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TREATISE 
 
 ON 
 
 ALGEBRA, 
 
 roR THE USX or 
 
 SCHOOLS AND COLLEGES, 
 
 ^yy-y : 
 
 WILLIAM SMYTH, A.M. ''^ ^ 
 
 PEOFESSOE OF MATHEMATICS IN BOWDOIN COLLEGE. 
 
 BOSTON: 
 
 SANBORN, CARTER & BAZIN, 
 
 PORTLAND: 
 
 BLAKE & CARTER. 
 
 1855. 
 
 A.^^ 
 
Entered according to Act of Congress, in the year 1862, 
 
 BY WILLIAM SMYTH, A.M. 
 
 In th« Clerk's Office of the Diatrict Court of the District of Maine. 
 
 Stereotyped by 
 
 ROBART k nOBBINS, 
 KNGLAUD TTPR AND STEREOTYrB FOCNDBBT, 
 BOSTON. 
 
Qk 
 
 ^3 
 
 Sicl 
 
 PREFACE 
 
 The present Treatise is composed substantially of the Elements of Alge' 
 bra of tlic author, -with such additional matter as fully to adapt it to the 
 advanced course of mathematics now generally pursued in the American 
 Colleges. In its preparation the object has been to give a clear view of 
 the nature and powers of Algebra. The analytic method is uniformly 
 pursued, and the topics are so presented as, in general, to lead the 
 student to feel the want of a new principle before proceeding to its investi- 
 gation. Thus, the work commences with the exposition of Algebra, as a 
 concise language adapted to facilitate the processes of reasoning required 
 in mathematical investigations. The operations of Algebra, therefore, with 
 which most treatises begin, are not introduced until, in the use of the 
 algebraic language in the solution of questions, the manner in which these 
 operations arise, and the reason for them, are seen. The same general 
 plan is pursued throughout. Much attention is paid to the Discussion of 
 Problems and Equations, a topic of the highest importance to the clear 
 understanding of the true nature of Algebra. A section is given on the 
 Indeterminate Analysis, a subject not usually introduced into our text 
 books, but of great value in itself and in its relation to Analytic Geometry. 
 A full view is given of the General Theory of Equations, and of the method 
 of solving Numerical Equations of any degree. The several subjects are 
 presented in the manner found by experience best adapted to the conven- 
 ience of recitations and the progress of the pupil. All needed help, it is 
 believed, is furnished, without that difFuseness of explanation which leaves 
 to the learner no room for the exercise of his own powers. The difficulties 
 to be encountered are such only as pertain of necessity to the subject, and 
 which serve to furnish a healthy stimulus to exertion and the njental 
 discipline necessary to the successful prosecution of more advanced studies. 
 
 The work in its present form is still better adapted, it is hoped, to the 
 use of Academies and High Schools, in which it has heretofore been exten- 
 sively used. For a younger class of pupils, the Elementaiy Algebra of 
 the author will be found sufficiently simple, and an easy introduction to 
 the present work. Wm Smyth 
 
 Bowdoin College, 1853 
 
TABLE OF CONTENTS. 
 
 BKCTION PAGB 
 
 1 . Explanation of Algebraic Signs 1 — 9 
 
 2. Equations 9—22 
 
 3. Algebraic Operations. Addition — Subtraction — Multi- 
 
 plication and Division of Algebraic Quantities . . . 22 — 44 
 
 4. Algebraic Fractions. Greatest Common Divisor .... 44 — 55 
 
 5. Equations of the First Degree. Elimination 55 — C9 
 
 6. Negative Quantities C9 — 7G 
 
 7. Indeterminate Analysis 7C — 87 
 
 8. Solution of Questions in a general manner 87 — 100 
 
 9. Discussion of Problems and Equations of the First Degree 100 — 111 
 
 10. Theory of Inequalities Ill — 115 
 
 11. Extraction of the Square Root 115 — 135 
 
 12. Equations of the Second Degree 135 — 149 
 
 13. Discussion of Equations and Problems of the Second Degree 149 — 1G5 
 
 14. Maxima and Minima 165 — 1C8 
 
 15. Powers and PkOots of Monomials 1G8 — 171 
 
 IG. Powers of Compound Quantities. Theory of Combinations. 
 
 Binomial Theorem 171—184 
 
 17. Roots of Compound Quantities 184—191 
 
 18. Calculus of Radical Expressions 191 — 197 
 
 19. Theory of Exponents 197—203 
 
 20. Proportions 203—210 
 
 21. Progressions 210—226 
 
 22. Theory of Continued Fractions 226—235 
 
 23. Exponential Equations. Logarithms 235 — 248 
 
 24. Application of Logarithms. Compound Interest. Annui- 
 
 ties 248—257 
 
 25. Praxis. Equations of the First and Second Degree . . . 257 — 264 
 
 26. General Theory of Equations. Detached Coefficients. 
 
 Synthetic Division. Divisibility of Equations. Num- 
 ber of the Roots. Coefficients. Form of the Roots. 
 Signs of the Roots. Limits of the Roots. Limiting 
 Equation. Equal Roots. Imaginary and Real Roots. 
 
 Sturm's Theorem 265 -309 
 
 27. Numerical Equations of any Degree. Commensurable 
 
 Roots. Incommensurable Roots. Homer's Method of 
 
 Approximation 309 -323 
 
 28. Elimination by means of the Greatest Common Divisor . 323- -326 
 
 29. Infinite Scrie?. Undetermined Coefficients. The Differ- 
 
 ential Method. Interpolation. Summation of Series 32G— 330 
 
ELEMENTS OF ALGEBRA. 
 
 SECTION I. — Explanation of Algebraic Signs. 
 
 1. Let it be proposed to divide the number 56 into two such 
 parts, that the greater may exceed the less by 12. 
 
 To resolve this question, we remark that, 
 
 1°. The greater part is equal to the less added to 12. 
 
 2°. The greater party added to the less part, is equal to 5Q. 
 
 It follows, therefore, that, 
 
 3°. The less part, added to 12, added also to the less part, is 
 iqual to 5Q. 
 
 But this language may be abridged, thus, 
 
 4°. Tioice the less part, added to 12, is equal to 56 ; whence, 
 
 5°. Twice the less part is equal to 56 diminished by 12. 
 
 Subtracting, therefore, 12 from 56, we have 
 
 6°. Twice the less part equal to 44 ; wherefore 
 
 7°. Once the less part is equal to 44 divided by 2, or perform- 
 ing the division, we have 
 
 8°. Once the less part equal to 22. 
 
 Adding 12 to 22 we have 34 for the greater part. The parts 
 required, therefore, ^re 22 and 34. 
 
 2. In the process of reasoning required in the solution of the 
 proposed question expressions, such as " added to," " diminished 
 by," ^^ eqical to," &c. are often repeated. These expressions 
 
6 ELE'tfSNTS OF ALGEBRA. 
 
 refer to the operations, by which the numbers given in the ques- 
 tion are connected among themselves, or to the relations which 
 they bear to each other. The reasoning, therefore, which per- 
 tams to the solution of the proposed, it is evident, may be rendered 
 much more concise, by representing each of these expressions by 
 a convenient sign. 
 
 It is agreed among mathematicians to represent the expression 
 " added to'^ by the sign -j-, read plus, the expression " diminished 
 }nf^ by the sign — , read minus, the expression " multiplied 3y" 
 by the sign X> that of " divided ly" by the sign -f-. Lastly, the 
 expression " equal to" is represented by the sign =. 
 
 3. By means of the above signs, the reasoning in the question 
 proposed may be much abridged ; still, however, we have frequent 
 occasion to repeat the expression " the less part." The reasoning, 
 therefore, may be still more abridged by representing this also 
 by a sign. 
 
 The less part is the unknown quantity sought directly by the 
 * reasoning pursued. It is agreed in general to represent the 
 unknown quantity or quantities sought in a question by some 
 one of the last letters of the alphabet, as, x, y, z. 
 
 4. Let us now resume the question proposed, and employ in 
 its solution the signs, which have been explained. 
 
 Let us represent by x the less of the two parts required, we 
 have then 
 
 a: -f- 12 = the greater part, 
 X + 12 + x==5Q 
 2 X ^ + 12 = 56 
 
 2 X a: = 56 — 12 
 2 X a: = 44 
 
 x = U-^2 
 .^^ X = 22. 
 
 The' multiplication of a; by 2 may be expressed more con- 
 cisely thus, 2.x, or still more concisely thus, 2x. Division 
 also is more commonly indicated by writing the number to be 
 diyided above a horizontal line, and the divisor beneath it in 
 
EQUATIONS OF THE FIRST DEGREE. 7 
 
 the form of a fraction ; 14 divided by 2, for example, is indicated 
 thus,-^-. 
 
 5. The question, which we have solved, is simple; it is 
 sufficient, however, to show the aid which may be derived 
 from convenient signs in facilitating the reasonings, that per- 
 tain to the solution of a question. Indeed in abstruse and 
 complicated questions, it would often be difficult, and sometimes 
 absolutely impossible to conduct, without such aid, the reasonings 
 required. 
 
 6. The signs which have been explained, ' together with 
 those which will hereafter be introduced, are called Algebraic 
 signs. It is from the use of these that the science of Algebra is 
 derived. 
 
 Let us now employ the signs already explained in the solution 
 of some questions. 
 
 1. Three men, A, B, and C trade in company and gain $405, 
 of which B has twice as much as A, and C three times as much 
 as B. Required the share of each. 
 
 Let X represent the share of A, then 2a; will represent the 
 
 share of B and 6a; the share of C. Then, since the shares added 
 
 together should be equal to the sum gained, we have 
 
 a;_|-2a; + 6a; = 405 
 
 9a; = 405 
 
 405 ._ 
 a;^_ = 45. 
 
 Thus we have A's share = $45; whence B's share is ^90 
 and C's $270. 
 
 2. A fortress is garrisoned by 2600 men ; and there are nine 
 times as many infantry, and three times as many artillery as 
 cavalry. How many are there of each ? 
 
 3. From two towns, which are 187 miles distant, two travel- 
 lers set out at the same time, with an intention of meeting. One 
 of them goes 8 miles, and the other 9 miles a day. In how 
 many days will they meet ^ 
 
m- 
 
 8 ELEMENTS OF ALGEBRA. 
 
 4. A gentleman meeting four poor persons distributed 5 shil- 
 lings among them; to the second lie gave twice, to the third 
 thrice, and to the fourth four times as much as to the first. What 
 did he give to each ? 
 
 5. Four persons, A, B, C and D made a joint stock; B puts' 
 in twice as much as A, C puts in three times as much as B, and 
 D puts in as much as the other three together. The whole stock 
 is $20,000. How much did each put in ? 
 
 6. To divide the number 230 into three such parts, that the 
 excess of the mean above the least may be 40, and the excess of 
 the greatest above the mean may be 60. 
 
 Let X represent the least part, then rr -[- 40 will be the mean, 
 and a; -j- 40 -|- 60 will be the greatest part ; we have therefore 
 2; -I- a: + 40 + a; + 40 + 60 = 230 
 8a; +140 = 230 
 3a: =90 
 a: = 30. 
 
 The parts will then be 30, 70 and 130 respectively. 
 
 7. A draper bought three pieces of cloth which together mea- 
 sured 159 yards. The second piece was 15 yds. longer than the 
 first, and the third 24 yds. longer than the second. What was 
 the length of each ? 
 
 8. Three men. A, B and C made a joint stock ; A puts in a 
 certain sum, B puts in SI 15 more than A, and C puts in $235 
 more that B ; the whole stock was $1753. What did each man 
 put in ? 
 
 9. A gentleman buys 4 horses, for the second of which he 
 gives £12 more than for the first, for the third £6 more than for 
 the second, and for the fourth £2 more than for the third. The 
 sum paid for all was £230. How much did each cost ? 
 
 10. A man leaves by will his property, amounting to $14000, 
 to his wife, two sons and three daughters ; each son is to receive 
 twice as much as a daughter, and the wife as much as all the 
 children together. What will each receive ? 
 
, EQUATIONS OF THE FIRST DEGREE. 9 
 
 11. An express sets out to travel 240 miles in 4 days, but in 
 consequence of the badness of the roads, he found he must go 5 
 miles the second day, 9 the third and 14 the fourth day less than 
 the first. How many miles must he travel each day ? 
 
 12; The sum of 8300 was divided among 4 persons; the 
 second received three times as much as the first, the third as 
 much as the first and second, and the fourth as much as the 
 second and third. What did each receive ? 
 
 13. A silversmith has 3 pieces of metal. The second weighs 
 6 oz. more than twice the first, and the third 9 oz. more than 
 three times the second. The weight of the whole being 52 oz., 
 what is the weight of each ? 
 
 SECTION II.— Equation^. 
 
 7. The difference between two numbers is 25 and the greater 
 is 4 times the less ; required the numbers. 
 
 Let X represent the less, then a:-f-25 will represent the 
 
 gTeater ; but since by the question the greater is four times the 
 
 less, 42; will also represent the greater; these two expressions 
 
 for the same thing will therefore be equal to each other, and we 
 
 have 
 
 a:4-25 = 4a:. 
 
 An expression for the equality of two things is called an 
 equation. The. two equal quantities, of which an equation is 
 composed, are called members of the equation ; the one on the 
 left of the sign of equality is called the first member and the 
 other the second. 
 
 If a member consists of parts separated by the signs -[- and — , 
 these parts are called terms. 
 
 Thus in the equation a; -|- 25 = 4 a:, the expression a: -|- 25 is 
 the first member and 4 a: the second. 
 
 The quantities x and 25 are the terms of the first member. 
 
10 ELEMENTS OF ALGEBRA. 
 
 A figure written before a letter, showing how many times the 
 letter is to be taken, is called the coefficient of that letter. In 
 the quantities 4 a:, 7 a;, 4 and 7 are the coefficients of x. 
 
 Equations are distinguished into different degrees. An equa* 
 tion, in which the unknown quantity is neither multiplied by 
 itself, nor by any other unknown quantity, is called an equation 
 of the first degree. 
 
 8. In the solution of a question by the aid of algebraic signs 
 there are, it is evident from the examples already performed, two 
 distinct parts. In the first, we form an equation by means of the 
 relations established by the nature of the question between the 
 known and unknown quantities. This is called putting the 
 question into an equation. 
 
 In the second part, from the equation, thus formed, we deduce 
 a series of other equations, the last of which gives the value 
 of the unknown quantity. This is called resolving or reducing 
 the equation. 
 
 9. No general and exact rule can be given for putting a ques- 
 tion into an equation When however the equation of a question 
 is formed, there are regular steps for its reduction, which we 
 shall now explain. 
 
 Since the two members of an equation are equal quantities, 
 it is evident, that, 1°. the same quantity may be added to both 
 sides of an equation ivithout destroying the equality ; 2°. the 
 same quantity may be subtracted from both sides of an equation 
 without destroying the equality; 3°. both sides of an equation 
 fnay be multiplied, or 4°. both sides may be divided by the same 
 ^quantity without destroying the equality. 
 
 10. Let it be proposed to resolve the equation derived trom 
 the following enunciation, viz. To find a number such that if 
 one half and one third of this number be added to itself the sum 
 will be equal to 30. 
 
 Let X represent the number, then one half of this number wiD 
 
EQUATIONS OF THE FIRST DEGREE. 11 
 
 1 X 1 X 
 
 be represented by ^ a; or - and one third by ^ a; or ^r, and we 
 have 
 
 .+1+5=30. 
 
 To resolve this equation we must free the fractional terms from 
 their denominators. In order to this we multiply both sides of 
 the equation first by 2, which gives 
 
 2a; + a: + ^| = 60; 
 
 multiplying next by three, we have 
 
 6a; + 3a; + 2a: =180, 
 an equation free from denominators. To free an equation there- 
 fore from denominators, multiply the equation by the denomina' 
 tors successively. 
 
 Ex. 1. Free from denominators the equation 
 
 - -4- - — - — 49 
 
 Ex. 2. Free from denominators the equation 
 
 3^7 12^11"" 
 Since in this equation the denominator 12 is a multiple of 3, 
 multiplying by 12, we have 
 
 4. + ~-x + ^=156. 
 
 Thus by multiplying first by 12, the number of multiplications 
 necessary to free the equation from denominators is diminished, 
 and the equation itself, when freed from denominators, is left m 
 * more simple state. 
 
 Ex. 3. Free from denominators the equation 
 
 -4-^ — — — — — 120 
 7^9 21 18~" • 
 
 Ex. 4. Free from denominators the equation 
 
 ± zj-Jl f_L in 
 
 6 4"^ 12 3'^2~^"' 
 
12 ELEMENTS OF ALGEBRA. 
 
 Ex. 5. Free from denominators the equation 
 
 The least number divisible by each one of the denominators 
 of the proposed, it is easy to see, is 20. Multiplying by 20, we 
 have 
 
 10a: + 2a: + 15a; — 4a:+ 120 = 180; 
 thus the proposed is freed at once from denominators, and the 
 equation which results, it is evident, is the most simple to which 
 it can be reduced free from denominators. 
 
 From what has been done, we have the following rule to free 
 an equation from denominators, viz. Find the least common 
 multiple of the denominators ; multiply each term hy this com- 
 mon multiple, observing to divide, as we proceed, the numerator 
 of each fractional term hy its denominator. 
 
 11. Let it be proposed to resolve the equation 
 
 3a: + 25 = 60 — 4a;. 
 To resolve this equation, it will be necessary to transfer tho 
 terms 25 and 4a; from the members, in which they now stand, to 
 the opposite. In order to this, let us first subtract 25 from both 
 members, we then have 
 
 3:^+25 — 25 = 60 — 4a; — 25. 
 or 3a; = 60 — 4a; — 25. 
 
 Adding next 4 a; to both sides of this last, we have 
 3z + 4a; = 60 + 4a; — 4a; — 25. 
 or 3a; + 4a; = 60 — 25. 
 
 Comparing the last equation with the proposed, the term 25 
 which is additive in the first member has, it is evident, passed 
 into the second member with the sign of subtraction, and the 
 term 4 a; which was subtractive in the second member has passed 
 into the first with the sign of addition. Whence the following 
 rule, for transposing a term from one member of an equation to 
 the other, will be readily inferred, viz. Efface the term in the 
 member in ivhich it stands, and write it in the other with the 
 contrary sign. 
 
EQUATIONS OF THE FIRST DEGREE. 13 
 
 12. Let it be proposed next to resolve the equation 
 
 5x 4x y. 7 13a; 
 
 IS — y"" 8 6"* 
 
 Freeing from denominators, we have 
 
 lOo; — S2x — 312 = 21 — 52a; ; 
 
 transposing and reducing, we have 
 30 a; = 333; 
 
 whence dividing both sides by 30 we obtain 
 a;=llyV- 
 
 The unknown quantity in equations of the first degree can be 
 combined with those which are known in four different ways 
 only, viz. by addition, subtraction, multiplication, and division. 
 From what has been done, we have therefore the following rule 
 for the resolution of equations of the first degree with one un- 
 known quantity, viz. 1°. Free the proposed equation from de- 
 nominators ; 2°. bring all the terms, which contain the unknown 
 quantity into the first member and all the knoion quantities into 
 the other ; 3°. unite in one term the terms which contain the un- 
 known quantity, and the knoion quantities in another ; 4°. divide 
 both sides by the coefficient of the unknown quantity. 
 
 13. Applying ^he above rule to the equation 
 
 ^_.| + 10 = |-|+ll,weobtaina;=12. 
 
 In order to verify this result we substitute 12 for z in the pro- 
 posed, it then becomes 
 
 lH_l?4-io-lH_l?+ii- 
 
 6 4+^"— 3 2+^^' 
 whence performing the operations indicated we obtain 
 
 9 = 9. 
 The value a; = 12 satisfies therefore the proposed equation. 
 
 In general, to verify the value of the unknown quantity de- 
 duced from an equation, we substitute this value for the unknown 
 quantity in the equation. If this renders the two members iden- 
 tically the same, the answer is correct. 
 
 B 
 
14 ELEMENTS OF ALGEBRA. 
 
 14. The following examples will serve as an exercise for the 
 learner in the reduction of equations. 
 
 1. 
 2. 
 3. 
 4. 
 
 2 + 3 + 5 — ^^- 
 
 Ans. a:=:30. 
 
 ^-'=5+3 
 
 
 a; =15. 
 
 X .X X 57 
 4"^6~"10"~T' 
 
 
 a: = 45. 
 
 3a: + 4 — ^ = 46 — 2a:. 
 
 
 a; = 9. 
 
 3- + 5a; + 3 = 28 + -- 
 
 6 
 7* 
 
 a: = 4. 
 
 |-- = 39-5. + |- 
 
 5 
 
 "8* 
 
 a:«=9. 
 
 | + 4 = - + 12-- 
 
 
 a:=13H. 
 
 ^x 7a: , 3a; ' 7a: 
 
 
 a: = 66|. 
 
 1-1+10=1-1 + 11. 
 
 
 a: =12. 
 
 4a: 3x 7a: 13a: ^x 
 
 . '5 
 
 
 5 "^ 4 3 "~ 10 2 
 
 +'!• 
 
 a: = 3. 
 
 7. 
 8. 
 9. 
 
 10. 
 
 The equations above have been taken at random. An equa- 
 tion, however, may always be considered as derived from the 
 enunciation of some question. Thus the first of the above equa- 
 tions may be considered as derived from the following enunci- 
 ation, viz., to find a number such that one half, one third, and 
 one fifth of this number may together be equal to 31. 
 
 15. Though no general and exact rule can be given for put- 
 ting a problem into an equation, yet the following precept will 
 be found very useful for this purpose, viz. : Indicate by the aid 
 of algebraic signs upon the unknown and known quantities the 
 same reasonings and the same operations, that it would be neces- 
 sary to perform in order to verify the answer, if it were known. 
 
 Let us illustrate this precept by some examples. 
 
EQUATIONS OF THE FIRST DEGREE. 15 
 
 1. A gentleman distributing money wanted 10 shillings to be 
 able to give 5 shillings to each person ; he therefore gave each 4 
 shillings only and found that he had 5 shillings left. Required 
 the number of persons. 
 
 In order to verify the answer if it were known, we should 
 multiply it first by 5 and from the product subtract 10; we 
 should next multiply it by 4 and add 5 to the product. The 
 results thus obtained would be equal to each other, if the answer 
 were correct. 
 
 Let us indicate the same operations by the aid of algebraic 
 signs. Putting x for the number of persons sought a-nd multi- 
 plying a: by o we have 5x, subtracting 10 from this we have 
 5z — ]j3 ; again x multiplied by 4 gives 4a;, adding 5 to this we 
 have 4:X-\-5. Then as these two results should be equal we 
 have for the equation of the problem 
 
 5x—10 = Ax-\-5, 
 which being resolved gives x = 15. 
 
 2. A person expends the third part of his income in board and 
 lodging, the eighth part in clothes and washing, the tenth part in 
 incidental expenses, and yet saves $318 yearly. What is his 
 yearly income ? Ans. $720. 
 
 3. A and B. began to play ; A with exactly four-ninths the 
 sum B had. After A had won $ 10, he found that they had each 
 the same sum. What had A at first ? Ans. $ 16. 
 
 4. A General having lost a battle found that he had only 
 3600 men more than half his army left, fit for action ; 600 more 
 than one-eighth of his men being wounded, and the rest, which 
 were one-fifth of the whole army, either slain, taken prisoners ox 
 missing. Of how many men did his army consist? 
 
 Ans. 24,000. 
 
 5. A sum of money was to be divided among six poor per- 
 sons; the second received lOd. the third IM. the fourth 25d. 
 the fifth 28i. and the sixth 23d. less than the first. Now the 
 Bum distributed was lOd. more than the treble of what the first 
 received. What money did the fijrst receive ? Ans. 40rf. 
 
16 ELEMENTS OF ALGEBRA. 
 
 6. A father intends by his will that his three sons should 
 share his property in the following manner. The eldest is to 
 receive 100 pounds less than half the whole property, the second 
 is to receive 80 pounds less than a third of the whole property, 
 and the third is to have 60 pounds less than a fourth of the 
 property. Required the amount of the whole property, and the 
 share of each son. 
 
 7. A cistern is supplied by two pipes, the first will fill it alone 
 in three hours, the second in four hours. In what time will the 
 cistern be filled if both run together ? 
 
 If the time were known, we should verify it by calculating 
 what part of the cistern would be filled by each pipe separately ; 
 these parts added toother would be equal to the whole cistern. 
 To indicate the same operations by the aid of algebraic signs, 
 let a; = the time, and let the capacity of the cistern be repre- 
 sented by 1. It is evident that if one of the pipes will fill the 
 
 cistern in three hours, in one hour it will fill - of it, in x hours it 
 
 X 
 
 will fill X times as much, that is, a part denoted by jr. In like 
 
 o 
 
 manner in the time x^ the second pipe will fill a part denoted 
 
 X 
 
 by -r ; since then these two parts should be equal to the whole 
 cistern, we have for the equation of the problem 
 
 -4-- — 1 
 
 from which we obtain x=.\\ hours. 
 
 8. A cistern is furnished with three cocks, the first will fill it 
 in 5 hours, the second in 13 hours, and by the third it would be 
 emptied in 9 hours. In what time will the cistern be filled if all 
 three run together ? Ans. 6^^j hours. 
 
 9. A gentleman having a piece of work to do hired three men 
 to do it ; the first could do it alone in 7 days, the second in 9, 
 tjie third in 15 days. How long would it take the three together 
 to do it ? Ans. S^Vx days. 
 
EQUATIONS OF THE FIRST DEGREE. 17 
 
 10. To divide the number 247 into three parts, which may be 
 to each other as the numbers 3, 5 and 11. 
 
 Two numbers are said to be to each other as 3 to 5, or in 
 proportion of 3 to 5, when the first is three-fifths of the second, 
 or which is the same thing, when the second is five-thirds of the 
 first. 
 
 If then one of the parts, the first for example, were known, we 
 should verify it thus. We should find a number, which would 
 be five-thirds of the first part; this would be the second part; 
 we should find also a number which would be eleven-thirds of 
 the first part; this would be the third part; the sum of these 
 parts would then be equal to 247. 
 
 To imitate this process let x = the first part, the second will 
 
 5x 11a: 
 
 then be — and the third —^. We have then for the equation 
 o o 
 
 of the question 
 
 ^ + f +^ = 247, 
 whence x = 39. 
 
 11. A sum of money is to be divided between two persons, 
 A and B, so that as often as A receives 9 pounds, B receives 4. 
 Now it happens that A receives 15 pounds more than B. What 
 are their respective shares? Ans. A £27, B £12. 
 
 12. A merchant bought a piece of cloth at the rate of 7 crowns 
 for 5 yards, which he sold again at the rate of 11 crowns for 7 
 yards, and gained 100 crowns by the traffic. How many yards 
 were there in the piece ? Ans. 583-^ yds. 
 
 13. On an approaching war 594 men are to be raised from 
 three towns A, B, C, in proportion to their population. Now 
 the population of A is to that of B as 3 to 5 ; whilst the popula- 
 tion of B is to that of C as 8 to 7. How many men must each 
 town furnish? Ans. A 144, B 240, C 210. 
 
 14. A gentleman employed two workmen at different times, 
 one for 3 shillings, and the other for 5 shillings a day. Now 
 
 2 
 
18 ELEMENTS OF ALGEBRA. 
 
 the number of days added together was 40 ; and they each re- 
 ceived the same sum. How many days was each employed ? 
 
 If the number of days one of the workmen was employed, 
 the second for example, were known, we should verify it thus, 
 we should subtract this number from 40, this would give the 
 number of days the first workman was employed; multiplying 
 next the number of days the first workman was employed by 
 3, and that of the second by 5, the two products would be 
 equal. 
 
 To indicate the same operations let x = the number of days 
 the second workman was employed, then 40 — x will be the 
 number of days the first was employed, and the product of 
 40 — X multiplied by 3 should be equal io x y^ 6. 
 
 The multiplication of 40 — a: by 3 is indicated by inclosing 
 this quantity in a parenthesis and writing the 3 outside, thus, 
 3 (40 — x) ; we have, therefore, for the equation of the ques- 
 tion 
 
 3 (40 — a;) = 52:. 
 
 With respect to the multiplication required in this equation, it 
 is evident, since 40 should be diminished by the number of units 
 in x, that 40 multiplied by 3 would be too great for the product 
 required, by the number of units in x multiplied by 3 ; to obtain 
 the true product therefore from 40 X 3, we must subtract a; X 3 ; 
 we have then 
 
 120 — 3a; = 5a:, 
 from which we obtain a: = 15. 
 
 15. Two workmen received the same sum for their labor; 
 but if one had received 15s. more, and the other 9s. less, then 
 one would have had just three times as much as the other. 
 What did they receive ? Ans. 21s. 
 
 16. A has three times as much money as B ; but if A gains 
 $ 50 and B loses S 93, then A will have five times as much money 
 as B. How much has each? Ans. A S772|, B $257|. 
 
 17. A and B engaged in trade, A with £240, and B with 
 £96. A lost twice as much as B, and upon settling their ac- 
 
EQUATIONS OF *THE FIRST DEGREE. 19 
 
 counts it appeared that A had three times as much remaining 
 as B. How much did each lose ? Ans. A £96, B £48. 
 
 18. Two merchants engage in trade, each with the same 
 sum ; A gains $ 150, B loses $ 63j when it appears that three 
 times A's money is equal to five times B's. What had each 
 at first? Ans. $382^. 
 
 19. A laborer was hired for 48 days ; for each day that he 
 wrought he was to receive 24 shillings, but for each day that he 
 was idle he was to forfeit 12 shillings. At the end of the time 
 he received 504 shillings. How many days did he work and 
 how many was he idle ? 
 
 To verify the numbers required in this problem we should 
 multiply them, if known, by 24 and 12 respectively ; subtracting 
 the last product from the first, the remainder would be 504. 
 To indicate these operations by the aid of algebraic signs let 
 X = the number of days in which the laborer wrought, then 
 48 — X will be the number of days, in which he was idle; 24a: 
 will be the sum due for the number of days in which he wrought, 
 and 576 — 12 a; will be the sum which he forfeited. 
 
 The subtraction of 576 — 12a; from 24a; is indicated by in- 
 closing this quantity in a parenthesis and writing the sign — 
 before it, thus, 24 a; — (576 — 12 a:); we have then for the 
 equation of the question 
 
 24a;— (576— 12a:) = 504. 
 
 To perform the subtraction required in this equation, it is 
 evident, since 576 should be diminished by 12 a; before subtrac- 
 tion, if we take 576 from 24 a; we subtract too much by 12 a;; 
 12a; must therefore be added to this result in order to have the 
 true remainder ; we have then 
 
 24a; — (576 — 12 i) = 24a; — 576 + 12a;, 
 the equation of the problem therefore becomes 
 
 24a; — 576+ 12a; = 504, 
 from which we deduce x = 30. 
 
 20. A father being questioned as to the age of his son replied, 
 
20 
 
 ELEMENTS OF ALGEBRA. 
 
 that if from double his present age, the triple of what it was six 
 years ago were subtracted, the remainder would be exactly his 
 present age. Kequired his age. Ans. 9 years. 
 
 21. Divide the number 68' into two such parts, that the dif- 
 ference between 84 and the greater may equal three times the 
 difference between 40 and the less. 
 
 Ans. The parts will be 26 and 42. 
 
 22. Two men commenced trade ; A had twice as much money 
 as B; A gained $50 and B lost $90; then if three times 
 B's money be subtracted from A^s, four times the remainder 
 will be exactly equal to A's money at first? What had each 
 at first? Ans. A $426f, B $213^. 
 
 23. A person at play won as much as he began with and 
 then lost 18 shillings ; after this he lost five-ninths of what re- 
 mained, and then counting his money, he found he had 14 shil- 
 lings less than at first. What had he at first ? 
 
 Let a: = the number of shillings he began with, then 2x will 
 be the sum he had after winning x, and 2x — 18 the sum re- 
 maining after the first loss, four-ninths of which will be the 
 sum remaining after the second loss. One-ninth of 2x — 18 
 
 is expressed thus, , ^ 
 
 y 
 
 four-ninths, therefore, will be , 
 
 y 
 
 and we have for the equation of the question 
 
 8a: — 72 ,^ 
 
 X 9— =14; 
 
 from which we obtain 9x — 8 a; -[- 72 = 126 ; 
 whence x = 54c. 
 
 24. Divide the number 96 into two such parts, that four-fifths 
 of the greater, diminished by three-fourths of the less, will be 
 equal to 15. Ans. The parts are 56^-j- aftgid 39f^. 
 
 25. It is required to divide 84 into two such parts, that if 
 one-half of the less be subtracted from the greater, and one- 
 
EQUATIONS OF THE FIRST DEGREE. 21 
 
 eighth of the greater be subtracted from the less, the remainders 
 shall be equal. Ans. The parts are 48 and 36. 
 
 26. A and B began to trade with equal sums of money. In 
 the first year A gained 40 pounds and B lost 40 ; but in the 
 second A lost one-third of what he then had and B gained a sum 
 less by 40 pounds than twice the sum A had lost; when it 
 appeared that B had twice as much money as A. What money 
 did each begin with ? Ans. £320. 
 
 27. What two numbers are as 3 to 5, to each of which if 4 be 
 added the sums will be as 5 to 7 ? 
 
 5x 
 
 Let X = the less number, then -k-= the greater ; adding 4 to 
 
 each, the first will be a: -| 4 and the second 
 
 5x , ^ 5x + 12 
 -3- + 4,or-:n_; 
 
 but by the question seven-fifths of the first should now be equal 
 to the second, we have therefore 
 
 7{x + 4) __ 5x+12 
 5 '^ 3 ' 
 
 28. Divide the number 49 into two such parts, that the greater 
 increased by 6 may be to the less diminished by 11 as 9 to 2. 
 
 ,. Ans. The parts are 30 and 19. 
 
 29. A and B begin trade, A with triple the stock of B. They 
 gain each $50, which makes their stocks in the proportion of 7 
 to 3. Kequired their original stocks. 
 
 Ans. A's S300, B's 100. 
 
 30. A, B and C make a joint stock. A puts in $60 less than 
 B, and $68 more than C, and the sum of the shares of A and B 
 is to the sum of the shares of B and C as 5 to 4. What did 
 each put in ? Ans. A $140, B $200, and C $72. 
 
 31. A man being at play lost one fourth of his money and then 
 won 3 shillings ; after which he lost one third of what he then 
 had and won 2 shillings ; lastly he lost one 7th of what he then 
 had ; this being done he had but 12 shillings left. What had he 
 at first? Ans. 20s. 
 
22 ELEMENTS OF ALGEBRA. 
 
 32. There are three pieces of cloth, whose lengths are in the 
 proportion of 3, 5 and 7 ; and 6 yards being cut off from each, 
 the whole quantity is diminished in the proportion of 20 to 17. 
 Required the length of each piece at first. 
 
 Ans. 24, 40 and 56 yds. 
 
 33. Two persons, A and B have both the same annual income. 
 A lays by one-fifth of his ; but B by spending £80 per annum 
 more than A, at the end of 4 years finds himself £220 in debt. 
 What did each receive and expend annually ? 
 
 Ans. Their income is £125. A spends £100, B £180. 
 
 34. A man bought a horse and chaise for S273. Now if 
 three fourths the price of the horse be subtracted from the price 
 of the chaise, the remainder will be equal to five-elevenths the 
 price of the chaise subtracted from four times the price of the 
 horse. Required the price of each. 
 
 SECTION III.— Algebraic Operations. 
 
 16. A quantity expressed by algebraic signs is called an alge- 
 braic of literal quantity. Thus, a-\-^lT-\-^x^al^xyz, are 
 algebraic or literal quantities. 
 
 From what has been done, it is easy to see that we shall 
 have frequent occasion to perform upon algebraic quantities 
 operations analogous to the fundamental operations of arith- 
 metic, viz. addition, subtraction, multiplication and division. 
 The operations upon algebraic quantities, differ however from 
 the corresponding ones in arithmetic in this respect, that the 
 results at which we arrive in the case of algebraic quantities are 
 for the most part only indications of operations to be performed. 
 All that we do is to transform the operations originally indicated 
 into others, which are more simple, or which become necessary 
 in order that the conditions of the question may be fulfilled. 
 Thus, in the equation a; 4-2 a; -f- 6a: = 405, given by the con- 
 
ADDITION OF ALGEBRAIC QUANTITIES. 23 
 
 ditions of question first art. 6, we simplify the operations 
 originally indicated by reducing the expressions x-\-2x'^6x 
 to one term, 9x, by an operation analogous to addition in arith- 
 metic, though not strictly the same. So likewise in qur,stion 
 nineteenth, art. 15, though we cannot, strictly speaking, subtract 
 .576 — 12 z from 24a:, yet, by an operation analogous to subtrac- 
 tion in arithmetic, we indicate upon these quantities operations, 
 which produce the same effect, as the subtraction which the 
 conditions of the question require. 
 
 17. Algebraic quantities consist'^ g only of one term are called 
 Tnonomiah, as 3 a, — 4^, &c. Those which consist of two 
 terms are called hinomials^ as a-\-h^ c — d. Those which con- 
 sist of three terms are called trinomials, &c. In general, 
 quantities consisting of more than one term are called poly- 
 nomials. Quantities consisting only of one term are also called 
 simple quantities, and those consisting of more than one term 
 are called compound quantities. 
 
 Quantities in algebra, which are composed of the same letters, 
 and in which the same letters are repeated the same number of 
 times, are called similar quantities, thus, 3 ah, 1 ab are similar 
 quantities, so also aab, 5aab. 
 
 ADDITION OF ALGEBRAIC QUANTITIES. 
 
 18. 1. Let it be required to add the monomials a, b, c, and 
 d ; the result, it is evident, will be a -\- b -\- c -\- d. 
 
 2. Let the quantities to be added be ab, c, ab, d Here we 
 have as before ab -\- c -{- ab -{- d ; but the quantities ab, ab in 
 this result are similar, they may therefore be united in one term, 
 thus, 2ab; whence the sum required will he 2ab -^ c -\- d. To 
 add monomials therefore, Write them one after the other with 
 the sign -\- between them, observing to simplify the result by 
 uniting in one, those lohich are similar. 
 
 3. Let it next be required to add the polynomials a-\-b and 
 c-\- d-\-e. The sum total of any number of quantities what- 
 
24 ELEMENTS OF ALGEBRA. 
 
 ever should be equal, it is .evident, to the sum of all the parts of 
 which these quantities are separately composed ; we have there- 
 fore for the sum required a-\-b-\-c-\-d-\-e. 
 
 Let the quantities proposed he a-\-b and c — d. If we 
 begin by adding c, the result a-\-b-\-c will, it is evident, be 
 too great by the quantity d, since it is not c, which we are to 
 add, but c diminished by d; to obtain the true result, therefore, 
 from a-\-b -{- c we must subtract d ; whence c — d added to 
 a-\- b gives 
 
 a-\-b -\- c — d. 
 
 To add polynomials therefore, Write in order one after the 
 other the quantities to be added with their proper signs, it being 
 observed that the terms, which have no signs before them, are 
 considered as having the sign -{-. 
 
 19. Let it riext be required to add the following quantities. 
 ^a-\-lib — 2c 
 2a — 5c 
 83 + c. 
 
 By the rule just given the sum required will be 
 Qa-{-7b — 2c-\-2a — 5c + 8b-{-c. 
 
 In this result the similar terms 9 a, 2 a may be united in one, 
 II a; also the terms 73 and 83 give 153. 
 
 The similar quantities — 2 c, — 5 c being both subtractive, 
 the effect will be the same, if we unite them in one sum 7c 
 and subtract this sum; and as there would still remain the 
 quantity c to be added, instead of first subtracting 7 c and then 
 adding c to the resuk, the effect will be the same if we subtract 
 only 6 c. 
 
 The sum of the expressions proposed will then be reduced to 
 .la-fl53 — 6c. 
 
 In order to verify this result, let us put numbers for the letters 
 a, 3, c, in the proposed • for example, the numbers, 10, 4, - 
 respectively, and we have 
 
ADDITION OF ALGEBRAIC QUANTITIES. X6 
 
 9a-{-7b'-2c=112 
 2a — 5c=5 
 8b + c= 25 
 
 9a_|-7i — 2c + 2a — 5c + 8^ + c== 152 
 Making the same substitution in the expression lid -{-15b 
 — t5c, we obtain the same result. 
 
 The operation, by which all similar terms are reduced to one, 
 whatever sign they may have, is called reduction. To perform 
 this operation, Take ike sum of similar quantities, which have 
 the sign -j- a7id that of those ivhich have the sign — ; subtract 
 the less of the two sums from the greater and give to the remain' 
 der the sign of the greater. 
 
 We have then the following general rule for the addition of 
 algebraic quantities, viz. Write the quantities in order one after 
 the other with their proper signs, observing to simplify the result 
 by reducing to one, terms which are similar. 
 
 EXAMPLES. 
 
 L To add the quantities 
 5x-{-3y — 4z 
 Qz-\-2x^5y-\-2t 
 35 — 42/ — 2z+a: 
 7a; — 3z + 4y — 65 
 
 Answer 15a; — 2?/- 3z4-2? — 35. 
 
 To verify this answer let the numbers 12, 5, 4, 3, 13, be put for 
 the letters x, y, z, t, s, respectively. 
 2. To add the-quantities 
 
 7^ + 3^_14;7 + 17r 
 3a4-9?i— ll7w + 2r 
 
 5y 4772 -f-8^ 
 
 lln — 2b — m — r-|-5 
 
 Answer 31%- 97/1- 9p+ 18r + 3a — 23 + 5. 
 c 
 
26 ELEMENTS OF ALGEBRA. 
 
 3. To add the quantities 
 
 llbc-{-4.ad — Sac-\-5cd 
 8ac-\-7hc — 2ad-}- 4:77171 
 2c d — 3ab-\- 5ac-\-am 
 '9 am — 2b c — 2ad-\-5cd 
 
 Answer Wbc-{- 5ac~{- l2cd-\-4mn — '3ab-\- IQam, 
 
 SUBTRACTION OF ALGEBRAIC QUANTITIES. 
 
 20. 1. To subtract a from b. Here the quantities being din 
 similar, the subtraction can only be expressed by the sign — 
 thus, b — a. 
 
 2. To subtract 5 a from 7«. The quantities in this case being 
 similar, the subtraction may be performed by means of the coef- 
 ficients, and the result, it is evident, will be 2 a. 
 
 3. To subtract 2 3 -}- 3 c from d. To subtract one quantity 
 from another, we must, it is evident, take from this other the 
 sum of all the parts, of which the quantity to be subtracted is 
 composed. The result required will therefore be 
 
 d — 2b'-3c. 
 
 4. To subtract a — b from c. If we begin by subtracting a 
 from c, it is evident, that we shall take away too much by the 
 quantity b, by which a should be diminished before its subtrac- 
 tion ; b should therefore be added to c — a to give the true result ; 
 whence a — b subtracted from c gives 
 
 c — a-\-b. 
 
 5. To subtract 5c + 2d — 4tb from 70 — 2d — 5b. The 
 result, it is easy to see, will be 
 
 70 — 2d — 5b — 5c — 3d + U, 
 which becomes by reduction 
 
 2c — 5d — b. 
 From what has been done the following rule for the subtrac- 
 tion of algebraic quantities will be readily inferred, viz. Change 
 
MULTIPLICATION OF ALGEBRAIC QUANTITIES. S7 
 
 the signs -\- into — , and the sig'os — into -[- in the quantities 
 to be subtracted, or suppose them to be changed^ and then proceed 
 as in addition. 
 
 EXAMPLES. 
 
 To subtract from 17a + 2w— -93— 4c + 23f^ 
 the quantity 51 « — 273 + llc — ^d 
 
 Answer 27/i — 34a + 18Z» — 15c + 27 d. 
 
 2. To subtract from 5 ac — Sab-\-^bc — 4am 
 the quantity 8am — 2ab-\-llac — 7cd 
 
 Answer 9bc — 6ac — 6ab — 12am-\-7cd 
 
 3. To subtract from 
 
 15abc—13xy-\-2lcd — 4.lx — 25 
 the quantity 75xy — 4:abc-\-16x — 5Scd — Slmc 
 
 Answer 19abc — 88xy — 57x-{-74:cd-{-Slmc — 25 
 
 MULTIPLICATION OF ALGEBRAIC QUANTITIES. 
 
 21. 1. The product of a quantity a by another quantity b is 
 expressed, as we have already seen, thus, ay,b, or in a more 
 simple manner, thus, ab. In like manner the product of abhy 
 cd is expressed thus, abXcd, or thus, abed. 
 
 2. The letters a and b are called factors of the product ab. 
 So also a, b, c and d are the factors of the product abed. The 
 value of a product, it is easy to see, does not depend at all upon 
 the order, in which its factors are arranged ; thus the value of 
 the product arising from the multiplication of a by ^ will evidently 
 be the same, whether we write ba or ab. 
 
 3. Let it be proposed to multiply Sab hj 5cd; by no. 1 we 
 have Sab 5cd, or by no. 2, 3 X 5abcd; but the factors 3 and 5 
 in this result may, it is evident, be reduced to one by multiplying 
 them together ; performing this operation, the product required 
 will be 15 abed. In like manner the product of the quantities 
 7ab,9cdf 13 e/ will be 
 
 Sldabcdef. 
 
28 ELEMENTS OF ALGEBRA. 
 
 4. Let it be required to multiply aahy a. According to no. 
 1 we have for the result aaa; but this expression for the product 
 required may, it is easy to see, be abridged by writing the letter 
 a but once only, and indicating by a figure the number of times 
 this letter enters into it as a factor. The figure which indicates 
 the number of times a given letter enters as a factor in a product 
 is called the exponent of that letter. And in order to distinguish 
 the exponent of a letter from a coefficient, we place the exponent 
 at the right hand of the letter and a little above it, the coefficient 
 being always placed before the letter, to which it belongs, and on 
 the same line with it. 
 
