x 
 
EVOLUTION. 123 
 
 4. What is the cube root of a,' 6 3^ 5 +5x 3 3^1 1 
 
 Ans. x z x 1 
 
 5. What is the cube root of a 3 Gcfb+lZab 2 Qb 3 ? 
 
 Am. a 26. 
 
 O I 
 
 6. What is the cube root of x*-\-3x-\ h- 
 
 * Extract the fourth root of 
 
 a 4 +8a 3 -f-24a 2 -f-32a-H 6(a-}-2 
 a 4 
 
 Ans. x-\ . 
 
 4a 3 ) 8a 3 , &c. 
 
 (Art. 77.) To apply this general rule to the extraction of the 
 cube root of numbers, we must first observe that the cube of 10 
 is 1000, of 100 is 1000000, &c.; ten times the root producing 
 1000 times the power, or one cipher in the root producing 3 in 
 the power ; hence any cube within 3 places of figures can have 
 only one in its root, any cube within 6 places can have only two 
 places in its root, &c. Therefore we must divide off the given 
 power into periods consisting of three places, commencing at the 
 unit. If the power contains decimals, commence at the unit 
 place, and count three places each way, and the number of pe- 
 riods will indicate the number of figures in the root.' 
 
 EXAMPLES. 
 
 1. Required the cube root of 12812904. 
 
 12-812-904(234 
 0=2 3 = 8 
 
 Divisor 3a 2 =12 )48 
 
 12167 = (23) 3 
 3(23) 2 = 1587) 6459 (4 
 
 12812904 = 
 
124 ELEMENTS OF ALGEBRA. 
 
 Here 12 is contained in 48, 4 times; but it must be remem- 
 bered that 12 is only a trial or partial divisor; when completed 
 it will exceed 12, and of course the next figure of the root can- 
 not exceed 3. 
 
 The first figure in the root was 2. Then we assumed a~2. 
 Afterwards we found the next figure must be 3. Then we as- 
 sumed a=23. To have found a succeeding figure, had there 
 been a remainder, we should have assumed a=234, &c., and 
 from it obtained a new partial divisor. 
 
 2. What is the cube root of 148877? Ans. 53. 
 
 3. What is the cube root of 571787? dns. 83. 
 
 4. What is the cube root of 1367631 ? rfns. 111. 
 
 5. What is the cube root of 2048383 ? rfns. 127. 
 
 6. What is the cube root of 16581375? dm. 255. 
 
 7. What is the cube root of 44361864 ? Ans. 354. 
 . What is the cube root of 100544625? Ans. 465. 
 
 (Art. 78.) When the power is not complete, and, of course, 
 its root surd, the methods of direct extraction are all too tedious 
 to be much used, and several eminent mathematicians have given 
 more brief and practical methods of approximation. 
 
 One of the most useful methods may be investigated as follows : 
 
 Suppose a and a-\-c two cube roots, c being very small in 
 relation to . a 3 and 3 +3a 2 c+3c 2 +c 3 are the cubes of the 
 supposed roots. 
 
 Now if we double the first cube (a 3 ), and add it to the second, 
 we shall have 3a >+3<?c+3ac>+c>. 
 
 
 
 If we double the second cube and add it to the first, we shall 
 have 
 
 As c is a very small fraction compared to , the terms con- 
 taining c 2 and c 3 are very small in relation to the others, and 
 the relation of these two sums will not be materially changed by 
 rejecting those terms containing c 2 and c 3 , and the same will 
 
 lhen be 3+3 etc 
 
 And 3o 3 +6a'c 
 
EVOLUTIOIS 7 125 
 
 The ratio of these terms is the same as the ratio* of a-\-c to 
 a-f2c. 
 
 Or the ratio is H ; 
 
 But the ratio of the roots a to +> is 1 -\ . 
 
 a 
 
 Observing again that c is supposed to be very small in rela- 
 tion to a. the fractional parts of the ratios ; and - are both 
 
 a-f-c a 
 
 small and very near in value to each other. Hence we have 
 found an operation on two cubes which are near each other in 
 magnitude, that will give a proportion very near in proportion to 
 their roots ; and by knowing the root of one of the cubes, by this 
 ratio we can find the other. 
 
 For example, let it be required to find the cube root of 28, true 
 to 4 or 5 places of decimals. As we wish to find the cube root 
 of 28, we may assume that 28 is a cube. 27 is a cube near in 
 value to 28, and the root of 27 we know to be 3. 
 
 Hence a, in our investigation, corresponds to 3 in this exam- 
 ple, and c is unknown; but the cube of a-\-c is 28, and a 8 
 is 27. 
 
 Then 27 28 
 
 2 2 
 
 54 56 
 
 Add 28 27 
 
 Sums 82 : 83 : : 3 : a-f-c very nearly. 
 
 Or (a-J-c)= 2 7 4 2 9 =3'03658+, which is the cube root of 28, 
 true to 5 places of decimals. 
 
 By the laws of proportion, which we hope more fully to in- 
 vestigate in a subsequent part of this work, the above propor- 
 tion, 82 : 83 : : a : a-}-c, may take this change . 
 82 : 1 : : a : c 
 
 Hence, c= T V c being a correction to the known root, 
 which, being applied, will give the unknown or sought root. 
 
 From what precedes, we may draw the following rule for find- 
 ing approximate cube roots : 
 
126 ELEMENTS OF ALGEBRA. 
 
 RULE. Take the nearest rational cube to the given number, 
 and call it an assumed cube ; or, assume a root to the given 
 number and cube it. Double the assumed cube and add the 
 given number to it ; also, double the given number and add 
 the assumed cube to it. Then, by proportion, as the first sum 
 is to the second, so is the known root to the required root. Or 
 take the difference of these sums, then say, as double of the 
 assumed cube, added to the number, is to this difference, so is 
 the assumed root to a correction. 
 
 This correction, added to or subtracted from the assumed root, 
 as the case may require, will give the cube root very nearly. 
 
 By repeating the operation with the root last found as an as- 
 sumed root, we may obtain results to any degree of exactness ; 
 one operation, however, is generally sufficient. 
 
 EXAMPLES. 
 
 1. What is the approximate cube root of 120? 
 
 tins. 4-93242-K 
 
 2. What is the approximate cube root of 8*5 ? 
 
 ns. 2-0408-f. 
 
 3. What is the approximate cube root of 63 ? 
 
 Jlns. 3-97905-}-. 
 
 4. What is the approximate cube root of 515 ? 
 
 tins. 8-01559-)-. 
 
 5. What is the approximate cube root of 16? 
 
 The cube root of 8 is 2, and of 27 is 3 ; therefore the cube 
 root of 16 is between 2 and 3. Suppose it 2-5. The cube of 
 this root is 15-625, which shows that the cube root of 16 is a 
 little more than 2-5, and by the rule 
 
 31-25 
 
 32 
 
 
 
 16 
 
 15-625 
 
 
 
 47-25 
 
 : 47-625 
 
 : : 2-5 : to the 
 
 required root. 
 
 47-25 
 
 : -375 
 
 :: 2-5 : -01984 
 
 
 
 Assumed root 
 
 2-50000 
 
 
 
 Correction 
 
 01984 
 
 Approximate root 2-51984. 
 
EVOLUTION. 127 
 
 We give the last as an example to be followed in most cases 
 where the root is about midway between two integer numbers. 
 
 This rule may be used with advantage to extract the root of 
 perfect cubes, when the powers are very large. 
 
 EXAMPLE. 
 
 The number 22-069'810'125 is a cube; required its root. 
 
 Dividing this cube into periods, we find that the root must 
 contain 4 figures, and the superior period is 22, and the cube root 
 of 22 is near 3, and of course th< whole root near 3000; but less 
 than 3000. Suppose it 2800 and cube this number. The 
 cube is 21952000000, which being less than the given number, 
 shows that our assumed root is not large enough. 
 
 To apply the rule, it will be sufficient to take six superior 
 figure^ of the given and assumed cubes. Then by the rule, 
 
 220698 
 2 
 
 439040 4413% 
 220698 219520 
 
 659738 : 660916 : : 2800 
 659738 
 
 659738 : 1178 :: 2800 
 2800 
 
 942400 
 2356 
 
 659738)3298400(5 
 3298690 
 
 Assumed root, 2800 
 Correction, 5 
 
 True root, 2805 
 
 The resuit of the last proportion is not exactly 5, as will be 
 seen by inspecting the work ; the slight imperfection arises from 
 the rule being approximate, not perfect. 
 
 When we have cubes, however, we can always decide the unit 
 figure by inspection, and, in the present example, the unit figure 
 
128 ELEMENTS OF ALGEBRA. 
 
 in the cube being 5, the unit figure in the root must be 5, as no 
 other figure when cubed will give 5 in the place of units. 
 
 [For several other abbreviations and expedients in extracting 
 cube root in numerals, see Robinson's Arithmetic.] 
 
 (Art. 79.) To obtain the 4th root, we may extract the square root 
 of the square root. To obtain the 6th root, we may take the 
 square root first, and then the cube root of that quantity. 
 
 To extract odd roots of high powers in numeral quantities is 
 very tedious and of no practical utility ; we therefore give no ex- 
 amples. 
 
 (Art. 80.) Some radical quantities may be simplified, and oth- 
 ers are not susceptible of simplification. For instance, the square 
 root of 75, written ^/75, is equal to 5 times the square root of 
 3; that is, ^/75=5^/3. But the square root of 71 can not be 
 fully expressed except by the sign, thus, ^/71, because 71 con- 
 tains no square factor. 
 
 It is obvious that *J4Q==*J8 :r 5==* i JS *^Ss=S(5)lj but 
 
 39 and 41 containing no cube factor, their cube roots can only 
 be indicated by the radical sign over them. 
 
 When simplification is possible, it can be effected by the 
 following 
 
 RULE. Separate the quantity into two factors, one of which 
 is a perfect power of the required root. Extract the root of 
 that factor and prefix the result as a coefficient to the other 
 factor placed under the radical sign. 
 
 We give the following examples for practice : 
 
 1. Reduce the square root of 75 to lower terms, or reduce 
 
 tins, 5^/3. 
 
 2. Reduce J98a 2 to lower terms. Jlns. 7 > /2. 
 
 3. Reduce jVZx 9 -y to lower terms. Jlns. 2x,j3y. 
 
 4. Reduce 3 /54.T 4 to lower terms. Jlns. 3x 
 
 5. Reduce 4 ^/T08 to lower terms. Jlns. 
 
 6. Reduce Jx? V to lower terms. Jlns. xjx a 2 . 
 
EVOLUTION. 20 129 
 
 7. Reduce 3 /32a 3 to lower terms. Ans. 2 3 > /4. 
 
 8. Reduce J'ZScfx? to lower terms. Ans. 2axj7a 
 
 9. Reduce 74 J to lower terms. 
 
 Where terms under the radical are fractional, it is expedient 
 to reduce the denominator to a power corresponding to the radi- 
 cal sign ; then by extracting the root there will be no fraction 
 under the radical. 
 
 The above example may be treated thus : 
 
 -y K'-J X33=ftV88. Aw. 
 
 We divided || into the factors and y ; the first factor is a 
 square ; the other factor, y , we multiply both numerator and 
 denominator by 3, to make the denominator a square. 
 
 In like manner reduce the following : 
 
 10. Reduce 3 7^ to more simple terms. Ans. %*J\Q. 
 
 11. Reduce 3 7V to more simple terms. Ans. 
 
 12. Reduce to more sim ^ e terms. Ans. 
 
 13. Reduce 3 f Ja^-\-a 3 b 2 to more simple terms. 
 
 Ans. a 
 
 14. Reduce ,J~ to more simple terms. Ans. 
 
 (Art. 81.) Radical quantities may be put into one sum, or the 
 difference of two may be determined, provided the parts essen- 
 tially radical are the same. 
 
 Thus the sum of ^/8 and ,/72 is 8^/2 and their 
 difference is 4^/2 
 
 For ^8= 
 
 And 72= 
 
 Sum 872 
 
 Difference 472" 
 
 When radical quantities are not and cannot be reduced to the 
 
130 ELEMENTS OF ALGEBRA. 
 
 same quantity under the sign, their sum and difference can only 
 tie taken by the signs plus and minus. 
 
 EXAMPLES. 
 
 1. Find the sum and difference of t J\.Qa 2 x and ,j4a 2 x. 
 
 Ans. Sum, Qajx ; difference, 2a t jjs. 
 
 2. Find the sum and difference of /128 and ^72. 
 
 Jlns. Sum, 14^/2 ; difference, 
 
 3. Find the sum and difference of 3 ^/135 and 3 N /40. 
 
 Am. Sum, 5 3 ,y5 ; difference, 
 
 4. Find the sum and difference of V 108 and 9 V 4. 
 
 Jins. Sum, 12 3 ^/4; difference, 6 3 > /4. 
 
 5. Find the sum and difference of ^/| and J'l. 
 
 Ans. Sum, f^2 ; difference, J^/2 U 
 
 6. Find the sum and difference of 3 ^/56 and 3 > /189. 
 
 fins. Sum, 5 3 ,y7; difference, 3 ^/?. 
 
 7. Find the sum and difference of 3Ja z b and 
 
 Am. Sum, 14-3a6; difference, 12^ 
 
 (Art. 82.) We multiply letters together by writing them one 
 after another, as abxy. If they are numeral quantities, their 
 product appears as a number ; if two or more of them are nu- 
 meral, the product of these quantities will appear as a number. 
 
 This fundamental principle of multiplication may be applied to 
 the multiplication of surds. Let it be required to multiply 
 5^/2 by 3^/7. Here suppose =5, 6=3, x=j2, y=,Jl. 
 Then the product of 5^/2T by S^/Y is abxy or 15^/2X7= 
 15^14. Hence, for the multiplication of quantities affected by 
 the same radical sign, we draw the following 
 
 RULE. Multiply the rational parts together for the rational 
 part of the product, and the radical parts together for the 
 radical part of the product. 
 
EVOLUTION. 131 
 
 EXAMPLES. 
 
 1. Required the product of 5^/5 and 3^/8. 
 
 Product reduced, tins. 
 
 2. Required the product of 4J12 and 3^/27 Am. 
 
 3. Required the product of 3,/2~ and 2J~8. .#ns. 24. 
 
 4. Required the product of 2yl4 and 3 3 ^/47 
 
 Ans. 12 V7. 
 
 5. Required the product of 2^5~and 2^/To. Jlns. 
 
 (Art. 83.) We multiply different powers of the same quantity 
 together by simply adding the exponents as, a 3 multiplied by a 3 
 is a 8 , and this holds when the exponents are fractional, as a 2 into 
 a 3 ; for the product we simply add J and -J, which make | to 
 write over a for the product of a 2 into a? which gives cfi. 
 
 When the radical quantities are different, as () 2 and 6^, and 
 the exponents different, are to be multiplied together, or one to be 
 divided by another, we better lay aside all rules and apply that 
 powerful engine, an equation. For illustration: 
 
 Required the product of a 2 into b^. 
 
 The product is P, then we have 
 
 Raise both members to the 6th power, then, 
 
 Conceiving the second member to be a single quantity, and 
 taking the 6th root, will give us 
 
 EXAMPLES. 
 
 1. Required the product of (+&)' and (a-\-b)*. 
 
 Jlns. 
 
 2. Required the product of JT~ and *JT, rfns. (7 5 ) 
 
132 ELEMENTS OF ALGEBRA. 
 
 (Art. 84.) If we divide one quantity by another, the product 
 of the quotient and the divisor must equal the dividend. 
 Divide the Q Jl2 by ^/2~: the quotient is Q. 
 Then, 
 
 By squaring, 
 
 Cubing 8 6=72. 
 
 Whence, Q*=9 Q 3 =3 
 
 3. Required the product of 2J3 and 3(4)^. 
 
 _ _ 
 4. Required the product of 3 ^/15 and ^10. 
 
 Am. 6^/225000. 
 
 EXAMPLES. 
 
 1. Divide 4^/50 by 2</57 Ans. 2JW. 
 
 2. Divide 6 yiOO by 3 y~5~. Am. 2 y20i 
 
 3. Divide Jl by B 4 J'7. 
 
 Let a=7. Then the example is, to divide the a 2 by a a , or 
 a by a ; quotient a?. As we always subtract the exponents 
 of like quantities to perform division (Art. 17) ; therefore 7 must 
 be the required quotient. 
 
 4. Divide 6^/54 by 3J2. Ans. 
 
 5. Divide (tfb z cl?Y by A -#rcs. 
 
 6. Divide (I6tfI2a 2 x)* by 2. ^?is. (4a 3xf. 
 (Art. 85.) In the course of algebraical investigations, we might 
 
 fall on the square root of a minus quantity, as J a, ,J 1, 
 J b, &c., and it is important that the pupil should readily 
 understand that such quantities have no real existence ; for no 
 quantity, either plus or minus, multiplied by itself will give 
 minus a, minus b, or minus any quantity whatever ; hence there 
 is no value to J a, &c., and such symbols are said to be im- 
 possible or imaginary. 
 
EVOLUTION. 133 
 
 (Art. 86.) The square and cube root of any quantity, as a, 
 being expressed by .Ja and 3 Ja, and as by involving the root 
 we obtain the power, hence the square of ,Ja, is a; and the 
 cube of I/a, is a. Hence, removing the sign involves to the 
 corresponding power. 
 
 EXAMPLES. 
 
 1. What is the square of J3ax 1 Jlns. Sax. 
 
 2. What is the cube of V% 2 ? J3ns. 6y 2 . 
 
 3. What is the square of Ja 2 x 2 ? JLns. a 2 x 2 . 
 
 4. What is the square of /vL_ ? rfns. - 
 
 6-f-z 6+0? 
 
 5. What is the cube of *Jl+a*. rfns. l+a. 
 
 (Art. 87.) When we have two or more quantities, and the 
 radical sign not extending over the whole, involution will not 
 remove the radical, but will change it from one term to another. 
 Thus, Jx-\-a the square will be x-{-2a,Jx-\-a* ; the radical 
 sign is still present, but not in the first term. 
 
 PURE EQUATIONS. 
 CHAPTER IV. 
 
 (Art. 88.) Pure equations in general are those in which a com- 
 plete power of the unknown quantity is contained, and no special 
 artifice is requisite to render the power complete. 
 
 The unknown quantity may appear in one or in several terms ; 
 when it appears in several, its exponents will be regular, descend- 
 ing from a higher to a lower value, or the reverse. 
 
 In such cases we must reduce the equation by evolution. 
 
 Like roots of equal quantities are equal. (Ax. 8.) 
 
 EXAMPLES. 
 
 1. Given 3x 2 9=66 to find the value of x. 
 Solution, 3,r 2 =^75 # 2 =25. Hence, #=5. 
 
134 ELEMENTS OF ALGEBRA. 
 
 We place the double sign before 5, as we cannot determine 
 \vhether25 was produced by the square of +5 or of 5. 
 
 In practical problems, the nature of the case will commonly 
 determine, but in every abstract problem we must take the 
 double sign. 
 
 S5. Given x z 6#-|-9=a 2 to find the value of x. 
 By evolution, x 3=a. Hence, a?=3#, Jlns. 
 
 3. Given x 3 x z -{-ix^=a s b 3 to find x. 
 
 Take cube root and x ^ab; and x=ab-\-\, Jlns. 
 
 4. Given a; 4 2^+1 = 16 to find the value of x. 
 
 Jlns. x=^ or 
 
 5. Given x z 4#+4=64 to find the value of x. 
 
 Jlns. #=10, or 6. 
 
 6. Given x 2 y 2 -}-2xy-}-l=4 : x 2 y 2 and x=2y to find the values 
 of x andy. Jlns. #==t,/2^ y 
 
 7. Given x 2 -\-y 2 =I3 and x 2 2/ 2 =5 to find the values of x 
 and y. Jlns. #=3 y=2. 
 
 (Art. 89.) The unknown quantity of an equation is as likely 
 to appear under a radical sign as to be involved to a power. In 
 such cases we free the unknown quantity from radicals by involu- 
 tion (Art. 86.), having previously transposed all the terms not 
 under the radical to one side of the equation, the radical being 
 on the other. 
 
 9. Given ,Jx-{-l=a 1 to find the value of x. 
 
 By involution, #-{-l=a 2 2a-r-l. Hence, xa 2 2a. 
 
 10. Given ,J 1 2 -f- x =3 -\-Jx to find the value of x. 
 Square both sides, and we have 
 
 Drop 9-}- x and 3=6^; or Jx~=s. x=%, Jim. 
 
 11. Given Jx 16=8 Jx to find x. Jlns. #=25. 
 
PURE EQUATIONS. 135 
 
 X ax J x 
 
 12. Given 1= = - to find x. 
 
 v# x 
 
 Multiply by ^/^ observing that x divided by x gives i ; and 
 we have x ax=l, 
 
 Or (I d)x=l, 
 
 Therefore x=- . 
 
 1 a 
 
 13. Given .= = r to find the value of x. 
 
 By clearing of fractions, we have 
 
 tf+34/H-168=;r 4-42^+152. 
 Reducing, 16=8^ 
 
 By division, 2= Jx, or 4=x. 
 
 To call out attention and cultivate tact, we give another solu- 
 tion. Divide each numerator by its denominator, and we have 
 
 Drop unity from both sides, and divide by 8 ; we then have 
 3 4 
 
 Clearing of fractions, 3 t Jx-\-l8=4 i Jx-{-lQ 
 
 Dropping equals, 2= ,Jx. Hence, x=4. 
 
 14. Given ,J x-{- ,J a-\-x= . = to find x. Jlns. x=\ 
 
 V a-\-x 
 
 2a z 
 
 15. Given x-}-Ja z -\-x 2 =j to find x. 
 
 Jlns. a?=a,u 
 
 . Given a?+a=Va 2 +a?V6 2 +ic 2 to find x. 
 
 b*4a 2 
 
 Jlns. x= 
 
 40 
 
136 ELEMENTS OF ALGEBRA. 
 
 Jftx 2 4tjQx 9 
 17. briven ^ = ^ to find a?.* Jlns. 
 
 18. Given V64-f- x 2 -^Sx= to find a:, dns. # 
 
 15 
 
 19. Given J&+x-}-Jx= - to find a?. rfns. #=4 
 
 V 5 +* 
 
 20. Given Jx+JxJx Jx=-( --^j- ) to find x. 
 
 2\#-rv#/ 
 
 25 
 Jin,. *. 
 
 Because a 2 6 2 =( a 
 
 We infer that 5z 9=(^/5a?-f3) (j5x3) 
 
 Therefore, =^5^ 3. The given equation then 
 
 -- 
 
 becomes .^53: 3=1-1^- -- . 
 2 
 
 Now assume 
 
 Then y=l+iy. Consequently, 2/=2. Returning to equa 
 tion (w5), we have ^5a; 3=2 
 
 t j5x=5. Therefore, x=5, Ans. 
 
 Jaxb 3j~ax2b Qb* 
 
 22. Given ^- ==-^ - to find x. Am. x= 
 Jax+b 3,Jax-{-5b 
 
 (Compare 22 with examples 13, and 17.) 
 
 23. Given (1+^7^+12) = l+a? to find x. rfns. 
 21. Given ^!!_ 1 + ^=9, to find x. 
 
 * See 2d solution to Equation 13. 
 
PURE EQUATIONS. 137 
 
 25. Given a 2 2ax-{-x z =b, to find x. ns. #= Jb 
 
 \ 1/3 
 
 26. Given =-^ r to find x. 
 
 Jlns. x=i. 
 
 4 
 
 27. Given r =|, to find x. Ans. x=5. 
 
 x 2 2#-{-l 
 
 28. Given -==l-\-. to find x. Aw. x= 
 
 Q/ 
 
 29. Given x--x 9= to find a?. 
 
 30 Given =- to find a?. ^ns. a=4 
 
 31. Given = L to find the value of a?. 
 
 Multiply the first member, numerator and denominator, by 
 x-\- ,Jx a), then both members by a, and extract square root. 
 
 32. Given ' q l^ =&> to find a-. 
 
 Assume a-{-#=2/. Then the equation becomes 
 
 '=&. Hence, v= 
 
 And 
 
 (Art. 90.) To resolve the following examples, requires a de- 
 
 gree of tact not to be learned from rules. Quickness of percep- 
 
 tion is requisite, as well as sound reasoning. Quickness to per- 
 
 ceive the form of binomial squares, and binomial cubes, and a 
 
 12 
 
138 ELEMENTS OF ALGEBRA. 
 
 readiness to resolve quantities into simple or compound factors, 
 as the case may require. 
 
 1. Given x*-}-2x=9-+- to find the value of x 
 
 Multiply by x, and x 3 +2x*=9x+18. 
 Separate into factors, thus : (a?-f-2)a? 2 =(#-{-2)9. 
 Divide by the common factor, tf-J-2, and x*=9, or #=3. 
 
 Given i a? 2 ,^~ 24 to find the values of x and y. 
 Add the two equations together, and we have 
 
 Extract square root, and x-\-y=6. (.#) 
 
 From the first equation we have (x-\-y)x=l.2. (B) 
 Divide equation (B} by (.#), and a?=2. 
 
 This example required perception to recognise the binomial 
 square, and also to separate into factors. 
 
 | Q / 
 
 3. Given x z -{-y z = - and xv= - to find the values 
 xy xy 
 
 of x and y. 
 
 From the first equation subtract twice the second, and 
 
 Therefore, (x #) 3 =1, and x y=l. 
 Continuing the operation, we shall find #=3, and y=2. 
 
 4. Given x 2 y-{-xy 2 =lSO and x s +y*= 189, to find the values 
 of x and y. J2ns. x=5 or 4 ; y=4 or 5. 
 
 To resolve this problem, requires the formation of a cube, or 
 to resolve quantities into factors. 
 
 5. Given x*-\-y*=(x-{-y}xy, and x-}-y=4, to find the 
 values of x and y. Jlns. x=2] 2/=2. 
 
 6. Given x-{-y : x :: 7 : 5, and xy+y*=I26i to find tha 
 values of x and y. Jlns. #=15, ?/=6 
 
PUKE EQUATIONS. 139 
 
 7. Given x y : y :: 4 : 5, and x 2 +4y*=l8], to find the 
 values of x and y. rfns. #=9, y=5. 
 
 8. Given Jx+Jy > Jx Jy :: 4 : 1, and x */=16, to 
 find the values of x and y. Ans. #=25, i/=9. 
 
 9 Given -^+|-f7=9 to find a?. Am x=7. 
 
 y o 4 
 
 10. Given x+y : x-y :: 3 = 1 ? to find x and 
 
 And a? 3 y =56 5 
 
 11. Given y?y-}-xy z =30 } 
 
 1 i 5 I to find x and 
 
 And + ' = 
 
 Observe that xy(x-\-y)=x?y-{-xy z . 
 
 Clear the 2d equation of fractions, and y-}-x or x-\-y= ~. 
 
 Now assume x-\-y=s, and xyp. Then the original equa- 
 tions become s/?=30 
 
 And 6s=5p 
 
 Equations which readily give s and j9, and from them we de- 
 termine x and y. 
 
 N. B. When two unknown quantities, as x and y, produce 
 equations in the form of 
 
 x+y=s (1) 
 
 And xy=p (2) 
 
 such equation can be resolved in the following manner : 
 
 Square (1), and x*-}-2xy-{-y 2 =s 2 
 Subtract 4 times (2) 4xy ==4p 
 
 I Diff. is x 2 2xy+y 2 =s 2 4p 
 
 By evolution x yJs* 4p (3) 
 
 Add equation (1) and (3), and we have, 
 
 (4) 
 
 Sub. (3) from (1), and 2y=sJs 2 4p (5) 
 
140 ELEMENTS OF ALGEBRA. 
 
 To verify equations (4) and (5), add them and divide by 2, 
 and we have x-\-y=s. Multiply (4) by (5), and divide by 4, 
 and we have xy=p. 
 
 (Art. 91.) No person can become very skilful in algebraic 
 operations as long as he feels averse to substitution; for judi- 
 cious substitution stands in the same relation to common algebra, 
 as algebra stands to arithmetic. The last example is an illustra- 
 tion of this remark. To acquire the habit of substituting, may 
 require some extra attention at first, but the power and advantage 
 gained will a thousand fold repay for all additional exertion. 
 
 When two equations involve quantities in the form (x-\-y) and 
 xy, or xy, and xy, and any product of these quantities, the 
 equations are easily solved. 
 
 The following examples will illustrate : 
 
 12. Given x++y= 197 find and 
 And x 2 -\-xy+y 2 =133$ 
 
 Put x -\-y =5, and 
 Then s+p= 19 () 
 
 And s 2 /=133 (E) 
 
 Divide (B) by (.#), and we have sp=7, &c. 
 
 Ans. #=9 or 4, y=4 or 9. 
 
 13. Given x 4 -}-2x z y 2 +y 4 =l2QQ-^4xy(x 2 -{-xy-{-y 2 ) and 
 x y=4, to find x and y. 
 
 Put x*-\-y 2 =s, and xy=p 
 
 Then the first equation becomes s 2 =1296 4p(s-\-p) 
 Multiply and transpose, and s 2 -\-4sp-\-4p 2 =l29Q 
 
 Square root +2p=36 
 
 But s-t-2/?=a: 2 -{-2^-f-2/ 2 ==36 
 
 Therefore x+= 6, or 
 
 Rejecting imaginary quantities, we find x=5 or 1, and 
 ?/=l or 5. 
 
 X 2 'U Z 
 
 14. Given -- =G, and x-{-y-\-xy=U, to find the values 
 x y 
 
 of x and y. Jlns. x=5 or 1, y=l or 5 
 
PURE EQUATIONS. HI 
 
 15. Given x*-{-y s =2xy(x-\-y], and xy=l6, to find the 
 values of x and y. Jlns. #=2^54-2, 7/^2^/5 2. 
 
 16. Given x 3 -\-y 3 , and x z y-}-xy*=a, to find the relative 
 values of x and y. tfns. xy. 
 
 1 7. Given x-\-y : x :: 5 : 3, and xy=Q, to find a? and y. 
 
 Ans. #=3, 2/=2. 
 
 18. Given #-f-?/ : a: :: 7 : 5, and a?7/-l-?/ 2 =126, to find a: 
 and y. Jlns. a? =15, i/=6. 
 
 19. Given ,T 2 -)-?/ 2 =, and xy=b, to find the values of 
 x and y. Ans. x=.* t Ja-}-2b-{-% l Ja < 2,b. 
 
 (Art. A)* Equations in the form of a? 4 2ax 2 -}-a z =b, require 
 for their complete solution, the square root of an expression in 
 the form of aJb ; for by extracting the square root of the 
 equation, we have 
 
 Hence x= \/ a 
 
 The right hand member of this equation is an expression well 
 known among mathematicians as 
 
 A BINOMIAL SURD. 
 
 Expressions in this form may or may not be complete powers ; 
 and it is very advantageous to extract the root of such as are 
 complete, for the roots will be smaller, and more simple quantities, 
 in the form of a'i^/6', or of .Ja'^jW. 
 
 Let us now investigate a method of extracting these roots ; and, 
 for the sake of simplicity, let us square 34-^7. 
 
 By the rule- of squaring a binomial, we have 94-6^/7-1-7, 
 Or, le+G/T; 
 
 Conversely, then, the square root of 164-6^/7, is 3+77. 
 
 * That the same Articles may number the same in both the School and Col- 
 lege Edition, we shall designate all additional Articles, in this volume, by 
 A, B, &c. 
 
142 ELEMENTS OF ALGEBRA. 
 
 But when a root consists of two parts, its square consists of tht 
 sum of the squares of the two parts, and twice the product of 
 the two parts. 
 
 Now we readily perceive that 16 is the sum of the squares of 
 the two parts expressing the root; and 6^/7 , the part containing 
 the radical, is twice the product of the two parts. 
 
 To find what this root must be, let a? represent one part of the 
 root, and y the other : 
 
 Then aHy=16, (1) 
 
 And 2xy=&J7, (2) 
 
 Add equations (1) and (2), and extract square root, and we 
 
 have 
 
 (3) 
 Subtract equation (2) from (1), and extract square root, and 
 
 we have xy=^lQ Gj7~. (4) 
 
 Multiply (3) and (4), and we have 
 
 a? 2 2/2=^/256 252= > /T=:2 ; (5) 
 Add (1) and (5), and we have 
 
 #=, or 
 Sub. (5) from (1), and 2# 2 =14, or y=, 
 
 Wberie x-\-y, or the square root of 16-|-6^7 is S-f-^/7. 
 f e shall now be more general. 
 
 Take two roots, one in the form of 
 and one in the form of 
 Square both, and we shall have 2 2a > /6-f-6, 
 
 And a2,Jab-\-b. 
 
 In numerals, and, in short, in all cases, the sum of the squares 
 of the two parts of the root, as ( 2 +6), in the first square, and 
 (a-f-6), in the second, contains noradical sign ; and the sum of 
 these rational parts may be represented by c and c', and the 
 squares represented in the form of 
 
PURE EQUATIONS. 143 
 
 c 2a/6, 
 or of c'2jab. 
 
 Hence, generally, if we represent the parts of the roots by y 
 and y, we shall have x?-}-y Z: = the sum of the rational parts, and 
 2xy= the term containing the radical. 
 
 The signs to x and y must correspond to the sign between 
 the terms in the power. If that sign is minus, one of the signs 
 of the root will be minus ; it is indifferent which one. 
 
 EXAMPLES. 
 
 1. What is the square root of ll-j-6^/2"? Jlns. 
 
 2. What is the square root of 7-f-4 A /3~7 Jlns. 
 
 3. What is the square root of 7 2^/10? 
 
 Jlns. 75 J2 or </2 J&1 
 
 4. What is the square root of 94+42^5'? Jlns. 7+3^5". 
 
 5. What is the square root of 28+1073? Jlns. 5-f-^ 
 
 6. What is the square root of 
 
 np-}-2m z 2m t Jnp-\-m 2 1 
 
 In this example put anp-{-m*, and x and y to represent 
 the two parts of the root, 
 
 Then x 2 +y 2 m 2 +a, 
 
 and 2xy = 2m Ja. 
 
 Jlns. db( Jnp-{-m z m) . 
 
 . What is the square root of bc-+-2bjbc b 2 1 
 
 Jlns. 
 
 8. What is the sum of 
 
 Jlns. 10. 
 
 9. What is the sum of ^/l 1+6^/2 and A /7 
 
 3+^/5*. 
 10 
 
 Am. S-\-2^5 or 4. 
 
144 ELEMENTS OF ALGEBRA. 
 
 In a similar manner we may extract the cube root of a bino- 
 mial surd, when the expression is a cube ; but the general solu- 
 tion involves the solution of a cubic equation, and, of course, 
 must be omitted at this place ? and, being of little practical utility, 
 we may omit it altogether. 
 
 (Art. 92.) Fractional exponents are at first very troublesome 
 to young algebraists ; but such exponents can always be ban- 
 ished from pure equations by substitution. For the exponents 
 of all such equations must be multiples of each other; otherwise 
 they would not be pure, but complex equations. 
 
 To make the proper substitution, place the unknown quanti- 
 ties having the lowest exponents equal to other simple quantities, 
 as P and Q ; and let this be a general direction. 
 
 EXAMPLES. 
 
 1 4. 2 
 
 1. Given x 3 -[-y 5 = G, and re 3 -}-;y s =20, to find the values 
 
 of x and y 
 
 2_ \_ 
 
 By the above direction, put x 3 =P, and y 5 Q. 
 Squaring these auxiliaries, or assumed equations, 
 
 And x^=P 2 , and y~* = Q\ 
 Now the original equations become 
 =Q (1) 
 =2Q (2) 
 
 By squaring equation (1), P 2 -f2P#-f-Q 2 =36. 
 Subtracting equation (2), we have 2PQ=IQ. 
 Subtracting this last from equation (2), and we have 
 
 By extracting square root P Q=2 
 But by equation (1), P J rQ= 6 
 
 Therefore, P=4 or 2, and Q=2 or 4, 
 
 fi 
 JL &VUV W, ~ =4 or 2, and?/ 5 =2 or 4. 
 
 Square root ^=2 or (2)* ) 
 
 ' 3 V y=32 or 1024 
 
 Cubing gives x=S or (2)* J 
 
PURE EQUATIONS. 145 
 
 2. Given xy z +y=2l, and a?y+2/ 2 =333, to find the values 
 of x and y. 
 
 By comparing exponents in the two equations, we perceive 
 that if we put xy 2 =P, and y=Q, the equations become 
 
 P+Q= 21 
 P*+ $2=333 
 
 Solved as the preceding, gives P=18, Q=3. 
 From which we obtain a?=2, or T ^ , y=3 or 18. 
 
 
 3. Given x z +x?y*= 208 
 
 24 to find the values of x and y. 
 
 And 
 
 
 2 
 
 Assume x*=P, and y* = Q. 
 
 By squaring and cubing these assumed auxiliary equations, 
 we have 4 
 
 Seek the common measure (if there be one) between 208 and 
 1053. 
 
 From the above substitution, the given equations become 
 P*-\-P 2 Q= 208=13.16 (1) 
 $3-}-$2p=:1053=13.81 (2) 
 Separate the left hand members into factors, and 
 \3.lG (3) 
 
 13.81 (4) 
 
 Divide equation (4) by (3), and we have 
 
 Q 2 81 & 9 
 
 -^2=. Extracting square root -^=- 
 
 9P 
 Or QT-- Substitute this value of Q in equation (1), 
 
 q/33 
 
 And / M -f-^-=13.16 
 4 
 
 Or 4P4-9P 2 =13.64 
 That is, 13P 3 =13.6i 
 Hence, P 3 =64 or P=4 
 
 13 
 
146 ELEMENTS OF ALGEBRA 
 
 But 2 =/ w =64. Therefore, #=8 
 
 As Q= and P=4, Q9 and #=27. 
 
 3 3 3 3_ 1 
 
 4. Given # 2 -f-# 4 i/ 4 -H/ 2 = 1009 =a 
 
 33 r to find a? and y. 
 
 And orM-a^v'-H/ 3 =582193=6 
 
 Put x*=P and #* = # 
 
 Then ar*=/ and y^=(f 
 And a- 3 = /* and y z = #* 
 Our equations then become 
 
 /_!_ PQ-L.Q* 1 Equations having no fractional exponents, 
 
 p\\piru\Qi t f an( l are of the same form as in Problem 
 
 J 12. (Art. 9 1.) 
 
 Jim. #=81 or 16, t/=16 or 8]. 
 
 5. Given x-{-x*y* = l2 } 
 
 j j r to find the values of x and y. 
 
 And y+x*y*= 4 J 
 
 ^ns. x=9, y=l 
 
 6. Given x-l-*V= | 
 
 j j r to find the values of x and y. 
 
 And y\-x*y*=zb } 
 
 a a* b* 
 
 rt jyns. x r^- v= ;-, 
 
 a+b 9 a-\-b 
 
 333 1 
 
 T. Given x 2 +x*y*=a [ 
 
 3 j- to find the values of x and y 
 
 And 1 3 
 
 8. Given Jx-\-Jy : Jx Jy :: 4 : I, and a: 1/=16, to 
 find the values of x and y. As. a?=25, y=9. 
 
 9. Given *+y~ 5 tQ find 
 
 And ar =13 
 
 fins. x=9 or 4, y=4 or 9 
 
PURE EQUATIONS. 147 
 
 JO. Given x-\-y : x y :: 3 : 1 } to find the values of 
 And * 3 */ 3 =56 $ * and y. 
 
 fins. #=4, 2/=2. 
 11. Given x -\-y =35 } 
 
 r to find the values of x and y. 
 And 3+7,3=5 J 
 
 CHAPTER V. 
 Problems producing Pure Equations. 
 
 (Art. 93.) We again caution the pupil, to be very careful not to 
 involve factors, but keep them separate as long as possible, for 
 greater simplicity and brevity. The solution of one or two of 
 the following problems will illustrate. 
 
 1. It is required to divide the number of 14 into two such parts, 
 that the quotient of the greater divided by the less, may be to the 
 quotient of the less divided by the greater, as 16 : 9. 
 
 fins. The parts are 8 and 6. 
 
 Let x= the greater part. Then 14 x= the less. 
 
 x 14 x 
 
 Per question, ~- - : -- : : 16:9. 
 14 x x 
 
 AT I ' 1 9X 16(14 X) 
 
 Multiply extremes and means, and - - = - - '- 
 
 14 x x 
 
 Clearing of fractions, we have 9# 2 =16(14 xf 
 By evolution, 3^=4(14 x) =4. 14 4x 
 
 By transposition, 7#=4.14 
 By division, #=4.2=8, the greater part. 
 
 14 "X 
 
 Had we actually multiplied - - by 16, in place of indi 
 
 rating it, the exact value and form of the factors would have been 
 lost to view, and the solution mighthave run into an adfected qua- 
 dratic equation. 
 
 The same remark may be applied to many other problems, 
 and many are put under the head of quadratics that may be re 
 duced by pure equations. 
 
148 ELEMENTS OF ALGEBRA. 
 
 2. Find two numbers, whose difference, multiplied by the 
 difference of their squares, is 32, and whose sum, multiplied by 
 the sum of their squares, gives 272. 
 
 If we put x= the greater, and y= the less, we shall have 
 
 (z-.yX^-yH 32 (i) 
 
 And (*+2/)(* 2 -f3/ 2 )=272 (2) 
 
 Multiply these factors together, as indicated, and add the equa- 
 tions together, and divide by 2, and we shall have 
 
 o; 3 +y 3 =152 (3) 
 
 If we take (1) from (2), after the factors are multiplied, we 
 shall have 2 t ri/ 2 -f 2arty=240, or xy(x+y} = \'2Q (4) 
 
 Three times equation (4) added to equation (3) will give a 
 cube, &c. A better solution is as follows : 
 
 Let x-}-y= the greater number, and x y the less. 
 Then 2x= their sum, and 2y= their difference. 
 Also, 4xy is equal the difference of their squares, and 
 2x 2 -\-2y z = the sum of their squares. 
 
 By the conditions, 2yX4xy= 32 
 
 And 2#(2or i 4-2*/ 2 )=272 
 
 By reduction, xy*= 4 
 And 
 
 By subtraction, x 3 =64 or #=4 
 
 Hence, y=l, and the numbers are 5 and 3. 
 
 We give these two methods of solution to show how much 
 depends on skill in taking first assumptions. 
 
 3. From two towns, 396 miles asunder, two persons, Ji and Z?, 
 set out at the same time, and met each other, after traveling as 
 many days as are equal to the difference of miles they traveled 
 per day, when it appeared that A had traveled 216 miles. How 
 many miles did each travel per day? Let x=JPs rate, nnd i/= 
 fl's rate. 
 
 Then x y= the days they traveled before meeting. 
 
 By question, (x y)x=21G, and (x y)y=lSQ. 
 
 216 180 6 5 
 
 vonsequently, -- = - or -=-. 
 y ' x y x y 
 
PURE EQUATIONS. 14 
 
 Therefore, ?/=|x, which substitute in the first equation, and 
 
 x 2 
 eve have (z f#)#=216, or --=216=6X6X6. 
 
 By evolution, #=36; therefore y= 30. 
 
 4. Two travelers, A and B, set out to meet each other, A 
 leaving the town C, at the same time that B left D. They traveled 
 the direct road between C and D ; and on meeting, it appeared 
 that A had traveled 18 miles more than B, and that A could have 
 gone 's distance in 1 5| days, but B would have been 28 days 
 in going ^'s distance. Required the distance between C 
 and D. 
 
 Let x= the number of miles Ji traveled. 
 Then x 18= the number B traveled. 
 
 x 18 
 
 =M s daily progress. 
 
 15 4 
 
 =J9's daily progress. 
 Sffo 
 
 Therefore, , : ,-18 : : f 
 
 if 4(x 
 
 Divide the denominators by 7, and extract square root, and we 
 
 Therefore, #=72 ; and the distance between the two towns 
 is 126 miles. 
 
 5. The difference of two numbers is 4, and their sum multi- 
 plied by the difference of their second powers, gives 1600. 
 What are the numbers 1 Jlns. 12 and 8. 
 
 6. What two numbers are those whose difference is to the 
 less as 4 to 3, and their product, multiplied by the less, is 
 equal to 504? dns. Hand 6. 
 
150 ELEMENTS OF ALGEBRA. 
 
 7. A man purchased a field, whose length was to its breadth 
 as 8 to 5. The number of dollars paid per acre was equal 
 to the number of rods in the length of the field; and the 
 number of dollars given for the whole was equal to 13 times 
 the number of rods round the field. Required the length and 
 breadth of the field. 
 
 Jlns. Length 104 rods, breadth 65 rods. 
 Put 8,r=the length of the field. 
 
 8. There is a stack of hay, whose length is to its breadth as 
 5 to 4, and whose height is to its breadth as 7 to 8. It is worth 
 as many cents per cubic foot as it is feet in breadth ; and the 
 whole is worth at that rate 224 times as many cents as there are 
 square feet on the bottom. Required the dimensions of the stack. 
 
 Put 5x = the length. 
 
 Jlns. Length 20 feet, breadth 16 feet; height 14 feet. 
 
 9. There is a number, to which if you add 7, and extract the 
 square root of the sum, and to which if you add 16 and extract 
 the square root of the sum, the sum of the two roots will be 9. 
 What is the number ? Jlns. 9. 
 
 Put x 2 7= the number. 
 
 10. Jl and B carried 100 eggs between them to market, and 
 each received the same sum. If A had carried as many as 
 BI he would have received 18 pence for them; and if B had 
 taken as many as .#, he would have received 8 pence. How 
 
 many had each ? Jlns. Jl 40, and B 60. 
 
 / 
 
 11. The sum of two numbers is 6, and the sum of their cubes 
 is 72. What are the numbers ? Jlns. 4 and 2. 
 
 12. One number is a 2 times as much as another, and the pro- 
 duct of the two is b 2 . What are the numbers ? 
 
 b 
 
 Jlns. and ab. 
 a 
 
 13. The sum of two numbers is 100, the difference of 
 their square roots is 2. What are the numbers? 
 
 Jlns. 36 and 64. 
 
PURE EQUATIONS. 151 
 
 Put x= the square root of the greater number, 
 And y= the square root of the less number ; or 
 Put x-\-y= the square root of the greater, &c. 
 
 14. It is required to divide the number 18 into two such parts, 
 that the squares of those parts may be to each other as 25 to 16 
 
 Let x= the greater part. Then 18 x= the less. 
 By the condition proposed, x*: (18 #) 2 ::25: 16. 
 Therefore, 16^=25(18 a?) 2 
 By evolution, 4a?= 5(18 a:) 
 
 If we take the plus sign, as we must do by the strict enuncia- 
 tion of the problem, we find #=10. Then 18 #=8. 
 
 And (10) 2 : (8) 2 ::25: 16 
 If we take the minus sign, we shall find a?=90. 
 
 Then 18 x= 18 90= 72. 
 
 And (90) 2 : ( 72) 2 ::25: 16; a true proportion, correspond- 
 ing to the enunciation ; but 18 in this case is not the number 
 divided, it is the difference between two numbers whose squares 
 are in proportion of 25 to 16. 
 
 15. It is required to divide the number a into two such 
 parts that the squares of those parts may be in proportion of b 
 to c. 
 
 Let a?= one part, then a x= the other. 
 By the condition, x 2 : (a x) 2 ::b:c 
 Therefore, cx 2 =b(a x) 2 
 
 By evolution, t jcx= l jb(a x) 
 
 Taking the plus sign, x=-~r- and a X- 
 
 ajb ajc 
 
 Faking the minus sign, x=- rr ^ -r- and a a?= ., *-. 
 
 Jb ,Jc 
 
 Prob. 14, is a particular case of this general problem, in which 
 a= 18, 6=25, and c=16; and substituting these values in the re- 
 sult, we find a?=10, and #=90, as before. 
 
 If we take b=c, the two divisions will be equal, each equal 
 to 5 a, when the plus sign is used ; but when the minus sign is 
 
152 ELEMENTS OF ALGEBRA. 
 
 ajb ajb 
 used, :r= r =i ~^~t a symbol of infinity, as the denomi- 
 
 A/^ - V ^ 
 
 nator is contained in the numerator an infinite number of times. 
 (Art. 58.) The other part, a x 
 
 symbol of infinity; and the two parts, 
 ajb ajc fl 
 
 Jb-Jc (Jb-Jc) 
 
 
 It may appear absurd, that the two parts, both infinite and 
 having a ratio of equality, (which they must have, if b=C) can 
 fetill have a difference of a. But this apparent absurdity will 
 vanish, when we consider that the two parts being infinite in 
 comparison to our standards of measure, can have a difference 
 of any finite quantity which may be great, compared with 
 small standards of measure, but becomes nothing in comparison 
 with infinite quantities. See (Art. 60.) 
 
 Application of the foregoing Problem. 
 
 (Art. 94.) It is a well established principle in physics that 
 light and gravity emanating from any body, diminish in inten- 
 sity as the square of the distance increases. 
 
 Two bodies at a distance from each other, and attracting at a 
 given point, their intensities of attraction will be to each other 
 as the masses of the bodies directly and the squares of their 
 distances inversely. Two lights, at a distance from each other, 
 illuminating at a given point, will illuminate in proportion to the 
 magnitudes of the lights directly, and the squares of their dis- 
 tances inversely. 
 
 These principles being admitted 
 
 16. Whereabouts on the line between the earth and the moon 
 will these two bodies attract equally, admitting the mass of the 
 earth to be 75 times that of the moon, and their distance asunder 
 30 diameters of the earth? 
 
 Represent the mass of the moon by c, 
 
 and the mass of the earth by 6, 
 
 their distance asunder by a. 
 
PURE EQUATIONS. 153 
 
 The distance of the required point from the earth's centre, 
 represent by x. Then the remaining distance will be (a x). 
 
 Now by the principle above cited, x 2 : (a xf ::b:c. 
 
 This proportion is the same as appears in the preceding gen- 
 eral problem ; except that we have here actually made the ap- 
 plication, and must give the definite values to a, b and c. 
 
 ajb ajc 
 
 As before, x = rrn j- and a x= y 
 
 fl =30, 6=75, c=l. 
 x= ^ =26&, nearly. Hence, a a?=3.1, nearly. 
 
 If we take the second values for the two distances, from the 
 
 ajb 
 
 neral result, namely, #=- v . a 
 V^ v c 
 
 give the numeral values, we shall have 
 
 ajb ajc 
 
 general result, namely, #=- v . and a x= . v . , and 
 V^ v c v v c 
 
 *= 3.9, nearly. 
 
 These values show that in a line beyond the moon, at a 
 distance of 3.9 the diameters of the earth, a body would be 
 attracted as much by the earth as by the moon, and the value 
 of (a x) being minus, shows that the distance is now counted 
 the other way from the moon, not as in the first case towards 
 the earth ; and the real distance, 30, corresponding to a in the 
 general problem, is now a difference. 
 
 We may make very many inquiries concerning the intensity 
 of attraction on this line, on the same general principle. 
 
 For example, we may inquire, whereabouts, on the line be- 
 tween the earth and moon will the attraction of the earth be 1 6 
 times the attraction of the moon? 
 
 Let x= the distance from the earth. 
 Then a x the distance from the moon. 
 
 The attraction of the earth will be represented by 
 
154 ELEMENTS OF ALGEBRA. 
 
 
 
 The attraction of the moon at the same point will be 
 By the 
 
 (-*)' 
 b 16c 
 
 3H=? 
 
 T> i Jb 4 V C 
 
 By evolution, ^_ = -t--^ 
 
 x a x 
 Clearing of fractions, ajb l jbx : =4 t jcx. 
 
 Using the plus sign, x= ^ . =20.5, nearly. 
 Using the minus sign, x ,, A , 55.7, nearly, or 
 
 25.7 diameters of the earth beyond the moon. 
 
 Observe that the 4 which stands as a factor to Jc is the 
 square root of 16, the number of times the intensity of the 
 earth's attraction was to exceed that of the moon. 
 
 If we propose any other number in the place of 16, its 
 square root will appear as a factor to /c; we may therefore 
 inquire at what distance the intensity of the earth's attraction 
 will be n times that of the moon, and the answer will be from 
 the earth in a line through the moon, 
 
 and a J b 
 
 Jb .Jnc 
 
 The same application that we have made of this general prob- 
 lem to the two bodies, the earth and the moon, may be made 
 to any two bodies in the solar system ; and the same application 
 we have made to attraction may be made to light, whenever 
 we can decide the relative intensity of any two lights at any 
 assumed unity of distance. 
 
 (Art. 95.) This problem may be varied in its application to 
 meet cases where the distances are given, and the compara- 
 tive intensities of light or attraction are required. 
 
 For example, the planet Mars and the moon both transmit 
 the sun's light to the earth by reflection, and we now inquire 
 
PURE EQUATIONS. 155 
 
 the relative intensities of their lights at given distances, and 
 in given positions. 
 
 If the surface of Mars and that of the moon were equal, 
 they would receive the same light from the sun at equal distance 
 from that luminary ; but at different distances equal surfaces 
 vould receive light reciprocally proportional to the squares of 
 their distances. 
 
 The surfaces of globular bodies are in proportion to the squares 
 of their diameters. Now let M represent the diameter of Mars 
 and m the diameter of the moon. Also, let R represent the dis- 
 tance of Mars from the sun, and r the distance of the moon from 
 the sun. 
 
 Then the quantity of light received by Mars may -be expressed 
 
 M 2 
 
 by -^j ; and the relative quantity received by the moon 
 
 m 2 
 by . But these lights, when reflected to the earth, must be 
 
 diminished by the squares of the distances of these two bodies 
 from the earth. Now if we put D to represent the distance of 
 Mars from the earth, and d the distance of the moon, we shall 
 
 M 2 
 have ~n2 for the relative illumination by Mars when the whole 
 
 m 
 
 enlightened face of that planet is towards the earth, and ~^-= for 
 
 the light of the full moon. 
 
 When the whole illuminated side of Mars is turned towards 
 the earth, which is the case under consideration, (if we take the 
 whole diameter of the body,) it is then in opposition to the sun, 
 and gives us light, we know not how much, as we have no 
 standard of measure for it ; but we can make a comparative mea- 
 sure of one by the other, and therefore the light of Mars in this 
 position may be taken as unity, and in comparison with this let 
 us call the light of the full moon x. 
 
 M 2 m 2 
 
 Theretor* *- 
 
156 ELEMENTS OF ALGEBRA. 
 
 As the value of a fraction depends only on the relation of the 
 numerator to the denominator, to find the numeral value of x, it 
 will be sufficient to seek the relation of m to M> of R to r, and 
 of D to d. 
 
 7)/=4000 miles nearly, and m=2150 ; hence, -r>=^- 
 
 Jfl 80 
 
 /?=H4000000, and r=95000000 ; or =^ 
 
 r 95 
 
 /)=144000000 95000000=49000000 or ; 
 
 a 24 
 
 That is, in round numbers, the light of the full moon is twenty- 
 seven thousand six hundred times the light of Mars, when that 
 planet is brightest, in its opposition to the sun. 
 
 We will add one more example by the way of farther illustra- 
 tion. 
 
 What comparative amount of solar light is reflected to the 
 earth by Jupiter and Saturn, when those planets are in opposi- 
 tion to the sun ; the relative diameter of Jupiter being to that 
 of Saturn as 111 to 83, and the relative distances of the Earth, 
 Jupiter and Saturn, from the sun, being as 10, 52 and 95, re- 
 spectively ? 
 
 .#iw. Taking the light reflected by Saturn for unity, that by 
 Jupiter will be expressed by 24. T \\ nearly. 
 
 The philosophical student will readily perceive a more ex- 
 tended application of these principles to computing the relative 
 light reflected to us by the different planets ; but we have gone to 
 the utmost limit of propriety, in an elementary work like this. 
 
 From Art. 94th to the end of this chapter can hardly be said 
 to be algebra ; it is natural philosophy, in which the science of 
 algebra is used ; however, we would offer no apology for thus 
 giving a glimpse of the utility, the cui bono, and the application 
 of algebraic science. 
 
QUADRATIC EQUATION. 157 
 
 SECTION IV. 
 
 QUADRATIC EQUATIONS. 
 CHAPTER I. 
 
 (Art. 96.) Quadratic equations are either simple or compound. 
 A simple quadratic is that which involves the square of the un- 
 known quantity only, as or*=; which is one form of pure equa- 
 tions, such as have been exhibited in the preceding chapter. 
 
 Compound quadratics, or, as most authors designate them, ad- 
 fectf.d quadratics, contain both the square and the first power 
 of the unknown quantity, and of course cannot be resolved as 
 simple equations. 
 
 All compound quadratic equations, when properly reduced, 
 may fall under one of the four following forms : 
 
 (1) x*+2ax=b 
 
 (2) x 2 2ax=b 
 
 (3) X s 2ox= -b 
 
 (4) x-f 2ax=b 
 
 If we take z-f-a and square the sum, we shall have 
 
 If we take x a and square, we shall have 
 
 If we reject the 3rd terms of these squares, we hare 
 3~-\-2ax, and x* 2ax 
 
 The same expressions that we find in the first members of the 
 four preceding theoretical equations. 
 
 It is therefore obvious that by adding a* to both sides of the 
 preceding equations, the first members become complete squares. 
 But in numeral quantities how shall we find the quantity corres- 
 ponding to cr ? We may obtain a* by the formal process of 
 taking half the coefficient of the first power of x, or the half of 
 2 or 2, which is a or a, the square of either being a*. 
 
 Hence, when any equation appears in the form of x 2 2x= 
 rt& ,we may render the first member a complete square, and effect 
 a solution by the following 
 
158 ELEMENTS OF ALGEBRA. 
 
 RULE. Add the square of half the coefficient of the lowest 
 power of the unknown quantity to the first member to complete 
 its square; add the same to the second member to preserve the 
 equality. 
 
 Then extract the square root of both members, and ive shall 
 have equations in the form of 
 
 Transposing the known quantity a and the solution is accom- 
 plished. 
 
 In this manner we find the values of x in the four preceding 
 equations, as follows : 
 
 (i) *=. 
 
 (2) *= 
 
 (3) *= 
 
 (4) *=. 
 
 When b is greater than a 2 equations (3) and (4) require the 
 square root of a negative quantity, and there being no roots to 
 negative quantities, the values of x in such cases are said to be 
 imaginary. 
 
 The double sign is given to the root, as both plus and minus 
 will give the same power, and this gives rise to two values 
 of the unknown quantity ; either of which substituted in the 
 original equation will verify it. 
 
 After we reduce an equation to one of the preceding forms, 
 the solution is only substituting particular values for a and b ; 
 but in many cases it is more easy to resolve the equation as an 
 original one, than to refer and substitute from the formula. 
 
 (Art. 97.) We may meet with many quadratic equations that 
 would be very inconvenient to reduce to the form of x 2 -{-2ax=b; 
 for when reduced to that form 2a and b may both .be 
 troublesome fractions. 
 
 Such equations may be left in the form of 
 
 An equation in which the known quantities, c, 6, and c, are all 
 whole numbers, and at least a and b prime to each other. 
 
QUADRATIC EQUATIONS. 159 
 
 We now desire to find some method of making the first mem- 
 ber of this equation a square, without making fractions. We 
 therefore cannot divide by 0, because b is not divisible by a, 
 the two letters being prime to each other by hypothesis. But 
 the first term of a binomial square is always a square. There- 
 fore, if we desire the first member of our equation to be convert- 
 ed into a binomial square, we must render the first term a 
 square, and we can accomplish that by multiplying every term 
 by a. 
 
 The equation then becomes 
 
 a 2 x 2 -}-bax=ca 
 Put y=ax. Then y z +by=ca 
 
 Complete the square by the preceding rule, and we have 
 
 .b 2 .b 2 
 y 4-%-f j=H-j 
 
 We are sure the first member is a square ; but one of the terms 
 is fractional, a condition we wished to avoid ; but the denomina- 
 tor of the fraction is 4, a square, and a square multiplied by a 
 square produces a square. 
 
 Therefore, multiply by 4, and we have the equation 
 
 4y 2 +4by+b 2 =4ca-{-b 2 
 
 An equation in which the first member is a binomial square and 
 not fractional. 
 
 If we return the values of y and y z this last equation becomes 
 4a 2 x 2 +4abx-\-b 2 =4ac-}-b 2 
 
 Compare this with the primitive equation 
 ax z -}-bx= c. 
 
 We multiplied this equation first by 0, then by 4, and in ad- 
 dition to this we find b 2 on both sides of the rectified equation, b 
 being the coefficient of the first power of the unknown quantity. 
 From this it is obvious that to convert the expression ax 2 -\-bx 
 into a binomial square, we may use the following 
 
 RULE 2. Multiply by four times the coefficient of x 2 , and 
 add the square of the coefficient of x. 
 
 To preserve equality, both sides of an equation must be mill- 
 
160 ELEMENTS OF ALGEBRA. 
 
 ti plied by the same factors, and the same additions to both sides. 
 We operate on the first member of an adfected equation to 
 make it a square, we operate on the second member to preserve 
 equality. 
 
 (Art. 98.) For the following method of avoiding fractions in 
 completing the square, the author is indebted to the late Professor 
 T. J. Matthews, of Ohio. 
 
 Resume the general equation ax 2 -\-bx=c 
 
 u , u* bit 
 
 Assume x=- Ihen aar= and bx= 
 
 a a a 
 
 The general equation becomes + =c 
 
 a a 
 
 Or u 2 +bu=ac 
 
 Now when b is even, we can complete the square by the first 
 rule without making a fraction. In such cases this transforma- 
 tion is very advantageous. 
 
 When b is not even, multiply the general equation by 2, and 
 the coefficient of x becomes even, and we have 
 
 2c (1) 
 
 Assume #= - Then 2ax 2 = and 2bx= 
 2a 2a 2a 
 
 With these terms, equation (1) becomes 
 
 u 2 2bu 
 2a + W 
 
 Or u*-\-2bu=4ac 
 Complete the square by the first rule, and we have 
 
 An equation essentially the same as that obtained by completing 
 the square by the rule under (Art. 97.) ; for we perceive the sec- 
 ond member is the same as would result from that rule ; hence 
 this method has no superior advantage except when b is even, in 
 the first instance. 
 
 (Art. 99.) The foregoing rules are all that are usually given 
 for the resolution of quadratic equations ; but there are some 
 
QUADRATIC EQUATIONS. 161 
 
 intricate cases in practice that we may meet with, where 
 neither of the preceding rules appear practical or convenient. 
 To master these with skill and dexterity, we must return to a 
 more general and comprehensive knowledge of binomial squares, 
 x* -\-2ax-\- a 2 is a simple and complete binomial square. Let 
 us strictly examine it, and we shall perceive, 
 1st. That it consists of three terms ; 
 
 2d. Two of its terms, i\\e first and the third, are squares; 
 3d. The middle term is twice the product of the square 
 roots of the first and last term. 
 
 Now let us suppose the third term, a\ to be lost, and we have 
 only x z -\-2ax. We know these two terms cannot make a square, 
 as a binomial square must consist of three terms.* 
 
 We know also that the last term must be a square. 
 Let it be represented by t*. 
 
 Then, by hypothesis, x z ^-2ax-^t* is a complete binomial 
 square. 
 
 It being so, 2xt2ax, by the third observation above. 
 
 Therefore, t=a and t*=a 2 
 Thus a z is brought back. 
 
 ! Again, 4a 2 -}-4ab are the first and second terms of a bi- 
 nomial square ; what is the 3rd term ? 
 
 Let t 2 represent the third term. 
 Then 4 2 -f 4ab-}-t 2 is a binomial square. 
 Hence, 4at4ab or t=b and / 2 =6 2 
 
 That is, t 2 represented the 3d term, and b 2 is the identical 3d 
 lerm, and 4a 2 -\-4ab-\-b 2 is the actual binomial square whose root 
 Is 2a+b. 
 
 2. 36?/ 2 -}-36?/ are the first and 2d terms of a binomial square, 
 what is the 3d term ? JLns. 9. 
 
 3. -\ \-9 are the 2d and 3d terms of a square, what is the 
 
 x ! 
 
 first? Ans. ? 
 
 x 2 - 
 
 * In binomial surds two terras may make a square, and this may condemn 
 the technicality here assumed ; but it is nothing against the spirit of this ar- 
 ticle. 
 
 14 
 
162 ELEMENTS OF ALGEBRA. 
 
 4Qx 2 
 
 4. -- 49 are the 1st and 2d terms of a binomial square, 
 
 49 
 
 what is the 3d 1 Jlns. . 
 
 tt 
 
 5. 9?/ 2 Qy are the 1st and 2d terms of a binomial square, 
 what is the 3d? rfns. 1. 
 
 6. tfx*-\-bx are the 1st and 2d terms of a binomial square, 
 
 b 2 
 
 what is the 3d 'I Jlns. 
 
 4a 
 ( 
 
 7. 8 la; 2 3 are the 1st and 3d terms of a binomial square, 
 what is the 2d or middle term ? fins. 18. 
 
 8. 2/ 2 Sx^y are the 1st and 2d terms of a binomial square, 
 what is the 3d ? rfns. I6x. 
 
 9. -- ?r+36 are the 2d and 3d terms of a binomial square, 
 
 1 7 
 
 or 2 
 
 what is the 1st ? Ans. - -. 
 
 36 1 
 
 77 2 
 
 10. ^--}-36 are the 1st and 3d terms of a binomial square, 
 301 
 
 what is the middle term ? rfns. 
 
 iy 
 
 11. If #-}--- are the 2d and 3d terms of a binomial square, 
 
 lo 
 
 what is the 1st term ? rfns. 4x*. 
 
 Qx* 
 
 12. The 1st term of a binomial square is the 2d term is 
 
 u 
 
 :12, what is the 3d term ? Ans. -^-. 
 
 (Art. 100.) Adfected quadratic equations, after being reduced 
 to the form of x?-\-2axb, can be resolved without any formality 
 of completing the square, by the following substitution : 
 
QUADRATIC EQUATIONS. 163 
 
 Assume xy o, 
 Then a^=?/ 2 2a?/+ 2 
 
 And 2ax= -\-2ay-2a* 
 By addition, x*+2ax=y z a z b 
 
 Hence, y= l Jb-\-a z 
 
 And x= a-i l Jb-^-a 2 j the same result as 
 may be found in equation (1), (Art. 96.) 
 
 RULE FOR SUBSTITUTION. Assume the value of the unknown 
 quantity equal to another unknown, annexed to half the coeffi- 
 cient of the inferior power with a contrary sign. 
 
 (Art. 101.) For further illustration of the nature of quadratic 
 equations, we shall work and discuss the following equation : 
 
 Given ^+4^=60, to find x. 
 
 Completing the square, (Rule 1st.) 2 +4aH-4=64. 
 
 Extracting square root, a?-f-2=rb8. 
 Hence, x=6 or x= 10. 
 
 That is, either plus 6, or minus 10, substituted for x in the 
 given equation, will verify it. 
 
 For 6H-4X 6=60. Also, ( 10) 2 4X 10=60 
 
 If x=6 then x 6=0 
 
 Ifa:= 10 then z+10=0 
 
 Multiply these equations together, and we have 
 x 6 
 x +10 
 
 Product, #M-4aN 60=0 
 
 Transpose, and 0^+42?= 60, the original equation. 
 
 Thus we perceive, that a quadratic equation maybe considered 
 as the product of two simple equations, and these values of x in 
 the simple equations are said to be roots of the quadratic, and this 
 view of the subject gives the rationale of the unknown quantity 
 having two values. 
 
164 ELEMENTS OF ALGEBRA. 
 
 In equations where but one value can be found, we infer that 
 the other value is the same, and the two roots equal, or one of 
 them a cipher. 
 
 EXAMPLES FOR PRACTICE. 
 
 1. Given x 2 6# 7=33, to find x. Jlns. #=10 or 4. 
 
 2. Given x 2 20#= 96, to find x. Jlns. 12 or 8. 
 
 3. Given x*-{-6x-{-l=92, to find x. Jlns. 7 or -13. 
 4. Given y 2 +12y=589, to find y. Jlns. 19 or 31. 
 
 5. Given t/ 2 6y-|- 10=65, to find y. Jlns. 1 1 or 5. 
 
 6. Given a?-}-12;c-}-2=110, to find x. Jlns. 6 or 18. 
 
 7. Given x 2 14z=51, to find x. Jlns. 17 or 3. 
 
 8. Given a? 2 -f6a:-|-6=9, to find x. Jlns. 32,y3. 
 
 9. Given ar4-8a?=12,to findtf. Jlns.. 
 
 10. Given a?-{-l2x=lQ, to find x. Jlns. 
 
 The reader will observe that the preceding examples are in, 
 or can be immediately reduced to the form of # 2 2a#=6, and of 
 course their solution is comparatively easy. The following are 
 mostly in the form of ax*-}-bx=c. 
 
 11. Given 5#*-j-4:e=204, to find x. 
 
 According to (Art. 98,) put x-. Then 5# 2 = and 
 
 4w w 2 4w 
 
 4x , and the equation becomes _-f ^-=204. 
 o o 5 
 
 Clearing of fractions, w 2 +4tt=1020. 
 
 Completing the square and extracting the root, we have, 
 w-f-2=rb32, or w=30 or 34 
 
 But x=~. Therefore, x 6 or , Jlns. 
 5 5 
 
 12. Given 5^+4#=273, to find x. Am. 7 or 7 4 S 
 
 13. Given 7a? 2 20z=32, to find x. Jlns. 4 or f . 
 
 14. Given 25# 2 20#= 3, to find x. Jlns. | or ^. 
 
 15. Given 21 x 2 292#= 500, to find x. 
 
 ins. Ilifor2 
 
QUADRATIC EQUATIONS. 165 
 
 16. Given 2x 2 5#=117, to find x. 
 
 Here, as 5 or b of the general equation is not even, we must 
 multiply the whole equation by 2, to apply the above principle ; 
 or we may take Rule 2. (Art. 97.) 
 
 Multiply by 8, and add 5 2 or 25 to both members. 
 
 Then 16^40a:4-25=961 
 
 Square root, 4x 5=31. Hence, x =9 or 65. 
 
 (Art. 102.) It should be observed that all quadratic equations 
 can be reduced to the form of x z .2ax=b t or, as most authors 
 give it, x z px=q; but when the terms would become fractional 
 by such reduction, we prefer the form ax*bx=c, for the sake 
 of practical convenience, as mentioned in (Art. 97.) 
 
 (Art. 103.) It is not essential that the unknown quantity 
 should be involved literally to its first and second powers ; it is 
 only essential that one index should be double that of the other. 
 In such cases the equations can be resolved as quadratics. For 
 example, a? 6 4^=621 is an impure equation of the sixth 
 degree, yet with a view to its solution, it may be called a quad- 
 ratic. For we can assume y=x 3 ; then y 2 =x 6 , and the equa- 
 tion becomes y z 4i/=621, a quadratic in relation to y, giving 
 y=27, or 23. 
 
 Therefore, #"=27 or 23 
 
 And a?=3 or V 23. 
 
 There are other values of x; but it would be improper to seek 
 for them now; such inquiries belong to the higher order of 
 equations. 
 
 3 
 
 For another example, take x 3 x 2 =56, to find the values 
 of x. 
 
 Here we perceive one exponent of x is double that of the other, 
 it is therefore essentially a quadratic. 
 
 Such cases can be made clear by assuming the lowest power 
 of the unknown quantity equal to any simple letter. In the 
 
 3 
 
 present case assume y=x 2 ; then y 2 =# 3 , and the equation is 
 
166 ELEMENTS OF ALGEBRA. 
 
 By Rule 3, 4y*4y+ 1=225 
 By evolution, 2y 1=15 
 Hence, y=& r 7 
 
 3 3 
 
 And by returning to the assumption y=x^ we find # 2 8, 
 or a? 4 =2. Hence, a?=4; or, by taking the minus value of y, 
 
 (Art. 104.) When a compound quantity appears under differ- 
 ent powers or fractional exponents, one exponent being double 
 that of the other, we may put the quantity equal to a single letter, 
 and make its quadratic form apparent and simple. For example, 
 suppose the values of x were required in the equation 
 
 Assume J2 
 Then by involution, 2# 2 -}-3#-f 9=?/ 2 (A] 
 And the equation becomes y 2 5y=6 (J9) 
 
 Which equation gives y=& or 1. These values of y, sub- 
 stituted fory in the equation (fi give 2,r 2 -j-3x-j-9=36 
 
 Or 
 From the first of tl^e we find a?=3 or 
 
 From the las* we find a?=|( 3dby 55,) imaginary quan- 
 tities. 
 
 EXAMPLES. 
 
 I. Given (#+12)*+ (a>H2)* = 6 to find the values of x. 
 
 rfns. x=4 or 6U 
 
 nd tlie values of or 
 Ans. xb*a or 81 b' 1 a.* 
 
 * It is proper to remark, that in many instances it would be difficult to verify 
 the equation by taking the second values of x, as by squaring, the minus quan- 
 tity becomes plus, and in returning the values, there is no method but trial to 
 decide whether we shall take a plus or a minus root. Hence, these second 
 answers are sometimes called roots of solution. In many instances hereafter, 
 we shall give the rational and positive root only. 
 
 Given (aH-</)~*-f 2&(#-f-a)* =3b 2 , to find the values of ar 
 
QUADRATIC EQUATIONa 107 
 
 3. Given 9;r+4+2,/9a'+4=15, to find the values of x. 
 
 Jlns. a?=f J. 
 
 4. Given (10+z)* (10+ar)*=2, to find x 
 
 Jlns. a?=6. 
 
 5. Given (# 5) 3 3(tf 5)2=40, to find x. 
 
 Ans. o?=9. 
 
 6. Given 2(1 -far a? 2 ) (l-f a; ^)^+^=0, to find a% 
 
 7. Given s+16 3(tf+16f=10, to find a;. Am. x=Q. 
 
 8. Given So.- 2 ' 1 2o? n =8, to find a*. #ns. o?=%/2. 
 
 9. Given 3^+^=756, to find x. Jlns. a?=243. 
 
 Q -I rt 
 
 1O. Given - -- r^ 1 "!" -- to ^ n( ^ ^ ^ n5 - a7 = 3 or 1 - 
 
 11. Given 4a?+=39, to find x. Am. a?=729. 
 
 12. Given ^2^+6(^2^+5)^ = 11, to find x. 
 
 Jlns. x=l 
 
 it 2 12ic 
 
 13. Given - T^ == 32, to find the value of #.. 
 
 rfns. a?=152 or 76. 
 
 If much difficulty is found in resolving this 13th example, the 
 pupil can observe the 9th example, (Art. 99). 
 
 14. Given 8 la^+ 17+^=99, to find the values of x. ] 
 
 Jlns. x=l, or 1, or |. 
 
 Observe that the 1st and 3d terms of the first number are 
 squares, see (Art. 99.) 
 
 1 R41 2^2 
 
 15. Given 81^+17+-,-=-^-+ +15, to find x. J 
 
 X XT X 
 
 rfns. x=2 or 1J. 
 
 16. Given 25# 2 +6+ '--, to find the values of x.~- 
 
 Jlns. a? 2, or 2, or 
 
 15 
 
168 ELEMENTS OF ALGEBRA. 
 
 IT. Given -^--j--|=6, to find the values of x. 
 
 Ans. x=7 or 11J. 
 
 (Art. 105.) Equations of the third, fourth, and higher degrees, 
 can be resolved as quadratics, provided we can find a compound 
 quantity in the given equation involved to its first and second 
 power, with known coefficients. 
 
 To determine in any particular case, whether such a com- 
 pound quantity is involved in the equation, we must transpose 
 all the terms to the first member, and if the highest power of 
 the unknown quantity is not even, multiply every term of the 
 equation by the unknown letter to make it even, and then extract 
 the square root, to two or three terms, as the case may require ; 
 and if we find a remainder to be any multiple or any aliquot 
 part of the terms of the root, a reduction to the quadratic form 
 is effected ; otherwise it is impossible, and the equation cannot 
 be resolved as a quadratic. 
 
 For example, reduce the following equation to the quadratic 
 form, if it be possible. 
 
 1. Given x 4 8a# 3 -f 8aV-f32a 3 a>- 9a 4 =0, to find the values 
 of x by quadratics. 
 
 OPERATION. 
 a- 4 Sax 3 +8aV+32a 3 *--9a 4 ==0 
 
 This remainder can be put into this form : 
 Now we observe the original equation can be written thus : 
 
 9a 4 =0 
 By putting # 2 4ax=y we have 
 
 y 2 8a 2 y=9a* a quadratic. 
 Completing the square?/ 2 8a 2 y-j-16a 4 =25a 4 
 By evolution y 4a 2 =5a 2 
 Hence y=9a 2 2 
 
QUADRATIC EQUATIONS. 109 
 
 Or x 2 4ax=Qa 2 or a 2 
 
 Completing the squares 2 4ax-\-4a 2 =l3a 2 or 3a 2 
 
 By evolution x 2a=a l jl3 or aj3 
 Hence x may have the four following 1 values (2-{->/iy) 
 (2aaJ~T3), (2-f aj3), (2aaj3). Either of which being 
 substituted in the original equation will verify it. 
 
 2. Reduce a7 3 -f-2ax 2 +5a 2 ^-f 4 3 =0 to a quadratic. 
 
 As ihe highest power of x is not ct'en, we must multiply by x 
 to make it even. Then 
 
 x 4 +2ax 3 +5aV-f-4tt 3 .r =0 
 
 By extracting two terms of the square root, and observing the 
 remainder, the part that will not come into the root, we find that 
 
 Divide by (x 2 -{-ax) and x z -\-ax-}-4a 2 =Q a quadratic. 
 3. Given x*+2x*7x 2 8#-hl2=0, to find the values of x. 
 This equation may be put in the following form : 
 
 Ans. x=l or 2, or 2 or 3. 
 
 4. Given x 3 8^ 2 -fl9 x 12=0, to find the values of x. 
 
 Am. x=l or 3 or 4. 
 
 5. Given a: 4 10^ 3 -|-35a; 2 50#-{-24=0, to find the values 
 of a?. JJns. x=l, 2, 3 or 4. 
 
 6. Given x 4 2x 3 +a?=132, to find the values of x. 
 
 Jjns. x=4 or 3. 
 
 7. Given i/ 4 2n/ 3 -f(c 2 2)t/ 2 -f 2cy=C", to find the values 
 
 f y- 
 
 (Art. 106.) The object of this article is to point out a few lit- 
 tle artifices in resolving quadratics, which apply in particular 
 cases only, but which at times may save much labor. It is there- 
 fore proper that they should be presented, though some minds 
 prefer uniformity to facility. 
 15 
 
J70 ELEMENTS OF ALGEBRA, 
 
 For example, take equation (J5) (Art. 104.) 
 
 1. y* -5y=6 Put 2a=5 
 Then y* 2ay=2a+l. 
 
 Add a 2 to both sides to complete the square, (Rule 1 .) 
 
 And i/ 2 2ay-f-a 2 = 2 +2a-}-l 
 By evolution, y fl=(a-f-l.) 
 Hence, y=2a-\-l=G or 1 
 
 2. Given y* 7i/=8, to find y. Jlns. y=8 or 1. 
 
 3. Given a: 2 -j-ll#=26, to find the values of x. 
 
 Assume 2a=ll ; then 4a-J-4=26. 
 
 Now put these values in place of the numerals, and complete 
 the square, and o; 2 -{-2rt#-J-a 2 =a 2 -|-4a+4. 
 
 By evolution, x -\-a=(a-\-2] Hence, x=Z or 13. 
 
 4. Given x 2 17^=60, to find the values of x. 
 
 Assume 2a=17; then 6a-f-9=60 
 And ** 2a?+a 2 =a 2 +6a-{-9. 
 By evolution, x a(a-\-3.) Hence, rr=20 or 3. 
 
 5. Given a^-f-19#=92, to find the values of x. 
 
 Assume 2a=19 ; then 8+ 16=92 
 Putting these values and completing the square we have 
 a?-}-2ax + 2 =a 2 +8a4- 1 6 
 ar-{-a=(a-r-4) or x=4 or 23. 
 
 Observe that in the preceding equations we invariably put the 
 coefficient of the first power of the unknown quantity equal 2a. 
 Then if we find the absolute term in the second member of the 
 equation equal to 2a-{- 1 
 
 or 4a-f 4 
 
 or 6a-f- 9 
 
 or 8a-}-16 
 
 Or, in general, m2a-\-m z . That is, any multiplier of 2 plus 
 the square of the same multiplier equal to the second member, 
 then the equation can be resolved in this manner ; for in fact one 
 
QUADRATIC EQUATIONS. 171 
 
 of the roots of the equation is this multiplier of 2a, and the other 
 root is (2a-j-77i), m being the multiplier, and it may represent 
 any number, integral or fractional ; but there is no utility in ope- 
 rating by this method unless m is an integer, and not very large 
 To present a case where m is fractional, we give tie following 
 equation: x z 9#=y, to find the value of x. 
 
 Put 20=9; then Ix2a+J= l 4 9 an( ^ ^ ie e q uat i n becomes 
 
 Therefore, x a=(fl-H). Hence, x= or 2a+s=9|. 
 
 (Art. 107.) When the roots of the equation are irrational or 
 surd, of course this method of operation will not apply ; but we 
 can readily determine whether the roots will be surd or not. For 
 example, take the equation # 2 -r"13#=40. 
 
 Put 2a=13; then 4a-|-4=30 And 60+9=48 
 From this, we observe that one of the roots of the equation 
 lies between 2 and 3. 
 
 (Art. 108.) When the roots of an equation are irrational or 
 surd, no artifice will avail us, and we must conform to set rules ; 
 but when the roots are small integers, we can frequently find 
 some method to avoid high numeral quantities ; but special artifi- 
 ces can only be taught by examples, not by precept. The follow- 
 ing are given as examples : 
 
 1. Given r j -f-9984#=l 60000, to find the values of x. 
 
 Observe that 9984=1000016 
 
 Put 2a=10000; then 320=160000 
 These substitutions transform the equation to 
 
 Completing the square by (Rule 1) and 
 
 a*+(2 16)a?-r-(a 8) 2 =a 2 -f-16a-f-64 
 By evolution, x-\-(a 8)=db(a-f-8) 
 Hence, #=16 or 2a= 10000. 
 
 2. Given # 2 +45#=9000, to find the values of x. 
 
 If we put 20=45, the multiplier and its square, requisite to 
 
172 ELEMENTS OF ALGEBRA. 
 
 produce 9000, is so large that it is not obvious, and of course 
 there will be no advantage in adopting this method ; at the same 
 time, we wish to avoid the high numerals we must encounter by 
 any set rule of solution; 
 
 We observe that 45 X 200=9000. Put a=45 
 
 Then x?+ax=2QOa 
 Complete the square by (Rule 2,) and 
 (13?+ 2 =a 2 +8000 
 
 By evolution, 2#+a=ya(a+800)= > /45X845 
 
 Multiply one of the factors, under the radical, by 5, and divide 
 the other by 5, and the equivalent factors will be 225 X 169, both 
 squares. Taking their root, resuming the value of a, and the 
 equation becomes 
 
 2;r+3.15=13.J5 
 Drop 3.15 from both sides 
 And 2a?=10.15 or #=75, rfns. 
 
 3. Given IQx 2 225#=225, to find the values of x. 
 
 This equation is found in many of the popular works on 
 algebra, and in several of them the common method of resolving 
 it may be seen. 
 
 Observe that 225=15X15. 
 
 Put a=15; then a+l = 16, and the equation becomes 
 
 Completing the square by (Rule 2), and 
 4(+])V 4(+l)a 2 a: +a 4 =a 4 
 By evolution, 2(a+l)a? 2 =<z 2 +2a 
 Transpose a 2 and divide by 2, and we have 
 
 Divide by (0+1) and x=a=\ 5, Jlns 
 
 We give one more example of the utility of representing nume- 
 rals, or numeral factors, by letters, in reducing the following 
 equation : 
 
QUADRATIC EQUATIONS. 173 
 
 18 , 81 rr 2 r 2 65 
 4. Given --[ ^ - 
 
 By examining the numerals, we find 9, and several multiples 
 of 9. Therefore, let a=9, and using a in the place of 9 the 
 equation becomes 
 
 2a , a 2 y?_x* 65 
 
 "P" 1 ^~ : : ~~&T~ 
 
 Clearing of fractions, we have 
 
 16 2 -f8a 2 ;r 8x*=x 4 65^ 
 
 Transposing all to one side, and arranging the terms according 
 to the powers of .T, we have 
 
 16a=0 
 
 ; 16a 2 
 Or 2 (x 2 4-8^+16) 
 
 Therefore, by (Art. 105,) the equation becomes 
 
 Or (ar+4)V= 2 (a:+4) 2 
 By division, ic 2 = 2 
 
 And a; ==9, 
 
 The preceding examples may be of service in reducing some 
 of the following 
 
 EXAMPLES. 
 1. Given a; 2 -}- 11 a? =80, to find a?. ^ns. a:=5, car 16 
 
 . 
 
 2. Given 5a? -- -=2a?-f - - , to find a?. 
 
 # 3 *4 
 
 rfns. ar=4, or 1 
 
 3. Given - ^+^ ~~"fi"' to fill(1 ^ *^ n5 ' a?=2 
 
174 ELEMENTS OF ALGEBRA. 
 
 72a? 250 
 
 4. Given 7x-{-- ^-=50, to find x. Ans. #=2, or - 
 
 10 3x 21 
 
 /6 V /6 \ 
 
 5. Given ( h*/)-H hy)=30, to find tne values of 
 
 Ans. y=3 or 2, or 
 
 _4 
 
 6. Given # 3 -|-7# 3 =44, to find the values of x. 
 
 Am. #=8 or ( 
 
 . Given ^-j-ll-j-^/i/ 2 -!-! 1+2=44, to find the values of y. 
 
 fins. 2/=5 or 
 it -1-7 
 
 it -- --# 
 
 8. Given 14-}-2.r -- =#4- - , to find the values of x. 
 
 Ans. #=28 or 9. 
 
 9. Given 3z? Qx 4=80, to find the values of x. 
 
 Am. x=7 or 4. 
 
 10. Given =T, to find x. Ans. x=4. 
 
 11. Given -{-x -2=24 3x, to find the values 
 
 x 3 
 
 of x. Ans. x=G or 5. 
 
 12. Given --- : = , to find the values of x. 
 
 x x* 9 
 
 A)is. a? 3 or y' f . 
 
 -^ _ l Ox 2 4-1 
 
 13. Given - ^ r-jr- =x 3, to find the values of x. 
 
 XT \)X~\~ J 
 
 Ans. x=l or 28. 
 
 14. Given ma? 2mxjn=nx* mn, to find x. 
 
 a Jmn 
 
 Ans. x= 
 
 15. Given # 4 H -=34a?+16, to find the values of x. 
 Z 
 
 Exms. Art. 99.) Ans ,r=2, or 2, or 8, or 5. 
 
QUADRATIC EQUATIONS. 175 
 
 16. Given ^(J- V 3 =^, to find the values of *. 
 
 N.B. Put -1-=. fins. x=S or 9. 
 
 17. Given y* ty 2 =y4-8y-f-12, to find the values of y. 
 (See Art. 99.) Jlns. y=3 or 2. 
 
 o 
 
 18. Given x 1=2+ , to find the values of a:. 
 
 Jx 
 
 Am. x=4 or 1 
 
 19. Given (-^~ -Y=# 2, to find the values of x 
 \zJx*9/ 
 
 tfns. x=5 or 3. 
 
 CHAPTER II. 
 
 Quadratic Equations, containing two or more unknown 
 quantities. 
 
 (Art. 108.) We have thus far, in quadratics, considered equa- 
 tions involving only one unknown quantity; but we are now 
 fully prepared to carry our investigations farther. 
 
 Two equations, essentially quadratic, involving two unknown 
 quantities, depend for their solution on a resulting equation of the 
 fourth degree. 
 
 This principle may be shown in the following manner : 
 
 Two equations, essentially quadratic, and in the most gencrul 
 form, involving two unknown quantities, may be represented 
 thus : 
 
 =0, 
 
 We do not represent the first terms with a coefficient, as any 
 coefficient may be reduced to unity by division ; and a, b, c, &c., 
 and a', b' t &c. may represent the result of such division ; and, 
 of course, may be of any value, whole or fractional, positive or 
 
 negative. 
 
176 ELEMENTS OF ALGEBRA. 
 
 Arranging the terms, in the above equations, in reference to 
 we have 
 
 a*+(m/+c)x+hf+dy+e =0, (1) 
 
 >=Q. (2) 
 
 13 y subtracting (2) from (1), we have 
 
 '}f+(dd>)y+e e'=0; 
 
 Therefore 
 
 Tliis expression for a?, substituted in either equation (1) or (2) 
 will give a final equation, involving only one unknown quan 
 tity, y. 
 
 But to effect this substitution would lead to a very complicated 
 result ; and as our object is only to show the degree to which 
 the resulting equation will rise, we may observe that the ex- 
 pression for the value of x is in the form of r -. This 
 
 ry+s 
 
 put in either of the equations (1) or (2), its square, or the ex- 
 pression for ar 2 , will be of the fourth degree ; and no term can 
 contain y of a higher degree than the fourth. 
 
 Therefore, in general, the resolution of two equations of the 
 second degree, involving two unknown quantities depends upon 
 that of an equation of the fourth degree involving one unknown 
 quantity. 
 
 (Art. 109.) Two or more equations, involving two or more 
 unknown quantities, can be resolved by quadratics, when they 
 fall under one of the following cases : 
 
 1st. One of the equations only may be quadratic; the other 
 must be simple, or capable of being reduced to a simple form. 
 
 2d. The equations must be similar in form, or the unknown 
 quantities similarly involved or combined in a similar manner, 
 as they combine in regular powers ; or, 
 
 3d. The equations must be homogeneous ; that is, the expo- 
 nents of the unknown quantities must make the same sum in 
 every term. 
 
 In the first and second oases, we eliminate one quantity in 
 
QUADRATIC EQUATIONS. J77 
 
 one equation and substitute its value in the other, or perform an 
 equivalent operation, by rules already explained. 
 
 In the third case, we throw in a factor to one unknown quan- 
 tity, to make it equal to the other, or assume it to be so ; but 
 these principles can only be explained by 
 
 EXAMPLES. 
 
 ax* -\-bxy -\- cy z =e, 
 
 These are homogeneous equations, for the exponents of the 
 unknown quantities make the same sum 2, in every term. In 
 such cases, assume x=vyi then the equations become 
 
 __ 
 
 or = 
 
 
 
 a'v*+b'v+c' 
 
 An equation involving the 1st and 2d powers of v, and of 
 course, a quadratic. 
 
 The solution of this equation will give v. Having v, we 
 have y 2 and y, and from vy we obtain x. 
 
 For a particular example, we give the following : 
 
 1. Given 4 ^-- 2; /= 12 I to find the values of x and y. 
 
 Put x=vy ; then the equations become 
 4v 2 y 2 2vy 2 =12, or y*=- 
 
 u 2 v)2 
 And 
 
 f* Q 
 
 Whence, = - =-. Dividing by 2, then clearing of 
 Zv v 2-\-3v 
 
 fractions, we have 
 
 v=8t> 2 4.v or 8u 2 
 12 
 
178 ELEMENTS OF ALGEBRA. 
 
 This last equation gives v=2 or |. 
 
 8 8 
 Omitting the negative value y*= -r- =-=l. 
 
 -- o 
 
 Therefore, 7/=l, and x=vy=2. 
 
 2. Given 2x 3y=l, and 2x*-}-xy 5y z =2Q, to find the 
 values of x and y. 
 
 These equations correspond to the first observation, one of 
 them only being quadratic, the other simple ; and the solution 
 is effected by finding the value of x in the first equation. Sub- 
 
 stituting that value - in the 2d, and reducing, we have 
 2?/ 2 -{-77/=39, which gives 2/=3. Hence, x=5. 
 
 3. Given & 2 -\-y z x #=78, and xy+x-\-y=39, to find 
 the values of x and y. 
 
 In these equations x and y are similarly involved, not equally 
 involved ; nor are the equations homogeneous. In cases of this 
 kind, as we have before remarked, a solution by a quadratic can 
 be effected, but no general or definite rule of operation can be 
 laid down ; the hitherto acquired skill of the learner, and his 
 power of comparison to discern the similarity, will do more than 
 any formal rules. 
 
 To resolve this example, we multiply the 2d equation by 2, 
 and add the product to the first ; we then have 
 l5G, or 
 
 Put x-}-y=s ;' then s 2 -f-s=156, a quadratic, which gives .9, 
 or x-{-y=l2. This value of or-J-y, taken in the second equa- 
 tion, gives xy=27. From this sum and product of x and y, 
 we find a?=9 or 3, and y=3 or 9. 
 
 Again, after we multiply the second equation by 2, if we sub- 
 tract it from the first, we shall have 
 
 or (xy) 2 (a?+y)=0 
 or (# y) 2 =3Xl2=36 
 or x yQ 
 But x+y=l$ 
 
 Hence, 2#=18 or 6, and #=9 or 3, as before. 
 
QUADRATIC EQUATIONS. 179 
 
 (Art. 110.) There are some equations to which the foregoing 
 observations do not immediately apply, or not until after reduc- 
 tions and changes take place. The following is one of them. 
 
 4. Given y > to find the values of x and y. 
 
 Here neither of the equations is simple, nor are both letters 
 similarly involved, nor are the equations homogeneous ; yet we 
 can find a solution by a quadratic, because the two equations 
 have a common compound factor, which taken away, will bring 
 the equations far within the limits or condition laid down ; and 
 this remark will apply to all problems that can be resolved by 
 quadratics not seemingly within the limits of the three con- 
 ditions. 
 
 12 
 
 From the first of these equations, we have y 
 
 From the second, - ... y= . 
 
 12 18 
 
 Hence, ~- = Divide the denominators by (#+1) 
 
 X"-\~X X + 1 
 
 find the numerators by 6, and we have 
 
 2 3 
 
 ~ == Z5 ITTT a Q ua ^ratic equation. 
 
 Clearing of fractions, and 2# 2 2x-{-2=3x 
 
 or 2.x 2 5#= -2. 
 (Rule 2.) I6x 2 .#+25=2516=9. 
 
 We write Jl to represent the second term. It is immaterial 
 what its numeral value may be, as it always disappears in taking 
 the root. 
 
 By evolution, 4# 5= 3 
 Hence, o?=2 or 5. 
 
 12 12 12 12X4 
 
 * r 
 
 The following is of a similar character : 
 
 
180 ELEMENTS OF ALGEBRA. 
 
 5 Given $ **"" 2/ 2 ~ (*+#) = 8 ? to find the values of 3, 
 
 W$ andy. 
 
 
 
 Divide the first equation by (x-\-y) and x y 1 -- (.#). 
 
 x-{-y 
 
 oo 
 
 Divide the second by (x-\-y) and (x 2/) 2 =-- (B). 
 Put x-\-y=s, and transpose minus 1, in equation (.#), and 
 =* + l. By squaring, (^-2/) 2 =^4-y-M. 
 
 00 
 
 Equation (5) gives (x y) 2 = 
 
 5 
 
 Therefore, V- 6 +l=??. 
 
 S' 2 5 5 
 
 Clearing of fractions, and transposing 32s, we have 
 
 64 -16s-|-s 2 =0 
 
 By evolution, 8 s=0 or s=8. That is, a:+y=8, which 
 value, put in equation (^j, gives x y=%. 
 Whence, x=5 and 2/=3. 
 
 MISCELLANEOUS EXAMPLES. 
 
 1. Given x=2y 2 and 5(0? 2/)=5, to find the values of x 
 and y. Jlns. x=l8 or 12^. 
 
 T/= 3 or 2j|. 
 
 2. Given 2^4-y=22, and a?i/+2i/ 2 =120, to find the values 
 of x and ?/. Jlns. x=8, y=G. 
 
 3. Given x+y : x y : : 13 : 5, and #-|-7/ 2 =25, to find 
 the values of x and y. Jlns. x=9, and 2/=4. 
 
 4. Given J^+S/ 3 ^ 8 ^ ? to find the values of # and y. 
 +y =12 5 ^5. a?=8 or 4, =4 or 8 
 
 5. Given z*-{-2xy-{-y 2 =l2Q 2x 2y, and xy / 2 =8, to 
 find the values of x and y. 
 
 x=Q or 9, or i 
 v=4 or 1, or 3 
 
6. Given 
 
 n 
 
 QUADRATIC EQUATIONS. 
 =56 
 
 =56 ^ 
 
 =60 S to 
 
 va * ues 
 
 or 10. 
 
 to find the values of x 
 and . 
 
 8. Given 
 
 C 
 
 s 4 2 i 3 ?/ 
 
 / = 68 ? 
 ?/== i60 S to 
 
 e va ^ ues 
 Jlns. ic= 
 
 In the first four examples, one of the equations is simple ; in 
 the 5th and 6th, x and y are similarly involved ; and the Oth, 
 7th and 8th are homogeneous. 
 
 (Art. 111.) When we have fractional exponents, we can re- 
 move them, as explained in (Art. 92.) ; but in some cases it may 
 not be important to do so. 
 
 EXAMPLES. 
 
 1. Given x*-}-y 3 '=3x and x*-\-y*=x, to find the values 
 of x and y. 
 
 Put a^=P; then x=P* and x^=P* 
 
 And^=#; then y=Q? and y%=Q* 
 
 Now the primitive equations become 
 
 P3+Q*=3P*, and P+Q=P* 
 From the 1st, # 2 =(3 P)P* 
 From the 2d, Q =(P1)P 
 By squaring, Q 2 =(P 1) 2 P 2 
 
 Put the two values of Q 2 equal to each other, rejecting or 
 dividing by the common factor P 2 , and we have 
 
182 ELEMENTS OF ALGEBRA. 
 
 or P 2 2/M-l=3P 
 or P*P=2 
 Hence, P =2 or 1, and #=4 or 1, and y=8. 
 
 2. Given #^+^+2#^+2y^=23, and #V 3 "=36, to find 
 the values of x and y. Jlns. #=27 or 8, y=8 or 27. 
 
 3. Given #"-i-i/ 5 =8, and # 3 -f-7/^+# 3 i/ 3 "=259, to find the 
 values of # and y. 
 
 ; #=125 or 27, 
 \ y= 27 or 125. 
 