 According to this method the product ca is expressed by a^, 
 aaa by a^, aaaa by a*, &c. 
 
 A letter, which is multiplied once by itself, or which has two 
 for an exponent, is said to be raised to the second power. A 
 letter which is multiplied twice successively by itself, or which 
 has 3 for an exponent is said to be raised to the third power. 
 In general, the power of a letter is designated according to the 
 figure, which it has for an exponent, thus a with 7 for an expo- 
 nent is called the seventh power of a. 
 
 A letter which has no exponent is considered as having unity 
 for its exponent, thus a is the same as a^ 
 
 From what has been said, it will be perceived, that in order to 
 raise a letter to a given power, it is necessary to multiply it suc- 
 cessively by itself as many times less one as there are units in the 
 exponent of this power. 
 
 5. Let it next be required to multiply a^ by a^. According 
 to no. 1 the product will be expressed by a^ a^. In this pro- 
 duct the letter a, it will be observed, occurs three times as a 
 factor, and also five times as a factor, whence on the whole it 
 is found eight times as . a factor. The product a^ a^ may there- 
 fore according to no. 4 be expressed more concisely, thus, a^. 
 In like manner the product of a' by a' will be <z'^ Whence, in 
 general, The product of two powers of the same letter will have 
 
MULTIPLICATION OF ALGEBRAIC QUANTITIES. 29 
 
 for an exponent the sum of the exponents of the multiplier and 
 multiplicand. 
 
 6. Let it be proposed next to multiply a^ h"^ c by a* b^ (? d. 
 According to no. 1 the product will be a^ h^ c «* b^ c^ d, or by 
 no. 2, a^ a* b^ b^ c c^ d ; but this expression may be reduced by 
 the rule just given to a^ b^ c^ d ; whence 
 
 a' b'' c X aHU^ d = an^ c^ d. 
 
 From what has been done we have the following rule for 
 the multiplication of simple quantities, viz. 1°. Multiply the 
 coefficients together; 2°. write in order in the product thus 
 obtained the letters ivhich are found at once in both the multiplier 
 and multiplicand, observing to give to each letter the sum of the 
 exponents, with which this letter is affected in the two factors ; 
 3°. if a letter is found in one of the factors only, write it in the 
 product with the exponent lohich it has in this factor. 
 
 EXAMPLES. 
 
 To multiply 1. SaHc'hy lahc^. Ans. Sea^iVdf. 
 
 2. 21tt3^2crfby8a3c^ Ans. 168a* J^cV. 
 
 3. MabUh^^df Ans. W^abUdf 
 22. Let us pass to the multiplication of polynomials. 
 
 To indicate that a polynomial a-\-b, for example, is multi- 
 plied by another c-\-d, we draw a vinculum over each and 
 connect them by the sign of multiplication, thus, 
 
 or, which is the better method, we inclose each of the quantities 
 in a parenthesis and write them in order one after the other, 
 either with or without a sign of multiplication, thus, 
 (« + *)X(c + e^),or(a + 3)(c + ^). 
 
 1. To multiply a-\-b\y^ c. To form the product required, it 
 is evident, that we must take c times each of the parts a and h 
 of which the quantity a-\-b\% composed. 
 
 The product of a-\-h 
 multiplied by c 
 
 is therefore ac-\-hc. 
 
30 ELEMENTS OF ALGEBRA. 
 
 In like manner 2a-\-Pc-\-d 
 
 multiplied by k 
 
 gives 2ah-\-b^ch-\-dk. 
 
 2. To multiply a — bhy c. Since a — 3 is smaller than a by 
 the quantity b, ac the product of a by c, it is evident, will be too 
 large for the product required by b times c or be; whence to 
 obtain the true result, from ac we must subtract be. 
 
 The product pf a — b 
 
 multiplied by c 
 
 is therefore ac — be 
 
 In like manner a^-^- c^ — dh — ef 
 
 multiplied hj ah 
 
 gives a^k-\-ahc^ — adk^ — ahef. 
 
 From what has been done, it is evident, that. If two terms 
 each affected with the sign -f- be multiplied together, the product 
 must have the sign -\- ; but if one of the terms be affected icith 
 the sign -\- and the other with the sign — , the product must have 
 the sign — . 
 
 3. Let it be proposed next to multiply a — 3 by c — d. In 
 this case, it is evident, that, if we take c times a — 5 the result 
 will be too great by d times a — b ; whence, to obtain the true 
 product, from c times a — b, or, ac — be, we must subtract d 
 times a — b ox ad — bd, 
 
 The product of a — b 
 
 multiplied by c — d 
 
 is therefore ac — bc' — ad-\-bd. 
 
 From this example it appears that. If two terms be affected 
 each with the sign — , the product of these terms should be affected 
 with the sign -|-. 
 
 If in the expression 'of a product there occur similar terms, 
 the expression may be abridged by uniting these terms into 
 one. 
 
MTTLTIPLICATION OF ALGEBRAIC QUANTITIES. 31 
 
 Thus 2a^.2 + a' — ( 
 
 multiplied by c^ — ah^ •\' & 
 
 
 gives aH^-\-a' — 2a!'b*-{-3ab'c' — c'. 
 
 To verify this result let « = 5, b = 2, c = 3. 
 
 From what has been done we have the following rule for the 
 multiplication of polynomials, viz. 1°. Multiply each term of the 
 multiplicand by each term of the multiplier, observing with 
 respect to the signs, that if tivo terms multiplied together have 
 each the same sign, the product must have the sign -\-, but if they 
 have different signs, the product must have the sign — ; 2°. Add 
 together the partial products thus obtai?ied, taking care to 
 unite in one, terms which are similar. 
 
 23. A polynomial is said to be arranged with reference to 
 some letter, when its terms are written in order according to the 
 powers of this letter. The polynomial 
 
 a^b^ + a'b — ab'+a'b^ 
 for example, arranged in descending powers of the letter a stands 
 thus, a* b"^ -\- a^ b -{- a^ P — ab^; arranged in ascending powers 
 of the letter b it stands thus, a^ b -{- a* b^ -\- a^b^ — abK 
 
 The letter with reference to which the arrangement is made is 
 called the principal letter. 
 
 To facilitate the multiplication of polynomials, it is usual, 
 1°. to arrange the quantities to be multiplied according to the 
 powers of the same letter; 2°. to dispose of the partial pro- 
 ducts in such a manner that those terms, which are similar, 
 shall fall under each other. Let it be proposed, for example, to 
 multiply 
 
 ^3 ^Pa-{-a' + banYU^ — 3ba-{- 2a\ 
 
 The multiplier and multiplicand being both arranged with 
 reference to the letter a, the work will be as follows : 
 
32 ELEMENTS OF ALGEBRA. 
 
 Sa' + Sba' + Sb^a'^-Sb'a' 
 
 — Sba' — Sb^a^ — Sb'a' — Sb'a 
 
 U^a' + Wa' + Wa + W 
 Sa'-\-4.b^a' + 4.b'a'-{-b'a + 4:b'. 
 
 24. The following examples will serve as an exercise in the 
 multiplication of polynomials. 
 
 To multiply 
 
 1. 5a^ — ^a'b + 5ab'' — 2b^ 
 by 4,a^^5ab + 2b^ ' 
 
 Answer 20a' — 4:la'b + 50a'b^ — A5a^b^'}-25ab' — 6b'. 
 
 2. a^ + Sa'b + Sab^ + b^ 
 
 by fl3_3^2^_j_0^^2_^3 
 
 Answer a^ — 2 a' b"" -\- 3 aH' — b\ 
 
 3. x^ + x^y + x'f + xy^ + ?/* 
 by a; — 7/ 
 
 Answer a:^ — 'if. 
 
 25. A term which contains one literal factor only, is said to 
 be of the first degree ; a term which contains two literal factors 
 only, is said to be of the second degree, &c. In general, the 
 degree of a term is marked by the number, which expresses the 
 sum of the exponents of the letters, which enter into this term. 
 The coefficient is npt reckoned in estimating the degree of the 
 term. Thus c^b^c is a term of the 6th degree, and 7«^^ is a 
 term of the fourth degree. 
 
 A polynomial is said to be homogeneous when all its terms are 
 of the same degree. Thus, 3c^ — ^ab,^d? -\-abc — W are 
 homogeneous polynomials. 
 
 26. From the rules for multiplication, which have been laid 
 down, it follows, 
 
 1°. If the polynomials proposed for multiplication are each 
 homogeneous, the product of these polynomials will also he ho* 
 
MULTIPLICATION OF ALGEBRAIC QUANTITIES. 33 
 
 mogeneouSi and the degree of each term of the product will be 
 equal to the sum of the degrees of any two terms whatever of the 
 multiplier and multiplicand. Thus in the first example, art. 24, 
 all the terms of the multiplicand being of the third degree and 
 those of the multiplier of the second degree, all the terms of the 
 product are of the fifth degree. When therefore the factors of a 
 product are homogeneous, we may readily detect by means of 
 this remark any error in regard to the exponents, which may 
 have occurred in the course of the work. 
 
 2°. In the multiplication of polynominals, if there be no re- 
 duction of similar terms, the number of terms in the 'product will 
 he equal to the number of terms in the multiplicand multiplied by 
 the number of terms in the multiplier. Thus if there be 5 terms 
 in the multiplicand and 4 in the multiplier, there will be 20 in 
 the product. 
 
 3". But if there be a reduction of similar terms, then the 
 number of terms in the product may be much less. It should 
 be observed, however, that among the different terms of the 
 product there will be two at least, which will not admit of 
 reduction with any other, viz. 1". The term arising from the 
 multiplication of the term in the multiplicand affected loith the 
 highest exponent of one of the letters, by the term in the multiplier 
 affected with the highest exponent of the same letter. 2°. The term 
 arising from the multiplication of the tioo terms affected luith the 
 lowest exponent of the same letter. 
 
 The manner in which an algebraic product is formed by means 
 of its factors is called the law of this product. This law, it will 
 readily be perceived, remains always the same, whatever may be 
 the values attributed to the letters which enter into the factors. 
 
 27. A product being given, we may sometimes by mere in- 
 spection decompose it into its factors, an operation which is 
 frequently useful. 
 
 Let there be the product a^h — a^b^. In the formation of 
 this product each term, it is evident, has been multiplied by a* 
 
34 ELEMENTS OF ALGEBRA. 
 
 and also by h, its factors therefore are c^, h and a — 3, and it 
 may be put under the form c^b {a — b). 
 
 In like manner the product ac-\-ad-\-bC'^bd may be 
 put under the form a{C'\- d) -\-b {c-\- d)^ or which is the 
 same thing {a -\-b) {c-\-d). 
 
 DIVISION OF ALGEBRAIC QUANTITIES. 
 
 28. 1. The object of division in algebra is the same as that 
 of division in arithmetic, viz. to find one of the factors of a given 
 product^ token the other is known. 
 
 According to this definition the divisor multiplied by the quo- 
 tient must produce anew the dividend; the dividend, therefore, 
 must contain all the factors both of the divisor and quotient; 
 whence the quotient is obtained by striking out of the dividend 
 the factors of the divisor. 
 
 Thus to divide abed by ac, we strike out of the dividend 
 the factors a and c of the divisor and obtain bd for the quo- 
 tient. 
 
 2. Let it be required to divide a^b by a^b. Decomposing a^ 
 into the two factors a^ and a^, the dividend may be put under the 
 form a^a^b; whence striking out of the dividend the factors 
 a^ and b of the divisor, the quotient will be a^. 
 
 From this example it appears that in order to find the quotient 
 of two powers of the same letter ; From the exponent of the divi' 
 dend lae subtract that of tke divisor, ike remainder loill be the 
 exponent of the quotient. 
 
 3. If it be required to divide 72 a J' c by 9^.^ we find that 72, 
 the coefficient of the dividend, may be decomposed into the two 
 factors 9 and 8 ; V may also be decomposed into the two factors 
 b^ and b"^', the dividend therefore may be put under the form 
 9 X 8ab^b^c; whence, suppressing 9 and P, the factors of the 
 divisor, we have 8ab^c for the quotient. 
 
 From what has been said we have the following rule for the 
 division of simple quantities, viz. P. Divide the coefficient of 
 
DIVISION OF ALGEBRAIC QUANTITIES. 35 
 
 the dividend by the coefficient of the divisor ; 2°. suppress in the 
 dividend the letters, ivhich are common to it and the divisor, when 
 they have the same exponent, and when the exponent is not the 
 same, subtract the exponent of the divisor from that of the divi- 
 dend and the remainder will be tlve exponent to be affixed to the 
 letter in the quotient ; 2°. write in the quotient the letters of tJte 
 dividend, which are not in the divisor. 
 
 EXAMPLES. 
 
 1. To divide 48an'c''d by I2ab^c. Ans. Aa'bcd. 
 
 2. To divide 150a'b'c^ by SOa'b'd!'. Ans. 5aH'cd. 
 
 29. From the preceding rule, it is evident, in order that the 
 division may be possible, 1°. that the coefficient of the divisor 
 should exactly divide the coefficient of the dividend; 2". the 
 exponent of a letter in the divisor should not exceed the expo- 
 nent of the same letter in the dividend ; 3°. that there should be 
 no letter in the divisor, which is not found in the dividend. 
 
 When these conditions do not exist, the division can only 
 be indicated by the usual sign. If it be required, for example, 
 to divide 12 a^b by 9cd, the division, it is easy to see, can- 
 not be performed; we therefore express the quotient by WTit- 
 ing the divisor under the dividend in the form of a fraction, 
 
 l2aH 
 thus, -Q— 7-. 
 9cd 
 
 30. The expression — — - is called an algebraic fraction. 
 
 9cd 
 
 Fractions of this species may be simplified, in the same manner 
 
 as those of arithmetic, by striking out the factors, which are 
 
 common to both terms, or which is the same thing, by dividing 
 
 both terms by the factors, which are common to them. 
 
 Let it be required, for example, to divide 48a^b^cd^ by 
 
 SQa^Pc^de; from what has been said, the most simple expres- 
 
 •n 1- 4a^ 
 sion for the quotient will be ^ — . 
 
 In like manner a^b divided by 5a^b gives ^ for the quo- 
 tient. 
 
38 t ELEMENTS OF ALGEBRA. 
 
 31. It scwne times ha,ppens, that the exponent of a letter ia 
 the same both in the divisor and dividend. The rule for ob- 
 taining the exponents of the letters of the quotient, art. 28, being 
 applied to a case of this kind, will give zero for the exponent 
 
 of the letter in the quotient. Thus, -^ according to this rule 
 gives aP for a quotient ; but -g, it is evident, is equal to unity ; 
 
 the expression a° may therefore be considered as equivalent 
 to unity. In general, a letter with zero for an exponent is to he 
 regarded as a symbol equivalent to unity. 
 
 This symbol, it is evident, will produce no effect upon the 
 value of the expression, in which it appears as a factor, since it 
 signifies nothing but unity. Its only use is to preserve in the 
 work the trace of a letter, which formed a part of the question 
 proposed, but which would otherwise disappear by the efTect of 
 division. Thus, if it be required to divide 24 a^i*^ by Sd^h^, the 
 quotient from what has been said may be put under the form 
 3<zi°. The symbol b° indicates that the letter b enters times 
 as a factor in this result, or in other words that it does not enter 
 into it as a factor, but at the same time it serves to show that 
 this letter belonged as a factor to the quantities, from which the 
 result 3 a is obtained by division. 
 
 32. We pass next to the division of polynominals. Since the 
 divisor multiplied by the quotient should produce anew the divi- 
 dend, it is evident, that the dividend must contain all the partial 
 products arising from the multiplication of each term of the 
 divisor by each term of the quotient. This being the case, it 
 is easy to see, that if we can find any one of these partial 
 products in the dividend, and the particular term of the divisor 
 upon which it depends is known, by dividing this term in the 
 dividend by the known term of the divisor, we shall obtain a 
 term of the quotient sought. 
 
 Let it be required to divide 
 
 50fl3*'' — 41a*i-|-20a5+10ai* — 33a»i' 
 by 5ah^ — ^a'b'\-5aK 
 
DIVISION OF ALGEBRAIC QUANTITIES. 37 
 
 It is evident from what has been said, art. 26, that the term 
 fl^, being affected with the highest exponent of the letter a in 
 the dividend, must have been formed without any reduction 
 from the multiplication of 5 a?, the term affected with the high- 
 est exponent of the letter a in the divisor, by the term affected 
 with the highest exponent of the same letter in the quotient; 
 that is, the term 20 a^ of the dividend is the product of 5 a? of the 
 divisor by a term of the quotient; whence, dividing 20 a^ by 6c^ 
 we obtain ^d? one of the terms of the quotient sought. Multi- 
 plying the divisor by 4a^, we produce anew all the terms of the 
 dividend, which depend upon 4fl^ viz. 20 a? b^ — \Q a^ b -\- 20 c^ \ 
 subtracting these from the dividend, the remainder 
 
 ^0a?b''-^25a'b-{-\0ab'-^^^aH^ 
 must contain all the partial products arising from the multipli- 
 cation of each one of the remaining terms of the quotient by 
 each term of the divisor. 
 
 Regarding this remainder as a new dividend, it is evident, 
 from what has been said, that the term — 25 a* b must have 
 arisen from the multiplication of 5(^ by the term affected with 
 the* highest exponent of the letter a in the remaining terms of 
 the quotient sought; whence," dividing — 25a^b by 5a^, we shall 
 be sure to obtain a new term of the quotient. 
 
 "With regard to the sign, which should be prefixed to this term 
 of the quotient, it is evident, that it should be the sign — ; since, 
 from the nature of multiplication, the divisor having the sign -)-, 
 the quotient must have the sign — in order that their product 
 may produce anew the dividend — 25a^b. 
 
 Performing the operation therefore, we have — 5ab for 
 another term of the quotient sought. Multiplying the divisor 
 by this term of the quotient, we obtain all the terms of the 
 dividend, which depend upon — 5ab, viz. 
 
 — 25aH^-\-20aH'' — 25a'b; 
 subtracting these from 30 a^ ^^ — 25 a* ^ + 10 a &* — 33 a^ b\ the 
 remainder lOa^i^^-j- lOai* — 8a^3^ will contain all the partial 
 products arising from the multiplication of each one of the 
 
 D 
 
38 ELEMENTS OF ALGEBRA. • 
 
 remaining terms of the quotient sought by each term of the 
 divisor; whence, for the same reasons as before, dividing 10 a^^* 
 by 5a^j we have 2b^ for a new term of the quotient; muhiplying 
 the divisor by this term and subtracting as before, nothing re- 
 mains; the division is therefore exact, and we have for the 
 quotient sought 4 a'^ — 5ab-]-2b^. 
 
 33. In the course of reasoning pursued above, we have been 
 obliged to seek in each of the partial operations the term in 
 the dividend affected with the highest exponent of one of the 
 letters, in order to divide it by the term of the divisor, affected 
 with the highest exponent of the same letter. We avoid this 
 research by arranging the dividend and divisor with reference 
 to the same letter; for, by means of this preparation, the first 
 term at the left of the dividend and the first term at the left of 
 the divisor will, in each of the partial operations, be the two 
 terms which must be divided, one by the other, in order to obtain 
 a term of the quotient. 
 
 The following is a table of the calculations in the preceding 
 example, the dividend and divisor being arranged with refer- 
 ence to the letter a, and placed one by the side of the other as 
 in arithmetic. 
 
 20a^ — 4U*H 
 
 -50an'' — Z'iaH^-^\0ab' 
 - 20 aH'' 
 
 5a'- 
 
 -.4.aH + 5ab^ 
 
 20a'—l6a'b-\ 
 
 ^a"- 
 
 -5ab'\-2b^ 
 
 — 25c'h- 
 
 — 25a-u- 
 
 -20aH^ — 25aH^ 
 
 
 
 
 
 -\Oab' 
 - lOab"^ 
 
 
 From what has been done we have the following rule for the 
 division of compound quantities, viz. 
 
 Having arranged the divisor and dividend with reference to 
 the powers of the same letter, P. Divide the first term of the 
 dividend by the first term of the divisor, the result will be the first 
 term of the quotient ; 2°. multiply the whole divisor by the term 
 of the quotient just found, and subtract the result from the divi- 
 dend; 3°. divide the first term of the remainder by the first term 
 
DIVISION OF ALGEBRAIC QUANTITIES. 39 
 
 of the divisor i the result will be the second term of the quotient ; 
 4°. multiply the lohole divisor by the second term of the quotient, 
 and subtract the product from the result of the first operation, 
 and continue the same course of operations until all the terms of 
 the dividend are exhausted. 
 
 Recollecting, that in multiplication the product of two terms 
 affected with the same sign should have the sign -[-» and that 
 the product of two terms affected with different signs should 
 have the sign — , we infer 1°. that if the two terms of the divi' 
 dend and divisor have each the same sign, the quotient arising 
 from their division should have the sign -\- ; but if they are 
 affected with co7itrary signs it should have the sign — .* This is 
 the rule for the signs. 
 
 EXAMPLES. 
 
 To divide « 
 
 by a^\-Ua\ 
 
 2. a:^ + 6:c^4- 92-^ 4-9ar^ + 4x4-1 
 
 by ar^4-:c+l. . ^ 
 
 3. ^2x' — IQx^y —l()xhf -\-\l x]^ -{'^y'^ 
 by I2x' — 5xy — ^y'. 
 
 4. x' — x^y — 132-y + x'f -J- \2xy'' 
 by y^J^^x'y — ^xf. 
 
 34. The dividend and divisor heing arranged with reference 
 to the powers of the same letter, if the first term of the divi- 
 dend is not divisible by the first term of the divisor, we infer 
 that the total division is impossible, or in other words, that 
 there is no polynominal, which multiplied by the divisor will 
 reproduce the dividend ; and, in general, we infer that the divis- 
 ion cannot be exactly performed, when the first term of any one 
 of the partial dividends is not divisible by the first term of the 
 divisor. 
 
 When the division cannot be exactly performed, in order to 
 complete the quotient, we write the remainder over the divisor 
 
40 ELEMENTS OF ALGEBRA. 
 
 in the form of a fraction and annex it to the quotient as in 
 arithmetic. 
 
 EXAMPLE. 
 
 To divide 
 
 hy 5a* — 2an + 4:a''b; 
 Answer a^ — 4:a^b'\-2b^ 
 
 5a' — 2a'b-\-4^d'b^ 
 35. We may remark in passing, that there is some analogy 
 between division in arithmetic and division in algebra with re- 
 gard to the manner in which the calculations are disposed and 
 performed; there is, however, this essential difference, that in 
 arithmetical division the iigures of the quotient are obtained by 
 trial; whereas, in algebraic division, we obtain with certainty a 
 term of the quotient sought, by dividing the first term of each 
 partial dividend by the first term of the divisor. In Algebraic 
 division, moreover, we may begin, as it will be easy to see from 
 the remarks, art. 26, at the right instead of the left of the divi- 
 dend, since, in this case, we shall have merely to operate upon 
 the terms affected with the lowest, instead of those affected with 
 the highest exponents of the letter, in reference to which the 
 arrangement is made ; whereas, in arithmetical division, we 
 must always begin at the left. Indeed, such is the indepen- 
 dence of the partial operations in algebraic division, that having 
 obtained one of the terms of the quotient and subtracted from 
 the dividend the product of this term by the divisor, we may in 
 the second partial operation divide, one by the other, the two 
 terms of the new dividend and the divisor affected with the 
 highest exponent of any other letter different from that, with 
 reference to which the arrangement is made, and thus obtain a 
 new term of the quotient. It is indeed only for the sake of 
 convenience, that we always regard the same letter in the course 
 of the partial operations necessary to obtain the quotient. 
 
DIVISION OF ALGEBRAIC QUANTITIES. 41 
 
 36. In the process of division, the muhiplication of the differ* 
 ent terms of the quotient by the divisor often produces terms, 
 which are not found in the dividend, and which it is necessary 
 to divide by the first term of the divisor. These terms are such, 
 as cancel each other in the process of forming the dividend by 
 the muhiplication of the divisor by the quotient. 
 
 To divide c^ — J^ by a — b, 
 
 c^ — b' 
 a' — aH 
 
 a — b 
 
 d^J^ab-^b^ 
 
 aH — P 
 a^b — ab^ 
 
 
 ab' — b' 
 ab' — b' 
 
 If we now multiply the divisor by the quotient in this exam- 
 ple, in order to produce anew the dividend, we shall find, that 
 the new, terms, which arise in the process of division, are those 
 which cancel each other in the result of multiplication. 
 
 EXAMPLES. 
 
 1. To divide 6a;* — 96 by 3a: — 6. 
 
 2. To divide a;' -j- ^ by a: + !/• 
 
 3. To divide a* — a;* by fl — x. 
 
 4. Todividea:' — a:* + a:^ — ar' + 2a: — Ibyar^ + a;— 1. 
 
 37. It sometimes happens, that one or both of the quantities, 
 proposed for division, contains several terms affected with the 
 same power of the letter, in reference to which the arrangement 
 is made. The following examples will exhibit the course to be 
 pursued in cases of this kind. 
 
 1. To divide 
 Ua'b — 19abc+ lOa' — 15a«c -f Sab^-{- ISbc" — 5Pc by 
 5fl' + 3ai — 53c. 
 
 The terms 11 a^b — I5a^c may be put under the form 
 
 (113 — 15 c) a^j or which is the more convenient method 
 
 lib 
 — 15c 
 
42 
 
 ELEMENTS OF ALGEBRA. 
 
 a vertical line being employed instead of a parenthesis to in- 
 dicate that the quantities lib, — 15c, placed one under the 
 other at the left hand, are multiplied each by a^. In like 
 manner the terms — 19abc-\-3ab^ may be put under the 
 
 form — 195 c a 
 + 3b^ 
 
 Arranging the quantities with reference to the letter a, the 
 
 calculations may be performed as follows. 
 
 5f+2ab--5bc 
 
 lOa' + llb 
 
 a'—ldbc 
 
 a — dPc+ldbc" 
 
 — 15c 
 
 + 3b' 
 
 
 10a' -{- 6b 
 
 a'—lObc 
 
 a 
 
 2a + b 
 
 Q 
 
 HC 
 
 1st Rem. 
 
 5b 
 
 — 15c 
 
 5b 
 
 — 15c 
 
 a' — dbc 
 -\-3b^ 
 
 a' — dbc 
 + 3P 
 
 a — 5b^c-\-15bc' 
 a — 5b'c-{-15bc' 
 
 2d Rem. 
 
 Dividing first 10 a^ by 5a^, we have 2a for the quotient; 
 subtracting the product of the divisor by 2 a from the dividend, 
 we obtain the first remainder ; dividing the part affected with a' 
 in this remainder by 5a^, we obtain b — 3c for the quotient; 
 multiplying successively each term in the divisor by 3 — 3 c, we 
 exhaust the dividend ; whence the quotient is 
 2a-\-b — 3c. 
 
 In like manner the following examples may be performed. 
 2. To divide 
 
 a^ — V" 
 + 2c2 
 
 a'-\-b' 
 
 b' 
 
 2b'(? 
 -\-b''c' 
 
 bya2_^2_g2 
 
 Answer — a* - 
 
 -24' 
 
 a 
 
 '' — b' 
 
 
 ■v<? 
 
 — v'e 
 
 
 3. To divide 
 
 • 
 
 f :^ — iyz3? — 'if 
 
 x + y' 
 
 -^ +^yz 
 
 — 1? 
 
 Answer y 
 
 x' + y 
 
 x — y 
 
 
 +' 
 
 
 ^ 
 
 + ^ 
 
 
 by 
 
 y\x — y 
 ■z I — z 
 
DIVISION OF ALGEBRAIC QUANTITIES. 
 
 43 
 
 38. When the dividend is not divisible by the divisor, we may 
 still attempt the division, according to the rules which have been 
 given, and continue it at pleasure. 
 
 Thus let it be required to divide xhy x-\' z. 
 
 X 
 
 x-^-z 
 
 X'\- z 
 
 a: ' ar x^ ^ x^ 
 
 — z 
 
 ^ 
 
 
 X 
 
 
 x^x' 
 
 
 z' 
 
 
 ■"? 
 
 
 2' 
 
 z' 
 
 x'" 
 
 -? 
 
 
 z' , 
 
 From the number of terms in the quotient already obtained in 
 the above example, the learner will readily infer a law, by 
 which the quotient may be continued at pleasure without per- 
 forming any more operations. 
 
 39. Miscellaneous examples in the division of algebraic quan- 
 tities. 
 
 1. To divide ar' + 1 by a: + 1. 
 
 2. Todivide 1 — 5a: +10ar^—10a;^ + 5a;^--a:« 
 byl — 2a; + ar». 
 
 3. To divide a^ -^ ct^^ hy a -\-x. 
 
 4. To divide a' -- 5 a' z -\- 10 a^ z"" — lOa'z' -\- 5az* — z" 
 by a^ — 2az-\- z^. 
 
 5. To divide 6x' — 5x'f + 2l3^y^ — 6x*y* + x'f+l5'/ 
 hy2:i^ — 3x'f + 5y\ 
 
44 ELEMENTS OF ALGEBRA. 
 
 6. To divide 1 by 1 — «. 
 
 7. To divide ahj a — ^. 
 
 8. To divide I ^ax -\-bx'-\' C3^ -{-dx* + &c., 
 by 1 — X. 
 
 Ans. 1 4- 1 1 a: 4- 1 1 a;*^ &c. 
 
 9. To divide a — bx-\-cx^ — da^ -{- &c., by 1 -f- a:. 
 
 SECTION IV.— Algebraic Fractions. 
 
 40. When the division of two algebraic quantities cannot be 
 exactly performed, the quotient, as we have seen, is expressed in 
 the form of a fraction, the dividend being taken for the numera- 
 tor and the divisor for the denominator. 
 
 A fraction in algebra has the same signification as a fraction 
 in arithmetic ; the denominator shows into how many parts unity 
 is divided, and the numerator how many of these parts are 
 taken. Thus, in the algebraic fraction -, unity is supposed to be 
 divided into b parts, and a number a of these parts is supposed 
 to be taken. 
 
 reduction of fractions to their lowest terms. 
 
 41. A fraction is said to be in its lowest terms, when there is 
 no quantity, that will divide both of its terms without a remain- 
 der. To reduce a fraction therefore to this state, we suppress in 
 the numerator and denominator the factors, which are common 
 to them. 
 
 The suppression of a factor is the same, it is evident, as divid- 
 ing by the factor, required to be suppressed. 
 
 When the two terms of an algebraic fraction are simple quan- 
 tities, it will be easy, from inspection, to determine the factors 
 
ALGEBRAIC FRACTIONS. 45' 
 
 common to them ; but if the terms of the fraction are polynomi- 
 als, this will not be so easy, and we must in this case have 
 recourse to the method of the greatest common divisor. 
 
 By the greatest common divisor of two algebraic quantities 
 we understand the greatest in regard to coefficients and expo- 
 nents, that ivill exactly divide these quantities. Its theory rests 
 upon the same two principles, as that of the greatest common 
 divisor in arithmetic, viz. 
 
 1°. The greatest divisor common to tivo quantities contains as 
 factors all the particular divisors common to these quantities and 
 does not contain any other factors. 2°. The greatest divisor 
 common to two quantities is the same with the greatest divisor 
 common to the less of these quantities and the remainder after 
 the division of the greater by the less. 
 
 42. A quantity is said to be prime in respect to another quan- 
 tity, when the two have no factor in common. 
 
 From the first of the preceding principles it follows that we 
 may multiply, or divide, either of the two quantities by any 
 quantity which is prime to the other, without affecting their 
 greatest common divisor ; for we shall not, by these operations, 
 either introduce or throw out a factor common to the two 
 quantities; the greatest common divisor, therefore, should re- 
 main the same. 
 
 This being premised, let it be proposed to find the greatest 
 common divisor of the" polynomials 
 
 (^^aH-\-^ab''—^b\ and a^ — 5ab-\-U\ 
 
 Pursuing the same general course as in arithmetic, we com- 
 mence by dividing the first of the proposed polynomials by the 
 second; we thus obtain a-\-^b for a quotient, with a remainder 
 lQab^—\^b\ 
 
 By the second of the above principles the question is now 
 reduced to finding the greatest common divisor to this remainder 
 and the divisor a^^5ab-\'W'. But 19a^2— 19i' may be 
 put under the form 19 5^ (a — h)\ and since the factor 193'^of 
 
46 ELEMENTS OF ALGEBRA. 
 
 this quantity is prime to a^ — 5ab-{-4Lp, it may, in virtue of the 
 first of the above principles, be suppressed; thus the question 
 will be still further reduced to finding the greatest common 
 divisor to a — b and a^ — 5a3-|-4Z»^ 
 
 Dividing the last of these two quantities by the first we obtain 
 an exact quotient a — 43; whence a — bis their greatest com- 
 mon divisor; and by consequence it is the greatest common 
 divisor of the polynomials proposed. 
 
 The following is a table of the calculations. 
 1st operation a^>—a'b-}-2ab''^3b^ ) a^ — 5ab + W^ 
 
 /^d'b — ah' — 3 // 
 /^aH — 2^ab''+\Qb^ 
 
 or l^b''{a — b) 
 
 a — b 
 
 2d operation c^ — 5ab-\-^h 
 c^ — ab 
 
 a — 4:b 
 
 
 
 
 2. To find the greatest common divisor of the polynomials 
 
 a^^5a'b + 3ab'' — b'\ . , , 
 
 a^J^2ab4b' 1 Ans.a + b. 
 
 3. To find the greatest common divisor of the polynomials 
 
 4. To find the greatest common divisor of the polynomials 
 
 a^ + ^x^+Sf ^ ^ 1 Ans. :. + 3y. 
 
 5. Let it be proposed, next, to find the greatest common divisor 
 of the polynomials 
 
 5b'—lSb^a + nba'--6a\sind7b^-^2Sba+6a^ 
 
 In this example 5b^, the fijst term of the dividend, is not 
 
 divisible by 7 b^, the first term of the divisor. It will be observed, 
 
 however, that 7, the coefiicient of the first term of the divisor, 
 
 will not divide the remaining terms of the divisor. We may, 
 
ALGEBRAIC FRACTIONS. 47 
 
 therefore, in virtue of the first principle, muhiply the dividend 
 by 7 without affecting the greatest common divisor sought. Per- 
 forming this operation, we have for the dividend 
 S5b^—126b^a + 77ba'''-A2a\ 
 
 Dividing next 35^^ by 7b'^, we obtain 53 for a quotient. Mul- 
 tiplying the whole divisor by db, and subtracting, we have for a 
 remainder — 11 b"^ a -^ 47 b a^ — 42 a^ 
 
 The exponent of b in this remainder, being equal to the expo- 
 nent of the same letter in the divisor, we continue the operation; 
 and in order to render the first term divisible by the first term 
 of the divisor, we multiply anew by 7, which gives — 77 b^ a 
 -{- 329 ba^ — 294 a^. Dividing this by the divisor, the quotient 
 is — 11a, which we separate froni the other by a comma, to show 
 that it has no connection with it, and the remainder is 76 ba^ 
 — 228a^or76a2(3 — 3 a). 
 
 Suppressing the factor 76 a*^, the question is reduced to finding 
 the greatest divisor common to b — 3 a and 7b^ — 23 3 a -|- 6 a^. 
 Dividing, therefore, the last of these quantities by the first, we 
 obtain an exact quotient 7b — 2a; whence b — 3a is the greatest 
 common divisor sought. 
 
 See a table of the calculations. 
 1st operation 
 
 25b'— 126Pa -^77 ba'' — 42a^ I 7b^ — 23ba + 6a' 
 35b'—115b^a-]-20ba' \ qJ^ —11a 
 
 2d operation 
 
 — 
 
 -lib' a 
 
 -_|-47^^2_ 
 
 42 a^ 
 
 — 
 
 -77 b'a 
 
 ',-\-329ba'- 
 
 -294 a' 
 
 — 
 
 -77b''a-\-2d3bd'- 
 
 -66 a' 
 
 
 
 7Ud' — 
 
 ■22Sa^ 
 
 
 or 
 
 7Qa^b- 
 
 -3a) 
 
 >n 
 
 7P- 
 7b'_- 
 
 — 23b. 
 
 — 21b 
 
 — 2ha 
 
 a + Qd?)^ b 
 + 6a'' 
 
 — 3a 
 
 — 2o 
 
 
 — 2ba 
 
 + 6a'^ 
 
 
48 ELEMENTS OF ALGEBRA. 
 
 6. To find the greatest common divisor of the polynomials 
 
 7. To find the greatest common divisor of the polynomials 
 
 4:r^_5:.^ + / ^J Ans. a:-y. 
 
 The suppression of a factor common to all the terms in the 
 first remainder in the preceding examples, serves not only to 
 simplify the calculations, but is also indispemable. Looking at 
 the first example, it is evident that unless the factor lO^'^in 
 the first remainder be suppressed, we must multiply all the 
 terms of the new dividend by 19 3*^, in order to render the first 
 term divisible by the first term of the divisor; we should thus 
 mtroducft into the dividend a factor, which is also contained in 
 the divisor, and by consequence we should introduce into the 
 greatest common divisor sought, a factor, which does not belong 
 to it. 
 
 8. Let it be proposed next to find the greatest common divisor 
 of the polynomials 
 
 \6a' -{-lOa'b + Mb'' -\-Q>aH^ ^^ab' 
 \2aH'' + 2QaH^ + lQab'—l(ib\ 
 Before proceeding to the division of the proposed polynomials, 
 we observe that the first contains the letter a as a factor common 
 to all its terms ; and since this letter does not enter as a factor 
 into the second polynomial, we may suppress it, as forming no 
 part of the greatest common divisor sought. 
 
 For a similar reason, the factor 2^^ may be suppressed in the 
 second polynomial. Thus the question is reduced to finding the 
 greatest common divisor of the polynomials 
 
 \5a' + lOa^i + Mb^ + 6 ab^^^h" 
 ^a^+l^aH^-Sab^ — ^bK 
 Pursuing with these polynomials the same course as in the 
 preceding examples, we should multiply the dividend by G, 
 
ALGEBRAIC FRACTIONS. 49 
 
 the coefficient of the first term of the divisor. But since 15 and 
 6 have a common factor 3, it will he sufficient to multiply hy 2 
 the other factor of 6, which does not enter into 15 ; multiplying 
 therefore by 2 and continuing the operations as above, we obtain 
 for the greatest common divisor, Sa'^ -|- 2a5 — b"^. 
 
 9. To find the greatest common divisor of the polynomials 
 
 c 3 /^ 2 o 1 ' t Ans. 2x — 1. 
 
 10. To find the greatest common divisor of the polynomials 
 
 ^x^y5a^x^ + 2Wx 1 Ans. 2.: + 3«. 
 
 From what has been done, we have the following rule, by 
 which to find the greatest common divisor of two polynomials, 
 viz. The polynomials proposed being arranged with reference 
 to the same letter, P. We suppress in each the monomial factors 
 which are not found in the other ; 2°. we divide one of the poly- 
 nomials by the other, and if the division cannot be exactly 
 performed, we divide the first divisor by the remainder, and so 
 on, observing to prepare each dividend when necessary in such a 
 manner, as to render the first term divisible by the first term of 
 the divisor, and to suppress in each remainder the monomial fac- 
 tors, which are n/)t contained in the preceding divisor ; and that 
 remainder, which will exactly divide the preceding, will be the 
 greatest common divisor sought. 
 
 43. The research for the greatest common divisor of two 
 polynomials admits, in certain cases, of simplifications which we 
 shall now explain. 
 
 1. Let it be proposed to find the greatest common divisor of 
 the polynomials 
 
 5a« + 10a^a: + 5aV 
 a^x-\-2a^x'+2a^x^-\-ax*. 
 The letter a, it will be perceived, enters as a factor into each 
 of the terms of the polynomials proposed. This letter will, 
 therefore, be a factor of the greatest common divisor sought. 
 4 
 
50 
 
 ELEMENTS OF ALGEBRA. 
 
 Suppressing a in the proposed, and applying the rule to the poly- 
 nomials which result, we obtain a -[- a; for their greatest common 
 divisor. ' The greatest common divisor sought will, therefore, be 
 a{a-\- a;), or c^ -\- ax. 
 
 2. To find the greatest common divisor of the polynomials 
 
 3. To find the greatest common divisor of the polynomials 
 
 6a*3 — 10 aH^ + 7 aH"^ — Sab' 
 Sa'b — 5a^b^ + 2ab' 
 
 4. Let it be required next to find the greatest common divisor 
 of the polynomials 
 
 > Ans. ab {a — b). 
 
 a' + b' 
 
 — b(? 
 
 a'-^b"" 
 
 — be 
 
 «5 + ^V 
 
 — b^c' 
 
 a^-^-b' 
 
 — b-'c 
 
 cS- 
 
 The proposed, it will readily be perceived, have a simple factor 
 c^ common to both ; recollecting that this will be a factor of the 
 greatest common divisor sought, we suppress it, and the polyno- 
 mials, which result, will be 
 
 a' + W 
 — b& 
 
 
 a^-\-b^ 
 — be 
 
 a^b\ 
 — bH 
 
 We may now commence the division of one of these polyno- 
 mials by the other according to the rule, in order to determine 
 their greatest common divisor. Before proceeding to this, how- 
 ever, let us see if there be not a polynomial divisor common to 
 the coefficients of the letter a, with reference to which the 
 arrangement is made. 
 
 Comparing for this purpose the two coefficients of the lowest 
 degree b'^ — (? and b — c, we find that b — c will divide both 
 without a remainder. We inquire next if i — c will divide the 
 remaining coefficients of a. This is the case; b — c, therefore, 
 is a divisor common to all the coefficients of the two last polyno- 
 mials. Recollecting that b — c will also be a factor of the 
 
ALGEBRAIC FRACTIONS. 
 
 51 
 
 greatest common divisor sought, we suppress b — c, and the 
 polynomials, which result, will be 
 
 b 
 
 "t 
 
 ^3 ^ 33.^2 _|. 32^^ and a^ + 3a + b^ 
 
 Applying the rule to these, the first, it will be perceived, 
 contains a factor b-^Cj which is not contained in the second. 
 Suppressing this, it remains to find the greatest common divisor 
 of the polynomials 
 
 a* + ba' + b'c\ and ^= + *a + b\ 
 These, it will be found, have no common divisor. The greatest 
 common divisor of the proposed will, therefore, be 
 a^ {b — c), or a^b — a^c. 
 7. To find the greatest common divisor of the polynomials 
 
 y 
 
 X 
 
 -1 
 
 2/5 — 3a; 
 + 3 
 
 + 3a; 
 — 2 
 
 y' + c^ 
 
 X 
 
 
 1 
 
 + 3 
 
 f-\-x\ 
 
 X 
 
 — 2 
 
 + x 
 
 y- 
 
 Ans. y{y—\) {x^l). 
 
 8. Let it be proposed next to find the greatest common divisor 
 of the polynomials 
 
 — yz' 
 
 fz 
 yz^ 
 
 x' + by' 
 
 — byz 
 
 — cyz" I 
 
 x^ -\-by^z 
 --dy^z 
 
 — by^ 
 
 — dyz" 
 
 :j^-\-bc'f 
 — bcyn^ 
 
 x-\-bd'ifz 
 — bdy^ 
 
 The simple quantity xy^ it will be perceived, will exactly 
 divide each of the terms of the first of the proposed polynomials, 
 and yz those of the second. The factor y common to these 
 quantities will be, it is evident, a factor of the greatest common 
 divisor sought. Setting apart the y therefore as such, and 
 
S^ ELEMENTS OF AteEBEA. 
 
 dividing the first polynomial hy xy and the second by yzy Uie 
 polynomials, which result, will be 
 
 
 — bcz" 
 
 x-^hdy 
 — bdz 
 
 y\x^-{-by 
 
 — z I -{-dy 
 
 — bz 
 
 — dz 
 
 The coefficients of the first of these aire divisible each by 
 ^ — z^ and those of the second by y — z; but y — 2, being 
 a factor common to y^ — :^ and y — z, will also, it is evident, 
 be a factor of the greatest common divisor sought; setting it 
 apart, therefore, as such, and dividing the first polynomial by 
 y* — 2^ and the second by 2/ — z, the polynomials, which result, 
 will be 
 
 -|-c| \d\ 
 
 Applying the rule to these last, we obtain x -J- 3 for their 
 greatest common divisor. The greatest common divisor of the 
 proposed will, therefore, be y{y — z) [x-\-b). 
 
 From what has been done, the following method for finding 
 the greatest common divisor of two polynomials will be readily 
 inferred, viz. 
 
 1*. Suppress in the polynomials proposed the greatest simple 
 divisors, which they respectively contain, observing to set aside as 
 a factor of the greatest common divisor sought, the greatest 
 factor, which these divisors have in common. 2°. Suppress in 
 the polynomials, which result, the greatest polynomial divisor, 
 independent of the principal letter, and set aside as a factor of 
 the greatest common divisor sought the greatest factor which 
 these divisors have in common. 3°. Find the greatest common 
 divisor of the polynomials which result, this will be the remain- 
 ing factor of the greatest common divisor sought, and the product 
 of the several factors, thus obtained, will be the greatest commoa 
 divisor sought. 
 
A](.aSBRAIC FBAOTIONS. -|3 
 
 44. To reduce a fraction to its lowest terms, we divide the two 
 terms of the fraction by their greatest common divisor. 
 
 EXAMPLES. 
 
 1. Reduce -= -r-^ to its lowest terms. Ans. — -f — . 
 
 ^4 x'^ 
 
 % Reduce -^—. — 5 5 = to its lowest terms. 
 