 4. Given x-}-y-\-x+y=2Q 1 and #=8, to find the 
 values of x and y. Jlns. #=8, i/=32. 
 
 # 4# 5 33 
 
 5. Given h f == -j- an( l # ^2/ =5 to ^ nt ^ ^ ie values of X 
 
 and y. *fl.ns. a?=9, y=4 
 
 to 
 
 6. Given \ to ^ n( e va l ues of ^ an(1 2/ 
 
 *=2|, y=16 
 
 7. Given #(y a + lJ-^# a + i;4-2;r 2 ?/ 2 =55, and #y a + 
 
 t^ v t >a =30, to find the values of # and y. ( #=4 or 9 
 
 < 3/=9 or 4 
 
 ?. 3. l 
 
 8. Given x 3 y z =2y z and 8# 3 1/^ = 14, to find the value* 
 of # and y. ^ m C #=2744 or 8 
 
 ^y=9604 or 4 
 
 (Art. 1 12.) No additional principles, to those already given, 
 are requisite for the solution of problems containing three or 
 more unknown quantities in quadratics. As in simple equations, 
 we must have as many independent equations as unknown 
 quantities. 
 
 As auxiliary to the solution of certain problems, particularly in 
 geometrical progression, we give the following problem : 
 
QUADRATIC EQUATIONS. 183 
 
 Given x+y=s, and xy=p, to find the values of x 2 -j-y 2 , 
 x'-J-y 5 ? x 4 +?/ 4 , and x 5 -J-y 5 , expressed in terms of s and p. 
 
 Squaring the first, x*-{-2xy-{-y 2 =s* 
 Subtract twice the second, 2xy 2p 
 
 1st result, x*+y 2 =s* 2p (A) 
 
 Again, 
 
 = P s 
 
 Subtract 
 
 2d result, 
 Square (A], and 
 Subtract 
 
 x*+y*=s* Bps 
 x 4 -\-2x 2 y z +y 4 =s* 4s*p+4p z 
 2x*y 2 = 2p z 
 
 3d result, 
 Multiply (A) by (#) and 
 
 x 4 +y*=s 4 4s z p-}-2p z 
 
 Subtract 
 
 x 2 y 2 (x-\-y) = 
 
 sp z 
 
 4th result, 
 
 (C) 
 
 x 5 +y 5 =s 5 5s 3 p-}-5sp 2 (D) 
 
 
 CHAPTER III. 
 Questions producing Quadratic Equations. 
 
 (Art. 113.) The method of proceeding to reduce the question 
 into equations, is the same as in simple equations ; and, in fact, 
 many problems which result in a quadratic may be brought out 
 by simple equations, by foresight and skill in notation. Others 
 again are so essentially quadratic, that no expedient can change 
 their form. 
 
 EXAMPLES. 
 
 1. A person bought a number of sheep for $240. If there 
 had been 8 more, they would have cost him $1 a-piece less. 
 What was the cost of a sheep, and how many did he purchase ? 
 
 240 
 
 Let x= the number of sheep j then =cost of one. 
 
184 ELEMENTS OF ALGEBRA. 
 
 If he had x-\-S sheep, _=:cost of one. 
 
 240 240 , 
 
 By the question, - : -- 1-1 
 x 3?-p8 
 
 Clearing of fractions, 240or-)-1920=240;r-{-a: 2 -f-8;c 
 Or 
 
 Resolving gives #=40 or 48 ; but a minus number will not 
 apply to sheep ; the other value only will apply to the problem 
 as enunciated. 
 
 This question can be brought into a simple equation thus : Let 
 x 4= the number of sheep, then 8 more would be expressed 
 by #4-4, and the equation would be 
 
 , ' +1. Put =240. 
 
 aH-4 
 
 Then " = +1 
 
 x 4 x -|-4 
 
 Clearing of fractions, ax-}-4aax 4a-\-x* 16. 
 
 Transposing, # 2 =8a-l-16=8(a+2)=16X 121 
 Extracting square root, #=4 X 1 1=44. Hence, a: 4=40, the 
 number of sheep. Divide 240 by 40, and we have $6 for the 
 price of one sheep. 
 
 (Art. 114.) In resolving problems, if the second member is 
 negative after completing the square, it indicates some impossi- 
 bility in the conditions from which the equation is derived, or an 
 error in forming the equation, and in such cases the values of the 
 unknown quantity are both imaginary. 
 
 2. For example, let it be required to divide 20 into two such 
 parts that their product shall be 140. 
 
 Let r one part, then 20 x= the other. 
 By conditions, 20# ^=140 
 
 Or, x* 20#=--140 
 
 Completing the square, x 2 20#-}-100= 40 
 
 By evolution, x 10=^=2^ 10 
 
 Or, x =10 
 
QUADRATIC EQUATIONS. 185 
 
 This result shows an impossibility ; there are no such parts 
 of 20 as here expressed. It is impossible to divide 20 into two 
 such parts that their product shall be over 100, the product of 
 10 by 10, and so on with any other number. The product of 
 two parts is the greatest possible, when the parts are equal. 
 
 3. Find two numbers, such that the sum of their squares being 
 subtracted from three times their product, 11 will remain; and 
 the difference of their squares being subtracted from twice their 
 product, the remainder will be 14. 
 
 Let x= the greater number, and i/=the less. 
 By the conditions, '3xy x 2 t/ 2 =ll 
 And 2xy # 2 -f-i/ 2 =14 
 
 These are homogeneous equations; therefore, put x=vy ; 
 Then 3w/ 2 ify 2 y 2 =l 1 (./?) 
 
 And 2vy*vy+y*=l4 (B) 
 
 Conceive (./?) divided by (B) and the fraction reduced, we have 
 3f v* 1^11 
 l 14 
 
 Clearing of fractions and reducing, we find 
 3u 2 20u = 25. 
 
 5 
 
 A solution gives one value of v, - 
 
 3 
 
 Put this value in equation (.#), and we have 
 
 5y 2 v 2 = ! 1 
 
 9 f 
 
 Multiply by 9. and 457/ 2 25^ 2 9y 2 =llX9, 
 Or, lli/ 2 =llX9, 
 
 y z = 9 or y=3. Hence, x5. 
 
 A. A company dining at a house of entertainment, had to pay 
 $3.50 ; but before the bill was presented two of them went away ; 
 in consequence of which, those who remained had to pay each 
 20 cents more than if all had been present. How many persons 
 dined ? Jlns. 7. 
 
 16 
 
186 ELEMENTS OF ALGEBRA. 
 
 5. There is a certain number, which being subtracted from 
 22, and the remainder multiplied by the number, the product will 
 be 117. What is the number ? Jlns. 13 or 9. 
 
 6. In a certain number of hours a man traveled 36 miles, but 
 if he had traveled one mile more per hour, he would have taken 
 3 hours less than he did to perform his journey. How many 
 miles did he travel per hour? Ans, 3 miles. 
 
 7. A person dies, leaving children and a fortune of $46,800, 
 which, by the will, is to be divided equally among them ; but it 
 happens that immediately after the death of the father, two of the 
 children also die ; and if, in consequence of this, each remaining 
 child receive $1950 more than he or she was entitled to by the 
 will, how many children were there? Jlns. 8 children. 
 
 8. A gentleman bought a number of pieces of cloth for 675 
 dollars, which he sold again at 48 dollars by the piece, and gain- 
 ed by the bargain as much as one piece cost him. What was 
 the number of pieces I Jlns. 15. 
 
 This problem produces one of the equations in (Art. 10.) 
 
 9. A merchant sends for a piece of goods and pays a certain 
 sum for it, besides 4 per cent, for carriage ; he sells it for $390, 
 and thus gains as much per cent, on the cost and carnage as the 
 12th part of the purchase money amounted to. For how much 
 did he buy it? Ans. $300. 
 
 10. Divide the number 60 into two such parts that their pro- 
 duce shall be 704. Jins. 44 and 16. 
 
 11. A merchant sold a piece of cloth for $39, and gained 
 as much per cent, as it cost him. What did he pay for it? 
 
 Jlns. $30. 
 
 12. Jl and B distributed 1200 dollars each, among a certain 
 number of persons. Jl relieved 40 persons more than B, and 
 B gave to each individual 5 dollars more than Jl. How many 
 were relieved by A and B? Jlns. 120 by Jl, and 80 by B. 
 
 This problem can be brought into a pure equation, in like man- 
 ner as (Problem 1.) 
 
QUADRATIC EQUATIONS. 187 
 
 13. A vintner sold 7 dozen of sherry and 12 dozen of claret 
 for 50, and finds that he has sold 3 dozen more of sherry for 
 10 than he has of claret for 6. Required the price of each? 
 
 Ans. Sherry, 2 per dozen; claret, 3. 
 
 14. Ji set out from C towards J9, and traveled 7 miles a 
 day. After he had gone 32 miles, B set out from D towards 
 C, and went every day ^ of the whole journey ; and after 
 he had traveled as many days as he went miles in a day, he met 
 Jl. Required the distance from C to 2). 
 
 J$ns. 76 or 152 miles ; both numbers will answer the con- 
 dition. 
 
 15. A farmer received $24 for a certain quantity of wheat, 
 and an equal sum at a price 25 cents less by the bushel for a 
 quantity of barley, which exceeded the quantity of wheat by 16 
 bushels. How many bushels were there of each ? 
 
 Jlns. 32 bushels of wheat, and 48 of barley. 
 
 16. A and B hired a pasture, into which Jl 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? 
 
 *flns. B had six horses in the pasture at first, and the price 
 of the whole pasture was 30 shillings per week. 
 
 I 1 ?. Find those two numbers whose sum, product, and dif- 
 ference of their squares, are all equal to each other. 
 Ans. i36, and 
 
 18. If a certain number be divided by the product of its 
 two digits, the quotient will be 2, and if 27 be added to the num- 
 ber, the digits will be inverted. What is the number? 
 
 JJns. 36. 
 
 19. It is required to find three numbers, whose sum is 33, 
 such that the difference of the first and second shall exceed the 
 difference of the second and third by 6, and the sum of whose 
 squares is 441. fins. 4, 13, and 16. 
 
188 ELEMENTS OF ALGEBRA. 
 
 2O. Find those two numeral quantities whose sum, product, 
 and sum of their squares, are all equal to each other. 
 
 Jlns. No such numeral quantities exist. In a strictly algebraic 
 sense, the quantities are 
 
 21. What two numbers are those whose product is 24, and 
 whose sum added to the sum of their squares is 62? 
 
 fins. 4 and G. 
 
 22. It is required to find two numbers, such that if their pro- 
 duct be added to their sum it shall make 47, and if their sum be 
 taken from the sum of their squares, the remainder shall be 62? 
 
 *ftns. 7 and 5. 
 
 23. The sum of two numbers is 27, and the sum of their 
 cubes 5103. What are their numbers? J%ns. 12 and 15 
 
 24. The sum of two numbers is 9, and the sum of their fourth 
 powers 2417. What are the numbers? Jlns. 7 and 2. 
 
 25. The product of two numbers multiplied by the sum of 
 their squares, is 1248, and the difference of their squares is 20 
 What are the numbers? Jlns. 6 and 4. 
 
 Let x-}-y=ihe greater, and x y=the less. 
 
 26. Two men are employed to do a piece of work, which 
 they can finish in 12 days. In how many days could each do 
 the \vork alone, provided it would take one 10 days longer than 
 the other? Ans. 20 and 30 days. 
 
 27. The joint stock of two partners, Jl and B, was $1000. 
 .tf's money was in trade 9 months, and J5's 6 months ; when 
 they shared stock and gain, Ji received $1,140 and B $640. 
 What was each man's stock? 
 
 Jlns. JFs stock was $600 ; B's $400. 
 
 28. A speculator from market, going out to buy cattle, met 
 with four droves. In the second were 4 more than 4 times the 
 square root of one half the number in the first. The third con 
 tained three times as many as the first and second. The fourth 
 was one half the number in the third and 10 more, and the whole 
 
ARITHMETICAL PROGRESSION. 199 
 
 number in the four droves was 1121. How many weie in each 
 drove? Ans. 1st, 162 ; 2d, 40 ; 3d, 606 ; 4th, 313 
 
 29. Divide the number 20 into two such parts, that the pro- 
 duct of their squares shall be 9216. Jlns. 12 and 8. 
 
 30. Divide the number a into two such parts that the product 
 of their squares shall be b. 
 
 rfns. Greater part -f- 
 
 Less part |- \(a*-4 jlty*. 
 
 31. Find two numbers, such that their product shall be equal 
 to the difference of their squares, and the sum of their squares 
 shall be equal to the difference of their cubes. 
 
 dns. ^5" and <l(5,/5) 
 
 SECTION V. 
 
 ARITHMETICAL PROGRESSION. 
 
 CHAPTER I. 
 
 A sertes of numbers or quantities, increasing or decreasing by 
 the same difference, from term to term, is called arithmetical pro- 
 gression. 
 
 Thus, 2, 4, 6, 8, 10, 12, &c., is an increasing or ascending 
 arithmetical series, having a common difference of 2 ; and 20, 
 17, 14, 11, 8, <fec., is a decreasing series, having a common dif- 
 ference of 3. 
 
 (Art. 1 15.) We can more readily investigate the properties of 
 an arithmetical series from literal than from numeral terms. Thus 
 let a represent the first term of a series, and d the common dif- 
 ference. 
 
190 ELEMENTS OF ALGEBRA. 
 
 a,(a-{-d),(a J r2d),(a-\-3d) 9 (a-\-'ld), &c., represent an ascend 
 ing series ; and 
 
 a, (a t?),(a 2d),(a 3d), (a 4cQ, &c., represent a descend- 
 ing series. 
 
 Observe that the coefficient of d, in any term is equal to the 
 number of the preceding terms. 
 
 The first term exists without the common difference. All 
 other terms consist of the first term and the common difference 
 multiplied by one less than the number of terms from the first. 
 
 Wherever the series is supposed to terminate, is the last term, 
 and if such term be designated by Z, and the number of terms 
 by n, the last term must be a-\-(n l)d, or a (n l)d, accord- 
 ing as the series may be ascending or descending, which we draw 
 from inspection. 
 
 Hence, L=a-(nl)d (A) 
 
 (Art. 116.) It is manifest that the sum of the terms will be the 
 same, in whatever order they are written. 
 
 Take, for instance, the series 3, 5, 7, 9, 11, 
 
 And the same inverted, 11, 9, 7, 5, 3. 
 
 The sums of the terms will be 14, 14, 14, 14, 14. 
 
 Take the series #, a-f- rf, a-{-2d, a-}-3d, a-\-4d t 
 
 Inverted, -j-4c/, a-{-3d, a-\-2d, a-}- d, a 
 
 Sums will be 2a+4rf, 2a+4d, 2a+4d, 2a-\-4d, 2a+4d. 
 
 Here we discover the important property, that, in an arithmeti- 
 cal progression, the sum of the extremes is equal to the sum of 
 any other two terms equally distant from the extremes. Also, 
 that twice the sum of any series is equal to the extremes, or 
 first and last term repeated as many times as the series contains 
 terms. 
 
 Hence, if S represents the sum of a series, and n the num- 
 ber of terms, a the first term, and L the last term, we shall 
 have 2S=n(a+L) 
 
 Or S=?(a+L) (B) 
 
 The two equations (.#) and (B] contain five quantities, a, d, 
 
ARITHMETICAL PROGRESSION. 191 
 
 Z, n, and S; any three of them being given, the other two can 
 be determined. 
 
 Two independent equations are sufficient to determine two un- 
 known quantities, (Art. 45,) and it is immaterial which two are 
 unknown if the other three are given. 
 
 By examining the two equations 
 
 Z=a4-(rc l)rf (JS) 
 
 S=%(a+L) (B) 
 
 We perceive that the value of any letter, L for example, can be 
 drawn from equation (B) as well as from (A). 
 
 It can also be drawn from either of the equations after n or a 
 is eliminated from them. Hence, the value of L may take/ot/r 
 different forms. The same may be said of the other letters, 
 and there being five quantities or letters and four different 
 forms to each, the subject of arithmetical progression may in- 
 clude twenty different equations. But we prefer to make no 
 display with these equations, believing they would add dark- 
 ness rather than light, as they are all essentially included in the 
 two equations, (A) and (B), and these can be remembered literal- 
 ly and philosophically, and the entire subject more surely under- 
 stood. 
 
 These two equations are sufficient for problems relating to 
 arithmetical series, and we may use them without modification 
 by putting in the given values just as they stand, and afterwards 
 reducing them as numeral equations. 
 
 EXAMPLES. 
 
 1 The sum of an arithmetical series is 1455, the first term 
 5, and the number of terms 30. What is the common difference! 
 
 Ans. 3. 
 
 Here =1455, a=5, n=W. L and d are sought. 
 Equation (B) 1455=(5-{-Z)15. Reduced Z=92 
 Equation (.#) 92=5-]-29<?. Reduced d=3, fins. 
 
192 ELEMENTS OF ALGEBRA. 
 
 2. The sum of an arithmetical series is 567, the first term 7, 
 and the common difference 2. What is the number of terms? 
 
 Am. 21 
 
 Here 5=567, a=7, d=2. L and n are sought. 
 Equation (A) Z=7-f-2n 2=5+2n 
 
 Equation (B) 567=(7+5-f-2n)~=6tt-|-n 2 
 
 Or n 2 -j-6/i-f-9=576 
 
 n-}-3=24, or n=2l,Ans. 
 
 3. Find seven arithmetical means between 1 and 49. 
 
 Observe that the series must consist of 9 terms. 
 Hence, o=l, Z=49, n=Q. 
 
 Ans. 7, 13, 19,25, 31, 37, 43. 
 
 4. The first term of an arithmetical series is 1, the sum of the 
 terms 280, the number of terms 32. What is the common dif- 
 ference, and the last term? Ans. d=z, Z=16j. 
 
 5. Insert three arithmetical means between | and 5* 
 
 Ans. The means are ~, T 5 2, \\ 
 
 6. Find nine arithmetical means between 9 and 109. 
 
 Ans. d--=lO. 
 
 "7. What debt can be discharged in a year by paying 1 cent 
 the first day, 3 cents the second, 5 cents the third, and so on, in- 
 creasing the payment each day by 2 cents? 
 
 Ans. 1332 dollars 25 cents. 
 
 8. A footman travels the first day 20 miles, 23 the second, 20 
 the third, and so on, increasing the distance each day 3 miles. 
 How many days must he travel at this rate to go 438 miles? 
 
 Ans. 12. 
 
 9. What is the sum of n terms of the progression of 1,2, 3, 
 
 1O. The sum of the terms of an arithmetical series is 950, 
 the common difference is 3, and the number of terms 25. What 
 is the first term? Ans. 2. 
 
ARITHMETICAL PROGRESSION. 193 
 
 11. A man bought a certain number of acres of land, paying 
 for the first, $5 ; for the second, $| ; and so on. When he came 
 to settle -he had to pay $3775. How many acres did he pur- 
 chase, and what did it average per acre ? 
 
 Ans. 150 acres at $25 per acre. 
 
 Problems in Arithmetical Progression to which the preceding 
 formulas, (A) and (B), do not immediately apply. 
 
 (Art. 117.) When three quantities are in arithmetical progres- 
 sion, it is evident that the middle one must be the exact mean 
 of the three, otherwise it would not be arithmetical progression ; 
 therefore the sum of the extremes must be double of the mean. 
 
 Take, for example, any three consecutive terms of a series, as 
 
 and we perceive by inspection that the sum of the extremes is 
 double the mean. 
 
 When there are four terms, the sum of the extremes is equal 
 to the sum of the means, by (Art. 116.) 
 
 To facilitate the solution of problems, when three terms are 
 in question, let them be represented by (x y), x, (x-\-y], y being 
 the common difference. 
 
 When four numbers are in question, let them be represented 
 by (x 3y), (x y), (aj+y), (x-\-3y)i 2y being the common dif- 
 ference. 
 
 So in general for any other number, assume such terms that 
 the common difference will disappear by addition. 
 
 1. There are five numbers in arithmetical progression, the 
 sum of these numbers is 65, and the sum of their squares is 
 1005. What are the numbers ? 
 
 Let x= the middle term, and y the common difference. Then 
 X 2y, xy, x, x-\-y, x-\-2y, will represent the numbers, 
 and their sum will be 5#=65, or #=13. Also, the sum of 
 their squares will be 
 
 5# 2 -f-10 < j/ 2 =1005 or ic 2 +2/ 2 =201. 
 
 But # 2 =169; therefore, 2?/ 2 ^32, i/ 2 =16 or y=i. 
 
 Hence, the numbers are 13 8=5, 9, 13, 17 and 21 
 17 
 
194 ELEMENTS OF ALGEBRA. 
 
 2. There are three numbers in arithmetical progression, their 
 sum is 18, and the sum of their squares 158. What are those 
 numbers? Jlns. 1, 6 and 11 
 
 3. It is required to find four numbers in arithmetical progres- 
 sion, the common difference of which shall be 4, and their coiv 
 tinued product 176985. Jlns. 15, 19, 23 and 27 
 
 4. There are four numbers in arithmetical progression, the 
 sum of the extremes is 8, and the product of the means 15 
 What are the numbers? Jlna. 1, 3, 5, 7. 
 
 *5. A person travels from a certain place, goes 1 mile the first 
 day, 2 the second, 3 the third, and so on ; and in six days after, 
 another sets out from the same place to overtake him, and travels 
 uniformly 15 miles a day. How many days must elapse after 
 the second starts before they come together? 
 
 Jlns. 3 days and 14 days. 
 Reconcile these two answers. 
 
 6. A man borrowed $60 ; what sum shall he pay daily to can- 
 cel the debt, principal and interest, in 60 days; interest at 10 per 
 cent, for 12 months, of 30 days each? 
 
 Jlns. $1 and f ^ of a cent. 
 
 7. There are four numbers in arithmetical progression, the 
 sum of the squares of the extremes is 50, the sum of the squares- 
 of means is 34 ; what are the numbers? J$ns. 1, 3, 5, 7 
 
 . The sum of four numbers in arithmetical progression is 
 24, their continued product is 945. What are the numbers '' 
 
 Jlns. 3, 5, 7, 9. 
 
 9. A certain number consists of three digits, which are in 
 arithmetical progression, and the number divided by the sum of 
 its digits is equal to 26 ; but if 198 be added to the number its- 
 digits will be inverted What is the number 1 fins. 234 
 

 GEOMETRICAL PROGRESSION. 195 
 
 CHAPTER II. 
 
 GEOMETRICAL PROGRESSION. 
 
 (Art. 118.) When numbers or quantities differ from each 
 othfr by a constant multiplier in regular succession, they consti- 
 tute a geometrical series, and if the multiplier be greater than 
 unity, the series is ascending ; if it be less than unity, the series 
 is descending. 
 
 Thus, 2 : 6 : 18 : 54 : 162 : 486, is an ascending series, the 
 multiplier, called the ratio, being three ; and 81 : 27 : 9 : 3 : 1 : 
 l : i, &c., is a descending series, the multiplier or ratio being |. 
 
 Hence, a : ar : ar* : r 3 : ar 4 : ar 5 : w 6 : &c., may represent 
 any geometrical series, and if r be greater than 1, the series is 
 ascending, if less than 1, it is descending. 
 
 (Art. 119.) Observe that the first power of r stands in the 
 2d term, the 2d power in the 3d term, the third power in the 
 4th term, and thus universally the power of the ratio in any 
 term is one less than the number of the term. 
 
 The, first term is a factor in every term. Hence the 10th 
 term of this general series is ar 9 . The 17th term would be ar 16 . 
 The rath term would be ar n ~ l . 
 
 Therefore, if n represent the number of terms in any series, 
 and L the last term, then L=ai**~ l (1) 
 
 (Art. 120.) If we represent the sum of any geometrical series 
 by 5, we have 
 
 5 =a-j-ar+cfr 2 -}-r 3 4- &c. . . ar n - 2 +ar n ~ l . 
 Multiply this equation by r, and we have 
 
 r 5 =ar-}-tfr 2 -f ar 3 -f- &c. ar n ~ l -\-ar n . 
 Subtract the upper from the lower, and observe thai 
 Zr=ar n ; then (r l)s=Lr a. 
 
 Therefore, 8 =- ( 2 ) 
 
 As these two equations are fundamental, and cover the whole 
 subject of geometrical progression, let them be brought together 
 for critical inspection. 
 
196 ELEMENTS OF ALGEBRA. 
 
 (1), S= (2). 
 
 These two equations furnish the rules given for the operations 
 in common arithmetic. 
 
 Here we perceive five quantities, 0, ?*, n, L and S, and any 
 three of them being given in any problem, the other two can be 
 determined from the equations. 
 
 To these equations we may apply the same remarks as were 
 made to the two equations in arithmetical progression (Art. 116.) 
 
 Equation (2), put in words, gives the following rule for the 
 sum of a geometrical series ; 
 
 RULE. Multiply the last term by the ratio, .and from the 
 vroduct subtract the first term, and divide the remainder by the 
 ratio less one. 
 
 EXAMPLES FOR THE APPLICATION OF EQUATIONS 
 (1) AND (2). 
 
 1. Required the sum of 9 terms of the series, 1, 2, 4, 8, 16, 
 &c. Ans. 511. 
 
 3. Required the 8th term of the progression, 2, 6, 18, 54, 
 &c. Ans. 4374. 
 
 3. What is the sum of ten terms of the series 1, f, J, &c. ? 
 
 * WAV 
 
 4. Required two geometrical means between 24 and 192. 
 
 N. B. When the two means are found, the series will consist 
 of four terms ; the first term 24 and the last term 192. 
 
 By equation (1) L=ar n ~\ 
 Here a 24, Z=192, n=4, and the equation becomes 
 
 192=24r 3 or r=2. 
 Hence, 48 and 96 are the means required. 
 
 5. Required 7 geometrical means between 3 and 768. 
 
 tins. 6, 12, 24, 48, 96, 192, 
 
 6. The first term of a geometrical series is 5, the last term 
 1215, and the number of terms 6. Wliat is the ratio 1 fins. 3. 
 
GEOMETRICAL PROGRESSION. J97 
 
 7". A man purchased a house, giving $1 for the fiist door, $2 
 for the second, $4 for the third, and so on, there being 10 doors. 
 What did the house cost him 1 tins. $1023. 
 
 (Art. 121.) By Equation (2), and the Rule subsequently given, 
 we perceive that the sum of a series depends on the first and last 
 terms and the ratio, and not on the number of terms ; and 
 whether the terms be many or few, there is no variation in the 
 rule. Hence, we may require the sum of any descending series, 
 as 1, a i, \, &c., to infinity, provided we determine the LAST 
 term. Now we perceive the magnitude of the terms decrease 
 as the series advances ; the hundredth term would be e'xtremely 
 small, the thousandth term very much less, and the infinite term 
 nothing ; not too small to be noted, as some tell us, but absolutely 
 nothing. 
 
 Hence, in any decreasing series, when the number of terms 
 is conceived to be infinite, the last term, Z, becomes 
 
 0, and Equation (2) becomes s=-- 
 
 By .change of signs s=- - 
 
 This gives the following rule for the sum of a decreasing infi- 
 nite series : 
 
 RULE. Divide the first term by the difference between unity 
 and the ratio. 
 
 EXAMPLES. 
 
 1. Find the value of 1, , T 9 , &c., to infinity. 
 
 o=l, r=\. JLns. 4 
 
 2. Find the exact value of the decimal .3333, &c., to infinity. 
 
 rfns. 5. 
 
 This may be expressed thus : T \+ T f , <fec. Hence. 
 
 3. Find the value of .323232, &c., to infinity. 
 
 =T 3 2 o r= T <nr 2 oo ; therefore r= T - . Ans. | 
 
 4. Find the value of .777, &c., to infinity. *ftns. . 
 
J98 ELEMENTS OF ALGEBRA. 
 
 5. Find the value of f : 1 : J : / 3 , &c. to infinity. 
 
 tins. 4 ' 
 
 6. Find tlie value of 5 : f : , &c. to infinity. Jlns. 7| 
 
 7. Find the value of the series 1, ^, &c., to infinity ? Am. |. 
 
 8. What is the value of the decimal .71333, &c., to infinity? 
 
 On*. Hi- 
 O. What is the value of the decimal .212121, &c., to infinity? 
 
 *>. A- 
 
 (Art. 122.) If we observe the general series, (Art. 11 8.) a: an 
 ar 2 : ar 3 : ar 4 , &c., we shall find, by taking three consecutive 
 terms anywhere along in the series, that the product of the ex- 
 tremes will equal the square of the mean. Hence, to find a 
 geometrical mean between two numbers, we must multiply them 
 together, and take the square root. If we take four consecutive 
 terms, theproduct of the extremes will be equal to the product 
 of the means. 
 
 (Art. 123.) This last property belongs equally to geometrical 
 proportion, as well as to a geometrical series, and the learner must 
 be careful not to confound proportion with a series. 
 
 a: ar : : b : br, is a geometrical proportion, not a continued 
 series. The ratio is the same in the two couplets, but the mag- 
 nitudes a and b, to which the ratio is applied, may be very dif- 
 ferent. 
 
 We may suppose a : ar two consecutive terms of one series, 
 and b : br any two consecutive terms of another series having 
 the same ratio as the first series, and being brought together they 
 form a geometrical proportion. Hence, the equality of the ra- 
 tio constitutes proportion. 
 
 To facilitate the solution of some difficult problems in geomet- 
 rical progression, it is desirable, if possible, to express several 
 terms by two letters only, and have them stand symmetrically. 
 
 Three terms maybe expressed by x : ,Jxy : y, or by x z : xy : 
 y\ as the product of the extremes is here evidently equal to the 
 square of the mean. 
 
 To express four terms with x and y symmetrically, we at first 
 
GEOMETRICAL PROGRESSION. 199 
 
 write P : x \ : y : Q. The firstthree being in geometrical progres- 
 
 y& y* 
 
 sion, gives Py=x 2 orP= . In the same manner, we find $= - 
 
 s? y 2 
 
 And taking these values of P and Q we have : x:: y: to ren- 
 
 y x 
 
 resent four numbers symmetrically with two letters. 
 
 Taking three numbers as above, and placing them between P 
 and Q, thus, P : x z : xy : y 2 : , we have five numbers ; and 
 
 3? 
 
 by reducing P and Q into functions of x and ?/, we have: #* : 
 
 V 3 
 xy : y z : y , for five terms symmetrically expressed. 
 
 X 
 
 3? z 2 y 2 y* 
 
 Six numbers thus, : : x : : y : : ^ 
 
 t/ 2 y xx* 
 
 Sometimes we may more advantageously express unknown 
 numbers in geometrical proportion by x, xy, xy 2 , <fec. ; x being 
 the first term, and y the ratio. 
 
 HARMONICAL PROPORTION. 
 
 (Art. 124.) When three magnitudes, a, ft, c, have the relation 
 of a : c : : a b : b c ; that is, the first is to the third as the dif- 
 ference between the first and second is to the difference between 
 the second and third, the quantities a, ft, c, are said to be in har- 
 monical proportion. 
 
 (Art. 125.) Four magnitudes 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, , ft, c, d, are in harmonical proportion when a : d : : 
 a ft : c d, or when a : d : : b a : d c. 
 
 An harmonical mean between two numbers is equal to twice 
 their product divided by their sum. For a : x : b representing 
 three numbers in harmonical proportion, we have by the definition, 
 (Art. 124.) a : ft : : a x : xb. 
 
 Therefore, ax ab=ab bx or x= r-y. 
 
200 ELEMENTS OF AUJEHKA. 
 
 1. Find the harmonica! mean between 6 and 12. JJns. 8. 
 
 2. Find the third harmonical number to 234 and 144. 
 
 rfns. 104. 
 
 3. Find the fourth harmonical proportion to the numbers 24 : 
 16 : 4. Am. 3, 
 
 4U There are four numbers in harmonical proportion, the firs' 
 is 16, the third 3, the fourth 2. The second is lost ; find it. 
 
 Ans. 8 
 
 PROBLEMS IN GEOMETRICAL PROGRESSION, AND 
 HARMONICAL PROPORTION. 
 
 1. The sum of three numbers in geometrical progression is 
 26, and the sum of their squares 364. What are the numbers ? 
 Let the numbers be represented by x : ,Jxy : y. 
 Then x+Jxy+y = 26=a (1) 
 
 And x 2 -!- ^4-2/2=364=6 (2) 
 
 Transpose, ,Jxy in Equation (I) and square, we have 
 
 a*+2xy+if=(fi2ajxy+xy (3) 
 Or x*-\- xy-{-y 2 =a~2a l Jxy (4) 
 
 The left hand members of Equations (2) and (4) are equal ; 
 
 hence, 
 
 a 2 2ajxy=b. 
 
 Therefore, 7^==6 (5) 
 
 This gives the second term of the progression, and now from 
 equations (1) and (5) we find #=2, y=I8, and the numbers 
 are 2, 6, 18. 
 
 2. The sum of four numbers in geometrical progression is 15, 
 (), and the sum of their squares 85 (6). What are the 
 numbers ? 
 
 Let the numbers be represented as in (Art. 123.) 
 
Then 
 
 GEOMETRICAL PROGRESSION. 201 
 
 Assume x-{-y=s 
 xy=p 
 
 Then by (Art. 112.) 
 
 X' "'"* 
 
 And 
 
 y 
 
 And x*-\-y*=s 8 3sp 
 
 Transposing (x-\-y) in Equation (1), and (x 2 -\-y 2 ) in Equation 
 (2), we have 
 
 =as (3) and +=b--#+2 p (4) 
 y *x y 2 x z 
 
 Square (3) and transpose 2xy or 2p and 
 
 |'+| 4 =(-*) 2 -2/> (5) 
 
 The left hand members of equations (4) and (5) are equal, there 
 fore, (ar-8)*2p=b s 2 +2p 
 
 Or a 2 2as-{-2s z 4p=b (6) 
 
 Clear equation (3) of fractions, and x 3J ry 3 ap ps. 
 
 s 3 
 
 That is, 6- 3 3span ps or = : - (7} 
 
 a+2s 
 
 Put this value of p in equation (6) and reduce, we have, 
 
 Or as*+bs=:- 
 
 2 
 
 Taking the given values of a and b we have 
 
 15s 2 -|-856'=70X15 
 
 Or 3* 2 -{-17s=210, an equation which gives s=6 
 Put the values of a and s in equation (7), and p=S 
 That is, x-\-y=Q, and xy=8, from which we find #=2, am! 
 =4 ; therefore the required numbers are 
 
 1,2,4 and 8, Ans 
 
 3. The arithmetical mean of two numbers exceeds the geo 
 
202 ELEMENTS OF ALGEBRA. 
 
 metrical mean by 13, and the geometrical mean exceeds the har- 
 monical mean by 12. What are the numbers? 
 
 Let x and y represent the numbers. 
 Then | (#+?/)= the arithmetical mean, > Jxi/= the geome- 
 
 trical mean, (Art. 112.) and r^-= the harmonical mean. 
 
 v+y 
 
 Let o=12; 
 
 Then, by the question, i(x-\-y) = i Jxy -\-a-\- 1 (1) 
 
 (2) 
 
 By our customary substitution, these equations become 
 5*=V/>+-fl (3) 
 
 And <//>=T+ a ( 4 ) 
 
 o 
 
 Take the ralue of s from equation (3) and put it into equation 
 (4), dividing me numerator and the denominator by 2, and we 
 have . p . 
 
 Clearing of fractions, we shall have 
 
 Drop equals, and Jp=(a-\-l}a (6) 
 
 Put this value of Jp in equation (3) and we have 
 
 Or 5=2(0+ 1) 2 (7) 
 
 For the sake of brevity, put (a-j-l)=6; squaring equation 
 (6) and restoring the values of s and p in equations (6) and (7), 
 and we have xy=a z b* (Jl) 
 
 x+y=2b* (JS) 
 Square (S) and 
 
 Subtract 4 
 
 / a\ 
 times ^ 
 
 And x z 2^-l-i/ 8 =4& 2 (6 2 2 )=46 2 (6+)(6 a) (C) 
 
GEOMETRICAL PROGRESSION. 803 
 
 As =12 and b=l3, b-\- a=25, and b a=l. 
 Therefore, (C) becomes (a? 2/) 2 =4& 2 X25X 1. 
 
 By evolution, a? y=2bX5 
 
 Equation () #-{-i/=2 2 
 
 By addition 2# 
 
 Or x= 6 2 -f- 5&=(6-f 5)6=18X13=234 
 
 By subtraction, 2y=2b 2 Wb 
 
 y= b 2 56=(6 5)6= 8X13=104. 
 A more brief solution is the following : 
 Let x y and x-\-y represent the numbers. 
 Then x= the arithmetical mean, >Jx 2 y z = the geometri- 
 cal mean, (Art. 112), and = the harmonical mean. By 
 
 the question, 
 
 x~ I3 = jx 2 y 2 (1), and - ^-4-12 = 7^ y 2 (2) 
 The right hand members of equations (1) and (2) being the 
 
 y2 yZ 
 
 same, therefore, ^--{-I2=x 13. 
 
 x 
 
 By reduction, y 2 =25x. 
 Put this value of y z in equation (1), and by squaring 
 
 # 2 26z-l-(13) 2 =z 2 25*, or a;=(13) 2 =169. 
 Hence, i/=65, and the numbers are 104 and 234. 
 
 4. Divide the number 210 into three parts, so that the last 
 shall exceed the first by 90, and the parts be in geometrical pro- 
 gression, dns. 30, 60, and 120. 
 
 5. The sum of four numbers in geometrical progression is 
 30 ; and the last term divided by the sum of the mean terms is 
 1|. What are the numbers ? Ans. 2, 4, 8, and 16. 
 
 6. The sum of the first and third of four numbers in geo- 
 metrical progression is 1 48, and the sum of the second and 
 fourth is 888. What are the numbers ? 
 
 Ans. 4, 24, 144, and 864. 
 
204 ELEMENTS OF ALGEBRA. 
 
 7". It is required to find three numbers in geometrical progres- 
 sion, such that their sum shall be 14, and the sum of their squares 
 84. Jlns. 2, 4, and 8. 
 
 8. There are four numbers in geometrical progression, the 
 second of which is less than the fourth by 24 ; and the sum of 
 the extremes is to the sum of the means, as 7 to 3. What are 
 the numbers ? Jlns. 1, 3, 9 and 27. 
 
 9. The sum of four numbers in geometrical progression is 
 equal to the common ratio -}-! and the first term is y 1 ^. What 
 are the numbers ? fins. T 1 ^, T \, T \, T . 
 
 10. The sum of three numbers in harmonical proportion is 
 26, and the product of the first and third is 72. What are the 
 numbers ? Jlns. 12, 8, and 6 
 
 11. The continued product of three numbers in geometrical 
 progression is 216, and the sum of the squares of the extremes 
 is 328. What are the numbers ? tins. 2, 6, 18. 
 
 12. The sum of three numbers in geometrical progression is 
 13, and the sum of the extremes being multiplied by the mean, 
 the product is 30. What are the numbers ? 
 
 1, 3, and 9 
 
 13. There are three numbers in harmonical proportion, the 
 sum of the first and third is 18, and the product of the three is 
 576. What are the numbers ? Jlns. 6, 8, 12. 
 
 14. There are three numbers in geometrical progression, the 
 difference of whose difference is 6, and their sum 42, What 
 are the numbers? Jlns. 6, 12,24. 
 
 15. There are three numbers in harmonical proportion, the 
 difference of whose difference is 2, and three times the product 
 of the first and third is 216. What are the numbers ? 
 
 Jlns. 6, 8, and 12. 
 
 16. Divide 120. dollars between four persons, in such a 
 way, that their shares may be in arithmetical progression ; and 
 if the second and third each receive 12 dollars less, and tho 
 
GEOMETRICAL PROPORTION. 205 
 
 fourth 24 dollars more, the shares would then be in geometri- 
 cal progression. Required each share. 
 
 Jlns. Their shares were 3, 21, 39, and 57, respectively. 
 
 17. There are three numbers in geometrical progression, 
 whose sum is 31, and the sum of the first and last is 26. What 
 are the numbers ? Jlns. 1, 5, and 25. 
 
 18 The sum of six numbers in geometrical progression is 
 189, and the sum of the second and fifth is 54. What are the 
 numbers ? rfns. 3, 6, 12, 24, 48 and 96. 
 
 19. The sum of six numbers in geometrical progression is 
 189, and the sum of the two means is 36. What are the num- 
 bers? Am. 3, 6, 12, 24, 48 and 96. 
 CHAPTER III. 
 PROPORTION. 
 
 (Art. 176.) We have given the definition of geometrical pro- 
 portion in (Art. 41.) and demonstrated the most essential prop- 
 erty, the equality of the products between extremes and means. 
 We now propose to extend our investigations a little farther. 
 
 Proportion can only exist between magnitudes of the same 
 kind, and the number of times and parts of a time, that one 
 measures another, is called the ratio. Ratio is always a num- 
 ber, and not a quantity. 
 
 (Theorem 1.) If two magnitudes have the same ratio as 
 two others, the first two as numerator and denominator may 
 form one member of an equation ; and the other two magnitudes 
 as numerator and denominator will form the other member. 
 
 Let A and B represent the first two magnitudes and r their ratio. 
 
 Also C and D the other two magnitudes, and r their ratio. 
 
 Then,^=r and =r Therefore, (Ax. 7) = 
 
 (Theorem 2.) Magnitudes which are proportional to tlie 
 same proportionals, are proportional to each other. 
 
 Suppose a : b : : P : Q \ Then we are to prove that 
 and c: d:: P: Q V a:b::c:d 
 
 and x : y : : P : Q ) and a : b : : x : y, &c. 
 
200 ELEMENTS OF ALGEBRA. 
 
 b Q 
 From the first proportion, -=73 
 
 From the second, = ~ 
 
 Hence, (Ax. 7) = or a : b : : c : d 
 
 In the same manner we prove a : b : : x : y 
 And c : d : : x : y 
 
 (Theorem 3.) If four magnitudes constitute a proportion, 
 the first will be to the sum of the first and second, as the 
 third is to the sum of the third and fourth. 
 
 By hypothesis, a : b : : c : d ;^then we are to prove that 
 a : a-\-b : : c : c-\-d. 
 
 By the given proportion, =-. Add unity to both members, 
 
 then reducing to the form of a fraction we have = . 
 
 a c 
 
 Throwing this equation into its equivalent proportional form, we 
 have ' a : a+b : : c : c+d. 
 
 N. B. In place of adding unity, subtract it, and we shall find 
 
 that a : a-b : : c : c-d. 
 
 or a : b a : : c : d c. 
 
 (Theorem 4.) If four magnitudes be proportional, the sum 
 of the first and second is to their difference, as the sum of 
 the third and fourth is to their difference. 
 
 Admitting that a : b : : c : d, we are to prove that 
 a-\-b : a b : : c-{-d : c d 
 
 X 
 
 From the same hypothesis, (Theorem 3.) gives 
 
 a : a-\-b : : c : c-\-d 
 And a : a b : : c : c d 
 
 Changing the means, (which will not affect the product of- the 
 extremes and means, and of course will not destroy proportion- 
 ality,) and we have, 
 
GEOMETRICAL PROPORTION. 207 
 
 arc:: -}-& : c-\-d 
 
 a i c : : a b : c d 
 
 Now by (Theorem 2.) a-{-b : c-\-d : : a b : c d 
 Changing the means, a-\-b : a b : : c-\-d : c d 
 
 (Theorem 5.) If four magnitudes be proportional, like 
 powers or rootsof the same, will be proportional. 
 
 Admitting a : b : : c : d, we are to show that 
 
 n n n n 
 
 a n : b n : : c n : er, and a : b : : c : d 
 
 By the hypothesis, ~j- = -7' Raising both members of this 
 equation to the nth power, and 
 
 a n c n 
 
 Changing this to the proportion a n : b n : : c n : d n 
 Resuming again the equation T=-y and taking the nth root 
 
 of each member, we have j- =7-. Converting this equation 
 
 r~**~ j"~ 
 
 b d 
 into its equivalent proportion, we have 
 
 JL _L i i 
 
 n , n n ,~n~ 
 
 a : b : : c id. 
 
 Now by the first part of this theorem, we have 
 
 m jrn m m 
 
 a n : b" : : c n : d" , m representing any 
 power whatever, and n representing any root. 
 
 (Theorem 6.) If four magnitudes be proportional, also 
 four others, their compound, or product of term by term, will 
 form a proportion. 
 
208 ELEMENTS OF ALGEBRA. 
 
 Admitting that a : b : : c . d 
 
 And x : y : : m : n 
 
 We are to show that, ax : by : : me : nd 
 From the first proportion, T = y 
 
 From the second, -= 
 
 y n 
 
 Multiply these equations, member by member, and 
 
 ax_mc 
 by nd 
 Or ax : by : : me : nd. 
 
 The same would be true in any number of proportions. 
 (Theorem 7.) Taking the same hypothesis as in (Theorem 
 6.) we propose to show, that a proportion may be formed by di- 
 viding one proportion by the other, term by term. 
 By hypothesis, a : b : : c : d 
 And x : y : : m : n 
 
 Multiply extremes and means, ad=bc (1) 
 And nx=my (2) 
 
 Divide (1) by (2), and -X -=- X - 
 x n m y 
 
 Convert these four terms, which make two equal products, into 
 a proportion, and we shall have 
 
 a ^ b c f d 
 x ' y m ' n 
 
 By comparing this with the given proportions, we find it com- 
 posed of the quotients of the several terms of the first propor- 
 tion divided by the corresponding term of the second. 
 
 (Theorem 8.) If four magnitudes be proportional, we may 
 multiply the first couplet or the second couplet, the antecedents 
 or the consequents, or divide them by the same factor, and the 
 results will be proportional in every case. 
 
GEOMETRICAL PROPORTION. 209 
 
 Suppose a : b : : c : d 
 
 Multiply ex. and means, and ad=bc (I) 
 
 Multiply this eq. by ?n, and .... madmbc 
 
 Now, in this last equation, ma may be considered as a single 
 term or factor, or md may be so considered. So, in the second 
 member, we may take mb as one factor, or me. Hence we may 
 convert this equation into a proportion in four different ways. 
 
 Thus, as . . ma : mb : : c : d 
 
 or as a : b :: me : md 
 
 or as ma : b : : me : d 
 
 or as a : mb :: c : md 
 
 If we resume the original equation (1), and divide it by any 
 number, m, in place of multiplying it, we can have, by the same 
 course of reasoning, 
 
 a 
 
 b 
 
 : : c 
 
 : d 
 
 m 
 
 m 
 
 
 
 a 
 
 : b 
 
 c 
 
 d 
 
 
 
 m 
 
 m 
 
 a 
 
 : b 
 
 c 
 
 : d 
 
 m 
 
 
 m 
 
 
 
 b 
 
 
 d 
 
 a 
 
 
 
 : : c 
 
 __ 
 
 
 m 
 
 
 771 
 
 The following examples are intended to illustrate the practi- 
 cal utility of the foregoing theorems : 
 
 EXAMPLES. 
 
 ! Find two numbers, the greater of which shall be to the 
 less, as their sum to 42 ; and as their difference is to 6. 
 
 Let a?=the greater, y=the less. 
 18 
 
210 ELEMENTS OF ALGEBRA. 
 
 ( x : y : : x-\-y :42 (1) 
 fhen.pcr quest, j x . * . . ^ : 6 fo 
 
 (Theorem 2.) x-\-y : 42 : : # y : 6 
 
 Changing the means x-}-y ' x y : : 42 : 6 
 (Theorem 4.) 2x : 2y : : 48 : 36 
 
 (Art. 42.) x : y : : 4 : 3 
 
 (Theorem 2.) 4:3:: xy : 6 
 
 And 4:3:: x+y : 42 
 
 From these last proportions, x y= 8 
 
 And x+y=5Q. Hence, x=32, i/=24. 
 
 2. Divide the number 14 into two such parts, that the 
 quotient of the greater divided by the less shall be to the quo 
 dent of the less divided by the greater, as 16 to 9. 
 
 Let #== the greater part, and 14 a?=the less. 
 
 x 14 x 
 
 By the conditions, - : -- : : 16:9 
 14 x x 
 
 Multiplying terms, x 2 : (14 a?) 2 : : 16:9 
 Extracting root, x : (14 x) : : 4:3 (Theor. 5.; 
 Adding terms, x : 14 : : 4:7 
 
 Dividing terms, x : 2 : : 4 : 1 
 
 Therefore, x=S. 
 
 3. There are three numbers in geometrical progression whose 
 sum is 52, and the sum of the extremes is to the mean as 10 to 
 3. What are the numbers ? Ans. 4, 12, and 36 
 
 Let x, xy, xy z represent the numbers. 
 Then, by the conditions, x-\-xy-\-xy z =52 (I) 
 
 And xy z -i-x : xy : : 10 : 3 
 (Art. 42.) yM-1 : y : : 10 : 3 
 
 Double 2d and 4th, ?/ 2 +l : 2y : : 10 : 6 
 
 Adding and sub. terms, 2 -f2?/4-l : y 2 2y-\-l : : 16 : 4 
 Extracting square root, 2/+1 : y 1 : : 4 : 2 
 Adding and sub. terms, y : 1 : : 3 : 1 Hence, y=3. 
 
 52 52 52 
 
 From equation (1), *=__=__- =-=4. 
 
GEOMETRICAL PROPORTION. *jll 
 
 4. The product of two numbers is 35, the difference of their 
 cubes, is to the cube of their difference as 109 to 4. What arc 
 the numbers ? Ans. 7 and 5 
 
 Let x and y represent the numbers. 
 
 Then, by the conditions, xy=35, and x 2 y* : (x y)* : : 109:4 
 Divide by (xy) (Art. 42.) and x 2 +xy-\-y* : (xy)* : : 109:4 
 Expanding, and # 2 -f xy -\-y z : x 2 2xy-{-y z : : 109:4 
 
 (Theorem 3.) 3xy : (z-^) ? ' : : 105:4 
 
 ttuiSxy, we know from the first equation, is equal to 105. 
 Therefore, (x y) 2 =4, or x y=2. 
 
 We can obtain a very good solution of this problem by putting 
 x-\-y= the greater, and x y= the less of the two numbers. 
 
 5. What two numbers are those, whose difference is to their 
 sum as 2 to 9, and whose sum is to their product as 18 to 77? 
 
 Jlns. 1 1 and 7 
 
 6. Two numbers have such a relation to each other, that if 4 
 be added to each, they will be in proportion as 3 to 4 ; and if 4 
 be subtracted from each, they will be to each other as 1 to 4 
 What are the numbers ? Ans. 5 and 8 
 
 7. Divide the number 16 into two such parts that their pro- 
 duct shall be to the sum of their squares as 15 to 34. 
 
 . 10, and 6. 
 
 8. In a mixture of rum and brandy, the difference between 
 the quantities of each, is to the quantity of brandy, as 100 is to 
 the number of gallons of rum; and the same difference is to the 
 quantity of rum, as 4 to the number of gallons of brandy. 
 How many gallons are there of each ? 
 
 tins. 25 of rum, and 5 of brandy. 
 
 9 There are two numbers whose product is 320 ; and the 
 difference of their cubes, is to the cube of their difference, as 61 
 to 1. What are the numbers? Am. 20 and 16. 
 
 1O. Divide 60 into two such parts, that their product shall be 
 to the sum of their squares as 2 to 5. Ans. 40 and 20 
 
 
21-2 ELEMENTS OF ALGEBRA. 
 
 11. There are two numbers which are to each other as 3 to 
 2. If 6 be added to the greater and subtracted from tie less, 
 the sum and the remainder will be to each other, as 3 to 1. 
 What are the numbers ? Jlns. 24 and 1 6. 
 
 12. There are two numbers, which are to each other, as 16 
 co 9, and 24 is a mean proportional between them. What are 
 the numbers 1 Ans. 32 and 18. 
 
 13. The sum of two numbers is to their difference as 4 to 1, 
 and the sum of their squares is to the greater as 102 to 5. 
 What are the numbers 1 Jlns. 15 and 9. 
 
 14. If the number 20 be divided into two parts, which are to 
 each other in the duplicate ratio of 3 to 1, what number is a 
 mean proportional between those parts ? 
 
 Jlns. 18 and 2 are the parts, and 6 is the mean proportion 
 between them. 
 
 15. There are two numbers in proportion of 3 to 2 ; and if 
 6 be added to the greater, and subtracted from the less, the results 
 will be as 3 to 1. What are the numbers ? Jins. 24 and 16. 
 
 16. There are three numbers in geometrical progression, the 
 product of the first and second, is to the product of the second 
 and third, as the first is to twice the second ; and the sum of 
 the first and third is 300. What are the numbers ? 
 
 Jlns. 60, 120, and 240. 
 
 17. The sum of the cubes of two numbers, is to the differ- 
 ence of their cubes, as 559 to 127 ; and the square of the first, 
 multiplied by the second, is equal to 294. What are the num- 
 bers ? rfns. 7 and 6. 
 
 18. There are two numbers, the cube of the first is to the 
 square of the second, as 3 to 1 ; and the cube of the second is 
 to the square of the first as 96 to 1. What are the numbers ? 
 
 tins. 12 and 24. 
 
BINOMIAL THEOREM. 213 
 
 SECTION VI. 
 
 CHAPTER I. 
 
 INVESTIGATION AND GENERAL APPLICATION OF 
 THE BINOMIAL THEOREM. 
 
 (Art. 127.) It may seem natural to continue right on to the 
 higher order of equations, but in the resolution of some cases in 
 cubics, we require the aid of the binomial theorem ; it is there- 
 fore requisite to investigate that subject now. 
 
 The just celebrity of this theorem, and its great utility in the 
 higher branches of analysis, induce the author to give a general 
 demonstration : and the pupil cannot be urged too strongly to 
 give it special attention. 
 
 In (Art. 67.) we have expanded a binomial to several powers 
 by actual multiplication, and in that case, derived a law for form- 
 ing exponents and coefficients when the power was a whole 
 positive number ; but the great value and importance of the 
 theorem arises from the fact that the general law drawn from that 
 case is equally true, when the exponent is fractional or negative, 
 and therefore it enables us to extract roots, as well as to ex- 
 pand powers. 
 
 (Art. 128.) Preparatory to our investigation, we must prove 
 the truth of the following theorem : 
 
 If there be two series arising from different modes of ex- 
 panding the same, or equal quantities, with a varying quan- 
 tity having regular powers in each series ; then the coefficients 
 of the same powers of the varying quantity in the two series 
 are equal. 
 
 For example, suppose 
 
 Jl+Bx+Cy?+nx*, &c. =a-\-bx+cx z +dx*, &c. 
 This equation is true by hypothesis, through all values of a?. 
 It is true then, when #=0. Make this supposition, and <ft=a 
 Now let these equal values be taken away, and the remainder 
 divided by x. Then again, suppose #=0, and we shall find 
 fi=b. In the same manner we find Cc, D=d, E=e, &c. 
 
214 ELEMENTS OF ALGEBRA. 
 
 (Art. 129.) A binomial in the form of a-\-x may be put in 
 the form of a X f 1 -j- - ) ; for we have only to perform 
 the multiplication here indicated to obtain a-\-x. Hence 
 
 (x\ m 
 1 + - ) > it will be sufficient to mul- 
 
 tiply every term of the expanded series by a m for the expansion 
 of (-j-.r) OT , but as every power or root of 1 is t, the first term 
 
 / x \ m 
 of the expansion of ( l-f~ J is 1, and this multiplied by a m 
 
 must give a m for the first term of the expansion of (a-\-x) m , what- 
 ever m may be, positive or negative, whole or fractional. 
 
 99 
 
 As we may put x in place of -, we perceive that any bino- 
 
 mial may be reduced to the form of (1-j-o?), which, for greater 
 facility, we shall operate upon. 
 
 (Art. 130.) Let it be required to expand (l-f-a;) m , when m is 
 a positive whole number. By actual multiplication, it can be 
 shown, as in (Art. 67.) that the first term will be 1, and the 
 second term mx. For if m=2, then 
 
 (l+a?) m =(l+a?) 8 =l+2ar, &c. 
 If m=3, (l-hr) ra =l+3x, &c. 
 And in general, (l-\-x) m =l+mx-\-tfx z ,+JBx*, &c. 
 
 The exponent of x increasing by unity every term, and .#, 
 It, C, &c., unknown coefficients, which have some law of de- 
 pendence on the exponent m, which it is the object of this investi- 
 gation to discover. 
 
 (Art. 131.) Now if m is supposed to be a fraction, or if m=-, 
 the expansion of (l-}-x) m will be a root in place of a power, and 
 
 we must expand (l-\-x) r . _i 
 
 For example, let us suppose r=2, then r 
 

 BINOMIAL THEOREM. 21 
 
 and to examine the form the series would take, let us actually 
 undertake to extract the square root of (l-\-x) by the common 
 rule. 
 
 OPERATION. 
 
 2-r-ar _i^ 
 
 Thus we perceive that in case of square root, the first term of 
 the series must be unity, and the coefficient of the second term 
 is the index of the binomial, and the powers of x increase by 
 unity from term to term. 
 
 We should find the same laws to govern the form of the series, 
 if we attempted to extract cube, or any other root; but, to be 
 general and scientific, we must return to the literal expression 
 
 Now as any root of 1 is 1, the first term of this root must be 
 1, and the second term will have some coefficient to x. Let 
 that coefficient be represented by p ; and as the powers of r will 
 increase by unity every term, we shall have 
 
 \ &c. 
 
 Take the r power of both members, and we shall have 
 l+x^^+px+dx 2 , &c.) r 
 
 As r is a whole number, we can expand this second mem- 
 ber by multiplication ; that is, by (Art. 130.), the second mem- 
 ber must take the following form 
 
 l+*=l-|-rjoa:-M'* 2 , &c. 
 
 Drop 1, and divide by x, and we have 
 
 Ky (Art. 128.) l=rp or p=- ; 
 
210 ELEMENTS OF ALGEBRA. 
 
 That is, the coefficient represented by p must be equal to tho 
 index of the binomial. 
 
 Therefore (\-\-x) r = \ + \.x-{- Ay?-\- Bx* -f, &c. ; the same 
 general form as when the exponent was considered a whole 
 number. 
 
 (Art. 132.) If we take m=- we have to expand the root of 
 a power. The first term must be 1, and the second term will 
 contain a?, with some coefficient, and the coefficients of x will 
 rise higher and higher every term. 
 
 n 
 
 That is, (l+x)^ = l+7?^+^^ 2 , &c. Take the r power of 
 both members, and (l-}-x) n =(l-\-px t &c.) r . 
 Expanding both members, as in (Art. 130), 
 
 l+nx-{-ax z , &c.=l+ rp.r-f-.tftf 2 , &c. 
 Now, by (Art. 128), 71=773 or p=^. 
 
 n 
 
 Therefore, (1-J-J?) r =l+ x+Jlx*+E3?, &c. ; the first two 
 
 terms following the same law, relative to the exponents, as in the 
 former cases. Now let us suppose m negative. Then 
 
 (1-Hr) 1 * will become (l-\-x)~ m = (Art. 18.) 
 
 Or by (Art. 130.) - * 2 
 
 ; l+maj-f-^a: 2 , &c. 
 
 By actual division, l+mx+rfx*, &c.) 1 (J mx-\-Jl'3f, &c. 
 
 mx 3? 
 
 mx m 2 x 2 
 
 That is, (\-\-x)~ m =lmx J rJl l yr 1 which shows that the same 
 general law governs the coefficient of the second term, as in the 
 former cases. 
 
 Hence it appears that whether the exponer* m of a binomial 
 
BINOMIAL THEOREM. 217 
 
 be positive or negative, whole or fractional, th same general 
 form of expression must be preserved. 
 
 That is, in all cases (l+ar) m =l+roar+-foM- Btf, &c. 
 
 (Art 133.) For clearness of conception, let the pupil bear in 
 mind that the coefficients of an expanded binomial quantity 
 depend not at all on the magnitudes of the quantities themselves, 
 but on the exponent. Thus, (a-\-b) to the 5th, or to any other 
 power, the coefficients will be the same, whether a and b are 
 great or small quantities, or whatever be their relation to each 
 other. 
 
 (Art. 134.) The equation (l+x^^l+mx+rftf+Ba*, &c., 
 is supposed to be true, therefore it must be true, if we square 
 both members. But we have only a portion of one member. 
 We have, however, as much as we please to assume, and suffi- 
 cient to determine the leading terms of its square, which is all 
 that we desire. Square both members, and (l-}-2a:-{-# 8 ) m = 
 (l+mH-^-l-^-f-Ck 4 , &c.) 2 . 
 
 By expanding the second member, and arranging the terms 
 according to the powers of tf, we shall have 
 
 
 Now if we assume yZx-}-^, the first member of this equa- 
 tion becomes (l-j-t/) m . If we expand this binomial into a series. 
 it must have the same coefficients as the expansion of (l-f-#) OT , 
 because the coefficients depend entirely on the exponent m. 
 (Art. 133.) 
 
 Therefore, l+*=l+rm/-Mi 2 -r- 8 -f &c - 
 
 Put the values of y, T/*, y 3 , &c., in this equation, and arrange 
 the terms according to the powers of #, and we have 
 
 SB 
 
 * t <fcc. (2) 
 
 The left hand members of equations (1) and (2), are identical, 
 19 
 
218 ELEMENTS OF ALGEBRA. 
 
 and the coefficients of like powers in the second member must 
 be equal. (Art. 128.) 
 
 Therefore, 4^-l-m=m 2 +2^, or 
 And 8 
 
 mu r r> - - - /o\ 
 
 Therefore B=~~- ^=m. - -- - (3) 
 
 o . o 
 
 By putting the coefficients of the fourth powers of x equal, 
 we have 
 
 To obtain the value of C from this equation, in the requisite 
 form, is somewhat difficult. 
 
 We must make free use of the preceding values of Jl and B, 
 which are alone sufficient ; but, to facilitate the operation, we 
 shall make use of the following auxiliary. 
 
 Assume P=m 2, then P-\-im 1, 
 
 m,P-\-m m 1 - , . 
 
 and o = m .-=rf (a) 
 
 Also, by inspecting equation (3), we perceive that 
 
 35=^P, and 2m=^^ (b) 
 
 3 
 
 By putting the values of 125 and 2m5 in the primitive equa- 
 tion, and dividing every term by .#, and in the second member 
 taking the value of Jl from equation (), and we shall have 
 
 _i_^ p ,, 
 _____ :__ __ 
 
 Multiply by 6, and substitute the value of 
 24F-}- 6 =24?7i 42, because P=m 2, and we have 
 
 84O 
 
 __-|-24m 42=4m/M-3mP-r-3ra. 
 Jl 
 
 Collecting terms, and dividing by 7, gives 
 
BINOMINAL THEOREM. 2-1-9 
 
 But 3m 6=3P, which substitute and transpose, and 
 
 ll:=mP_3P=P(m 3) 
 yl 
 
 Or C^ f ( m ~ 3 )=m m ~ 1 m ~~ m ~ 3 
 
 ~~12 2 ~~3~" 4 
 
 Therefore the development of (l-\-x) m must be 
 
 2 2 
 
 234 
 
 whatever be the value of m, positive or negative, whole or 
 fractional, and thus far the law of development has been demon- 
 strated, and we infer, and can only infer, that this law will 
 continue true, whatever be the number of terms. Hence, the 
 demonstration is complete, only so far as we extend it. 
 
 This series will terminate when (771) is a whole positive 
 uumber, but in all other cases it will be infinite. 
 
 Few pupils, who pay attention, find difficulty in compre- 
 hending the preceding method of demonstration, and for that 
 reason we preferred it ; but to the following demonstration, no 
 objection as to completeness or perfection can be found. 
 
 SECOND DEMONSTRATION. 
 Assume (l+ x )*==l+Ax+x a + Cx*+2)x*-}- &c. (1) 
 
 Then (\J ry y=\J r Ay+By* + Cy*-\-Dy* + & c > W 
 By subtracting (2) from (1) and dividing the result by 
 (ff y) we obtain 
 
 x y xy xy xy 
 
 + <&c. (3) 
 
 Place (l+z)=P, and(l+y) = . (a) 
 
 Then xy=PQ. Whence (3) becomes 
 
 ^U c 
 
 y 
 
 If we divide (x 2 y 2 ) by (x y) the quotient will term- 
 inate with the second term, and (x 3 y 3 ) divided by (x y) the 
 quotient will terminate with its third term, and so on. 
 
220 ELEMENTS OF ALGEEBRA. 
 
 Then if n is a whole number, the division of the first member 
 as indicated will terminate with the n th term of the quotient. 
 
 Now, divide m the last equation as indicated, and after- 
 ward suppose xy. It then follows that P= Q, and we have 
 p*-i_^p*-a g_|_p- 3_|_ < c> to n terms. And if P= Q the 
 whole quotient must be nP D ~ l for the first member. 
 
 Whence, nP*- l =A+2Bx-{-3Cx s -{-4I>x 3 +5Ex* &c. 
 
 But P = l+x. 
 
 Multiply these two equations, member by member, and 
 we shall have, 
 
 But we perceive, by inspecting equation (1), that 
 
 By equating the right hand members of these two equa- 
 tions, and observing that the coefficients of the like powers of 
 x, must be equal, we shall have 
 
 or = 
 
 or C 
 
 or D= 
 
 
 <KC. 
 
 Whence A=n 
 
 2 
 
 n 1 n2 
 _- 
 
 n 1 n 2 n 3 
 
 234 
 
 and so on, the law of continuation being obvious from the 
 equations, to whatever term it may be extended. 
 
 Thus the demonstration is complete when n is a whole posi- 
 tive number. 
 
 Now let n be an irreducible fraction, as ~, 
 
BINOMIAL THEOREM. 221 
 
 1 ,.x 
 
 Then (l+*) t= =: 
 
 And 
 
 j. i^ 
 
 Now assume ( 1 -f-#) l =P and ( 1 +y ) l = 
 
 8 
 
 8 
 
 Then, (J+*) 1 =P 
 And l+a= 
 
 Whence zy=P i (>* (c) 
 
 Now if we subtract (b) from (), and divide the remainder 
 by x y, we shall have 
 
 Now divide numerator and denominator of the first member 
 by (P Q), and it will become 
 
 , fcc. 
 
 Now, because 5 and J are whole numbers, the division in both 
 numerator and denominator will be complete, and if we sup- 
 pose x=y, it follows that P= Q, and the results of the several 
 divisions will be 
 
 +, &c. (e) 
 
 But JP* =!+. 
 
 Multiplying these equations, member by member, will 
 produce 
 
 , 3 
 
 But by inspecting equation (a), and observing that (l-J-#)r 
 
222 ELEMENTS OF ALGEBRA. 
 
 s 
 
 -P =-+-.Ax+-Bx*-\--Cx*+, Ac. (g) 
 
 t t t ' t t 
 
 Equating (/) and (g), and operating as before, we find 
 
 A=f- 
 
 3 * 2 3~' 
 
 &c. &c. 
 
 The same law of development as before, and this completes the 
 demonstration for all positive fractional exponents. 
 
 Now, let the exponent be a negative fraction, 
 
 Then, (l~}-3;)-^=l+Ax-{-x a -\-Cx 3 -^, &c. (a') 
 
 * 
 
 g j 
 
 Now assume, (l-fz)-7= />" Then, (l+#)-T= p~ 4 
 
 Or, Pt = i_|^. 
 
 In like manner, Q l = 
 
 By subtraction, ^P l Q l x y 
 
 Subtracting (V) from (a') and dividing by (x y), as in the 
 other cases, we shall obtain 
 
 Ac. 
 
 The first member of this equation is 
 1 1 
 
 - 
 
BINOMIAL THEOREM. 223 
 
 But _=- 
 t 
 
 When the division is performed, and P supposed equal to Q. 
 hence, * P~ 8 " 1 =A+2 
 
 But, P i =l+x. 
 
 Whence, * P~ 8 " 1 =A+2x+3 Cx 2 +, &c. 
 
 Therefore, *-P * = 
 
 A 
 
 -, <fec. 
 
 But, - P =-JLLAa;-ix* , 
 
 t t t t 
 
 Whence, A= 8 - J5= l.lil!, &c. 
 t t 2 
 
 and thus the demonstration is shown to be complete, under 
 all possible circumstances. 
 
 We may use either of the three following forms :* 
 
 Ac. (1) 
 ^+,&c.(2) 
 
 2 a 2 2 3 a 
 
 2 3 a 
 
 &c. (3) 
 _ 1 ; !!Z^a-3 ;z .3 +> &c . 
 
 
 
 APPLICATION". 
 
 I. Required the square root of (l-|-a?). 
 Apply formula (1), making w*=-|- ; then the development 
 
 2.4 2.4.6 2.4.6.8 
 The law of the series, in almost every case, will become 
 apparent, after expanding three or four terms, provided we 
 keep the factors separate. 
 
 *In practical examples we may be required to expand (1 x) as well as 
 (1-t-tf), and the exponent may be negative as well as positive. But in all 
 cases the products in the second member must conform to the rules of mul- 
 tiplication. 
 
224 ELEMENTS OF ALBKBRA. 
 
 The above will be the coefficients for any binomial square 
 root (Art. 133 ); hence the square root of 2 is actually expressed 
 in the preceding series, if we make a?=l. 
 
 Then (1-f 1) 5 ==1-K_-L, & c . The square root of 3 w 
 expressed by the same series, when we make #=2, &c. 
 
 2. Required the cube root of (a-\-b) or its equal, af H V 
 Formula (2) gives 
 
 This expresses the cube root of any binomial quantity, or any 
 quantity that we can put into a binomial form, by giving the 
 proper values to a and b. For example, required the cube root 
 
 of 10, or its equal, 8+2. Here a=8, 6=2. Then ^=2, 
 
 and b -=$. 
 a 
 
 1 2 25 
 
 Therefore, 2(1+-^ - z +^ Q ' Q ^ &c.) is the cube root 
 
 of 10, and so for any other number. 
 
 S. Expand - -r into a series, or its equal, (a b)~ l . 
 
 1 b .b*b* 
 
 b* 6 4 . b* 
 
 4. Expand (a 2 -h&*) into a series, or its equal, M-| 2 
 
 Expand d(c*-}-x?) * into a series. 
 d a? 
 
BINOMIAL THEOREM. 225 
 
 6. Expand (a 2 #*)* into a series. 
 
 Ans. 
 
 f 
 
 2 5 a 2 2 7 a 4 
 
 
 7. Expand 
 
 8. Find the cube root of 
 
 a 3 +b 
 
 9. Find the cube root of 31. 
 
 31=27+4=27/^1+^ 
 
 277- 
 
 , . 
 
 . 1 - + -- - , &c 
 3a 3 9a 6 8 la 9 
 
 Place-i=a 
 
 27/ 27 
 
 Ans. 3l+_-. +.-- 3 +,&c. =3.156066 
 V 327 36 (27)^369 (27) 3 / 
 
 1O. Find the seventh root of 2245, or any other number, by 
 the binomial theorem. 
 
 N. B. Find by trial the greatest seventh power, less than the 
 given number, (2245), and divide the given number by it. 
 
 Thus, 3 7 =2187 
 
 2187 2187 2187 
 
 Ans. sf 1+1. Jl? __ A YJLY+&0. ^=3. 01 13574 
 \ ' 7 2187 7.14 \2187/ / 
 
 11. Find the cube root of 9, by the binomial series. 
 9=8(1+|). Cube root, 
 
 
 Ans. 2.080084. 
 19.' Find the cube root of 7, by the binomial series. 
 
 7=(8 1)=8(1 ). Cube root, 2(1 J)* 
 
 Ans. 1.912933. 
 13. Expand (x -y)" 1 , into a series. 
 
 n 1 
 Ans. x* tt B 
 
226 ELEMENTS OF ALGEBRA. 
 
 When n = 1, this series becomes 
 
 V^JJ/ 3 , y 3 , 
 SF?FiTF^, &c., Ac. 
 
 14. Find an approximate cube root of 100, by the binomial 
 series. 
 
 N. B. The nearest cube number to 100, is 125. 
 
 / 25 \ 
 Therefore, we place 100=125 25=125^1 ~\%). 
 
 Or, 100=125(l). Whence (100)^=5(1 $)%. 
 
 Ans. 4.64159, nearly. 
 
 Thus, by a little artifice, this series can be used to extract 
 the cube (or any other root), of any number. 
 
 15. Expand - into a series. 
 
 a+cx 
 
 N. B. It is much easier to expand all such expressions as 
 this, by actual division, than by the binomial theorem, and 
 we take the example, merely to show that it can be expanded 
 by the binomial theorem. 
 
 The above expression is the same as 1 ( ~r c ) x t 
 
 a-\-cx 
 
 Place -J-c=w, then we are required to expand 
 
 / M, nix/* , cx\ 
 mx(a+cx) n or, _(!+ ) 
 a \ a/ 
 
 because n= 1. 
 
 Because n= 1, this series becomes 
 
 IJL-i, Ac. 
 a 5 
 
 
 this multiplied by , and the product subtracted from 1, pro- 
 a 
 
 , mx , mcx 2 me 2 x 3 . mc 3 x* mc*x 5 ., u 
 
 duces 1 _ -4- ___ - -- -t- --- - , &c. the result 
 a a 2 a 2 a 4 a 5 
 
 sought, when (b-\-c) is written in place of m. 
 
INFINITE SERIES. 227 
 
 16. Expand ' , or its equal 14- _ , or its equal 
 x 1 x 1 
 
 -^. 
 1 x 
 
 Ans. 1 2* 2s 3 2* 3 2# 4 , &c., &o. 
 17 Expand z(l a?)-" 3 into a series. 
 
 Ans. *+3:r 2 +6z 3 -f-10z 4 -f. 
 
 18. Expand (64+1)*, or its equal, 8(1+^)*, into a series. 
 N. B. This is the same as demanding the cube root of 65. 
 
 CHAPTER II. 
 
 OF INFINITE SERIES. 
 
 (Art 136.) An infinite series is a continued rank, or progres- 
 sion of quantities in regular order, in respect both to magnitudes 
 and signs, and they usually arise from the division of one quantity 
 by another. 
 
 The roots of imperfect powers, as shown by the examples in 
 the last article, produce one class of infinite series. Some of 
 the examples under (Art. 121.) show the geometrical infinite 
 series. 
 
 Examples in common division may produce infinite series for 
 quotients ; or, in other words, we may say the division is con- 
 tinuous. Thus, 10 divided by 3, and carried out in decimals, 
 gives 3.3333, &c., without end, and the sum of such a series is 
 3|. (Art. 121.) 
 
 (Art. 137.) Two series may appear very different, which arise 
 from the same source ; thus 1, divided by 1-f-a, gives, as we may 
 bee, by actual division, as follows: 
 
 l-f-a)l (1 a-j-a 2 a 3 -f-a 4 , <fcc. without end. 
 
228 ELEMENTS OF ALGEBRA. 
 
 Also, +!)! ( -+- j, &c. without end. 
 
 ' \ a a* a* a* 
 
 a 
 
 These two quotients appear very different, and in respect to 
 blngle terms are so ; but in these divisions there is always a re- 
 mainder, and either quotient is incomplete without the remain- 
 der for a numerator and the divisor for a denominator, and when 
 these are taken into consideration the two quotients will be 
 equal. 
 
 We may clearly illustrate this by the following example : 
 Divide 3 by l-f-2, the quotient is manifestly 1 ; but suppose them 
 literal quantities, and the division would appear thus : 
 
 (3 6+12, &c. 
 3+6 
 
 6 
 612 
 
 12 
 12+24 
 
 24 
 
 Again, divide the same, having the 2 stand first. 
 
 |+f,&c. 
 
 Now let us take either quotient, with the real value of its re- 
 mainder, and we shall have the same result. 
 
 Thus, 3+12=15; and 6, and the remainder 24 divided 
 
INFINITE SERIES. 229 
 
 by 3, gives 8, which make 14 ; hence, the whole quotient 
 is 1. 
 
 Again, H-t=V, and |i=l. 
 
 Hence, y |=f =1 the proper quotient. 
 
 If we more closely examine the terms of these quotients, we 
 shall discover that one is diverging, the other converging, and 
 by the same ratio 2, and in general this is all a series can show, 
 the degree of convergency. 
 
 (Art. 138.) We convert quantities into series by extracting 
 the roots of imperfect powers, as by the binomial, and by actual 
 division, thus : 
 
 1. Convert -. into an infinite series. 
 a-\-x 
 
 Thus, +*, (l_+_ + , & c . 
 
 2. Convert - into an infinite series. 
 a x 
 
 Observe that these two examples are the same, except the 
 signs of x : when that sign is plus the signs in the series will be 
 alternately plus and minus ; when minus, all will be plus. 
 
 3. What series will arise from - ? 
 
 1 x 
 
 Jlns. 1+2*+ 2*M-2# 3 , &c. 
 Observe that in this case the series commences with 2x. The 
 
 unit is a proper quotient, and the series arises alone from r -- 
 
 1 z, 
 
 the remainder after the quotient 1 is obtained. 
 
230 ELEMENTS OF ALQEBRA. 
 
 4. What series will arise from -r- r ^ ? 
 
 ,,,2 ^4 
 _ f___^ 
 
 a a 4 
 
 Observe, in this example, the term x, in the numerator does 
 not find a place in the operation ; it will be always in the re- 
 
 mainder : therefore,^-; - will give the same series. 
 
 2 -- z 
 
 5. What series will arise from dividing 1 by 1 a-}- a 2 , or 
 
 from 
 
 - - -, -? rfns. 1-l-a a 3 a 4 -}-a 6 +a 7 a 9 a 10 , 
 1 a-j-cr 
 
 In this example, observe that the signs are not alternately plus 
 and minus, but two terms in succession plus, then two minus ; 
 this arises from there being two terms in place of one after the 
 minus sign in the divisor. 
 
 6. What series will arise from y~~ ? 
 
 fins. a-j-ar-r-ar 2 -|-ar 3 -l-ar 4 , &c. 
 
 Observe that this is the regular geometrical series, as appears 
 in (Art. 118.) 
 
 7. What series will arise from - - ? 
 
 rfns. 1-f-l-f-l-fl, &c 
 
 That is - is 1 repeated an infinite number of times, or infinity, 
 a re/ U corresponding to observations under (Art. 60.) 
 
SUMMATION OF SERIES. 231 
 
 8. What series will arise from the fraction ? 
 
 If =&, this series will be -, repeated an infinite number of 
 
 times. 
 
 This series can also be expanded by the binomial theorem, 
 
 for . j=h(a 6) . This observation is applicable to seve- 
 ral other examples. 
 
 (Art. 139.) A fraction of a complex nature, or having com- 
 
 j / 
 
 pound terms, such as 2 , may give rise to an infinite 
 
 1 &XdX 
 
 series, but there will be no obvious ratio between the terms. 
 Some general relation, however, will exist between any one term 
 and several preceding terms, which is called the scale of rela- 
 tion, and such a series is called a recurring series. Thus 
 the preceding fraction, by actual division, gives l-}-x-{-5x c *-\r 
 ISa^-f-^la^+lSltf 5 , &c., a recurring series, which, when carried 
 to infinity, will be equal to the fraction from which it is derived. 
 
 l+2a? 
 Expand -^ into a series 
 
 CHAPTER III. 
 
 SUMMATION OF SERIES. 
 
 (Art, 140.) We have partially treated of this subject in geo- 
 metrical progression, in (Art. 121) ; the investigation is now 
 more general and comprehensive, and the object in some respects 
 different. There we required the actual sum of a given number 
 of terms, or the sum of a converging infinite series. Here the 
 series may not be in the strictest sense geometrical, and we may 
 not require the sum of the series, but what terms or fractional 
 quantities will produce a series of a given convergent]/. 
 
232 ELEMENTS OF ALGEBRA. 
 
 The object then, is the converse of the last chapter ; and for 
 every geometrical series, our rule will be drawn from the sixth 
 
 example in that chapter ; that is, - - , a being the first term 
 
 of any series, and r the ratio. We find the ratio by dividing 
 any term by its preceding term. 
 
 Hence, to find what fraction may have produced any geomet- 
 rical series, we have the following rule: 
 
 RULE. Divide the first term of the series by the alge- 
 braic difference between unity and the ratio. 
 
 EXAMPLES. 
 
 1. What fraction will produce the series 2, 4, 8, 16, &c? 
 
 2 
 
 Here a=2, and r=2 ; therefore, - - Jlns 
 
 1 Z 
 
 2. What fraction will produce the series 3 9-J-27 81, &c.? 
 Here 0=3, and r= 3; then r=3, 
 
 Hence, -fL- 
 
 3. What fraction will produce the series ^, T 7 , &c. 
 Here = T V and r= T V; therefore, 
 10 3 31 
 
 4. What fraction will produce the series, 
 _+__ 3 , & c . ? [See example 1, (Art. 138.)] 
 
 x 1 a 
 
 Here 0=1, and r= - ; then - = = , 
 a .x a-i-x 
 
 a 
 
 5. What fraction will produce the series 
 &c. ? [See example 3, (Art. 138.) and the observation in con- 
 nection.] 
 
SUMMATION OF SERIES. 233 
 
 2# 
 
 Here a=2x, r=x\ then - - , will give all after the first 
 
 2iX 1 l % 
 term: therefore, 1+- -- =- - , flns. 
 
 , w . e d db . db 3 (/6 s 
 6. What fraction will produce the series - H F -- 4 
 
 c er c* c* 
 
 &c. ? AM. 
 
 What fraction will produce the series 7+- r^ i rr~ ' 
 
 'b ax' 
 8. What fraction will produce the series 
 
 2ab 
 
 3 ,- ' FT' 
 
 a a 2 a 3 a-J-o 
 
 9. What fraction will produce the series 
 
 !-}- a 3 a 4 4-a 6 +a 7 a 9 a 10 , &c. ? 
 See example 5, (Art. 138.) 
 
 Put l+a=6; then a 3 -J-a 4 =a 3 6, and a e + 7 =a 6 6, and the 
 scries becomes b a?b-\-a*b , <fec. 
 
 b 1-fa 1 
 
 Hence, the fraction required is = 3=^-: 5 =- : 5. 
 
 1-j-a 3 1+a 3 1 +a 2 
 
 10. What fraction will produce the series a?+a^-f-a^, &c. ? 
 
 Ana. .. 
 1 x 
 
 If a?s=l f this expression becomes j f~o' a S 7 m bol of in 
 fmity 
 
 11. What fraction will produce the series 1-f-f+JL, &c. ? 
 
 1 5 
 
 53 
 
 Hence, the sum of this series, carried to infinity, is 2. 
 20 
 
234 ELEMENTS OF ALGEBRA. 
 
 In the same manner, we may resolve every question in (Art. 
 
 tin.) 
 
 12. What is the sum oi the series, or what fraction will pro- 
 duce the series 1 a-\-a? 4 -f-a 6 a'-\-cP , &c. ? 
 
 Jlns. 
 
 13. What is the value of the infinite series 
 
 RECURRING SERIES. 
 
 (Art. 141.) We have explained recurring series in (Art. 139.) 
 and it is evident that we cannot find their equivalent fractions 
 by the operation which belongs to the geometrical order, as no 
 common relation exists between the single terms. The fraction 
 
 |_l_2<>' 
 
 --- 2> by actual division, gives the series l-|~3#-f-4#M-7a: 8 
 
 L^~~3C' i 3C 
 
 -\-llx* -\-\8x*, &c., without termination ; or, in other words, the 
 division would continue to infinity. 
 
 Now, having a few of these terms, it is desirable to find a 
 method of deducing the fraction. 
 
 There is no such thing as deducing the fraction, or in fact no 
 fraction could exist corresponding to the given series, unless 
 order or a law of dependence exists among the terms ; therefore 
 some order must exist, but that order is not apparent. 
 Let the given series be represented by 
 
 d+B+C+D+E+F, &c. 
 
 Two or three of these terms must be given, and then each sue 
 oeeding term may depend on two or three or more of its pre- 
 ceding terms. 
 
 In cases where the terms depend on two preceding terms we 
 may have 
 
 C=mBx-}-nJlx* 
 J)=mCx-\-nBx* 
 
RECURRING SERIES. 236 
 
 In cases where the terms, or law of progression depend on 
 three preceding terms we may have 
 
 fi=m 
 
 E=mDx+n 
 F=m 
 &c.=&c. 
 
 The reason of the regular powers of x coming in as factors, 
 will be perfectly obvious, by inspecting any series. 
 
 The values of m, n and r express the unknown relation, or law 
 that governs the progression, and are called the scale of relation 
 We shall show how to obtain the values of these quantities in a 
 subsequent article. 
 
 (Art. 142.) Let us suppose the series of equations (1), to be 
 extended indefinitely, or, as we may express it, to infinity, and 
 add them together, representing the entire sum of Jl-\-B-\- 
 C-\-D, <fec., to infinity, by S ; then the first member of the 
 resulting equation must be (S J? B), and the other member 
 is equally obvious, giving 
 
 S A 
 
 A+BmJlx 
 
 Hence, S=-l -- r (a) 
 
 1 mx no? 
 
 In the same manner, from equations (2), we may find 
 
 s _J+B+G-(Jl+B)mx--Jlna* , 
 1 mx nx* rx* * ' 
 
 (Art. 143.) The/orw of a series does not depend on the 
 value of x, and any series is true for all values of x. Equations 
 (1) then, will be true, if we make 2=1. 
 
 Making this supposition, and taking the first two equations of 
 the series (1), we have 
 
 C^mB+nA ) (c) 
 
 And D=mC+nB \ 
 
 In these equations, ,/?, B, (7, 7J, are known, and m and n un- 
 known ; but two unknown quantities can be determined from two 
 equations ; hence m and n can be determined 
 
236 ELEMENTS OF ALGEBRA. 
 
 When the scale of relation depends upon three terms, wt 
 take three of the equations (2), making x=l, and we determine 
 m, n, and r, as in simple equations. 
 
 EXAMPLES. 
 1. What fraction will produce the series 
 
 To determine the scale of relation, we have w#=l, .#=3, 
 =4, and D=7. Then from equations (c), we have 
 n+3m=4 
 3n+4m=7 
 
 These equations give m=l, and 72= 1. 
 Now to apply the general equation (a), we have Jj=\ t 23=3x. 
 
 A+B-mAx 1+3*-* 1+2* 
 1 hen o= - 5- = -; -- ?=; -- a t * 
 1 mx nx* 1 x ar 1 x or 
 
 2. What fractional quantity will produce the series 
 
 l + 6z+12arM-48z 3 -f 120;r 4 , &c., to infinity? 
 Here .#=1, B=Qx, m=l, n=6. 
 
 Hence, by applying equation (a), we find - - j for the 
 
 1 ~^x o* , 
 
 expression required. 
 
 3. What quantity will produce the series 
 
 l_L3;r-f Sa^-fTar'-i-Qtf 4 , to infinity ? l-|-ar 
 
 4. What quantity will produce the infinite series 
 
 1-f 4^+6^+11^+ 28^ 4 +63^, <fec. ? in which m=2, n= 1, 
 r=3. 
 
 [Apply equation (6).] ^ns. 
 
 5. What fraction will produce the series 
 
 , &c.? 
 
REVERSION OF A SERIES. 
 6. What fraction will produce the series 
 
 , &c. 
 
 287 
 
 REVERSION OF A SERIES- 
 
 (Art. B.) To revert a series is to express the value of the un- 
 known quantity in it, (which appears in the several terms under 
 regular powers,) by means of another series involving the powers 
 of some other quantity. 
 
 Thus, let x and y represent two indeterminate quantities, and 
 the value of y be expressed by a series involving the regular 
 powers of x. That is 
 
 &c. 
 
 To revert this, is to obtain the simple value of ar, by means 
 of another series containing only the known quantities, a, b, c, 
 d, &c. and the powers of y. 
 To accomplish this,asswme 
 
 *, &c.* 
 
 Substitute the assumed value of x for x, x*, a? 8 , <fcc., in the pro- 
 posed series, and transposing y, we shall have 
 
 1 
 
 y-\-aB 
 
 +3 cA*B 
 
 &c. 
 
 * By examining previous authors, we have found none that explain the 
 rationale of such assumptions : but these are points on which the learner 
 requires the greatest profusion of light. We explain thus : the term x must 
 have some value, positive, negative, or zero, and the series Ay-\-By 2 -\- Cy*, 
 fcc., can be positive and of any magnitude. It can be negative, and of any 
 magnitude, by giving the coefficients A, B and C, negative values. It can bo 
 zero by making ,y=0. Therefore it can express the value of or, whatever that 
 may be ; and because the powers of y are regular, the substitution of this 
 value of x for the several powers of x, in the primitive series, will convert 
 that series into regular powers of y, which was the object to be accomplished. 
 
238 
 
 ELEMENTS OF ALGEBRA. 
 
 As every term contains y, \ve can reduce the equation by 
 dividing by y ; and afterwards, in consideration that the equation 
 must be true for all values of y : making y=0, we shall have 
 
 I 1=0 or 
 
 .*-! 
 
 a 
 
 Continuing the same operation and mode of reasoning as in (Art. 
 128.), and we shall find, in succession, that 
 
 n 
 
 a 
 
 2b z ac 
 
 a 1 
 
 = &c. 
 
 Substituting these values of .#, B, C, &c., in the assumed 
 equation and we have the value of x in terms of , b, c, &c., 
 and the powers of y as required, or a complete reversion of the 
 series. 
 
 EXAMPLES. 
 
 1. Revert the series i/=x-}-x' 2 -}-j: 3 , &c. 
 Here a=l 6=1 e=l, &c. 
 Assume x=^.y-f-% 2 +Oj/ 3 -[-&c. 
 
 Result, ^= +i/ 3 i/ 4 + 5 &c. 
 
 ft. Revert the series x=y J - -\-- - &c. 
 
 234 
 
 Here a I b=% c=| J= | &c. 
 
 Assume y^Ax+Btf+Ctf+Dx*-^ &;c., and find 
 the values of rf, B, C, &c., from the formulas (F), 
 
 Result, ^+l. + _+.-. & c 
 
LOGARITHMS. 
 
 3. Revert the series y=rc+3a? 2 +5a >3 +7a? 4 +9a ;5 , &c. 
 
 Result x= 
 
 239 
 
 In case the given series is of the form of xay-\- by*-}-cy 5 , 
 &c., the powers of y varying by 2, the equations (F] will not 
 apply, and we must assume y=J2x-\-J3x 3 -\-Cx 5 , &c., and sub- 
 stitute as before, and we shall find 
 
 (0) 
 
 
 a 10 
 
 In common cases, after the coefficients, as far as D, are deter- 
 mined, the law of continuation will become apparent, especially 
 if the factors are kept separate. 
 
 CHAPTER IV. 
 EXPONENTIAL EQUATIONS AND LOGARITHMS, 
 
 (Art. 144.) We have thus far used exponents only as known 
 quantities ; but an exponent, as well as any other quantity, may 
 be variable and unknown, and we may have an equation in the 
 form of a*=b. 
 
 This is called an exponential equation, and the value of x can 
 only be determined by successive approximations, or by making 
 use of a table of logarithms already determined. 
 
 (Art. 145.) Logarithms are exponents. A given constant 
 number may be conceived to be raised to all possible powers, and 
 thus produce all possible numbers ; the exponents of such pow- 
 ers are logarithms, each corresponding to the number produced. 
 
 Thus, in the equation a*=6, x is the logarithm of the number 
 b ; and to every variation of x, there will be a corresponding 
 
240 ELEMENTS OF ALGEBRA. 
 
 variation to b a is constant, and is called the base of the sys 
 tern, and differs only in different systems. 
 
 The constant a cannot be 1, for every power of 1 is 1, and the 
 variation of x in that case would give no variation to b ; hence, 
 the base of a system cannot be unity ; in the common system it 
 is 10. 
 
 In the equation 10 Z =2, x is, in value, a small fraction, and 
 is the logarithm of the abstract number 2. 
 
 In the equation a x =b, if we suppose x=l, the equation becomes 
 a l =a; that is, the logarithm of the base of any system is unity. 
 
 If we suppose 37=0, the equation becomes a=l ; hence, the 
 logarithm of 1 is 0, in every system of logarithms. 
 
 (Art. 146.) The logarithms of two or more numbers added 
 together give the logarithm of the product of those numbers, 
 and conversely the difference of two logarithms gives the lo- 
 garithm of the quotient of one number divided by the other. 
 
 For we may have the equations z =&, a v =b', and a*=b". 
 
 Multiply these equations together, and as we multiply powers 
 ly adding the exponents the product will be 
 
 Hence, by the definition of logarithms, x-\-y-\-z is the loga- 
 rithm for the number represented by the product bb'b". Again. 
 
 divide the first equation by the second, and we have a x ~ y =j- i ; 
 
 and from these results we find that by means of a table of loga- 
 rithms multiplication may be practically performed by addition, 
 and division by subtraction, and in this consists the great utility 
 of logarithms. 
 
 (Art. 147.) In the equation a x =b, take a=10, and x succes- 
 sively equal to 0, 1, 2, 3, 4, &c. 
 
 Then 10=1, 10^10, lO^lOO, 10 3 =1000, &c. 
 Therefore, for the numbers 
 
 1, 10, 100, 1000, 10000, 100000, &c., we have for corres- 
 ponding logarithms 
 
 0, 1, 2, 3, 4, 5, <fcc. 
 
LOGARITHMS. 241 
 
 Here it may be observed that the numbers increase in geometri- 
 cal progression, and their logarithms in arithmetical progression. 
 
 Hence the number which is the geometrical mean between two 
 given numbers must have the arithmetical mean of their lo- 
 garithms, for its logarithm. 
 
 On this principle we may approximate to the logarithm of any 
 proposed number. For example, we propose to find the /a- 
 garithm of 2. 
 
 This number is between 1 and 10, and the geometrical mean 
 between these two numbers, (Art. 122.), is 3.16227766. The 
 arithmetical mean between and 1 is 0.5; therefore, the number 
 3.16227766 has 0.5 for its logarithm. 
 
 Now the proposed number 2 is between 1 and 3.162, &c.,and 
 the geometrical mean between these two numbers is 1.778279, 
 and the arithmetical mean between 0. and 0.5 is 0.25 ; therefore, 
 the logarithm of 1.778279 is 0.25. 
 
 Now the proposed number 2 lies between 1.778279, and 
 3.16227766, and the geometrical mean between them will fall 
 near 2, a little over, and its logarithm will be 0.375. Continuing 
 the approximations, we shall at length find the logarithm of 2 to 
 be 0.301030, and in the same manner we may approximate to 
 the logarithm of any other number, but the operation would be 
 very tedious. 
 
 (Art. 148.) We may take a reverse operation, and propose a 
 logarithm to find its corresponding number; thus, in the general 
 equation a z =n; x may be assumed, and the corresponding value 
 of n computed. 
 
 Thus suppose ar= T 3 ff ; then (10) T =n, or 10 3 =n 10 . 
 
 Hence, n=J 1000=1. 9952623 15. 
 
 That is, the number 1.9952, <fcc., (nearly 2) has 0.3 for its 
 logarithm. In the same way we may compute the numbers cor- 
 responding to the logarithms 0.4, 0.5, 0.6, 0.7, &c. 
 
 (Art. 149.) We may take another method of operation to find 
 the logarithm of a number. 
 21 
 
242 ELEMENTS OF ALGEBRA. 
 
 Let the logarithm of 3 be required. 
 The equation is 10 Z =3, the object is to find x. 
 It is obvious that x must be a fraction, for 10 1= =10 ; there 
 fore, 
 
 Let #= . Then 10 Z '=3, or 10=3*. Here, we perceive, 
 that x' must be a little more than 2. Ma.te x'=2-{- p Then 
 
 Hit x 1 
 
 3 2 X3 =10; or 3 = . 
 
 7 
 
 Or ( ~) =3. 
 
 Here we find by trial that x" is between 10 and 1 1 ; take it 10, 
 and a?'=2-f-y^ ; hence, a?=!,j=0. 476, for a rough approxima- 
 tion for the logarithm of 3. A more exact computation gives 
 0.4771213; but all these operations are exceedingly tedious, and 
 to avoid them, mathematicians have devised a more expeditious 
 method by means of a converging series ; which we shall inves- 
 tigate in a subsequent article. 
 
 (Art. 150.) It is only necessary to calculate directly the lo- 
 garithms of prime numbers, as the logarithms of all others may 
 be derived from these. Thus, if we would find the logarithm 
 of 4, we have only to double that of 2; for, taking the equation 
 (10) z =2, and squaring both members we have, (10) 2z =4 ; or 
 taking the same equation, and cubing both members, we have 
 (10) 3z =8 ; which shows that twice the logarithm of 2 is the 
 logarithm of 4, and three times the logarithm of 2 is the loga- 
 rithm of 8 ; and, in short, the sum of two or more logarithms 
 corresponds to the logarithm of the product of their numbers, 
 (Art. 142.), and the difference of two logarithms corresponds to 
 the logarithm of the quotient, which is produced from dividing 
 one number by the other. 
 
 Thus, the logarithm of j log. a log. b. Tne abbrevia- 
 tion, (log. a), is the symbol for the logarithm of a. 
 
LOGARITHMS. 243 
 
 (Art. 151.) As the logarithm of 1 is 0, 10 is 1, 100 is 2, &c., 
 we may observe that the whole number in the logarithm is one 
 less than the number of places in the number. 
 
 The whole number in a logarithm is called its characteristic, 
 and is not given in the tables, as it is easily supplied. For ex- 
 ample, the integral part of the logarithm of the number 67430 
 must be 4, as the number has 5 places. The same figures will 
 have the same decimal part for the logarithm when a portion of 
 them become decimal. 
 
 Thus, 67430 logarithm 4.82885 
 
 6743.0 3.82885 
 
 674.30 2.82885 
 
 67.430 1.82885 
 
 6.7430 0.82885 
 
 .67430 1.82885 
 
 For every division by 10 of the number, we must diminish 
 the characteristic of the logarithm by unity. 
 
 The decimal part of a logarithm is always positive ; the index 
 or characteristic becomes negative, when the number becomes 
 less than unity. 
 
 By reference to (Art. 18.), we find that =10-', = = 
 
 10~ 2 , &c. That is : fractions may be considered as numbers 
 with negative exponents, and logarithms are exponents ; there- 
 fore the logarithm of , or .1, is 1 ; of , or .01, is 2, 
 10 100 
 
 &c. If any addition is made to .01 the logarithm must be more 
 than 2 ; but, for convenience, we still let the index remain 
 2, and make the decimal part plus. Hence the index alone 
 must be considered as minus. 
 
 Negative numbers have no logarithms; for, in fact, as tot 
 have before observed, there are no such numbers. 
 