 Ans. — — . 
 a-J-a: 
 
 ^ _, , Qa7?-\-a3^ — \2ax . . 1 /. 
 
 3. Reduce -^ ^ to its most simple form. 
 
 Qax — 8a ^ 
 
 23?4-2x 
 Ans. J , 
 
 4. Reduce -5 — . . , ^ » — T 1 ^ to its lowest terms.' 
 
 x^ — 4a:^-j-6ar* — ^x-\-2 
 
 Ans. "■ 
 
 45. Algebraic fractions being of the same nature as fractions 
 in arithmetic, the rules for the fundamental operations are the 
 same. We shall merely subjoin these rules, with some exam- 
 ples und^ each, the results being reduced to their lowest terms. 
 
 MULTIPLICATION OF ALGEBRAIC FRACTIONS. 
 
 Rule. — Multiply the numerators together for a new numerator, 
 and the denominators for a new denominator. 
 
 EXAMPLES. 
 
 1. Multiply —5 by - — 5- Ans. -„. 
 
 2. Multiply — !-^ 1 — by — r -. Ans. • ^ '" -^. 
 
 cd — d* a-\-o c — d 
 
 e n/r u' 1 «'* + «a: , a^ — a^ . (^4-a^x4-aa* 
 
 3. Multiply o o by -^. Ans. — ?^ -\= — . 
 
 a^_l.i X 1 
 
 4. Multiply 3:p, -^^~ and — t— 7 together. 
 2a a-j-o 
 
 Sa:' — 3a: 
 Ans. 
 
 2a^-^-2ab 
 
54 ELEMENTS OF ALGEBRA. 
 
 6, Multiply — .—r-t r — -2 and a-{ together. 
 
 ^•' a-{-b ax-\-x^ 'a — x ^ 
 
 Ans. ^(^i). 
 
 X 
 DIVISION OF ALGEBRAIC FRACTIONS. 
 
 Rule. — Invert the divisor ^ and then proceed as in multiplica- 
 tion. 
 
 EXAMPLES. 
 
 1. Divide i+* by i+|. Ans. ^*'. 
 
 X — y a — b ar — if 
 
 2. Divide ^^ by ^. Ans. -^+^. 
 
 a^ — x^ ^ a — x a^J^ax-{-3? 
 
 3. Divide -2 — r—r^ by '—r-* Ans. — ' — . 
 
 a^ — 2bx-\-b^ ^ X — b x 
 
 4. To divide 12 by ^-^i^ — a. Ans. "^^^ 
 
 X c^-^-ax-^-a^' 
 
 ADDITION OF ALGEBRAIC FRACTIONS. 
 
 Rule. — Reduce the fractions to a common denomincUor ; then 
 add the numerators together, and place their sum over the common 
 denominator. 
 
 EXAMPLES. 
 
 1. Add together ^-±1 and ^. Ans. %±^. 
 
 x — y x-f-y a^ — f 
 
 2. Add together ^\nd^^:=^. Ans. ^i^+^V + f), 
 
 x-j-y x — y x-j-y 
 
 3.jj^ , a a — 3^ . a^ — b^ — ab 
 . Add together 7, — , and ; — ; . ♦ 
 
 o cd bed 
 
 . acd — 4tb^4-a^ 
 
 Ans. i— 7—^ • 
 
 bed 
 
 4. Add together - — r-— 5, — - — 5-— -„, and 
 
 {a + bY' {a + bf a-\-b 
 
 i 
 
 la + bY ' 
 
 a^A.ab^ + b* 
 Ans. — * — ' 
 
EQUATIONS OF THE FIRST DEGREE. 55 
 
 5. Add together ^r-r:^, - — j — ^7" ^ n v » and j — . 
 
 • (a — 2xy {a-{-x) {a — 2x) a-\-x 
 
 20ax — 223^ 
 
 Ans. 
 
 {a-^-x) {a — 2xf 
 
 SUBTRACTION OF ALGEBRAIC QUANTITIES. 
 
 Rule. — Reduce the fractiom to a common denominator ; then 
 place the difference of their numerators over the denominator ^ and 
 it will be the difference required. 
 
 EXAMPLES. 
 
 , ^ 5a: — 3 ^ ^x-\-2 , Sar*— 13a:+l 
 
 1. From ;-— - subtract ^;-. Ans. 5 — ^ — . 
 
 x-\-\ X — 1 ar — 1 
 
 « T. 1 , 1 * 2v 
 
 2. Ironx subtract — ; — . Ans. „ „ • 
 
 X — y ^-tV ^ — 2r 
 
 - _ az , a — z . 2az — a^ — z^ 
 
 3. From -:; 5 subtract — -, — . Ans. 
 
 a^ — z^ a-\-z ^ d^ — ^ 
 
 "T ^ ' subtract ^~~ . Ans. ^ , . 
 
 V —y y y—i 
 
 SECTION V. — Equations of the First Degree. 
 
 46. The rules obtained in the preceding sections, are suf- 
 ficient for the solution of all equations of the first degree, 
 however complicated. We place below a few examples, in- 
 volving operations a little more complicated than those, which 
 have been previously introduced. 
 
 , _. 7x — S.l5x-{-S _ 31— a: , . , ,, 
 
 1. Given — — 1 Y^ — = 3 a: -^ — , to find the 
 
 value of a:. Ans. a; = 9. 
 
 ^ „. 2a:+l 402 — 3a: _ 471 — 6a: ^ . , 
 
 2. Given -^ 12— = ^ 2"' '' ^^^ 
 
 the value of x. Ans. x = 72. 
 
66 ELEMENTS OF ALGEBRA. 
 
 3. Given — ^ — = — -+- -, to find the value of x. 
 
 36 5a; — 4 ' 4 
 
 Ans. x = S. 
 
 . ^. 10a:+17 12a: + 2 5a: — 4 , . , . 
 
 4. Given _^-^^_ = ^-^, to find the 
 
 value of z. Ans. a: = 4. 
 
 / 18a:— 19 , lla: + 21 9a:+15 , . , . 
 
 5. Given _^^ + _^_ = _^, to find the 
 
 value of X. Ans. a: = 7. 
 
 PROBLEMS AND EQUATIONS OF THE FIRST DEGREE WITH TWO 
 UNKNOWN QUANTITIES. 
 
 47. Most of the questions we have hitherto considered, in- 
 volve more than one unknown quantity. We have been able 
 to solve them, however, by representing one of the unknown 
 quantities only by a letter, since, by means of this, it has been 
 easy, from the conditions of the question, to express the other 
 unknown quantity. In many questions the solution becomes 
 more simple by representing more than one of the unknown 
 quantities by a letter, and in complicated questions, it is fre- 
 quently necessary to do this. ' 
 
 The question, art. 1. viz. To divide the number 56 into two 
 suck parts that the greater may exceed the less by 12, presents 
 itself naturally with two unknown quantities. Thus, denoting 
 the less part by x and the greater by y, we have by the con- 
 ditions of the question 
 
 x~\"y = 5Q 
 y-.x=\2. 
 
 Deducing the value of y from the second equation, we have 
 y=zx-\-12; substituting for y in the first equation its value 
 X -j- 12, we have x -\- x -\- 12 z= 56, an equation, which con- 
 tains only one unknown quantity, and from which we obtain 
 a: = 22. 
 
EQUATIONS OF THE FIRST DEGREE. W 
 
 2. A person has two horses and a saddle, which of itself is 
 worth $ 10. If the saddle be put upon the first horse, his value 
 will be twice the second; but if the saddle be put upon the 
 second, his value will be three times the first. What is the 
 value of each ? 
 
 Let X = the value of the first horse, and y that of the second, 
 we have by the question 
 
 a:-[-10 = 2y 
 y4-10 = ^«. 
 
 Deducing the value of y from the second of diese equations, 
 and substituting it for y in the first, we have 
 
 a;-f.l0 = 6a: — 20; 
 whence a: = 6. ' 
 
 Substituting next for x its value 6 in the second equaticm, 
 we have z/ -[" 1^ = 1® ' 
 
 whence 2/ = S* 
 
 The process by which one of the unknown quantities in an 
 equation is made to disappear, is called elimination. The 
 method of eliminating one of the unknown quantities, pursued 
 above, is called elimination by substitution. 
 
 48. Since the two members of an equation are equal quanti- 
 ties, it is evident, P. that we may add two equations, Tnember to 
 member, without destroying the equality ; 2°. we may subtract 
 the members of one equation from those of another without destroy- 
 ing the equality. 
 
 Taking advantage of this remark, we may frequently elimi- 
 nate one of the unknown quantities in a more simple manner, 
 than by the process of substitution. 
 
 Let there be proposed, for example, the equations 
 5a; + 7y = 43 
 lla: + 9y = 69. 
 
 If either of the unknown quantities in these equations were 
 aflfected with the same coefficient, we might, it is evident, 
 eliminate this unknown quantity by a simple subtraction. But 
 
'58 ELEMENTS OF ALGEBRA. 
 
 if the first equation be multiplied by 9, the coefficient of y in 
 the second, and the second by 7, the coefficient of y in the 
 first, we shall obtain two new equations, which may be sub- 
 stituted for the proposed, and in which the coefficient of y will 
 be equal, viz. 
 
 45a; + 63?/ = 387 
 77 a: + 63?/ = 483. 
 Subtracting then the first of these equations from the second, 
 we have 32 a; = 96, from which we obtain a; = 3. Substi- 
 tuting this value of z in either of the proposed we obtain the 
 value of y. 
 
 In like manner, if we wish first to eliminate a:, we multiply 
 the first of the proposed equations by 11, the coefficient of x 
 in the second, and the second by 5, the coefficient of x in the 
 first ; we thus obtain two new equations, which may be substi- 
 tuted for the proposed, and in which the coefficients of x will 
 be equal, viz. 
 
 55 a: + 77^ = 473 
 55 a: +45?/ = 345. 
 Subtracting therefore the second of these equations from the 
 first, we have 32?/ = 128 ; whence ?/ = 4. 
 Let us take as a second example the equations 
 8a: — 21?/ = 33 
 6a: + 35?/ = 177. 
 The coefficients of x in these equations have, it will be per- 
 ceived, a common factor 2. It will be sufficient therefore, in 
 order to render these coefficients equal, to multiply the first 
 equation by 3 and the second by 4. Performing the operations 
 we have 
 
 24a: — 63 2/ = 99 
 24a; +140?/ = 708; 
 whence, subtracting the first of these equations from the second 
 
 we obtain 
 
 203y = 609; 
 therefore 2/ = 3. 
 
EQUATIONS OF THE FIRST DEGREE. 59 
 
 In like manner, since the coefficients of y contain the common 
 factor 7, in order to render the coefficients of y equal we multi- 
 ply the first of the proposed equations by 5 and the second by 3, 
 which gives two new equations, 
 
 40 a;— 1052/ =165 
 18a: +1052/ = 531; 
 whence by addition we obtain 
 
 58 a: = 696; 
 therefore x = 12. 
 
 49. The method of elimination, which we have now ex- 
 plained, is called elimination by addition and subtraction, since, 
 the equations being properly prepared, we cause one of the 
 unknown quantities to disappear by addition or subtraction. 
 
 In the use of this method, it is important to ascertain whether 
 the coefficients have common factors, since, if this be the case, 
 by omitting the common factors in the multiplications required, 
 the calculations to be performed become more simple. The 
 equations, moreover, should be reduced to the form of the pre- 
 ceding examples, — that is, they should be freed from denomi- 
 nators, the unknown quantities collected each into one term on 
 one side of the sign of equality, and the known quantities col- 
 lected in one term on the other. 
 
 EXAMPLES. 
 
 1. To find the values of x and y in the equations 
 
 4a: — 3t/=l 
 3a:-j-42/ = ^. 
 
 2. To find the values of x and y in the equations 
 
 4a: — 9?/ = 51 
 8a: + 132/ = 191. 
 3» To find the values of x and y in the equations 
 
 82/ — 3a: = 29 
 
 67y — 4a; = 20 
 
•0 »I,EM(ENTS Q? AW31SBRA. 
 
 4. To find the values of ^ and y in the equatious 
 
 Ans. x=l2,y=il6, 
 
 5. To find the values of x and y in the equations 
 
 a; + 2 
 
 3 
 
 y + 5 
 
 82/ = 31 
 
 ^ + 10a; =192. 
 4 ' 
 
 Ana. a:=3sI9, ysBB^k 
 
 6. To find the values of x and y in the equations 
 
 2^ + 32,-4=15 
 
 Ans. 3; = 7, 1/ = 5. 
 
 7. To find the values of x and y in the equations 
 
 Ans. a; = 3, y = 2, 
 
 8. To find the values of x and y in the equations 
 
 37/ + 4a: _ 9y + 33 
 '*'"+"^ 7 ~ 14 
 
 „ 5a: — 4y llw— 19 
 
 Ans. a; 5=? 6> 2^ == 5. 
 
 50. We pass next to the solution of some questions producing 
 equations involving two unknown quantities. 
 
 1. A number consisting of two figures when divided by 4, 
 gives a certain quotient, and a remainder of 3; when divided 
 
eqitatkWS of the first begree. 61 
 
 by 9 gives another qaolient and a remainder of 8. The value 
 of the figure on the left hiand is equal to the quotient obtained, 
 when the number was divided by 9, and the other figure is equal 
 to tV of ^^® quotient obtained, when the number was divided 
 by 4. Required the number. 
 
 Let a: = the figure in the place of tens, y that in the place 
 of units; then 10^-{"2/ = *^6 number, and we have by the 
 question 
 
 Deducing the values of x and y from these equations, we 
 obtain a: == 7, y = 1. The number required is therefore 71. 
 
 2. A purse holds 19 crowns and 6 guineas. Now 4 crowns 
 and 5 guineas fill ^^ of it. How many will it hold of each ? 
 
 Let z = the number of crowns and y = the number of 
 
 guineas, then ~ = the space occupied by a crown, and - = 
 
 the space occupied by a guinea, we have therefore by the ques*- 
 tion 
 
 19 ,6 ^ ,4,5 17 
 
 f- - = 1, and - + - = ^. 
 
 sc ' y z y o3 
 
 Multiplying the first equation by 5 and the second by 6, subtract- 
 ing the second from the first and reducing, we obtain z = 21, 
 whence y = 63. 
 
 3. What fraction is that, whose numerator being doubled, and 
 denominator increased by 7, the value becomes §j but the de- 
 nominator being doubled, and the numerator increased by 2, the 
 value becomes f ? Ans. f. 
 
 4. A owes $1200, B $2500; but neither has enough to pay 
 his debts. Lend me, said A to B, the eighth part of your 
 fortune, and I shall be enabled to pay my debts. B answered, 
 I cin discharge my debts, if you will lend me the 9th part of 
 youjfs. What was the fortune of each? 
 
 Ans. A's $900, B's $2400. 
 
 F 
 
62 ELEMENTS OF ALGEBRA. 
 
 5. A farmer with 28 bushels of barley at 2s. M. per bushel, 
 would mix rye at 3 shillings per bushel, and wheat at 4 shillings 
 per bushel, so that the whole mixture may consist of 100 bushels, 
 and be worth 35. 4^. per bushel. How many bushels of rye, and 
 how many of wheat must he mix with the barley ? 
 
 Ans. 20 bushels of rye and 52 bushels of wheat. 
 
 6. A and B speculate with different sums; A gains $150, 
 B loses $50, and now A's stock is to B's as 3 to 2. But if A 
 had lost $50 and B gained $100, then A's stock would have 
 been to B's as 5 to 9. What was the stock of each ? 
 
 Ans. A's $300 and B's $350. 
 
 7. A rectangular bowling green having been measured, it 
 was observed, that if it were 5 feet broader, and 4 feet longer, it 
 would contain 116 feet more; but if it were 4 feet broader, and 
 5 feet longer, it would contain 113 feet more. Required the 
 length and breadth. 
 
 Let a; = the length, ?/ = the breadth, then xy=zth.e content, 
 and by the first condition (x -|- 4) {y -\- 5) = xy -^ 116, &c. 
 Ans. The length was 12 and the breadth 9 feet. 
 
 8. There is a number consisting of two figures, the figure in 
 the place of units being the greater ; if the number be divided by 
 the sum of its figures, the quotient is 4; but if the figures be 
 inverted, and the number which results be divided by a number 
 greater by 2 than the difference of the figures, the quotient 
 becomes 14. Required the number. Ans. 48. 
 
 9. A person has two horses and two saddles, one of which 
 cost $50, the other $2. If he places the best upon the first 
 horse, and the worst upon the second, then the latter is worth 
 $8 less than the other; but if he puts the worst saddle upon 
 the first, and the best upon the second horse, then the latter is 
 worth 3f times as much as the former. What is the value of 
 each horse ? Ans. The first $30, the second $70. 
 
 10. A cistern containing 210 buckets, may be filled by 2 
 
EQUATIONS OF THE FIRST DEGREE. 63 
 
 pipes. By an experiment, in which the first was open 4 and 
 the second 5 hours, 90 buckets of water were obtained. By 
 another experiment, when the first was open 7, and the other 3^ 
 hours, 126 buckets were obtained. How many buckets does 
 each pipe discharge in an hour ? 
 
 Ans. The first 15, and the second 6 buckets. 
 
 11. A person having laid out a rectangular bowling green, 
 observed that if each side had been 4 yards longer, the adjacent 
 sides would have been in the proportion of 5 to 4, but if each 
 had been 4 yards shorter, the proportion would have been 4 to 3. 
 What are the lengths of the sides ? Ans. 36 and 28 yds. 
 
 12. A vintner has two casks of wine, from the greater of 
 which he draws 15 gallons, and from the less 11 ; and finds the 
 quantities remaining in the proportion of 8 to 3. After the casks 
 become half empty, he puts 10 gallons of water into each, and 
 finds that the quantities of liquor now in them are as 9 to 5. 
 How many gallons will each hold ? 
 
 Ans. The larger 79 and the smaller 35 gallons. 
 
 13. Two persons, A and B, can perform a piece of work in 
 16 days. They work together for 4 days, when A being called 
 oflf, B is left to finish it, which he does in 36 days more. In 
 what time would each do it separately ? 
 
 Ans. A in 24 and B in 48 days. 
 
 14. A work is to be printed, so that each page may contain 
 a certain number of lines, and each line a certain number of 
 letters. If we wished each page to contain 3 lines more, and 
 each line 4 letters more, then there would be 224 letters more in 
 each page ; but if we wished to have 2 lines less in a page, and 
 3 letters less in each line, then each page would contain 145 
 letters less. How many lines are there in each page ? and how 
 many letters in each line ? 
 
 Ans. There are 29 lines in a page and 32 letters in a line. 
 
 15. There is a number consisting of two digits, which is 
 
■9fm ELEMENTS OF ALGEBRA. 
 
 equal to four times the sum of those digits ; and if 18 be added 
 to it, the digits will be inverted. What is the number? 
 
 Ans. 24. 
 
 16. To find a fraction such, that if 3 be subtracted from the 
 numerator and denominator, it is changed into ^^ but if 5 be 
 added to the numerator and denominator it becomes J. What 
 is the fraction? Ans. -f^. 
 
 17. There is a cistern, into which water is admitted by three 
 cocks, two of which are of exactly the same dimensions. When 
 they are all open, five-twelfths of the cistern is filled in four 
 hours; and if one of the equal cocks be stopped, seven-ninths 
 of the cistern is filled in ten and two-thirds hours. In how 
 many hours would each cock fill the cistern? 
 
 Ans. Each of the equal ones in 32 
 hours and the other in 24. 
 
 18. A person owes a certain sum to two creditors. At one 
 time he pays them $53, giving to one four-elevenths of the 
 sum due to him, and to the other $3 more than one-sixth of 
 his debt to him. At a second time he pays them $42, giving 
 to the first three-sevenths of what remains due to him, and to 
 the other one-third of what is due to him. What were the 
 debts? Ans. $121 and $36. 
 
 PROBLEMS AND EQUATIONS OF THE FIRST DEGREE WITH THREE 
 OR MORE UNKNOWN QUANTITIES. 
 
 51. Let now the following question be proposed, viz. 
 
 There are three persons. A, B, and C, whose ages are as 
 follows ; If from 4 times A's age added to 5 times B's age, we 
 subtract three times C's age, the remainder will be 70; if from 3 
 times A's age we subtract 4 times B's age, and to the remainder 
 add twice C's age, the sum will be 25; and if twice A's age, 3 
 times B's, and 5 times C's age be added together, the sum will 
 be 240 What is the age of each ? 
 
EQUATIONS OF THE FIRST DEGREE. 0S 
 
 This question presents itself naturally with three unknown 
 quantities. Thus denoting A's age by a:, B's age by y, and C's 
 by Zy we have by the question 
 
 4a; + 5y — 32r = 70 
 3a; — 4y + 2z = 25 
 2a; + 3y + 5z = 240. 
 
 Multiplying the first equation by 2, and the second by 3, and 
 adding the results, we obtain 
 
 17x — 2y=:2l5. 
 
 Again, multiplying the second equation by 5, and the third by 
 2, and subtracting, we obtain, 
 
 — 11a: + 262/ = 355. 
 
 We have now two equations with two unknown quantities only. 
 Deducing next the values of x and y from these, in the same 
 manner as in the preceding equations with two unknown quanti- 
 ties, we have a; = 15, y = 20 ; substituting these values in the 
 first of the proposed equations, we obtain z = 30. 
 
 52. In the same manner, if there be four equations, with four 
 unknown quantities, we combine the equations two by two, until 
 one of the unknown quantities is eliminated from the whole; 
 we then have three equations with three unknoAvn quantities. 
 Combining next these three, two by two, until one of the un- 
 known quantities is eliminated, we obtain two equations with 
 two unknown quantities, and so on. The process is altogether 
 similar for five or more equations with the same number of 
 unknown quantities. 
 
 EXAMPLES. 
 
 1. To find the values of x, y, and z in the equations 
 5a; — 6y + 42r=15 
 7a;-|-4y — 32=19 
 2x-f- y+62:r=:46. 
 , Ans. a: r= 3, y s= 4, z = 6. 
 
 6 
 
66 ELEMENTS OF ALGEBRA. 
 
 % To find the values of x, y, and z in the equations 
 2x-]-4:y — 3z=^22 
 4.x^2y + 5z=18 
 ex + 7y— z = 63. 
 
 Ans. a; = 3, 2/ = 7, z = 4. 
 
 3. To find the values of x, y and z in the equations 
 
 3x + 5y + 7z=179 
 Sx-\'3y — 2z= 64 
 5x— y-\-3z=. 75. 
 
 Ans. a; = 8, 2/ = 10, 2 = 15. 
 
 4. To find the values of x, y and z in the equations 
 
 3x-\-2y — Az=m 8 
 5x — 3y-\-3z = 33 
 7x+ y + 5z = 65. 
 
 Ans. a; = 6, 2/ = 3, 2: = 4. 
 
 5. To find the values of a:, y, 2r, and u in the equations 
 
 a:-(~ y~{~ ^ — u = 5 
 2x-\-3y — 2z-^ M==2 
 5x—2y-\- z — 2u = 9 
 3X+ y — 2z-\- u = 2. 
 
 Ans. a; = 2, 2/=l, z = 3, M=l. 
 53. It sometimes happens, that all the unknown quantities are 
 not found in each of the equations. In this case, the elimination 
 may, with a little attention, he very readily performed. 
 
 1. Let there be proposed, for example, the four following^ 
 equations, with four unknown quantities, viz. 
 2x — 3y-\-2z=\3 
 4w — 2a: = 30 
 42/ + 2z=14 
 5y+ 32^=32. 
 With a little examination we see, that the elimination of z 
 from the first and third equations will give an equation in z and 
 
EQUATIONS OF THE FIRST DEGREE. 6Jf 
 
 y, and that the elimination of u from the second and fourth equa- 
 tions will also give an equation in x and y. From these last the 
 values of z and y may be readily found. Performing the neces- 
 sary operations we obtain a: = 3, y = 1. Substituting next for 
 z its value in the second equation, we have w = 9, and substi- 
 tuting for y its value in the third, we have 2; = 5. 
 
 2. To find the values of a:, y, z and u in the following equa- 
 tions. 
 
 3a:— y + 2z = 7 
 5x-\-2y — u==5 
 2z—Sy-{-2z = 2 
 
 7y— 3w = 2. 
 
 Ans. a:=l, y = 2, z=3, M = 4. 
 
 3. To find the values of the unknown quantities in the follow- 
 ing equations 
 
 5z— y — 3z = 8 
 Sy — 2z-\- t = 
 z-\-2y—r z = 3 
 5y — u-{'5t = 5 
 4?^+ M = 9. 
 Ans. a: = 3, 2/= 1, z=2, w = 5, ^=1. 
 54. We pass next to some questions producing three equations 
 with three unknown quantities. 
 
 1. Three laborers are employed in a certain work. A and B 
 together can perform it in 8 days,' A and C together in 9 days, 
 and B and C together in 10 days. In how many days can each 
 alone perform the same work ? 
 
 Let a;, y and z represent the number of days respectively, 
 then, in one day A will do one-a:th part of it, B one-yth part, 
 and C one-zth part of it, and we shall have for the equations of 
 the question, 
 
 z ^ y X ^ z y z 
 
6P ELEMENTS OF ALGEBRA. 
 
 Dividing the first of these equations by 8, the second by 9, 
 and the third by 10, we have 
 
 1 , i__i 1 i_i 1 , i_2.. 
 
 a;~^y"~"8' x'^z^d' y"^z""10' 
 subtracting the second equation from the first, and adding the 
 third to the remainder, and reducing, we obtain y == 17|f ; and 
 in like manner we find x = 14f |, z = 23/i-. 
 
 2. It is required to find three numbers, such, that one-half of 
 the first, one-third of the second, and one-fourth of the third shall 
 together make 46 ; one-third of the first, one-fourth of the second 
 and one-fifth of the third shall together make 35 ; and one-fourth 
 of the first, one-fifth of the second and one-sixth of the third, 
 shall together make 28^. 
 
 Ans. 12, 60, and 80. 
 
 3. Three brothers purchased an estate for $15,000, and the 
 first wanted, in order to complete his part of the payment, half 
 of the property of the second ; the second would have paid his 
 share with the help of a third of what the first owned ; and the 
 third required, to make the same payment, in addition to what 
 he had, a fourth part of what the first possessed ; what was the 
 amount of each one's property ? 
 
 Ans. S3,000, $4,000, and $4,250, respectively., 
 
 4. Three persons, A, B, and C, compare their fortunes, A 
 says to B, give me $700 of your money, and I shall have twice 
 as much as you retain ; B says to C, give me $1400, and I shall 
 have thrice as much as you have remaining ; C says to A, give 
 me $420, and then I shall have 5 times as much as you retain. 
 How much has each ? 
 
 Ans. A $980, B $1540, C $2380. 
 
 5. Three men, A, B, C, driving their sheep to market, says A 
 to B and C, if each of you will give me 5 of your sheep, I shall 
 have just half as many as both of you will have left. Says B 
 to A and C, if each of you will give me 5 of yours I shall have 
 just as many as both of you will have left. Says C to A and 
 
EQUATIONS OF THE FIRST DEGREE. ^ 
 
 B, if each of you will give me 5 of yours I shall have just twice 
 as many as both of you will have left. How many had each ? 
 
 Ans. 10, 20, and 30 respectively. 
 
 6. A cistern is furnished with three pipes, A, B and C. By 
 the pipes A and B it can be filled in 12 minutes, by the pipes B 
 ^nd C in 20 minutes, and by A and C in 15 minutes. In what 
 time will each fill the cistern alone, and in what time will it be 
 filled if all three are open together ? 
 
 Ans. A will fill it in 20, B in 30, and C in 60 minutes, 
 and the three together in 10 minutes. 
 
 7. It is required to divide the number 72 into four such parts, 
 that if the first part be increased by 5, the second part diminished 
 by 5, the third part multiplied by 5, and the fourth part divided 
 by 5, the sum, difference, product and quotient shall all be 
 e(^ual. Ans. The parts are 5, 15,. 2 and 50. 
 
 SECTION VI.— Negative Quantities. Questions producing 
 Negative Results. 
 
 55. The length of a certain field is eight rods and the breadth 
 five rods, how much must be added to the length, that the field 
 may contain 30 square rods ? 
 
 Let z = the quantity to be added, then by the question 
 40 + 5a; =30, 
 and 5a; = 30 — 40, * 
 
 or dividing by 5 a; =; 6 — 8. 
 
 In this result 8, the quantity to be subtracted, is greater than 
 that, from which it is required to be taken ; the subtraction there- 
 fore cannot be performed. We may, however, decompose 8 into 
 
7P ELEMENTS OF ALGEBRA. 
 
 two parts 6 and 2, the successive subtraction of which will be 
 equal to that of 8, and we shall have for 6 — 8 the equivalent 
 expression, 6 — 6 — 2, which is reduced to — 2 or more simply 
 — 2, the sign — being retained before the 2 to show that it 
 remains to be subtracted. 
 
 A monomial with the sign — prefixed is called a Tiega- 
 five quantity, thus, — 2, — 3 a, — 5 a 3, are negative quanti- 
 ties. 
 
 Monomials with the sign -f- either prefixed or understood 
 are called positive quantities. Thus, 2, 3 a, 5 ah are positive 
 quantities. 
 
 Negative quantities, it will be perceived, differ in nothing 
 from positive quantities except in their sign. They are derived 
 from endeavoring to subtract a larger quantity from one that is 
 smaller, and are to be regarded merely as positive quantities to 
 be subtracted. 
 
 56. If it now be asked what is the sum of the monomials 
 -j- a, — 3, -|- c, the question, from what has been said, is reduced 
 to this, what change will be produced in the quantity a, if the 
 quantity b be subtracted from it and the quantity c be added to 
 the remainder. Indicating the operations required to obtain the 
 answer to the question thus proposed, the result will be 
 
 a — b-\-c. 
 In order then to add monomials aflfected with the signs -f" 
 and — , it will be sufficient to write them one after the other 
 with the signs with which they are affected^ when they stand 
 alon£. 
 
 57. If we now add the quantities -(- i, — by the result b — bf 
 it is evident, will be equal to zero. If then the expression b — b 
 be added to a, it will not affect the value of a ; and «--[-& — b 
 will only be a different form of expression for the same quantity 
 a. If it now be proposed to subtract -|- b from a, it will be 
 sufficient, it is evident, to efface -}- ^ in the equivalent expression 
 a-j-i — b, and the result will be a — b. Ag-ain, if it be 
 
EQUATIONS OF THE FIRST DEGREE. 71 
 
 required to subtract — * h from a, it will be sufficient to efface — b 
 m the same expression, and we shall have for the result a-\-h. 
 Thus, to subtract a positive quantity is the same as to add an 
 equal negative quantity, and to subtract a negative quantity is 
 the same as to add an equal positive quantity. To subtract mo- 
 nomials therefore of whatever sign, we change the signs^ and 
 then proceed as in addition. 
 
 58. If we multiply b — bhj a, the product must he ab — ab^ 
 because the multiplicand being equal to zero, the product must 
 be zero. Since then the product of 3 by a is evidently ab^ 
 that of — bhj a must be — a 3, in order that the second term 
 may destroy the first. For a similar reason the product of a by 
 b — b will be ab — ab. Whence if a negative quantity be mul' 
 tiplied by a positive, or a positive by a negative, the product wilt 
 be negative. 
 
 Again, if we multiply — a by 3 — b, from what has been 
 proved above, the product of — ahj b will be — ab, the product 
 of — 3 by — a must therefore be -\- ab, in order that the result 
 may be zero, as it should be, when the multiplier is zero. 
 Whence, the product of a negative quantity by a negative 
 quantity will be positive. 
 
 The rules for division follow necessarily from those for multi- 
 plication. We have therefore the same rules for the signs in 
 the multiplication and division of isolated simple quantities, as 
 are applied to these quantities, when they make a part of poly- 
 nomials ; and in general, monomials, when they are isolated are 
 combined in the same manner with respect to their signs, as when 
 they make a part of polynomials. 
 
 59. From what has been said, it will be perceived, that the 
 term addition does not in algebra, as in arithmetic, always imply 
 augmentation. Thus, the sum of a and — b is, strictly speaking, 
 the difference between a and b ; it will therefore be less than a. 
 To distinguish this from an arithmetical sum, we call it an 
 algebraic sum. Thus the polynomial 
 
 Sab — 5bc.-\-cd — ef, 
 
72 ELEMENTS OF ALGEBRA. 
 
 considered as formed by uniting the quantities 
 Sab, — 53c, -|- cd, — ef 
 with their respective signs, is called an algebraic sum. lu 
 proper acceptation is the arithmetical difference between the 
 sum of the units contained in the terms, which are additive, 
 and the sum of those contained in the terms, which are sub- 
 tractive. 
 
 In like manner the term subtraction in algebra does not 
 always imply diminution. Thus — b subtracted from a gives 
 a-\-b, which is greater than a. This result may, however, be 
 called an algebraic difference, since it may be put under the form 
 a-{-b). 
 
 60. Resuming now the question proposed, art. 55, we havQ 
 for the answer x = — 2. In order to interpret this negativf 
 result, we return to the equation of the question 40 -|- 52; = 30 
 Here, the addition intended in the enunciation of the questioij 
 being arithmetical, it is evidently absurd to require that some 
 thing should be added to 40 in order to make 30, since 40 iar 
 already greater than 30. The negative result indicates, there- 
 fore, that the question is arithmetically impossible, or in other 
 words, that it cannot be solved in the exact sense of the enuncia- 
 tion. If, however, in the equation 40 -f- 5 a: = 30, we substitute 
 — 2 for z, we have 40 — 10 = 30, an equation which is exact. 
 In order then that the result may be positive, or which is the 
 same thing, that the question may be arithmetically possible, the 
 enunciation should be modified thus. 
 
 The length of a certain field is eight rods, and its breadth five 
 rods ; how much must be subtracted from the length, that the 
 field may contain 30 square rods ? 
 
 Putting X for the quantity to be subtracted, we have by this 
 new enunciation 40 — 5 2: = 30, from which we obtain a: = 2. 
 
 2. The length of a certain field is 11 rods and its breadth 7 
 rods ; how much must be subtracted from the length, that the 
 field may contain 98 square rods ? 
 
EQUATIONS OF THE FIRST DEGREE. tJI 
 
 Let a;= the quantity to be subtracted ; then by the question 
 77 — 7a: = 98; 
 whence a: = — 3. 
 
 To interpret this resuh, we return to the equation of the 
 question. Here, as an arithmetical subtraction is intended in 
 the enunciation, it is evidently absurd to require, that some- 
 thing should be subtracted from 77 to make 98, since 77 is 
 already less than 98. The question therefore cannot be solved 
 in the exact sense of the enunciation. If, however, instead 
 of X in the equation of the question, we substitute — 3, we 
 have 77 -|- 21 = 98, an equation which is exact. In order 
 then that the result may be positive, the question should be 
 modified, thus, 
 
 The length of a certain field is 11 rods and the breadth 7 rods; 
 how much must be aMedi to the length in order that the field 
 may contain 98 square rods ? 
 
 Resolving the question according to this new enunciation, we 
 obtain a; = 3. 
 
 Let us take as a third example the following question. 
 
 3. A laborer wrought for a person 12 days and had his wife 
 and son with him 7 days, and received 46 shillings. He after- 
 wards wrought 8 days, having his wife and son with him 5 days, 
 and received 30 shillings ; how much did he earn per day him- 
 self, and how much did his wife and son earn ? 
 
 Let X a= the daily wages of the man, and y that of his wife 
 and son ; we have by the question 
 
 12a; + 72/ = 46 
 8a;-J-oy = 30. 
 
 Resolving these equations, we obtain a; = 5, y = — 2. 
 
 In order to interpret this negative result, we substitute 5 for x 
 in the equations above, by which we have 
 60 + 7?/== 46 
 40 + 5^ = 30, 
 
 G 
 
74 ELEMENTS OF ALGEBRA. 
 
 equations which are evidently absurd, since it is required to 
 add something to 60 in order to make 46, and to 40 in order to 
 make 30. If, however, we substitute — 2 for y in these last we 
 have 
 
 60—14 = 46 
 40 — 10 = 30, 
 equations which are exact. The negative value therefore obtained 
 for y, shows that the allowance made for the wife and son instead 
 of augmenting the pay of the laborer, should be regarded as a 
 charge placed to his account. The question therefore should be 
 modified, thus, 
 
 A laborer wrought for a person 12 days and had his wife and 
 son with Jiim 7 days at a certain expense^ and received 46 shil- 
 lings. He afterwards wrought 8 days, having his wife and son 
 with him 5 days at expense as before, and received 30 shillings. 
 How much did the laborer earn per day, and how much was 
 charged him per day on account of his wife and son ? 
 
 Resolving the question, thus stated, we have 
 a:=5,2/ = 2. 
 
 61. From what has been done, it will be perceived, that in 
 problems of the first degree, a negative result indicates some 
 inconsistency in the enunciation of the question, arithmetically 
 considered, and at the same time shows how this inconsistency 
 may be reconciled by rendering subtractive certain quantities, 
 which had been regarded as additive, or additive certain quanti- 
 ties, which had been regarded as subtractive. 
 
 Negative results, however, in the extended sense, in which 
 the terms addition and subtraction are used in algebra, may 
 be regarded as answers to questions. Thus, in the equation 
 40 -[- 5a: == 30, the negative result — 2 shows that it is necessa- 
 ry to add — 10 to 40 to obtain 30. By means of this exten- 
 sion of the meaning of the terms, addition and subtraction, 
 we may regard as one single question, those, the enunciations 
 of which are such, that the solution, which satisfies one of 
 
EQUATIONS OF THE FIRST DEGREE. 75 
 
 the enunciations, will by a mere change of sign satisfy the other 
 also. 
 
 62. The following examples will serve as an exercise in the 
 interpretation of negative resuUs. 
 
 1. A father is 55 years old, and his son is 16. In how many 
 years will the son be one-fourth as old as the father ? 
 
 2. What number is that, whose fourth part exceeds its third 
 part by 12 ? 
 
 3. There are two numbers such, that if twice the second be 
 added to the first, the sum will be 20, but if 3 times the second 
 be subtracted from the first, the difference will be 45. What are 
 the numbers ? 
 
 4. To divide the number 40 into two such parts, that if the 
 first be multiplied by 7 and the second by 5, the sum of the 
 products will be 90. 
 
 5. What number must be subtracted from the numbers 70 
 and 50 respectively in order that their differences may be as 4 
 to 3? 
 
 6. Three persons comparing their property, it is found, that 
 A's and B's together amount to $1000, A's and C's to $480, and 
 B's and C's to $400. What amount of property has each ? 
 
 7. A laborer wrought for a gentleman 10 days, having his 
 wife with him 5 days and his son 4 days, and received $14,25. 
 At another time he wrought 8 days, having his wife with him 6 
 days and his son 3 days, and received $13. At a third time he 
 wrought 6 days, having his wife with him 4 days and his son 5 
 days, and received $8,00. How much did he receive a day 
 himself, and how much for his wife and son severally ? 
 
76 ELEMENTS OF ALGEBRA. 
 
 SECTION VII.— Indeterminate Analysis. 
 
 63. Let it be proposed to find two numbers such, that the first 
 added to three times the second shall be equal to 15. 
 
 Putting z and y for the numbers sought, we have by the 
 question 
 
 a: + 32/=15; 
 here as we have two unknown quantities and but one equation, 
 the particular numbers intended in the question proposed cannot 
 be determined. Deducing, however, from the equation the value 
 of one of the unknown quantities, x for example, we have 
 
 a:=15 — Sy. 
 If we now assume arbitrarily any values whatever for y, we shall 
 obtain values for a:, which will satisfy the equation, 
 
 thus, let 2/== 1, IJ, 2, SJ 
 
 webave a:= 12, lOJ, 9, 8 
 
 or otherwise t/ = — 1, — 1 J, — 2, — 2| 
 
 wehave a;= 18, 19J, 21, 22 
 
 pairs of values for x and y^ which, it is easy to see, will satisfy 
 the equation, and the number of w^hich may be increased without 
 limit. 
 
 In general, if the conditions of a problem furnish fewer 
 equations, than there are unknown quantities to be determined, 
 the equations of the problem will admit of an infinity of systems 
 of values for the unknown quantities, if we understand by these 
 any quantities whatever, entire or fractional, positive or negative. 
 It is frequently the case, however, that the nature of the question 
 requires, that the values of the unknown quantities should be 
 entire numbers. This circumstance, it is evident, will very 
 much restrict the number of solutions, especially if we reckon 
 the direct solutions only, that is to say, solutions in entire and 
 positive numbers. 
 
 Thus, if in the question proposed the numbers sought are 
 
INDETERMINATE ANALYSIS. 77 
 
 required to be entire and positive, the value of y, it is evident, 
 must not exceed 5 ; if then we put successively for y 
 
 2^ = 0, 1,2,3,4,5 
 we have x = 15, 12, 9, 6, 3, 0, 
 
 and the question admits of six different solutions only, the 
 solution in which is reckoned as a value of one of the unknown 
 quantities being included. 
 
 Problems of the kind, which we are here considering, are 
 called indeterminate problems^ and that part of algebra, which 
 relates to the solution of indeterminate problems, is called inde- 
 terminate analysis. 
 
 64. The preceding question, in which the coefficient of one 
 of the unknown quantities is equal to unity, presents no diffi- 
 culty. We shall now show, that whatever the coefficients of the 
 unknown quantities, the solution of the question proposed may 
 be made to depend upon the solution of an equation, in which 
 the coefficient of one of the unknown quantities is equal to 
 unity. 
 
 Let it be proposed then to find the entire values of x and y m 
 the equation 17 a: = 542 — \\y. 
 
 Deducing from this equation the value of y, we have 
 
 542— 17 a: 
 
 or performing the division as far as possible, we have 
 
 ._ , 3 — 6a; 
 
 y==49~a:+-yp-. 
 
 But, by the question the values of x and y should be entire 
 
 numbers, it is necessary^ therefore, and it is sufficient ^ that — rrr— 
 
 should be equal to an entire number. Let t be this number 
 {t is called an indeterminate,) we have 
 y = 49 — a; + ? 
 1U= 3 — 6ar. (2) 
 
78 ELEMENTS OF ALGEBRA. 
 
 and the question is now reduced to resolve in entire numbers 
 the equation (2), the coefficients of which are more simple 
 than those of the proposed. Deducing from this equation the 
 value of X and performing the division as far as possible, we 
 have 
 
 Here, since x and t are entire numbers — - — must be equal 
 
 to an entire number ; let t' be this number, the letter t being 
 marked with an accent to show that it represents a quantity 
 different from that before represented by it, we have 
 
 6f = 3 — 5t, (3) 
 And the question is still further reduced to resolve in entire 
 numbers the equation (3), the coefficients of which are more 
 simple than those of equation (2). Deducing from this equation 
 the value of t, we have 
 
 2 f 
 
 but t and f in this equation are entire numbers ; — ^ — must 
 
 therefore be equal to an entire number ; let f be this number, 
 we have t = —t.'-^t" 
 
 or f = S — 5t", (4) 
 
 and the question is now reduced to resolve in entire numbers 
 the equation (4), in which the coefficient of one of the unknown 
 quantities t' is equal to unity. Indeed, the two principal unknown 
 quantities and the several indeterminates employed are, it is 
 evident, connected together by the equations 
 y = 49 — a; + ^ 
 
 X=:'—t']-f 
 
INDETERMINATE ANALYSIS. 79 
 
 if then we give any entire value whatever to t" and return to 
 the values of x and y corresponding, the values thus found, it is 
 evident, will be entire numbers, and will satisfy the equation 
 proposed. Thus, let ^" = 1, we have a: = — 5, 2/ = 57, values 
 which, it is easy to see, will satisfy the equation proposed. 
 
 To determine with more facility the values of t" which will 
 give entire values for x and y, we express x and y immediately 
 in teirms of t". In order to this we substitute for t' its value in 
 the equation for ?, which gives 
 
 ? = — (3 — 50 + ^" = 6r — 3. 
 
 Substituting next for t and f their values in the equation for a:, 
 and for x and t their values in the equation for y, we obtain 
 finally a; = 6— llif" 
 
 2^ = 40-(-17r. 
 If then we make successively ^"=0, 1, 2, 3, . . . or other- 
 wise, ?"=0, — 1, — 2, — 3, ... in the above, we shall ob- 
 tain all the entire values of x and y proper to satisfy the equa- 
 tion proposed. But if entire and positive values only are re- 
 quired for X and y, it will be necessary to give to t" such values 
 only, as will render 6 — 11 if", 40 -j- 17 if" positive. It is evi- 
 dent, that ^" = 0, t" = — 1, ^" = — 2, are the only values of 
 t", that will fulfil this condition ; for, every positive value of 
 t" will render x negative, and every negative value of t" nu- 
 merically greater than 2, willvrender y negative. Putting there- 
 fore if"= 0, — 1, — 2 successively, we have 
 
 a; = 6, 17,28 
 2, = 40, 23, 6. 
 
 The proposed therefore admits of three different solutions in 
 entire and positive numbers, and of three only. 
 
 2. Let it be proposed, as a second example, to divide the 
 number 159 into two such parts, that the first may be divisible 
 by 8, and the second by 13. 
 
 Designating by x and y the quotients, arising from the division 
 
80 ELEMENTS OF ALGEBRA. 
 
 of the parts sought by the numbers 8 and 13 respectively, we 
 have by the question 8a; -f- 137/= 159. 
 
 Pursuing with this equation the same process, as in the pre- 
 ceding example, we have the five equations 
 
 a; = 19 — 2/ + ^ 
 y= l — t^-f 
 
 Expressing x and y in terms of t"\ we have 
 
 a:=15+13r' 
 
 y= 3 — Sr'. 
 