 (Art. C.) We now design to show a practical method of com- 
 puting logarithms. The methods hitherto touched upon were 
 only designed to be explanatory of the nature of logarithms ; but, 
 to calculate a table, by either of those methods, would exhaust the 
 
244 ELEMENTS OF ALGEBRA. 
 
 patience of the most indefatigable. To arrive at easy practical 
 results requires the clearest theoretical knowledge, and we must 
 therefore frequently call the attention of students to first prin- 
 ciples. 
 
 The fundamental equation is a*=b, in which a is the con 
 slant base and x is the logarithm of the number b ; and b may 
 be of any magnitude or in any form, if it be essentially positive. 
 
 Now we may take a=l+c and 6=1 -j 
 
 Then the fundamental equation becomes (l-f-c) z =(l-| J (1) 
 
 Here ar=log. (l+i)=log.(^i)=log.(/i+l)-]og./i (2) 
 
 Raise both members of equation (1) to the nth power, then 
 we shall have 
 
 Expand both members into a series by the Binomial Theorem, 
 
 Then l+nxc+nx. ^^c^+nx . ^^ . ^=^c 3 4- 
 
 I 
 
 nx 1 nx 2 nx 3 4 1 . n 1 1 . 
 
 nX 2 8 4-'+' &C " = l+n -j +n -2-p* + 
 
 n 1 n 2 1 , n 1 n 2 n 3 1 
 n. 
 
 a 3 
 
 Drop unity, from both members, and divide by n, then we have 
 
 <, nx 1 , . nx 1 nx 2 . nx 1 nx 2 nx 3 . 
 c+- a -c'+- . ^-c'+- 2 - . -jr- f-S 
 
 \ In 1 1 , n 1 n 2 I n 1 n2 n 3 1 
 
 "*" ) ~~ p 2""'/" 1 2~'~3~y~ h ~2~'~3~~T~'^ 
 
 +, &c. 
 
 This equation is true for all values of n. It is, therefore, true 
 
LOGARITHMS. 945 
 
 when =0. Making this supposition and the above equation 
 reduces to 
 
 c 2 c 3 c 4 c 5 \ 1 1,1 1.1 
 
 - 2 - +rr +5 ^rrw^w^W'^' w 
 
 As c remains a constant quantity for all variations of x and /), 
 the series in the vinculum may be represented by the symbol AI 
 
 Then xM=- -- - 2 -f -L - L+J_ & c . (3) 
 jo 2/> 2 3/ 4jo 4 5/ 
 
 Take the value of x from equation (2), and substitute it in 
 equation (3), 
 
 Then [\ og .(p+l)-l os .p-]M= l ~;L+;L-+, &c. (4) 
 
 Again, we may have the fundamental equation 
 ( I - Y in which y is the logarithm of ( I -- Y the same as 
 
 of(l+l). 
 
 W as 
 
 Or 2/=log.(- ) = log.(j9 1) lg P" (5) 
 
 Operating on the equation (l-j-c) 1/ =l , the same as w* 
 
 did on equation (1), we shall find 
 
 Subtract equation (6) from equation (4), and we obtain 
 
 /I 1 1 1 \ 
 
 [1og.(;;-fl) log.(p 1)]M=2( ^~5~i~^"K~iH ^ & Ct ) P; 
 
 Dividing by M, and considering that log/^- - )=log.(p-f-l). 
 log. ( p 1), the equation can take this form 
 
246 ELEMENTS OF ALGEBRA. 
 
 As p may be any positive number, greater than 1, make 
 " -=1-J . Then j0=2*4-l, and equation (8) becomes 
 
 log \~~F L ) = 
 
 + 
 
 3(2H-1)* ' 5(2z-r-l) 5 ' 
 
 By this last equation we perceive that the logarithm of (z-f-1) 
 will become known when the log. of z is known, and some 
 value assigned for the constant M. Baron Napier, the first dis- 
 coverer of logarithms, gave M the arbitrary value of unity, for 
 the sake of convenience. 
 
 Then, as in every system of logarithms, the logarithm of 1 is 
 0, make z=l, in equation (9), and we shall have 
 
 h^Lrh &cA=.O.C9314718. 
 
 )= 
 
 This is called the Napierian logarithm of 2, because its magni- 
 tude depends on Napier's base, or on the particular value of M 
 being unity. 
 
 Having now the Napierian logarithm of 2, equation (9) will 
 give us that of 3. Double the log. of 2 will give the logarithm 
 of 4. Then, with the log. of 4, equation (9) will give the loga- 
 rithm of 5, and the log. of 5 added to the log. of 2, will give the 
 logarithm of 10. 
 
 Thus the Napierian logarithm of 10 has been found to great 
 exactness, and is 2.302585093. 
 
 The Napierian logarithms are not convenient for arithmetical 
 computation, and Mr. Briggs converted them into the common 
 logarithms, of which the base is made equal to 10. 
 
 To convert logarithms from one system into another, we 
 may proceed as follows : 
 
 Let e represent the Napierian base, and a the base of the 
 common system, and N any number. 
 
LOGARITHMS. 247 
 
 Also, let 3: represent the logarithm of JV, corresponding to 
 the base , and y the logarithm of N 9 corresponding to the 
 base c. 
 
 Then a x =N t 
 
 and & ~N. 
 
 Now, by inspecting these equations, it is apparent that if the 
 base a is greater than the base e, the log. x will be less than the 
 log. y. 
 
 These equations give a T e y . 
 
 Taking the logarithms of both members, observing that x and 
 y are logarithms already, we have 
 
 a? log. a=y \og.e. 
 
 This equation is true, whether we consider the logarithms 
 taken on the one base or on the other. Conceive them taken on 
 the common base, then 
 
 log. 0=1, and x=y\Q<r*e (10) 
 
 x 
 or log.e=-. 
 
 y 
 
 In this equation x and y must be logarithms of the same num- 
 ber, and therefore if we take a?=l, which is the logarithm of 
 10, in the common system, y must be 2.302585093, as pre- 
 viously determined. 
 
 Hence log. f = 2 L- =0.434394482. . . . (11) 
 
 This last decimal is called the modulus of the common sys- 
 tem ; for by equation (10) we perceive that it is the constant 
 multiplier to convert Napierian or hyperbolic logarithms into 
 common logarithms. 
 
 But equation (9) gives Napierian logarithms when M=\\ 
 therefore the same equation will give the common logarithms by 
 causing M to disappear, and putting in this decimal as a factor. 
 
 Equation (9) becomes the following formula for computing 
 common logarithms : 
 
ELEMENTS OF ALG'KBKA 
 log.(z-r-l) log.*= 
 
 0.868588961^ 
 
 (-L_+ ^+^-1--,+, Ac.) (F) 
 
 To apply this formula, assume z= 10. Then 
 log. *=1, and 2z-f-l=21. 
 
 21 
 
 441 
 
 0.86858896 
 
 0.0413613791 = 0.041361379 
 
 93792 3 = 31264 
 
 2125 = 42 
 
 .041392685=sum of series. 
 log.lO= 1.0 
 
 log.(*+l) = 1.041 392685 =log.ll. 
 
 If we make 2=99, then (z-H) = 100, and Iog.(*-H)=s2, 
 and 224-1=199. 
 
 199 
 39601 
 
 0.86858896 
 
 436477-H =0.00436477 
 11-A_3= 4 
 
 0.00436481 =sum of series. 
 
 Hence 2.00000log.99-=0.00436481 
 By transposition log.99=l. 99563519 
 
 Subtract log.l l=1.04139269=log.l 1 
 
 log. 9=0.95424234 
 log.9=!og. 3=0.47712 117=log.3. 
 
 Thus we may compute logarithms with great accuracy ami 
 rapidity, using the formula for the prime numbers only. 
 
 By equation (11) we perceive that the logarithm of the Na- 
 pierian base is 0.434294482 ; and this logarithm corresponds to 
 the number 2.7182818, which must be the base itself. 
 
 We may also determine this base directly : 
 
 In the fundamental equation (1), the base is represented by 
 (1+c). In equation (A), c must be taken of such a value as 
 
LOGARITHMS. 
 
 249 
 
 c 8 c 3 c 4 
 shall make the series c +- --f, &c., equal to 1. But to 
 
 determine what that value shall be, in the first place, put 
 c 2 , c 3 c 4 
 
 ^-sf+s-T &c ' 
 
 Now by reverting the series (Art. B.), we find that 
 
 But, by hypothesis, the series involving c equals unity that is, 
 y=l. Therefore 
 
 By taking 12 terms of this series, we find (l-j-c)=2.7182818, 
 the same as before. 
 
 TABLE I. LOGARITHMS FROM J TO 100. 
 
 N. 
 
 Log. 
 
 N. 
 
 Log. 
 
 N. 
 
 Log. 
 
 N. 
 
 Log. 
 
 1 
 
 000000 
 
 26 
 
 1 414973 
 
 51 
 
 1 707570 
 
 76 
 
 1 880814 
 
 2 
 
 301030 
 
 27 
 
 1 431364 
 
 62 
 
 1 716003 
 
 77 
 
 1 886491 
 
 3 
 
 477121 
 
 28 
 
 1 447158 
 
 53 
 
 1 724276 
 
 78 
 
 1 892095 
 
 4 
 
 602060 
 
 29 
 
 1 462398 
 
 54 
 
 1 732394 
 
 79 
 
 1 897627 
 
 5 
 
 698970 
 
 30 
 
 1 477121 
 
 55 
 
 1 740363 
 
 80 
 
 1 903090 
 
 6 
 
 778151 
 
 31 
 
 491362 
 
 56 
 
 1 748188 
 
 81 
 
 1 908485 
 
 7 
 
 845098 
 
 32 
 
 505150 
 
 57 
 
 1 755875 
 
 82 
 
 1 913814 
 
 8 
 
 903090 
 
 33 
 
 518514 
 
 58 
 
 1 763428 
 
 83 
 
 1 919078 
 
 9 
 
 954243 
 
 34 
 
 531479 
 
 59 
 
 1 770852 
 
 84 
 
 1 924279 
 
 10 
 
 1 000000 
 
 35 
 
 544068 
 
 60 
 
 1 778151 
 
 85 
 
 1 929419 
 
 11 
 
 1 041393 
 
 36 
 
 556303 
 
 61 
 
 1 785330 
 
 86 
 
 1 934498 
 
 12 
 
 1 079181 
 
 37 
 
 568202 
 
 62 
 
 1 792392 
 
 87 
 
 1 939519 
 
 13 
 
 113943 
 
 38 
 
 579784 
 
 63 
 
 1 799341 
 
 88 
 
 1 944483 
 
 14 
 
 146128 
 
 39 
 
 591065 
 
 64 
 
 1 806180 
 
 89 
 
 1 949390 
 
 15 
 
 176091 
 
 40 
 
 602060 
 
 65 
 
 1 812913 
 
 90 
 
 1 954243 
 
 16 
 
 204120 
 
 41 
 
 612784 
 
 66 
 
 1 819544 
 
 91 
 
 1 959041 
 
 17 
 
 230449 
 
 42 
 
 623249 
 
 67 
 
 1 826075 
 
 92 
 
 1 963788 
 
 18 
 
 255273 
 
 43 
 
 633468 
 
 68 
 
 1 832509 
 
 93 
 
 1 968483 
 
 19 
 
 278754 
 
 44 
 
 643453 
 
 69 
 
 1 838849 
 
 94 
 
 1 973128 
 
 20 
 
 301030 
 
 45 
 
 653213 
 
 70 
 
 1 845098 
 
 95 
 
 1 977724 
 
 21 
 
 322219 
 
 46 
 
 662578 
 
 71 
 
 1 851258 
 
 96 
 
 1 982271 
 
 22 
 
 342423 
 
 47 
 
 672098 
 
 72 
 
 1 857333 
 
 97 
 
 1 986772 
 
 23 
 
 361728 
 
 48 
 
 681241 
 
 73 
 
 1 863323 
 
 98 
 
 1 991226 
 
 24 
 
 380211 
 
 49 
 
 690196 
 
 74 
 
 1 869232 
 
 99 
 
 1 995635 
 
 25 
 
 397940 
 
 50 
 
 698970 
 
 75 
 
 1 875061 
 
 100 
 
 2 000000 
 
250 
 
 ELEMENTS OF ALGEBRA. 
 
 TABLE II. LOGARITHMS OF LEADING NUMBERS WITHOUT INDICES. 
 
 
 
 1 
 
 2 
 3 
 4 
 6 
 6 
 7 
 8 
 9 
 
 N. 10Q. N. 101. 
 
 N. 1(W ! N. 10^. N. 104. 
 
 N. 105. 
 
 N. U)6. 
 
 N. 107. 
 
 N. 108.! N. 109. 
 
 000000 
 000434 
 000867 
 001301 
 001734 
 002166 
 002598 
 003029 
 003461 
 003891 
 
 0043-21 
 004750 
 005881 
 006609 
 006038 
 006466 
 006894 
 007321 
 007748 
 008174 
 
 003600 
 009026 
 009451 
 009876 
 010300 
 010724 
 011147 
 011570 
 011993 
 012415 
 
 012837 
 013259 
 013680 
 014100 
 014521 
 014940 
 015360 
 015779 
 016197 
 016616 
 
 017033 
 017451 
 017868 
 018284 
 018700 
 019116 
 019532 
 019947 
 020361 
 020776 
 
 021189 
 021603 
 022016 
 022428 
 022841 
 023252 
 023664 
 024075 
 024486 
 024896 
 
 025306 
 025715 
 026125 
 026653 
 026942 
 027350 
 027757 
 028164 
 028571 
 J028978 
 
 029384 
 029789 
 030195 
 030600 
 031004 
 031408 
 031812 
 032216 
 032619 
 1033021 
 
 033424 
 033826 
 034227 
 034628 
 035029 
 035430 
 035830 
 036230 
 036629 
 037028 
 
 0374-26 
 037825 
 038223 
 038620 
 039017 
 039414 
 039811 
 040207 
 040602 
 040998 
 
 (Art. 15la.) The preceding tables are sufficient to give us 
 the logarithms of any number whatever, provided we under- 
 stand the theory of logarithms, and are able to use a few prac- 
 tical artifices. 
 
 If to this knowledge, and these artifices, we add the formula, 
 we can with ease find the log. of any number, however large, 
 or however small, to any required degree of accuracy.* 
 
 We can, in fact, find the logarithm of any number, very 
 nearly, by Table I. 
 
 ILLUSTRATIONS. 
 
 1. We require the logarithm of 520; but 520 is not in the 
 table, 52 and 10 are. 
 
 Log. of 52 is 
 Log. of 10 is 
 
 Log. of 520, therefore, is 
 
 2 The logarithm of 5.2 is, 
 And of .52 is 
 
 1.716003. 
 1.000000. 
 
 2.716003. 
 
 0.716003, 
 1.716003, (fee., <fec. 
 
 3. What is the logarithm of 146 ? 
 
 146=73X2. Whence log. 73, 1.863323 
 log. 2, 301030 
 
 Log of 146 is 2.164353 
 
 *It is very bad policy to give a learner the use of voluminous tables ; 
 such tables are made to supply the place of theory, and of thought. Con- 
 tracted tables, like these, are the best educators. 
 
LOGARITHMS. 251 
 
 4. What is the logarithm of 3244 ? 
 
 3240=60 54. 3250=65 50. 
 
 Log. 60, 1.778151 Log. 65, 1.812913 
 
 Log. 54, 1.732394 Log. 50, 1.698970 
 
 Log. 3240, is 3.510545. Log. 3250, is 3.511883. 
 
 (Art. 1515.) Here we have the logarithm of 3240, and of 
 3250 ; and the number 3244 is intermediate, therefore its loga- 
 rithm must be intermediate between 3.510545 and 3.511883. 
 The difference between the two extreme numbers is 10, and 
 the difference of their logarithms is .001338. 
 
 Therefore, by proportion, 10 : .001338 : : 4. 
 
 That is -^ths of the difference of the known logarithms is 
 the correction to add to the less logarithm to obtain the loga- 
 rithm sought. 
 
 Whence to log. 3240 3.510545 
 
 Add .0001338X4 635 
 
 Log. of 3244 is 3.511080 
 This example shows that Table I., and the true theory of 
 
 logarithms, are sufficient to find the logarithm of any number 
 
 whatever. 
 
 ANOTHER METHOD. 
 (Art. 151c.) When the number is large, as in the present 
 
 example, the first term of the series will be the increase of the 
 
 logarithm corresponding to one ; four times this would be the 
 
 increase corresponding to four, and so on. 
 
 The logarithm of 3240 is 3.510545, as before found, and 
 
 here z=3240, and 22+1=6481, 
 
 .86858896 
 Whence, T- - =0.000134 
 
 Whence to losr. 3.540545 
 
 Add 0.000134X4 536 
 
 Log. 3244 3.511081 
 
 Ar.d the two methods agree the first may be clearest to a 
 learner, but the second is more brief and scientific. 
 
252 ELEMENTS OF ALGEBRA. 
 
 (Art. 151d.) Table II. requires some explanation. The 
 numbers are at the top and side, and only the decimal part of 
 the logarithm is contained in the table. The operator, in each 
 particular case, will supply the index, according to the theory 
 
 of logarithms. 
 
 EXAMPLES. 
 
 1. Find the logarithm of 1013. Ans. 3.005609. 
 We find 101 at the top, and 3 at the side. Under the former, 
 
 and opposite the latter, we find the decimal .005609, and this 
 is the decimal part of the logarithms of all numbers expressed 
 by the figures 1013, as they here stand. 
 
 Thus, it is the decimal logarithm of 1013, 101.3, 10.13, 1.013, 
 or of the decimals .1013, .01013, &c. 
 
 It is also the decimal logarithm of 10130, 101300, <fcc., but 
 this was sufficiently illustrated in (Art. 151). 
 
 2. Find the logarithm of 1 .075. Ans. 0.031408. 
 
 3. Find the logarithm of 10940. Ans. 4.039017. 
 Here, under 109, at the top, and opposite 4, at the side, we 
 
 find the decimal, and we take 4 for the index, because there 
 are five places of figures, in whole numbers. 
 
 4. Find the logarithm of 101 .5. Ans. 2.031408. 
 
 5. Find the logarithm o/. 001043. Ans. 3.018284. 
 
 (Art. 1510.) Table II. will be very useful in solving prob- 
 lems in compound interest, and annuities ; but its greatest 
 utility is in solving the following problems : 
 
 6. Find the logarithm of 5&1521. Ans. 5.753977. 
 Divide by the two superior figures (in this example it is 56), 
 
 and the quotient is 10134^|. 
 
 The divisor, consisting of two figures, its logarithm can 
 always be found in Table I. 
 
 And the four superior figures of the quotient (any quotient) 
 will be found in Table II. 
 
 The quotient in this example may be written thus, 
 10130+41!-. 
 
 The logarithm of 10130 (the decimal part of it), can be 
 taken directly out of Table II. It is .005609 ; and the next 
 
LOGARITHMS. 253 
 
 greater (the number below) is .006038 ; the difference between 
 the two is .000429, and the correction for 4\% must be 4 
 times one tenth of .000429, which is .000180. 
 
 Whence log. 10130 - 4.005609 Table II. 
 
 Correction for 4}|, - - .000180 
 
 Log. of quotient, 10134^, 4.005789 
 
 Log. of 56, - - 1.748188 Table I. 
 
 Log. of 567521, 5.753977. 
 
 Thus we may find the logarithm of any number by the use of 
 these two tables. We give one more example. 
 
 7. Find the logarithm of 365. 25638 the days and parts of 
 a day in a siderial year. 
 
 (N. B. As three figures are whole numbers, the index must 
 be two. Divide by 36, and taking five figures in the quotient 
 as whole numbers, we shall have 10146.01 for a quotient.) 
 This quotient may be written in the following form : 
 
 10140+6.01. 
 
 The Log. of 36, Table I., - 0.556303 
 
 Log. of 10140, Table II., - 0.006038 
 
 6.01 
 Correction for -y^- of diff.* .000257 
 
 Sum +2.=log. of 365.25638, 2.562598 
 
 (Art. 151/.) To find the number corresponding to a given 
 logarithm we take the converse operation. 
 
 For example, we take the logarithm just found, and demand 
 its corresponding number. 
 
 Log. (omit index), 0.562598 
 
 Next less in Table I. 36, - .556303 
 
 .006295 
 Next less in Table II. 10140, .006038 
 
 .000257 
 
 *The difference in the table is the difference between .006038 and the 
 next greater .006466, which is .000428, one-tenth of this .000042.8, multi- 
 ply by (6.01) and the product is .000257. 
 
254 ELEMENTS OF ALGEBRA. 
 
 Divide this difference by the tabular difference, which is 
 .000428, and we obtain 6, and a very small remainder. 
 
 Whence, to 10140, add 6, and we have 10146, which being 
 multiplied by 36, produces 365.256, for the number sought. 
 
 We take three of the figures for whole numbers, because 
 the index of the given logarithm is 2. Had it been 3, we 
 should have taken four figures for whole numbers. 
 
 1. What number corresponds to the logarithm 0.317879 ? 
 
 Ans. 2.079118. 
 
 From the given log. 0.317879 
 
 Sub. next less. Table I., 2, 0.301030 
 
 0.016849 
 Next less. Table II., 1039, 0.016616 
 
 0.000233 
 
 Divide 233 by 417 (tab. diff.), and we obtain 559, nearly, 
 which we annex to 1039, making 1039559, which, multiplied 
 by 2, produces 2.079118, the answer. 
 
 2. What number corresponds to the logarithm 4.031718? 
 
 Ans. 10757.673. 
 
 As the first figure in the decimal is 0, we had better use 
 Table II. at once. 
 
 The number next less in the table is the log. of 1075. As 
 the index is 4, we must have another figure for whole numbers. 
 Hence, from the given log., 4.031718 
 
 Subtract log. of 10750 (Table II.), which is 4.031408 
 
 .000310 
 
 This difference is less than any other logarithm in Table II. 
 Therefore, we can obtain no other factor, but we can correct 
 10750. 
 
 The next greater logarithm in the table is that of 10760, 
 which is .031812 ; the tablular diff. is .000404. 
 
 Therefore, .000404 : 000310 : : 10 : correction. 
 
 Or, 404 : 310 : : 10 : correction = 7.673. 
 
LOGARITHMS. 255 
 
 3. The logarithm of the British gallon is 2.442909; a cubic 
 inch being the unit. How many cubic inches does it contain ? 
 
 Ans. 277.27+ 
 
 (Art. 151^.) Logarithms are exponents; and as powers and 
 roots may be expressed by exponents, therefore, logarithms 
 may be used for finding powers and roots. 
 
 To extract the square root of any number, a, we divide its 
 
 exponent (one understood), by 2, and the result is a? . To 
 
 extract the third root, fourth root, &c., we divide the exponent 
 by 3, 4, &c., as the case may be. 
 
 The exponent of a number is the logarithm of that number. 
 Hence, to extract roots by logarithms, we have the following 
 rule : 
 
 RULE. Take the logarithm of the number and divide it by 2 for 
 the square root, by 3 for the third root, by 4 for the fourth root, 
 and so on. 
 
 The quotient will be the logarithm of the root sought, and the 
 number corresponding to this logarithm will be the root itself. * 
 
 (Art. 15U.) If x=a 3 Then log. x2> log. a. 
 
 By hypothesis, x=a.a.a. 
 
 By taking the logarithm of each member, we have 
 
 log. x= log. a + log. a -j- log. a = 3 log. a. ( 1 ) 
 
 Therefore, in general terms, 
 
 If #=a n log. x=n log. ff.(2) 
 
 Now, let n be a fraction, then xcfo. 
 Cube both members ; then x 3 =a. That is, x.x.x=a. 
 By the property of logarithms we have 
 
 log. x + log. x + log. x = log. a. 
 Or, 3 log. x = log. a. 
 
 Or, log. x = i log. a. (3) 
 
 All these equations conform to the rule just given. 
 
 * Examples can be taken out of any Arithmetic. 
 
256 ELEMENTS OF ALGEBRA. 
 
 USE AND APPLICATION OF LOGARITHMS. 
 
 (Art. 152.) The sciences of trigonometry, mensuration, and 
 astronomy alone, can develop the entire practical utility of lo- 
 garithms. The science of algebra can only point out their 
 nature, and the first principles on which they are founded. To 
 explain their utility, we must suppose a table of logarithms formed, 
 corresponding to all possible numbers, and by them we may re- 
 solve such equations as the following : 
 
 1. Given 2*= 10 to find the value of x. 
 
 If the two members of the equation are equal, the logarithms 
 of the two members will be equal, therefore take the logarithm 
 of each member ; but as a? is a logarithm already, we shall have 
 v log. 2=log. 10. 
 
 2. Given (729)* =3, to find the value of x. 
 
 Raise both members to the x power, and 3 X =729=9 3 , 
 Or 3 Z =3 6 . Hence, #=6. 
 
 3. Given r +&*=c, and a x b v =d, to find the values of x 
 and y. 
 
 By addition, 2a x =c-\-d. Put c-\-d=2m ; 
 
 Then a z =m. Take the logarithm of each mem- 
 
 loff. m 
 her, and x log. =log. m, or x=^ - . 
 
 By subtracting the second equation from the first and 
 making c rf=2n, we shall find 2/ == f"^"~l* 
 
 3 
 
 4. Given (2 1 6) x =18, to find the value of x. 
 
 9 log. 6 
 
 Ans. x=- r -^~- 
 log. 12 
 
 9. Given ^=e, to find the value of x. 
 
 losf. m log. a /fix 
 
 Jtns. x=2 p.- m being equal to (rfe-f c] 
 
LOGARITHMS. 257 
 
 
 
 6. Given 4 3 =16, to find the value of x. rftis. x=Q. 
 
 7 Given 6*= - to find the value of x. 
 
 18 log.24-{-log.l7--3 log.71 
 
 Ans. x ^ . 
 
 3 log. 6 
 
 8. Required the result of 23.46 multiplied by 7.218, and the 
 product divided by 11.17. 
 
 OPERATION. 
 
 23.46 log. 1.37033 
 
 7.218 log. 0.85842 
 
 Sum 2.22875 
 
 11.17 Subtr log. 1.04805 
 
 Result,. . . . 15.16 .log. 1.18070. 
 
 N". B. The log. of a vulgar fraction is found by subtracting 
 the log. of its denominator from the log. of its numerator. 
 
 For instance, T T is simply 9 divided by 11, and division in 
 logarithms is performed by subtraction. 
 
 Thus, from the log. of 9, .... 0.954343 
 
 Subtract log. of 11, .. . 1.041393 
 
 Log. of T T therefore is .... 1 .912850 
 
 The decimal part of a logarithm is never minus, but the 
 index is always minus when the number is less than unity. 
 Hence, the logarithm of a very small fraction has a large nega- 
 tive index, and the logarithm of is minus infinity. The loga- 
 rithm of 10 is 1, of 1 is 0, of .1 is I, of .01 it is 2, of .001 
 it is 3, and so on. As the decimal number goes down to 
 zero, the index of the logarithm goes up to minus infinity. 
 
 The previous example was nothing more than obtaining the 
 logarithm of a common fraction, where the numerator con- 
 sisted of two factors. That example might have been stated 
 thus : 
 
 Find the log. of the fraction, ( 23 - 46 ) ( 7 - 2 * 8 ) 
 
 11.17 
 
 22 Ans. 1.18070. 
 
258 ELEMENTS OF ALGEBRA. 
 
 CHAPTER V. 
 
 COMPOUND INTEREST. 
 
 (Art. 153.) Logarithms are of great utility in resolving some 
 questions in relation to compound interest and annuities ; but for 
 a lull understanding of the subject, the pupil must pass through 
 the following investigation : 
 
 Let p represent any principal, and r the interest of a unit of 
 this principal for one year. Then 1 -\-r would be the amouni 
 of $1, or J61. Put^=l-fr. 
 
 Now as two dollars will amount to twice as much as one dol 
 lar, three dollars to three times as much as one dollar, &c. 
 Therefore, 1 : Ji : : A : JP=the amount in 2 years, 
 And 1 : Ji : : JP : .# 3 =the amount in 3 years, 
 
 <fec. &c. 
 
 Therefore, /?" is the amount of one dollar or one unit of the 
 principal in n years, and p times this sum will be the amount for 
 p dollars. Let a represent this amount ; then we have this gen- 
 eral equation, 
 
 pA n =a. 
 
 In questions where n, the number of years, is an unknown 
 term, or very large, the aid of logarithms is very essential to a 
 quick and easy solution. 
 
 For example, what time is required for any sum of money 
 to double itself, at three per cent, compound interest? 
 
 Here a=2/?,and .#=(1.03), and the general equation becomes 
 
 />(1.03) n =2/> 
 Or (1.03)"=2. Taking the logarithms 
 
 nk*. (1.03)=log. 2, or n= 
 years nearly. 
 
 2. A bottle of wine that originally cost 20 cents was put away 
 for two hundred years: what would it be worth at the end of 
 that time, allowing 5 percent, compound interest? 
 
ANNUITIES. 259 
 
 This question makes the general equation staml thus : 
 
 (20 cts. being l of a dollar) i(1.05) 200 = a 
 Therefore (1.05) 200 =5 
 
 Taking the logarithms 200 log. (1.05)=log.5-{-log. a 
 Hence log. 0=200 log. (1.05) log. 5. JJns. $3458.10 
 
 3. A capital of $5000 stands at 4 per cent, compound interest; 
 what will it amount to in 40 years ? rfns. $24005.10. 
 
 4. In what time will $5 amount to $9, at 5 per cent, com- 
 pound interest ? Jlns. 12.04 years. 
 
 5. A capital of $1000 in 6 years, at compound interest, 
 amounted to $1800; what was the rate per cent? 
 
 Ans. log. (l+r)= log * L8 or 10 T \ nearly 
 
 6. A certain sum of money at compound interest, at 4 per 
 cent, for four years, amounted to $350. 95| ; what was the sum? 
 
 fins. $300. 
 
 7. How long must $3600 remain, at 5 per cent, compound in- 
 terest to amount to as much as $5000, at 4 per cent, for 12 years 1 
 
 Am. 16 years, nearly. 
 
 ANNUITIES. 
 
 (Art. 154.) An annuity is a sum of money payable peri 
 odically, for some specified time, or during the life of the re 
 ceiver. If the payments are not made, the annuity is said to h 
 in arrear, and the receiver is entitled to interest on the several 
 payments in arr3ar. 
 
 The worth of an annuity in arrear, is the sum of the several 
 payments, together with compound interest on every payment 
 after it became due. 
 
 On this definition we proceed to investigate a formula to be 
 applied to calculations respecting annuities. 
 
 Let p represent the annual principal or annuity to be 
 
2GO ELEMENTS OF ALGEBRA. 
 
 paid, and \-\-r=*fl, the amount of annuity of principal for one 
 year, at the given rate r. 
 
 Let n represent the number of years, and put A' to represent 
 the entire amount of the annuity in arrear. 
 
 It is evident, that on the last payment due, no interest could 
 accrue, and therefore the sum will be p. The preceding pay- 
 ment will have one year's interest ; it will therefore be pJl ; the 
 payment preceding that will have two years' compound interest ; 
 and, of course, will be represented by pJP. (Art. 153.) Hence 
 the whole amount of A' will be 
 
 , &c., to 
 This is a geometrical series, and its sum (Art. 120.) is 
 
 This general equation contains four quantities, .#', p, r, and n , 
 any three of them being given in any question, the others can be 
 found, except r. 
 
 EXAMPLES. 
 
 1. An annuity of $50 has remained unpaid for 6 years, at 
 compound interest on the sums due, at 6 per cent., what sum is 
 now due ? 
 
 By the general equation, 
 
 50[(t.06)'-l] 
 
 ~:o6~ 
 
 Faking the log. of both members, we have 
 
 log. A'= log. 50-flog. [(1.06) 6 1] log. .06. 
 
 The value of (1.Q6) 6 , as found by logarithms, is 1.41852, from 
 which subtract 1, as indicated, and take the log. of the decimal 
 number .41852, we then have 
 
 log. ^'=1. 69897 +( 1.62172) ( 2.778151)=2.54218, 
 From which we find, .#'=$348.56 Jim. 
 
 2. In what time will an annuity of $20 amount to $1000, at 
 4 per cent., compound interest ? 
 
ANNUITIES. 261 
 
 The equation applied, we have 
 
 Dividing by 20, and multiplying by .04, we have 
 2=(1.04) 1 or (1.04)=3. 
 
 log. 3 .477121 
 
 3. What will an annuity of $50 amount to, if suffered to 
 remain unpaid for twenty years, at 3| per cent, compound in- 
 terest? Ans. $1413.98. 
 
 4. What is the present value of an annuity or rental of $50 
 a year, to continue 20 years, discounting at the rate of 3 per 
 cent., compound interest ? 
 
 N. B. By question 3d, we find that if the annuity be not paid 
 until the end of 20 years, the amount then due would be $1413.98. 
 If paid now, such a sum must be paid as, put out at compound 
 interest for the given rate and time, will amount to $1413.98. 
 
 Now if we had the amount of $1 at compound interest for 20 
 years, at 3j per cent., that sum would be to $1 as $1413.98 is 
 to the required sum, $710.62. 
 
 (Art. 1 55.) To be more general, let us represent the present 
 worth of an annuity by P. By (Art. 153.) the amount of one 
 dollar for any given rate and time, is Ji n ; A being \-\-r and n 
 the number of years. By (Art. 154.) the value of any annuity 
 p remaining unpaid for any given time, n years, at any rate of 
 
 pA n p 
 compound interest r, is -- or Jl'. 
 
 Now by the preceding explanation we may have this propor- 
 tion : 
 
 Jl* :l::J':P, or P=~- ...... (1) 
 
 Hence, to find the present worth of an annuity, we have this 
 RULE. Divide the amount of the annuity supposed unpaid 
 
 for the given number of years, by the amount of one dollar for 
 
 the same number of years. 
 
2 ELEMKXTS OF ALCEBRA. 
 
 If in equation (1) we put the value of .#, we shall have 
 
 ^nE L m Divide both members by .#*, and we have 
 
 (2) 
 
 r 
 
 This last equation will apply to the following problems : 
 
 5. The annual rent of a freehold estate is p pounds or dollars, 
 to continue forever. What is the present value of the estate, 
 money being worth 5 per cent., compound interest ? 
 
 Here, as n is infinite, the term, becomes 0, and equation 
 
 (2) becomes P=LJL^=2Qp that is, the present value of the 
 estate is worth 20 years' rent. 
 
 6. The rent of an estate is $3000 a year ; what sum could 
 purchase such an estate, money being worth 3 per cent., com- 
 pound interest ? Jim. $100000. 
 
 7. What is the present value of an annuity of $350, assigned 
 for 8 years, at 4 per cent. ? Jlns. $2356.46. 
 
 8. A debt due at this time, amounting to $1200, is to be dis- 
 charged in seven annual and equal payments ; what is the 
 amount of these payments, if interest be computed at 4 per cent. ? 
 
 Ans. $200, nearly. 
 
 9. The rent of a farm is $250 per year, with a perpetual lease. 
 How much ready money will purchase said farm, money being 
 worth 7 per cent, per annum 1 rfns. $3571 7 
 
 10. An annuity of $50 was suffered to remain unpaid for 
 20 years, and then amounted to $1413.98; what was the rate 
 per cent., at compound interest ? 
 
 N. B. This question is the converse of problem 3, and, ol 
 course, the answer must be 3i per cent. But the general equa- 
 tion gives us 
 
GENERAL THEORY OF EQUATIONS. % 63 
 
 Or 28.2796 JH^ 2 ; 
 
 an equation from which it is practically impossible to obtain r, 
 except by successive approximations. 
 
 SECTION VII. 
 
 CHAPTER I. 
 GENERAL THEORY OF EQUATIONS. 
 
 (Art. 156.) In (Art. 101.) we have shown that a quadratic 
 equation, or an equation of the second degree, may be conceived 
 to have arisen from the product of two equations of the first de- 
 gree. Thus, if x=a, in one equation, and x=b in another 
 equation, we then have 
 
 x a=0, 
 and x &=0; 
 
 By multiplication, x 2 (a-{-b}x-{-ab=Q. 
 
 This product presents a quadratic equation, and its two roots 
 are a and b. 
 
 If one of the roots be negative, as x= , and x=b, the 
 resulting quadratic is 
 
 x z -\-(ab)xab=0. 
 
 If both roots be negative, then we shall have 
 
 Now let the pupil observe that the exponent of the highest 
 power of the unknown quantity is 2 ; and there are two roofs. 
 The, coefficient of the first power of the unknown quantity is 
 the algebraic sum of the two roots, with their signs changed ; 
 and the absolute, term, independent of the unknown quantity, 
 is the product of the roots (the sign conforming to the rules of 
 multiplication}. 
 
264 ELEMENTS OF ALGEBRA. 
 
 When the coefficients and absolute term of a quadratic are not 
 largo, and not fractional, we may determine its roots by inspec- 
 tion, by a careful application of these principles 
 
 EXAMPLES. 
 
 Given #* 20#-f96=0, to find x. 
 
 The roots must be 12 and 8, for no other numbers will make 
 20, signs changed, and product 96. 
 
 Given y z 6y 55=0 to find y. Roots 11 and 5. 
 
 Given a? 6# 40=0 to find x. Roots 10 and 4. 
 
 Given s?+x 91=0 to find x. Roots 7 and 13. 
 
 Given y* 5y 6=0 to find y. Roots 6 and 1. 
 
 Given y z -\-\'2y 589=0 to find y. 
 
 Here it is not to be supposed that we can decide the values of 
 the roots by inspection; the absolute term is too large; but, 
 nevertheless, the equation has two roots. 
 Let the roots be represented by P and Q. 
 From the preceding investigation 
 
 P+$=- 12 (1) 
 
 And PQ= 589 (2) 
 
 By squaring eq. (1) P 2 -}-2PQ-{-Q 2 = 144 
 4 times eq. . . . (2) 4PQ =2356 
 
 By subtraction, p22PQ+Q 2 = 2500 
 
 By evolution, P #=50 
 But P-f-= .12 
 
 Hence P=19 or 31, and Q=3l or +19, the true 
 roots of the primitive equation ; and thus we have anothci 
 method of resolving quadratics. 
 
 (Art. 157.) In the same manner we can show that the product 
 of three simple equations produce a cubic equation, or an equa- 
 tion of the third degree. Conversely, then, an equation of the 
 third degree has three roots. 
 
 The three simple equations, #=, x=b, x=c,* may be put 
 
 * Of course, x cannot equal different quantities at one and the same time 
 and these equations must not be thus understood. 
 
GENERAL THEORY OF EQUATIONS. 2 65 
 
 in the form of x a=Q, x &=0, and x c=0, and the pro- 
 duct of these three give 
 
 (x a}(x b] (x c)=0 ; 
 and by actual multiplication, we have 
 x 3 a 
 
 If one of the roots be negative, as x= c, or a?-j-c=0, the 
 product or resulting cubic will be 
 
 If two of them be negative, as x= b and #== e, the 
 resulting cubic will be 
 
 a^+C^+c a)x 2 -f(6c ab ac}x abc=0. 
 If all the roots be negative, the resulting cubic will be 
 
 Every cubic equation may be reduced to this form, and con- 
 ceived to be formed by such a combination of the unknown term 
 and its roots. 
 
 By inspecting the above equations, we may observe 
 
 1st. The first term is the third power of the unknown 
 quantity. 
 
 2d. The second term is the second power of the unknown 
 quantity, with a coefficient equal to the algebraic sum of the 
 roots, with the contrary sign. 
 
 3d. The third term is the first power of the unknown 
 quantity, with a coefficient equal to the sum of all the products 
 which can be made, by taking the roots two by two. 
 
 4th. The fourth term is the continued product of all the roots, 
 with the contrary sign. 
 
 It is easy, then, to form a cubic equation which shall have 
 any three given numbers for its roots. 
 
 Assuming x for the unknown quantity, find an equation which 
 shall have 1, 2 and 3 for its roots. 
 
 tins. a*(l+2+3)3M-(2+ 3+6)* 6=0 ; 
 Or 
 
 Find the equation which shall have 2, 3, and 4 for its roots. 
 23 Ans. x* 9? 14a?+24=0. 
 
266 ELEMENTS OF ALGEBKA. 
 
 Find the equation which shall have 3, 4 y and +7 for its 
 roots. Am. x 3 0a; 2 37z 84=0, 
 
 Or x* 37# 84=0. 
 
 These four general cases of cubic equations may all be repre- 
 sented by the general form. 
 
 Thus: x s -\-px z +qx+r=0, ......... (1) 
 
 (Art. 158.) When the algebraic sum of three roots is equal U 
 zero, equation (1) takes the form of 
 
 x*+qx+r=0 ............. (2) 
 
 Equation (1) is a regular cubic, and is not susceptible of a 
 direct solution, by Cardan's rule, until it is transformed into 
 another wanting the second term, thus making it take the form 
 of equation (2). To make this transformation, conceive one of 
 the roots, or x, in equation (1), represented by u-{-v* 
 
 Then x*=u*+3u z v+3uv z -f-u 3 
 
 px*= pu 2 -\-2puv-\-pv 2 
 qx = qu -\-qv 
 
 r = r 
 
 By addition, and uniting the second member according to tho 
 powers of u, we shall have 
 
 for the transformed equation. But the object was to make such 
 a transformation that the resulting equation should be deprived 
 of its second power ; and to effect this, it is obvious that we 
 must make the coefficient of u 2 equal zero, or 3v-}-p=Q. 
 Therefore, v= ip. 
 
 Hence, we perceive that if x, in the general equation (1), be 
 
 put equal to u ~, there will result an equation in the form of 
 o 
 
 u*-\-qu-\-r=0, or the form of equation (2). 
 
 As x=u ^, and if , 6, and c represent the roots of equa- 
 o 
 
 tion (1), or the values of x, the roots of (2), or values of u will be 
 
 and 
 
GENERAL THEORY OF EQUATIONS. 267 
 
 EXAMPLES. 
 
 1. Transform the equation ar 3 9x?-{-26x 30=0, into another 
 wanting the second term. 
 
 By the preceding investigation, we must assume 
 
 xu-}-3. Here p= 9 ; therefore, sp=3. 
 
 = 26w-f78 
 
 30 = 30 
 
 Sum, =w 3 u 6=0, the equation required. 
 
 2. Transform the equation a? 3 6r J +10a> 8=0, into another 
 not containing the square of the unknown quantity. 
 
 Put x=u+2. Result, u 3 2w 4=0. 
 
 3. Transform x s 3x z -}-6x 12=0, into another equation, 
 wanting the second power of the unknown quantity. 
 
 Put a?=w+l. Result, w 3 +3w 8=0. 
 
 (Art. 159. We have shown, in the last article, that any icgular 
 cubic equation containing all the powers of the unknown quan- 
 tity can be transformed into another equation deficient of the 
 second power ; and hence all cubic equations can be reduced to 
 the form of 
 
 We represent the coefficient of x by 3p, and the absolute term 
 by 2</, in place of single letters, to avoid fractions, in the course 
 of the following investigation. 
 
 Now, if we can find a direct solution to this general equation, 
 it will be a solution of cubic equations generally. 
 
 The" value of x must be some quantity ; and let that quantity, 
 whatever it is, be represented by two parts, v-\-y, or let 
 x=v-\-y. Then the equation becomes 
 
 By expanding and reducing, we have 
 
 Now as we have made an arbitrary division of x into two parts, 
 v and y we can so divide it, that 
 
268 ELEMENTS OF ALGEBRA. 
 
 This hypothesis gives 
 
 v 3 +y*= 2q, (.#) 
 
 And V y=p, (J9) 
 
 Here we have two equations, (A) and (/?), containing two un- 
 known quantities, similarly involved, which admit of a solution 
 by quadratics. (Art. 108.) Hence we obtain v and y, and their 
 algebraic sum is x. 
 
 From equation (B], 
 
 This substituted in equation ($), gives 
 
 Or, V Q 2yu 3 =p 3 , a quadratic. 
 
 Hence v- 
 
 And y. 
 
 Or, as y- 
 
 Therefore x=(q+ jf+fi}* +(q 
 
 Or , 
 
 These formulas are familiarly known, among mathematicians, aa 
 Cardan's rule. 
 
 (Art. 160.) When p is negative, in the general equation, and 
 its cube greater than <? 2 , the expression Jq* p s becomes imagi-' 
 nary ; but we must not conclude that the value of # is therefore 
 imaginary ; for, admitting the expression ,J(f -jo 3 imaginary, it 
 
 can be represented by aj 1, and the value of a?, in equation 
 (C), will be 
 
(JENERAL THEORY OF EQUATIONS. 269 
 
 Now by actually expanding the roots of these binomials by the 
 binomial theorem, and adding their results together, the terms 
 containing J 1 will destroy each other, and their sum will be 
 a real quantity ; and, of course, the value of x will become real. 
 If in any particular case it becomes necessary to make the series 
 
 converge, change the terms of the binomial, and make J 1 
 stand first, and 1 second. 
 
 EXAMPLES. 
 Given x 9 6a?=5.6, to find the value of x. 
 
 Here, 3p= 6, and 2? 5.6, or p= 2, and #=2.8. 
 
 Then 
 
 *=(2.8+/7.84 8) +(2.877.848) , by equation (C) 
 
 Or tf=(2.8+.4,/ 1) +(2.8 .47 1) 
 
 Or " 
 
 Expand the binomials by the binomial theorem, (Art. 135.), 
 i d for the sake of brevity, represent 4 J 1 by b; 
 
 Then &*= , and fl='x- 
 
270 ELEMENTS OF ALGEBRA 
 
 =24-0.0045350.000034=2.0045. 
 Therefore, 
 
 
 3 -=2.0045 or a?=(2.0045)(2.8)=2.8256, nearly 
 (Art. 161.) Every cubic equation of the form of 
 
 has three roots, and their algebraic sum is 0, because the equa- 
 tion is wanting its second term. (Art. 157.) 
 
 If the roots be represented by , b, and c, we shall have 
 
 If any two of these roots are equal, as b=c, then a= 2b (1), 
 and ab 2 =q (2). Putting the value of a taken from equation 
 (1), into equation (2), and we have 26 3 =g r . 
 
 Hence, in case of there being two equal roots, such roots must 
 each equal the. cube root of one half the quantity represented 
 byq. 
 
 EXAMPLES. 
 
 The equation re 3 48^=128 has two equal roots; what are 
 .he roots ? 
 
 Here, 26 3 ,=128, or b s = 64; therefore, b= 4. 
 
 Two of the roots are each equal to 4, and as the sum of the 
 three roots must be 0, therefore 4, 4, -\-8, must be the 
 three roots. 
 
 If the equation a: 3 27#=54 have two equal roots, what are 
 the roots ? Am. 3, 3, and +6. 
 
 Either of these roots can be taken to verify the equation ; and 
 if they do not verify it, the equation has not equal roots. 
 
 (Art. 162.) If a cubic equation in the form of 
 
GENERAL THEORY OF EQUATIONS. 271 
 
 have two equal roots, each one of the equal roots will bo 
 equal to 
 
 The other root will be twice this quantity subtracted from p, 
 because the sum of the three roots equal p. (Art. 157.) 
 
 This expression is obtained from the consideration that the 
 tfirce roots represented by , 6, and c, must form the folloAving 
 equations: (Art. 157.) 
 
 a+b+c=p ......... (1) 
 
 ao-{-ac-\-bc=q ......... (2) 
 
 abcr ......... (3) 
 
 On the assumption that two of these roots are equal, tha* is, 
 a=b, equations (1) and (2) become 
 
 2a-\-c=p ........ (4) 
 
 And a 2 -{-2ac=q ........ (5) 
 
 Multiply equation (4) by 2, and we have 
 
 4a 2 -\-2ac=2ap ....... (7) 
 
 Subtract (5) 2 +2oc= q ....... (8) 
 
 And we have 3a 2 =2ap q. 
 
 This equation is a quadratic, in relation to the root or, and a 
 solution gives a=i(p.Jp 2 3q). 
 
 (Art. 163.) A cubic equation in the form of x s px=q can 
 be resolved as a quadratic, in all cases in which q can be resolved 
 into two factors, m and rc, of such a magnitude that m 2 -\-p=n. 
 
 For the values of p and </, in the general equation, put the 
 assumed values, mn=q, and p=n m z . 
 
 Then we have x 3 -{-nx m 2 x=mn. 
 
 Transpose m z x, and then multiply both members by x, and 
 = m z x z -{-mnx. 
 
 Add -- to both members, and extract square root ; 
 
 Then x*-\~-mx-\--. Drop -, and divide by a?, and x-=m. 
 
272 ELEMENTS OF ALGEBRA. 
 
 Therefore, if such factors of q can be found, the equation is 
 already resolved, as x will be equal to the factor m. 
 
 EXAMPLES. 
 
 1. Given x?+6x=S&, to find the values of x 
 
 Here, wm=88=4X22, 4 2 -j-6=22. Hence, x=4 
 
 2. Given x 3 -\-3x=l4, to find one value of x. 
 
 2X7=14, 2 2 +3=7. Hence, x 2. 
 
 U. Given # 3 -|-6#=45, to find one value of x. *fl.ns. x=3. 
 
 4. Given x 3 13;r= 12, to find x. rfns. x=3. 
 
 5. Given 7/ 3 +48?/=104, to find y. Jlns. y=2. 
 
 In the above examples we have given only one answer, or one 
 root ; but we have more than once observed, that every equation 
 of the third degree must have three roots. Take, for example, 
 the 4th equation. We have found, as above, one of its roots to 
 be 3. Now we may conceive the equation to have been the 
 product of three factors, one of which was (x 3) ; therefore the 
 equation must be divisible by x 3 without a remainder, (other- 
 wise 3 cannot be a root) ; and if we divide the equation by x 3, 
 the quotient must be the product of the other two factors. 
 
 Thus, a: 3)* 3 I3x+12(x z -\-3x 4 
 
 2 I3x 
 
 4H-12 
 
 By putting x*-\-3x 4=0, and resolving the equation, we find 
 a?=l, or 4, and the three roots are 1, 3, 4. Their sum is 
 0, as it should be, as the equation is deficient of its second term. 
 In the general equation 
 
 If p and q are each equal to 0, at the same time, the equation 
 becomes a i3 -j-r=0, a binomial equation. 
 
GENERAL THEORY OF EQUATIONS. 373 
 
 Every binomial equation has as many roots as there are 
 units in the exponent of the unknown quantity. 
 
 Thus a; 3 +8=0, and a 3 8=0, or o; 3 +l=0, and a; 3 1=0, 
 &c., are equations which apparently have but one root, but a 
 full solution will develop three. 
 
 Take, for example, a? =8 
 
 By evolution, x =2 
 
 4xS 
 4xS 
 
 Now by putting a; 2 +2^-}-4=0, and resolving the equation, 
 
 we find x= 1 + V ** an( ^ x ~ * >/"~~^ anc ^ tne tnree 
 roots of the equation X s 8 = 0, are 2, 1 + ^/ 3, and 
 1 J 3, two of them imaginary, but either one, cubed, will 
 give 8. 
 
 The three roots of the equation ic 3 +l=0, are 
 
 1 , 1 . 1 1 
 
 CHAPTER II. 
 GENERAL THEORY OF EQUATIONS CONTINUED. 
 
 (Art. 164.) In the last Chapter we confined our investigations 
 to equations of the second and third degrees ; and if they are 
 well understood by the pupil, there will be little difficulty, in 
 future, as many of the general properties belong to equations of 
 every degree. 
 
 All the higher equations may be conceived to have been formed 
 by the multiplication of the unknown quantity joined to each 
 of the roots of the equation with a contrary sign, as shown in 
 (Art. 157.). 
 
274 ELEMENTS OF ALGEBRA. 
 
 Let a, b, c, d, e, &c., be roots of an equation, and a; 
 its unknown quantity, then the equation may be formed by the 
 product of (x a}(x b)(x c), &c., which product we may 
 represent by 
 
 Now it being admitted that equations ran be thus formed by 
 the multiplication of the unknown quantity joined to its roots, 
 conversely, when any of its roots can be found, such root, with 
 its contrary sign joined to the unknown term, will form a com- 
 plete divisor for the equation ; and by the division the equation 
 will be reduced one degree, and conversely. 
 
 If any quantity, connected to the unknown quantity by the 
 sign plus or minus, divide an equation without a remainder, 
 such a quantity may be regarded as one of the roots of the 
 equation. 
 
 The product of all the roots form the absolute term U. 
 
 (Art. 165.) Every equation having unity for the coefficient 
 of the first term, and all the other coefficients, whole numbers, 
 can have only whole numbers for its commensurable* roots. 
 
 This being one of the most important principles in the theory 
 of equations, its enunciation should be most clearly and distinctly 
 understood. Such equations may have other roots than whole 
 numbers ; but its roots cannot be among the definite and irre- 
 ducible fractions, such as f , J, U , &e. Its other roots must be 
 
 among the incommensurable quantities, such as J2, (3)% &c., 
 i. e., surds, indeterminate decimals, or imaginary quantities. 
 
 To prove the proposition, let us suppose j- a commensurable 
 but irreducible fraction, to be a root of the equation 
 
 A, B, fec., being whole numbers. 
 Substituting this supposed value of x, we have 
 
 * Commensurable numbers are all those that measure or can be measured by 
 unity ; hence, all whole numbers and definite fractions are commensurable. 
 Surds, and imaginary quantities, are incommensurable. 
 
GENERAL THEORY OF EQUATIONS. 275 
 
 Transpose all the terms but the first, and multiply by b m ~ } , and 
 we have 
 
 a m 
 
 y- 
 
 Now, as a and b are prime to each other, b cannot divide a, 
 or any number of times that a may be taken as a factor, for j 
 
 being irreducible, - X a is also irreducible, as the multiplier a 
 
 a? 
 will not be measured by the divisor b ; therefore y cannot be 
 
 expressed in whole numbers. Continuing the same mode of 
 
 a m 
 reasoning, y cannot express whole numbers, but every term in 
 
 the other member of the equation expresses whole numbers. 
 Hence, this supposition that the irreducible fraction = is a root 
 
 of the equation, leads to this absurdity, that a series of whole 
 numbers is equal to another quantity that must contain a fraction. 
 
 Therefore, we conclude that any equation corresponding to 
 these conditions cannot have a definite commensurable fraction 
 among its roots. 
 
 (Art. 166.) Any equation having fractional coefficients, can be 
 transformed to another in which the coefficients are all whole 
 numbers, and that of the first term unity. 
 For example, take the equation 
 
 m n p 
 
 Assume ar=--, and put this value of x in the equation, 
 mnp 
 
 And _ 
 
 _.. 
 mn*p p 
 
 Multiply every term by m*n*p*, and we have 
 
 
276 ELEMENTS OF ALGEBRA. 
 
 When m, n, and p have common factors, we may put x equal 
 to y divided by the least common multiple of these quantities, 
 as in the following examples: 
 
 Transform the equation y*-\ -- .+ --- h-=0, into another 
 
 pm m p 
 
 which shall have no fractional coefficients, and that of the first 
 term unity. 
 
 To effect this, it is sufficient to put x=-^. With this value 
 
 pm 
 
 of x the equation becomes 
 
 __. = . 
 
 pV p*m 3 pm 2 p 
 Multiplying every term by j9 3 m 3 , we obtain 
 
 for the transformed equation required. 
 Transform the equation x*-\ - 1 'T"i"nA~^"Tn == ^9 into an- 
 
 other, having no fractional coefficients. 
 
 Result, y+20?/ 3 +18.24y4-7(24) 2 y-}-2(24) 3 =0. 
 
 (Art. 167.) Now as every commensurable root consists of 
 whole numbers, and as the coefficients are all whole numbers, 
 each term of itself consists of whole numbers, and the commen- 
 surable roots are all found among the whole number divisors of 
 the last term ; and if these divisors are few and obvious, those 
 answering to the roots of the equation may be found by trial. If 
 the factors are numerous, we must have some systematic method 
 of examining them, such as is pointed out by the following rea- 
 soning : 
 
 Take the equation * 4 +^ 3 +jar z +Ca?-j-Z)==(). 
 
 Let a represent one of its commensurable roots. Transpose 
 all the terms but the last, and divide every term by a. 
 
 But, since a is a root of the equation, it divides D without a 
 
GENERAL THEORY OF EQUATIONS. 277 
 
 remainder, the left hand member of this last equation is therefore 
 a whole number, to which transpose C, also a whole number, 
 
 and represent \-C by N. 
 
 Then N=a?rfa?a. 
 
 Divide each term by a, and transpose J3, and we have 
 
 -+= a 2 da. 
 a 
 
 The right hand member of this equation is an entire quantity, 
 (not fractional), therefore the other member is also an entire 
 quantity ; let it be represented by N 1 , and the equation again 
 divided by a. 
 
 Then = a A. 
 
 a 
 
 Transpose A ; reasoning the same as before, we can repre- 
 sent the first member by JV", and we then have 
 
 N" 
 Divide by a, and = 1. This must be the final result, in 
 
 case a is a root. 
 
 From these operations we draw the following rule for deciding 
 what divisors of the last term are roots of an equation. 
 
 RULE. Divide the last term by the several divisors, (each 
 designated by ,) and add to the quotient the coefficient of the 
 term involving x. 
 
 Divide this sum by the divisors (a), and add to the quotient 
 the coefficient of the term involving #*. 
 
 Divide this sum by the divisors (a), and add to the quotient 
 the coefficient of the term involving x?. 
 
 And thus continue until the first coefficient, .#, is transposed, 
 and the sum divided by a ; the last quotient will be minus one, 
 if a is in fact a root 
 
278 
 
 ELEMENTS OF ALGEBRA. 
 
 EXAMPLES. 
 
 1. Required the commensurable roots (if any) of the equation 
 
 =0. 
 
 N 
 
 
 
 2 1 
 
 (M 
 
 
 
 7 
 
 
 1 1 
 
 
 
 CD 
 
 <N 
 
 _u "T3 
 
 O 
 
 1 
 
 
 7 i 
 
 1 
 
 7 
 
 CO 
 
 i 
 2 1 
 
 0* 
 
 o> 
 
 7 
 
 * 
 
 7 
 
 7 +7 
 
 i 
 
 0) 
 
 1 
 
 CO 
 
 ^5 ift 
 
 i-H 
 
 ,| , 7 ,_ 
 
 1 
 
 7 
 
 c* 
 
 14 
 
 7 ' 
 
 CO CO '-''-* 
 
 1 + 1 a 
 
 1 
 
 
 
 
 "c 
 
 
 ^4 
 
 
 00 QO 
 
 05 05 .22 
 
 ij 
 
 
 7 
 
 
 tl I 
 
 bo 
 
 c* 
 
 CD 
 
 <N f-t 
 
 ^UT 7 | 
 
 .s 
 
 CO 
 
 7 
 
 
 
 ? 
 
 3 
 
 !i 
 
 
 
 1 
 
 o 
 
 a 2 
 
 * 
 
 1 
 
 05 
 
 7 
 
 1 
 
 |l 
 
 
 
 
 o 
 
 O^ "~^ 
 
 CO 
 
 T 
 
 777 
 
 1 1 
 
 1 i 
 
 e* <M 
 
 r-l CD 
 
 ^ ^ 
 
 C | 
 
 1 
 
 1 7 
 
 i e 
 
 CD 
 
 " oT 
 
 a ^ 
 il 1 Is 
 
 it 
 
 . 1 1 1 1 t 
 
 1 || 11 11 & | 
 
 _>: rJ r^ ^ 
 
 o -5 
 
 O M 
 
 sl 
 
 CO 
 
EQUAL ROOTS. 279 
 
 2. Required the commensurable roots of the equation 
 ar 5 * 6r4-lla: 6=0. dns. 1,2,3. 
 
 3. Required the commensurable roots of the equation 
 s_6;e 2 16^+21=0. Am. 3 and 1 
 
 Here the student might hesitate, as one regular term of the 
 equation is wanting, or rather the coefficient of x* is . hence, 
 the equation is # 4 dbOa; 3 6cr 2 16H-21=0. 
 Go through the form of adding 0. 
 
 4. Required the commensurable roots of the equation 
 a> 4 6a? 3 4-5^+2# 10=0. tins. 1, +5. 
 
 As the commensurable roots are only two, there must be two 
 incommensurable roots ; and they can be found by dividing the 
 given equation by a?-j-l, and that quotient by x 5, and resolv- 
 ing the last quotient as a quadratic. 
 
 EQUAL ROOTS. 
 
 (Art. 168.) In any equation, as 
 
 the roots may be represented by , &, c, d, e, and either one, put 
 in the place of x, will verify the equation. 
 
 Now, let y represent the difference between any two roots, as 
 a b ; then y=a &, and by transposition b-\-y=a. But as a 
 will verify the equation, it being a root, its equal, (b-\-y), sub- 
 stituted for x, will verify it also. That is, 
 
 By expanding the powers, and arranging the terms according 
 to the powers of y, we have 
 +106V 
 
 Cb 2 +2Cby 
 
 Db+Vy 
 
 E 
 
 =0. 
 
 We might have been more general, and have taken x m -{-Ax m 1 , &c., for 
 the equation ; but, in our opinion, we shall be better comprehended by taking 
 an equation definite in degree : the reasoning is readily understood as general 
 
280 ELEMENTS OF ALGEBRA. 
 
 Now, as I is a root of the equation, the first column of this 
 transformation is identical with the proposed equation, on sub- 
 stituting the root b for x. Hence, the first column is equal to 
 xero ; therefore, let it be suppressed, and the remainder divided 
 byy. 
 
 We then have 
 
 =0. 
 
 2Cb 
 D 
 
 On the supposition that the two roots a and b are equal, y 
 becomes nothing, and this last equation becomes 
 
 As b is a root of the original equation, x may be written in 
 place of b ; then this last equation is 
 
 5zM-4^e 3 +3^+2Ca:+#=0. . . . . . (2) 
 
 This equation can be derived from the primitive equation by 
 the following 
 
 RULE. Multiply each coefficient by the exponent of x, and 
 diminish the exponent by unity. 
 
 Equation (2) being derived from equation (1), by the above 
 rule, may be called a derived polynomial. 
 
 (Art. 169.) We again remind the reader that b will verify the 
 primitive equation (1), it being a root, and it must also verify 
 equation (2) ; hence, b at the same time must verify the two 
 equations (1) and (2). 
 
 But if b will verify equation (1), that equation is divisible by 
 (x 6), (Art. 164.), and if it will verify equation (2), that equa- 
 tion also, is divisible by (x b), and (x 6) must be a common 
 measure of the two equations (1) and (2). That is, in case 
 the primitive equation has two roots equal to b. 
 
 (Art. 170.) To determine whether any equation contains equal 
 roots, take its derived polynomial by the rule in (Art. 168.), and 
 seek the greatest common divisor (Art. 27.), [which designate by 
 
GENERA L THEORY OF EQUATIONS. ogj 
 
 (/)),] of the given equation and its first derived polynomial ; 
 and if the divisor D is of the first degree, or of the form of x h, 
 then the equation has two roots each equal to h. 
 
 If no common measure can be found, the equation contains no 
 equal roots. If D is of the second degree, with reference to x, 
 put Z)=0, and resolve the equation ; and if D is found to be in 
 the form of (x A) 2 ; then the given equation has three roots 
 equal to h. 
 
 If D be found of the form of (x h)(x h') t then the given 
 equation has two roots equal to h, and two equal to h'. 
 
 Let D be of any degree whatever ; put Z)=0, and, if possible, 
 completely resolve the equation ; and every simple root of D is 
 twice a root in the given equation ; every double root of D will 
 be three times a root in the given equation, and so on. 
 
 EXAMPLES. 
 
 1. Does the equation x 4 2x s 7x 2 -}-2Qx 12=0 contain 
 any equal roots, and if so, find them ? 
 
 Its derived polynomial is 4x 3 6x 2 14#-f20. 
 
 The common divisor, by (Art. 27.), is found to be x 2 
 therefore, the equation has two roots, equal to 2. 
 
 The equation can then be divided twice by x 2, or once by 
 (x 2) 2 , or by fl 2 4#-{-4. Performing the division, we find 
 the quotient to be x z -^-2x 3, and the original equation is now 
 separated into the two factors, 
 
 (xtlx+^tf+Zx 3) =0. 
 
 The equation can now be verified by putting each of these 
 factors equal to zero. From the first we have already x=2 r 
 and 2, and from the second we may find x=\ or 3; hence, 
 the entire solution of the equation gives 1, 2, 2, 3 for the four 
 roots. 
 
 2. The equation arH-2;r 4 lla? 8x*-{-2Qx+lG=Q has two 
 equal roots ; find them. Jlns. 2 and 2. 
 
 3. Does the equation a? 2# 4 +3r } 1x*+8x 3=0 contain 
 equal roots, and how many ? 
 
 Ans. It contains three equal roots, each equal to 1 
 
 24 
 
282 ELEMENTS OF ALGEBRA. 
 
 4L Find the equal roots, if any, of the equation 
 
 x 3 ~\-tf 16#-f 20=0. tins* 2 and 2. 
 
 5. Find the equal roots of the equation 
 
 X 4_|_ 2 ^ 3 3x 2 4^+4=0. 
 Jlns. Two roots equal to 1, and two roots equal 2. 
 
 6. Find the equal roots of the equation 
 
 a? 53H-10a:--8==0. 
 
 JJns. It contains no equal roots. 
 
 (Art. 171.) Equations which have no commensurable roots, or 
 those factors of equations which remain after all the commensu- 
 rable and equal roots are taken away by division, can be resol- 
 ved only by some method of approximation, if they exceed the 
 third or fourth degree. It is possible to give a direct solution in 
 cases of cubics and in many cases of the fourth degree ; but, in 
 practice, approximate methods are less tedious and more conve- 
 nient. 
 
 We may transform any equation into another whose roots 
 shall be greater or less than the roots of the given equation by 
 a given quantity. 
 
 Suppose we have the equation 
 
 a*+Jla*+a*+Ca*+Dx+JE=0, ..... (1) 
 
 and require another equation, whose roots shall be less than those 
 of the above by a. 
 
 Put x=a-\-y, and, of course, the equation involving y will 
 have roots less than that involving x, by a, because y=x , or 
 ?/ is less than x, by a. 
 
 In place of x t in the above equation, write its equal (a~\-y) 
 and we have 
 
 By expanding and arranging the terms according to the powers 
 of t/, we shall have, as in (Art. 168.), 
 
EQUAL ROOTS. 
 
 E 
 
 (2) 
 
 After a little observation, these transformations may be made 
 very expeditiously, for the first perpendicular column may be 
 written out by merely changing x to a, in the original equation, 
 and then, each horizontal column run out by the law of the 
 binomial theorem. 
 
 Thus a 5 becomes 5a 4 , and this, again, 10a 3 , &c. 
 
 Now, the first column of the right hand member of this equa- 
 tion consists entirely of known quantities ; and the coefficients 
 of the different powers of y are known ; hence we have an 
 equation, involving the several powers of y, in form of equa- 
 tion (1), 
 
 Or, ^+^y+j&y+Cy+/)'y-H5;-==0; the equation 
 required; .#', B', &c., representing the known coefficients of 
 the different powers of y. 
 
 In commencing this subject, we took an equation definite in 
 degree, for the purpose of giving the pupil more definite ideas ; 
 but it is now proper to show the form of transforming an equation 
 of the most general character. 
 
 For this purpose, let us take the equation 
 
 1 
 
 Now let it be required to transform this equation into another 
 whose roots shall be less than the roots of this equation by a. 
 
 Put x=a-{-y, as before ; then 
 
284 
 
 ELEMENTS OF ALGEBRA. 
 
 % .2 
 
 rd ^ 
 
 CL> cC 
 
 +j 9 
 
 
 .f^f 
 
 
 rl ^ 
 
 
 <; 
 
 
 'f3 TJ 
 
 
 eS **** 
 
 
 o . . g j> 
 
 
 CM 5^ 
 
 
 
 
 1 ? 
 
 
 , . * 3 2 
 %> 1 
 
 
 2 ^53 
 
 
 
 
 s ^ 
 
 
 I 2 
 
 
 u ?T 
 
 
 __. r^3 HH 
 
 
 55 ^N 
 
 
 1 
 
 *s " s 
 
 o s| 
 
 - : : : i! 1 
 
 ^^ ^ i^ 
 
 T 
 
 if 
 
 i 
 
 1 
 
 f 
 
 ( t,-. ^ 
 
 
 OD S 
 
 CO f+z 
 
 '~-< 
 
 i 1 I 
 
 
 
 & 
 
 ^3 03 C m ' 
 
 cT ^ ^ C 
 
 p 
 
 C/) ep*i 
 
 .f% 
 
 >C o 
 
 I 
 
 i I : ts * o 
 
 ?3 ff M 03 O 
 
 7 >:N : o 
 
 M 7 'Jo & * 
 
 s 1 i * 
 
 ^s J, - >. ^^ : 
 
 heretofore, p 
 
 !l 
 
 I i 
 
 ^ 
 
 f 
 
 4- 4- * o ^ o 
 
 ' ' 03 rt S3 || 
 
 IB 
 
 C o 
 
 .2 
 
 o ^ hn n^ 
 
 CJ 
 
 8 ^ 
 
 03 
 
 III; s -^ + 
 : ^ 5, ^ -t'l 1 
 
 a 
 
 II 
 (i 
 
 S .| 
 
 2 .S 
 1 2 
 
 *^ c^i 
 
 ^S *Ki ^ 
 
 1 
 
 1 1 11 i|J 
 
 j^ ^T^ j^ ^ ri 
 
 I 
 1 
 
 :|i| 
 
 55 s ^ 
 
 S SH g, 
 
 C. tf *? 
 
 03 
 
 1 II ll'i = 
 
 *8 
 
 ^^^ 
 
 ^ "^ 
 
 
 e S353C3 S3 *>H ^ ^ PH n . 
 ^^ tt5 C ^ C** ^^ ^^ pj ^i. FH 
 
 ^* * ' c3 c3 
 
 i 
 
 
 u 
 
 ^T* i3 .2 
 
 3 
 
 "*** S <u 
 o ^ ^ 
 
 1 
 
 S o "^ 
 
 
 * . fc 
 
 
 "H "5 S 
 
 .S 
 
 . l 64 
 
 p. 
 
 f) , ^ 
 
 u? 
 
 ^ "^ "53 
 
 tfi 
 
 "t^ O +* 
 
 *.-< 
 
 s p ^ 
 
 
 ^j *- ^* 
 
 -*-> 
 
 ^~ S ^^ 
 
 ^ 
 
EQUAL ROOTS 285 
 
 If we should desire to make the third term (counting from the 
 highest power of y] of equation (2) to disappear, we must 
 
 Put 10rt 2 -Htf0+-#=0 ; and this involves the 
 solution of an equation of the second degree, to find the definite 
 value of a. To make the fourth term disappear would require 
 the solution of an equation of the third degree ; and so on. 
 
 If a is really a root of the primitive equation, then x=a, <y=0, 
 and each perpendicular column of the transformed equation is 0. 
 
 If we designate the first perpendicular column of the general 
 transformed equation by X, and the coefficients of the succeeding 
 columns by 
 
 X' X" X"' 
 
 The coefficients of the different powers of y, as X', X", X'", &c. 
 arc called derived polynomials, because each term of X' can be 
 derived from the corresponding term of X ; and each term of X" 
 can be derived from the corresponding terms of X', by the law of 
 the binomial theorem, as observed in the first part of this article. 
 But, to recapitulate : 
 
 X is derived from the given equation by simply changing x 
 to a. 
 
 X' is derived from X by multiplying each of the terms of X 
 by the exponent of a, in that term, and diminishing that expo- 
 nent by unity, and dividing by the exponent of y increased by 1. 
 
 X" is derived from X', in the same manner as X' is derived 
 from X ; and so on. 
 
 X' is called the first derived polynomial ; X" the second, &c. 
 
 To show the utility of this theorem, we propose to transform 
 the following equations : 
 
 1. Transform the equation 
 
 x 4 l2x 3 +l7x 2 9*4-7=0, 
 
 into another, which shall not contain the 3d power of the un- 
 known quantity. 
 
286 ELEMENTS OF ALGEBRA. 
 
 By (Art. 172 ), put .r=7/+^ ..... or- ... x=3+y 
 
 Here =3 and m=4. 
 
 X =(3) 4 12(3) 3 +17(3) 2 9(3)-}-7 . or . . . X =110 
 X' =4(3) 3 36(3) 2 4-34(3) - 9 ...... or ... X' =123 
 
 ~=6(3) 2 36(3)4-17 .......... or. . . = 37 
 
 z <& 
 
 Y"' 
 12 .............. m ---M 
 
 Therefore the transformed equation must be 
 2. Transform the equation 
 
 into another wanting its second term. Put x=2-{-y. 
 
 X =(2) 3 6(2) 2 +13(2) 12 ....... or. . .X =2 
 
 X' =3(2) 2 12(2)+13 .......... or. - X' =+1 
 
 6 ............... or...= 
 
 X'" X' 
 
 21T 1 ................... or ---2 
 
 Therefore the transformed equation must be 
 2/ 3 + 
 
 3. Transform the equation 
 
 into another whose roots shall be less by 2. 
 
 Put a?=24-y. Result, ?/ 4 4-4?/ 3 24y=0, 
 
 As this transformed equation has no term independent of y, 
 y=0 will verify the equation ; and x=2 will verify the original 
 equation, and, of course, is a root of that equation. 
 
EQUAL ROOTS. 28 7 
 
 4. Transform the equation 
 
 into av.other whose roots shall be greater by 3. 
 
 Put x =3-}-y. Result, 2/ 4 +4?/ 3 -r-9i/ 2 42y=0. 
 
 5 Transform the equation 
 
 x 4 8z 3 -{-ar J +82;r 60=0, 
 into another wanting its second term. 
 
 Result, y 4 23*/ 2 +22i/-f-60=0. 
 
 (Art. 173.) We may transform an equation by division, as 
 well as by substitution, as the following investigation will show. 
 
 Take the equation 
 
 * 4 4-^ 3 4- Bx>+Cx+D=Q ........ (1) 
 
 If we put x=a-}-y, in the above equation, it will be trans- 
 formed (Art. D.) into 
 
 As x a-\-y, therefore T/ # a ; and put this value of y in 
 equation (2), we have 
 
 (*-)'+ J^-) 3 +^V-) 2 +X '(*-)+X=0. . .(3) 
 
 Now it is manifest that equation (3) is identical with equation 
 (1), for we formed equation (2) by transforming equation (l),and 
 from (2) to (3) we only reversed the operation. 
 
 Now we can divide equation (3), or in fact equation (1), by 
 (x a). and it is obvious that the first remainder will be X. 
 
 Divide the quotient, thus obtained, by the same divisor, (x a). 
 and the second remainder must be X'. 
 
 Divide the second quotient by (x a), and the third remaindei 
 
 nmst be -. 
 
 m 
 
 X'" 
 
 The next remainder mus', be , <fec., <fec., according to the 
 
 2.3 
 
 degrpe of the equation 
 
238 ELEMENTS OF ALGEBRA. 
 
 Now if we reserve these remainders, it is manifest that they 
 may form the coefficients of the required transformed equation ; 
 taking the last remainder for the first coefficient ; and so on, in 
 reverse order. 
 
 For illustration, let us take the third example of the last article. 
 
 01 y=x 2. 
 16 
 
 2x* Sx 
 
 
 16H-32 4x 16 
 
 16#-{-32 4x+ 8 
 
 0=X 24=X 
 
 2)a:+2(l 
 a: 2 
 
 X'" 
 ~ 
 
 Hence the transformed equation is 
 
 0; 
 or, t/ 4 -f-4?/ 3 24i/=0, as before. 
 
 For a further illustration of this method, we will again operate 
 on the first example of the last article. 
 
EQUAL ROOTS. 
 
 10a>-39 
 
 lOic 2 9x- 
 
 39a'-f- 7 
 _39.r-|-117 
 
 110 X. 1st Remainder. 
 ;_3U- 3 9^ 1 O.r 39(0.- Gx 28 
 
 28^39 
 
 123 = X'. 2d Remainder. 
 2S(x 3 a> 3).r 3(1 
 
 3x28 X'" 
 
 __. 3d Remainder. 
 
 Hence 7/ 4 ()j/ 3 3 7/ 123^110 = 0, must be the trans- 
 formed equation. 
 
 We shall have a 4th remainder, if we operate on an equation 
 of the 4th degree; a 5th remainder with an equation of the f>tli 
 degree; and, in general, n number of remainders with an equa- 
 tion of the nth degree. 
 25 
 
290 ELEMENTS OF ALGEBRA. 
 
 But to make this method sufficiently practical, the operator 
 must understand 
 
 SYNTHETIC DIVISION. 
 
 (Art. 174.) Multiplication and division are so intimately blended 
 that they must be explained in connection. For a particular 
 purpose ve wish to introduce a particular practical form of per- 
 forming certain divisions ; and to arrive at this end, we commence 
 with multiplication. 
 
 Algebraic quantities, containing regular powers, may be 
 multiplied together by using detached coefficients, and annexing 
 the proper literal powers afterwards. 
 
 EXAMPLES. 
 
 1 Multiply a 2 +2^+^ by a+x. 
 Take the coefficients. Thus 
 1+2+1 
 1 + 1 
 
 1+2+1 
 1+2+1 
 
 Product, ... 1+3+3+1 
 By annexing the powers, we have 
 
 J. Multiply 3t*+xy+y* by tfxy+if. 
 
 As the literal quantities are regular, we may take detached 
 coefficients, thus : 
 
 1 + 1 + 1 
 
 1 + 1 + 1 
 __!_ i i 
 
 1 + 1 + 1 
 
 Product,. 1+0+1+0+1 
 
SYNTHETIC DIVISION. 291 
 
 Hero the second and fourth coefficients are ; therefore the 
 terms themselves will vanish ; and, annexing the powers, we shall 
 have for the full product 
 
 3. Multiply 3,r 2 2.r 1 by 
 321 
 
 4+2 
 
 1284 
 
 642 
 
 12282 
 Product, . . . 12^2^ Sx 2. 
 
 . Multiply x 4 aar'+flrV 3 #+a 4 by 
 1 1 + 1 1 + 1 
 
 11 + 1 1+ 1 
 + 1 1 + 11 + 1 
 
 1+0+0+0+0+1 
 
 As all the coefFjcients are zero except the first and last, there- 
 fore the product must be 
 
 (Art. 175.) Now if we can multiply by means of detached 
 coefficients, in like cases we can divide by means of them. 
 
 Take the last example in multiplication, and reverse it, that is, 
 divide a^+a 5 by a-+a. 
 
 Here we must suppose all the inferior powers of x 5 and a 5 
 really exist in the dividend, but disappear in consequence of their 
 coefficients being zero ; we therefore write all the coefficients of 
 the regular powers thus : 
 
292 ELEMENTS OF ALGEBRA. 
 
 Divisor. Dividend. Quotient. 
 
 Annexing the regular powers to the quotient, we have x* ax 3 -^- 
 a z x 2 3 vT+ a 4 , for the full quotient. 
 
 2. Divide a 5 5a 4 6+lO 3 6 2 10 2 6 3 +5a& 4 tf by a 8 
 2ab+b z . 
 
 12+1)15+1010+51(13+31 
 12+ 1 
 34-910 
 34-6 3 
 
 " 37+5 
 3_6+3 
 
 1+2-1 
 1+21 
 
 These coefficients are manifestly the coefficients of a cube, 
 therefore the powers are readily supplied, and are 
 
 N. B. If we change the signs of the coefficients in the divisor, 
 except the first, and then add the product of those changed terms, 
 we shall arrive at the same result. 
 
 Perform the last example over again, after changing the signs 
 of the second and third terms of the divisor. Thus, 
 
SYNTHETIC DIVISION. 093 
 
 l-f2__l)l_5_}-10 10+5 1(1 3+ 31 
 l-j-2 1 
 
 Sum * 3-f- 910 
 _3_ 6+ 3 
 
 Sum . . . . ; 3 7+5 
 
 3-f 63 
 
 Sum ....... * J+2 J 
 
 12+1 
 
 Sum ....... ~~*~0+0 
 
 3. Divide or 3 Gr'+l 1# 6 by x2. 
 Change the sign of the second term of the divisor. 
 1+2)16+116(14+3 
 1+2 
 4+11 
 4 8 
 
 3+6 
 
 Let the reader observe, that when the first figure of the divisor 
 is 1, the first figure of the quotient will be the same as the first 
 figure of the dividend ; and the succeeding figures of the quotient 
 are the same as the first figures of the partial dividends. 
 
 Now this last operation can be contracted. 
 
 AY rite down the figures of the dividend with their proper signs, 
 and the second figure of the divisor, with its sign changed, on 
 the right. Thus 
 
 16+1 16(2.= Divisor 
 2 8+6 
 
 (144-3) o 
 
 The first figure, 1, is brought down for the first figure of 
 the quotient. 
 
 The divisor, 2, is put under 6 ; their sum is 4, which, 
 multiplied by 2, and the product 8 put under the next term, 
 
294 ELEMENTS OF ALGEBRA. 
 
 the sum of +11 8 is 3, which multiplied by 2, gives 6, and the 
 sum of the last addition is 0, which shows that there is no 
 remainder. 
 
 The numbers in the lower line show the quotient, except the 
 last ; that shows the remainder, if any. 
 
 This last operation is called synthetic division. 
 
 4. Divide a,' 3 - r -2r ! -Sx 24 by rr 3. 
 
 COMMON METHOD. 
 
 a- 3 3.r 2 
 
 SYNTHETIC METHOD. 
 
 14-2 824(3 
 3+15-1-21 
 
 (14-54-7) 3 
 
 Now we are prepared to work the examples in (Art. E.) in a 
 more expeditious manner. 
 
 Transform again, the equation # 4 4x 3 8#4-32=0, to an 
 other, whose roots shall be less by 2. 
 
 This equation has no term containing a, 12 , therefore the coefli 
 cient of a? must be taken =0, if we use Synthetic Division. 
 
 FIRST OPERATION. 
 
 1 40 84-32 (2 
 24832 
 
 (12416), 0=X. 
 
 SECOND OPERATION. 
 
 12416(2 
 20 8 
 
 (l-f-0 4),-24=X'. 
 
SYNTHETIC DIVISION. 295 
 
 THIRD OPERATION. 
 IrbO 4 (2 
 
 2+4 
 
 Y" 
 (1+2) 0=-. 
 
 FOURTH OPERATION. 
 
 1+2(2 
 2 
 
 (fence oui transformed equation is y 4 -f-4t/ 3 24*/=0, as 
 before. 
 
 To transform an equation of the fourth degree, we must have 
 four operations in division ; an equation of the nth degree n 
 opeiations, as before observed. 
 
 But these operations may be all blended in one. Thus 
 
 j __4 o 8 32 (2 
 
 248 32 
 
 0"=:X 
 
 -2 
 2 
 
 4 
 
 
 16 
 
 8 
 
 =X 
 
 
 2 
 
 2 
 2 
 
 4 
 
 +4 
 
 
 
 24 
 
 X" 
 
 = 
 2 
 
 2.3* 
 
 We omit the first column, except in the first line, as there aro 
 no operations with it. 
 
 The pupil should observe the structure of this operation. It 
 is an equation of the 4th degree, and there are four sums in nd- 
 dition, in the 2d column; three in the next; two in the next, 
 &.C., giving the whole a diagonal shape. 
 
29 U ELEMENTS OF ALGEBRA. 
 
 Transform the equation x* l2x 3 -\-l7x 2 O.r+7 0, into an- 
 other whose root shall be 3 less. 
 
 OPERATION. 
 
 1 12 +17 9 _J- 7 (3 
 4. 3 27 30 117 
 
 _ y io 39 110=X 
 
 -f 3 18 84 
 
 6 28 123=X' 
 4.39 
 
 - 3 -37=^ , 
 3 
 
 ^51; 
 
 Hence the transformed equation is 
 
 10=0. 
 
 Transform tlie equation x 3 l'2x 28 = 0, into another whose 
 roots shall be 4 less. 
 
 
 
 12 
 
 28 ( 
 
 4 
 
 + 10 
 
 + 10 
 
 4 
 
 4 
 
 12 = ] 
 
 4 
 
 32 
 
 
 IF 
 
 ~~36 = : 
 
 V' 
 
 4 
 
 
 
 Hence the transformed equation must be ?/ 3 +12?/ s +30i/ 
 12 0, on the supposition that we put y=x 4. 
 
 Transform the equation # 3 10a^+3a: 0946=0, into another 
 whose roots shall be less by 20. 
 Put #=20+?/. 
 
SYNTHETIC DIVISION. 29? 
 
 10 
 
 3 
 
 6946 
 
 (20 
 
 20 
 
 200 
 
 4060 
 
 
 10 
 
 203 
 
 2886 
 
 
 20 
 30 
 
 600 
 
 803 
 
 
 
 20 
 
 = 
 
 
 
 50 
 
 The three remainders are the numbers just above the doulde 
 lines, which give the following transformed equation : 
 y+50?/ 2 +803?/ 2886=0. 
 
 Transform this equation into another, whose roots shall be 
 less by 3. Put y=3-\-z. 
 
 50 
 3 
 
 53 
 3 
 
 803 
 159 
 
 962 
 
 168 
 
 2886 
 +2886 
 
 (3 
 
 
 
 
 56 
 
 a 
 
 1130 
 
 
 
 59 
 Hence the second transformed equation is 
 
 This equation may be verified by making 2=0 ; which gives 
 y3 and #=20+3=23. 
 
 Thus we have found the exact root of the original equation by 
 successive transformations ; and on this principle we shall here- 
 after give a general rule to approximate to incommensuralle roots 
 of equations of any degree ; but before the pupil can be prepared 
 to comprehend and surmount every difficulty, he must pay more 
 attention to general theory, as developed in the following 
 Chapter 
 
298 ELEMENTS OF ALGEBRA. 
 
 CHAPTER III. 
 
 GENERAL PROPERTIES OF EQUATIONS. 
 
 (Art. 170.) Jlny equation, having only negative roots, will 
 have all its signs positive. 
 
 If wo take a, b, c, &c., to represent the roots of an 
 equation, the equation itself will be the product of the factors ; 
 
 (z-r-a), (x+b), Or-fc), &c., =0 : 
 
 and it is obvious that all its signs must be positive. 
 
 From this, we decide at once, that the equation a? 4 -f-3a^-f- 
 6a?+6=0 ; or any other numeral equation, having all its signs 
 plus, can have no rational positive roots. 
 
 (Art. 177.) Surds, and imaginary roots, enter equations by 
 pairs. 
 
 Take any equation, as 
 
 and suppose (a-f-^/6) to be one of its roots, then ( Jb) must 
 be another. 
 
 In place of x, in the equation, write its equal, and we have 
 
 By expanding the powers of the binomial, we shall find some 
 terms rational and some surd. The terms in which the odd 
 powers of ,Jb are contained will be surd ; the other terms 
 rational ; and if we put R to represent the rational part of this 
 equation, and SJb to represent the surd part, then we must have 
 
 But these terms not having a common factor throughout, cannot 
 equal 0, unless we have separately /?=0, and #=0; and if this 
 be the case we may have 
 
 This last equation, then, is one of the results of 
 being a root of the equation. 
 
GENERAL PROPERTIES OF EQUATIONS. 299 
 
 Now if we write (a Jb) in place of a?, in the original equ* 
 tion, and expand the binomials, using the same notation as 
 before, we shall find 
 
 But we have previously shown that this equation must be true ; 
 and any quantity, which, substituted for x, reduces an equation 
 to zero, is said to be a root of the equation ; therefore (a Jit) 
 is a root. 
 
 The same demonstration will apply to (-f~V a ) ( >/ a ) to 
 
 , J a, and to imaginary roots in the form of 
 
 (a+bjl), 
 
 (Art. 178.) If we change the signs of the alternate terms of 
 an equation, it will change the signs of all its roots. 
 At first, we will take an equation of an even degree. 
 If a is a root to the equation 
 
 x*+J33*+a*+Cx+D=0 ....... (1) 
 
 then will a be a root to the equation 
 
 x*x*+Bx*Cx+I)=Q ....... (2) 
 
 Write a for x, in equation (I), and we have 
 
 Now write a for re, in equation (2), and we have 
 
 a 4 rfa*+Ea 2 -}-Ca-{-I}=() ....... (4) 
 
 Equations (3) and (4) are identical ; therefore if a, put for .c in 
 equation (1), gives a true result, a put for x in equation (2), 
 gives a result equally true. 
 
 We will now take an equation of an odd degree. 
 
 If the equation x s -\-^x 2 -\-Bx-\-C=O t 
 have a for a root, then will the equation 
 
 have a for a root. 
 From the first 
 From the second a 3 4#a 2 /?a 6V.-0. 
 
300 ELEMENTS OF ALGEBRA. 
 
 This second equation is identical with the first, if \ve change 
 all its signs, which does not essentially change an equation. 
 
 The equation x 4 -f a: 3 l9x 2 -{-l l;r-{-30=0, has 1, 2, 3, 
 and -5, for its roots ; then from the preceding investigation 
 \vc learn that the equation 
 
 must have 1 2, 3, and +5 for its roots. 
 
 (Art. 179.) If we introduce one positive root into an equa- 
 tion, it will produce at least one variation in the signs of its 
 term; if two positive roots, at least two variations. 
 
 The equation x 2 -\-x-\-l.=Q, having no variation of signs, can 
 have no positive roots. (Art. 176.) Now if we introduce the 
 root +2, or which is the same thing, multiply by the factor x 2, 
 
 x*+ a?+l 
 x 2 
 
 x-- x x 
 Zx 2 2x 2 
 
 Then a; 3 - a*- a: 2=0; 
 
 and here we find one variation of signs from -\-x* to 2 , and 
 one permanence of signs through the rest of the equation, 
 
 If we take this last equation and introduce another positive 
 root, say -|-5, or multiply it by x 5, we shall then have 
 
 11 1 2 
 
 5 -[-5 +5 +10 
 
 Here are two variations of signs, one from 4-a? 4 to O.r 3 , and 
 another from Gx 3 to -f-4;r 2 . 
 
 And thus we might continue to show that every positive root, 
 introduced into an equation, will produce at least one variation 
 of signs. But we must not conclude that the converse of this 
 proposition is true. 
 
GENERAL PROPERTIES OF EQUATIONS. 3QJ 
 
 Every positive root will give one variation of signs ; but 
 every variation of signs does not necessarily show the existence 
 of a positive root. 
 
 For an equation may have 
 
 (a+bjV), ( V^f), c, d, 
 
 for roots ; then the equation will be expressed by the product of 
 the factors 
 
 (a* 2ax+a z +b*) (x+c) (x+d)=Q. 
 
 As one of these terms, ( 2ax), has the minus sign, it will 
 produce some minus terms in the product ; and there must neces- 
 sarily be variations of signs ; yet there is no positive root. At 
 the same time, the whole factor in which the minus term is found, 
 must be plus, whatever value be given to x, as it is evidently 
 equal to (x a) 2 -\-b*, the sum of two squares. 
 
 The equation 
 
 ^ 2x 3 x 2 -\-2 x +10=0, 
 
 has two variations of signs, and two permanences, but the roots 
 are all imaginary, viz., 
 
 T and -.1-=T. 
 
 If it were not for imaginary roots, the number of variations 
 among the signs of an equation would indicate the number of 
 plus roots : and this number, taken from the degree of the equa- 
 tion, would leave the number of negative roots; or the number 
 of permanences of signs would at once show the number of 
 negative roots. 
 
 To determine a priori the number of real roots contained in 
 any equation, has long bafHed the investigations of mathemati- 
 cians ; and the difficulty was not entirely overcome until 1829, 
 when M. Sturm sent a complete solution to the French Academy. 
 The investigation is known as Sturm's Theorem, and will be pre- 
 sented in the following Chapter. 
 
 LIMITS TO ROOTS. 
 
 (Art. 180.) All positive roots of an equation are comprised 
 between zero and infinity ; and all negative roots between zera 
 
302 ELEMENTS OF ALGEBRA. 
 
 and minus infinity ; but it is important to be able at once to 
 assign much narrower limits. 
 
 We have seen, (Art. 179.), that every equation, having a posi- 
 tive root, must have at least one variation among its signs, and 
 at least one minus sign. 
 
 If the highest power is minus, change all the signs in the 
 equation. 
 
 Now we propose to show that the greatest positive root must 
 be less than the greatest negative coefficient plus one. 
 
 Take the equation 
 
 =Q. 
 
 It is evident, that as the first term must be positive for all de- 
 grees, x must be greater and greater, as more of the other terms 
 are minus : then x must be greatest of all when all the othei 
 terms are minus, and each equal to the greatest coefficient, (D 
 being considered the coefficient of a?). 
 
 Now as A, B, &c., are supposed equal, and all minus, we shall 
 have 
 
 For the first trial take x=J2, and transpose the minus quantity, 
 and we have 
 
 Divide by ^ 4 , and we have 
 
 1=1+ xh/F*~;#~ 3 * 
 
 Now we perceive that the second member of the equation is 
 greater than the first, and the expression is not, in fact, an equa- 
 tion. x=rf proves x not to be large enough. 
 
 For a second tiial put x= 
 
 Then 
 
 Dividing by (^-fl)' 1 , we have 
 
 A Jl A A 
 
 -* 8 - 4 ' 
 
 We retain the sign of equality for convenience, though the 
 
GENERAL PROPERTIES OF EQUATIONS. 303 
 
 members are not equal. The second member consists of terms 
 in geometrical progression, and their sum, (Art. 120), is 
 
 1 T~aTT~\\' Hence tne fi rst member is greater than the second, 
 
 which shows that (.#+1) substituted for x, is too great. But Jl 
 was too small, therefore the real value of x, in the case under 
 consideration, must be more than Jl and less than (.#+1). 
 
 77m/ is, the greatest positive root of an equation, in the 
 most extreme case, must be less than the greatest negative 
 coefficient plus one. 
 
 In common cases the limit is much less. 
 
 From this, we at once decide that the greatest positive root of 
 the equation x 5 S-r'+To: 3 8x z 9a? 12=0, is less than 13. 
 
 Now change the second, and every alternate sign, and we 
 have the equation 
 
 ar 5 +3.r 4 +7r 3 +8;r 2 9^+12=0. 
 
 The greatest positive root, in this equation, is less than 10 ; but, 
 by (Art. 178.), the greatest positive root of this equation is the 
 greatest negative root of the preceding equation; therefore 10 is 
 the greatest limit of the negative roots of the first equation ; and 
 all its roots must be comprised between +13 and 10 ; but as 
 this equation does not present an extreme case, the coefficients 
 after the first are not all minus, nor equal to each other ; there- 
 fore the real limits of its roots must be much within +13 and 
 -10. In fact, the greatest positive root is between 3 and 4, and 
 the greatest negative root less than 1 . 
 
 If it were desirable to find the limits of the least root, put 
 
 0,=-, and transform the equation accordingly. Then find, as 
 
 y 
 
 just directed, the greatest limit of y, in its equation ; which will, 
 of course, correspond to the least value of x in its equation. 
 
 (Art. 181.) If we substitute any number less than the least 
 root, for the unknown quantity, in any equation of an even 
 degree, THE RESULT WILL BE POSITIVE. *ftnd if the degree of the 
 equation be odd, THE RESULT WILL BE NEGATIVE. 
 
 Let , b, c, &c., be roots of an equation, and x the unknown 
 
3J4 ELEMENTS OF ALGEBRA. 
 
 quantity. Ayso, conceive a to be the least root, /; the next 
 greater, and so on. Then the equation will be represented by 
 (x a)(x b)(x c)(a? rf), &c., =0. 
 
 Now in the place of x substitute any number h less than a, and 
 the above factors will become 
 
 (ha)(hb)(hc}(hd), &c. 
 
 Each factor essentially negative, and the product of an even 
 number of negative factors, is positive ; and the product of an 
 odd number is negative ; therefore our proposition is proved. 
 
 Scholium. If we conceive h to increase continuously, until it 
 becomes equal to a, the first factor will be zero ; and the product 
 of them all, whether odd or even, will be zero, and the equa- 
 tion will be zero, as it should be when A becomes a root. 
 
 If h increases and becomes greater than a, without being 
 iqual to I), the result of substituting it for x will be NEGATIVE, 
 in an equation of an even degree, and positive in an equation 
 of an odd degree. 
 
 For in that case the first factor will be positive, and all the 
 other factors negative ; and, of course, the signs of their product 
 will be alternately minus and plus, according as an even or odd 
 number of them are taken. 
 
 If h is conceived to increase until it is equal to b, then the 
 second factor is zero, and its substitution for x will verify the 
 equation. If h becomes greater than b, and not equal to c, then 
 the first two factors will be positive ; the rest negative ; and the 
 result of substituting h for x will give a positive or negative 
 result, according as the degree of the equation is even or odd. 
 
 If we conceive h to become greater than the greatest root, 
 then all the factors will be positive, and, of course, their product 
 positive. 
 
 For example, let us form an equation with the four roots 5, 
 2, 6, 8, and then the equation will be 
 
 (^+5)(x 2) (a? 6)(ar 8)=0, 
 Or. ... # 4 11^ 4z 2 +284z 480=0. 
 (The greater a negative number is, the less it is considered.) 
 
GENERAL PROPERTIES OF EQUATIONS. 305 
 
 Now if we substitute 6 for x, in the equation, the result must 
 be positive. Let 6 increase to 5, and tho result will be 0. 
 Let it still increase, and the result will be negative, until it has 
 increased to +2, at which point the result will again be 0. 
 
 If we substitute a number greater than 2, and less than 6, for 
 #, in the equation, the result will again be positive. A number 
 between 6 and 8, put for x, will render the equation negative ; 
 and a number more than 8 will render the equation positive ; 
 and if the number is still conceived to increase, there will be no 
 more change of signs, because we have passed all the roots. 
 
 If in any equation we substitute numbers for the unknown 
 quantity, which differ from each other by a less number than 
 the difference between any two roots, and commence with a 
 number less than the least root, and continue to a number greater 
 than the greatest root, we shall have as many changes of signs 
 in the results of the substitution as the equation has real roots. 
 
 If one real root lies between two numbers substituted for the 
 unknown quantity, in any equation, the results will necessarily 
 show a change of signs. 
 
 If one, or three, or any odd number of roots, lie between the 
 two numbers substituted, the results will show a change of signs 
 
 If an even number of roots lie between the two numbers sub- 
 stituted, the results will show no change of signs. 
 
 In the last equation, if we substitute 6 for a?, the result 
 will be plus. 
 
 If we substitute -J-3, the result will also be plus, and give no 
 indication of the two root? 5 and -f-2, which lie between. 
 