 Here it is evident, that t'" = 0, and t'" = — 1 are the only 
 values of t"\ which will give entire and positive values for z 
 and y. Making successively it"'= 0, t'" = — 1, we have 
 
 a:=15,2 
 2^ = 3,11. 
 
 Since then 82; and \2y represent the parts required, the 
 proposed admits of two solutions, viz. 120 and 39 for the first 
 solution, and 16 and 143 for the second. 
 
 The expressions, a:= 15+ 13^', 2/ = 3 — 8^', are called 
 formulas for x and ?/, since they indicate the manner in which 
 the values of x and y are obtained. In the use of these formulas 
 the accents, it is evident, may be omitted, as we have now no 
 further occasion to distinguish, one from the other, the indeter- 
 minates which have been employed. 
 
 3. It is required to divide 25 into two parts, one of which may 
 be divisible by 2, and the other by 3. 
 
 Ans. The parts are 16 and 9, 10 and 15, or 4 and 21. 
 
 4. A person has in his pocket pieces of 5 shillings and 3 
 shillings only, and wishes to pay a bill of 58 shillings. How 
 many pieces must he give of each ? 
 
 Ans. 2 of the first and 16 of the secondj or &c« 
 
INDETERMINATE ANALYSIS. 81 
 
 5. A sum of $81 was distributed among some poor persons, 
 men and women; each woman received $5, and each man S7. 
 How many men and women were there ? 
 
 Ans. There were 3 men and 12 women, or &c. 
 
 65. Let it be required next to solve in entire numbers the 
 equation 49a; — S5y =11. 
 
 Here it will be observed, that the coefficients of x and y 
 have a common factor 7; dividing therefore both members by 
 
 7, we have 7x — 6y=. — ^ an equation which is evidently im- 
 possible in entire numbers ; the proposed therefore does not 
 admit of entire and positive values for x and y. In general, 
 the proposed equation being reduced to the form of the pre- 
 ceding, if the coefficients of x and y have a common factor ^ which 
 does not enter into the second member ^ the equation is impossible 
 in entire numbers. 
 
 If there be a factor, common to the coefficients of x and y, 
 which does not enter into the second member, and this factor 
 be not perceived at first, the course of the calculation will make 
 known, sooner or later, the impossibility of solving the question 
 in entire numbers. 
 
 Applying the process, explained above, to the equations 
 49 X — 35?/ =11, we obtain finally the equation 
 
 an equation, which is evidently impossible in entire numbers 
 for t and t\ from which it is readily inferred, that the proposed 
 . will not admit of entire solutions. 
 
 If the equation of the proposed question has therefore a factor 
 common to both members, we suppress it; the coefficients of 
 z and y will then be prime to each other, if the question admits 
 of solution in entire and positive numbers. This being the case, 
 the process explained above will always lead to a final equation, 
 in which the coefficient of one of the indeterminates is equal to 
 6 
 
82 ELEMENTS OF ALGEBRA. 
 
 unity. Indeed, it will readily be perceived, that in the course 
 pursued we apply to the coefficients of x and y in the proposed 
 the process of the greatest common divisor; since then these 
 coefficients are by hypothesis prime to each other, we arrive 
 finally at a remainder equal to unity, which will be the coefficient 
 of the last but one of the indeterminates introduced in the course 
 of the calculation. 
 
 66. In certain cases the preceding process admits of simpli- 
 fications, which it is important to introduce in practice. 
 
 1. Let it be required to solve in entire numbers the equa- 
 tion 5a: + 3?/ = 49. 
 
 Proceeding as before, we have 
 
 32/ = 49 — 5a; 
 y=^lQ^x 3—;^ 
 
 but the quotient on dividing 5 a; by 3 being nearer 2a;, we put 
 the equation under the form 
 
 3y = 49 — 6a; + a;; 
 
 x-\- 1 
 whence y=16 — 2x-\ ?— J 
 
 from which we obtain a; = 3 ^ — 1 
 y=18 — 5^. 
 
 By means of the simplification, here introduced, the number 
 of indeterminates, employed in the calculation, is one less, than 
 would otherwise be necessary. 
 
 2. A person purchases wheat at I65. and barley at 9*. a 
 bushel, and pays in all 167s. How many bushels of each did 
 he purchase ? Ans. 2 of wheat and 15 of barley. 
 
 3. To find two numbers such, that if the first be multiplied 
 by 7 and the second by 13, the sum of the products will be 
 128. Ans. The numbers are 9 and 5, or &c. 
 
 4. Let it be proposed next to resolve in entire numbers the 
 equation 13 a; -— 57y = 101. 
 
INDETERMINATE ANALYSIS. 83 
 
 Deducing the value of z, we have 
 
 a: = 4y + 7+-^^^ = 4y + 7+ ^J ^ 
 In order that a; and y in this equation may be entire num- 
 bers, ^7^ must be equal to an entire number; but since 
 5 and 13 are prime to each other, it is necessary in order to 
 
 y j o 
 
 this, that •^ Jl should be an entire number ; putting t for this 
 
 number, we have 
 
 a; = 4y-|-7 + 5^ 
 13^= y + 2. 
 from which we obtain 
 
 a: = 57^— 1 
 
 Here, every entire and positive value for t will give similar 
 values for x and y ; but if we suppose ^ = 0, or to be negative, 
 the values of x and y will be negative. The number of entire 
 and positive solutions of the proposed is therefore infinite^ and 
 the smallest system of values for x and y is 
 a; = 56, 2/=ll. 
 
 By means of the simplifications, here introduced, one inde- 
 terminate only is employed, instead of three, which would other- 
 wise be necessary. 
 
 5. To divide 100 into two such parts, that if the first be 
 divided by 5, the remainder will be 2; and if the second be 
 divided by 7 the remainder will be 4. 
 
 Ans. The parts are 47 and 53, or 12 and 88. 
 
 6. To find two numbers such, that 11 times the fitst dimin- 
 ished by 7 times the second, may be equal to 53. 
 
 Ans. 8 and 5, or &c. 
 
 7. A person purchases some horses and oxen; he pays $30 
 for each horse, and $23 for each ox; and he finds, that the oxen 
 cost him $12 more than the horses. How many horses and 
 oxen did he buy ? Ans. 18 horses and 24 oxen, or &c. 
 
Bi ELEMENTS OF ALGEBRA. 
 
 8. To find two numbers such, that if 8 be added to 17 tirKies 
 the first, the sum will be equal to 49 times the second. 
 
 Ans. 37 and 13, or^&c. 
 #, Let it be proposed next to resolve the equation 
 
 39 a: — 56 2/ =11. 
 JDeducing from this equation the value of x, we have 
 
 ^^y^ 39 * 
 
 Here, yi the expression — ~J^ — , it will be observed that the 
 
 difference between 17, the coefficient of y, and the divisor 39, 
 contains the other term 11 as a factor; on this account we take 
 the quotient 56 y divided by 39 in excess, which gives 
 ^_g^ 22y-ll _ ll(2y-l). 
 
 ^ — '^y 39^ """^^ 39 ' 
 
 from which we readily obtain 
 
 a: = 56?' — 27 
 y = 39j^'— 19. 
 
 10. To find two numbers such, that if the first be multiplied 
 by 11 and the second by 17, the first product is 5 greater than 
 the second. Ans: 19 and 12, or &c. 
 
 11. In how many ways can a debt of 546 livres be paid, by 
 paying pieces of 15 livres, and receiving in exchange pieces of 
 11 livres? Ans. The number of ways is infinite. For the 
 
 first we have 43 of the one, and 9 of the other. 
 
 12. The difference between two numbers is 309, and if the 
 greater be divided by 37 the remainder will be 5, and if the 
 less be divided by 54 the remainder will be 2; what are the 
 numbers ? Ans. 1337 and 1028, or &c. 
 
 67. From what has been done, it will be perceived, that if the 
 equation proposed be of the form 2a;-f-3y = 10, the number 
 of solutions in entire and positive numbers will be limited; but 
 if the equation be of the form 2 a: — 3y=10, 10 being either 
 positire or negative, the number of solutions will be infinite. 
 
INDETERMINATE ANALYSIS. 05 
 
 If moreover we compare the formulas for x and y with the 
 equations from which they are derived, the coefficient of the 
 indeterminate in the formula for x is the same, it will be ob- 
 served, with the coefficient of y in the equation; and the co- 
 efficient of the indeterminate in the formula foT y is the same 
 with the coefficient of x in the equation, taken with the con- 
 trary sign, or the converse, as it respects the signs of the co- 
 efficients. Having obtained then a first solution of the ques- 
 tion proposed, those which follow will be found by adding 
 successively to the values of x the coefficient of y in the equa- 
 tion, and subtracting successively from the values of y tlie co- 
 efficient of X in the equation, or the converse, the coefficients 
 of X and y being taken with the signs, which they have in the 
 equation. 
 
 68. We pass next to the solution of problems and equations 
 with three or more unknown quantities. 
 
 1. Let it be proposed to pay 741 livres with 41 pieces of 
 money of three different species, viz. pieces of 24 livres, 19 
 livres, and 10 livres. 
 
 Let X, y and z represent respectively the number of pieces of 
 each kind, we have by the question 
 
 a: + 2/-|-^ = 41 
 24a: + 192/ + 10z = 74L 
 
 Eliminating one of the unknown quantities, x for example, we 
 have 5y+14z = 243. 
 
 Deducing from this equation formulas for entire values for z 
 and y, according to the method explained above, we have, 
 omitting the accents, 
 
 2 = 5^ — 3 
 
 y = 57 — UU 
 
 Substituting next, in the first of the equations of the proposed 
 the expressions for z and y just obtained, and deducing the 
 ▼alue of X, we have x = 9t — 13. 
 
86 ELEMENTS OF ALGEBRA. 
 
 If we now put for f, in the above formulas for ar, y, and «, 
 any entire values whatever, we shall obtain entire values for 
 Xt y, and z, which will satisfy the equations of the proposed. 
 But to obtain the entire and positive values only, as the nature 
 of the question requires, it is evident, 1°. that 9t must be 
 greater than 13, or which is the same thing, that t must be 
 greater than IJ; 2**. that 14^ must be less than 57, or which 
 is the same thing, that t must be less than 4^ ; t can therefore 
 have only the values of 2, 3, 4. 
 
 Putting in the formulas above t equal to 2, 3, and 4 succes- 
 sively, we have 
 
 x= 5,14,23 
 3^ = 29,15, 1 
 z= 7, 12,17. 
 
 The proposed, therefore, admits of three different solutions 
 and of three only. 
 
 2. Thirty persons, men, women, and children, spend 80 
 crowns in a tavern. The share of a man is 7 crowns, that of 
 a woman 5 crowns, and that of a child is 2 crowns. How 
 many persons were there of each class ? 
 
 Ans. For a first answer we have, 1 man, 
 5 women, and 24 children. 
 
 3. The sum of three entire numbers is 15, and if the first 
 be multiplied by 2, the second by 3 and the third by 7, the sum 
 of the products will be 65. What are the numbers ? 
 
 Ans. 4, 5, and 6. 
 
 From what has been done, it will be easy to see how we are 
 to proceed, in the case of three equations with four unknown 
 quantities, and so on. 
 
SOLUTION OF QUESTIONS IN A GENERAL MANNER. 87 
 
 SECTION VIII. — Solution of Questions in a General 
 Manner. 
 
 69. In the solution of a question m numbers, there are, it 
 must have been perceived, two distinct things which require 
 attention. 1°. To determine by a process of reasoning what 
 operations must be performed upon the numbers given in the 
 question in order to obtain the answer sought; 2°. to perform 
 these operations. In the questions which have been solved 
 thus far, the operations have each been performed as soon as 
 determined. Let us now resume the question, art. 1, and in- 
 stead of performing the operations, as we proceed, let us retain 
 them by means of the proper signs. 
 
 Representing as before the less part by x, the greater will 
 be a; -j- 12, and we have 
 
 x-\-x+l2 = 5Q 
 
 2x=:56 — l2 
 56 12 
 ^=2—2- 
 Here the process of reasoning required in the solution of 
 the proposed has been conducted by itself; the expression, at 
 which we arrive, is not the answer sought, but the result of the 
 reasoning pursued ; it shows what operations must be performed 
 in order to obtain the answer, viz. that from one half of 56, the 
 number given to be divided, there must be subtracted one half 
 of 12 the given excess. Performing next the operations thus 
 determined, we have 22 for the less part as before. 
 
 Let us next resume the sixth question, art. 6; representing 
 again the least part by x, the mean will he x -\- 40, the great- 
 est a; -|- 40 -[- 60, and we shall have 
 
 a: + z + 40 + a: + 40 + 60 = 230 
 
 8a: = 230 — 40 — 40 — 60 
 
 3a: = 230— 2 x40 — 60 
 
 230—2 X 40 — 60 
 
S3 ELEMENTS OF ALGEBRA. 
 
 Here also the result shows the operations to be performed, 
 according to which to find the least part sought, from 230, the 
 number given to be divided, we subtract twice 40, or twice the 
 excess of the mean part above the least, and also 60, the excess 
 of the greatest part above the mean, and take one third of the 
 remainder. 
 
 70. If the reasoning pursued in the solution of the preceding 
 questions be examined with attention, it will be perceived, that 
 it does not depend upon the particular numbers given in these 
 questions. It will be precisely the same for any other numbers. 
 The same operations will therefore be necessary to obtain the 
 parts sought, whatever the given number may be. 
 
 By preserving the operations, therefore, we resolve the pro- 
 posed in a general Tnanner^ that is, we determine once for all 
 what operations are necessary for all questions, which differ 
 from the proposed only in the particular numbers, which are 
 given. 
 
 Let it be proposed next to find a number such that the differ- 
 ence between one-ninth and one-seventh of this number shall be 
 equal to 10. 
 
 Putting X for the number, we obtain 
 
 9a; — 7a; = 7X 9X 10. 
 
 Here it will be observed that x is taken 9 times minus seven 
 times, or 9 — 7 times ; 9 — 7 will, therefore, be the coefficient 
 of a;, and the above equation may be written thus : 
 
 (9 — 7) a: =.7X9X10 
 
 , 7 X 9 X 10 
 
 whence x = — 5- — = — . 
 
 Let the following questions be now resolved in a general 
 manner. 
 
 L A company settling their reckoning at a tavern, pay 8s. 
 each ; but if there had been 4 persons more, they should only 
 have paid 7*. each. How many were there ? 
 
SOLUTION OF QUESTIONS IN A GENERAL MANNER. 89 
 
 . 2. Divide the number 91 into two such parts, that 6 times the 
 first, diminished by 5 times the second, may be equal to 40. 
 
 3. Divide the number 56 into two such parts, that one part 
 being divided by 7 and the other by 3, the quotients may together 
 be equal to 10. 
 
 71. In the solution of questions in a general manner, accord- 
 ing to the method above explained, we should be liable, through 
 inadvertence, to perform some of ihe operations as we proceed ; 
 thus the result would not show how the answer is to be 
 found by means of the numbers originally given in the question. 
 To avoid this inconvenience and at the same time to render the 
 solution more concise, it is usual to represent the given things 
 in a question by signs, which will stand indifferently for the 
 particular numbers given in the question, or for any other 
 numbers whatever. 
 
 It is agreed to represent known quantities, or those which 
 are supposed to be given in a question, by the first letters of the 
 alphabet, as a, i, c. 
 
 The given things in the question, art. 1, are the number to be 
 divided, and the excess of the greater part aboVe the less ; rep 
 resenting these by a and b respectively, the question may be 
 presented generally, thus ; To divide a number a into two stcch 
 parts that the greater may exceed the less by b. 
 
 To resolve the question, thus stated, we denote still the less 
 part by x ; the greater will then be a; -{- ^> and we have 
 x-{-x-\-b = a 
 2x-\'b = a 
 
 2x = a — b 
 
 a b 
 
 ^=2-2 
 
 The translation of a formula into common language is called 
 a rule. Thus we have the following rule, by which to obtain 
 the less of the parts required according to the question pro- 
 
90 ELEMENTS OF ALGEBRA. 
 
 posed, viz. From half the number to he divided^ subtract half 
 the given excess, the remainder will be the answer. 
 
 Knowing the less part, we obtain the greater by adding to the 
 less the given excess. We may, however, easily obtain a rule 
 for calculating the greater part without the aid of the less. 
 Indeed since the less part is equal to 
 
 - — -, if we add b to this, we have ^ — o~t~^ equal to the 
 
 greater. But this expression may, it is easy to see, be reduced 
 
 to Q + o 5 whence we have the following rule, by which to find 
 
 the greater part, viz. To half the number to be divided, add 
 half the given excess, the result will be the answer. 
 
 To apply these rules, let it be required to divide $1753 
 between two men in such a manner, that the first may have 
 $325 more than the second. 
 
 72. The 6th question, art. 6, may be presented in a general 
 manner, thus ; To divide a number a into three such parts, that 
 the excess of the mean above the least may be b, and the excess 
 of the greatest above the mean may be c. 
 
 Let X = the least part ; 
 
 then x-\-b=- the mean, 
 
 and x-\-b-\-c = the greatest, 
 
 therefore x-\-x-\-b-\-x-\-b-\-c = a, 
 
 or transposing and reducing 2x = a — 2b — c, 
 
 a — 2b — c 
 whence x == ^ . 
 
 Translating the above formula into common language, we have 
 the following rule, by which to find the least part, viz. From the 
 number to be divided, subtract twice the excess of the mean part 
 above the least, and also the excess of' the greatest above the mean 
 and take a third of the remainder. 
 
 To obtain a formula for the mean part, we add b, the excess 
 
SOLUTION OF QUESTIONS IN A GENERAL MANNBR. 91 
 
 of the mean above the least, to the above expression for the leas» 
 
 part, which gives for the mean 
 
 a — 2b — c , . 
 3- + *. 
 
 or reducing to a common denominator 
 g — 23 — c , 3^ 
 3 +"3 ' 
 
 whence we obtain for the mean part 
 a-\'b — c 
 3 * 
 
 In like manner the following formula will readily be obtained 
 for the greatest part, viz. 
 
 g-|-3 + 2c 
 /^ 3" 
 
 Translating these formulas into common language we obtain 
 rules also for the mean and for the greatest part. 
 
 1. To apply these rules let it be required to divide $973 among 
 three men, so that the second shall have $69 more than the first, 
 and the third $43 more than the second. 
 
 2! A father, who has three sons, leaves them his property 
 amounting to $15730. The will specifies, that the second shall 
 have $2320 more than the third, and the eldest shall have $3575 
 more than the second. What is the share of each ? 
 
 73. The operations necessary for the solution of this last ques- 
 tion are, it is easy to see, the same with those for the preceding. 
 It may therefore be solved by the same formulas. In like man- 
 ner the seventh and eighth questions, art. 6, may be solved by 
 the same formulas. This circumstance is worthy attention, since 
 we are thus enabled to comprehend in one the solution of a multi- 
 tude of questions, differing from each other not only in the par- 
 ticular numbers, which are given, but also in the language, in 
 which they are expressed. 
 
 Let now the following questions be generalized. 
 
 1. The sum of $3753 is to be divided among 4 men, in such 
 a manner, that the second will have $159 more than the first, the 
 
92 ELEMENTS OF ALGEBRA. 
 
 third $275 more than the second, and the fourth $389 more than 
 the third. Wliat is the share of each ? 
 
 2. Three men share a certain sum in the following manner ; 
 the sum of A's and B's shares is $123, that of A's and C's $110, 
 and that of B's and C's $83. What is the whole sum and the 
 share of each ? 
 
 Let X = the whole sum, a, h, and c the sum of the shares of 
 A and B, A and C, B and C, respectively; then x — a=C's 
 share, &c., and we have 
 
 « + h-\-c 
 
 74. The seventh question, art. 15, may be stated generally, 
 thus. A cistern is supplied by two pipes ; the first will fill itUn 
 a hours, the second in h hours. In what time will the cistern he 
 filled if both run together ? 
 
 Let X = the time ; the capacity of the cistern being supposed 
 equal to unity, we have 
 
 a^b — ^' 
 whence freeing from denominators 
 
 ax-{-bx = ab. 
 Here it will be observed, that x is taken a times and also b 
 times; whence on the whole it is taken a-\-b times; a-{-b 
 is then the coefficient of x, and the above equation may be 
 written thus, 
 
 {a-{-h) x = ab ; 
 
 . ab 
 
 whence x = — r—: . 
 
 a-f-b 
 
 Translating this formula. We have the following rule foi 
 every case of the proposed question, viz. Divide the product 
 of the numbers, which denote the times employed by each pipe in 
 flUing the cistern, by the sum of these numbers; the quotient loill 
 he the time required by both the pipes running together to fill 
 the cistern. 
 
SOLUTION OF QUESTIONS IN A GENERAL MANNEE. VtS 
 
 Example. Suppose one pipe will fill the cistern in 5 J hours, 
 and the other in 9 hours ; in what time will it be filled if both 
 run together ? 
 
 75. The 4th question, art. 6, may be thus generalized. A 
 gentleman meeting four poor persons distributed a shillings ammig 
 them ; to the second he gave b times^ to the third c times ^ and to 
 the fourth d times as much as to the first. What did he give to 
 each"^ 
 
 Let X repiresent what he gave to the first, we then have 
 z-\'bx-\'CX'\- dx = a, 
 or {1 -\- b -{- c '\- d) x = a ; 
 
 whence a;=: ., , , , f— 3. 
 
 \ -\- b -\- c -\- d 
 
 Let next the following questions be generalized. 
 
 1. A bankrupt wishing to distribute his remaining property 
 among his creditors, finds, that in order to pay them $175 apiece, 
 he should want $30, but if he pays them $168 apiece he will 
 have $40 left. How many creditors had he ? 
 
 2. It is required to divide the number 91 into two such parts, 
 that the greater being divided by their difference the quotient 
 may be 7. 
 
 3. Divide the number 138 into two such parts, that 5 times 
 the first part diminished by 4 times the second will be equal 
 to 85. 
 
 4. Three men, A, B, and C, engage in trade and gain $500, 
 of which C is to have twice as much as B, and B $50 less than 
 4 times as much as A. How much will each receive ? 
 
 5. A trader having gained $3450 by his business, and lost 
 $2375 by bad debts, found, that | of what he had left equalled 
 the capital with which he commenced trade. What was his 
 capital? 
 
 6. In a certain school \ of the pupils lean nayigation, | leaia 
 
94 ELEMENTS OF ALGEBRA. 
 
 geometry, ^ learn algebra, and the rest, 23 in number, leam 
 arithmetic. How many pupils are there in all ? 
 
 76. The nineteenth question, art. 15, may be presented in a 
 general manner, thus. A laborer was hired for a certain number 
 a of days; for each day that he wrought he loas to receive b shil' 
 lings, but for each day that he was idle, he was to forfeit c 
 shillings. At the end of the time he received d shillings. How 
 many days did he loorh, and how many was he idle ? 
 
 Putting X = the number of days, in which he wrought, and 
 resolving the question, we obtain 
 
 d-\-ac 
 b-j-c 
 
 Example. A laborer was hired for 75 days ; for each day that 
 he wrought he was to receive $3, but for each day that he was 
 idle, he was to forfeit $7. At the end of the time he received 
 $125. To determine by the above formula the number of days 
 in which the laborer wrought. 
 
 The two following questions may also be solved by the same 
 formula. Why is this the case ? 
 
 1. A man agreed to carry 20 earthen vessels to a certain 
 place on this condition; that for every one delivered safe he 
 should receive 11 cents, and for every one he broke, he should 
 forfeit 13 cents; he received 124 cents. How many did he 
 break ? 
 
 2. A fisherman to encourage his son promises him 9 cents 
 for each throw of the net in which he should take any fish, 
 but the son, on the other hand, is to forfeit 5 cents for each 
 unsuccessful throw. After 37 throws the son receives from the 
 father 235 cents. What was the number of successful throws 
 of the net? 
 
 77. Lei it be proposed next to make a rule for Fellowship, and 
 in order to this, let us take the following example. 
 
 Three men, A, B, and C commence trade together, and fur- 
 nish money in proportion to the numbers w, n and p respectively; 
 
T 
 
 SOLUTION OF QUESTIONS IN A GENERAL MANNER. 95 
 
 they gain a certain sum a. What is each man's share of the 
 gain ? 
 
 Let X = A's share ; 
 
 7tX 7) X 
 
 then — = B's, and — = C's share. 
 m m 
 
 By the question, therefore, 
 
 . nx , px 
 x-\ \-'— = a. 
 
 Freeing from denominators, we have 
 
 mx-\-nx-\-'px-=.ma^ 
 or, which is the same thing 
 
 {m -\- n -\- "p) x=^ma; 
 
 whence x = ; ; — = A's share. 
 
 m-f-n-f-p 
 
 Muhiplying next the value of x by w, and dividing by tw, we 
 obtain 
 
 na 
 
 m ■\'n'\-p 
 In like manner, we find 
 
 pa 
 
 = B's share. 
 
 = C's share. 
 
 m-\-n-\-p 
 
 To find a share of the gain therefore; Multiply the corre- 
 sponding proportion of the stock into the whole gaiuy and divide 
 the product by the sum of the proportions. 
 
 78. Let now the following questions be generalized. 
 
 1. Three merchants, A, B, C, enter into partnership. A ad- 
 vances $750, B $1300 and C $825. A leaves his money 9 
 months, B 13 months, and C 15 months in the business. They 
 gain $830. What is the share of each ? 
 
 Since A advances $750 for 9 months, he advances what is 
 equivalent to $750 X 9 for 1 month. In like manner B advances 
 what is equivalent to $1300 X 13 for one month, &c. 
 
 Let p, p\ p" represent respectively the sums advanced by, 
 each, and j, j', g", the times in which these sums were severally 
 
96 ELEMENTS OF ALGEBRA. 
 
 employed ; putting a for the sum gained, and z for A's share of 
 the gain, we have • 
 
 3. A bankrupt leaves $18000 to be divided among three 
 creditors, in proportion to their claims. Now A's claim is to B's 
 as 2 to 3, and B's claim to C's as 4 to 5. How much does each 
 creditor receive ? 
 
 3. A gentleman hired three men to perform a certain piece of 
 work ; the first working 9 hours a day would perform the work 
 in 10 days, the second working 7 hours a day, in 15, and the 
 third, working 12 hours a day, in 6 days. How long will it take 
 them together to perform the work ? 
 
 4. A merchant purchased 24 yards of cloth of two different 
 kinds for $408. The first cost $18, the second $15 a yard. 
 How many yards were there of each kind ? 
 
 5. A gentleman hired two workmen for 50 days ; to the first 
 he gave $3, to the second $2 a day. On settling with them he 
 paid both together $130. How many days did each work ? 
 
 What have these last two questions in common, and what 
 general statement will comprehend both ? 
 
 79. Thus far we have employed the first letters of the alphabet 
 to represent known quantities, and the last to denote those which 
 are unknown. In some cases it is more convenient to represent 
 tlie quantities, whether known or unknown, by the initials of the 
 words for which they stand. 
 
 Let it be proposed to determine what sum of money must be 
 put at interest, "at a given rate, in order to amount to a given sum 
 in a given time. • 
 
 Let p == the principal, or sum put at interest, 
 r = the rate, 
 fl = the given amount, 
 ;( = the given time. 
 
SOLUTION OF QUESTIONS IN A GENERAL MANNER. 97 
 
 By the question, we have p-\-trp = a^ 
 or {l-\-tr)p = a; 
 
 whence p = - — ; . 
 
 ^ l-\-tr 
 
 We have therefore the following rule, by which to find the 
 principal required, viz. Multiply the rate by the time and add 1 
 to the product ; the amount divided by the sum thus obtained will 
 give the principal. 
 
 Examples. 1. What sum of money must be put at interest 
 at 6 per cent., in order that the principal and interest may, at the 
 end of 5 years, amount to $748,80 ? 
 
 Six per cent, will be $6 on $100, or, $.06 on one dollar ; r in 
 the formula will be then, for this case, .06 and we obtain $576 
 for the answer. 
 
 2. A man lent a certain sum of money at 5 per cent. ; at the 
 end of 7 years he received for principal and interest $1237.47, 
 W]>at was the sum lent ? Ans. $916.65. 
 
 3. A merchant finds that by a fortuiaate speculation with his 
 floating capital, he has gained 15 per cent., and that by this 
 means it has increased to S15571. What was his capital? 
 
 Ans. $13540. 
 
 80. The equation p-\-trp=^a, contains, it will be perceived, 
 four different things, any one of which may be determined, when 
 the others are known. Deducing, for example, the value of ty 
 we have 
 
 a — p 
 rp 
 
 Whence to find the time, when the amount, principal and 
 rate are given ; From the amount subtract the principal^ and 
 divide the remainder by the product of the rate multiplied by the 
 principal. 
 
 Examples. 1. A man put at inJerest $345 at 4 per cent.; 
 at the end of a certain time he received for principal and interest 
 
 7 
 
^ ELEMENTS OF ALGEBRA. 
 
 $483. Required the time for which the money was lent. 
 
 Ans. 10 years. 
 
 2. A merchant lets out his floating capital, amounting- to $5873 
 at 10 per cent, interest. At the end of a certain number of years 
 he finds that he has received $3523,80 interest. For how many 
 years was his capital let out ? Ans. 6. 
 
 3. Let the learner prepare the formula and solve the following 
 example. 
 
 A gentleman put at interest $6840, and at the end of 5 years 
 received for capital and interest $8208. What rate per cent, did 
 he receive ? Ans. 4 per cent. 
 
 81. In the preceding questions the object has been to deter- 
 mine certain unknown numbers by means of others, which 
 are known, and which have relations to the unknown num- 
 bers established by the enunciation of the question. We shall 
 now show the aid derived from the same signs in demon- 
 strating certain properties in relation to known and given 
 numbers. 
 
 1. To demonstrate that if both terms of a fraction be multi- 
 plied by the same number, the value of the fraction will not be 
 changed. 
 
 Let the proposed fraction be designated by -, and let n be any 
 
 number whatever. 
 
 Putting - = m, we have a = bm; 
 
 multiplying both sides of this last by n, we have 
 
 na=inbm, 
 from which we deduce 
 
 na 
 
 no 
 
 , na a 
 
 whence -—=-. 
 
 no h 
 
 2. Let the same number be added to both terms of a proper 
 
SOLUTION OF QUESTIONS IN A GENERAL MANNER. 99 
 
 fraction; to determine what effect this will produce upon the 
 value of the fraction. 
 
 Let us designate the fraction by -. Let m he the number 
 
 added to both terms of this fraction; it then becomes 
 
 To compare the two fractions, it is necessary to reduce them 
 to the same denominator. Performing this operation, we have 
 for the first 
 
 ab-\-am 
 
 and for the second 
 
 bb -\-bm 
 ab-\- bm 
 
 bb-\-bm 
 
 Here the two numerators have the part ab common to both; 
 but the part bm oi the second is greater than the part am oi the 
 first, since b is greater than a ; the second fraction is therefore 
 greater than the first ; whence, If the same number be added to 
 both terms of a proper fraction^ the value of the fraction will be 
 increased* 
 
 3. It has been shown in arithmetic that, Every divisor common 
 to tivo numbers must divide the remainder after the division of 
 the greater of these numbers by the less. Let us now demonstrate 
 this property by the aid of algebraic symbols. 
 
 Let D be the divisor common to the two numbers ; let A D 
 represent the greater of the two numbers and B D the less ; let 
 Q be the entire' quotient arising from the division of the greater 
 by the less, and let R be the remainder ; we have then 
 
 AD = BDxQ+.R; 
 
 dividing both sides by D, we have 
 
 A = BxQ+5. 
 
 Here the first member of the equation is an entire numli^r, 
 
100 ELEMENTS OF ALGEBRA. 
 
 the second must, therefore, be equal to an entire number; but 
 of this member the term BQ is an entire number; whence 
 
 _ must be an entire number, that is, R must be exactly divisible 
 
 by D. The proposition above is, therefore, demonstrated. 
 The following propositions may now be demonstrated. 
 
 1. If the sum of any two quantities be added to their differ- 
 ence, the sum will be twice the greater. 
 
 2. If the difference of any two quantities be taken from their 
 sum, the remainder will be twice the less. 
 
 3. The second power of the sum of two quantities contains 
 the second power of the first quantity, plus double the pro- 
 duct of the first by the second, plus the second power of the 
 second. 
 
 4. The second power of the difference of two quantities is 
 composed of the second power of the first quantity, minus the 
 double product of the first by the second, plus the second power 
 of the second. 
 
 5. The product of the sum and difference of two quantities is 
 equal to the difference of their second powers. 
 
 The questions, art. 15, will furnish additional exercises for 
 the learner in stating and resolving questions in a general 
 manner. * 
 
 SECTION IX. — Discussion of Problems and Equations op 
 THE First Degree. 
 
 82. When a problem has been solved in^ a general manner, 
 it may be proposed to determine what values the unknown 
 quantities will take for particular hypotheses made upon the 
 known quantities. The determination of these different values, 
 
DISCUSSION OF FORMI^LAS^ ] ', ' ',' ; I.; > ,101' 
 
 and the interpretation of the results to which we arrive, form 
 what is called the discussion of the problem. 
 
 The discussion of the following problem presents nearly all 
 the circumstances, that can ever occur in equations of the first 
 degree. 
 
 Two couriers set out, at the same time, from two different 
 points A and B, in the line E D, and travel towards D until 
 they meet ; the courier, who sets out from the point A, travels 
 at the rate of m miles an hour, the other travels at the rate of 
 n miles an hour ; the distance between the points A and B is 
 a miles; at what distance from the points A and B will they 
 meet ? 
 
 I I 
 
 E C A B C D 
 
 Suppose C to be the point in which they meet ; let a; = the 
 distance A C, y = the distance B C. We have for the first 
 equation x — y = a. 
 
 The first courier, travelling at the rate of m miles an hour, 
 
 cc 
 will be — hours in passing over the distance x; the second, 
 m 
 
 travelling at the rate of n miles an hour, will be - hours in 
 
 n 
 
 passing over the distance y; and since these distances must 
 
 each be passed over in the same time, we shall have for the 
 
 second equation 
 
 m n 
 
 Resolving these two equations, we have 
 
 am an 
 
 m — n m — n 
 
 Discussion, 
 
 1. Let m be greater than n. In this case the values of x and 
 y will be positive, and the problem will be solved in the exact 
 of the enunciation j for, it is evident, that if the courier, 
 I* 
 
.tte : ^'l"*,.^' '; ; ' p'^Eli'E^TS OF ALGEBRA. 
 
 who sets out from A, travels fasti^r tlian the other, they will 
 meet somewhere in the direction A D. 
 
 2. Let n be greater than m. This being the case we shall 
 have 
 
 am an 
 
 y 
 
 n — m n — m 
 
 Here the values of x and y are negative. In order to interpret 
 this result, we observe that the courier from B travelling faster 
 than the courier from A, the interval between them must in- 
 crease continually. It is absurd therefore to require that they 
 should meet in the direction A D. The negative values for 
 X and y indicate, then, an absurdity in the conditions of the 
 question. To show how this absurdity may be done away, let 
 us substitute in the equations of the problem — x, — y instead 
 of X and y, we shall then have 
 
 — x-{-y=^a>i (y — x = a 
 
 — — = — ^{ ^^ ]-=t 
 m n) (m n 
 
 The second equation is not affected by the change of sign, as 
 indeed it ought not to be, since it only expresses the equal- 
 ity of the times. In regard to the first, however, we have 
 y — a: = «, instead of x — y=.a. This shows that the point, 
 in which the couriers meet, must be nearer to A than to B by 
 the distance A B; it must, therefore, be at some point C on 
 the other side of A with respect to B. In order then to remove 
 the absurdity in the enunciation of the question, it is necessary 
 to suppose the couriers, instead of travelling in the direction 
 A D, to travel in the opposite direction B E. Indeed, if we 
 resolve the equations 
 
 y — x = a 
 
 we have a: = , y = , values which are positive, 
 
 n — m n — m 
 
 and which answer the conditions of the probletri modified, thus, 
 
 Two couriers set out at the same time from two points, A dfid 
 
DISCUSSION OF FORMULAS. 103 
 
 B, in the line E D, and travel towards E ; the courier, who sets 
 out from the point B, travels at the rate of n miles an hour, the 
 other travels at the rate of m miles an hour; the distance be- 
 tween the points B and A is a miles ; at what distance from the 
 points B and A will they meet ? 
 
 3. Let m=.n. In this case we have m — n=0, and the 
 values of x and y become 
 
 am an 
 
 But how shall we interpret this result ? Returning to the 
 question, we perceive it to be absolutely impossible to satisfy 
 the enunciation; for, the couriers travelling equally fast, the 
 interval between them must always continue the same, how- 
 ever far they may travel in either direction. It is impossible, 
 then, that they should meet, and no change in the enunciation, 
 so long as we have m = n, can make it possible. Indeed, 
 the equations of the problem on the hypothesis m=.n become 
 
 X — y = a 
 
 x — yz=0, 
 equations, which are evidently incompatible. Zero being a 
 divisor is, then, a sign of impossibility. 
 
 The expressions -^, -^ are considered, however, by mathe- 
 maticians as forming a species of value for x and y, to which 
 they give the name of infinite value. To show the reason for 
 this, let us suppose that the difference between m and 7i with- 
 out being absolutely nothing is very small; in this case, it is 
 evident that the values of x and y will be very largo. Let, for 
 example, m = 3, m — tz = 0. 01, we shall then have n = 2. 99, 
 whence 
 
 am 3 a _-^_ an 
 
 m — n .01 ' m 
 
 Again let m — 7z = .0001, m being equal to 3, n will then 
 = 2. 9999, whence 
 
 -^^ = 30000a, — — = 29999a. 
 m — n m — n 
 
104 ELEMENTS OF ALGEBRA. 
 
 In a word, so long as there is any difference, however small, 
 
 between m and n, the couriers will meet in one direction or 
 
 the other; but the distance of the point, in which they meet, 
 
 from the points A and B will be greater in proportion as the 
 
 difference between m and n is less. If then the difference he- 
 
 tween m and n is less than any assignable quantity, the distances 
 
 am an .^^ . ^ . ,, 
 , will be greater than any assignable quantity or 
 
 infinite. Since then is less than any assignable quantity, we 
 may employ this character to represent the ultimate state of a 
 quantity which may be decreased at pleasure ; and since the 
 value of a fractional quantity is greater, in proportion as its 
 
 denominator is less, the expression — r-, and in general, any 
 
 quantity with zero for a denominator may be considered as the 
 symbol of an infinite quantity, that is, a quantity greater than 
 any, which can be assigned. 
 
 We say then that the values x=--^, y=.-— are infinite. 
 
 To show how the notion indicated by the expression -r-- does 
 
 away the absurdity of the equations 
 
 X — y=-a, X — 2/ = ^' 
 from the second of these equations, we deduce the value of y 
 and substitute it in the first, we then have x — x = a; dividing 
 both sides of this last by x, we have 
 
 l_l=f,.or? = 0. 
 x x 
 
 Here, as we put for x values greater and greater, the fraction 
 - will differ less and less from 0, and the equation will approach 
 nearer and nearer to being exact. If then x be greater than any 
 assignable quantity, - will be less than any assignable quantity 
 or zero. 
 
DISCUSSION OF FORMULAS. 105 
 
 4. Let us suppose next m=:n, and at the same time a = 0, 
 we shall then have 
 
 
 
 ^ = 0' 2^ = 0- 
 
 But how shall we interpret this new result? Returning to 
 the enunciation, we perceive, that if the couriers set out each 
 from the same point and travel equally fast, there is no par- 
 ticular point in which they can be said to meet, since in this 
 case, they will be together through the whole extent of their 
 route. Indeed, on this hypothesis the equations of the problem 
 become 
 
 a: — y = 0, 
 
 a; — y = 0, 
 equations which are identical ; the problem is tnerefore indetev' 
 mijiate, since we have in fact but one equation with two un- 
 known quantities. The expression - is therefore a sign of inde-- 
 
 termination in the enunciation of the problem. 
 
 The preceding hypotheses are the only ones, which lead to 
 remarkable results. They are sufficient to show the manner 
 in which algebra corresponds to all the circumstances m the 
 enunciation of a problem. 
 
 GENERAL FORMULAS FOR EQUATIONS OF THE FIRST DEGREE WITH 
 ONE OR TWO UNKNOWN QUANTITIES. . 
 
 83. Every equation of the first degree with one unknown 
 quantity may, by collecting all the terms which involve a:, into 
 one member and the known quantities into the other, be re- 
 duced to an equation of the form Aa:= B, A and B denoting 
 any quantities whatever, positive or negative. 
 
 Let there be, for example, the equation 
 
 mz 
 
 p = x — a. 
 
 n ^ ^ 
 
 Freeing from denominators, transposing and upiting terms, we 
 
 have {m — n)x=.n{p — q). 
 
106 ELEMENTS OF ALGEBRA. 
 
 Comparing this equation with the general formula, we have 
 7» — 7^ = A, n {p — q) = 'B. 
 
 ■p 
 84. Resolving the equation Aa; = B, we have « = -t-. This 
 
 A. 
 
 is a general solution for equations of the first degree, with one 
 unknown quantity. 
 
 Discussion, 
 
 1. Let it be supposed, that in consequence of a particular 
 hypothesis made upon the known quantities, we have A = 0, 
 
 T> 
 
 the value of z will then be -jr. But the equation Aa; = B on 
 
 this hypothesis becomes "^ x=B, an equation which, it is 
 
 evident, cannot be satisfied by any determinate value for x. 
 
 The equation X ^ = B may, however, be put under the 
 
 ■p 
 form — ■ = 0. Here, if we consider x greater than any assign- 
 
 ■p 
 
 able quantity, the fraction — will be less than any assignable 
 
 quantity or zero. On this account we say that infinity in this 
 case satisfies the equation. It is evident, at least, that the equa- 
 tion cannot be satisfied by any finite value for x. 
 
 2. Let us suppose next A = 0, and at the same time B = 0, 
 
 the value of x will then take the form ^. In this case the equa- 
 tion becomes X ^ == 0? an equation which may be satisfied by 
 any finite quantity whatever, positive or negative. Thus the 
 equation, or the 'problem, of which it is the algebraic translation, 
 is indeterminate. 
 
 It should be observed, however, that the symbol * does not 
 
 always indicate that the problem is indeterminate. 
 
 Let, for example, the value of x derived from the solution of a 
 problem be 
 
DISCUSSION OF FORMULAS. 107 
 
 If we put a = b in this formula, it will, under its present 
 form, be reduced to ^ ; but this value for x may be put Tinder 
 the form 
 
 ^"" {a — b){a-^b) ' 
 If then, before making the hypothesis a = i, we suppress the 
 factor a — bj the value of x becomes 
 
 a^ + ab-^b\ 
 a + b ' 
 
 from which we obtain x = —, on the hypothesis a = b. 
 
 We conclude therefore that the symbol j- is sometimes in alge- 
 bra the sign of the existence of a factor common to the two terms 
 of a fraction, ivhich in consequence of a particular hypothesis be- 
 comes 0, and reduces the fraction to this form. 
 
 Before deciding then, that the result tt is a sign that the 
 
 problem is indeterminate, we must examine whether the ex- 
 pressions for the unknown quantities, which in consequence of 
 a particular hypothesis are reduced to this form, are in their 
 lowest terms, if not, they must be reduced to this state ; the par- 
 ticular hypothesis being then made anew,, the result ^ shows 
 
 that the problem is really indeterminate. 
 
 85. Every equation of the first degree with two unknown 
 quantities may be reduced to an equation of the form 
 
 A, B, and C denoting any quantities whatever, positive or nega- 
 tive. It is evident, that all equations of the first degree with 
 two unknown quantities niay be reduced to this state, P. by 
 freeing the equation from denominators ; 2°. by collecting into 
 one member all the terms, which involve x and y, and the 
 known quantities into the other; 3**. by uniting the terms 
 
108 ELEMENTS OF ALGEBRA. 
 
 which contain x into one term, and those which contain y into 
 another. 
 
 Let us tak^ the equations 
 
 A'a;-j-BV = C'. 
 Resolving these equations we have 
 
 _ CB^ — BC^ _AC^--CA/ 
 "^""AB' — BA" ^""AB— BA'* 
 
 This is a general solution for all equations of the first 
 degree with two unknown quantities. 
 
 To show the use which may be made of these formulas in the 
 solution of equations, let there be the two equations, 
 5a;4-32/=19, 4a: + 7y = 29. 
 Comparing these with the general equations, we have 
 A = 5, B = 3, C = 19, A' = 4, B' = 7, C = 29, 
 whence, by substitution in the formulas for x and y^ we have 
 19X7 — 8X29 133 — 87 46 ^ 
 
 5X7 — 3X4 35—12 
 
 -23*-"'^ 
 
 5X29 — 19X4 145 — 76 
 5X 7— 3X4 35 — 12 
 
 -23-^' 
 
 Discussion. 
 
 
 In the above formulas for x and ?/, let A B' — B A' = 0, 
 CB' — BC and AC — CA' being each different from zero, 
 we shall then have 
 
 CB' — BC AC — CA' 
 
 ^ = "0 ' 2/= . 
 
 To interpret these results, we observe that the equation 
 
 AB' 
 
 A B' — B A' = gives A' = -r^r- ; substituting this value in 
 
 B 
 
 the equation A'a: -|- B'y = C, we have 
 A VJ 
 
DISCUSSION OF FORMULAS. 109 
 
 from which we obtain Aa;-|-By = -^; comparing this with 
 
 the equation Aa; + By = C, the left hand members, it will bo 
 perceived, are identical, while the right are essentially different ; 
 for if in the numerator CB'— BC, CB' be greater than BC, 
 
 C will be greater than -^^; and if C B' be less than B C, C will 
 
 be less than -^57- We conclude, therefore, that the two equations 
 B 
 
 proposed cannot in this case he satisfied^ at the same time, by any 
 
 system whatever of finite valiies for x and y. The question 
 
 therefore in ihis case is impossible. 
 