 (Art. 182.) If an equation contains imaginary roots, the 
 factors pertaining to such roots will be either in the form of 
 (a-f-) or in the form of [(a? a) 2 -{-& 2 ], both positive, whatever 
 numbers may be substituted for x, either positive or negative ; 
 hence, if no other than imaginary roots enter the equation, all 
 substitutions for x will give positive results, and of course, no 
 changes of sign. It is only when the substitutions for x pass 
 real roots that we shall find a change of signs. 
 
 26 
 
306 ELEMENTS OF ALGEBRA. 
 
 CHAPTER IV. 
 GENERAL PROPERTIES OF EQUATIONS CONTINUED. 
 
 (Art. 183.) If we take any equation which has all its roots 
 real and unequal, and make an equation of its first derived poly- 
 nomial, the least root of this derived equation will be greater than 
 the least root of the primitive equation, and less than the next 
 greater. 
 
 If the primitive equation have equal roots, the same root will 
 verify the derived equation.* 
 
 We sha)l form our equations from known positive roots. 
 
 Let <i, b, c, d, &c., represent roots ; and suppose a less than 
 &, b less than c, <fec., and x the unknown quantity. An equation 
 of the second degree is 
 
 x 2 ( 
 
 Its first derived polynomial is 
 
 If we make an equation of this, that is, put it equal to 0, we 
 shall have 
 
 Now if b is greater than , the value of x is more than a, 
 and less than 6, and proves our proposition for all equations of 
 the second degree. If we suppose a=6, then x=a, in both 
 equations. 
 
 An equation of the third degree is 
 
 ^(a+b+c^+^b+ac+bclxabc^ ..... (i) 
 Its first derived polynomial is 
 
 This equation, being of the second degree, has two roots, and 
 only two. 
 
 To ensure perspicuity and avoid too abstruse generality, we ope-ate on 
 equations definite in degree ; the result will be equally satisfactory to the 
 learner, and occupy, comparatively, but little space. 
 
GENERAL PROPERTIES OF EQUATIONS. 30? 
 
 Now if we can find a quantity which, put for x, will verify 
 equation (2), that quantity must be one of its roots. If we try 
 two quantities, and find a change of signs in the results, we are 
 sure a root lies between such quantities. (Art. 181.) Therefore 
 we will try a, or write a in place of x. As b and c are each 
 greater than a, we will suppose that 
 
 With these substitutions, equation (2) becomes 
 
 Reduced, gives ...... -j-/j/i'=0. 
 
 Therefore a cannot be a root; if it were we should have 0=0. 
 If we now make a substitution of b for a*, or rather (a-\-li) 
 for a*, and reduce the equation, we shall find 
 
 It is apparent that this quantity is essentially minus, as h' 
 is greater than h. Hence, as substituting a for a?, in the equa- 
 tion, gives a small plus quantity, and b for x gives a small 
 minus quantity, therefore one value of a?, to verify equation (2), 
 must lie between a and b. 
 
 This proves the proposition for equations of the third degree : 
 and in this manner we may prove it for any degree; but the labor 
 of substituting for a high equation would be very tedious. 
 
 If we suppose =6, and put c=a-\-h', and then substitute a 
 in place of a?, we shall find equations (1) and (2) will be verified. 
 
 Therefore in the case of equal roots, the equation and its 
 first derived polynomial will have a common measure, as before 
 shown in (Art. 168). 
 
 If all the roots of an equation are equal, the equation itself 
 i:.5ay be expressed in the form of 
 
 (x a) m =0. 
 
 Its first derived polynomial, put into an equation, will be 
 m ( x a) m ~'=0. 
 
 It is apparent that the primitive equation has m roots equal to 
 a; and the derived equation, (ra 1), roots also equal to a. 
 
308 ELEMENTS OF ALGEBRA. 
 
 Lastly, take a general equation, as 
 
 Its first derived polynomial, taken for an equation, will be 
 
 mx m - l +(m l)rfx m ~* ..... 7?=0. 
 
 We may suppose this general equation composed of the 
 factors 
 
 (x a)(x b) (x c), &c., =0, 
 
 and also suppose b greater, but insensibly greater, than a ; c 
 insensibly greater than b, &c. Then the equation will be nearly 
 
 (x ) OT =0 ; 
 and its derived polynomial, 
 
 m(x )" l ~ 1 =0, 
 
 cannot have a root less than (a), the least root of the primitive 
 equation ; but its root cannot equal a, unless the primitive equa- 
 tion have equal roots ; therefore it must be greater. 
 
 By the same mode of reasoning we can show that the greatest 
 root of an equation is greater than the greatest root of its derived 
 equation ; hence the roots of the derived equation are interme- 
 diate, in value, to the roots of the primitive equation, or contained 
 within narrower limits. 
 
 (Art. 184.) If we. take any equation, not having equal roots, 
 and consider its first derived polynomial also an equation, 
 and then substitute any quantity less than the least root of 
 either equation, for the unknown quantity, the result of such 
 mbstilution will necessarily give opposite signs. 
 
 Let a, b, c, &c., represent the roots of a primitive equation, 
 and a', b', &c., roots of its first derived polynomial ; x the un- 
 known quantity. Then the equation will be 
 
 (x a)(x b)(x c), &c., to m factors =0 ; 
 the derived equation will be 
 
 (x a')(x b')(x c'), &c., to (m 1) factors =0. 
 
 Now if we substitute h for x, and suppose h less than either 
 root, then every factor, in both equations, will be negative. 
 
 The product of an even number of negative factors is positive 
 and the product of an odd number is negative ; and if the factor?, 
 
GENERAL THEORY OF EQUATIONS 309 
 
 in the primitive equation are even, those in the der ved equation 
 must be odd. 
 
 Hence any quantity less than any root of either equation, will 
 necessarily give to these functions opposite signs. 
 
 (Art. 185.) Now if we conceive h to increase until it becomes 
 equal to a, the least root, the factor (x ) will be 0, and reduce 
 the whole equation to 0. Let h still increase and become 
 greater than a, and not equal to a', (which is necessarily greater 
 than a, (Art. 184.), and the factor (x a) will become plus, while 
 till the other factors, in both equations, will be minus, and, of 
 course, leave the same number of minus factors in both functions, 
 which must give them the same sign. Consequently, in passing 
 the least root of the primitive equation a variation is changed into 
 a permanence. 
 
 Sturm's Theorem. 
 
 (Art. 186.) If we take any equation not having equal roots, and its 
 first derived polynomial, and operate with these functions as though 
 their common measure was desired, reserving the several remain- 
 ders with their signs changed, and make equations of these func- 
 tions, namely, the primitive equation, its first derived polynomial 
 and the several remainders with their signs changed, and then 
 substitute any assumed quantity, h, for x, in the several functions, 
 noting the variation of signs in the result ; afterwards substitute 
 another quantity, h', for x, and again note the variation of signs ; 
 the difference in the number of variation of signs, resulting 
 from the two substitutions, will give the number of real roots 
 between the limits h and h'. 
 
 If * QQ and +00 are taken for h and h', we shall have the 
 whole number of real roots ; which number, subtracted from the 
 degree of the equation, will give the number of imaginary roots. 
 
 DEMONSTRATION. 
 
 Let X represent an equation, and X' its first derived polyno- 
 mial. 
 
 In operating as for common measure, denote the several quo- 
 
 * Symbols of infinity. 
 
310 ELEMENTS OF ALGEBRA. 
 
 tients by Q, Q' Q", &c., and the several remainders, with their 
 sign$ changed, by R, R', R", &c. 
 
 In these operations, be careful not to strike out or introduce 
 minus factors, as they change the signs of the terms ; then a 
 re-change of signs in the remainder would be erroneous. 
 
 From the manner of deriving these functions, we must have 
 the following equations 
 
 X R 
 
 R R 
 
 IF* -IF 
 
 <fcc. &c. 
 Clearing these equations of fractions, we have 
 
 X =X' Q R 
 
 X'=RQ' R' 
 R =nR'Q" R" 
 R'=R"Q"' R'" 
 
 As the equation X=0 must have no equal roots, the functions 
 X and X' can have no common measure (Art. 168.), and we shall 
 arrive at a final remainder, independent of the unknown quantity, 
 and not zero. 
 
 Proposition 1. No two consecutive functions, in equations (A), 
 can become zero at the same time. 
 
 For, if possible, let such a value of h be substituted for x, as 
 to render both X' and R zero at the same time ; then the second 
 equation of (A) will give R'=0. Tracing the equations, we must 
 finally have the last lemainder R TO =0; but this is inadmis- 
 sible ; therefore the proposition is proved. 
 
 Prop. 2. If lien one of the functions becomes zero, by giving 
 
GENERAL PROPERTIES OF EQUATIONS. 311 
 
 a particular value to x, the adjacent functions between which it 
 is placed must have opposite signs. 
 
 Suppose R' in the third equation, (A), to become 0, then the 
 equation still existing, we must have R= R". 
 
 The truth of Sturm's Theorem rests on the facts demonstrated 
 in Arts. 184, 185 and in the two foregoing propositions. 
 
 If we put the functions X, X', R, R', &c., each equal to ; 
 that is, make equations of them, and afterwards substitute any 
 quantity for #, in these functions, less than any root, the firs 
 and second functions, X and X', will have opposite signs, (Art. 
 184.) ; and the last function will have a sign independent of x, 
 and, of course, invariable for all changes in that quantity. The 
 other functions may have either plus or minuSj and the signs 
 have a certain number of variations. 
 
 Now all changes in the number of these variations must 
 come through the variations of the signs in the primitive func- 
 tion X. A change of sign in any other function will produce 
 no change in the number of variations in the series. 
 
 For, conceive the following equations to exist: 
 
 (B) 
 
 Now take x=h, yet h really less than any root of the equa- 
 tions, (B), and we may have the following series of signs : 
 
 X = 1 
 
 X' = 4- 
 
 R = 
 R' = 
 R"= 
 R'"= 4- 
 
 Or we mav have any other order of signs, restricted only to the 
 fact that the signs of the two first functions must be opposite, and 
 the last invariable, or unaffected by all future substitutions. 
 are three variations of signs. 
 
312 ELEMENTS OF ALGEBRA. 
 
 Now conceive h to increase. No change of signs can take 
 place in any of the equations, unless h becomes equal and passes 
 a root of that equation ; and as there are no equal roots, no two 
 of these functions can become at the same time (Prop. 1.) ; 
 hence a change of sign of one function does not permit a change 
 in another ; therefore by the increase of A, one of the functions, 
 (C), will become 0, and a further increase of h will change its 
 sign. 
 
 In the series of signs as here represented, X' cannot be the 
 first to change signs, for that would leave the adjacent functions, 
 X and R, of the same sign, contrary to Prop. 2 ; nor can the 
 function R' be the first to change sign, for the same reason. 
 
 Hence X or R or R" must be first to change sign. 
 
 If we suppose X to change sign, the other signs remaining the 
 same, the number of variations of signs is reduced by unity. 
 
 If R or R" change sign, the number of variations cannot be 
 changed ; a permanence may be made or reduced, and all cases 
 that can happen with three consecutive functions may be ex- 
 pressed by the following combinations of signs ; 
 
 + ) 
 Or + j 
 
 either of which gives one variation and one permanence. 
 
 Now as no increase or decrease in the number of variations of 
 signs can be produced by any of the functions changing signs, 
 except the first, and as that changes as many times as it has real 
 roots, therefore the changes in the number of variations of signs 
 show the number of real roots comprised between h and h'. 
 
 If h and h' are taken at once at the widest limits of possibility, 
 from infinity to -1- infinity, the number of variations of signs 
 will indicate the number of real roots; and this number, 
 taken from the degree of the equation, will give the number of 
 the imaginary roots. 
 
 (Art. 187.) The foregoing is a full theoretical demonstration 
 of the theorem ; but the subject itself being a little abstruse, 
 some minds may require the following practical elucidation. 
 
GENERAL PROPERTIES OF EQUATIONS. 3J3 
 
 Form an equation with the four assumed roots, 1, 3, 4, 6. 
 
 The equation will be 
 
 or X = a? 4 14# 3 +67# 2 126aH-72=0 Roots 1, 3, 4, 6. 
 X = 4x s 42r4-J34# 126 =0. . Jfoofa 2, 3.3, 5 nearly. 
 R =13y?9lx +153 ......... Moots 2.8, 4.1 nearly. 
 
 R'=72# 252 ............. Hoot 3.5 
 
 R"= + 
 
 \ 
 
 Let the pupil observe these functions, and their roots, and see 
 that they correspond with theory. The least root of X is less 
 than the least root of X'. (Art. 183.) The roots of any func- 
 tion are intermediate between the roots of the adjacent functions. 
 This corresponds with (Prop. 2.) ; for if three consecutive func- 
 tions have the same sign as , -, , or -f-, +, -K the middle 
 one cannot change first and correspond to (Prop. 2.) ; but signs 
 change only by the increasing quantity passing a root, and it must 
 pass a root of one of the extreme functions first ; therefore the roots 
 of X' must be intermediate in value between the roots of X and 
 R ; and the roots of R intermediate in value between the roots 
 of X' and R' ; and so on. But the roots of X' are within nar 
 rower limits than the roots of X (Art. 183.) ; therefore the roots 
 of all the functions are within the limits of the roots of X. 
 
 We will now trace all the changes of signs in passing all the 
 roots of all the functions. 
 
 We will first suppose x or h = ; which is less than any 
 root ; then as we increase h above any root, we must change the 
 sign of that function, and that sign only 
 
 We represent these changes thus : 
 
 27 
 
314 ELEMENTS OF ALGEBRA. 
 
 X X' R R' R" 
 
 * When xQ + + -{-.... 4 variations 
 
 #=1.1 * 4- 4-. . . .3 
 
 " #=2.1 4-* 4. 4- .... 3 " 
 
 " #=2.9 4- * _)_.... 3 
 
 " #=3.1 4-* + + .... 2 
 
 #=3.4 4" * 4" 2 " 
 
 " #=3.7 4- -p _{-.... 2 
 
 " #=4.1 * + 4. . . . . i 
 
 " #=4.11 4-* 4- 4-....1 " 
 
 #=5.i 4-* 4- 4- -i- .... i 
 #=6.1 4-* 4- 4- 4- 4 o 
 
 We commenced with 4 variations of signs, and end with 
 variations, after we have passed all the roots ; therefore the real 
 roots, in the primitive equation, must be 4 0=4. 
 
 By this it can be clearly seen, by inspection, that the changes 
 of sign in all the functions, except the first, produce no change 
 in the number of variations. 
 
 In making use of this theorem we do not go through the inter- 
 mediate steps, unless we wish to learn the locality of the roots as 
 well as their number. We may discover their number by sub- 
 stituting a number for x less than any root, and then one greater ; 
 tne difference of the variations of signs will be equal to the num- 
 ber of real roots. 
 
 If we take oo and 4-QD, the sign of any whole function 
 will be the same as that of its first term. 
 
 * In making this table, we did not really substitute the numbers assumed for 
 x, as we previously determined the roots ; and as passing any root changes the 
 sign in that function, we write a star against that sign which has just 
 changed. 
 
APPLICATION OF STURM'S THEOREM. 315 
 
 CHAPTER V. 
 
 APPLICATION OF STURM'S THEOREM. 
 (Art. 188.) In preparing the functions, remember that we are 
 at liberty to suppress positive numeral factors. 
 
 EXAMPLES. 
 
 ! How many real roots has the equation x 3 -}-9x=Q ? 
 Here X = 
 
 X'= 
 R= ar+1 
 R'= 
 Now for x substitute co or 100000, and we see at once 
 
 ^ X X' R R' 
 
 -j- ~J- 2 variations. 
 
 Again, for x put -f-co or +100000, and the resulting signs 
 
 must be , , 
 
 f- + 1 variation. 
 
 Hence the above equation has but one real root ; and, of course, 
 two imaginary roots. 
 
 To find a near locality of this root, suppose x=Q, and the 
 
 signs will be , 
 
 4- =t 2 variations. 
 
 x=l -|- 4~ rb .1 variation. 
 
 Hence the real root is between and 1 
 
 Now as we have found x, in the equation x 3 -}-Qx 6=0, to 
 be less than 1, X s may be disregarded, and 9x 6=0, will give 
 us the first approximate value of a?; that is, a?=.6, nearly. 
 
 2. How many real roots has the equation a; 4 3x z 4=0 ? 
 X= a; 4 Zx 2 4 
 X'=4a: 3 Qx 
 R=+25 
 
 If XCQ -f- +2 variations. 
 
 #=-1-03 -|- -}- + variation. 
 Hence there must be 2 real roots, and 2 imaginary roots. 
 
316 ELEMENTS OF ALGEBRA. 
 
 3. How many real roots has the equation x" 4# 3 621=0 1 
 (See Art. 103.) 
 
 X=z 6 4^621 
 
 X'=x 5 23? 
 
 R =+625 
 
 When x= co + +2 variations. 
 
 When #=-fco + + +0 variation. 
 Hence there are two real roots and four imaginary roots. 
 
 4. How many real roots has the equation x 9 15#-|-21=0? 
 
 Arts. 3. 
 
 5. How many real roots are contained in the equation 
 
 6. How many real roots are contained in the equation 
 
 =0? dm. 2. 
 
 7. Find the number and situation of the roots of the equation 
 
 =0. 
 
 X= 
 
 X'=3^ 2 +22a? -102 
 R=122o? 393 
 R=+ 
 
 Putting x= co H- +3 variations. 
 
 ar=-|-co -t- + + + variation. 
 Hence all the roots are real. 
 
 To obtain the locality of these roots there are several principles 
 to guide us, there is (Art. 180), but the real limits are much 
 narrower than that article would indicate, unless all the coeffi- 
 cients after the first are minus, and equal to the greatest. 
 
 A practised eye will decide nearly the value of a positive root 
 by inspection ; but by (Art. 183.) we learn that the root of R, 
 or 122# 393=0, must give a value to x intermediate between 
 the roots of the primitive equation. 
 
 From this we should conclude at once that there must be a 
 root between 3 and 4. 
 
NEWTON'S METHOD OF APPROXIMATION. 317 
 
 Substituting 3 for #, in the above functions, we have 
 
 + 4-2 variations. 
 #=4 -h + 4- -{- variation. 
 Hence there are two roots between 3 and 4. 
 
 As the sum of the roots must be 11, and the two positive 
 roots are more than 6, there must be a root near 17. 
 
 As there are two roots between 3 and 4, we will transform 
 the equation, (Art. 175), into another, whose roots shall be less 
 by 3; or put a?=3-J-y. Then we shall have 
 X= 
 X'= 
 R I22y 27 
 
 R'=+ 
 
 The value of y, in this transformed equation, must be near 
 the value of y in the equation 122y=27, (Art. 183.) ; that is, y 
 is between .2 and .3 
 
 2/=.2 gives 4~ + 2 variations. 
 7/=.3 gives 4-4~4-4-0 variation. 
 Hence there are two values of y between .2 and .3 ; and, of 
 course, two values of x between 3.2 and 3.3. 
 
 We may now transform this last equation into another whose 
 roots shall be .2 less, and further approximate to the true values 
 of x r in the original equation. 
 
 Having thus explained the foregoing principles, and, in our 
 view, been sufficiently elaborate in theory, we shall now apply it 
 to the solution of equations, commencing with 
 
 NEWTON'S METHOD OF APPROXIMATION. 
 
 (Art. 189.) We have seenf m (Art. 175.), that if we have any 
 equation involving x, and put x=a-{-y, and with this value 
 transform the equation into another involving y, the equation 
 will be 
 
 If a is the real value of x. then i/=0, and X=0. 
 
318 
 
 ELEMENTS OF ALGEBRA. 
 
 If a is a very near value to a?, and consequently y very 
 small, the terms containing y 2 , i/ 3 , and all the higher powers of 
 T/, become very small, and may be neglected in finding the op- 
 proximate value of y. 
 
 Neglecting these terms, we have 
 
 X-f-X'2/=0, 
 Or ..... y=~ ......... (1) 
 
 In the equation x=a-{-y, if a is less than x, y must be posi- 
 tive ; and if y is positive in the last equation, X and X' must 
 have opposite signs, corresponding to (Art. 184.). 
 
 Following formula (1), we have an approximate value of y ; 
 and, of course, of x. The value of x, thus corrected, again call 
 , and find a correction as before ; and thus approximate to any 
 required degree of exactness. 
 
 EXAMPLES. 
 
 1. Given 3a? 5 +4a? 3 5x 140=0, to find one of the approxi 
 mate values of x. 
 
 By trial we find that x must be a little more than 2. 
 Therefore, put x=2+y. 
 
 X = 3(2) 5 -{- 4(2) 3 5(2) 140. . . or . . . X = 22 
 X'=15(2) 4 +12(2) 2 5 ........ or. . . X'= 283. 
 
 X 22 
 
 Whence y= _= =0.07 nearly. 
 
 For a second operation, we have 
 
 X= 3(2.07) 5 -{- 4(2.07) 3 5(2.07) 140...By log. X = -0.854 
 X'=15(2.07) 4 +12(2.07) 2 5 ....... By log. X'= 321.82 
 
 Q &A 
 
 Hence the second value, or 2/=xHT^-r=0.00265-}- 
 
 o21o.2 
 
 And ............... #=2.07265+ 
 
 2. Given x*-}-2x 2 23#=70, to find an approximate value 
 of x. Am. 5.1 3454-. 
 
HORNER'S METHOD OF APPROXIMATION. 319 
 
 3. Given x 4 '3x z -\-75x=lQQQQ, to find an approximate 
 value of x. Ans. 9.886+ 
 
 4. Given 3jJ* 35# 3 llx 2 14rr-f30=0, to find an approxi- 
 mate value of x. dns. 11.998-}-. 
 
 5. Given 5x 3 3x* 2x=1560, to find an approximate value 
 of x rfns. 7.00867+. 
 
 CHAPTER VI. 
 HORNER'S METHOD OF APPROXIMATION. 
 
 (Art. 190.) In the year 1819, Mr. W. G. Homer, of Bath, 
 England, published to the world the most elegant and concise 
 method of approximating to roots of any then known. 
 
 The parallel between Newton's and Homer's method, is 
 this ; both methods commence by finding, by trial, a near value 
 to a root. 
 
 In using Homer's method, care must be taken that the number, 
 found by trial, be less than the real root. Following Newton's 
 method we need not be particular in this respect. 
 
 In both methods we transform the original equation involving 
 #, into another involving y, by putting xr-{-y, as in (Art. 175), 
 r being a rough approximate value of #, found by trial. 
 
 The transformed equation enables us to find an approximate 
 value of y, (Art. 189.). 
 
 Newton's method puts, this approximate value of y to r, and 
 uses their algebraic sum as r was used in the first place ; again 
 and again transforming the same equation, after each successive 
 correction of r. 
 
 Homer's method transforms the transformed equation into 
 another whose roots are less by the approximate value of y ; and 
 again transforms that equation into another whose roots are less, 
 and so on, as far as desired. 
 
 By continuing similar notation through the several transforma- 
 tions we may have 
 
320 ELEMENTS OF ALGEBRA. 
 
 x=r+y 
 
 y=s-\-z 
 z = t+z' 
 z'=u+z" 
 &c. &c. 
 
 Hence x=r-\-s-\-t, &c. ; r, s, t, &c., being successive figures 
 of the root. Thus if a root be 325, r=300, s=20, and /=5. 
 
 On the principle of successive transformations is founded the 
 following 
 
 RULE for approximating to the. true value of a real root 
 of an equation. 
 
 1st. find by Sturm 1 s Theorem, or otherwise, the value of 
 the first one or two figures of the root, which designate by r. 
 
 2d. Transform the equation (Art. 175), into another whose 
 roots shall be less by r. 
 
 3d. With the absolute term of this transformed equation for a 
 dividend, and the coefficient of y for a divisor, find the next 
 figure of the root. 
 
 4th. Transform the last equation into another, whose roots 
 shall be less, by the value of the last figure determined; and 
 so proceed until the whole root is determined, or sufficiently 
 approximated to, if incommensurable. 
 
 NOTE 1. In any transformed equation, X is a general symbol 
 to represent the absolute term, and X' represents the coefficient 
 of the first power of the unknown quantity. If X and X' be- 
 come of the same sign, the last root figure is not the true one, 
 and must be diminished. 
 
 NOTE 2. To find negative roots, change the sign of every alter- 
 nate term, (Art. 178.) : find the positive roots of that equation, 
 and change their signs. 
 
 (Art. 191.) We shall apply this principle, at first, to the solu- 
 tion of equations of the second degree ; and for such equations 
 as have large coefficients and incommensurable roots, it will fur- 
 nish by far the best practical rule. 
 
HORNER'S METHOD OF APPROXIMATION. 321 
 
 EXAMPLES. 
 1. Find an approximate root of the equation 
 
 x*-\-x 60=0. 
 
 We readily perceive that x must be more than 7, and less 
 than 8, therefore r=7. 
 
 Now transform this equation into another whose roots shall be 
 less by 7. 
 
 Operate as in (Art. 175.), synthetic division 
 
 i i .60 (7 
 
 7 56 
 
 8 4 
 7 
 
 15 
 
 Trans., eq. y* -f- 15y 4=0 
 
 Here we find that y cannot be far from T 4 ^, or between .2 and 
 .3 ; therefore transform the last equation into another whose 
 
 roots shall be .2 less ; thus, 
 
 s 
 
 1 15 4 (.2 
 
 0.2 3.04 
 
 15.2 0.96 
 
 2 
 
 15.4 
 
 The second transformed equation, therefore, is 
 z 2 -f- 15.4* 0.96=0 
 
 .96 
 
 To obtain an approximate value of z, we have ^ or 0.06. 
 
 15.4 
 
 In being thus formal, we spread the work over too large a 
 space, and must inevitably become tedious. To avoid these diffi- 
 culties, we must make a few practical modifications. 
 
 1st. We will consider the absolute term as constituting the 
 second member of the equation ; and, in place of taking the 
 algebraic sum of it, and the number placed under it, we will take 
 their difference. 
 
322 ELEMENTS OF ALGEBRA. 
 
 2d. We will not write out the transformed equations ; that is, 
 not attach the letters to the coefficients ; we can then unite the 
 whole in one operation. 
 
 3d. Consider the root a quotient ; the absolute term a dividend, 
 and, corresponding with these terms, we must have divisors. 
 
 In the example under consideration, 8 is the first divisor ; 15 
 .s thejirst trial divisor ; 15.2 is the second divisor, and 15.4 is 
 the second trial divisor ; 15.46 is the third divisor, &c. 
 
 Let us now generalize the operation. The equation may be 
 represented by 
 
 Transform this into another whose roots shall be less by r ; that 
 equation into another whose roots shall be less by s, &c., &c. 
 
 SYNTHETIC DIVISION. 
 
 1 a n ( r-j-s 
 
 r ar-f-r 8 
 
 1st divisor, .... a-\-r n' 
 
 r (a-\-2r-}-s}s 
 
 1st trial divisor, . . a-f-2r n" 
 
 0-t-s &c. 
 
 2d divisor, . . . .a-\-2r-}-s 
 
 s 
 
 2d trial divisor, . .a+2r+2 
 &c. 
 
 In the above we have represented the difference between n 
 and (ar-{-' 2 ) by n', &c. As n', n", n'", &c,, with their corre- 
 sponding trial divisors, will give 5, t, u, &c., the following for- 
 mula will represent the complete divisors for the solution of all 
 equations in the form of 
 
IIORJNER'S METHOD OF APPROXIMATION 3^3 
 
 1st divisor, .... a+ r 
 
 add .... r-\- s 
 
 2d divisor, 
 add 
 
 3d divisor, .... a+2r+2*-f- t 
 
 add .... t-\-u 
 
 4th divisor, .... 
 
 &c. &c. &c. 
 
 Equations which have expressed coefficients of the highest 
 power, as 
 
 the formulas will be : 
 
 1st divisor, .... a-h cr 
 
 add .... cr-f- cs 
 
 2d divisor, .... a-f 2cr+ cs 
 
 add .... cs-\-ct 
 
 3d divisor, .... a-f 2cr-j-2cs-t-c* 
 &c. &c. 
 
 To obtain trial divisors we would add cr only, in place of 
 (cr+cs), &c. 
 
 We will now resume our equation for a more concise solution. 
 
 
 1 
 
 7 
 
 n r 
 60 ( 7. 
 56 
 
 stu 
 262 
 
 1st divisor, . . 
 add . . 
 
 8 
 7.2 
 
 4 
 
 304 
 
 
 2d divisor, . . 
 add . . 
 
 15.2 
 26 
 
 96 
 9276 
 
 
 3d divisor, . . 
 add . . 
 
 15.46 
 
 62 
 
 324 
 
 31044 
 
 
 4th divisor, . . 15.522 1356 
 
324 ELEMENTS OF ALGEBRA. 
 
 We can now divide as in simple division, and annex the quo- 
 tient figures to the root, thus : 
 
 15522)1356 (08734 
 124176 
 
 11424 
 10865 
 
 559 
 
 465 
 
 94 
 Hence #=7.2620873-f- 
 
 2. Find #, from the equation x z 700#=59829. 
 On trial, we find x cannot exceed 800 ; therefore, r 700. 
 
 n rst 
 
 -h r -700+700= 59829(777 
 
 r+ s 700-H 70=770 00)000 
 
 =770 770)59829=n' 
 
 77 5390 
 
 =847 847)5929=n" 
 
 5929 
 
 Hence, #=777. 
 
 3 Find # from the equation x 2 1283#= 16848. 
 
 By trial, we find that # must be more than 1000, and 
 
 than 2000; therefore, r=1000. 
 
 n rstu 
 
 4- r = 283 16848(1296 
 
 H- 5= 1200 283 
 
 a-r-2r4- ~s= 917 917)2998 
 s+t 290 1834 
 
 = 1207 1207)11644 
 96 10863 
 
 1303 1303) 
 Hence, 
 
HORNER'S METHOD OF APPROXIMATION. 305 
 
 4. Given x 2 5;r=8366, to find x. 
 
 By trial, we find x must be more than 90, and less than 100 
 Therefore, a-f- r = 85 ) 8366 ( Q4=x 
 
 r+s . . 94 
 +2H-s=179 
 
 765 
 
 716 
 716 
 
 5. Find a?, from the equation x 2 375#+ 1904=0. 
 Here the first figure of the root is 5. 
 
 5 
 375 
 
 1st divisor, 370. 
 5.1 
 
 2d divisor, 364.9 
 .14 
 
 1904(5.1480052207 
 1850 
 
 5400 
 3649 
 
 3d divisor, 364.76 
 48 
 
 4th divisor, 364.712 
 8 
 
 5th divisor, 364.7039 
 
 175100 
 145904 
 
 2919600 
 2917696 
 
 190400 
 1823519 
 
 80491 
 72941 
 
 7550 
 7294 
 
 256 
 255 
 
 1 
 
 6. Given ^+7^1194=0, to find x. Ans. 31.2311099. 
 
 7. Given x 3 21^=214591760730, to find x. rfns. 463251. 
 
 It might be difficult for the pupil to decide the value of r, as 
 applied to the last example, without a word of explanation : x 
 must be more than the square root of the absolute term, that is, 
 
326 ELEMENTS OF ALGEBRA. 
 
 more than 400000 ; then try 500000, which will be found too 
 great. 
 
 (Art. 192.) When the coefficient of the highest power is not 
 unity, we may (if we prefer it to using the last formulas for di- 
 visors), transform the equation into another, (Art. 166.), which 
 shall have unity for the coefficient of the first term, and all the 
 other coefficients whole numbers. 
 
 8. Given 7x* 3#=375. 
 
 Put #=- and we shall have y 2 3^=262 5. 
 
 One root of this equation is found to be 52.7567068+, one- 
 seventh of which is 7.536672+ ; the approximate root of the 
 original equation. 
 
 9. Given 7# 2 83#+ 187=0, to find one value of a?. 
 
 Jlns. 3.024492664 
 
 1O. Given x* T 3 T a?=8, to find one value of x. 
 
 Am. 2.96807600231 
 
 11. Given 4# 2 +J#=^, to find one value of x. 
 
 Jlns. Jlns. .14660+ 
 
 12. Given \x*-\-\x T 7 T , to find one value of x. 
 
 Jlns. .6042334631 
 
 13. Given 115 3x* 7#=0, to find one value of x. 
 
 Jlns. 5.13368606 
 
 (Art. 193.) We now apply the same principle of transforma- 
 tion to the solution of equations of the third degree. 
 
 EXAMPLES. 
 1. Find one root of the equation 
 
 We find, by trial, that one root must be between 3 and 4. 
 
HORNER'S METHOD OF APPROXIMATION. 
 
 327 
 
 1st Transformation. 
 
 1 1 
 3 
 
 3 
 
 ~5 
 3 
 
 ~S 
 
 70 
 
 *76 
 15 J 
 
 91 =X' 
 
 r 
 300 (3. 
 
 228 
 
 72=X 
 
 .... _- 72 _A 
 
 to 
 
 o. 
 
 H 
 1 
 
 I 8 
 0.7 
 
 91 ' 
 6.09-| 
 
 s 
 72 (0.7 
 67.963 
 
 8.7 
 
 7 
 
 *97.09 , 
 6.58J 
 
 4.037=X 
 
 9.4 
 
 7 
 
 To.7 
 
 103.67=X' 
 
 
 1 10.1 
 .03 
 
 103.67 
 .3039-} 
 
 t 
 4.037 (0.03 
 3.119217 
 
 10.13 
 3 
 
 *103.9739 
 .3048J 
 
 0.917783=X 
 
 10.16 
 3 
 
 HU9" 
 
 104.2787=X' 
 
 
 1 10.19 
 .008 
 
 104.2787 
 .081584 
 
 u 
 0.917783(0.008 
 0.834882272 
 
 10.198 
 8 
 
 10.206 
 
 8 
 
 10.214 
 
 *104.360284 
 81648 
 
 104.441932 
 
 .082900728 
 
328 ELEMENTS OF ALGEBRA. 
 
 The terms here marked X are trial divisors ; we have pro 
 fixed stars to the numbers that we may call complete divisors. 
 We rest here with the equation 
 
 (z") 3 +10.214(z // ) 2 4-104.4419z" 0.0829=0. 
 
 The value of z" is so small that we may neglect all its powers, 
 
 except the first, and obtain several figures by division, thus : 
 
 104 ) 829 ( 797 
 
 728 
 
 74 
 
 r s tu 
 
 Hence, one approximate value of # is .... 3.738797+. 
 (Art. 193.) We may make the same remarks here as in (Art. 
 191.), and, as in that article, generalize the operation. 
 
 Let x s -\-J2z? -\-Bx-N represent any equation of the third 
 degree, and transform it into another whose roots shall be r less ; 
 thus, 
 
 1 A B = N ( r 
 
 r 
 
 N' 
 
 The transformed equation is 
 
 If we put ( 
 and 
 
 we shall have . . . y*-\-*ft'y z -{-B'y=N t , an equation similar 
 to the primitive equation. 
 
 If we transform this equation into another whose roots shall be 
 less by s, we shall have 
 
 Or, . . z*-\-"z*-\-B"z=N" ; an equation also similar to 
 
HORNER'S METHOD OF APPROXIMATION. 329 
 
 the first equation. And thus we may go on forming equation 
 after equation, similar to the first, whose roots are less and less. 
 
 The quantities N', N", &c., are the same as X in our previous 
 notation, and the quantities B', B", &c., are the same as the 
 general symbol, X' ; but we have adopted this last method of 
 notation to preserve similarity. 
 
 Observe, that as (r-{-JT)r-{-JB is the first complete divisor, N 
 is the number considered as a dividend and r the quotient ; and 
 
 therefore... r^-^, nearly. 
 
 The next equation gives us 
 
 ' ' 
 
 ( ' " 
 
 JV N" very nearly. 
 
 And the next, /= 
 
 N'" N" 
 
 &c. = &c. &c. 
 
 The denominators of these fractions are considered complete 
 divisors, and the quantities, B', B", B'", are considered trial 
 divisors. The further we advance in the operation the nearer 
 will the trial and true divisors agree. 
 
 Before the operation is considered as commenced, we must 
 find the first figure of the root (r) by trial. Then the operator 
 can experience no serious difficulty, provided he has in his mind 
 a clear and distinct method of forming the divisors ; and these 
 may be found by the following 
 
 RULE. 1st. Write the number represented by B, and directly 
 under it, write the value of r(r+A) ; the algebraic sum of these 
 two numbers is the first complete divisor. 
 
 2d. Directly under the first divisor write the value of r 2 , and 
 the sum of the last three numbers is B 1 , or the first trial divisor. 
 
 3d. Find by trial, as in simple division, how often B' is con- 
 tained in N, calling the first figure s, (making some allowance 
 for the augmentation o/'B'), and s will be a portion of the 
 root under trial. 
 28 
 
330 ELEMENTS OF ALGEBRA. 
 
 4th. Take the value of the expression (3r+s-[-A) and add 
 it to the first trial divisor ; the sum is the second divisor (if 
 we have really the true value of s). 
 
 IN GENERAL TERMS; 
 
 Under any complete divisor, write the square of the last figure of 
 the root ; add together the three last columns, and their sum is 
 the next trial divisor. With this trial divisor decide the next 
 figure of the root. 
 
 Take the algebraic sum, of three times the root previously 
 found, the present figure under trial, and the coefficient Jl, and 
 multiply this sum by the figure under trial, and this product, 
 added to the last trial divisor, gives the next complete divisor. 
 
 EXAMPLES. 
 
 1. Given a? 3 +2a^+3a?= 13089030, to find one value of x. 
 By trial, we soon find that x must be more than 200 and less 
 than 300 ; therefore we have 
 
 r=200, .#=2, .5=3. 
 By the rule, 
 
 N rs 
 B ....... 3 40403 ) 13089030 ( 235 
 
 40400 - 80806 
 
 -j 
 
 1st divisor ....... 40403 f 139763 ) 500843 
 
 r 2 ........ 40000J 419289 
 
 1st trial divisor . . B' =120803 163108 ) 815540=^ 
 (3H-s-M)s 18960^ 815540 
 
 2d divisor ....... 139763 f 
 
 s 2 ........ 900J 
 
 2d trial divisor JB"=I 59623 
 3485 
 
 3d divisor ....... 163108 
 
 Hence, ........ . ............... z=235, 
 
HORNER'S METHOD OF APPROXIMATION. 331 
 
 3. Given ^-(-173^=14760638046, to find one value of x 
 Here .#=0, I?=173, and we find, by trial, that x must be 
 more than 2000, and less than 3000 ; therefore r=2000. 
 B ........ 173 
 
 4000000 
 
 1st divisor ......... 4000173 f 
 
 r 2 ......... 4000000J 
 
 B' ........ 12000173 
 
 2560000 
 
 2d divisor ......... 14560173 f 
 
 s 2 ......... leooooJ 
 
 B" ....... 17280173 
 
 . 362500> 
 
 3d divisor ......... 17642673 f 
 
 2500 J 
 
 B'" ....... 18007673 
 
 .... 22059 
 
 4th divisor ........ 18029732 
 
 4000173 ) 14760638046 ( 2453=z 
 8000346 
 
 14560173 ) 67602920 =N' 
 58240692 
 
 17642673 ) 93622284 =N" 
 88213365 
 
 18029732 ) 54089196=^'" 
 54089196 
 
 3. Given a? 3 +2a^ 23#=70, to find an approximate value 
 of x. rfns. #=5.134578-K 
 
 4. Given x* 17a^-i-42iC=185, to find an approximate value 
 of x. Jns. r== 15.02407. 
 
332 ELEMENTS OF ALGEBRA. 
 
 (Art. 194.) When the coefficient of the highest power is not 
 unity, we may transform the equation into another, (Art. 166.), 
 in which the coefficient of the first term is unity, and all the 
 other coefficients whole numbers ; but it is more direct and con- 
 cise to modify the rule to suit the case. 
 
 If the coefficient of the first power is c, the first divisor will 
 be (cr-f^)r-f-^, in place of (r+A)r-\-B. 
 
 In place of (3r-{-s-\-JI)s, to correct the first trial divisor, we 
 must have (3cr-\-cs-\-tf)s ; and, in genera], in place of using 3 
 times the root already found, we must use 3c times the root ; and, 
 in place of the square of any figure, as r 2 , s 2 , &c., we must use 
 cr 2 , cs 2 , &c. 
 
 EXAMPLES. 
 
 5. Find one root of the equation, 3:e 3 -{-2;r 2 -}-4;r=75, 
 
 By trial, we find that x must be more than 2, and less than 3; 
 
 therefore , a A 
 
 r=2, c=3, #=2, /j4. 
 
 N r stu 
 B ........ 4. 75 ( 2.577 
 
 16. > 
 
 1st divisor 20. 
 
 cr 2 12. J 
 
 40 
 
 35=7V' 
 29375 
 
 B 1 ....... 48. 5625=^" 
 
 . .10.75 5038579 
 
 2d divisor ...... 58.75 f . 586421 =N'" 
 
 cs 
 
 75J .517301099 
 
 B" ....... 70.25 69119901 
 
 3d divisor ..... 71.9797 [ Continue, by simple division, thus 
 
 ct * ....... _ Jfy 739 ) 6911 ( 935 
 
 B>" ....... 73.7241 6651 
 
 176057 ~ 
 
 4th divisor ..... 73.900157 221 
 
 Hence, ................. z =2. 577935 -f. 
 
 * R is a symbol to represent the entire root, as far as determined. 
 
HORNER'S METHOD OF APPROXIMATION. 333 
 
 6. Find one root of the equation, 5# 3 Qx?-\-3x 85. 
 
 rfns. x= 2.16399-. 
 
 7. Find one root of the equation, \2x*-\-x* 5a?=330. 
 
 dns. #=3.0364754- 
 
 8 Find one root of the equation, 5# 3 -f 9# 2 7#=2200. 
 
 Ans. #=7.107353G-f-. 
 
 9. Find one root of the equation, 5# 3 3a? 2 2.r=1560. 
 
 Ans. #=7.0086719-1- . 
 
 (Art. 195.) This principle of resolving cubic equations ma)' 
 be applied to the extraction of the cube root of numbers, and 
 indeed gives one of the best practical rules now known. 
 
 For instance, we may require the cube root of 100. This 
 gives rise to the equation 
 
 in which .#=0, and 5=0, and the value of x is the root 
 sought. 
 
 As Ji and B are each equal to zero, the rule under (Art. 193.) 
 may be thus modified. 
 
 1st. Keeping the symbols as in (Art. 193.), and finding r by 
 trial, r 2 will be the first divisor, and 3r 2 is B', or the first TRIAL 
 divisor. 
 
 2d. By means of the dividend (so called), and the first trial 
 divisor, we decide s the next figure of the root. 
 
 3d. Then (3r-{-s)s ; that is, three times the portion of the 
 root already found, with the figure under trial annexed, and 
 the sum multiplied by the figure under trial, will give a sum, 
 which, if written two places to the right, under the last trial 
 divisor, and added, will give the next complete divisor. 
 
 4th. After we have made use of any complete divisor, write 
 the square of the last quotient figure under it ; the sum of the 
 three preceding columns is the next trial divisor ; which use, 
 and render complete, as above directed, and so continue as far as 
 necessary.* 
 
 * In case of approximate roots after three or four divisors are found, we 
 may find two or three more figures of the root, with accuracy, by simple division. 
 
334 ELEMENTS OF ALGEBRA. 
 
 We may now resolve the equation 
 
 1st divisor 
 
 (3r-f-s> . 
 2d divisor 
 
 . 16 
 
 . .48 
 . . 756 
 
 . . 5556 
 
 r stu 
 
 100 ( 4.6415889-t- 
 64 
 
 36 
 33336 
 
 B" 6348 
 
 (3R+t)t . 5536] 
 
 3d divisor .... 640336 f 
 
 ? 16 J 
 
 B"' 645888 
 
 . . 13921 
 
 4th divisor. .64602721 
 
 64616643 
 646166 
 
 2. Extract the cube root of 
 1st divisor * * 64 
 B' 192 
 
 (3r-f-s)s 1729 
 
 2d divisor 
 
 B" . 
 
 2664000 
 2561344 
 
 102656 
 64602721 
 
 38053279 
 32308321 
 
 5744958 
 5169331 
 
 575627 
 516933 
 
 58694 
 58154 
 
 673373097125. 
 
 N rstu 
 673373097125 (8765 
 512 
 
 3d divisor 
 
 B'" 
 
 20929 \ 
 49 j 
 
 161373 
 
 146503 rNO TE.-To deter- 
 
 22707 
 15696] 
 
 mine s, we have 
 
 14870097 1 i92)1613( 
 
 13718376 | Some allowance 
 
 2286396 \ 
 36J 
 
 11 51721 1251 crease of 192. 
 
 1151721125 
 
 2302128 
 131425 
 
 230344225 
 
HORNER'S METHOD OF APPROXIMATION. 335 
 
 3. Extract the cube root of 1352605460594688. 
 
 tins. 110592. 
 
 4. Extract the cube root of 5382674. tins. 175.25322796. 
 
 5 Extract the cube root of 15926.972504. 
 
 Ans. 25.16002549. 
 
 6. Extract the cube root of 91632508641. 
 
 Ans. 4508.33859058. 
 
 7. Extract the cube root of 483249. Am. 78.4736142 
 
 (Art. 196.) The method of transforming an equation into an- 
 other, whose roots shall be less by a given quantity, will resolve 
 equations of any degree ; and for all equations of higher degrees 
 than the third, we had better use the original operation, as in 
 (Art. 192.), and attempt no other modification than conceiving the 
 absolute term to constitute the second member of the equation ; 
 and the difference of the numbers taken in the last column in 
 place of their algebraic sum. 
 
 The following operation will sufficiently explain : 
 1. Find one value of x from the equation 
 
 
 
 
 r 
 
 1 
 
 3 
 
 75 
 
 = 10000 (9 
 
 9 
 
 81 
 
 702 
 
 6993 
 
 9 
 
 78 
 
 777 
 
 3007=^V 
 
 9 
 
 162 
 
 2160 
 
 
 18 
 
 240 
 
 2937 
 
 
 9 
 
 243 
 
 
 
 27 
 
 483 
 
 
 
 9 
 
 
 
 
 36 
 
 
 
 
 (Continued on the next page.) 
 
336 
 
 ELEMENTS OF ALGEBRA. 
 
 
 1 36 
 
 483. 
 
 2937. 
 
 9 
 
 =3007. ( 0.8 
 
 
 to 
 
 c- 
 
 t -^ 
 
 .8 
 
 29.44 
 
 409.952 
 
 2677.5616 
 
 
 36.8 
 
 512.44 
 
 3346.952 
 
 329.4384=^" 
 
 ansformation. 
 
 8 
 
 3776 
 
 8 
 
 38.4 
 
 30.08 
 
 542.52 
 30.72 
 
 573.24 
 
 434.016 
 
 coefficients to their nearest 
 
 3780.968 
 Take the 
 
 
 8 
 
 
 unit. 
 
 
 
 
 39.2 
 
 
 
 
 
 
 
 
 t 
 
 
 1 39 
 
 573 
 
 3781 
 
 = 329.4384 ( 0.08 
 
 
 
 + 
 
 3 
 
 46 
 
 306.16 
 
 
 39 
 
 576 
 
 3827 
 
 23.2784=^'" 
 
 + 
 
 3 
 
 46 
 
 
 
 39 
 
 579 
 
 3873 
 
 
 
 
 
 
 u 
 
 
 1 39 
 
 579 
 
 3873 
 
 = 23.2784(0.00600-1- 
 
 
 
 3 
 
 23.256 
 
 
 
 
 3876 
 
 224 
 
 
 
 
 3 
 
 
 
 3879 
 Hence, a-=9.88600+. 
 
 N. B. We went through the first and second transformations in full. Had 
 
 we been exact, in the third, we should have added .08 to 39.2, and multiplied 
 
 their sum, (39.28), by .08, giving 3.1424 ; we reserve 3. only to add to the 
 
 next column. By a similar operation we obtain 46. to add to the next column. 
 
 EXAMPLES. 
 
 1. Given x & x 3 x 4 -J-500=0, to find one value of a;. Ans. 4.46041671 
 
 2. Given a; 4 5.r 3 -}-9;r=2.8, to find one value of x. Ans. .32971055072 
 
 3. Given 20#-{-ll;r 2 -j-9z 3 ^=4, to find one value ofx. 
 
 Ans. .17968402502 
 
 4. Required the 5th root of 5000; or, in other terms, find one root of the 
 equation ^=5000. Ans. 5.49280-}- 
 
 5. Given 2?= 
 
 to find one value of x. 
 
 Ans. 2.120003355 
 
THEORY OF EQUATIONS-SUM OF COEFFICIENTS. 337 
 
 In following out general principles, we have, thus far, omit- 
 ted some artifices which apply in particular cases, or depend 
 on particular circumstances, which we will now set forth, as 
 they will enlarge the views of all who pay attention to them. 
 
 The first circumstance to be observed is the sum of the 
 numerical coefficients. 
 
 (Art. 197.) Transpose every term to the first member of the 
 equation, and if the sum of the coefficients is zero, we are sure 
 that unity is one of the roots of the equation. 
 
 For example, we are sure that the equation 
 a;4+7a; 3 133 2 -79*+84==0, 
 
 has unity for one of its roots, because l-J-7 13 79+84=0. 
 Assume x=l, and substitute 1 for x in the equation, and the 
 equation will be verified. But assume x, a little greater or a 
 little less than 1, and substitute, and the equation cannot be 
 verified ; therefore the following principle is established : 
 
 In any equation having numerical coefficients, if the sum of the 
 coefficients is zero, -\-l is one of the roots of the equation. 
 
 We can now divide the given equation by (x 1) and thus 
 depress it one degree. 
 
 For another example we will recall the 4th, on page 266, 
 which is 
 
 Here 1 13-(-12=0, whence x=\. The answer on page 
 266 is 3 ; that is another root. The third root must, therefore, 
 be 4 ; the sum of all the roots must be 0, because the second 
 term of the equation is zero. (Art. 157, second observation.) 
 
 We perceive at once, that unity is one root in each of the 
 following equations : 
 
 2. x 3 Qx^ + Ux 6=0. 
 
 , 3. x *Q x 3J r \4 x 28 x }5Q. 
 
 4. X 3_ i3^2_j_3 9a ._ 27=0. 
 
 (Art. 198.) When the absolute term of an equation is 
 numerically too large (whether it be plus or minus), to make 
 29 
 
338 ELEMENTS OF ALGEBRA. 
 
 the sum of the coefficients zero, we can diminish it, as well as 
 other coefficients, by the following transformation. 
 For instance : In the equation 
 
 the sum of the coefficient is obviously not equal to zero ; 70 is 
 too large, but we can diminish it by placing x=2y, or x=ny, 
 and substituting this value of x in the equation, we have 
 
 H 3y3 iOw 2 
 Dividing by n 3 we obtain 
 
 n n 
 
 Now, we can assume n equal to 2, 3, 4, or any other number 
 whatever. 
 
 Let w=2, and the preceding equation becomes 
 
 y'-5y s +"H-=o. (i) 
 
 let=-2, 
 
 Let 7i=5, and y 3 -^+y +=0. (3) 
 
 Zo ~o 
 
 The sum of the coefficients in each of the equations (2) and 
 (3), is zero. Therefore, y equals 1 in each of them. 
 
 But x=ny, therefore z 2, and #=5, 
 
 which are two of the three roots in the original equation. 
 
 But the sum of the three roots must be 10, by the genera] 
 theory of equations ; therefore,, the third root must be 7, 
 
 Because, 2+5-f 7= 1 0. 
 
 If we take the first equation in (Art. 197), and depress it 
 one degree, by dividing by (x 1), we shall obtain the equation 
 
 z*--8x 2 oxU=Q. (a) 
 
 Here it is obvious that the coefficients require depressing to 
 make their sum zero. 
 
 Therefore, place x=nP, and 
 
 p 3 i 8 j p 2 _l p __84 =a ^ 
 
 w n 2 n 3 v ' 
 
THEORY OF EQUATIONS SUM OF COEFFICIENTS. 339 
 
 Making n=3, this equation becomes 
 
 2_5 p _28 =0 
 99 
 
 Here ?+?i - =0. Whence P=l, but =3, 
 99 9 9 
 
 and z=nP=3, another of the roots of the original equation. 
 
 We may now depress equation (a) one degree, by dividing 
 by (x 3), and then find the other two roots. 
 
 Or, we may assume n 4, and substitute this value for n 
 in (b), then that equation becomes 
 
 Here, because H~ff r 5 6-h?6=- ^=1, and x= 4, 
 which is a third root of the original equation. x= 7 is the 
 fourth root. 
 
 Thus we can find all the roots in the equation contained in 
 the last article. 
 
 If it were our object to transform an equation so as to 
 
 increase the coefficients in place of diminishing them, as 
 
 p 
 in this article, we should place #= , but this has already 
 
 n 
 
 been shown in (Art. 166). 
 
 This method of solution may be considered a sequence to 
 Newton's method of dividers as shown in (Art. 147). 
 
 (Art. 199.) An equation which has one for a root, can bo 
 changed into another, which will have minus one for a root, by 
 changing its second, and every alternate sign. (Art. 178.) 
 
 Consequently, then, an equation which has minus one for 
 one of its roots, can be changed into another, which will have 
 plus one for one of its roots, by changing the sign of its second, 
 and every alternate term. 
 
 Therefore, if changing the sign of the second, and every alter- 
 nate term in any equation ivill make the sum of the coefficients 
 ZERO, one root of that equation must be 1. 
 
 For example, the sum of the coefficients of the equation 
 4.c 3 20;r 2 -{-ll.r+35=0, is not zero; 
 
340 ELEMENTS OF ALGEBRA. 
 
 but change the signs as above directed, and we have the 
 equation, 
 
 4z 8 -f20# 2 -)-l 1^35=0, 
 
 and the sum of the coefficients is now zero, and consequently 
 unity is one root of this equation, and therefore 1 is a root 
 of the first mentioned equation. 
 
 Hence, in seeking to solve any equation of this kind, if the 
 sum of the coefficients is not zero, we may change the sign of 
 the second, and each alternate term, and if the sum is then 
 zero, 1 may be taken as one root of the equation ; and then 
 we may depress the equation one degree by dividing by (x-\-\). 
 
 (Art. 200.) Another circumstance to be observed in rela- 
 tion to coefficients, is that of their recurrence, producing 
 what is called 
 
 RECURRING EQUATIONS. 
 
 In a recurring equation the coefficients of those terms which are 
 equally distant from the extreme terms, are numerically equal. 
 Thus x 5 3x*+7x 3 +7x 2 3a?-fl=0, (1) 
 
 are all recurring equations of an odd degree. 
 
 If we substitute 1 for x, in each of the equations (1) and 
 (2), they will be verified; and -f- 1 put for x will verify (3) 
 and (4). Therefore, any equation of an odd degree has either 
 1 or _j_ i for one of its roots, and the equation itself is redu- 
 cible to an even degree, by dividing by (x-\-\ ) or by ( x ] ). 
 
 Minus one is a roof, when the like coefficients have the same sign; 
 and plus one is a root when the like coefficients have opposite signs. 
 
 (Art. 201.) Recurring equations, of an even degree, in ivhich 
 the like coefficients have opposite signs, and ivhoss middle term is 
 ivanting, are divisible by (x 2 1), and therefore +1, and 1, 
 are both roots of every such equation. 
 
THEORY OF EQUATIONS SUM OF COEFFICIENTS. 341 
 The following is an equation of this description : 
 
 which can be put in the following form : 
 
 (x*l)+8x(x*l)-\-llx 2 (x 2 l)=Q. 
 
 and this is obviously divisible by (x 2 1). Whence we may 
 place x 2 1=0, and x=l, or 1. 
 
 And thus the principle enunciated in this article is estab- 
 lished. Dividing the equation by (x 2 1), will reduce it two 
 degrees lower, and leave it an equation of an even degree. 
 
 (Art. 202.) Every recurring equation of an even degree, above 
 the second, can be reduced to an equation of half that degree, by the 
 following artifice. 
 
 We take the following equation for an example : 
 
 #6_7 a .5_|_8 2; 4 9z 3 -far 2 72+1=0. (1) 
 
 Divide each term by the square root of the highest power of 
 the unknown quantity. In this case divide by x 3 , and we have 
 
 . (2) 
 
 Which can be put in the following form : 
 
 Now assume, /Vfi'Wz. Then, fx 2 -\-^\=z 2 2. 
 
 And Sx 3 -\-- )=z 3 3z. Whence (3) becomes 
 
 v as 3 / 
 
 ( 2 3_3 2 )7(s 2 2)4-8^ 9=0, (4) 
 
 an equation of the third degree. 
 
 Thus (1), an equation of the 6th degree is reduced to (4), an 
 equation of half that degree ; and this process will reduce any 
 other recurring equation of an even degree to another of half 
 that degree. 
 
 (Art. 203.) A recurring equation of the 4th degree can be 
 reduced to a quadratic by the following rule. 
 
 Divide by the coefficient of x 4 , and transpose the term containing 
 -x 1 . Add to each member +2#-. Use the plus sign when the signs 
 
342 ELEMENTS OF ALGEBRA. 
 
 of the second and fourth term are alike, and use the minus sign 
 when tJtey are unlike. Also add to each member the square of 
 half the term containing x. Extract the square root of each mem- 
 ber, and the result is an equation of the second degree. 
 
 EXAMPLES. 
 
 a a 
 
 Add 
 
 .x, a quadratic. 
 
 We are indebted to Professor Stevens, of Greenmount Col- 
 lege, Indiana, for this article. It is original with him. 
 
 BINOMIAL EQUATIONS. 
 
 (Art. 204.) Binomial equations have but two terms, such as 
 
 a; 1=0, z 2 1=0, z 3 1=0, &c. 
 Or, such as #a=0, #"a n =0, &c., &c. 
 
 Every binominal equation is also a recurring equation, for 
 the coefficient certainly recurs. 
 
 A binomial equation has apparently but one root, but a full 
 investigation will discover as many roots as there are units 
 in the exponent of the unknown quantity, as is explained by an 
 example in (Art. 163). We here give one more practical 
 illustration. 
 
 The equation x 5 1=0, is a binomial equation of an odd 
 degree, and it is also a recurring equation. 
 
 It is also divisible by (x 1), having apparently but one 
 root, -f-1. The division produces 
 
 
THEORY OF EQUATIONS SUM OF COEFFICIENTS. 343 
 
 O. (1) 
 
 a recurring equation of the 4th degree. 
 Dividing again by x 2 , we obtain 
 
 l=0. (2) 
 
 Assuming fx+-\=P. Then z 2 +-L=P 2 2, 
 \ x / x 
 
 And (2) becomes P 2 +P 1=0, a quadratic. 
 
 Again: Operating under (Art. 203), we obtain 
 
 Add * 
 
 x 
 Square root, 
 
 Or x 2 4-1(1^:^/5)2:= 1, a quadratic. 
 
 Whence the five roots in (x 5 1)=0, or the five-fifths roots 
 of unity are 
 
 x=\, or x= 
 
 or s=i(,y5 1 V 
 
 or x= 
 
 or a?= 1(^5+1 +V 10+2 V 6 )- 
 
 EXAMPLES. 
 1. Find the roots of the equation 
 
 Ans. Three of its roots, are each equal 1, and f. _=h- 75 j 
 
 2. 
 
 3. 
 
 See (Art. 201). Jws. a?=l, or 
 4. a; 4_|, r 3_j_22; 2 --fa;+l=0. 
 
 r=2, or , or db^/ 
 
344 ELEMENTS OF ALGEBRA. 
 
 5. 5x<+8x*+9x*+8x+5=Q. 
 
 (Art. 203) reduces this to the following quadratic : 
 
 6. 4* 24* 5 -f-57z 4 73z 3 -f57# 2 24*+4=0. 
 
 Roots, 2 ,^l^, 
 
 7. 4# 4 4-3* 3 8z 2 3^+4=0. 
 
 Ans. #=rhl, or ~ 
 
 o 
 
 8. 
 
 ;=, or 
 
 Some one, or more of the foregoing principles will apply to 
 the solution of the following examples. 
 
 9. a? 4 2s 3 7a a 824-16=0. 
 
 Ans. ar=l, or 4, or ^f 
 
 10. x 4 +2a: 3 3^ 2 _4a:+4=0. 
 
 ^4ws. ar=l, 1, or 2, 2. 
 
 11. x* 2x 3 25a: 2 4-26a:+120=0. 
 
 Ans. #=3 or 5, or 2, or 4. 
 
 12. x* 
 
 3. 
 . #=0, or 1, or 
 
 13. 
 
 . .r=0, or 4, or db^/ 8. 
 
 14. o; 3 +5a: 2 -f3^ 9=0. 
 
 ^Iw5. =!, or 3, or 3. 
 
 15. a^Gz 2 7# 60=0. 
 
 ^4tts. ar=3, 4, or 5. 
 
 16. 2 3 +8.r'-fl7;r-f-10=0. 
 
 Ans. x= 1, or 2, or 5. 
 
 17. x 3 29# 3 -|-198# 360=0. 
 
 Ans. x=3, 6, or 20. 
 
THEORY OF EQUATIONS SUM OF COEFFICIENTS. 345 
 
 27=0. 
 
 Ans. #=-, -J, or 27. 
 
 37=0. 
 
 Ans. x=%, -J, or 37. 
 20. 4x*-{-3x 3 -\-ttx 2 3.r-[-4=0. 
 
 Ans. x=aj+a 2 and2a=(dr N / 247). 
 21. *5 13z 4 +67z 3 171* 2 +216z 108=0. 
 
 (See Art. 198.) Ans. 3, 3, 3, or 2, 2. 
 
 REMARK. This treatise has hitherto omitted one process 
 of elimination, which Bourdon, and some others, have intro- 
 duced, and which, some teachers regard as very scientific and 
 interesting. We cannot quite agree with this class of teachers, 
 for whatever is most simple, and most practical, we regard as 
 most scientific, and this method is neither very simple, nor 
 practical. Still, in a work like this, it is worthy of a passing 
 investigation like the following. 
 
 GENERAL METHOD OF ELIMINATION AMONG EQUATIONS 
 ABOVE THE FIRST DEGREE. 
 
 (Art. 205.) Suppose we have two equations, each contain- 
 ing x and y, like the following : 
 
 x+y-6=0. (1)) Ans x=4, or 2. 
 #3+2/3 72=0. (2)j ' y=2, or 4. 
 
 To make the illustration clear, we give the values of # and 
 y. Now by the theory of equations (Art. 156), if x=4, equa- 
 tion (1) is divisible by (# 4), without a remainder. 
 
 That is, whatever the remainder appears to be, it is in value 
 zero. 
 
 Dividing (x-\-y 6) by (x 4) the quotient is 1, and the 
 apparent remainder is (y 2), but this must be zero ; therefore 
 y=2, corresponding to x=4. 
 
 But x=4, corresponds to both the given equations ; there- 
 fore equation (2) is divisible by (x 4), and (x 4) is- a 
 common measure to the two equations (1) and (2). 
 
316 ELEMENTS OF ALGEBRA. 
 
 Now, therefore, if we take equations (1) and (2), and ope- 
 rate with them for common measure, as taught in (Art. 27), 
 the last remainder must be zero ; and in fact we shall show, 
 that each and every remainder, throughout the whole operation, must 
 be zero. And if we take care that only one letter shall appear 
 in the last remainder that is, operate until one letter only 
 appears in the remainder that remainder can be put equal to 
 zero, and the desired elimination is effected. 
 
 OPERATION". 
 
 (A) 
 
 y 3 72 (B) remains. 
 
 Observe that the divisor is in value zero, and it being mul- 
 tiplied by x 2 , the product (A) must be zero in value ; and that 
 product, subtracted from the dividend (zero), the remainder 
 (B) must be zero. 
 
 Thus we might establish the fact, that each remainder (and 
 in fact each product also), throughout the operation, must be 
 zero. 
 
 Continuing the operation we obtain 
 
 x*y xy 2 -\-6x 2 +6y* 72. 2d remainder. 
 Or, xy(x-\-y)-\-Q(x 2 -\-y 2 ) 72 =2d remainder. 
 But we learn by equation (1) that (x-{-y)=6, therefore 
 we can divide this second remainder by 6, and we have 
 
 Qx 
 
 6x 2xy-\-y 2 12 3d remainder. 
 
 Or, (6 2y)z-f-y 2 12 3d remainder. 
 
 (6 2y )#-]-( 6 2y)(y 6) 
 
 y 2 12 (6 2y)(y 6). 4th remainder. 
 This last remainder does not contain x, therefore that letter 
 is completely eliminated, and the remainder put equal to zero 
 will give the values of y.- 
 
THEORY OF EQUATIONS SUM OF COEFFICIENTS. 347 
 
 N. B. When this method is clearly analyzed, it will be 
 found to be nothing more, than elimination by addition and 
 subtraction. 
 
 Here we have two equations, (1) and (2), each equal to 
 zero, and we multiply one of them, and subtract the product 
 from the other. The remainder is an equation equal to zero, 
 and if its terms have a common divisor, we can divide the 
 equation by it. Hence we have no new principle, nothing to 
 gain by keeping these forms in view. The common method 
 gives us much more freedom of action. 
 
 We give one more example. 
 
 Given, ( f *+ f *)(*- y )-1220=0> d and 
 
 And, x 2 -\-y 2 -^-x y 132=0[ 
 
 Assume, % 2 -}-y 2 =P, and x y=Q 
 Then, P+Q 132=0, 
 
 And, PQ 1220=0 (Q, 
 
 2 -f 1320 1220=0, a quadratic. 
 
 We extract the following from Prof. Perkin's Algebra. 
 Indeed, we have observed it in several other works, and it is 
 given as an appropriate example to illustrate this method of 
 elimination. 
 
 We give it to show the utter inutility of this mode of ope- 
 ration, in a practical point of view. 
 
 Given, a.*+ X y+y*l=0, (1) 
 
 And, .^+y 3 =0, (2) 
 
 to find a single equation in terms of y. 
 
 FIRST OPERATION. 
 1) x*+y*(x y 
 
 x 2 y y 2 x -f- yy 
 
 y=first remainder. 
 
348 ELEMENTS OF ALGEBRA. 
 
 SECOND OPERATION. 
 
 / 2 ' !(* (2y 3 2y) 
 
 _(2t/ 
 
 second remainder. 
 
 Whence, 4y 6 6y 4 -|-3y 2 1=0, is the equation sought. 
 As the sum of these coefficients equal zero, therefore, y=l, 
 and if it were not for this lucky circumstance the values of x 
 and y, in (1) and (2) would still be far away. 
 
 We hare observed, that all remainders in these operations, 
 must be zero ; therefore, our first remainder must be zero, and 
 xy 2y 3 . 
 
 Or, x=(l 2y 2 )y. This value of x put in (2) and we 
 have (l-f2y 2 ) 3 2/ 3 -f?/ 3 =0. Or, (1 22/ 2 ) 3 +l=0. 
 
 Expanding and dividing by 2, produces 
 
 4y6_6 2/ 4_|_3 y 2_ 1 _o. The same as before. 
 But if the object is to solve equations (1) and (2), and find 
 the values of x and y, the common method is incomparably 
 the best. 
 
 s a -H*y+y a = i. (1) 
 
 Equation (2) is divisible by (#-f-y), and the other factor is 
 x 2 xy-\-y 2 . Therefore we have 
 
 x+y=o. (2) 
 
 Or, xtxy+yV^Q, (3) 
 
 Subtracting (3) from (1) and we dbtain 
 2ary=l, or, xy=^. 
 
 This last added to (1), and subtracted from (3) will produce 
 
 And, 
 
 A.lso, x= 1, andy=l. 
 
APPENDIX. 
 
 DESIGNED TO SHOTf THE UTILITY OF ALGEBRA IN PHILOSOPHICAL 
 INVESTIGATIONS. 
 
 INEQUALITY. 
 
 Expressions of inequality sometimes occur in the mathematics, but we 
 can always make equations of them by adding a symbol to the smaller 
 quantity. Then we can reduce the expressions the same as equations. 
 without danger of confusion. 
 
 EXAMPLES. 
 1. Find the limit of the value of x in the inequation 
 
 Let h represent the difference between these two expressions, 
 
 en, 
 
 Whence, 
 
 then, ..... lx 8?=?f+5+A, is a perfect equation 
 o o 
 
 or 
 That is, x is greater than the number 2. 
 
 2. The double of a number diminished by 5 is greater than 25, and triple of 
 the number diminished by 1 is less than the double increased by 13. What 
 numbers will satisfy the conditions? 
 
 Let ar= the number, and h and k the difference of the expressions. 
 
 Then, ...... 2;c 5=25-^. (1) 
 
 3x 7-t-& 2*4-13. (2) 
 
 From .... (1) *=15-H/i. From (2) *=20 k. 
 
 That is, x must be more than 15 and less than 20, consequently, 16, 17, 
 18 and 19 will each answer the conditions. 
 
 3. The sum of two numbers is 32, and if the greater be divided by the less, 
 the quotient will be less than 5, but greater than 2. What are the numbers? 
 
 Let x= the greater, y= the less, then ar-f-y=z32. (1) 
 
 And .... f=5-nft. (2) f=3+Jfc. (3) 
 
 y y 
 
 From (2) x=5yhy. This put in (1) gives 6yhy=32. 
 
 on 
 
 Therefore, y=_. From (3) x=2y+ky; this, put in (1), 
 
 32 
 gives JF=f 
 
 From these last equations we perceive that 32 divided by less than 6 will 
 give the limit of y in one direction, and divided by more than 3 will give 
 its limit in the opposite direction; that is, y must be more than 5 and less 
 than 10 ; consequently x must be more than 22 and less than 27. 
 
 349 
 
350 - APPENDIX. 
 
 ' DIFFERENTIAL METHOD OF SERIES. 
 
 A series purely of the arithmetical order has been fully treated, as also 
 those of the geometrical order; but a series in general exist, when each of 
 the terms is derived from one or more of the preceding terms, according to 
 some definite law. (Art. 141 ) 
 
 Therefore, a variety of series may exist, neither arithmetical nor geo- 
 metrical, and these can be investigated by means of finite differences; and 
 the word differential at the head of these remarks, is not identical with the 
 same word applied to the differential calculus. For example, 
 
 1, 5, 15, 35, 70, 126, 
 
 and so on, is a regular series, and if we wished to find its 10th, 12th, or 
 nth terms, it would be too unscientific and tedious to lind it through the 
 succession of terms ; we must, therefore, investigate for general principles. 
 
 The series is 1, 5, 15, 35, 70, 126, 210, 
 
 1st order of diff. 4, 10, 20, 35, 56, 84, 
 
 2d 6, 10, 15, 21, 28, 
 
 3d " " " 4, 5, 6, 7, 
 
 4th " " " 1, 1, 1, 
 
 5th " " 0, 0, 
 
 Here it will be observed that the first term of the series is 1. The first 
 term of the first order of differences is 4. The first term of the 2d order 
 of differences is 6. The first term of the 3d order of differences is 4. The 
 first term of the 4th order is 1 j and lastly, the first term and all the terms 
 of the 5th order are 0. 
 
 To obtain general principles, however, we must take general notation 
 as follows : (The symbol a 4 is read a sub. 4, and so on for other like 
 symbols.) 
 
 Let ..... a,, a 2 , a 3 , <* 4 , a^ a a , be a series. 
 
 ij, 6 2 , 6 3 , & 4 , 6 a , 1st differences 
 c i C 2 c s' C 4' 2< * differences. 
 
 d lt d 2 , cZ 3 , 3d differences. 
 
 &c. &c. 
 
 By this notation we perceive that d^ would represent the 4th term of 
 the third order of differences, d being the third letter of the alphabet. 
 
 From the foregoing notation, we readily derive the following equations : 
 
 and so on, and so on, and so on. 
 
 By transposition (writing the letters in alphabetical order), we have 
 
 and so on, and so on, and so on* 
 
 In the 2d equation of the first column we will substitute the values of 
 an I 6^, then we shall have fl 3 =ff 1 4~'W>i~h c i t (1) 
 
DIFFERENTIAL SERIES. 351 
 
 By the same process we find 
 
 ^i-H'i-M- (2) 
 
 In the 3d equation of the first column we find 
 
 Substituting the values of a 3 , 6 3 from equations (1) and (2) gives us 
 
 a 4 =a 1 -f36 1 -|-3c 1 -t-rf 1 . (3) 
 
 By the same process, 
 
 6 4= 6 ,-i- 3c rt-3<VK ( 4 ) 
 
 Also by referring back we find 
 
 a 5= a 4-H4- 
 Substituting the values of a 4 and b 4 from (3) and (4), we find 
 
 (5) 
 
 By inspecting equations (1), (3) and (5), we perceive that the coefficients 
 in theSd members correspond to binomial coefficients, and if we should con- 
 tinue the process to any degree, they would still correspond, and the letters 
 represent the first terms of the several orders of differences. 
 
 Therefore, in general, 
 
 And . . 6 n+l =H-nc-f-n . ^~d-\- n . "f . ?=-V}- &c. 
 
 We drop the sub. ones in the second members. 
 
 The series terminates with that order of difference which becomes con 
 etant ; for the succeeding order becomes zero. 
 
 EXAMPLES. 
 
 1. Required the 12th term of the series. 
 
 1, 5, 15, 35, 70, 126, 
 4, 10, 20, 35, 56, 
 6, 10, 15, 21, 
 4, 5, 6.. 
 
 1, 1. 
 Here a=l, 6=4, c==6, rf=4, =1, /=0, n=ll. 
 
 Whence, a 12 =l-{-ll . 4+11 . . 6+11 . {? . L . 4-J-ll . 5 |.f 
 Or, ... a l2 =l-f-44-j-330-f-660-t-330=1365, Ans 
 
 2. Required the 15th term of the series 
 
 1, 4, 10, 20, 35, &c. Ans. 670 
 3 Required the 20th term of the series 
 
 6, 10, 15, 21, Ac. Ans. 253. 
 
 4. Required the nth term of the series 
 
 1, 3, 6, 10, Ac. AntJ 
 
 5. Required the nth term of the series 
 
 2, 6, 12, 20, 30, <fe. Ans. n"-\-n. 
 N. B. To solve the 4th and 5th examples, write (n 1) for n in the 
 
 formula. 
 
352 APPENDIX 
 
 6. Find the 20th term of the series 1, 8, 27, 64, 125, Ac. Ans. 8000. 
 
 7. Required the 50th term of the series 
 
 1, 3, 6, 10, 15, Ac. Ans. 1275. 
 
 Compare examples 4 and 7. 
 
 Our investigation thus far has been limited to finding a particular term 
 of a series. 
 
 We now propose to find the sum of any proposed number of consecutive 
 terms. 
 
 Let . . . Oj, a^, a 3 , a 4 , a., . . . a n , <fec., represent a series. 
 
 With the terms of this series form a new series commencing with 0, 
 for its first term ; adding the succeeding term of the proposed series to the 
 preceding term of the new series, thus, 
 
 0, (0+flj), (0+ ar f-a 2 ), (0+ flr f-a 2 -{-a 3 ), and so on. 
 
 Here it is obvious that the (n-{-l) th term of this new series is the sum 
 of n terms of the proposed series. Therefore, we will find the (/i-|-l/ lh 
 term of the new series by the formula, and that will be n terms of the 
 proposed series as required. 
 
 EXAMPLES. 
 
 1. Find the sum of 10 terms of the series 3, 5, 7, 9, 11, Ac. 
 The new series to be formed is 
 
 0, (O-j-3), (0-f3-l-5), (0-f3-}-5-|-7), (0-j-3-{-5-f 7+9), Ac 
 That is, .... 0, 3, 8, 15, 24, Ac. 
 
 We must now find the llth term of this series, which will be the sum 
 of 10 terms of the proposed series as required. Ans. 120. 
 
 2. Find the sum of 20 terms : also the sum of n terms of the series 
 
 1, 3, 6, 10, 15, Ac. 
 
 Ans. 20 terms =1540, n 
 
 3. Find n terms of the series 2, 6, 12, 20, 30, Ac. 
 
 Ans. 
 
 4. Find the sum of n terms of the series of cubes 
 
 P, 23, 33, 43, & c . Ans. 
 
 5. Find the sum of 10 terms, also the sum of n terms of the series 
 
 1, 4, 9, 16, 25, 36, Ac. 
 N. B. The new series is 0, 1, 5, 14, 30, 55, 91, &c 
 
 Ans, Sum of 10 terms =385. Sum of n terms =L 2re2 + 3ra + 1 ) ra . 
 
 6 
 
 The series of numbers in example 2 are called triangular numbers, be- 
 cause they may be represented by points forming equilateral triangles, thus: 
 
 t * 
 
 36 10 
 
DIFF1 RENTIAL SERIES. 353 
 
 In like ma.iner the square numbers in example 5 can be represented by 
 points, thus: 
 
 In forts and armories shot and shells are piled on triangular, square, or 
 
 oblong bases. 
 
 When the base is a triangle, and the pile complete, the number of balls 
 
 in the pile is expressed by the answer to example 2, in which re represents 
 
 the number of layers in the pile. 
 
 When the base is a square, and the pile complete, the number of shot in 
 
 the pile is expressed by the answer to example 5, in which n represents the 
 
 number of layers in the pile. 
 
 When the pile is an oblong and complete, the upper layer is a single row. 
 
 The next larger has two rows and one more ball in each row. The third 
 
 layer consists of three rows, and each row one more ball than the preceding, 
 Now if the top layer consists of ..... m-j-1 balls, 
 The 2d must consist of ... 2(m+2) or 2ro+4 " 
 The 3d must consist of ... 3(m+3) or 3m+9 " 
 The 4th must consist of . . . 4(m+4) or 4m-4-l6 " 
 The 5th must consist of ... 5(m+5) or 5m+25 " <fcc. 
 
 The number of balls in the whole pile must then consist of the sum of 
 the series (m+1), (2w+4), (3m-J-9), <fec., to n terms n representing the 
 number of layers in any pile. 
 
 We therefore require the sum of this series to n terms. 
 
 The new series to be formed is 
 
 0, (0+m+l), (0+3771+5), (0+6/71+14), <fcc. 
 
 1st diff. (m+1), (2771+4), (3m+9), (4771+16), &c. 
 
 2d diff. (>H-3)> (+5), (wi+7). 
 
 3d diff. 2 2 
 
 4th diff. 
 
 Now by the formula, the (n+l) th term of this series is 
 
 a n+1 =0+n(m+l)+n . ^.(w+^+n . n ~ l . n -^X2. 
 
 The right hand member of this equation is the general formula for an 
 oblong p'le of shot or shells, in which (wi+1) represents the number in the 
 top row, and n the number of layers. 
 
 We have previously found that rciw+I) (M+^) re p resen t s a triangular 
 
 pyramidical pile and, n ( n "f" w +_) represents a square pyramidical pile 
 6 
 
 EXAMPLES. 
 
 1 . How many shot are in a triangular pyramidical pile consisting of 1 8 
 layers? Ans. 1140. 
 
 2. How many shot or shells in a square pile consisting of 15 layers ? 
 
 Ans. 1240. 
 
 3. The top row of an oblong pile of shot or shells consists of 31 balls, and 
 the number of layers is 30. How many balls are there in the whole pile ? 
 
 30 An 
 
354 APPENDIX. 
 
 4. There is an oblong pile of balls consisting of 20 layers, and 644* 
 balls ; how many balls are in its base ? Ans. 740. 
 
 5. A square pile of shells consists of 12 layers, the upper layer has 8 
 shells on a side ; how many shells are in the pile ? Ans. 2330. 
 
 6. The number of balls in a triangular pile is to the number in a square 
 pile (having the same number of layers) as 5 to 9 j required the number 
 in each pile. Ans. 455 and 819. 
 
 Our object is now to express the several orders of differences by the terms 
 of the series, 
 
 a i a , a 3 4 fl 5 6 &C ' 
 
 1st order, (a aj, (a 3 2 ), (a 3 ), (a, a 4 ) &c. 
 
 2d order, (a 3 2^+^), (a 2 3 +a 2 ), (a 2a 4 +a 3 ). 
 
 3d order, ( 4 3a 3 +3a 2 a,), (a^^a.^aa ) 
 
 4th order, (a. 4a^- r -6a 3 ia-j-Cj). 
 
 Here it can be observed that the coefficients correspond to those of the powers 
 of a binomial. 
 
 Also observe that to compose thejirst term of the 2d order of differences, 
 we must use the first three terms of the series. To compose the first term 
 of the 3d order, we must use the first four terms of the series. To com- 
 pose the first term of the 4th order of differences, we must use the first 
 five terms of the series : and in general, to compose the first term of the 
 n th order of differences, we must use the first (n-f-1) terms of the series. 
 Observing these facts, the first term of the 5th order of differences of the 
 preceding general series, must be expressed thus : 
 
 o 6 5a.+lOa 4 10a 3 +5 2 aj. 
 
 If the first term of any order vanishes, or becomes very small, the 
 expression for it may be put equal to zero, and any term of the series com- 
 prised in it, can be found, provided the other terms are given. 
 
 For example, suppose the 4th order of differences in a series becomes 0, 
 then 
 
 Now suppose the 3d term of the series lost or unknown, the others 
 being given, it is found thus : 
 
 __ 
 
 The series 1, 8, a, G4, 125, has lost its third term, its 4th order of dif- 
 
 ference vanishes ; what is that lost term ? 
 
 Ans. 
 
 6 6 
 
 One of the most important applications of this calculus of finite differ- 
 ences is, that of inserting between the terms of a series some new term 
 or terms subject to the same law. This is called 
 
 INTERPOLATION. 
 
 We have already seen that the general equation is 
 
 n 1. , _ n 1 n 2 
 
 T 
 This series will apply to any term beyond a. If we required the 3d 
 
INTERPOLATION. 
 
 355 
 
 term beyond a, we must put n=3. If the 4th beyond a, we put n=4, 
 and so on. 
 
 If part of a term beyond a, we put w= to that fractional part. 
 
 EXAMPLE . 
 
 Given the logarithms of 102, 103, 104, and 105, to find the logarithm 
 of 103.6, which is the 1.6th term beyond a. 
 
 NO. 
 
 LOGARI'MS. 
 
 1ST DIFF. 
 
 2o DIFF. 
 
 3D DIFF. 
 
 102 
 103 
 104 
 105 
 
 2. 0086002 
 2. 0128372 
 2. 0170333 
 2. 0211893 
 
 .0042370 
 .0041961 
 .0041560 
 
 .0000409 
 .0000401 
 
 .0000008 
 
 a=2.0086002, 6= .0042370, c .0000409, d .0000008, n=1.6 
 
 a 2. 0086002 
 
 nb +0. 0067792 
 
 1 
 
 c 0. 0000196 
 
 Ld ....... +0. 00000005 
 
 log. 103.6 2 01535985 
 
 In most cases the 3d difference may be omitted : in this case it only in- 
 fluenced the 8th decimal place. If we were to require the logarithm of 
 102.3, then n=0.3. If 104.7, then n=2.7, and so on. 
 
 Interpolation is much used in astronomy, as the following example will show: 
 
 At noon (Greenwich time), May 10th, 1846, the moon's declination was 
 3 27' 17.5" north; at midnight following it was 5 46' 38" north; on 
 the llth, at noon, 7 38' 57" north, and the following midnight 10 2' 
 31.7" north. What was the declination on the 10th of May at 3h P. M? 
 
 Here a=3 27' 17.5", 6=2 19' 20.5", c= T 1.5", d 1' 43.2", and 
 the interval of 12 hours must be considered as the unit of time : 3 hours 
 is, therefore, J of this unit. Hence, n=0.25, and the required result must 
 be the sum of the following series : 
 
 The 1st term a =3 27' 17.5". 
 2d termn6 34' 50.1" 
 3d term 39.5". 
 
 4th term 5.6".' 
 
 Sum~42' 41.5"T Ans. 
 To show tne utility and application of this formula, we give the following 
 
 EXAMPLE S. 
 
 Given the logarithmic sine of 5, 5 12', 5 24', 5 36', 5 48' and 6, ft 
 find the sine of any or all intermediate arcs. 
 
 a. 
 
 Sine 5 0'=8. 9402360 
 Sine 50 12'=8. 9572843 
 Sine524'=8 9736280 
 Sine5036'=8 9893737 
 Sine548'=9 0045634 
 Sine GO 0'=9 0192346 
 
 +6. 
 1st diff. 
 .0169883 
 .0163437 
 .0157457 
 .0151897 
 .0146712 
 
 2d diff. 
 
 6446 
 
 5980 
 5560 
 5185 
 
 -M. 
 
 3d diff. 
 
 466 
 420 
 375 
 
 4th diff 
 
 46 
 45 
 
356 
 
 APPENDIX. 
 
 In applying the formula to this example, 12' must be taken as the unit 
 of space between the terms. Then if we require the sine of 5 3', we 
 must put n=y 3 .2=2, and the formula gives 
 
 Sine53'=8.9402960- r -2(.Ol69883)-H . tz? 
 
 4G6=8.9402960+.0042471+604.3-f 25.6=8.9446061. 
 Sine54'=8.9402960+i(.0169883)-H . tz? 
 
 456=8.9402960+.0056628+716+29=8.9460333. 
 For the sine 5 13', we put n= T 1 2 and 
 fin. 50 13'=8.9572843+ T 1 3 (.0163437)+ T 1 2 . j 
 
 jti(420)=8.9572843+.0013619+S29+13=8.9586704. 
 
 O 
 
 To find the log. sine of 5 14' 13", we could put n=?I?=!??. ' 
 
 12 720 
 
 The work of finding sines, cosines, and tangents, &c., is already done; 
 and, therefore, in respect to this application, the formula for interpolation 
 has comparatively lost its importance ; but with the practical astronomer, 
 this formula is of the greatest importance, and must forever be in use. 
 
 We give the following 
 
 EXAMPLE. 
 
 For the English Nautical Almanac the moon's declination is required 
 for every hour of Greenwich time; but to compute it directly from the lunar 
 tables and trigonometry, would present an apalling amount of labor, and to 
 surmount this difficulty, the declination is computed from the lunar tables 
 for each noon and midnight of Greenwich time ; and the declination in- 
 serted for the intermediate hours by interpolation. 
 
 Opening the English Nautical Almanac for 1854, at page 91, I find the 
 moon's declination as follows : 
 
 3) dec. N. +6 
 
 May 23, mean moon, 6 13' 14.4" 
 
 May 23, midnight, 8 55' 6.4" 2 41' 52.0" 
 
 May 24, noon, 11 30' 26.2" 2 35' 19.8" 
 
 May 24, midnight, 13 57' 44.4" 2 27' 16.2" 
 
 May 25, noon, 16 15' 33.6" 2 17' 49.2" 
 
 6' 32.2" 
 8' 3.6" 
 9' 27.0" 
 
 1' 31.4" 
 1' 23.4' 
 
 From the above data we require the moon's declination at the commence- 
 ment of every hour between noon and noon of May 23d and 24th. 
 For the first hour, or 1, afternoon, we must put n=J . 
 
 The D's dec. at 1 P. M. =6 13' 14.4''+^ (2 41' 52.0")+T2 . itzl X 
 
 14.9" 2.2"r=6 26' 56.4" North. 
 
SPECIFIC GRAVITY. 
 
 357 
 
 The following are the results taken from the Nautical Almanac which 
 serve as answers to twenty-two examples in the application oi this formu-la 
 
 Hours. 
 
 f)Dec. 
 
 Hours. 
 
 Dec. 
 
 1 
 
 6 C 
 
 Ofi' 
 
 565" 
 
 13 
 
 9 
 
 8' 19.3" 
 
 2 
 
 6 
 
 40 
 
 365 
 
 14 
 
 9 
 
 21 29.4 
 
 3 
 
 6 
 
 
 143 
 
 15 
 
 9 
 
 34 36.6 
 
 4 
 
 7 
 
 7 
 
 498 
 
 16 
 
 9 
 
 47 41.0 
 
 5 
 
 7 
 
 91 
 
 23.0 
 
 17 
 
 10 
 
 42.5 
 
 6 
 
 7 
 
 'M 
 
 53.9 
 
 18 
 
 10 
 
 13 40 9 
 
 7 
 
 7 
 
 48 
 
 22.3 
 
 19 
 
 10 
 
 26 36.4 
 
 8 . . . 
 
 8 
 
 1 
 
 48.3 
 
 20 
 
 10 
 
 39 28.7 
 
 9 . 
 
 8 
 
 15 
 
 11.7 
 
 21 
 
 10 
 
 52 17.9 
 
 10 
 
 8 
 
 
 32.6 
 
 22 
 
 11 
 
 5 3.9 
 
 11 
 
 8 
 
 -11 
 
 50.8 
 
 23 
 
 11 
 
 17 46.7 
 
 12 
 
 8 
 
 rr 
 
 64 
 
 24 
 
 .. 11 
 
 30 26.2 
 
 
 
 
 
 
 
 
 SPECIFIC GEAVITY. 
 
 Gravity is weight. Specific gravity is the specified weight of one body, 
 compared with the specified weight of another body (of the same bulk), taken 
 as a standard. 
 
 Pure water, at the common temperature of 60 Fahrenheit, is the standard 
 for solids and liquids ; common air is the standard for gases. 
 
 Water will buoy up its own weight. If a body is lighter than water, it 
 will float ; if heavier than water, it will sink in water. 
 
 If a body weighs 16 pounds, in air, and when suspended in water weighs 
 only 14 pounds, it is clear that its bulk of water weighs 2 pounds ; and the 
 body is 8 times heavier than water ; therefore the specific gravity of this body 
 is 8, water being 1. 
 
 If the specific gravity of a body is n, it means that it is n times heavier 
 than its bulk of water. Therefore 
 
 If we divide the weight of any body by its specific gravity, the quotient 
 will be, the weight of its bulk of water. 
 
 On this fact alone we may resolve all questions pertaining to specific gravity 
 
 EXAMPLES. 
 
 1. Two bodies, whose weights were A and B, and specific gravities a and b, 
 were put together in such proportions as to make the specific gravity of the 
 compound mass c. What proportions of A and B were taken? 
 
 A quantity of water, equal in bulk to A, must weigh 
 
 A quantity " 
 
 A+B 
 c 
 
 A quantity of water, equal in bulk to (A-^-B), will weigh 
 
 A B A4-B 
 Therefore, - ) = ! ; Or, bcA-\-acB=abA-\-abB ; 
 
 Or, b(o a)A=a(b c}B. 
 
3-5?; AITENDIX. 
 
 Hence the quantities of each must be reciprocal to these coefficients; or if we 
 
 a /b c\ 
 take one, or unity of B, we must take j ( - J units of A. 
 
 U \,C"~~flyX 
 
 2. Hiero, king ef Sicily, sent gold to his jeweler to make him a crown ; he 
 afterwards suspected that the jeweler had retained a portion of the gold, and 
 substituted the same weight of silver, and he employed Archiwedes to ascer- 
 tain the fact, who, after due reflection, hit upon the expedient of specific gravity 
 
 He found, by accurately weighing the bodies both in and out of water, that 
 the specific gravity of gold was 19, of silver 10.5, and of the crown 16.5. 
 Fromthesedata he found what portion of the king's gold was purloined. Re- 
 peat the process. 
 
 The preceding problem is the abstract of this, in which A may represent the 
 weight of the gold in the crown, B the weight of the silver, and (./i-J-fi) the 
 weight of the crown ; a=19, 6=10$, c=16. 
 
 Then if we take J5=l, one pound, one ounce, or any unity of weight, of 
 
 silver, the comparative weight of the gold will be expressed by -j( - J. 
 
 That is, for every ounce of silver in the crown, there were 4 ^ 2 j ounces of 
 gold. If clearer to the pupil, he may resolve this problem as an original one, 
 without substituting from the abstract problem. 
 
 3. I wish to obtain the specific gravity of a piece of wood that weighs 10 
 pounds ; and as it will float on water, I attach 21 pounds of copper to it, of a 
 specific gravity of 9. The whole mass, 31 pounds, when weighed in water, 
 weighs only 4 pounds ; hence 27 pounds of the 31 were buoyed up by the 
 water ; or we may say, the same bulk of water weighed 27 pounds. Require-1 
 the specific gravity of the wood. 
 
 Let s represent the specific gravity of the wood. 
 
 10 
 Then = the weight of the same bulk of water. 
 
 21 
 
 And = the weight of water of the same bulk as the copper. 
 y 
 
 10 7 30 
 
 Hence, ........ -{--=27, Or, s==^=.405, nearly , 
 
 4. Granite rock has a specific gravity of 3. A piece that weighs 30 ounces, 
 being weighed in a fluid, was found to weigh only 21.5 ounces. What waa 
 the specific gravity of that fluid ? 
 
 The weight of the fluid, of the same bulk as the piece of granite, was evi- 
 dently 8.5 ounces. Let a represent its specific gravity. 
 
 8.5 80 
 
 Then = the weight of the same bulk of water; also - Q =10= tlio 
 
 3 o 
 
 weight of the same bulk of water. 
 
 8.5 
 Hence =10, Or, s=.85 Ans. t which indicates impure alcohol 
 
MAXIMA AND MINIMA. 
 
 5. The specific gravity of pure alcohol is .797 ; a quantity is offered of the 
 specific gravity of .85, what Proportion of water does it contain? 
 
 Let A= the pure alcohol, and W = the water. 
 
 Then 
 
 Ans. The resolution of this equation shows 1 portion of water to 2.255-j- 
 portions of alcohol. 
 
 6. There is a block of marble, in the walls of Balbeck, 63 feet long, 12 wide, 
 and 12 high. What is the weight of it in tons, the specific gravity of marble 
 being 2.7 and a cubic foot of water 62^ pounds, Ans. 683 y\ tons. 
 
 7. The specific gravity of dry oak is 0.925 ; what, then, is the weight of a 
 dry oak log, 20 feet in length, 3 feet broad, and 2 feet deep ? Ans. 867 if Ibs. 
 
 We may now change the subject, to make a little examination into maxima 
 and minima. For this purpose, let us examine Problem 2 (Art. 114). 
 
 1. Divide 20 into two such parts that their product shall be 140. It may 
 be impossible to fulfil this requisition, therefore we will change it as follows : 
 
 Divide 20 into two such parts, that their product will be the greatest possible. 
 
 Let x-\-y one part, and x y= the other part. 
 
 Then 2.r=20, and #=10, and the product, x 2 y*, is evidently the 
 greatest possible when y=Q. Hence the two parts are equal, and the greatest 
 product is 100, or the square of one half the given number. 
 
 2. Given the base, m, and the perpendicular, n, of a plane triangle, to find 
 the greatest possible rectangle that can be inscribed in the triangle.* 
 
 Let ABC be the triangle, BC=m, A 
 
 AF=n, AD=x, and AE=x', De be a 
 very small distance, so that x' is but insen- 
 sibly greater than x. 
 
 As Z), comparatively, is not far from the 
 vertex, it is visible, that the rectangle 
 a'b'c'd' is greater than the rectangle abed. 
 
 If we conceive the upper side of the 
 rectangle to pass through D 1 , in place of 
 D, and we represent AD' by x, and At 
 by or', it is visible that the rectangle 
 e'fg'h' is less than the rectangle efg h. 
 
 If we subtract the rectangle abed from the rectangle a'b'c'd', we shall have 
 a positive remainder. 
 
 If we subtract the rectangle efg h from the rectangle t'f'g'h', we shall have 
 a negative remainder. 
 
 * We do not introduce this problem to show its solution ; it bt.ongs to the calculus, 
 and, in its place, is extremely simple ; we introduce it to show a p-inciple of reasoning 
 extensively used in the higher mathematics, and, perchance our illustration may uj] 
 a pupil in his progress in the calculus. 
 
 A A 
 
360 APPENDIX. 
 
 The rectangle abed cannot be the greatest possible, so long as we can have 
 a positive remainder by subtracting it from the next consecutive rectangle 
 immediately below. 
 
 After we pass the point on the line AF where the greatest possible rectangle 
 comes in, the next consecutive rectangle immediately below, will become less ; 
 and by subtracting the upper from it, the difference will be negative. 
 
 Hence, when abed becomes the greatest possible rectangle, the difference 
 between it and its next consecutive rectangle can be neither plus nor minus 
 but must be zero. 
 
 Therefore, it is manifest, that if we obtain two algebraical expressions for tho 
 two rectangles abed and a'b'c'd', and put their difference equal to 0, a resolu 
 lion of the equation will point out the position and magnitude of the maximum 
 rectangle required. 
 
 Put the line ab=y, and a'b'=y'. As AD=x and AE=x l , DF=n x 
 Wid EF=n x f . The rectangle abcd=y(n x), and a'b'dd'=y'(n a:/). 
 
 From the consideration just given, the maximum must give 
 
 y'(n -a;') y(n #)=0. 
 
 mx 
 
 By proportional triangles, we have x : y : : n : m or. y =- 
 
 n 
 
 By a like proportion, we have y=x'. 
 
 Put these values of y and yi in the above equation, and, dividing by , 
 
 we have x'(n x'}=x(n x} ; 
 
 Or, x 2 x'* =n(xx'). 
 
 By division, x -{- x' =n 
 
 As x' is but insensibly greater than x, 2x=n ; which shows that AD is one 
 half AF, and the greatest rectangle must have just half the altitude of the triangle. 
 
 3. Required the greatest possible cylinder that can be cut from a right cone. 
 
 Conceive the triangle (of Prob. 2.) to show the vertical plane cut through 
 the vertex of the cone, and ab=y the diameter of the required cylinder. Then, 
 the end of the cylinder is .7854?/ 2 , and its solidity is .7854# 2 (n *). The next 
 consecutive cylinder is .7854i/'2(?i x"). Hence y' 2 (n a:')=;y 2 (n ;r). 
 
 By similar triangles x : y : : n : m, Or, y 2 = ^ and y' 2 = -z' 9 . 
 
 Hence, *' 2 (n a:')=r 2 (n *), 3r, x 3 *'3=n(z 2 z' 2 ) ; 
 Divide both members by (z z'), and a -f zz4-z' 2 ==n(z-|-z'). 
 As x=x' infinitely near, 3z 2 =2wz, or, x=|n; which shows tnat the 
 altitude of the maximum cylinder is 5 the altitude of the cone. 
 
 In this way all problems pertaining to maxima and minima can be resolved ; bui 
 the notation and language of the calculus, in all its bearings, is preferable to this. We 
 had but a single object in view that of showing the power of algebra. 
 
 (r , 
 

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XC 49563 
 
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