 Again, let us suppose A B' — B A'= 0, and at the same time, 
 
 C B' — BC'= ; the value of x in this case is reduced to jr. 
 
 To interpret this result, we remark that the equations proposed 
 may, in consequence of the relation A B' — B A'= 0, be put 
 ffider the form 
 
 Aa;4-By==C 
 
 Aa;+By = ^g7-, 
 
 equations which are identical, since from the relation C B' — BC 
 
 = 0, we have -^^j- = C. 
 
 JD 
 
 In order then to resolve the problem, we have in fact but one 
 equation with two unknown quantities ; the question therefore is 
 
 indeterminate. 
 
 B A' 
 
 Since the equation A B' — B A' = gives B'= —r— we have 
 
 A. 
 
 by substitution in the equation C B' — BC' = 0. 
 
 £J^-BC'=0, 
 
 or reducing, AC — C A' = ; we infer ^ therefore, that if the 
 
 valice of X be of the form ^ the. valtie of y will he of the sarM 
 form and the converse. 
 
 J 
 
tlO ELEMENTS OF ALGEBRA. 
 
 PROBLEMS FOR SOLUTION AND DISCUSSION. 
 
 1. A mercharkt has two sorts of wine, one of which costs a, 
 the other b shillings a gallon; from these he would make a 
 mixture of c gallons to be worth d shillings a gallon. How 
 much of each must he take ? 
 
 Let a: = the number of gallons of the first, y of the second, 
 
 . cid — b) cia — d) 
 
 we nave ' x = — ~^y = — — i. 
 
 a — b ^ a — b 
 
 How shall we interpret these results 1°. when i or a is equal 
 to d; 2°. when a^^b ; 2°. when az=b, and at the same time, 
 h=-d ; 4°. what condition is necessary in order that the question 
 may be solved in the exact sense of the enunciation? 
 
 Ans. In the first case, the quantity of one of the ingredients 
 v/ill be 0, as it should be, since, if the price of one of the ingre- 
 dients is equal to that of the mixture, none of the other will be 
 needed to make the mixture of the required price. In the second, 
 since the prices of the ingredients are both the same, a mixture 
 of a different price cannot, it is evident, be made from them ; the 
 question, therefore, requires an impossibility. In the third case, 
 the price of the ingredients and that of the mixture being each 
 the same, whatever number of gallons be taken of either, the 
 mixture will be of the required price ; the question is, therefore, 
 indeterminate. The number of solutions is, however, limited by 
 the circumstance that the number of gallons of both ingredients 
 taken together must be equal to c. Finally, in order that the 
 question may be answered in the exact sense of the enunciation, 
 the price of the mixture must be comprised between the prices 
 of the ingredients. 
 
 2. To find a number such, that if it be added to the num- 
 bers a and b respectively, the first sum will be m times the 
 second. 
 
 Putting z for the number, we have x = -:; . 
 
 1 — m 
 
DISCUSSION OF FORMULAS. Ill 
 
 'How shall we interpret this result when 7w= 1? How when 
 m=l, and at the same time a = b? How when m is greater 
 than 1, and mb greater than a? What conditions are necessary, 
 in order that the question may be solved in the exact sense of the 
 enunciation ? 
 
 3. The sum of two numbers is a, and the sum of their pro- 
 ducts by the numbers m and n respectively is b. "WTiat are the 
 numbers ? 
 
 Putting X and y for the numbers, we have 
 
 b — na ma — b 
 
 x = , 7J. 
 
 Tfi —•— n 
 
 How shall we interpret these results, when w is greater than 
 
 n and na greater than b? How when m = n? How Avhen 
 
 m = n, and at the same time na = b? What conditions are 
 
 necessary in order that the question may be solved in the exact 
 
 sense of the enunciation ? 
 
 4. Two numbers are in proportion of fl to i ; but if c be added 
 
 to both, they will then be in proportion of m to n ; what are the 
 
 numbers ? ^ 
 
 . acim — n) . bcim — n) 
 
 Ans. — ^^ ; — - and — ^ -, — -. 
 
 an — bm an — om 
 
 SECTION X.— Theory of Inequalities. 
 
 86. In the reasonings, which relate to the discussion of a 
 problem, we have frequent occasion to make use of the expres- 
 sions '^greater ihari^'' ^^less than.''' In such cases we shall attain 
 a greater degree of conciseness, by representing each of these 
 expressions by a convenient sign. It is agreed to represent the 
 expression " greater tkarC' by the sign ]> ; thus, a greater than h 
 is expressed hy g.^b. The same sign by a change of position 
 
112 ELEMENTS OF ALGEBRA. 
 
 is made to represent the phrase "less than;" thus, 2 less than h 
 is expressed by a <^ b. 
 
 An equation of the form a=za is called an eqality. An 
 expression of the form a ]>• 3, or a <^ ^ is called an inequality. 
 
 The principles established for equations apply in general to 
 inequalities. As there are some exceptions, however, we shall 
 state the principal transformatioi^s, which may be made upon 
 inequalities, together with the exceptions which occur. 
 
 1°. We may always add the same quantity to both members of 
 an inequality, or subtract the same quantity from both members^ 
 and the inequality will continue in the same sense. 
 
 Thus-, let 3 <^ 5 ; adding 8 to both sides, we have 
 8-|-3<5 + 8, or 11<13. 
 
 Again let — 3 ^ — 5 ; adding 8 to both sides we have 
 8_3>8 — 5, or5>3. 
 
 This principle enables us, as in the case of equations, to 
 transpose a term from one member of an inequality to the 
 other; thus, from the inequality a'^-)- ^^^^^^^^ — aSwe obtain 
 2a« + i'^>3c* 
 
 2°. We may in all cases add, member to merriber, two or more 
 inequalities established in the same sense, and the in£qualityt 
 which results, will exist in the sense of the proposed. 
 
 Thus, let there he a'^ b, c"^ d, e'^f; we have 
 a + c + e>b + d+f 
 
 But ifive subtract, member from member, two or more inequal' 
 ities established in the same sense, the inequality, which results^ 
 will not always exist in the sense of the proposed. 
 
 Let there be the inequalities 4 <^ 7, 2 <] 3, we have bysub- 
 traction4 — 2<7 — 3, or2<4. 
 
 But let there be the inequalities 9 <^ 10 and 6 < 8, subtract- 
 ing the latter from the former, we have 
 
 9 — 6>10 — 8, or3>2. 
 
 3*. We may multiply or divide the two members of an inequal' 
 
THEORY OP INEQTTALITIES. 113 
 
 ity by any positive or absolute number, and the inequality ^ which 
 results, ivill exist in the sense of the proposed. 
 
 Thus, if we have a<^b, multiplying both sides by 5, we have 
 5a<:^5b. 
 
 By means of this principle, we may free an inequality from 
 its denominators Thus, let there be 
 
 a^ — b^a' — b^ 
 2d ^ 3a ' 
 we have by multiplication (a'' — b^)2a'^{a^ — b^)2d, and by 
 division 3a ^2^. 
 
 But if we multiply or divide the two members of an inequality 
 by a negative quaiitity, the inequality, which results, will exist in 
 the contrary sense. 
 
 Thus, let 8 ^ 7 ; multiplying both sides by — 3, we have 
 _24< — 21. 
 
 From this it follows, that if we change the sign of each term 
 of an inequality, the inequality, which results, will exist in a sense 
 contrary to that of the proposed ; for this transformation will be 
 equivalent to nntUiplying both members by — 1, 
 
 87. Let there now be proposed the inequality 
 
 Multiplying both sides by 3, we have 
 
 21a: — 23>2a:+15; 
 whence transposing and reducing, we have 
 
 a:>2. 
 
 Here 2 is the limit to the value of x, that is, if we substitute 
 for X in the proposed any value greater than 2, the inequality 
 will be satisfied. The process, by which the limit to the value 
 of the unknown quantity is determined, is called resolving the 
 inequality. 
 
 8 
 
114 ELEMENTS OF ALGEBRA. 
 
 EXAMPLES. 
 
 1. To find the limit to the value of x in the inequalities 
 
 2. To find the limit to the value of x in the inequalities 
 
 X X ^6 X 
 
 3. To find the limit to the value of x in the inequalities 
 
 ax-\- ah <C—, 
 
 88. The theory of inequalities may be applied to the solution 
 of certain problems. 
 
 1. The double of a number diminished by 5 is greater than 
 25, and triple the number diminished by 7 is less than double the 
 number increased by 13. Kequired a number that shall possess 
 tliese properties. 
 
 By the question, we have 
 
 2a: — 5>25 
 3a; — 7<2a; + 13. 
 Resolving these inequalities, we have a;^ 15, a;<^20. Any 
 number J^herefore, entire or fractional, comprised between 15 and 
 20 will satisfy the conditions of the question. 
 
 2. A shepherd being asked the number of his sheep re- 
 plied, that double their number diminished by 7 is greater 
 than 29, and triple their number diminished by 5 is less than 
 double their number increased by 16. Required the number of 
 sheep. 
 
EXTRACTION OF THE SQUARE ROOT. 115 
 
 Resolving the question, we have x^ 18, and a;<^21. Here 
 all the numbers, comprised between 18 and 21, will satisfy the 
 inequalities ; but since the nature of the question requires that 
 the answer should be an entire number, the number of solutions 
 is limited to 2, viz. a; = 19, a; = 20. 
 
 3. A market woman has a number of oranges, such, that 
 triple the number increased by 2, exceeds double the number 
 increased by 61 ; and 5 times the number diminished by 70 is 
 less than 4 times the number diminished by 9. How many 
 oranges had she ? 
 
 4. 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 ? 
 
 5. The sum of two numbers is 25 ; if the greater be divided 
 by the less, the quotient will be greater than 3, and if the less 
 be divided by the greater the quotient will be greater than \. 
 What are the numbers ? 
 
 SECTION XI. — Extraction of the Square Root. 
 
 89. Let it now be proposed to find a number, which multiplied 
 by five times itself, will give a product equal to te5. 
 
 Putting X for the number required, we have by the question 
 5r*=«:125, from which we obtain x^ = 25. This equation is 
 essentially different from any, which we have hitherto considered. 
 It is called an equation of the second degree, because it contains 
 X raised to the second power. To find the value of x, we must 
 see what number ihultiplied by itself will give 25. It is obvious, 
 that the number 5 will fulfil this condition ; we have therefore 
 « = 5. 
 
 The value of x is easily found in the present example, but 
 
116 ELEMENTS OF ALGEBRA. 
 
 in Others it will be more difficult. Hence arises this new arith- 
 metical question, viz. To find a number, which multiplied by 
 itself will give a product equal to a proposed number, or which 
 is the same thing, from the second power of a number to deter- 
 mine the first. 
 
 A number, which multiplied by itself will produce a given 
 number, is called the square or second root of this number. The • 
 process for finding the second root is called extract^g the square 
 or second root. 
 
 In the following table, we have the nine primitive numbers 
 with their squares written under them respectively. 
 1, 2, 3, 4, 5, 6, 7, 8, 9. 
 1, 4, 9, 16, 25, 36, 49, 64, 81. 
 
 By inspection of this table, it will be perceived, that among 
 entire numbers consisting of one or two figures, there are nine 
 only, which are squares of other entire numbers. The remain- 
 der have for a root an entire number plus a fraction. Thus 53, 
 which is comprised between 49 and 64, has for its square root 7 
 plus a fraction. 
 
 The numbers in the second line of this table being the squares 
 of those in the first, conversely, the numbers in the first line 
 are the square roots of those in the second. If, therefore, 
 the number, the square root of which is required, consists of 
 one or two figures only, its root will be readily found by means 
 of the table. 
 
 Let it be propose^ to find the root of a number consisting of 
 more than two figures, 6084, for example. 
 
 The square of 9, the largest number consisting of one figure, 
 is 81, and the square of 100, the smallest number consisting of 
 three figures, is 10000 ; the square root of 6084 will, therefore, 
 consist of two places, viz. units and tens. 
 
 To determine then a method, by which to return from the 
 proposed number to its root, let us observe the manner, in 
 which the different parts of a number consisting of two places, 
 47, for example, are employed in forming the square of this 
 
EXTRACTION OF THE SQUARE ROOT. 117 
 
 number. For this purpose we decompose 47 into two parts, 
 viz. 40 and 7, or 4 tens and 7 units. Designating the tens by 
 a and the units by 3, we have a -}" ^ = ^7, and squaring both 
 sides a^-f-2a3-f-Z''^ = 2209. Thus the square of a number, 
 consisting of units and tens, is composed of three parts, viz. ike 
 square of the tens, plus twice the product of the tens multiplied, 
 by the units, plus the square of the units. Thus in 2209, the 
 square of 47, we have 
 
 The square of the tens {c^) = 1600 
 
 Tioice the tens by the units (2 a 3) = 560 
 The square of the units {b^) = 49 
 
 2209 
 Considering, then, the proposed number 6084 as composed 
 of the square of the tens of the root sought, twice the product 
 of the tens by the units, and the square of the units, if we can 
 discover in this number the first of these parts, viz. the square 
 of the tens, the tens of the root will be readily found. The 
 square of an exact number of tens, it is evident, can have no 
 figure inferior to hundreds. Separating then the two last figures 
 of the proposed from the rest by a comma, the square of the 
 tens will be found in 60, the part at the left of the comma, 
 which, in addition to the hundreds in the square of the tens, 
 will also contain those, which arise from the other parts of the 
 square. 60 is comprised between 49 and 64, the roots of 
 which are 7 and 8 respectively ; 7 will, therefore, be the figure 
 denoting the tens in the root sought. Indeed 60 00 is com- 
 prised between 49 00, and 64 00, the squares of 70 and 80 
 respectively ; the same is the case with 60 84 ; the root required 
 will therefore consist of 7 tens and a certain number of units 
 less than ten. 
 
 The figure 7 being thus obtained, we place it at the right 
 of the proposed, taking care to separate them by a vertical 
 line; we then subtract 49, the square of 7, from 60, and to 
 the remainder 11 we bring down 84, the two other figures of 
 
118 ELEMENTS OF ALGEBKA. 
 
 the proposed. The result 1184 of this operation will then con- 
 tain twice the product of the tens of "the root hy the units, plus 
 th6 square of the units. Twice the product of the tens by the 
 units will, it is evident, contain no figure inferior to tens. Sep- 
 arating then 4, the right hand figure of the remainder 1184, 
 from the rest by a comma, the part 118 of this remainder, at the 
 left of the comma, must contain the double product of the tens 
 by the units, together with the tens arismg from the square of 
 the units. 
 
 The double product of the tens is 14; dividing, therefore, 
 118 by 14, the quotient 8 will be the unit figure exactly, or in 
 consequence of the tens arising from the square of the units, 
 it may be too large by 1 or 2. To determine whether 8 be the 
 right figure for the units of the root, we multiply twice the tens 
 by 8 and subtract the result from 1184, the remainder 64 being 
 equal to the square bf 8, shows that 8 is the unit figure sought. 
 We have 78, therefore, for the root required. The operation 
 will stand thus, 
 
 60,84 
 49 
 
 118,4 
 112 
 
 14 
 
 8 
 
 64 
 64 
 
 To complete the root, we place 8, the unit figure, at the right 
 of 7, the figure for the tens. The work, moreover, may be 
 abridged by writing the 8 at the right of the divisor, and 
 multiplying 148 the number thus formed by 8. We thus ob- 
 tain in one expression twice the tens by the units and the 
 square of the units; this being equal to the remainder 1184 
 proves, as before, that 8 is the right figure for the units of the 
 root. 
 
EXTRACTION OF THE SQUARE ROOT. 119 
 
 With this modification, the work will stand thus, 
 
 60, 84 I 78 
 49 
 
 118,4 148 
 118 4 
 
 Let us take, as a second example, the number 841. Pursuing 
 the same course as in the preceding example, we find 2 for 
 the tens of the root ; subtracting the square of the tens, the 
 remainder will be 441. Separating the unit figure in this re- 
 mainder from the rest by a comma, and dividing the part at the 
 left by double the tens, in order to obtain the unit figure of the 
 root, we have 11 for the result. This is evidently too much. 
 Indeed, we cannot have more than 9 for the units ; we therefore 
 try 9. This proves to be the correct figure. The root sought is 
 therefore 29. 
 
 The operation will be as follows : 
 
 8,41 
 4 
 
 29 
 
 44,1 I 49 
 441 
 
 90. Any number however large may be considered as com- 
 posed of units and tens ; 345, for example, may be considered as 
 composed of 34 tens and 5 units. 
 
 Let it now be proposed to find the second root of 190969. 
 This number exceeds 10 000 and is less than 1000 000 ; its 
 root will therefore consist of three places. But from what has 
 been said, the root may be considered as composed of two 
 parts, units and tens. The proposed will, therefore, consist 
 of three parts, viz. the square of the tens of the root, twice 
 the tens by the units and the square of the units. The square 
 of the tens will have no figure inferior to hundreds. Separating, 
 therefore, the last two figures from the rest by a comma, the 
 tens of the root will be found by extracting the square root 
 
Sfe 
 
 ELEMENTS OF ALGEBRA. 
 
 of 1909, the part of the proposed at the left of the comma. 
 Regarding 1909 for the moment as a separate number, its root 
 will evidently consist of two places, units and tens. The method 
 of finding the root will, therefore, be the same as in the pre- 
 ceding examples. Performing the necossary operations we obtain 
 43 for the root and a remainder of 60. There will therefore, 
 be 43 tens in the root of the proposed, and bringing down the 
 last two figures of the proposed by the side of 60, the result 
 6069 will contain twice the product of the tens of the root 
 sought by the units, plus the square of the units. Separating, 
 therefore, the right hand figure from the rest by a comma, we 
 divide 606, the part on the left of the comma, by 86, twice the 
 tens ; this gives 7 for the unit figure. Placing the 7, therefore, 
 at the right of 43, the part of the root already found, and also 
 at the right of 86, and multiplying this last by 7, we have 6069 
 for the result. 7 is, therefore, the right unit figure, and the root 
 of the proposed is 437. 
 
 The following is a table of the operations. 
 
 19,09,69 I 437 
 16 
 
 309 I 83 
 249 
 
 606,9 
 606 9 
 
 867 
 
 The same process, it is easy to see, may be extended to any 
 number however large. From what has been done, therefore, 
 the following rule for the extraction of the second root will 
 be readily inferred, viz. 1°. Separate the number into parts 
 of two figures each, beginning at the right. 2°. Find the great- 
 est second power in the left hand part ; write the root as a quO' 
 tient in divisio?i, and subtract the second power from the left hand 
 part. 3°. Bring down the two next figures at the right of the 
 remainder for a dividend and double the root already found for 
 a divisor. See how many times the divisor is contained in the 
 
EXTRACTION OF THE SQUARE ROOT. 121 
 
 dividend, neglecting the right hand figure. Write the result in 
 the root at the right of the figure previously found, and also at 
 the right of the divisor. 4°. Multiply the divisor, thus aug- 
 mented, by the last figure of the root and subtract the product 
 from the whole dividend. 5°. Bring down the next two figures 
 as before, to form a neio dividend, and double the root already 
 found for a divisor, and proceed as before. The root will be 
 doubled, if the right hand figure of the last divisor be doubled. 
 
 91. If the number proposed be not a perfect square, we shall 
 obtain by the above rule, the root of the greatest square number 
 contained in the proposed. Thus, let it be required to find the 
 square root of 1287. Applying the rule to this number, we ob- 
 tain 35 for the root with a remainder 62. This remainder shows 
 that 1287 is not a perfect square. The square of 35 is 1225, 
 that of 36 is 1296 ; whence 35 is the root of the greatest square 
 contained in the proposed. 
 
 92. When the proposed number is not a perfect square a 
 doubt may sometimes arise, whether the root found be that of 
 the greatest square contained in this number. This may be 
 readily determined by the following rule. The square oi a-\-\ 
 h c^ -\- 2a -{-!', whence the square of a quantity greater by 
 unity than a exceeds the square oi a hj 2a-\-\. From this it 
 follows, that if the root obtained should be augmented by unity 
 or more than unity, the remainder after the operation must be at 
 least equal to twice the root plus unity. When this is not the 
 case, the root obtained is that of the greatest square contained in 
 the proposed. 
 
 EXAMPLES. 
 
 1. To find the square root of 56821444. Ans. 7538. 
 
 2. To find the square root of 17698849. Ans. 4207. 
 
 3. To find the square root of 1607448649. Ans. 40093. 
 
 4. To find the square root of 12103441. Ans. 3479. 
 
 5. To find the square root of 48303584206084. Ans. 6950078. 
 
122 ELEMENTS OF ALGEBRA. 
 
 93. From what has heen done, it will be perceived, that there 
 are many whole numbers, the' roots of which are not whole 
 numbers. What is remarkable in regard to these numbers is, 
 that they will have no assignable roots. Thus the numbers 
 3, 7, 11 have no assignable roots, that is, no number can be 
 found either among whole or fractional numbers, which multi- 
 plied by itself will produce either of these numbers. The proof 
 of this depends upon the following proposition, which, we shall 
 now demonstrate, viz. 
 
 Every number P, which luill exactly divide the product A B q/* 
 two numbers A and B, and which is prime to one of these num- 
 bers must necessarily divide the other number. 
 
 Let us suppose that P will not divide A, and that A is greater 
 than P. Let us apply to A and P the process of the great- 
 est common divisor, designating the quotients, which arise, by 
 Q> Q'> Q" • • • and the remainders by R, R', R" . . . respec- 
 tively. It is evident, that if the operation be pursued sufficiently 
 far, we shall obtain a remainder equal to unity, since by hypothe- 
 sis A and P are prime to each other. . This being premised we 
 have the following equations 
 
 A = PQ + R 
 P = RQ'4-R' 
 R = R'Q"+R" 
 
 Multiplying the first of these equations by B, and dividing 
 by P, we have 
 
 -p- = BQ+-p- 
 
 AB 
 
 By hypothesis -p- is an entire number, and since B and Q 
 
 are each entire numbers the product BQ is an entire number. 
 
 BR 
 
 It follows therefore, that -p- must be an entire number ; whence 
 
 B multiplied by the remainder R is divisible by P. 
 
EXTRACTION OF THE SQUARE ROOT. 123 
 
 Again, multiplying the second of the ahove equations by B 
 and dividing by P, we have 
 
 BRQ' , BR' 
 
 B 
 
 p -T- p 
 
 "RR 
 
 But we have already shown that -5- is an entire number, 
 
 whence — p — is an entire number. This being the case, -p- 
 
 must be an entire number ; whence B multiplied by the remain- 
 der R' must be divisible by P. 
 
 If then the remainder R' is equal to unity, the proposition 
 is demonstrated, since in this case we shall have B X 1 or B 
 divisible by P. But if the remainder R' is not equal to 
 unity, it is evident, that if the process of the greatest common 
 divisor be applied to the quantities A and P until a remain- 
 der is obtained equal to unity, we may in the same manner 
 as above, prove that B multiplied by this remainder will be 
 divisible by P. 
 
 We conclude, therefore, that if P, which we have supposed 
 not to divide A, will not divide B, it will not divide AB the 
 product of A by B. 
 
 Returning now to our purpose, it is evident, in order that a 
 
 fractional number - may be the root of an entire number c, we 
 must have 
 
 1^ = '' 
 But if c be not a perfect square, its root will not be an en- 
 tire number, that is, a will not be divisible by h; but from 
 what has just been demonstrated, if a is not divisible by 
 b, ay^a or c^ will not be divisible by 3, and by consequence 
 
 c^ will not be divisible by b^ ; whence —^ cannot be equal to an 
 
 entire number c. 
 
 94. Though the roots of numbers, which are not perfect 
 squares cannot be assigned either among whole or fractional 
 
124 ELEMENTS OF ALGEBRA. 
 
 numbers, yet, it is evident, there must be a quantity, which mul- 
 tiplied by itself will produce any number whatever. Thus the 
 root of 53 cannot be assigned; yet there must be a quantity, 
 which multiplied by itself will produce 53. This quantity, it is 
 evident, lies between the numbers 7 and 8, for the square of 7 is 
 49, and the square of 8 is 64. If then we divide the difference 
 between 7 and 8 by means of fractions, we shall obtain numbers, 
 the squares of which will be greater than 49 and less than 64, 
 and which will approach nearer and nearer to 53. 
 
 95. All numbers, whether entire or fractional, have a common 
 measure with unity; on this account they are said to be cotw- 
 mensurable ; and since the ratio of these numbers to unity may 
 always be expressed by entire numbers, they are on this account 
 called rational numbers. 
 
 The root of a number which is not a perfect square can have 
 no common measure with unity; for, since it is impossible to 
 express this root by any fraction, into how many parts soever we 
 conceive unity to be divided, no fraction can be assigned suffi- 
 ciently small to measure at the same time this root and unity. 
 The roots of numbers, which are not perfect squares, are on this 
 account called incommensurable or irrational quantities. They 
 are sometimes also called surds. 
 
 To indicate that the square root of a quantity is to be taken, 
 we use the character /^, which is called a radical sign. Thus 
 y\/l6 is equivalent to 4. /k^2 is an iiicommensurable or surd 
 quantity. 
 
 EXTRACTION OF THE SQUARE ROOT OF FRACTIONS. 
 
 96. Since a fraction is raised to the second power by rais- 
 ing the numerator to the second power, and the denominator 
 to the second power, it follows that the square root of a frac- 
 tion will be found by extracting the square root of the numera- 
 
 9 3 
 tor, and of the denominator. Thus, the square root of -r-r is -. 
 
EXTRACTION OF THE SQUARE ROOT. 125 
 
 If either the numerator or denominator of the fraction is 
 
 not a perfect square, the root of the fraction cannot be found 
 
 exactly. We may, however, always render the denominator 
 
 of a fraction a perfect square by multiplying both terms of 
 
 the fraction by the denominator. This will not alter the 
 
 / value of the fraction. The root of the denominator may then 
 
 be found, and for that of the numerator, we must take the 
 
 number nearest the root. Thus, if it be required to extract 
 3 
 
 the square root of -, multiplying both terms by 5, the fraction 
 o 
 
 becomes 7^, the root of which is nearest -, accurate to within 
 
 less than -. 
 o 
 
 If the denominator of the fraction contain a factor, which 
 is a perfect square, it will be sufficient to multiply both terms by 
 the other factor of the denominator. Thus, let. it be required 
 
 Q 
 
 to find the square root of — ; multiplying both terms by 6, the 
 
 48 . 7 
 
 fraction becomes k— 7, the root of which is —^, accurate to within 
 
 oJ4 Jo 
 
 less than —^. 
 10 
 
 If a** greater degree of accuracy is required, we convert the 
 fraction into another, the denominator of which is a perfect 
 square, but greater than that obtained by the method above. 
 
 3 1 
 
 To find, for example, the square root of - to within -r^, the 
 
 o 15 
 
 fraction must be converted into 225ths. This is done by multi- 
 
 3 135 
 
 plying both terms by 45. Thus we have - = — — , the root of 
 
 5 225 
 
 which is nearest v?» accurate to withm less than — -. 
 15 15 
 
 After making the denominator a perfect square, we may mul- 
 tiply both terms of the proposed fraction by any number, 
 which is a perfect square, and thus approximate the root 
 more nearly. If, for example, we multiply both terms of 
 
126 ELEMENTS OF ALGEBRA. 
 
 -^ by 144, the square of 12, we obtain 75^7^, the root of which 
 
 4-fi *^ 
 
 is nearest 7^7:. Thus, we have the root of - to within less 
 60 
 
 thanl. 
 
 97. We may in this way approximate the roots of whole 
 numbers, the roots of which cannot be exactly assigned. 
 
 If it be required, for example, to find the square root of 2 ; 
 we convert it into a fraction the denominator of which will be 
 
 a perfect square. Thus, if we put 2 = ^— -, we have for the 
 
 root — or 1^^, accurate to within less than —=. 
 lo 10 
 
 In general, to find the square root of a number accurate to 
 within a given fraction, we multiply the proposed number by the 
 square of the denominator of the giveii fraction ; we then fina 
 the entire part of the square root of this product, and divide the 
 result by the denominator of the given fraction. 
 
 This rule may be demonstrated as follows. Let a be the num- 
 ber proposed, and let it be required to find the root of a to within 
 
 less than!. 
 n 
 
 a 7l 
 We shall have, it is evident, a = —^ ; let r be the entire part 
 
 of the root of the numerator of an^; an^ will be comprised between 
 r^ and (r -f- 1 Y, and by consequence the square root of a will 
 
 be comprised between those of — ^ and - — -^ — , that is to say, 
 
 r r I 1 7* 
 
 between - and "^ ; whence - will be the root of a to within 
 n n n 
 
 less than l 
 n 
 
 98. To approximate the root of a number, which is not a 
 perfect square, it will be most convenient to employ some power 
 oT 10 as the multiplier of the proposed, or which is the same 
 thing, to convert the proposed into a fraction, the denominator of 
 
EXTRACTION OF THE SQUARE ROOT. 127 
 
 which shall be some power of 10. Thus, to approximate 
 the root of 2, let us put 2 = -r^ or 2.00, the approximate root 
 
 will be 1.4. Again, let 2 = ^ or 2.0000, the approximate 
 
 root will be 1.41. 
 
 99. From w^hat has been done, and indeed from tl^e nature 
 of multiplication it follows, that the number of decimal places 
 in the power will be double the decimal places in the root. To 
 find the approximate root of an entire number by the aid of 
 decimals therefore, we must annex to this number twice as many- 
 zeros as there are decimal places Avanted in the root. Thus, 
 if 5 places are required in the root, ten zeros must be annexed. 
 The zeros may be annexed as we proceed, it being observed, 
 that two zeros must be annexed for every new figure placed in 
 the root. 
 
 The root of 7, to three places, will be found as follows. 
 7 (2.645 
 4 
 
 300 
 276 
 
 2400 
 2096 
 
 30400 
 26425 
 
 3975 
 
 If the proposed be already a decimal, the number, of decimal 
 places must be made even by annexing a zero, if necessary. If 
 the root of the number, thus prepared, is not sufficiently exact, 
 two zeros must be annexed for every new figure required in \he 
 root. 
 
 100. To find the root of a vulgar fraction by the aid of deci 
 
128 ELEMENTS OF ALGEBRA. 
 
 mals, we convert this fraction into a decimal and then extract the 
 root. 
 
 If the proposed consist of an entire part and a fraction, we 
 convert the fraction into a decimal, annex it to the entire part, 
 and then extract the root. 
 
 In converting the fraction into a decimal, it will be necessary 
 to pursue the operation, until twice as many decimals are obtained, 
 as are wanted in the root. 
 
 EXAMPLES. 
 
 1. Find the square root of 11 to within less than — 
 
 It 
 
 Ans. 3 
 
 2. Find the square root of 223 to within less than —- 
 
 15' 
 
 15' 
 
 40* 
 Ans. 14^. 
 
 3. Find the square root of 7 to within .01. Ans. 2.64. 
 
 4. Find the square root of 227 to within .0001. 
 
 Ans. 15.0665. 
 
 5. Find the square root of j^ to 3 places of decimals. 
 
 Ans. 0.645. 
 3 
 
 6. Finu the approximate square root of 1~. Ans. 1.32 -[-• 
 
 13 
 
 7. Find the approximate square root of 2 —=. 
 
 Ans. 1.6931 +. 
 
 8. Find the approximate square root of 31.027. 
 
 Ans. 5.57+. 
 
 9. Find the approximate square root of 0.01001. 
 
 Ans. 0.10004 +. 
 
 10. Find the approximate square root of 3271.4707. 
 
 Ans. 57.19+. 
 
SQUARE ROOT OF ALGEBRAIC QUANTITIES. 129 
 
 EXTRACTION OF THE SQUARE ROOT OF ALGEBRAIC QUANTITIES. 
 
 101. By the rule for multiplication we have 
 
 A monomial is therefore raised to the square by squaring the 
 coefficient and doubling the exponent of each of the letters. 
 Whence to extract the square root of a monomial, it is necessary 
 1°. to extract the root of the coefficient ; 2°. to divide the expo- 
 nents of each of the letters by 2. 
 
 According to this rule, we have 
 
 V6^7¥=San\ * 
 
 ^/625^¥? = 25abU\ 
 
 In order that a monomial may be a perfect square, its coeffi- 
 cient, it is evident from the preceding rule, must be a perfect 
 square and the exponent of each of the letters must be an even 
 number. 
 
 Thus 98a 5* is not a perfect square. Its root can, therefore, 
 be only indicated by means of the radical sig-n, thus ^98 a b*. 
 Expressions of this kind are called irrational quantities of the 
 second degree, or more simply radicals of the second degree. 
 
 102. The second power of a product, it is easy to see, is the 
 same as the product of the second powers of all its factors. It 
 follows, therefore, that the square root of a product will be the 
 same as the product of the square root of all its factors. 
 
 By means of this principle, we may frequently reduce to a 
 more simple form expressions of the kind, which we are here 
 considering. Thus, the above expression w98ab* may be put 
 under the form \/Wb*X^^; but ^/49^=7Z>^ whence 
 ^/98^'=7bW2^. 
 
 In like manner, we have 
 V864a'^'c" = Vl44a^i*c'»X6^c= 12aiV V6bl 
 
130 ELEMENTS OF ALGEBRA. 
 
 In the expression 1b^ w2aj \2ab^c^ ^/Qbc, the quantities 73', 
 \2ab'^c^ placed without the radical sign are called the coefficients 
 of the radical. The expressions themselves are said to be 
 reduced to their most simple form. 
 
 From what has been done, we have the following rule for 
 reducing irrational quantities, consisting of one term, to their 
 most simple form, viz. Separate the quantity proposed into two 
 parts, one of lohich shall contain all the factors, which are perfect 
 squares, and the other those lohich are not ; write the roots of the 
 factors, which are perfect squares, without the radical sign as 
 multipliers of the radical quantity, and retain under the radical 
 sign the factors, which are not perfect squares. 
 
 1. To reduce ^/ida^bc to its most simple form. 
 
 Ans. 5a^3abc. 
 
 2. To reduce ^S2a^b^c to its most simple form. 
 
 Ans. 4.aHW2c'. 
 
 3. To reduce ^ 175 a^b^c''d to its most simple form. 
 
 Ans. 5aHcWTbcd, 
 
 4. To reduce ^^Oda^b^c^de to its most simple form. 
 
 Ans. 9ab^c\^5ade, 
 
 5. To reduce \f 29^.c^ b'^ & d^ ^ to its most simple form. 
 
 Ans. 1 d^b^cde'^^abce. 
 
 6. To reduce VsiTo^^V^ to its most simple form. 
 
 Ans. Wd^b^edfsfr^lxd, 
 
 7. To reduce 's/Vd\^:(£'b^(^d to its most simple form. 
 
 Ans. Via^b'^c's/^abcd. 
 
 103. The square of — a, it will be observed, is c^, as well 
 as that of A- a; the root therefore of c?, may be either -^-aon 
 — a. Both of these roots may be comprehended in one expres- 
 sion by means of the double sign ±. Thus 
 
SQUARE ROOT OF ALGEBRAIC QUANTITIES. 131' 
 
 The double sign, it is evident, should be considered as ajSect- 
 ing the square root of all quantities whatever. 
 
 If, the monomial proposed be negative, the square root i»' 
 impossible ; since there is no quantity, positive or negative, 
 which multiplied by itself will produce a negative quantity. 
 Thus, a/ — a, /v/ — 2b^ are impossible or imaginary quanti- 
 ties. 
 
 Expressions of this kind may be simplified in the same man- 
 ner as radical expressions, which are real. Thus V — 9 may 
 be put under the form \^ — 1 X 9 ; whence 
 
 In like manner \/ — 4<z^ = 2flV — 1. 
 
 104. We proceed to the extraction of the square root of 
 polynomials. 
 
 A quantity consisting of two terms cannot, it is evident, be a 
 perfect square, for the square of a simple quantity will be a sim- 
 ple quantity, and the square of a binomial consists always of 
 three terms. 
 
 This being premised, let the proposed be a trmomial, its root, 
 it is evident, will consist of at least two terms. Let w -|- w be 
 the root, we have {m -\- rif = 7r? -\- 2m7i -J- n?. 
 
 This shows, that if the proposed be arranged with reference to 
 the powers of some letter that, P. the first term of the proposed 
 will be the square of the first term of the root sought; 2°. the 
 second term of the proposed will be equal to twice the first 
 term of the root multiplied by the second; 3°. the third term 
 of the proposed will be the square of the second term of the 
 root. 
 
 Let it be proposed to extract the root of the trinomial 
 2^a^b^c+\Qa^(?-\-^b\ 
 
 Arranging with reference to the letter a, the proposed be^ 
 comes 16a*c= + 2^aH^c + 9^»^ 
 
 In order to obtain the root, we extract according to what 
 has been said the root of the first term 16 aV, which gives 
 
132 ELEMENTS OF ALGEBBA. 
 
 4fl'c. This is the first term of the root. Dividing next the 
 second term 2^(^b^c by Sc^c, twice the term of the root already 
 found, we have 3^' for the second term of the root, and since 
 the square of this is equal to 93^ the remaining term of the 
 proposed, the proposed is a perfect square, the root of which 
 IS ^0^0 + W. ^ f 
 
 Again, let the proposed consist of more than three terms, 
 its root will consist of more than two terms. Let it consist 
 of three and let m-\-n-\-p be the root. The expression 
 m-\-n-\-p may be put under the form {m •\- n) -\- p ; forming 
 the square after the manner of a binomial, we have for the re- 
 sult {m -f- Tif -\- 2 [m -\- n) p -\- p^, or developing [m -\- rif the 
 result will be tt^ -\- ''Zmn -\~ rl^ ~\- 2 {m '\- n) p -^ p^. The pro- 
 posed, therefore, being arranged with reference to the powers 
 of some letter, it is evident, that the first term of the root will 
 be found by extracting the root of the first term of the proposed, 
 and that the second term of the root will be found by dividing 
 the second term of the proposed by twice the first term of the 
 root already found. If, then, we subtract from the proposed 
 the square of the two terms of the root already obtained, the 
 remainder will be equal to twice the first two terms of the 
 root multiplied by the third plus the square of the third. Divid- 
 ing this remainder, therefore, by twice the terms of the root 
 already found, or which is the same thing, dividing the first 
 term of the remainder by twice the first term of the root, we 
 shall obtain the third term sought. Subtracting from the first 
 remainder twice the product of the first two terms of the root 
 by the third, together with the square of the third, if the result 
 be 0, the proposed is a perfect square, and the root is exactly 
 obtained. 
 
 Let it be proposed to find the square root of the polynomial 
 
 The proposed being arranged with reference to the letter 0, 
 the work will be as follows : 
 
SQUARE ROOT OF ALGEBRAIC QUANTITIES. 133 
 
 25a*^20aH4-4:9aH^ — 24:ab'+16b*)5^-^^^nb+A^ 
 
 A0a'b^ — 24:ab^ + l6b* 
 maH^ — 24:ab^ + l6b* 
 
 
 
 We begin by extracting the root of 25 a*, this gives 5a^ for 
 the first term of the root sought, which we place at the right 
 of the proposed and on the same line with it ; we then multiply 
 this term of the root by 2 and write the result 10 a^ under the 
 root. Dividing next the second term of the proposed by 10 a^ we 
 obtain — Sab for the second term of the root sought. Squaring 
 the part of the root already found, viz. 5a^^-3ab, and subtract- 
 ing the square from the proposed, we have for the first term of 
 the remainder 4.0 a^ P. Dividing this last by 10 a* the double of 
 5 a',. we obtain 4^^ for the quotient. 
 
 This is the third term of the root sought; forming next the 
 double product of 5a* — 2ab by 45*, and subtracting the result 
 together with the square of 45* from the first remainder the re- 
 sult is 0. The proposed is, therefore, a perfect square, and we 
 have for the root required 
 
 5a* — 3a5 + 45*. 
 
 The calculations in the above example may be performed with 
 more facility as follows. 
 
 25 a'- 
 25 a* 
 
 -30a^5 + 49a*5* — 24a53+165* 
 
 5a* — 3a5 + 45» 
 lOa^ — 3a5 
 
 " 
 
 -30a35 + 49a*5* 
 -SOaH+ 9a* 5* 
 
 10^2 — 6a5 + 45* 
 
 
 40a*5* — 24a53+165* 
 40a*5* — 24a53-j-165* 
 
 
 
 
 Having found the first term 5c^ of the rooot, we subtract its 
 square from the first term of the proposed, and bring down the 
 next two terms for a dividend. Dividing the first term of th« 
 
 L 
 
434 ELEMENTS OF ALGEBRA. 
 
 dividend by 10 a^, we obtain — 2ab, the second term of the root; 
 this we place by the side of 10 a^; we then muhiply the whole, 
 viz. 10 a^ — 3a^ by this second term and subtract the result from 
 the dividend, which gives a remainder 40a^b^; to this remainder 
 we bring down the two remaining terms of the proposed for a 
 new dividend. Doubling the two terms of the root already 
 'found for a new divisor, we write the result under 10 a^; divid- 
 'ing next the first term of the new dividend by the first term of 
 the divisor, we obtain 4:b^ the third term of the root, which we 
 place by the side of the last divisor ; we then multiply the whole 
 by this last term of the root, and subtracting the result from the 
 ilast dividend, remains. 
 
 105. The same process, it is easy to see, may be extended to 
 a polynomial of any number of terms whatever. 
 
 EXAMPLES. 
 
 1. To find the square toot of 
 
 4a^ + 12ffl3 :» + 13aV + 6fla:3 -f- a^. 
 
 Ans. 2a^-\-2aX'\- a?, 
 
 2. To find the square root of 
 
 9x*—12x^ + 16x^ — Sx + 4.. 
 
 Ans. Saf*-— 2a; + 2. 
 
 3. To find the square root of 
 
 4x' ^16x^-{- 24ar^ — 16a: + 4. 
 
 Ans. 2a^ — 4:X + 2. 
 
 4. To find the square root of 
 
 a;« _[_ 4 a:5 + lOz^ + SOa:^ + 25r^ + 24:r + 16. 
 
 Ans. a;3 + 2ar^ + 3a; + 4. 
 6. To find the square root of 
 
 4a:'' + 12a:5 + 5a:* — 2a;^ + 7ar^ — 2a: + 1. 
 
 Ans. 2a,-^ + 3ar^ — a:+l. 
 
 106. The polynomial proposed being arranged with refer- 
 ence to the powers of some letter, if the first term of the 
 
SQUARE ROOT OF ALGEBRAIC QUANTITIES. 135 
 
 proposed is not a perfect square, or if in the course of the 
 operation we arrive at a remainder, the first term of which is 
 not divisible by twice the first term of the root, the proposed 
 is not a perfect square, and the root cannot be exactly as- 
 signed. 
 
 The polynomial a^b-{- 4:0,^5"^ -\- 4:ab^, for example, is not, it 
 is easy to see, a perfect square ; the root therefore can only 
 be indicated thus, ^/a■^^ -\- 4.d^b'^ -\- 4^ab'^. We may, however, 
 apply to expressions of this kind the same simplifications, that 
 have already been applied to monomials. The proposed in- 
 deed may be put under the form w{d^-\-4:ab-\-4:b^)ab; 
 but the root of a^-\- Aab -\- 4tb^ is evidently a-\-2b, whence 
 
 ^/a?b-\-4aH^-{-^ab^ z={a-\-2b) s/'ab. 
 
 EXAMPLES. 
 
 1. To find the square root of 3 a* & — 6 a^ i'^ + 3 a'^ i^ . 
 
 Ans. a{a^b)^W. 
 
 2. To find the square root oi5a^b — mab^ + 46 b\ 
 
 Ans. (a — 3Z>)\/5^ 
 
 3. To find the square root of 12 aV/ + \2aH^ -\-2ab\ 
 
 Ans. b[2a-\-b)^^. 
 
 4. To find the square root o^ a? ~\-^aH ^^ah" -[-b\ 
 
 Ans. [a + b) Va + b. 
 
 5. To find the square root of a^ -{-a^b — ab"^ — P. 
 
 Ans. [a -)- b) w a — b. 
 
 SECTION XII. — EQUATIONS of the Second Degree. 
 
 107. An equation is said to be of the second degree, when it 
 contains the second power of the unknown quantity, without any 
 of the higher powers. 
 
136 ELEMENTS OF ALGEBRA. 
 
 In an equation of the second degree there can be, therefore, 
 three kinds of terms only, viz. 1". terms, which involve the 
 second power of the unknown quantity, 2°. terms, which in- 
 volve the first power of the unknown quantity, 3°. terms con- 
 sisting entirely of known quantities. 
 
 An equation, which contains all three of these different kinds 
 of terms is called a complete equation of the second degree. 
 
 If the second of these different kinds of terms be wanting, the 
 equation is then called an incoviplete equation of the second 
 degree. 
 
 A complete equation of the second degree is sometimes called 
 an affected equation, and an incomplete equation is sometimes 
 called a pure equation of the second degree. 
 
 108. We are now prepared for the solution of incomplete 
 equations of the second degree. 
 
 Let there be proposed, for example, the equation 
 
 3r^_29 = ^ + 510. 
 
 Freeing from denominators, we have 
 
 12ar^— 116 = ar^ + 2040; 
 transposing and uniting terms 
 
 llar^ = 2156, 
 or r^=196, 
 
 whence, extracting the foot of both members 
 
 :i=:14. 
 Equations of the second degree, it should be observed, admit 
 of two values for the unknown quantity, while those of the first 
 degree admit of but one only. This arises from the circum- 
 stance, that the second power of a quantity will be positive, 
 whether the quantity itself be positive or negative. 
 
 Thus we have x in the preceding example equal -f" 14 or 
 — 14, or, uniting both values in one expression, we have 
 2: = ±14. 
 
EQUATIONS OF THE SECOND DEGREE. 137 
 
 Let US talce, as a second example, the equation 
 
 Freeing from denominators, transposing and reducing, we 
 have 
 
 ar = -^, whence x 
 
 Y 29' 
 
 252 
 
 In this example -^ is not a perfect square ; we can therefore 
 
 obtain only an approximate value for x. 
 
 Let us take, as a third example, the equation 
 ar^+25 = 9. 
 
 Deducing the value of x from this equation, we have 
 • x=^^f'^^^^. 
 
 To find the value of ar, we are here required to extract the 
 square root of — 16. But this is impossible ; for, as there is no 
 quantity positive or negative, which multiplied by itself will 
 produce a negative quantity, -^ 16, it is evident, cannot have 
 a square root either exact or approximate. — 16 may indeed be 
 considered as arising from the multiplication of -f-4 by — 4; 
 but -\- 4 and — 4 are different quantities ; their product therefore 
 is not a square. 
 
 The result a: = \/ — 16 shows then,- that it is impossible to 
 resolve the equation, from which it is derived. In general, an 
 expression for the square root of a negative quantity is to be 
 regarded as a symbol of impossibility. 
 
 109. Equations of the kind, which we are here considering, 
 may always be reduced to an equation of the form ax'^ = b^ a and 
 h denoting any known quantities whatever, positive or negative. 
 It is evident, that they may be reduced to this state, by collecting 
 into one member the tervis, ivhich involve y^ and reducing them 
 to one term, and collecting the known terms into the other 
 member. 
 
13S ELEMENTS OF ALGEBRA. 
 
 v/ 
 
 Resolving the equation ax^ = b^ we have 
 
 T 
 a 
 
 This is a general solution for incomplete equations of the 
 second degree. 
 
 If - be a perfect square, the value of x may be obtained ex- 
 a 
 
 actly, if not, it may be found with such degree of approximation 
 as we please. If - be negative, we shall have! y^ a sym- 
 bol of impossibility. 
 
 From what has been done, we have the following rule for the 
 solution of incomplete equations of the second degree, viz. Col- 
 lect into one member all the terms, which involve the square of the 
 unknown quantity, and the known quantities into the other 4 free 
 the square of the unknown quantity from the quantities, by which 
 it is multiplied or divided ; the value of the unknown quantity 
 will then be obtained by extracting the square root of each memler. 
 
 QUESTIONS PRODUCING INCOMPLETE EQUATIONS OF THE 
 SECOND DEGREE. 
 
 1. What two numbers are those, whose difference is to the 
 greater as 2 to 9, and the difference of whose squares is 128 ? 
 
 Let 9 a: = the greater and 2 a: = the difference, then, &c. 
 
 Ans. 18 and 14. 
 
 2. It is required to divide the number 14 into two such parts, 
 that the quotient of the greater part divided by the less may be 
 to the quotient of the less divided by the greater as 48 to 27. 
 
 Let X = the greater, then 14 — x= the less, and we have 
 
 14 — X X 
 
 or 272r^ = 48(14 — a:)'^; , 
 
 dividing by 3 to make the coefficients perfect squares 
 
 9a:'=16(14 — a:)=^; 
 whence 8a: = 4 (14 — x). Ans. 8 and 6. 
 
EQUATIONS OF THE SECOND DEGREE. 100 
 
 3. It is required to divide the number 18 into two such parts, 
 4hat the squares of these parts may be in the proportion of 25 to 
 16. Ans. 10 and 8. 
 
 4. In a court there are two square grass plots ; a side of one 
 of which is 10 yards longer than the side of the other ; and their 
 areas are as 25 to 9. What are the lengths of the sides ? 
 
 Ans. 25 and 15 yards. 
 
 5. A person bought two pieces of linen, which together mea- 
 sured 36 yards. Each of them cost as many shillings a yard as 
 there were yards in the piece ; and their whole prices were in 
 the proportion of 4 to 1. What were the lengths of the pieces ? 
 
 Ans. 24 and 12 yards. 
 
 6. There is a rectangular field, whose length is to the breadth 
 in the proportion of 6 to 5. A part of this equal to |- of the 
 whole being planted, there remain for ploughing 625 square 
 yards. What are the dimensions of the field ? 
 
 Ans. The sides are 30 and 25 yards. 
 
 7. Two workmen, A and B, were engaged to work for a 
 certain number of days at different rates. At the end of the 
 time, A who had played 4 of the days, received 75 shillings, but 
 B who had played 7 of the days, received only 48 shillings. 
 Now had B played 4 days, and A played 7 days, they would 
 have received exactly alike. For how many days were they 
 engaged ; how many did each work, and what had each per day ? 
 
 Ans. 19 days ; A worked 15, and B 12 dajj's, and 
 A received 5s. and B 4s. a day. 
 
 8. Two travellers, A and B, set out to meet each other, A 
 leaving the town C at the same time that B left D. They 
 travelled the direct road C D, and on meeting, it appeared that 
 A had travelled IS miles more than B ; and that A could have 
 gone B's journey in 15J days, but B would have been 28 days 
 in performing A's journey. What was the distance between C 
 and D ? Ans. 126 miles. 
 
 9. A and B carried 100 eggs between them to market and 
 each received the same sum. If A had carried as many as B 
 
140 ELEMENTS OF ALGEBRA. 
 
 he would have received 18 pence for them, and if B had carried 
 only as many as A, he would have received only 8 pence. How 
 many had each ? Ans. A 40, B 60. 
 
 10. What two numbers are those, whose sum is to the greater 
 as 11 to 7, the difference of their squares being 132? 
 
 Ans. 8 and 14. 
 
 11. A merchant sold for $960 a certain number of pieces of 
 silk, for which he paid four-fifths as many dollars a piece as the're 
 were pieces. He gained $1000 by the sale, how many pieces 
 did he sell ? Ans. The question is impossible. 
 
 COMPLETE EQUATIONS OF THE SECOND DEGREE. 
 
 110. Let us take next the equation a;'-}- 8a; = 209. This 
 is a complete equation of the second degree. The solution of 
 this equation, it is evident, would present no difficulty, if the left 
 hand member v/ere a perfect square. But this is not the case ; 
 for the square of a quantity -consisting of one term will consist 
 of one term, and the square of a quantity consisting of two 
 terms will contain three terms. Let us then see if a^-\-8x can 
 be made a perfect square ; for this purpose, it will be recollected, 
 that the three parts which compose the square of a binomial are 
 P. the square of the first term of the binomial^ 2°. tioice the first 
 term multiplied by the second, 3°. the square of the second term. 
 Thus, 
 
 [x-\-af = x' + 2ax-\-a\ 
 
 If, then, we compare a;*^ + 8a: with x^ -\-2ax-\- a^/\i is evi- 
 dent that af^-f-' 8 a: may Se considered the first and second terms 
 in the square of a binomial. The first term of this binomial 
 will evidently be x; then as 8 a; must contain twice the first term 
 by the second, the second will be found by dividing 8a; by 2a:, 
 which gives 4 for the quotient, ar^-f- 8a: is, therefore, the first 
 two terms in the square of the binomial x-\-^. If? then, we 
 add 16, the square of 4, to a;*^ + 8a:, the left hand member of the 
 proposed, the result ar^ + 8a: + 16 will be a perfect square. 
 
EQUATIONS OF THE SECOND DEGREE. 141 
 
 But if 16 be added to the left hand member, it must also bo 
 added to the right in order to preserve the equality ; the proposed 
 will then become 
 
 a^-^Sx-^ 16 = 225. 
 Extracting the root of each member of this last, we have 
 a: + 4 = ±15, 
 whence a;=ll, a; = — 19. 
 
 Let us take, as a second example, the equation 
 
 2 
 
 Comparing a^ — -^x with the square of the binomial x — a, 
 3 
 
 2 
 
 viz. ar* — 2ax-\-a^,itis evident, that a^ — ^x may be considered 
 
 the first two terms of the square of a binomial. By the same 
 course of reasoning as in the preceding example, we find this 
 
 binomial to be a: — ^r. If, then, the square of ^ be added to 
 o o 
 
 both sides, the left hand member will be a perfect square, and 
 we have 
 
 Extracting the root of each member, we have 
 x-i = =fc4; 
 
 whence x= 4| , a: = — 3|. 
 
 Let us take, as a third example, the equation a^-\'px = q. 
 
 Comparing the left hand member of this equation with 
 a?-^-2aa:-}-a^ it is evident, that it may be considered as the 
 
 first two terms in the square of the binomial a: + ^ ; whence, if 
 
 the square of ^ be added to both sides, the left hand member will 
 become a perfect square, and we shall have 
 
142 ELEMENTS OF ALGEBEA. 
 
 Extracting the root of each member 
 
 Making the left hand member a perfect square is called com- 
 pleting the square. This is done, as will readily be inferred 
 from the preceding examples, hy adding to both sides the square 
 of one half the coefficient of x in the second term. 
 Let us take for a fourth example, the equation 
 3 61— ar^ 
 
 5'^~'4x — 2* 
 Freeing from denominators, we have 
 
 140 a; — 70 — 12ar' + 6a; = 305 — 5a?. 
 Transposing and uniting terms, we have 
 146 a; — Tar' = 375. 
 Or, changing the signs of each term and dividing by the 
 coefficient of a;^ 
 
 146a; _ 375 
 7 7 ' 
 
 Completing the square, we have 
 
 146a: 5329 _ 375 5329 _ 2704 
 ~7~ ' 49 ~" 7 ' 49 ~" 49 * 
 Whence, extracting the root of each member 
 __73__ 52 
 
 a;=17f, a; = 3. 
 
 111. The rule for completing the square applies only, it is 
 evident, to equations of the form :i^ -\-'px-=q, p and q denoting 
 any quantities whatever, positive or negative. 
 
 If not already of the form x^ -\-px = q^ equations of the 
 kind, which we are here considering, must always be reduced 
 to this form, before completing the square. Thus, in the pre- 
 ceding example, the given equation was reduced, before com* 
 
EQUATIONS OF THE SECOND DEGKEE. 
 
 1^ 
 
 , . ^ / 146 375 . ^ V 
 
 pletmg the square, to ar —x = =-, an equation of the 
 
 form required. 
 
 It is evident, that all complete equations of the second degree 
 may be reduced to the form a? -\-pz-=-q, 1°. by collecting 
 all the terms which involve x into the first member and uniting 
 the terms, which contain ar^, into one term, and those which 
 contam x into another, 2°. by changing the signs of each 
 term, if necessary, in order to render that of 3? positive, 3°. by 
 dividing all the terms by the multiplier of x^^ if it have a mul- 
 tiplier, and multiplying all the terms by the divisor of 3?, if it 
 have a divisor. 
 
 CLX C X 
 
 Let the equation ~ 5ar^ = — -j-ae be reduced to tho 
 
 form x^-\-px = q. 
 
 Freeing from denominators, we have 
 
 5ax-^20bx^ = 'icx-\'20ae 
 By transposition — 20bx^ -\-5ax — 4:cx==20ae 
 Changing signs 20bx^ — 5ax-\-4:Cx = — 20ae 
 
 Uniting terms 20b x^ — {5a — 4c) a: = — 20ae 
 
 Dividing by 20 A :^ - ^^^^ ^ = " y • 
 
 Comparing this equation with the general formula, we have 
 
 {5a — 4c) ae 
 
 P= 203—' «=-T- 
 
 From what has been done, we have the following rule for 
 the solution of complete equations of the second degree, viz. 
 ]°. The equation being reduced to the form x'^-f-pxsrrq, add 
 to both members the square of half the coefficient ofx in the second 
 term; 2°. extract the square root of both members, taking care 
 to give to the root of the second member the double sign ± ; 
 3°. deduce the value of :s.from the equationt which arises from 
 the last operation. 
 
y^fk ELEMENTS OF ALGEBBA. 
 
 EXAMPLES. 
 
 2a^ X 
 
 1. Given -o- + ^s- == « + 8, to find the values of x. 
 
 O <6 
 
 2. Given 4a; = 46, to find the values of x. 
 
 Ans. a; = 3, or — 2J. 
 I the values of x. 
 Ans. X = 12, or — .75. 
 
 40 27 
 
 3. Given ^r -I =13, to find the values of x. 
 
 X — 5 ' X 
 
 14 X 
 
 4. Given 4 a; r— r- = 14, to find the values of x. 
 
 x-f- 1 
 
 Ans. a; = 9, or ly^. 
 he values of x. 
 Ans. a; = 4, or — If. 
 
 X 7 
 
 5, Given — t-ttr = 7i ^j to find the values of x. 
 
 a;-|-60 Sx — o 
 
 Ans. X = 14, or — 10. 
 
 , =.'?JL. 
 
 2 ' 2a; — 5 
 
 6. Given — J^ \- -^ ~ = 5|, to find the values of x, 
 
 Ans. x = 5, or 6.9. 
 
 '• a^d7+ t/lT, + 11 i '° fi°d 'he values of . and ,. 
 Ans. a; = — 46, or 2 ; y=15, or 3. 
 
 ^* ^ind^a^^'^-lt^V^I^ 62 I ^° ^^^ ^^^ ^^^"^' ^^^ ^"^ 2/. 
 
 Ans. a; = 5, or — 36^y ;^y = 3, or 30^^^ . 
 
 9.Give„?£±^^ = 2,_51±^l 
 4a; ^ 10 
 
 , 4a; + 3y 
 and — _L_^ = 2/ — 2 
 
 to find the values of 
 
 a; and y. 
 
 16 
 
 Ans. a; = 5, or — 2^\ ; y = 4, or Ifff. 
 112. "We pass next to the solution of some questions. 
 1. To find a number such, that if three times this number be 
 added to twice its square, the sum will be 65. 
 
EQUATIONS OF THE SECOND DEGREE. 145 
 
 Putting X for the number sought, we have by the question 
 
 Dividing by 2, we have oi? -\--x=--^. 
 
 3 9 65 9 
 
 Completing the square, ^ + 2^ + 3;q = "2"^T6* 
 
 T. , .3 23 
 
 bxtractmg the root a: -j- ^ = db -j-, 
 
 whence a: = 5, a; = ^. 
 
 The first value of x satisfies the question in the sense, in 
 which it is enunciated. In order to interpret the second, it 
 will be observed, that if we put — x instead of x in the equa- 
 tion 23? -\'^x = Q5, it becomes 2r^ — 3 a; = 65. Resolving 
 
 13 
 this equation, we obtain a; = -^, x = — 5, values of a:, which 
 
 13 
 
 differ from the preceding only in the signs. The number -^ will, 
 
 tit 
 
 therefore, satisfy the conditions of the question modified thus. 
 To find a number such, that if three times this number be 
 
 subtracted from twice its square, the remainder will be 65. 
 
 2. A person bought some sheep for £ 72 ; and found if he had 
 
 bought 6 more for the same money, he would have paid £ 1 less 
 
 for each. How many did he buy ? 
 Let X = the number, we have 
 
 Z?__2?_ — 1 
 
 X x-\-6 
 from which we obtain x = 18, or — 24. To interpret the nega- 
 tive result, we write — x for x in the equation, which becomes 
 72 72 
 
 __a; —a:-f 6~ ' ' 
 or which is the same thins: 
 
 •& 
 
 72 _72_ 
 
 X — 6 X 
 
 an equation, which corresponds to the following enunciation 
 10 
 
146 ELEMENTS OF ALGEBRA. 
 
 A person bought some sheep for £72, and found if he had 
 bought 6 less for the same money, he would have paid £ 1 more 
 for each. How many did he buy ? 
 
 The negative values here modify the proposed questions, in a 
 mariner analogous to what takes place, as we have already seen, 
 in equations of the first degree. 
 
 3. To find a number such, that if 15 be added to its square, 
 the sum will be equal to eight times this number. 
 
 Putting X for the number sought, we have by the question 
 
 a,-2-f 15==8a:. 
 Resolving this equation, we have 
 
 re = 5, 2; = 3.- 
 In this example both values of x are positive, and ansfver 
 directly the conditions of the question, in the sense in which it 
 is enunciated. 
 
 4. To find a number such, that if the square of this number 
 be augmented by 5 times the number and also by 6, the result 
 will be 2. 
 
 Putting X for the number sought, we have by the question 
 
 ar^4-5a; + 6 = 2. 
 Whence, resolving the equation we have r^ 
 
 The values of x in this example are both negative ; the ques- 
 tion, therefore, as is evident from inspection, cannot be solved in 
 the sense, in which it is enunciated. 
 
 If instead of x we write — a; in the equation of the proposed 
 it becomes z^ — 5a: -[-6 = 2, from which we obtain a:=l, 
 a: = 4. The numbers 1 and 4 will, therefore, satisfy the con- 
 ditions of the proposed modified thus, 
 
 To find a number such, that if five times this number be 
 subtracted from its square, and 6 be added to the remainder, 
 the result will be 2. 
 
 5. To divide the number 10 into two such parts, that the 
 product of these parts will be 30. 
 
EQUATIONS OF THE SECOND DEGREE. 147 
 
 Putting X for one of the parts, 10 — x will be the other ; w« 
 have therefore by the question 
 
 10a; — a:' =30. 
 Resolving this equation, we obtain 
 
 a; = 5 -f V — 5, x = 5-r-^^5. 
 
 This result indicates, that there is some absurdity in the con- 
 ditions of the question proposed, since in order to obtain the 
 value of X, we must extract the root of a negative quantity, which 
 is impossible. 
 
 In order to see in what this absurdity consists, let us exam- 
 ine into what two parts a given number should be divided, in 
 order that the product of these parts may be the greatest pos- 
 sible. 
 
 Let us represent the given number by p, the product of the 
 two parts by q, and the difference of the two parts by d; the 
 
 greater part will then be f + ^j and the less ^ — -, and we 
 
 shall have 
 
 (1+1) (1-1)=^- 
 
 p" (f 
 
 4 4^ 
 
 Here the value of q, it is evident, will be greater as that of 
 d is less ; the value of q will, therefore, be the greatest possible 
 when d is zero, that is, the product will be the greatest possible, 
 when the difference between the two parts is zero, or in other 
 words, lohen the two parts are equal. 
 
 The greatest possible product, which can be obtained by di- 
 viding 10 into two parts and taking their product will be 25. 
 The absurdity of the question above consists, therefore, in re- 
 quiring, that the product of the two parts, into which 10 is to be 
 divided, should be greater than 25. 
 
 113. The following questions will serve as an exercise for the 
 learner. 
 
148 ELEMENTS OF ALGEBRA. 
 
 ■ 1. There is a field in the form of a rectang-ular parallelogram, 
 whose length exceeds the hreadth by 16 yards, and it contains 
 960 square yards. Required the length and breadth. 
 
 Ans. 40 and 24 yards. 
 
 2. There are two numbers, whose difference is 9, and their 
 sum multiplied by the greater, produces 266. What are those 
 mumbers ? Ans. 14 and 5. 
 
 ^. A regiment of soldiers, consisting of 1066 men, is formed 
 into two squares, one of which has four men more in a side than 
 the other. What number of men are in a side of each of the 
 squares ? Ans. 21 and 25. 
 
 4. Two partners, A and B, gained £ 18 by trade. A's money 
 was in trade 12 months, and he received for his principal and 
 gain £26. Also B's money, which was £30, was in trade 
 16 months. What money did A put into trade? 
 
 Ans. £20. 
 
 5. The plate of a looking glass is 18 inches by 12, and is to 
 ,be framed with a frame of equal width, whose area is to be equal 
 to that of the glass. Required the width of the frame. 
 
 Ans. Jpnches. 
 
 6. A grazier bought as many sheep as cost him £60 ; out of 
 which he reserved 15, and sold the remainder for £54, gaining 
 2 shillings a head by them. How many sheep did he buy, and 
 what was the price of each ? ^• 
 
 Ans. 75 sheep, and t^e price was 16^. 
 
 7. A person bought two pieces of clotK' of different sorts; 
 whereof the finer cost 4 shillings a yard more than the other ; 
 for the finer he paid £18; but the coarser, which exceeded 
 the finer in length by 2 yards, cost only £16. How many 
 yards were there in each piece, and what was the price of a 
 yard of each? 
 
 Ans. 18 yards of the finer, and 20 of the coarser, and 
 the prices were £ 1 and 16s. respectively. 
 
 8. Three merchants, A, B, and C, made a joint stock, by 
 
GENERAL EQUATION OF THE SECOND DEGREE. l49 
 
 which they gained a sum less than that stock by £80 ; A^s share 
 of the gain was £60, and his contribution to the stock was £17 
 more than B's. Also B and C contributed together £325. How 
 much did each contribute ? 
 
 Ans. 75, 58, and 267 pounds respectively. 
 
 9. Two messengers, A and B, were dispatched at the same 
 time to a place 90 miles distant ; the former of whom riding one 
 mile an hour more than the other, arrived at the end of his 
 journey an hour before him. At what rate did each travel per 
 hour ? Ans. A 10 miles, B 9. 
 
 10. The joint stock of two partners, A and B, was $416. A's 
 money was in trade 9 months and B's sii» months ; on dividing 
 their stock and gain, A received $228, and B $252. What was 
 each man's stock ? Ans. A's $192, B's $224. 
 
 11. A and B sold 130 ells of silk, of which 40 ells were A's 
 and 90 B's, for $42. Now A sold for a dollar J of an ell more 
 than B did. How many ells did each sell for a dollar ? 
 
 Ans. B sold 3 ells, and A 3J for a dollar. 
 
 12. A square court-yard has a rectangular gravel walk round 
 it. The side of the court wants 2 yards of being 6 times the 
 breadth of the gravel-walk ; and the number of square yards in 
 the walk exceeds the number of yards in the periphery of the 
 court by 164. Required the area of the court. 
 
 Ans. 256 yards. 
 
 SECTION XIII. — Discussion of the General Equation and 
 OF Problems of the Second Degree. 
 
 114. All complete equations of the second degree may, as 
 we have already seen, be reduced to an equation of the form 
 ar^-|- Pa;=; Q, P and Q denoting any known quantities whatever, 
 positive or negative. Resolving this equation, we have 
 
150 ELEMENTS OF ALGEBRA. 
 
 This is a general solution for equations of the second degree. 
 We shall now examine the circumstances, which result from 
 the different hypotheses, which may be made upon the known 
 quantities P and Q. This is the object of the discussion of the 
 general equation of the second degree. 
 
 115. Any quantity, which substituted for the unknown quantity 
 in an equation of the second degree will satisfy it, is called a root 
 of the equation. 
 
 Before proceeding to the proposed discussion, we shall show, 
 that every equation of the second degree admits of two values 
 for the unknown quantity, or in other words of two roots, and of 
 two only. In order lo this we take the general equation 
 
 x' + Vx^Q,; (1) 
 Completing the square, we have 
 
 p2 F / P\2 P 
 
 p2 
 
 Let Q -|- — - == M', we shall then have 
 
 (=" + 1) = ^'' "' G + 1)' - "^^ °- 
 
 But the first member of this equation being the difference 
 between two squares, it may be put under the form 
 
 (:. + ?+ m)(:. + |-m):^0. (2) 
 
 This fast equation is, it is evident, a necessary consequence of 
 equation (1) and the converse. Each of these equations will be 
 satisfied therefore by the values of x, which satisfy the other, 
 and by these only. But since the left hand member of equation 
 (2) is composed of two factors, this member will become zero 
 if either of its factors is equal to zero, and thus the equation will 
 be satisfied. 
 
 P P 
 
 If we suppose a; -|- n — M=:0, we shall harve x = — ^--j-M. 
 
 P P 
 
 If we suppose a: -f- ^ + M = 0, we shall have z = — ^ — IML 
 
GENERAL EQUATION OF THE SECOND DEGREE. 151 
 
 Or substituting for M its value, we have 
 
 
 Since equation (2) can be satisfied only by putting for x a 
 value which will reduce to zero one or the other of the two fac- 
 tors, of which the left hand member is composed, it follows, that 
 every equation of the second degree admits of tivo roots or values 
 for the unknown quantity and of two only. 
 
 It follows also from what has been done, that every equation 
 of the second degree may be decomposed into two binomial factors 
 of the first degree with respect to x, having a for a common term^ 
 and the two roots taken with their signs changed, for the second 
 terms. ^ 
 
 Resolving the equation 3? -\-Sx — 209 = 0, for example, we 
 have a: =11, x = — 19. Either of these values will satisfy 
 the equation. We have also 
 
 (a;— 11) (2: + 19) = 2-=' + 8a: — 209 = 0. 
 
 If we add together the two general values for a:, found above, 
 the sum, it is evident, will be — P; if we multiply them together 
 the product will be — Q ; whence 1°. The algebraic sum of 
 the two roots is equal to the coefficient of the second term taken with 
 the contrary sign. 2°. The product of the two roots% equal to 
 the second member of the equation, taken also ivith a contrary 
 sign. 
 
 116. Let us new proceed to Uie discussion proposed. Re- 
 suming the value of a;, obtained from the general equation 
 ar* -j- Pa; = Q, we have 
 
 In order to find the value of this expression, which contains a 
 radical, that is a quantity the root of which is to be extracted, 
 we must be able to extract the root either exactly or by approxi- 
 
152 ELEMENTS OF ALGEBRA. 
 
 F . 
 
 mation ; Q -f- -r- the quantity placed under the radical sign must, 
 
 therefore, be positive. But -r- will necessarily be positive, what- 
 
 P 
 
 ever the sign of P may be ; the sign of the quantity Q,-\--t- will, 
 
 therefore, depend principally upon that of Q or the quantity in 
 the equation altogether known. 
 
 1. This being premised, let Q in the first place be positive. 
 In this case P may be either positive or negative, and the general 
 equation may be written under the two forms ^ 
 
 a^-^Yx = +Q, x' — 'Px = + Q, 
 or uniting both in one 
 
 x':i.Vx = + Q; 
 from which we have 
 
 I±v/q+?- 
 
 p2 
 
 Here Q-\--t- will evidently be positive ; the value of x may, 
 
 therefore, be obtained, either exactly or with such degree of 
 approximation as we please. 
 
 With respect to the two values of x, the first, viz. 
 
 ==p|+\/q+? 
 
 F . P 
 
 will be positive, for the square root of -r- alone being — , the square 
 
 P2 P 
 
 root of Q -f- — will be greater than — -, the value of x will, there- 
 
 fore, have the same sign with the radical and will by consequence 
 be positive. This value will answer directly the conditions of 
 the equation, or the problem of which the equation is the algebraic 
 translation. 
 
 P / F 
 
 The second value of x, viz. x = ^ — — Xy^ ^"^"47' ^^^^S 
 
 also necessarily of the same sign with the radical, will be essen- 
 
GENERAL EQUATION OF THE SECOND DEGREE. 163 
 
 tially 7iegative. This value, though it satisfies the equation, 
 will not answer the conditions of the question, from which the 
 equation is derived. It belongs to an analogous question corre 
 spending to the equation, after — x has been introduced instead 
 of X, that is, to ar^ ^ Pa:= Q. Indeed, from this last equation 
 
 P / W 
 
 we deduce z = ± ^ ± i / ^ "f" X ^^^^^^ which do not dif- 
 fer from the preceding, except in the sign. Thus the same 
 equation connects together two questions, which differ from each 
 other only in the sense of certain conditions. 
 
 2. Again, let Q be negative. The equation will then be of 
 the form 2r^ db Pa; = — Q, and we have 
 
 Here, in or(Jer that the root of the quantity placed under the 
 radical sign may be taken, or in other words, that the value of x 
 
 F 
 
 may be real, it is evident that Q must not exceed -^ . 
 
 /p p 
 
 Since moreover \y/ -r Q is numerically less than — it 
 
 follows, that the values of x will both be negative, if P is posi- 
 tive in the equation, that is, if the equation is of the form 
 3?-i^Vx = — Q, and that they wdll both be positive, if P is 
 negative in the equation, that is, if the equation is of the form 
 ar^ — Pa; = — Q. 
 
 Indeed, it may be shown a priori, that always tvJieii Q is 
 negative, in the second memher and P negative in the first, the 
 problem will admit of tivo direct solutions, provided that Q does 
 
 not exceed -v-. 
 4 
 
 The equation x^ — Pa: = — Q, may, by changing the signs 
 of all the terms, be put under the form 
 
 Pa: — r' = Q, or a: (P — a:) = Q. 
 
 But the equation a: (P — a:) = Q is evidently the algebraic 
 translation of the following enunciation, viz. To divide a number 
 
154 ELEMENTS OF ALGEBRA. 
 
 P into iiDo parts, the product of which shall he equal to a given 
 number Q. For if we put x for one of the parts, the other 
 part will be P — x, and the product of the two parts, will be 
 x{V — x). 
 
 This being premised, the enunciation of the problem admits, 
 it is evident, of two direct solutions; for the equation of the 
 problem will be the same, whether x be put for one or the other 
 of the parts ; there is no reason then, why the equation, when 
 resolved, should give one of the parts rather than the other ; it 
 should therefore give both at the same time. 
 
 Moreover, in order that the problem may be possible, it is 
 
 p2 
 
 necessary, that Q should not exceed ~ ; for the greatest possible 
 product of the parts, into which the number P may be divided 
 
 P2 , . 
 
 being equal only to — it is absurd to require that their product, 
 
 P 
 
 which we have represented by Q, should be greater than — . We 
 
 conclude therefore that, in all cases when the hnoion quantity 
 is negative in the second member, but numerically greater than 
 the square of half the coefficient of the second term, the question 
 proposed is impossible. 
 
 117. The following examples will serve as an exercise upon 
 the different cases, which we have here been considering. What 
 change must be made in the enunciations of the first four ques- 
 tions respectively, in order that the negative solutions may become 
 positive ? How must the fifth question be modified, so that the 
 answers shall become positive ? In what does the absurdity in 
 the seventh question consist ? 
 
 1. A company at a tavern had £8, 15^. to pay, but two of 
 them having left before the bill was settled, those who remained 
 had each in consequence IO5. more to pay. How many were in 
 the company at first ? 
 
 2. A man travelled 105 miles, and then found that if he had 
 not travelled so fast by 2 miles an hour, he should have been 
 
GENERAL EQUATION OF THE SECOND DEGREE. 155 
 
 6 hours longer in performing the same journey. How many- 
 miles did he go per hour ? 
 
 3. A regiment of foot was ordered to send 216 men on gar- 
 rison duty, each company being to furnish an equal number; 
 but before the detachment marched, 3 of the companies were sent 
 on another service, when it was found that each company that 
 remained was obliged to furnish 12 additional men, in order to 
 make up the complement 216. How many companies were 
 there in the regiment, and what number of men was each ordered 
 to send at first ? 
 
 4. A and B set out from two towns, which were distant 247 
 miles, and travelled the direct road till they met. A went 9 
 miles a day ; and the number of days at the end of which they 
 met, was greater by 3, than the number of miles, which B went 
 in a day. Where between A and B did they meet ? 
 
 On substituting — x for x in the equations, which pertain 
 respectively to the preceding questions, it will be easy to trans- 
 late these equations into enunciations analogous to those of the 
 questions proposed ; there are questions however, in which it will 
 be very difficult to do this, and the negative solutions in such 
 cases are to be regarded merely as connected with the first in the 
 same equation of the second degree. 
 
 5. A gentleman counting the guineas, which he had in his 
 purse, finds that if 24 be added to their square, and 8 times their 
 number be subtracted from 17, the sum and remainder will be 
 equal. How many guineas had he in his purse ? 
 
 6. A set out from C towards D, and travelled 7 miles a day. 
 After he had gone 32 miles, B set out from D towards C, and 
 went every day one-nineteenth of the whole journey ; and after 
 he had travelled as many days as he went miles in one day, he 
 met A. Required the distance of the places C and D ? 
 
 7. The difference of two numbers is 7, and the square of the 
 greater is equal to 25 times the less. What are the numbers f 
 
156 ELEMENTS OF ALGEBRA. 
 
 EXAMINATION OF PARTICULAR CASES. 
 
 1. In the general equation let Q be negative, that is, let the 
 equation be of the form x^-\-Fx = — Q, P being of any sign 
 
 whatever; if we suppose Q = — , the radical 
 
 \/ 
 
 pa 
 
 will be reduced to 0, and the values of x will be equal each to 
 
 P F 
 
 — — . Thus if Q be negative in the equation and equal to -^, 
 
 the values of x will be equal, and will both be positive if P is 
 negative, or both negative if P is positive. 
 
 2. In the general formula 
 
 let Q = 0, the values of x will then be a: = 0, a; = — P. 
 
 3. In the same formula let P = 0, we have then 
 
 that is to say, the values of x will in this case be equal, but of 
 contrary signs, real if Q is positive, and imaginary if Q is 
 negative. 
 
 4. Let P = 0, Q = 0, the values of x will then be each equal 
 to 0. 
 
 5. We have next to examine a remarkable case which fre- 
 quently occurs in the solution of problems of the second degree. 
 For this purpose, let us take the equation 
 
 Aar^+Ba:=C. 
 
 This equation being resolved, gives 
 
 — B±VB'^ + 4AC 
 
 ^= ^ ' 
 
 Let it now be supposed, that in consequence of a particular 
 
 hypothesis made upon the given things in the question, we have 
 
 A= 0, the values of x then become 
 
 _0 _ 2B 
 
DISCUSSION OF PROBLEMS OF THE SECOND DEGREE. 157 
 
 The second value of x here presents itself under the form of 
 infinity, and may be regarded as a true answer, when the ques- 
 tion is susceptible of infinite solutions. In order to interpret the 
 first, if we return to the equation, we see that the hypothesis 
 
 Q 
 
 A = reduces it to Ba: = C, from which we deduce x = ^, an 
 
 expression Jinite and determinate, and which must be regarded 
 
 as the true value of ^ in the present case. 
 
 6. Let it be supposed finally, that we have at the same time 
 A = 0, B = 0, C = 0. The equation will then be altogether 
 indeterminate. This is the only case of indetermination, which 
 the equation of the second degree presents. 
 
 DISCUSSION OF PROBLEMS. 
 
 118. The following problems offer all the circumstances, which 
 usually occur in problems of the second degree. 
 
 1. To find on the line A B, which joins two luminous bodies 
 A and B, the point where these bodies shine with equal light. 
 
 C" A C B C 
 
 The solution of this problem depends upon the following prin- 
 ciple in physics, viz. The intensity of light from the same 
 luminous body will be, at different distances, in the inverse ratio 
 of the square of the distance. 
 
 This being premised, let <z = A B, the distance between the 
 two bodies ; let b= the intensity of A at the unit of distance, 
 c = the intensity of B at the same distance ; let C be the point 
 required, and let A C = x. 
 
 The intensity of A at the distance 1 being b, its intensity 
 
 &t the distance 2, 3, 4, ... . will ^ t> q> t^ • • . • • . and by 
 consequence, at the distance x, it will be -5. For the same 
 
158 
 
 ELEMENTS OF ALGEBRA. 
 
 reason, the intensity of B at the distance a — x will be 
 ; whence, by the question, we have i 
 
 {a — x)' 
 
 From which, we obtain 
 
 {a^xf 
 
 
 a'b 
 
 cY 
 
 b — c 
 
 or reducing 
 
 a{b±^bc) 
 
 But^±/v/^c may, it will be observed, be put under the 
 form /^ b (^ b zLa/ c), and b — c may be put under the form 
 {A/bf-i^cf, or {^b + ^c) i^b-^c). 
 
 Taking advantage of this remark, the value of x may be 
 expressed more simply, thus 
 
 X = — J-. J-, whence a — x= —— — ^'^- 
 
 or x = 
 
 a \f b 
 
 V^ + Vc 
 a s/ b 
 
 ■ and 
 
 a /s/ c 
 
 
 DISCUSSION. 
 
 1. Let b be greater than c. 
 
 The first value of x is positive and less than <z, since 
 
 . , - — J- is a fraction. The point sought, therefore, accord- 
 
 ing to this value of x, is situated between A and B. It is 
 moreover nearer B than A ; for, in consequence of 3 ^ c, 
 we have ^ b -{- /s/ b, or '^ h/ b^ s/ h -\- s/ c, whence 
 
 aJ b 1 a/s/ b ^ a 
 
 17b + ^0 > 2' ''"'* ^y consequence ^^^^^ >^. 
 
 This, indeed, should be the case, since we have supposed the 
 intensity of A greater than that of B. 
 
 The corresponding value of a — x is also positive and less 
 as it will be easy to see, than -. 
 
DISCUSSION OF PROBLEMS OF THE SECOND DEGREE. 159 
 
 The second value of x is positive, but greater that c, since 
 
 we have . , -z- "> 1. This value of x gives, therefore, a 
 
 V — f^ c 
 
 second point C situated upon AB produced and at the right 
 
 of A and B. Indeed, since the light from A and B expands 
 
 itself in all directions, there should be, it is easy to see, on AB 
 
 produced a second point where A and B shine with equal light. 
 
 This point moreover should be nearer the body, the light of 
 
 which is least intense. 
 
 The second value oi a — a: is negative : this should be the case, 
 
 since we have x^a. 
 
 2. Let b be less than c. 
 
 a 
 The first value of x is positive, but less than -. The corre- 
 sponding value of a — a; is also positive and greater than -. 
 
 Thus on the present hypothesis the point C, situated between 
 A and B, should be nearer A than to B. 
 
 The second value of x is essentially negative. In order to 
 
 interpret it, we return to the equation, which becomes by substi- 
 
 b c 
 
 tuting — a: for a:, -^ = -— -j— — -^. But a — x expressing in the 
 
 X ( Q* ~T~ Xj 
 
 first instance the distance of the point sought from B,a-\- x must 
 in the present case express the same distance. Thus the point 
 sought should be at the left of A, in C" for example. Indeed, 
 since by hypothesis the intensity of B is greater than that of A, 
 the second point sought should be nearer A than to B. 
 
 3. Let b = c. 
 
 The first value of x, and also that of a — z is reduced in this 
 
 case to -. Thus we have the middle of AB for the point sought. 
 This result conforms to the hypothesis. 
 
 The remaining values are reduced to -^ — or become infinite, 
 that is, the second point where the bodies shine with equal .light, 
 
160 ELEMENTS OF ALGEBRA. 
 
 is situated at a distance from A and B greater than any assign- 
 able quantity. This result corresponds perfectly with the present 
 hypothesis; for, if we suppose the difference b — c, instead 
 of being absolutely nothing, to be very small, the second point 
 will exist, but at a very great distance from A and B. If then 
 b = c or 
 
 the point required must cease to exist, or be placed at an infinite 
 distance. 
 
 4. Let b = c and a = 0. 
 
 The first system of values of x and a — x reduce themselves 
 
 in this cat* to 0, and the second system to ^. This last charac- 
 ter is here the symbol of indetermination ; for, on returning to 
 the equation of the problem, 
 
 [b — c) a^ — 2abx = — fl^3, • 
 
 this equation becomes on the present hypothesis 
 
 0.2^—0. ar=-0, 
 an equation which may be satisfild by any number whatever 
 taken for x. Indeed, since the two bodies have the same intensity 
 and are placed at the same point, they should shine with equal 
 light upon any point whatever in the line A B. 
 
 5. Finally let a = 0, b being different from c. 
 
 Both systems in this case will be reduced to 0, which in- 
 dicates, that there is but one point, where the bodies shine 
 with equal light, viz. the point, in which the two bodies are 
 situated. 
 
 2. To find two numbers such, that the diflierence of their 
 products by the numbers a and b respectively may be equal to a 
 given number s, and the difference of their squares equal to 
 another given number q. 
 
 Denoting by x and y the numbers sought, we have by the 
 question ax — by = s 
 
 x' — f^g. 
 
DISCUSSION OF PROBLEMS OF THE SECOND DEGREE. 161 
 
 Resolving these equations, we have for the first system of 
 values for x and y 
 
 as-\-l 
 
 ^V^- 
 
 -Q{^- 
 
 -b-) 
 
 
 a^- 
 
 -b^ 
 
 
 _bs + a^ ^- 
 
 -q{d^- 
 
 -f>^) 
 
 
 d'- 
 
 -h' 
 
 
 and for the second system, we have 
 
 as- 
 
 -bVs"- 
 
 -q{d'- 
 
 -b^) 
 
 
 d"- 
 
 -b^ 
 
 
 bs- 
 
 -a\f s"- 
 
 -q{d^- 
 
 -*'1 
 
 ^"^ d'—b^ 
 
 DISCUSSION. 
 
 1. Let a be greater than b, and by consequence a* — b"^ posi- 
 tive 
 
 In order that the values of x and y may be real, it is necessary 
 that we have 
 
 q {c^ ' — b^) <^ s^, and therefore, q <^ -^ jz. 
 
 This condition being fulfilled, the values of x arid y in the first 
 system will be necessarily positive, and will, by consequence, 
 form a direct solution of the problem in the sense, in which it is 
 enunciated. 
 
 In the second system the value of x will be essentially posi- 
 tive; fora]>^ gixes as'^bs, and for a still stronger reason, 
 as^bs/s'—q{d^ — b''). 
 
 With respect to the value of y^ it may be either positive or 
 negative. In order that it may be positive, we must have 
 
 bs^aVs' — qid'—b^) 
 or, squaring both sides, 
 
 11 
 
162 ELEMENTS OF ALGEBRA. 
 
 or, adding c^ q [a^ — //) to both sides of this last, and subtracting 
 b^s' from both sides 
 
 or, by division S'^l- 
 
 Thus, in order that the second system maybe a real and direct 
 solution, we must have 
 
 If, then, we take for a, b, and 5 any absolute numbers what- 
 ever, provided that we have a'^b and that we take for q a 
 
 number comprised between the two limits -^ and -^ r^, we shall 
 
 (T (T — b^ 
 
 be certain of obtaining two direct solutions. 
 
 Thus, let a = 6, ^ = 4, 5 = 15 ; we have 
 
 i 225 „, , / 225 „. 
 
 if then we take q == 10, for example, we shall have 
 6X 15±4V225— 20X 10 11 7 
 
 ^- 20 
 
 -2 ^^2 
 
 4Xl5±6/v/225- 
 
 -20X10 9 3 
 
 ^— 20 
 
 2' °' 2' 
 
 If on the present hypothesis, we have ? <C^ — > and for a still 
 
 stronger reason, q <C —^ JIS' ^^® value of y in the second sys- 
 tem will be negative. This system, therefore, will not be a 
 solution of the proposed problem in the sense, in which it 
 is enunciated, but of an analogous problem, the equations of 
 which are ax-{-by = s 
 ^ 3? — f = q 
 
 and which will differ from the proposed in this respect only, that 
 s will express an arithmetical sum instead of difference. 
 2. Let a be less than b and therefore c^ — b"^ negative. 
 
DISCUSSION OF PROBLEMS OF THE SECOND DEGREE. 163 
 
 In this case the expressions for x and y in the first system 
 may be put under the form 
 
 _ — as — hs/^-\-q{}i'^d') 
 ''"" h' — d' 
 
 __ — bs — a^/s- -j-q{l?^-— a') 
 and in the second 
 
 '^~" b-' — a' 
 — bs^a s/s" -\-q{b' — a') 
 
 y— ^537^ • 
 
 The values of x ,and y in both systems, it is evident, will be 
 real, since the quantity placed under the radical is essentially 
 positive. 
 
 In the first system the values of x and y are essentially nega- 
 the; in the second the value of^a:, it is easy to see, is neces- 
 sarily positive, but the value of y may be either positive or nega- 
 
 tive ; in order that it may be positive, we must have q^—^. 
 
 3. Let a-=hj and therefore d^ — b'^=. 0. 
 On this hypothesis, we have for the first system of values 
 for x and y 
 
 2as 2as 
 
 ^=-0-. y=^; 
 
 and for the second 
 
 
 
 Returning to the equations of the proposed in order to inter- 
 pret these last, we obtain for x and y on the present hypothesis 
 
 ?q-\-s^ ^^Q — ^ 
 
 2as ' ^~" 2as ' 
 
 PROBLEMS FOR SOLUTION AND DISCUSSION. 
 
 1. There are two numbers, whose sum 'is a and the sum of 
 whose second powers is b. Required the numbers. 
 
164 ELEMENTS OF ALGEBRA. 
 
 futtijig z and y for the numbers, we have 
 
 a^s/2h- 
 
 -d' 
 
 ~~ 2 
 
 azh^2b- 
 
 -a^ 
 
 What conditions are necessary in order that the values of x 
 and y may be real ? When will the values of x both be posi- 
 tive? Can both be negative? When will one of them be 
 positive and the other negative, and to what question does the 
 negative value belong ? 
 
 2. To find two numbers such, that the sum of their products 
 by the numbers a and b respectively may be equal to 25, and 
 their product equal to p. 
 
 Putting X and y for the numbers, we have 
 
 szh^s^ — abp 
 
 x=. — 
 
 a 
 
 szh^s^ — abp 
 
 y = -^ . 
 
 What conditions are necessary in order that the values of x 
 and y may be real ? What is the greatest value of which p ad- 
 mits ? Can either of the values of a: or y be negative ? 
 
 8. To find two numbers such, that the sum of their products 
 by the numbers a and b respectively may be equal to a given 
 number s, and the sum of their squares equal to another given 
 number q. 
 
 Putting X and y for the numbers respectively, we have 
 __ as±b ^{d' + b') q — s" 
 
 _ bszt^^{d' + l>')q — s' 
 y— a^J^b^ 
 
 What conditions are necessary in order that the values of x 
 and y may be real ? Within what limits must q be comprised 
 \\i, order that both values of x may be positive ? Within what 
 limits must q be comprised in order that both values of y may 
 
JiAXlMA AND IttlNIMA. 165 
 
 he positive? In the second system of values for x and y, 
 when will the value of x be positive, and that of y negative, 
 and what is the analogous problem, to which this system be- 
 longs ? 
 
 4. To find a number such, that its square may be to the 
 product of the differences between this number and two other 
 numbers a and b in the ratio of q to p. 
 
 Putting X for the number sought, we have 
 
 q{a + b) ± ^q' {a — bf + 4:pqab 
 '^- 2{q-^p) 
 
 Let this formula be examined on the different hypotheses 
 9<P^ S=P, Q>P' 
 
 SECTION XIV.— Maxima and Minima. 
 
 119. In several of the preceding questions, the given things, 
 we have seen, are so connected among themselves, that one is 
 determined by the others to be comprised within certain limits, 
 or to have a greatest or least possible value. 
 
 A quantity, the value of which may be made to vary, is called 
 a variable quantity; the greatest value of which is called a 
 viaximum and the least a minimum. 
 
 Questions frequently occur, in v/hich it is required to deter- 
 mine under what circumstances the result of certain arithmet- 
 ical operations performed upon numbers will be the greatest or 
 least possible. We shall resolve a few questions of this kind 
 the solutions of which depend upon equations of the second 
 degree. 
 
 1. To divide .a number, 2a, into two parts such, that the 
 product of these parts may be a maximum. 
 
 Let X be one of the parts, then 2a — x will be the other, and 
 their product will be x{2a — x). By assigning different values 
 
166 ELEMENTS OF ALGEBRA. 
 
 to x^ the product xi^a — x) will vary in magnitude, and the 
 question is to assign to :?: a value such, that this product may be 
 the greatest possible. Let m be the maximum sought, we have 
 by the question 
 
 Eegarding for the moment m as known, and deducing from 
 this equation the value of x^ we have 
 
 a: = a ± w of- — m. * 
 
 From this result it appears, that in order that x may be real, 
 m must not exceed c? ; the greatest value of m will therefore be 
 c^^ in which case we have x = a. Thus to obtain the greatest 
 possible product, the proposed must be divided into two equal 
 parts, and the maximum obtained ivill be equal to the square of 
 one of these parts. 
 
 In the equation x{2a — x)=^m, the expressions x{2a — x) 
 is called a function of x. This function is itself a variable, 
 the value of which depends upon that given to the first variable 
 ox X. 
 
 2. To divide a number, 2a, into two parts such, that the sum 
 of the square roots of these parts may be a maximum. 
 
 Let 7? be one of the parts, then 2 a — x^ will be the other, and 
 the sum of the square roots will be x-\r ^ 2a — x^. Let m be 
 the maximum sought, we have by the question 
 
 x-\-^2a — x^=-m; ^'—<^Jn ^ ^' 
 
 2a 
 
 from which we obtain .'^- '^ '' f s •''^^'^ — • - ■ ^^'^^ 
 
 ^ , 4 / "^ 
 ''=2^V ^ 
 
 or simplifying a; = ^ ± _ ^-^^ZT^, 
 
 In order that the values of a: may be real, the value of m^ must 
 not exceed 4a; 2 /^ a is therefore the greatest value, which wi 
 can receive. 
 
 Let us put m==-2 s/ a, we have x==- tsj a and 3? = a, whence 
 
 i 
 
 
 -f- 4./ 
 
MAXIMA AND MINIMA. 167 
 
 2a — x^ = a. Thus, the proposed must be divided into two equal 
 parts in order that the sum of the square roots of the parts may 
 be a maximuyn. This maximum moreover will be equal to twice 
 the square root of one of the parts. 
 
 From what has been done, the following rule for the solution 
 of questions of the kind which we are here considering will 
 readily be inferred^yi-z. HarAng formed the algebraic expression 
 of the quantity f^jeptible of becoming a maximum or minimum, 
 make this expression equal to any quantity m. If the equation 
 thus made is of the second degree in x, x designating the variable 
 quantity^ lohich enters into the algebraic expression^ resolve this 
 equation in relation to x; make next the quantity under the 
 radicaLequal to zero, and deduce from this last equation the value 
 ofm; this ivill be the maximum or minimum sought. Substi- 
 tuting finally the value of m in the expressioii for x, lue obtain 
 the value of x proper to satisfy the enunciation proposed. 
 
 If the quantity placed under the radical remains essentially 
 positive, whatever the value of m, Vv^e infer that the expression 
 proposed may be of any assignable magnitude whatever, or in 
 other words, that it will have infinity for a maximum and zero 
 for a minimum. 
 
 42-4- 4 :r 3 
 
 Thus let there be proposed the expression c/o. i ~T\ — ' 
 
 to determine v/hether this expression is susceptible of a maxi- 
 mum or minimum. 
 
 4^2 I 4^. 3 
 
 Putting i^i^' , I - >^ = ^ and deducing the value of a:, we 
 
 v) I ^ *// ' I ■ Jl I 
 
 3^ — 1 , 1 , 
 
 have X = g ^-2^9m'-\-^. Here, whatever value we 
 
 give to m, the quantity placed under the radical will be positive ; 
 the proposed therefore may be of any magnitude whatever. 
 
 EXAMPLF'-S FOR PRACTICE. 
 
 1. To divide a given number a into two factors, the sum of 
 which shall be a minimum. 
 
 Ans. The two factors should be equal. 
 
168 ELEMENTS OF ALGEBKA. 
 
 2. Let d be the difference between two numbers; required 
 that the square of the greater divided by the less may be a 
 minimum. Ans. The minimum required is 4^ and the value 
 
 of the greater part corresponding is 2d. 
 
 3. Let a and b be two numbers of w^hich a is the greater, to 
 find a number such, that if a be added "to this number, and b be 
 subtracted from it, the product of the suny-^nd difference thus 
 obtained being divided by the square of the number, the quotient 
 will be a maximum. 
 
 nnn thc» m«svinrmm J ' 
 
 Ans. The number =: '^'^'^, , and the maximum . , 
 
 a — b ^ab 
 
 4. To divide a number 2 a into two parts such, that the sum 
 of the quotients obtained by dividing the parts mutually, one by 
 the other, may be a minimum. ' 
 
 Ans. The number should be divided into two 
 equal parts, and the minimum is 2. 
 
 5. To find a number such, that if a and b be added to this 
 number respectively, the product of the two sums thus obtained, 
 divided by the number, may be a minimum. 
 
 Ans. The number = /^y ab, and the minimum 
 
 SECTION XV. — Powers and Roots of Monomials. 
 
 120. AVhen a quantity is multiplied into itself, the product, 
 we have seen, is called a poiver, the degree of which is marked 
 by the exponent of tne product, thus aaaaa or a^ is called the 
 f^fih power of a; in like manner o/' is called the mih. power 
 of a. 
 
 The original quantity, from which a power is derived, is 
 called the root of this power. The degree of the root is de- 
 termined by the number of times the root is found as a factor 
 in the poorer ; thus a is the fifth root of a^ ; in like manner a 
 
POWERS AND ROOTS OF MONOMIALS. 169 
 
 is the ?7zth root of a"*. The number which marks the degree of 
 the root is called the index of the root. 
 
 121. Let it be proposed to find the fifth power of 2a^3'; 
 this power is indicated thus, {2a?b'^f, and we have, it is evi- 
 dent, 
 
 {2a'by = 2a'b^ X ^a'b^ X ^a^b^ X ^ciH'' X 2(^b\ 
 
 Here, it is evident, 1°. that the coefficient 2 must be multiplied 
 into itself four times or raised to the fifth power ; 2°. that each 
 one of the exponents of the letters must be added, until it is 
 taken as many times as there are units in the exponent of the 
 power, or in other words, multiplied by 5 ; we have therefore 
 {2a^b''f=:32a''b'', 
 
 In like manner {8aH'cY = 5l2a^b'c\ 
 
 To raise a monomial therefore to any given power, we raise 
 the coefficient to this potver, and multiply each one of the expo^ 
 nenis of the letters by the exponent of the povjer. 
 
 With respect to the sign, with which the powers of a mono- 
 mial should be affected, it is evident, that whatever be the sign 
 of the quantity itself, its second power will be positive. More- 
 over if the exponent of the power of a monomial .,]be an even 
 number, it is easy to see, that this power may be considered 
 as a power of the square of the proposed quantity. Thus a^, it 
 is evident, may be considered as the fourth power of a^ ; in like 
 manner a^"", any even power of a, may be considered the mih 
 power of a^. It follows, therefore, that lohatever be the sign of a 
 monomial, any poiver of it, the exponent of ivhich is an even 
 number, is positive. 
 
 Again, since the power of a simple quantity, the exponent 
 of which is an odd number, is equal to a power of this quan- 
 tity of an even degree multiplied by the first power, it follows, 
 that every power of a monomial, the exponent of which is an odd 
 number, loill have the same sign as the quantity from ivhich it is 
 formed. 
 
 122. Let it now be proposed to find the third root of 64a'i'c'. 
 
 o 
 
170 ELEMENTS OF ALGEBRA. 
 
 The root required is indicated thus, wQ'la'^b^c^; the sign a/ 
 being employed to denote in general that a root is to be taken, 
 and the index 3. placed above the radical sign to denote the 
 particular root required. 
 
 Since the root of a quantity must evidently be sought by a 
 process the reverse of that, by which it is raised to a power, in 
 order to extract the root of a monomial, 1°. we extract the root 
 of the coefficient, 2°. we divide the exponent of each of the letters 
 hy the index of the root. 
 
 According to this rule, the third root of the proposed will 
 be ^c^h^c. In like manner the seventh root of a^^b^^c^ is 
 aH'c\ 
 
 With respect to the signs, with which the roots of monomials 
 should be affected, it is an evident consequence of the principles 
 already established that, 
 
 1°. Every root of an even degree of a positive monomial may 
 have indifferently either the sign -j- or — . Thus, the sixth root 
 of 64«^Ms±2a^ 
 
 2°. Every root, the degree of which is expressed by dn odd 
 number, will have the same sig7i as the quantity proposed. Thus 
 the fifth roet of —32a'' b' is —2a'b. 
 
 3°. Every root of an even degree of a negative monomial is an 
 impossible or imaginary root. For there is no quantity, which 
 raised to a power of an even degree can give a negative result. 
 
 Thus s/ — a, \/ — b denote impossible or imaginary quanti- 
 ties, in the same manner as 'v — a, V — b. 
 
 123. From what has been said, it is evident in order that a 
 root may be extracted, 1°. that the coefficient of the proposed 
 must be a perfect power of the degree marked by the index of 
 the root to be extracted ; 2°. that the exponents of each of the 
 letters must be divisible by the index of the root. 
 
 When this is not the case the root can only be indicated. It 
 should be observed, however, that radical expressions, of any 
 degree whatever admit of the same simplifications as those of the 
 
THEORY OF COMBINATIONS. . 171 
 
 second degree. These simplifications are founded upon the prin- 
 ciple, that any root whatever of a product is equal to the product 
 of the same root of the several factors. 
 
 Thus let it be proposed to find the third root of SAa^Pc^. 
 The third root, it is evident, cannot be taken ; for 54 is not a 
 perfect third power, and the exponents of the letters a and c 
 are not divisible by 3. We therefore indicate the root, thus, 
 w54.a*l)^c^; but this expression may be put under the form 
 ^27^^^*^ X 2<zr; whence taking the third root of the factor 
 
 27an\ we have W5UF¥?=SabW2^. 
 Let it be proposed next to find the third root of 
 l25a'b-{-S75a^c. 
 This expression, which it is easy to see is not a perfect third 
 power, may be put under the form 125 a^ [a^ b -\- 2 c) ; whence 
 extracting the third root of the first factor, we have for the root 
 
 sought 5 a w ctb -j- 3 c. 
 
 EXAMPLES. 
 
 1. To reduce ^/S^a^b'' c to its most simple form. 
 
 2. To reduce ^/l^Sx^ifz^ to its most simple form. 
 
 3 - • 
 
 3. To reduce wax^ -\-bx^ io its most simple form. 
 
 4. To reduce ^\/21 d^ -\- SI a^b to its most simple form. 
 
 5. To reduce V96a^^'c^^ to its most simple form. 
 
 ^ rp T 'i /md' — \Qa'b' . . , ^ 
 
 b. lo reduce 1 / r-—- ^^ ^ ^ , , to its most simple form. 
 
 Y SW — 324a* ^ ^ 
 
 SECTION XVI. — Powers of Compound Quantities, Theory 
 OF Combinations, Binomial Theorebl 
 
 124. Powers of compound quantities are found like those of 
 monomials by the continued multiplication of the quantity into 
 itself. They are indicated by inclosing the quantity in a paren- 
 
172 ELEMENTS OF ALGEBRA. 
 
 thesis, to which is annexed the exponent of the power. The 
 third power of a^-\-5ab — bc^, for example, is indicated, thus, 
 {a^-\-5ab — bc^f. This same power may also he indicated 
 thus, d^-\-5ab — bc^- 
 
 Next to monomials, binomials are those, which are the least 
 complicated. We begin therefore with these. 
 
 Below are several of the first powers of the binomial a: -f- «> 
 viz. 
 
 {x-^ay = x-\-a 
 
 {x + aY = x'-{-2ax-{-a' 
 
 {x-\'af = x'-{-2ax^-\-3a''x + a^ 
 
 {x + ay = x*-\- 4:ax^ + 6aV + U^x + a\ 
 
 We have formed the different powers of x~\-am this table by 
 the continued multiplication of a: -|- ^ into itself. In this way 
 we arrive only at particular results. To form any of the higher 
 powers, the process of multiplication must still be continued. 
 This would be tedious, especially, as the power to which the 
 binomial is to be raised, becomes more and more elevated. We 
 proceed, therefore, to investigate a method, by which a binomial 
 may be raised to any power whatever, without the necessity of 
 forming the inferior powers. This method was discovered by 
 Newton. The principle on which it is founded is called the 
 Binomial Theorem. The most simple and elementary demon- 
 stration of this theorem depends upon the theory of combinations, 
 10 which we shall first attend. 
 
 THEORY OF COMBINATIONS. 
 
 125. The results, obtained by writing one after the other, in 
 every possible way, a given number of letters, in such a manner, 
 that all the letters will enter into each result, are called permu- 
 tations. 
 
 Let there be, for example, two letters, a and b. These give, it 
 is evident, two permutations, ab^ ha. 
 
THEORY OF COMBINATIONS. 173 
 
 Again, let there be three letters, «, h and c. If we set apart 
 one of the letters, a for example, the remaining letters give 
 two permutations, viz. 3c, ch; placing next the a at the right 
 of each of these, we have two permutations of three letters, 
 viz. hca^ cha; but each of the remaining letters h and c, being 
 set apart in the same manner, will also furnish each two 
 permutations of three lettters ; ivhence the permutations of three 
 letters will be equal to the permutations of two letters^ multiplied 
 by three. 
 
 In like manner the permutations of four letters will be found 
 equal to the permutations of three letters multiplied by four. 
 
 And in general, the permutations of any number whatever n 
 of letters^ will be equals it is evident, to the permutations of n — 1 
 letters, multiplied by n the number of letters employed. 
 
 Let Q represent the permutations of ?i — 1 letters, then Qn 
 will represent the permutations of n letters; thus Qn will be a 
 general formula for permutations. 
 
 In the general formula Q n, let n = 2, then Q will be 1 ; 
 whence 1 X 2 will be the permutations of two letters. Again 
 let 71 = 3, then Q will be 1X2; whence 1X^X3 will be 
 the permutations of three letters. In like manner the permuta- 
 tions of 4 letters will be 1 X 2 X 3 X 4. The following rule 
 for permutations will, therefore, be readily inferred, viz. Multi- 
 ply in order the natural numbers, 1, 2, 3, 4, ^c. to the number 
 denoting the letters employed inclusive; the result ivill be the 
 permutations of the given number of letters. 
 
 126. When a given number of letters are disposed in order 
 one after the other in every possible way, 2 and 2, 3 and 3, and, 
 in general, n and ti at a time, the number of letters taken at a 
 time being always less than the given number of letters, the 
 results obtained are called arrangements, 
 
 0* 
 
174 ELEMENTS OF ALGEBRA. 
 
 Let it be required to form the arrangements of three letters, 
 ay b, and c, two and two at a time. 
 
 a, ab 
 , ac 
 
 b, ba 
 
 be 
 
 c, ca 
 cb 
 
 Setting apart first one of the letters, a for example, we write 
 after this letter each one of the reserved letters b and c ; we 
 thus form two of the arrangements sought, viz. ab, ac ; setting 
 apart next the letter Z*, and writing by its side each one of the 
 reserved letters a and c, we form two more of the arrangements 
 sought, viz. hu, he; pursuing the same course with the re- 
 maining letter c, we have in the result, it is plain, all the ar- 
 rangements required and no more ; lohence the arrangements of 
 three letters 2 and 2 at a time, will le equal to the arrangements 
 of the same letters one at a time, multiplied by the number of 
 letters reserved. 
 
 Let it be required next to form the arrangements of four letters 
 a, b, c, d, three and three at a time. 
 
 ab, ab c c a, cab 
 
 a bd cad 
 
 ac, 
 
 acb 
 acd 
 
 cb, 
 cd, 
 da, 
 db, 
 dc, 
 
 cb a 
 cbd 
 
 ad, 
 
 adb 
 adc 
 
 cda 
 cdb 
 
 ba. 
 
 b ac 
 bad 
 
 dab 
 dac 
 
 he, 
 
 bca 
 bed 
 
 dba 
 dbc 
 
 bd, 
 
 bda 
 bdc 
 
 dca 
 deb 
 
 Having formed the arrangements of the given letters, 2 and 
 
THEORY OF COMBINATIONS. 176 
 
 2 at a time, we set apart one of these, ab for example, and write 
 successively by its side each one of the reserved letters c and dy 
 we thus form two of the arrangements sought, viz. abc, abd. 
 The same being done with each one of the remaining arrange- 
 ments of the given letters, 2 and 2 at a time, we obtain, it is 
 evident, all the arrangements required and no more. Thus, the 
 arrangements of 4 letters taken 3 and ^ at a time, ivill be equal 
 to the arrangements of the same letters, taken 2 and 2 at a timet 
 multiplied by the number of letters reserved. 
 
 In like manner, understanding by letters reserved those which 
 remain, when the given letters are taken one less than the re- 
 quired number at a time, we have the arrangements of any 
 number m of letters, taken n arid n at a time, equal to the 
 arrangeinents of the same letters, n — 1 at a time, multiplied by 
 the number of letters reserved. 
 
 Let P represent the arrangements of m letters n — 1 at a 
 time ; it being required to take the letters n and n at a time, the 
 reserved letters will be 77i — {n — 1), or m — 7i-\-l; thus the 
 arrangements of m letters, 7i and ?i at a time, will be expressed 
 by the formula, 
 
 ^{m — n+l). 
 
 This will be the general formula for arrangements. 
 * In the general formula 'P [m — n -\- 1), let n equal 2. ' In this 
 case P will represent the arrangements of m letters 1 at a time ; 
 thus P will equal m; whence m {m — 1), will express the 
 arrangements of m letters 2 and 2 at a time. 
 
 Again, in the general formula P (?7^ — n-^l), let n==3. 
 In this case P will represent the arrangements of ?ra letters 2 
 and 2 at a time; P will therefore equal m {m — 1) ; whence 
 m{?n — 1) {m — 2) will express the arrangements of m letters 
 
 3 and 3 at a time. 
 
 In like manner the arrangements of m. letters 4 and 4 at a 
 time, will be expressed hy m {m — I) {m — 2) {m — 3). 
 
 From inspection of the above formulas the following rule for 
 arrangements will be readily inferred, viz. From the number 
 
176 ELEMENTS OF ALGEBRA. 
 
 denoting the given letters subtract successively the natural num.* 
 lers 1, 2, 3, ^c. to the number which denotes the letters to he 
 taken at a time ; multiply these several remainders and the num" 
 her denoting the given letters together ; the product will he the 
 arrangements required. 
 
 127. Arrangements, any two of which differ at least by one of 
 the letters, which enter into them, are called combinations. 
 
 Let it he proposed to determine the number of combinations of 
 three letters, a, b, and c taken two and two at a time. 
 
 The arrangements of these letters, two and two at a time, 
 will be 
 
 ah 
 ha 
 
 ac 
 ca 
 
 he 
 ch 
 Among these arrangements we have, it is evident, but three 
 combinations, viz. ab, ac, he, each one of which is repeated as 
 many times as there are permutations of two letters. Hence the 
 combinations of three letters taken 2 and 2 at a time, will he equal 
 to the arrangements of three letters 2 and 2 at a time, divided hy" 
 the permutations of two letters. 
 
 In like manner, it will be seen, that the combinations of 4 
 letters, 3 and 3 at a time, are equal to the arrangements of 4 
 letters, 3 and 3 at a time, divided by the permutations of three 
 letters. 
 
 And in general the combinations of m letters, n and n at a 
 time, will be equal, it is evident, to the arrangements ofm. letters^ 
 n and n at a time, divided by the permutations of n letters. 
 
 From what has been done, we have therefore the following 
 general formula for combinations, viz. 
 
 ^{m — n+iy 
 Qn 
 
BINOMIAL THEOREM. 177 
 
 In the general formula let w = 2, the formula which results 
 
 will be ^^^^. 
 
 This will give the combinations of m letters 2 and 2 at a 
 time. 
 
 Again, let % = 3 ; the formula which results will be 
 m{7n — 1) {m — 2) 
 
 \7%7z * 
 
 This will give the combinations of m letters 3 and 3 at a time. 
 
 In like manner we obtain — ^ , ' ^ — -r^ -. a 
 
 1.2.3.4 
 
 formula which gives the combinations of m letters 4 and 4 at 
 a time. 
 
 From inspection of the formulas obtained by making % = 2, 3, 4, 
 &c. in the general expression, we may infer a general rule for 
 combinations as has been done already with respect to permuta- 
 tions and arrangements. 
 
 1. For how many days can 7 persons be placed in a different 
 position at dinner ? 
 
 2. How many words can be made with 5 letters of the alpha- 
 bet, it being admitted that a number of consonants may make a 
 word? 
 
 3. How many combinations can be made of 24 letters of the 
 alphabet taken two and two at a time ? 
 
 4. A general was asked by his king, what reward he should 
 confer on him for his services ; the general only desired a farthing 
 for every file of ten men in a file, which he could make with a 
 body of 100 men. At this rate what would he receive ? 
 
 BINOMIAL THEOREM. 
 
 128. If we examine with attention the different powers of 
 x-\-a^ art. 124, it will be easy to fix upon the law, according 
 to which the exponents of x and a proceed. But it will not 
 be so easy to determine the law for the numerical coefficients. 
 y 12 
 
178 
 
 ELEMENTS OF ALGEBEA. 
 
 If we observe, however, the manner in which the different 
 terms which compose a power are formed, we shall perceive 
 that the numerical coefficients are occasioned by the reduction 
 of several similar terms into one, and that these similar terms 
 arise from the equality of the factors which compose a power. 
 These reductions, it is easy to see, will not take place, if the 
 second terms of the binomials are different. AVe begin there- 
 fore by investigating a law for the formation of the product of 
 any number of binomials x-\-a, x~\-b, x-\-c. . . , the first 
 terms of which are the same in each, while the second are 
 different. 
 
 {x + a){x + b) 
 
 x^ -\-a 
 b 
 
 + a5 
 
 {x-\-h) (x + b) {x-{-c)=x^ + a 
 
 b 
 
 x + a){x + b){x + c){x + d) 
 
 ac 
 be 
 
 = x'-\-a 
 
 x' + ab 
 
 x'-^abc 
 
 b 
 
 ac 
 
 abd 
 
 c 
 
 ad 
 
 acd 
 
 d 
 
 be 
 
 bed 
 
 bd 
 
 
 
 cd 
 
 
 x-\-abe 
 
 x-\-abcd. 
 
 From inspection of the above products, which we have formed 
 by the common rules of multiplication, it will be observed, 
 
 1°. hi each product there is one term more than there are units 
 in the number of factors. 
 
 2°. The exponent of ^ in the first term is the same as the num*- 
 ber of factors, and goes on decreasing by unity in each of the foU 
 lowing terms. 
 
 3°. The coefficient of the first term is unity. The coefficient 
 of the second term is equal to the sum of the second terms of the 
 binomials ; that of the third term is equal to the sum of the differ- 
 ent combinations or products of the second terms of the binomials 
 taken two and two ; that of the fourth is equal to the sum of the 
 products of the second terms of the binomials taken three and, 
 
BINOMIAL THEOREM. 179»' 
 
 three, and so on. The last term is equal to the product of the 
 second terms of the binomials. 
 
 129. A\re readily infer from analogy, that the same law wilF 
 obtain, whatever be the number of factors employed. This law 
 may, however, readily be shown to be general. In order to this, 
 it will be sufficient to show, that if the Jaw be true for the 
 product of any number m of binomials, it will also be true for the 
 product of 772 -|~ 1 binomials. 
 
 The number of binomial factors being represented by w, 
 the different powers of x will be a;*", a:*""*, a;'""^ &c. Let 
 A, B, C, . . . U denote the quantities, by which these powers 
 beginning with z'""^ are to be multiplied; but as the number 
 of terms must remain indeterminate, until m receives a particular 
 value, we can write only a few of the fifst and last terms of the 
 expression, designating the intermediate terms by a series of 
 points. 
 
 The product of any number m of factors will then be repre- 
 sented by the expression 
 
 a:'" + Aa;'"-^ + Bx"'-- 4 Ca:'"-^ . . U. 
 
 Multiplying this expression by a new factor a: -[- K, it becomes 
 
 2:—=^..,. UK. 
 
 + ^ + A 
 K 
 
 a:'" + B 
 AK 
 
 BK 
 
 Here the law for the exponents is evidently the same, as in 
 the first expression. With respect to the coefficients, it is 
 evident, 1°. that the coefficient of the first term is unity. 2°* 
 A -(- K, the coefficient of the second term, is equal to the sum 
 of the second terms of the w, -(- 1 binomials. 3°. Since B by 
 hypothesis expresses the sum of the second terms of the m 
 binomials taken two and two, and AK expresses the sum of the 
 second terms of the m binomials multiplied each by the new 
 second term K, B -f- AK, the coefficient of the third term, will 
 be the sum of the products two and two of the second terms of 
 the m-\'\ binomials. 
 
 In the same manner C -|- BK, it is easy to see, will be tlw 
 
180 
 
 ELEMENTS OF ALGEBRA. 
 
 sum of the products three and three of the second terms of the 
 m-\-l hinomials, and so on. 4°. The last term UK it is evi- 
 dent, is the product of the m-\- 1 second terms. 
 
 The law laid down, art. 128, being true therefore for expres- 
 sions of the fourth degree will, from what has just been de- 
 monstrated, be true for those of the fifth; and being true for 
 expressions of the fifth degree, it will be true for those of the 
 sixth and so on ; thus it is general. 
 
 130. If in the different products which have been formed, 
 art. 128, we make b, c and d each equal to a, these products will 
 be converted into powers of x -\- a, thus 
 
 {x + a) {x-{-b) = {x-\-af=^x'-{-a 
 
 a 
 
 x' + a' 
 
 = {x + a] 
 
 ' 
 
 a^ + a' 
 
 x' + a' 
 
 a' 
 
 a^ 
 
 a' 
 
 a' 
 
 a' 
 
 (^ 
 
 a' 
 
 
 a' 
 
 
 +^- 
 
 x + a'. 
 
 {x-\-a){x-]-b) {x-\-c) = {x-{-aY = a^-}-a 
 
 a 
 a 
 
 x-\-d){x-\-b){x^ c) {x + d): 
 
 = x' + ^ 
 a 
 a 
 a 
 
 Comparing these expressions with the different products, 
 from which they are derived, we perceive 1°. that the multi- 
 plier of X in the second term has been converted into the first 
 power of a, repeated as many times as there are units in the 
 number of binomials employed, or which is the same thing, 
 as there are units in the exponent of x in the first term. 2°. 
 That the multiplier of the third term has been converted into 
 the second power of c, repeated as many times, as there can 
 be formed different products from a number of letters equal to 
 the number of binomials employed, taken two and two at a 
 time. 3**. That the multiplier of the fourth term has been con- 
 verted into the third power of a, repeated as many times as 
 
BINOMIAL THEOREM. 181 
 
 there can be formed different products from a number of letters, 
 equal to the number of binomials employed, taken three and 
 three at a time, and so on. 
 
 131. From what has been done, it is evident, therefore, that 
 whatever be the power to which a binomial a: -|" ^ is to be raised, 
 1°. the exponent of x in the first term will be equal to the expo- 
 nent of the power, and that it will go on decreasing by unity in 
 each of the following terms to the last, in which it will be 0. 
 2°. That the exponent of a in the first term will be 0, in the 
 second unity, and that it will go on increasing by unity, until it 
 becomes equal to the exponent of the power to be formed. 
 3°. The numerical coefficient of x in the first term will be unity, 
 in the second it will be equal to the exponent of x in the first 
 term, in the third it will be equal to the number of products, 
 which may be formed from a number of letters, equal to the 
 exponent of x in the first term, taken two and two at a time, in 
 the fourth it will be equal to the number of products, which may 
 be formed from the same number of letters, taken three and three 
 at a time and so on. 
 
 Let it be required to form the 5th power Qi x-\- a. The dif- 
 ferent terms, without the numerical coefficients, will be by the 
 preceding rule x" -\- ax^ -\- c^ :»? -\- c^ :i^ •\- a^ x -\- a^. 
 
 With respect to the numerical coefficients, that of the first 
 term will be 1, that of the second will be 5, that of the third will 
 be equal to the number of products, which may be formed of 5 
 letters taken 2 and 2, that of the fourth will be equal to the num- 
 ber of products, which may be formed of 5 letters taken 3 and 3, 
 and so on. Thus the numerical coefficients will be 
 
 1, 5, 10, 10, 5, 1 ; 
 whence 
 
 {x + af=^7?^5ax' + lOa^x' + lOa^r'-f- 5a'x + a'. 
 
 Let it next be required to raise x-\- aio the vTzth power, we 
 shall have, according to the preceding rule, for a few of the first 
 terms without the numerical coefficients 
 
 x"^ -\- ax"^-' -{- a^x"^-^ -\- a^x"^-^ -\- . ., 
 p 
 
182 ELEMENTS OF ALGEBRA. 
 
 . Here the numerical coefficients cannot be determined until we 
 
 assign a particular value to m ; by the preceding rule, however, 
 
 the numerical coefficient of the second term will be equal to wz, 
 
 whatever the value of m may be. In the development therefore 
 
 of [x -|- cl)"" we write m for the coefficient of the second term. 
 
 With respect to the third term the numerical coefficient will 
 
 be equal to the number of products, which may be formed of 
 
 m letters 2 and 2 at a time ; this is expressed by the formula 
 
 m{vi — 1) . , - mim — 1) . , zr • . r.-u 
 
 — - — we write theretore — - — - — lor the coemcient oi the 
 
 ... „ ... m{7n — \) [m — 2) .„ , 
 third term, bor a similar reason — ^ = — s^^i ^^"^^ "® 
 
 the coefficient of the fourth term, and so on. We have then 
 
 .... a'". 
 
 From inspection of the different terms of this development 
 it will be perceived, that the coefficient of the fourth, for example, 
 
 is formed by multiplying — ^r— ? the coefficient of the third 
 
 1 . o 
 
 term, by 7n — 2 the exponent of x in this term, and dividing 
 by 3 the number, which marks the place of this term. It 
 will be perceived, also, that the coefficient of the third term is 
 formed in the same manner by means of the second term, and 
 that of the second by means of the first. We readily infer, 
 therefore, the following rule, by which to form the coefficient 
 of any term v/hatever, viz. Multiply the coefficient of the pre' 
 ceding term by the exponent of x in that term, arid divide the 
 product by the number which marks the place of that term from 
 the left. 
 
 From what has been done, we have therefore the following 
 rule, by which to raise a binomial to any power whatever, viz. 
 P. The coefficient of x in the first term is unity, and its exponent 
 is equal to the number of units in the degree of the power to which 
 the binomial is to be raised. 2°. To pass from any term to the 
 
BINOMIAL THEOREM. 1S3 
 
 folloioing, we multiply the numerical coefficient by the exponent 
 of X in that term^ dioide by the number which marks the place of 
 that term from the left, increase by unity the exponent of a and 
 diminish by unity the exponent of x. 
 
 According to this rule 
 {x-\-af=^x' + Qax'+\5d'x' + 2Qa^x^-]r^5a'x'-\-Qa'x-{-a\ 
 
 132. It sometimes happens, that the terms of the proposed 
 binomial are affected with coefficients and exponents. The 
 following example will exhibit the course to be pursued in cases 
 of this kind. 
 
 Let it be proposed to raise the binomial ^c^b — 3aic to the 
 fourth power. 
 
 Putting ^a^b = x, and — 2abc = y, we have 
 
 {X + yY = x' + ^x'y + Qx'f + ^xf -\- y\ 
 Substituting next for x and y their values, we have 
 
 (4^2^ _ ^abcY = {^aHf + 4 {^aHf (— 3aic) + . .. . 
 6 {^aHf {—^abcf^^ (4a^i) { — ^abcf + {—^abc)\ 
 or performing the operations indicated, we have 
 
 {^aH — ^abcy=25QaH' — ima'b'c-l-SUa%*(?. . 
 — ^^2a'b'c^-{-Sla'b'c\ ■ 
 The terms produced by this development are alternately- 
 positive and negative. This, it is evident, should always be 
 the case, when the second term of the proposed binomial has the 
 sign — . 
 
 133. The powers of any polynomial whatever may be found 
 by the binomial theorem. Let it be proposed to find for example, 
 the third power of the trinomial a-^b '\- c. 
 
 In order to apply the rule to this case, we put a-\-b=:m; 
 the proposed is then reduced to the binomial m -\- c, and we have 
 
 {m + cf = 771^ + 3';^^'c -\-'^me -\- & 
 whence, restoring the value of m, we have 
 
 (a + ^ + cf == fl3 + 3a=^ 4- 3^^=^ + b^ 
 
 + 3a=c + 6a^c + 33'^c 
 + 3ac2 + 33c^ 
 + c3. 
 
184 ELEMENTS OF ALGEBRA. 
 
 The same process, it is easy to see, may be applied to any 
 polynomial whatever. 
 
 MISCELLANEOUS EXAMPLES. 
 
 1. To find the third power of 2 a — ^ -[- c^ 
 
 2. To find the seventh power of Sa^ — 2 a*. 
 
 3. To find the fifth power oi a' — c — 2d. 
 
 4. To find the third power of 2^^ — 4a^ + S^*^. 
 
 SECTION XVII.— Roots of Compound Quantities. 
 
 134. We pass next to the extraction of the roots of com- 
 pound quantities, beginning with the third or cube root of num- 
 bers. 
 
 In the following table, we have the nine first numbers, with 
 their third powers or cubes written under them respectively. 
 1, 2, 3, 4, 5, 6, 7, 8, 9, 
 1, 8, 27, 64, 125, 216, 343, 512, 729. 
 
 By inspection of this table, it will be perceived, that among 
 numbers consisting of two or three figures, there are nine only, 
 which are perfect third powers, the others have each for a root 
 an entire number plus a fraction. 
 
 If the proposed number consists of not more than three figures, 
 its third root or that of the greatest third power contained in it, 
 may be found immediately by the above table. 
 
 Let it be proposed to extract the third root of a number, con- 
 sisting of more than three figures, 103823, for example. 
 
 The proposed being comprised between 1000, the third power 
 of 10, and 1000000, the third power of 100, its root will 
 consist of two places, units and tens. To return therefore from 
 the proposed to its root, let us observe the manner, in which the 
 
ROOTS OF COMPOUND QUANTITIES. 185 
 
 units and tens of a number are employed in forming the third 
 power of this number. For tliis purpose designating the tens by 
 a and the units by 3, we have 
 
 From this we learn, that the third power of a number consisting 
 of units and tens, is composed of the third poioe?' of the tens, the 
 triple product of the square of the tens by the units, the triple 
 product of the tens by the square of the units, a?id the third power 
 of the units. 
 
 If then we CE^n determine in the proposed the third power of 
 the tens, the tens of the root will be found by extracting the third 
 root of this part. The third power of the tens, it is evident, can 
 have no significant figure below the fourth place, the three 
 figures on the right will, therefore, form no part of the third 
 power of the tens, and may on this account be separated from 
 the rest by a comma. The third power of the tens will then be 
 contained in 103, the part at the left of the comma. The great- 
 est third power contained in 103 is 64, the root of which is 4; 4 
 is, therefore, the significant figure in the tens of the root sought. 
 Indeed, the proposed is evidently comprised between 64000, the 
 third power of 40 or 4 tens, and 125000 the third power of 50 
 or 5 tens. The root sought is, therefore, composed of 4 tens and 
 a certain number of units less than ten. 
 
 The tens of the root being thus obtained, we subtract the third 
 power 64 from 103, the part of the proposed at the left of the 
 comma, and to the remainder bring down the figures at the right. 
 The result of this operation, 39823, must contain, from what has 
 been said, the triple product of the square of the tens by the 
 units, together with the two remaining parts in the third power 
 of the root sought. 
 
 The square of the tens, it is evident, will contain no signifi- 
 cant figure less than hundreds ; on this account we separate 23, 
 the two figures on the right of the remainder 39823, from the 
 rest by a comma; 398, the figures on the left of the comma, will 
 then contain the triple product of the square of the tens of the 
 
186 ELEMENTS OF ALGEBRA. 
 
 root sought by the units and something more, in consequence of 
 the hundreds arising from the two remaining parts of the third 
 power of the root sought. Dividing therefore 398 by 48, the 
 triple product of the square of the tens, already found, the 
 quotient 8 will be the unit figure sought, or, from what has been 
 said, it may be too large by 1 or 2. 
 
 To determine whether 8 be the right unit figure we raise 48 to 
 the third power. This gives 110592, a number greater than the 
 proposed ; 8 is, therefore, too large for the unit figure. We 
 next try 7 ; 47 raised to the third power gives 108823. The 
 proposed is, therefore, a perfect third power, the root of which 
 is 47. 
 
 The operation may be exhibited as follows. 
 
 103,823 I 47 
 64 
 
 398,23 I 48 
 
 103,823 
 
 Any number however large may be considered as composed 
 of units and tens ; the process for finding the root may therefore 
 be reduced to that of the preceding example. 
 
 Let it be proposed, for example, to find the third root of. 
 43725678. Considering the root of this number as composed 
 of units and tens, 678 the three right hand figures, it is evident, 
 will form no part of the third . power of the tens. On this ac- 
 count we separate them from the rest by a comma. The third 
 power of the tens being contained, then, in the part at the left 
 of the comma, we obtain the tens of the root sought by ex- 
 tracting the third root of this part. Considering therefore, 
 for the moment, the part of the proposed 43725 as a separate 
 number, its third root, it is evident, may be found as in the 
 preceding example. Performing the operations, we have 35 
 for the root and a remainder of 850. There will therefore be 
 
ROOTS OF COMPOUND QUANTITIES. 187 
 
 3«) tens in the root of the proposed, and in order to find the units, 
 we bring- down the three right hand figures 678 by the side of 
 850, which gives 850678. Separating next the two right hand 
 figures of this last from the rest by a comma, and dividing the 
 part on the left by the triple square of the tens already found, 
 we obtain 2 for the unit figure of the root sought. To determine 
 whether this is the right figure, we raise 352 to the third power, 
 which gives 43614208, a result less than the proposed. 352 is, 
 therefore, the root of the- proposed to within less than a unit. 
 The operation may be exhibited as follows: 
 
 43,725,676 
 27 
 
 352 
 
 Is Dividend 167,25 [ 27 1st Divisor 
 
 Third power of 35 42875 
 
 2d Dividend 8506,78 | 3675, 2d Divisor 
 
 Third power of 352 43614208 
 
 Remainder 111470 
 
 The same process, it is easy to see, may be extended to any 
 number however large. The rule, therefore, for the extraction 
 of the third root will be readily inferred. 
 
 If it happens, that the divisor is not contamed in the dividend 
 prepared as above, a zero must be placed in the root, and the 
 next figure brought down to form the dividend. 
 
 EXAMPLES. 
 
 1. Find the third root of 150568768. Ans. 532. 
 
 2. Find the third root of 205483447701. Ans. 5901. 
 
 3. Find the third root of 32977340218432. Ans. 32068. 
 135. If the proposed be a fraction its third root is found by 
 
 extractinf? the third root of the numerator and denominator. 
 
 lus y/ , 
 
 Thus I / |,is|: 
 
188 ELEMENTS OF ALGEBRA. 
 
 If the denominator is not a perfect third power, it may be 
 
 made so, by multiplying both terms by the square of the denom- 
 
 3 
 inator ; thus if the proposed be -, we multiply both terms by 49 ; 
 
 147 
 
 the fraction then becomes ^-r^, the root of which is nearest 
 
 t>4«3 
 
 -, accurate to within less than -. 
 7 7 
 
 136. We have seen, art. 94, that the square root of an entire 
 number, which is not a perfect square, cannot be exactly assigned. 
 The same is true with respect to the roots of all entire numbers, 
 which are not perfect powers of a degree denoted by the index 
 of the root. 
 
 The third root of a number which is not a perfect third power 
 
 may be approximated by converting the number into a fraction, 
 
 the denominator of which is a perfect third power. Thus let it 
 
 be required to find the approximate third root of 15. This num- 
 
 1 r 15 X 12' 25920 , ^. J 
 ber may be put under the form — —3 — ==———, the third 
 
 29 5 1 
 
 root of which is j^ , or 2 — , accurate to within less than — . If 
 
 a greater degree of accuracy were required, we should convert 
 the proposed into a fraction, the denominator of which is the 
 third power of some number greater than 12. 
 
 In such cases it is most convenient to convert the proposed 
 number into a fraction, the denominator of which shall be the 
 third power of 10, 100, 1000, &c. 
 
 Thus if it be required to find the third root of 25 to within 
 .001, we convert the proposed into a decimal, the denominator 
 of which is the third power of 1000, viz. 25.00Q000000, the 
 third root of which is 2.920 to within .001 ; we have then 
 >v/25=: 2.920, accurate to within less than .001. 
 
 To approximate therefore the third root of an entire number by 
 means of decimals, we annex to the 'proposed three times as many 
 zeros as there are decimal places required in the rout, we then 
 
Hoots of compound quantities. 189 
 
 extract the root of the number thus prepared to within a unit, 
 and point off for decimals, as mamj places as there are decimal 
 figures required in the root. 
 
 187. If 'the proposed number contain decimals, beginning at 
 the place of units, we separate the number both to the right and 
 left into periods of three figures each, annexing zeros, if neces- 
 sary, to complete the right hand period in the decimal part. We 
 then extract the root, and point off for decimals in the root 
 as many places as there are periods in the decimal part of the 
 power. 
 
 If the proposed be a vulgar fraction, the most simple method 
 of finding the third root is to convert the proposed into a decimal, 
 the number of places in which shall be equal to three times the 
 number of decimal figiires required in the root. The question is 
 thus reduced to extract the third root of a decimal fraction. 
 
 EXAMPLES. 
 
 1. Fnd the approximate third root of 79. Ans. 4.2908. 
 
 2. Find the approximate third root of ^J. Ans. 0.824. 
 
 3. Find the approximate third root of 3.00415. 
 
 Ans. 1.4429. 
 
 4. Find the approximate third root of 15f . Ans. 2.502. 
 138. By processes altogether similar to that, which we have 
 
 employed in tlm extraction of the third root of numbers, we 
 may extract the root of any degree whatever. The method of 
 extracting the root of any degree whatever, in the case of alge- 
 braic quantities, is also founded upon the same principles. The 
 following example will be sufficient to illustrate the course 
 to be pursued, whatever the degree of the root required may be. 
 
 Let it be proposed to extract the fifth root of the polynomial 
 32a'' — 80aH' + S0an'^^0a*b'-\-l0a'b'^ — b'^. 
 
 The proposed being arranged with reference to the powers 
 of the letter a, we seek the fifth root of the first term 32 a" 
 
19(^ ELEMENTS OF ALGEBRA. 
 
 Its root 2 a' will be the first term of the root sought. "We 
 write, therefore, 2d^ in the place of the quotient in division, and 
 subtracting its fifth power from the whole quantity, we have for 
 a remainder 
 
 The second term of the binomial [a -|- if is 5 a^ 3 ; this shows, 
 that in order to obtain the second term of the root, we must 
 divide — S^a^b^, the second term of the proposed, by five times 
 the 4th power of ^c^, the term of the root already found. Per- 
 forming the operation we obtain — b^. This will be, therefore, 
 the second term of the root. Raising 2c^ — b^ to the fifth power, 
 it produces the quantity proposed. The root is therefore obtained 
 exactly. If the root contained more than two terms, it would be 
 necessary to subtract the fifih power of 2a^ — b^ from the pro- 
 posed quantity, and then in order to find the next term of the 
 root, to divide the first term of the remainder by five times the 
 4th power of 2a^ — b^. In this case, however, only the first term 
 of the divisor would be used ; we should have therefore the same 
 divisor, that was used the first time. 
 
 139. When the index of the root has divisors the root may be 
 found more readily than by the general method. Thus the 
 fourth root may be found by extracting the square root twice 
 successively ; for the square root of a^ is a^^ and that of c^ is a, 
 the fourth root of a*. In general, all roots of a degree marked 
 by 4, 8 or any power of 2 may be found by si^cessive extrac- 
 tions of the square root. Roots, the indices oT which are not 
 prime numbers, may be reduced to others of a degree less 
 elevated. The 6th root, for example, may be found by first 
 extracting the square and then the third root ; for the square root 
 of c^ is a^ and the third root of a^ is a, 
 
 EXAMPLES. 
 
 1. To find the third root of Qs^ + ZQj? + 54a: + 27. 
 
 Ans. 22: + 3. 
 
CALCULUS OF RADICAL EXPRESSIONS. 191 
 
 2. To find the third root of x' + 6x' — 4.0x' + 96x — 64 
 
 Ans. 3^-\-2x — 4. 
 
 3. To find the third root oi x' — Qa^-{- \5x' — 20a:^ + \5a? 
 ^Qx-\ \. Ans. x' — 2x-\-\. 
 
 4. To find the third root oi 21 x^ — 5^x^ -\-Q'^x' ^^^x^ + 
 21.£2 — e^r+l. Ans. ^x' — 2x-\-\, 
 
 5. To find the fourth root of 16 a* — 96 a? x -{- 216 a" x" — 
 \n6ax'^-^8lx*. Ans. 2a— 3a:. 
 
 SECTION XVIII. — Calculus of Radical Expressions. 
 
 140. Radical expressions, the roots of which cannot be found 
 exactly, frequently occur in the solution of questions. On this 
 account mathematicians have been led to investigate rules for 
 performing upon quantities subjected to the radical sign, the 
 operations designed to be performed upon their roots. In this 
 way the calculations required in the solution of a question are 
 frequently rendered more simple, and the extraction of the root 
 is left to be performed at last, when the radical expression is 
 reduced to the most simple form, which the nature of the question 
 will allow. 
 
 ^DDITION AND SUBTRACTION. 
 
 141. Radical expressions of the same degree, and which have 
 the quantities placed under the radical sign also the same, are 
 said to be similar. 
 
 The addition and subtraction of similar radicals is performed 
 upon the coefiicients. Thus the sum of the radicals 
 34/*, 94/3, is 12^3; the sum o( a4/W, b4/¥7,—cA/l^ 
 is {a + b — c)4/¥7. 
 
 In like manner 9\/a*c, subtracted from 12 \^a^c gives 
 
192 ELEMENTS OF ALGEBRA. 
 
 Vc, and bwab^ subtracted from a^/ab^ gives 
 
 Radical expressions, which are at first dissimilar, frequently 
 become similar when reduced to their most simple form. Thus, 
 let it be required to add 5 \/2a?b^ and a s/'E\cFW. These ex- 
 pressions, reduced to their most simple form, become 5aw2c^b^, 
 ^a^2a^b'^; their sum is therefore 
 
 Sa^/2^. 
 
 The addition and subtraction of dissimilar radicals can be 
 effected only by means of the signs -\- and — . 
 
 EXAMPLES. 
 
 1. To find the sum of ^a'f^bl, and a^W^b^. 
 
 2. To find the sum of ac? s/T^Wc, and 3 Va^^V. 
 
 3. To find the sum of ^/^WJb?, and ^^sflE^Tc. 
 
 4. To find the sum of 2 V8, — 7 Vl8, 5^72, and — V^O. 
 
 Ans. 8V2. 
 
 5. To find the sum of 8 Vi —J'^IS, 4 a/27, and — 2 Vi^g- 
 
 Ans. VV3. 
 
 6. To find the sum of 2^1^ V^O, — Vl5, and Vf- 
 
 Ans. ffVlSr 
 
 7. To find the sum of ^/lQa'b\ and VdOa'bK 
 
 Ans. {3aH-{-5ab)^2ab, 
 
 8. To subtract sT^^Wc from 1 as/¥c. 
 
 9. To subtract ^24 from \/l92. Ans. 24/3. 
 
 10. To subtract l/^^ from l]/^?^. 
 
 Ans, (3a- 1)^^. 
 
CALCULUS OF RADICAL EXPRESSIONS. 199 
 
 MULTIPLICATION AND DIVISION. 
 
 142. Let it be required to multiply /^ a by iHJh^ we have 
 /i/ay^^b = ^ab; for ^/fl X \/^ raised to the seventh 
 power gives ab for the resuh, and wab raised to the seventh 
 power gives also ab for the result; whence the seventh powers 
 of these expressions being equal, the expressions themselves 
 must be equal. 
 
 The same reasoning may be applied to all similar cases ; we 
 have, therefore, the following rule for the multiplication of radi- 
 cal expressions of the same degree, viz. Take the product of the 
 quantities under the radical sign, observing to place the result 
 under a sign of the same degree. 
 
 Let it next be required to divide /^a hy A^b. In this 
 
 , /^a J /a ^ j^a 
 
 case we have -57-7 = 1 / - ; lor the expressions -^77, 
 A/ o Y ^ w^ 
 
 \X - being raised to the fifth power give each -\ these 
 
 expressions are therefore equal. 
 
 We have then the following rule for the division of one radical 
 quantity by another of the same degree, viz. Take the quotient 
 arising from the division of the quantities under the radical 
 sign, recollecting to place it under a sign of the sam£ degree. 
 
 • 
 
 EXAMPLES. 
 
 1. Multiply 4/4, 74/6, and ^4/5 together. 
 
 Ans. l^/^^. 
 
 2. Multiply 5^3, 7;^ f, and a/ 2 together. Ans. 140. 
 
 3. Multiply7 + 2V6by9 — 5V6. Ans. 3—17^6. 
 
 4. Multiply 8 + 2V7by8 — 2V7. Ans. 36. 
 
 5. Multiply5V3 — 7V6by2V8 — 3. 
 
 Ans. 41V6 — 71V3. 
 
 13 n^ ^ 
 
194 ELEMENTS OF ALGEBRA. 
 
 6. Divide V243 by VlS. Ans. SJ. 
 
 7. Divide /v/24^^ by ^8TJb. Ans. |a^3. 
 
 8. Divide 1 by r-|-^2. Expressing the quotient in the 
 form of a fraction, and multiplying both terms by 1 — >^2 
 we have Ans. ^2 — 1. 
 
 9. Divide l + V6by2V2 — V-*^- Ans. a/^+V^. 
 
 FORMATION OF POWERS AND EXTRACTION OF ROOTS. 
 
 143. Let it be required to raise the radical ^a^b to the third 
 power ; we have 
 
 {\/^f = ^^b X /^^ X ^^ = 4/^3 
 according to the rule established for multiplication. 
 
 Whence to raise a radical quantity to any power ; we raise ike 
 quantity placed under the radical sign to the power required^ 
 observing to place the result under the same radical sign. 
 
 When the index of the radical is a multiple of the exponent 
 of the power to which the radical is to be raised, it may be 
 raised to the power required in a more simple manner than by 
 the preceding rule. 
 
 Thus let it be required to raise ^2 a to the second power. 
 The proposed from what has been said, art. 139, may be put 
 
 under the form | / sf2a; but to raise this expression to the 
 
 second power, it is sufficient to suppress the first radical sign; 
 whence (^/2 of = \/2 a. 
 
 Again, let it be required to raise \/5b io the third power. 
 The proposed maybe put under the form | / wdb ; whence 
 
 Whence if the index of the radical is divisible by the exponent 
 
FORMATION OF POWERS AND EXTRACTION OF ROOTS. 1^5 
 
 of the poiuer, to ivkich the proposed quantity is to be raised, the 
 operation is performed by dividing the index of the radical by the 
 exponent of the power. 
 
 144. With respect to the extraction of roots, it is evident, from 
 the preceding rules, that to extract the root of a radical, we may 
 extract the root of the qicaniity placed under the radical sign, 
 the result being left under the same radical sign, or we may 
 multiply the index of the radical by the index of the root to be 
 extracted. 
 
 3 
 
 Thus, 
 
 ^y^y^^=4/3^. ^y^yrc^f^sF. 
 
 EXAMPLES. 
 
 1. To raise ^a*b^ to the fourth power. 
 
 2. To raise i>/ dh^c to the sixth power. 
 
 3. To find the fourth root of sfZ2a'b\ 
 
 4. To find the fifth root of >v/243^^ 
 
 REDUCTION OF RADICAL EXPRESSIONS TO THE SAME 
 INDEX. 
 
 145. It follows from the principles established above, that 
 if we multiply at the same time the index of the radical and the 
 exponents of the quantity placed under the radical sign by the 
 same number, the value of the radical remains the same. 
 
 Thus if we multiply the index of the radical s/c^b by 3, 
 we have wc^b, the third root of the proposed; if then we 
 multiply the exponent of the quantity placed under the radi- 
 cal sign by 3, we have />/a^b^ the third power of i^c^b; the 
 second operation, therefore, restores the expression to its original 
 value. 
 
 146. By means of this last principle, we may reduce two or 
 
196 ELEMENTS OF ALGEBRA» 
 
 more radicals of different indices to the same index. Thus let 
 there be the two radicals s/^a^ s/h^c. Multiplying the index 
 and also the exponents of the quantities placed under the radical 
 sign in the first by 4, and in the second by 3, we have for the 
 first v'SV or f^\^a\ and for the second ^¥J. The pro- 
 posed are, therefore, reduced to equivalent expressions having a 
 common index 12. 
 
 In like manner, the three quantities 
 
 f^TW, ^'dfW, V7W, 
 become respectively 
 
 105 105 105 
 
 ^/a'^V\ s/a'^y^ s/c^iS^ 
 having a common index 105. 
 
 From what has been done we have the following rule for re- 
 ducing radical expressions to the same index, viz. Multiply at 
 the same time the index belonging to each radical sign, and the 
 exponents of the quantities placed under this sign, by the product 
 of the indices belonging to all the other radical signs. 
 
 If the indices of the radicals have common factors, the calcula- 
 tions are rendered more simple, by taking for the common index 
 the least number exactly divisible by each of the indices. 
 
 A quantity, which has no radical sign, may on the same prin- 
 ciples be placed under a radical sign ; for this purpose, we raise 
 the quantity proposed to the power denoted by the index of the 
 radical sign, under which it is to be placed. 
 - Thus if it be required to put the quantity o^ under the sign 
 
 ^, we have for the result Va^". 
 
 147. Radical expressions having different indices must be re- 
 duced to the same index before applying to them the rules for 
 multiplication and division laid down above. The following 
 examples will serve as an additional exercise in the multiplica- 
 tion and division of radical quantities. 
 
THEORY OF EXPONENTS. 197 
 
 1. Multiply ^2 by >,J/3. Ans. ^2592. 
 
 2. Multiply V« by ^b. Ans. \/'^¥'. 
 
 3. Multiply ^ahj ^b. Ans, \^'^. 
 ' 4. Multiply 3 a a/8^ by 2 bs/TH. Ans, 12 a'^ i \/2Z 
 
 5. Multiply V 2, yv^ 3, and ^ 5 together. Ans. a^648000. 
 
 6. Multiply ^f, ;^J, and ^6 together, Ans. ^/y. 
 
 7. Multiply 2V6 — 3V5 by 4 VS — '^IO^ 
 
 Ans. 39V2— 16V15. 
 
 a Multiply 4 V^ + 5 V^ by Vi + 2a/^- _« 
 
 Ans. '^j^ + V^ V42. 
 
 9. Divide ></6 by ^2. Ans. J/v'55296: 
 
 10. Divide \/l35 by V^. Ans. a/ts: 
 
 11. Divide aX/b by ^^. Ans. a^^b, 
 
 12. Divide ^« by s/b ■\- s/c. Ans. . 
 
 — c 
 
 13. Divide V 3 by 1 + V 2. Ans. V 3 (V 2 — 1). 
 
 14. Divide 1 Vi by V 2 + 3 Vi- Ans. ^n- 
 
 15. Divide 1 by Va -j- V** Ans 
 
 16. Divide 
 
 Y a^ — b ' 
 
 a/s/^^ + s/^ by ^/s/a — s/b, 
 
 A„s. »/^±ll2V5 
 
 r a — b 
 
 SECTION XIX.— Theory of Exponents. 
 
 148. We have seen, art. 28, that with respect to the same 
 letter, division is performed by subtracting the exponent of 
 
198 ELEMENTS OF ALGEBRA. 
 
 the divisor from that of the dividend. The application of this 
 rule to the case, in which the exponent of the divisor is equal to 
 that of the dividend, gives rise to the exponent 0. An expres- 
 sion a", in v/hich this exponent is found, is to he regarded, 
 art. 31, as a symbol equivalent to unity, 
 
 149. The application of the same rule to the case, in which 
 the exponent of the divisor exceeds that of the dividend, gives 
 rise to negative exponents. Thus let it be required to divide 
 a^ by a^. Subtracting the exponent of the latter from that of the 
 former, we have a~^ for the result. But a^ divided by c^ is 
 
 expressed by the fraction -g ; reducing this fraction to its lowest 
 terms, we have —^. The expression a~'^ must therefore be re- 
 garded as equivalent to —^. 
 
 In like manner ^^_^_,^ gives by subtracting the exponent of 
 the divisor from that of the dividend a~"; but the fraction 
 fifives when reduced to its lowest terms — ; whence a"" 
 
 is equivalent to — . 
 
 The expression «"" is, therefore, the symbol of a division, 
 which cannot be performed. Its true value is the quotient of 
 unity divided by a raised to a -power denoted by the negative 
 exponent n. 
 
 150. To find the roots of monomials, we divide, art. 122, the 
 exponents of the proposed by the index of the root required. 
 The application of this rule to the case, in which the exponents 
 of the proposed are not divisible by the index of the root, gives 
 rise to fractional exponents. Thus let the third root of a be 
 required. Indicating upon the exponent of a the operation re- 
 quired in order to obtain the third root, we have for the re- 
 sult a^. But we have agreed to indicate the third root by >^; 
 
THEORY OF EXPONENTS. 199 
 
 the expressions /^ a, a^ are, therefore, to be regarded as equiva- 
 lent. In like manner, we have 
 
 m 
 
 The expression a" is, therefore, to be regarded, as a symbol 
 equivalent to the nth root of the mth power of a. 
 
 151. The two preceding cases sometimes meet in the same 
 expression. This gives rise to negative fractional exponents. 
 Thus let it be required to extract the seventh root of a? di- 
 
 vided by a^ ; we have -g = a~^ the seventh root of which is 
 a ~ T. In like manner the wth root of 
 
 The expression a " is, therefore, the symbol of a division 
 which cannot be performed, combined with the extraction of a 
 root. Its true value is the nth root of the quotient of unity di- 
 vided by a raised to the mth power. 
 
 m m 
 
 The expressions a", «"'", a", a ", derived in the manner 
 above explained from rules previously established, have be- 
 come by agreement notations equivalent respectively to 1, 
 
 1 "" y I 
 
 — , /^(f"^ t / — ; we may, therefore, at pleasure substitute 
 
 the former of these expressions for the latter, and the converse. 
 152. We proceed to show, that the rules already established 
 for performing the operations of arithmetic upon quantities 
 affected with entire and positive exponents are sufficient for 
 these operations, whatever the exponents may be, with which 
 the quantities are affected. 
 
200 ELEMENTS OF ALGEBRA 
 
 MULTIPLICATION. 
 
 Let it be required to multiply a^ hy a . To perform the 
 Operation required, it is sufficient to add the exponents. 
 
 Indeed J = ^^, J = a/^, whence 
 But adding the exponents, we have 
 
 ,fxJ = .i + i = e« 
 
 the same result as before. 
 
 Again, let it be required to multiply a~^ by a*; we have 
 
 whence a~^ X «^ = l/^^3 X ^7^^= tV'^ X )^a}' . . 
 
 But adding the exponents of the proposed, we have 
 the same result as by the former operation. 
 
 m p 
 
 Let it be required next to multiply a " by <z' ; 
 we have a '^ = \X '^t> a? = t^oF ; 
 
 whence 
 
 nq /■ 
 
 We arrive at the same result by adding the exponents of the 
 proposed. 
 
 _ 2! I Z. np — mq 
 
 Indeed a "" ' =a '"' . 
 
THEORY OF EXPONENTS. 201 
 
 To multiply two monomials therefore, it is sufficient, whatever 
 the exponents, to add the exponents of the letters^ which are the 
 same in each. 
 
 EXAMPLES. 
 
 1. Multiply a* c*, a "" ^ 3, and c^ i "" ^ together. 
 
 Ans. a"^^^, 
 
 2! Multiply -y^ by ^. Ans. a^^bK^^. 
 
 3. Multiply a^ + b^ by a^ — b^. Ans. a^ — b^, 
 
 4. Multiply 3 + 52 by 2 — 5*. Ans. 1 — sK 
 
 DIVISION. 
 
 153. Whatever the exponents may be, in order to divide one 
 monomial by another, we subtract for each letter the expon£nt 
 of the divisor from that of the dividend. 
 
 Indeed, since the exponent of each letter in the quotient should 
 be such, that when added to the exponent of the same letter 
 in the divisor, the sum will be equal to the exponent of the divi- 
 dend, it follows, that the exponent of the quotient should be equal 
 to the difference between that of the divisor and the dividend. 
 
 EXAMPLES. 
 
 1. Divide a' by a"*. 
 
 Ans. a^. 
 
 2. Divide a* by a^. 
 
 Ans. a"^\ 
 
 3. Divide a^b^ by a''^b^. 
 
 Ans. a"^^*"*. 
 
 4. Divide a^ — b^ by a* — i*. Ans. a^ + ah^ + b^. 
 
 5. Divide 5a^ — 41a^i + 42fl^i« by 5a^ — eah. 
 
 Axis, fli — y-""?'- 
 
ELEMENTS OF ALGEBRA. 
 FORMATION OF POWERS AND EXTRACTION OF ROOTS. 
 
 154. From the rule for multiplication, it follows, that to raise 
 a monomial to any power, it is necessary whatever the exponents 
 of the letters, to multiply the exponent of each letter by the 
 exponent of the power required. 
 
 _g 
 Thus a raised to the third power 
 
 _,-i+(-i)^(-i)_,-«_„-'. 
 
 Conversely, to extract the root of a monomial, we dividb the 
 exponent of each letter by the index of the root. 
 
 Thus ^a-2 = a"^ 
 
 The utility of exponents, of the kind which we are here con- 
 sidering, consists principally in this, that the calculation of quan- 
 tities affected with these exponents is performed by the rules 
 already established for quantities affected with entire and positive 
 exponents. The calculation is moreover reduced to operations 
 upon fractions, with which we are already familiar. 
 
 155. By means of negative exponents we may give an entire 
 form to fractional expressions. Thus, let there be the fraction 
 
 -5, this is the same as a; X -5 J t)ut — = 2/ ~ ^ ; whence 
 r f f 
 
 ^ -a 
 
 156. Fractional and negative exponents enable us to arrange 
 polynomials, which contain radical terms. Thus let it be re- 
 quired to arrange the polynomial 
 
 according to the descending powers of the letter a. 
 
 To perform the operation required, 1°. we give to the radical 
 quantities fractional exponents ; 2^. we reduce to an entire 
 form, terms which have denominators; 3°. we reduce all the 
 exponents of the letter, according to which the arrangement is to 
 be made, to their least common denominator. The proposed 
 
PROPORTION BY DIFFERENCE. 269 
 
 may then be arranged according to the powers of the letter 
 required. 
 
 In the preceding example we have for the result 
 
 SECTION XX.— Proportions. 
 
 157. When two quantities are compared with respect to their 
 magnitude, the result of the comparison is called their ratio. In 
 general, there are two different ways, in which the magnitude 
 of two quantities may be compared; P. we may wish to de- 
 termine how much the greater exceeds the less; the result is 
 then obtained by subtraction, and is called the ratio of the 
 quantities by difference; 2°. we may wish to determine how 
 often one of the quantities is contained in the other ; the result is 
 then found by division and is called the ratio of the quantities 
 by quotient. 
 
 Thus, the ratio by difference of the quantities a and b is 
 
 a — b, and the ratio by quotient is -; a and b are the terTns 
 
 of the ratio. 
 
 The same quantity may be added to, or subtracted from, both 
 terms of a ratio by difference loithout changing the ratio, for 
 a — b={a-\-c) — {b-\-c) = {a — c) — [b — c). 
 TJie two terms of a ratio by quotient may be multiplied of 
 divided by the same quantity without changing the ratio, for 
 a am 
 b bm' 
 Ratios by difference are sometimes called arithmetical ratios 
 and those by quotient geometrical ratios. 
 
 158. An expression for two equal ratios is called a proportion* 
 
204 ELEMENTS OF ALGEBRA. 
 
 If the ratios are by difference, the proportion is called a propor 
 tion by difference. Thus the equality 
 
 b — a = d — c, 
 is a proportion by difference, and is usually written thus, 
 a .b:c . d. 
 If the ratios are by quotient, the proportion is called a propor- 
 
 CL C 
 
 tion by quotient. Thus, the equality j = - is a proportion by 
 
 CL 
 
 quotient, and is usually written 
 
 a:b::c:d. 
 
 The proportions above are read thus, a is to i as c to d. The 
 first and last terms are called the extremes of the proportion; 
 the second and third are called the means ; a is called the ante- 
 cedent, b the consequent of the first ratio ; c the antecedent, d the 
 consequent of the second ratio. 
 
 Proportion by difference is sometimes called arithmetical pro- 
 portion, that by quotient geometrical proportion. Proportion by 
 difference is now, however, more commonly called equidiffer- 
 ence, while the term proportion is limited to proportions by 
 quotient. 
 
 EQUIDIFFERENCES. 
 
 159. Let there be the equidifference a.b:c.d\ this is the 
 same with the equation b — a = d — c, from which we deduce 
 a -\- d =z b -\- c. 
 
 Thus in an equidifference the sum of the extremes is equal to 
 the sum of the means. This is the leading property of equi- 
 differences. 
 
 Reciprocally, let there be four quantities a, b, c, d, such that 
 a-\- d = b -^-c. From this equation we obtain 
 b — a = d — c, or a . b : c . d. 
 
 Thus, if there be four quantities such, that any two of them 
 give the same sum with the other two, the first are the extremes ^ 
 the second the mearis, or the converse, of an equidifference. 
 
PROPORTION BY QUOTIENT. 205 
 
 Any three terms of an equidifference are sufficient to deter- 
 mine the fourth ; thus, from the equidifference, a .b :c , dj we 
 deduce a = b — d -{- c, b==.a-\- d — c. 
 
 In the equidifference a .b: c . d, let c = 3 ; we have 
 a .b :b . d. 
 
 This is called a continued equidifference, and b is called an 
 arithmetical mean between a and d. 
 
 From the equidifference a .b :b . d, we deduce 
 b = \{a + d); 
 the arithmetical mean betioeen tivo quantities is, therefore, equal 
 to half their sum. 
 
 160. In order that an equidifference may exist, it is sufficient, 
 that the sum of the extremes should be equal to the sum of the 
 means ; we may, therefore, make any transposition of the terms, 
 of an equidifference, which will not alter the equality between 
 whe sum of the extremes and that of the means. The equation 
 a — b = c — d furnishes the eight following equidifferences, 
 
 a . b : c . d, a . c: b , d, d , b : c . a, d . c: b . a, 
 h . a: d . c, b . d: a . c, c . d: a .b, c . a: d .b. 
 
 PROPORTION BY QUOTIENT. 
 
 161. Let us take the proportion a:b: : c:d; this returns 
 o -=i-, an equation, which gives 
 
 adi=bc. 
 
 In a proportion by quotient, therefore, the product of the eX' 
 tremes is equal to the product of the means. This is the funda- 
 mental property of proportions. 
 
 Reciprocally, let there be four quantities a, b, c, d, such, that 
 
 ad = bc; this leads to the equation - = -, or 
 
 a c 
 
 a: bit: d. 
 
 Whence if four quantities be such, that any two of them give the 
 
 tame product as the remaining two, the first will form the ex* 
 
 iremes and the second the means^ or the converse, of a proportion. 
 
206 ELEMENTS OF ALGEBRA. 
 
 Three terms of a proportion are sufficient to determine the 
 
 fourth ; thus from the proportion a:h '.: c: d^ we deduce 
 
 he ^ ad ^ 
 a=^-rr, b = — , &c. 
 a c 
 
 The proportion a: b : : b: d, in which the two mean terms are 
 the same, is called a continued proportion, and b is called a rnean 
 proportional between a and d. 
 
 From the continued proportion a: b : : b : d,vfe deduce P=:ad, 
 
 whence br=W ad. Thus to find a mean proportional betioeen 
 two quantities, we take the square root of their product. 
 
 162. In order that a proportion may exist, it is sufficient, that 
 the product of the extremes should be equal to that of the means. 
 We may, therefore, make any transposition in the terms of a 
 proportion, which will leave the product of the extremes equal to 
 
 that of the means. Thus the equation - = - gives the eight 
 
 following proportions 
 
 a : b : : c : d, a : c : : b : d, b : d : : a : c, d : c : : b : a, 
 b : a : : d : c, c : a: : d : b, d : b : : c : a, c : d : : a : b. 
 
 163. The same quantity m, it is evident, may be added to or 
 
 subtracted from the equation - = -, so that we have 
 
 a c 
 
 b^ d^ 
 
 . a c 
 
 , b-:^ma d-±:mc 
 whence = . 
 
 a c 
 
 but this last may assume the form 
 
 c d-^mc 
 
 a b z^ma' 
 
 from which we have the proportion 
 
 b :^ma\ d-:^mc\ : a:c; 
 
 c d 
 *ut since - = T> we have also 
 a 
 
 d d-±z "fnc 
 b' h-^ma 
 
PROPORTION BY QUOTIENT. 207 
 
 from which we have the proportion 
 
 h -i^ma : d ± ^c : : b : d. 
 
 These two proportions may be enunciated thus ; The first 
 
 consequent plus or minus its antecedent taken a given number of 
 
 times, is to the second consequent plus or minus its antecedent 
 
 taken the same number of times, as the first term is to the third, 
 
 or as the second is to the fourth. 
 
 ^aA m. . d±:mc c 
 
 lo4. 1 he expression ~ = - returns to 
 
 ± 'ma a 
 
 d-\-mc c d 
 
 c 
 
 b-\-ma a b — ma a 
 
 . d~\-mc d — mc 
 
 whence ,— y = -, , 
 
 b-\-ma b — ma 
 
 or b -\- ma: d-\-mc::b — ma:d — mc, 
 
 or, changing the relative places of the means, 
 
 b -\-ma:b — ma'. '. d-\-mc: d — mc] 
 
 whence making m=^\, we have 
 
 b -\- a:b — a:: d-\- c d — c, 
 
 a proportion which may be enunciated thus. 
 
 The sum of the first two terms is to their difference, as the sum 
 
 of the last two is to their difference. 
 
 165. The proportion a : b : : c: d may be written thus, 
 
 a : c : : b : d, 
 
 we have then 
 
 c d 
 
 a b 
 
 from which we obtain 
 
 ^ c-^zma-.d^mb:: a-.b, OT :: c: d; 
 
 whence, the second antecedent plus or minus the first taken a 
 
 given number of times, is to the second consequent plus or minus 
 
 the first taken the same number of tim£s, as any one of the ante' 
 
 cedents whatever is to its consequent. 
 
 If in the above proportion we make m= 1, we have 
 
 c ':^ a : d :^b : : a : b, ox '. : c : d ', 
 
 whence c-\-a:c — a::d-\-b'.d — h. 
 
SOS ELEMENTS OF ALGEBRA. 
 
 Therefore, the sum or difference of the antecedents is to the sum 
 or difference of the consequents, as one antecedent is to its consc' 
 quent ; and the sum of the antecedents is to their difference^ as the 
 sum of the consequents is to their difference. 
 
 166. Let there be the series of equal ratios 
 
 a: b : : c: d: : e :/: : g : h , . . , 
 
 ^__d__f_h 
 a c e g' 
 
 Making — = q; we have 
 
 b d f h 
 
 - = q,- = q,- = q,-=zq; 
 
 a c e g 
 
 whence bz=aq, d = cq, fz=eq, h = gq; 
 
 adding these equations member to member, we have 
 
 b + d+f+h = {a + c + e + g)q; 
 
 , • b+d+f+h b 
 
 whence — ' r^ — = <7 = -» 
 
 a-f-c + e + ^ a 
 
 or a-{-c-\-e-^ g: b-{-d-{-f-^h::a:b; 
 
 whence in a series of equal ratios, the sum of any number what' 
 
 ever of antecedents, is to the sum of the like number of consC' 
 
 quents, as one antecedent is to its consequent. 
 
 167. Let there be the two equations - = -,--=-; muhiply- 
 
 a c e g 
 
 ing these equations, member to member, we have 
 bj__dh 
 ae cg^ 
 or ae : bf: : eg : dh. 
 
 We obtain the same result by multiplying, term by term, the 
 proportions a:b : : c: d, e:f::g:h; this is called multiplying 
 the proportions in order ; it follows then, that if two proportions 
 be multiplied in order, the results will be proportional. 
 
 It will be seen also, that if two proportions be divided, term by 
 term, or in order^ the quotients will be proportional. 
 
PROPORTION BY QUOTIENT. 209 
 
 If in the equation - = - we raise both members to the mth 
 a c 
 
 power, we have 
 
 3"* dr 
 
 which gives aP" : b^ : : c^ : dl^. 
 
 It follows, therefore, that the second^ thirds and in general the 
 similar powers of four proportional quantities are also propor- 
 tional. 
 
 In like manner, it may be shown that the roots of the same 
 degree of four proportional quantities are also proportional. 
 
 QUESTIONS IN WHICH PROPORTIONS ARE CONCERNED. 
 
 1. The sum of the squares of two numbers is to the difference 
 of their squares as 17 to 8, and their product is 15. What are 
 the numbers ? 
 
 Putting X and y for the numbers, we have by the first con- 
 
 dition 
 
 ^ + f 
 
 7?- 
 
 -7/::17:8; 
 
 whence art. 164, 
 
 22? 
 
 
 22/^:: 25: 9, 
 
 or 
 
 a? 
 
 
 2/^::25:9; 
 
 whence art. 167, 
 
 X 
 
 
 y '.: 5:3, 
 
 wherefore 
 
 
 
 3z=5v. 
 
 By the second condition we have xy=\5', comparing this 
 w:th the equation just found, we readily obtain a: = 5, ?/ = 3. 
 
 2. The product of two numbers is 24, and the difference of 
 their third powers is to the third power of their difference as 
 19 to 1. Required the numbers. 
 
 Let X and y be the numbers, we have by the question 
 
 a:y = 24 
 7? — 'if:{x--yf:'.\^'.\. 
 
 Transposing terms and developing {x — y)^ in this last, we 
 
 have 7?-'f:\^:'.7?—^7?y-{-^xif — f'.\\ 
 
 whence art. 165, ^y?y — 2xy^ : 18 : : (a: — yf : 1, 
 
 or dividing by a: — y and substituting for xy its value from the 
 
 first equation 72 : 18 : : (a: — y)' : 1 ; 
 
 whence z — v = 2. 
 
 U ^ 
 
210 ELEMENTS OF ALGEBRA. 
 
 Comparing this last with the first equation we obtain 
 a; = 6, ?/ = 4. 
 
 3. The sum of two numbers is to their difference as 3 to 1, 
 and the difference of their third powers is 56; what are the 
 numbers ? Ans. 4 and 2. 
 
 4. There are two numbers, whose product is 135, and the 
 difference of their squares is to the square of their difference 
 as 4 to 1. What are the numbers ? Ans. 15 and 9. 
 
 5. A merchant mixes wheat, which cost him 10 shillings a 
 bushel, with barley which cost him 4 shillings a bushel, in such 
 proportion as to gain 43J per cent, by selling the mixture at 
 11 shillings a bushel. Required the proportion. 
 
 Ans. 14 bushels of wheat to 9 of barley. 
 
 6. The product of two numbers is 63, and the square of their 
 sum is to the square of their difference as 64 to 1. What are 
 the numbers ? Ans. 9 and 7. 
 
 SECTION XXL— Progressions. 
 
 168. A series of quantities increasing or decreasing by a con- 
 stant difference, is called an arithmetical progression or prO' 
 gression by difference. The constant difference is called the 
 ratio of the progression. 
 
 Thus let there be the two following series 
 1, 4, 7, 10, 13, 16 
 60,56,52,48,44,40. 
 the first is called an increasing progression, the ratio of which 
 is 3 ; the second is called a decreasing progression, the ratio of 
 which is 4. 
 
 To indicate that the quantities c, h, c, d . . . form a pro- 
 gression by difference, we write them thus 
 -^ a .b ,c .d . . , , 
 A progression by difference, it will be readily perceived, is 
 
PEOGRESSION BY DIFFERENCE. 5^| 
 
 simply a series of continued equidifferences. Each term there- 
 fore is at once antecedent and consequent, with the exception of 
 the first term, which is only an antecedent, and of the last, which 
 is only a consequent. The progression -f- a . i . c . d is enun- 
 ciated thus, a is to i as 3 to c, as c to d^ &cc. 
 
 169. Let us take the increasing progression 
 
 •7-a.b.c.k.h.., 
 and let d represent the ratio. 
 
 From the nature of the progression, we have, it is evident, 
 b=a -|- d 
 c==a-\-2d 
 k = a-!rSd, 
 from which it is readily inferred, that a term of any rank lohat- 
 ever is equal to the first term plus as many times the ratio, as 
 there are units in the number of the preceding terms. 
 
 Let L represent a term of any rank whatever, and let n denote 
 the number, which marks the place of this term ; we have from 
 what has been said 
 
 L = a-\- {n — l)d. 
 This expression for L is called the general term of the series. 
 If the series were decreasing, we should have, as it is easy to 
 see, for the general term 
 
 L = a — {n — l)d. 
 By means of the above formulas we may find any term of a 
 progression by difTerence, when the first term, the number of the 
 term required, and the ratio are given. 
 
 Thus, let it be required to find the 50th term of the pro- 
 gression -T- 1 . 4 . 7 , we have by the first formula 
 
 I<=1 + (50 — 1)3=148. 
 Again let it be required to find the 40th term of the pro- 
 gression -t- 5 . 3 . 1 . . ., we have by the second formula 
 L==5— (40 — 1)2 = — 73. 
 
 170. The first and last terms of a progression are called the 
 extremes; if the number of terms be odd, the middle term is 
 called the mean; if the number of terms be even, the two 
 
212 ELEMENTS OF ALGEBRA. 
 
 terms having the same number of terms on each opposite side 
 are called the means. 
 
 Let us take the general progression -^ a .h .c . , , ,h .h .1, 
 from the nature of the progression, we have 
 
 h — a = l — A 
 whence b -\-k-=a-\-l 
 
 so also c — b = k — n 
 
 whence c-|-A = i-|-Ar = «4-Z, 
 
 from which we infer that in a progression by difference, the sum 
 of any two terms taken at equal distances from the extremes is 
 equal to the sum of the extremes. 
 
 Let S represent the sum of all the terms in the progression 
 ■^ a .b . c . . . ,k . k .1,. Writing this progression in an inverse 
 order below itself, we have 
 
 S = a-{-b + c .... k + k-^l 
 S=l-\-k + h .... c-i-b-i-a; 
 adding these equations member to member and uniting the 
 corresponding terms, we have 
 
 2S={a + l) + {b + k).... + {h + c) + {k + b) + {l + a) 
 but the parts b-{-kf k-j-c . . . are equal each to a + Z ; the 
 number of these parts moreover is the same, it is evident, with 
 the number of terms in the progression; designating then this 
 number by «, we have 
 
 2S = n{a-{-l); 
 
 whence S^^^pl. 
 
 By means of this formula we find the sum of all the terms, 
 when the first term, the last term and the number of the terms 
 are given. 
 
 EXAMPLES. 
 
 1. What is the sum of the natural series of numbers 1, 2, 
 3,4, &c. up to 1000? 
 
 2. The last term in a progression by difference is 60, the first 
 term 12, and the number of terms 5. Wliat is the sum of all 
 the terms ? 
 
PROGRESSION BY DIFFERENCE. 
 
 213 
 
 3. What is the sum of the uneven numbers 1, 3, 5, 7, &c. 
 up to 93 ? 
 
 171. The equations L = a-{- {n—l)d, g^ ^(g + [ ) f^^. 
 
 nish us with the means of resolving the following general 
 problem, viz. Any three of the five quantities, a, d, n, 1 and s, 
 which enter into a progression by difference, being given, to dc' 
 termin£ the remaining two. 
 
 This general problem resolves itself into as many particular 
 problems as there are combinations of 5 letters taken 2 and 2, or 
 3 and 3 at a time. The number will therefore be 10. See the 
 enunciations below. 
 
 Let there be given 1°. a, d, n to find I and * ^ ^ ■' I 
 
 2°. a, d, I ... n and s -^ 
 
 S°. a, d, s . . . n and I 
 
 4°. a,n, I . . . d and s 
 
 5°. a, n, s . . . d and I 
 
 6°. a, I, s . . . d and n 
 
 7°. dfU, I . . . a and i 
 
 B°. d, n, s . . .a and I 
 
 9°. d, I, s . . . «and» 
 10°. n,l,s... «andrf^i>^^ . - 
 
 172. Let it now be proposed to find the number of terms in 
 the progression by difference, the sum of which is 145, the first 
 term 1, and the ratio 3. ^ . — 
 
 This example is a particular case of the third general prob- 
 lem ; to prepare a formula for its solution, we eliminate L from 
 
 the equations 
 
 71 {a -\- 1) 
 2 ' 
 
 L = a-\'{n—\)d, S = 
 
 by which we obtain for the equation for n 
 2 {d — 2a) 
 
 d 
 
 
 2£ 
 
 d' 
 
 resolving this equation we have 
 
 __ d — 2a±^/{d — 2af-{-Sds 
 "*"" 2d 
 
214 ELEMENTS OF ALGEBRA. 
 
 from which we obtain 10 for the number of terms sought. This 
 being done we readily obtain by means of the formula for L the 
 last term required. 
 
 Let the learner prepare the formulas and solve the following 
 particular cases. 
 
 1. To find the first and last terms in a progression by differ- 
 ence, the sum of which is 567, the number of terms 21 and the 
 ratio 2. 
 
 2. The sum of a progression by difference is 1455, the first 
 term 5, and the number of terms 80. What is the last term and 
 the ratio ? 
 
 3. The first term in a progression by difference is 5, the last 
 term 185, and the ratio 6, to find the number of terms, and the 
 sum of all the terms. 
 
 4.- The first term of a progression by difference is 3, the last 
 term 41, and the sum of all the terms 440. What is the num 
 ber of terms and the ratio ? 
 
 173. The formula L = a-\- {n — 1)^ gives d = =■ ; this 
 
 expression for d enables us to resolve the following problem, viz. 
 to insert between two quantities, b and c, m arithmetical means, 
 that is to say, quantities, which comprised between b and c, will 
 form with them a progression by difference. 
 
 To resolve this problem, it will be sufficient to ^determine the 
 ratio of the progression required. For this we have given the 
 first term b, the last term c, and the number of terms m -{- 2. 
 Substituting therefore c, b, and m -|- 2 for I, a, and n in the 
 
 above expression for d, viz. d = -, the ratio required will 
 
 be r— -= — -r-— ., that is, to find the ratio sought, we 
 
 m-\-2 — I m-f-1 
 
 divide the difference of the two numbers b and c by the number of 
 
 terms to be inserted plus 1. 
 
 Let it be required, for example, to insert 11 arithmetical 
 
 means between 17 and 77. 
 
 77 17 
 
 Here d = — — — = 5. 
 
PROGRESSION BY DIFFERENCE. ^ISj 
 
 The progression required will therefore be 
 
 -r 17 . 22 . 27 . 32 72 . 77. 
 
 Let it be required, as a second example, to insert 9 arithmet- 
 ical means between each antecedent and consequent of the pro- 
 gression -J-2.5.8.11.14.... 
 
 It will readily be inferred from what has been done, that, if 
 between the terms of a progression by difference^ taken two and 
 two^ we insert the same number of arithmetical means^ the terms 
 of this progression together with the arithmetical means iriserted 
 win form a progression by difference, 
 
 QUESTIONS INVOLVING PROGRESSIONS BY DIFFERENCE. 
 
 1. A number consisting of 3 digits, which are in arithmetical 
 progression, being divided by the sum of its digits gives a quo- 
 tient 48; and if 198 be subtracted from it the digits will be 
 inverted. Required the number. 
 
 Let X = the second digit and y the common difference, the 
 three digits will then be expressed hy x ■-\- y, x, x — y. 
 
 Resolving the question we obtain a: = 3 ; and the number re- 
 quired is 432. 
 
 2. Four numbers are in arithmetical progression. The sum 
 of their squares is equal to 276, and the sum of the numbers 
 themselves is equal to 32. What are the numbers ? 
 
 Let 2?/= the common difference and let x-\-Sy, x-\-y, 
 X — y,x — 3 y be the numbers. 
 
 Resolving the question, we obtain for the numbers sought, 
 11, 9, 7 and 5. 
 
 3. A traveller sets out for k certain place and travels 1 mile 
 the first day, 2 the second and so on. In 5 days afterwards 
 another sets out and travels 12 miles a day. How long and how 
 far must he travel to overtake the first ? 
 
 Let X = the number of days ; then a; -|- 5 = the number of 
 
 a; 4-5 
 days the first travels, and {x-\- 6) — ^ — = the distance he 
 
 travels. 
 
216 ELEMENTS OF ALGEBRA. 
 
 Resolving the question, we obtain a; = 3, or 10. 
 
 4. There are three numbers in arithmetical progression, 
 whose sum is 21 ; and the sum of the first and second is to 
 the sum of the second and third as 3 to 4. Required the 
 numbers. . Ans. 5, 7, 9. 
 
 5. From two towns, which were 168 miles distant, two per- 
 sons, A and B, set out to meet each other; A went 3 miles the 
 first day, 5 the next, and so on ; B went 4 miles the first day, 
 6 the next, and so on. In how many days did they meet ? 
 
 Ans. a 
 
 6. There are four numbers in arithmetical progression, whose 
 sum is 28, and their continued product is 585. Required the 
 numbers. Ans. 1, 5, 9, 13. 
 
 7. A and B, 165 miles distant from each other, set out with a 
 design to meet ; A travels 1 mile the first day, 2 the second, and 
 so on ; B travels 20 miles the first day, 18 the second, and so on. 
 How soon will they meet ? 
 
 Ans. They will meet in 10, and also in 33 days. 
 
 8. The sum of the squares of the extremes of four numbers in 
 arithmetical progression is 200, and the sum of the squares of the 
 means is 136. What are the numbers ? 
 
 Ans. 14, 10, 6, 2, or — 14, — 10, — 6, — 2. 
 
 9. A regiment of men was just sufiicient to form an equilateral 
 wedge. It was afterwards doubled, but was still found to want 
 385 men to complete a square containing 5 more men in a side, 
 than in a side of the wedge. How many did the regiment at 
 first contain ? Ans. 820. 
 
 PROGRESSION BY QUOTIENT. 
 
 174. A series of quantities such, that if any term be divided by 
 the one v/hich precedes it, the quotient is the same in whatever 
 part of the series the two terms are taken, is called a geometricai 
 progression or progression by quotient. 
 
 The constant quotient is called the ratio of the progression, 
 
PROGRESSION BY QUOTIENT. iflf 
 
 If the series is increasing, the ratio will be greater than unity ; 
 if decreasing, the ratio will be less than unity. 
 
 The following series are examples of this kind of progres- 
 sion, 3 . 6 . 12 . 24 . 48 . 96 
 64 . 16 . 4 . 1 . J . tV 
 
 In the first the ratio is 2, in the second J. A progression by 
 quotient, it will readily be perceived, is simply a series of equal 
 ratios by quotient, in which each term is at once antecedent and 
 consequent, ivith the exception of the first, which is only an ante- 
 cedent, and of the last, which is only a consequent. 
 
 To indicate that the quantities a,b,c,d... form a progres- 
 sion by quotient, we write them thus 
 
 ^ a'.h '. c '. d'. . . ». 
 
 The progression is enunciated thus, a to ^ as i to c as c, &c. 
 
 175. Let us take the general progression 
 ^ a'.b:c:d . . . , 
 and let the ratio be represented by q; from the nature of the 
 progression, we have, it is evident, 
 
 b=iaq, c=bq^=-aq^, d = cq-=a(f 
 
 from which it will be readily inferred, that a term of any rank 
 whatever is equal to the first term multiplied by the ratio raised 
 to a power, the exponent of luhich is one less than the number, 
 which marks the place of this term. 
 
 Let L designate any term whate^^er of the progression, and 
 let n represent the number of this term; from what has been 
 said, we have 
 
 This is called the general term of the progression. By means 
 of it we may find any term required, when the first term and the 
 ratio are given. 
 
 Thus let it be required to find the 8th term of the progression 
 •fr 2 : 6 : 18 . . . in this case, we have 
 
 i = 2x3' = 4374. 
 
218 ELEMENTS OF ALGEBRA. 
 
 In like manner if it be required to find the 12th term of the 
 progression -ff 64 : 16 : 4 : 1 : J . ., we have 
 
 ^ = «K5)" = 6i6- 
 
 176 Resuming the general progression 
 
 we have from the nature of the progression the series of equa- 
 tions, 
 
 b = aq, c = hq, d = cq . . . l:=kq; 
 adding these equations member to member, we have 
 
 b-\-c-{-d-\-. , .l=[a-^rh + c-\-. , .k)q. (1) 
 Let S represent the sum of all the terms, we have 
 
 * + c + ^ + l=zS — a 
 
 a-\-b'-\- c-\- . . . . k= S — I; 
 whence by substitution in equation (1), we have 
 S — a = q{S — l), 
 
 and by consequence S = =-. 
 
 By means of this formula we may obtain the sum of any num- 
 ber of the terms of a progression by quotient ; for this purpose, we 
 multiply the last term by the ratio, subtract the first term from 
 this product, and divide the remainder by the ratio diminished by 
 unity. 
 
 Let it be required to find the sum of the first 4 terms of the 
 progression -tt 2 : 6 : IS : 54 : 162, we have 
 
 o — 1 
 
 When the progression is decreasing, that is, when q is less 
 than 1, it will be more convenient to put the above expression 
 
 for S under the form <S = ; since in this case the two 
 
 l — q 
 
 terms of the fraction will be positive. 
 
 Let it be required to find the sum of the 12 first terms of the 
 
 progression^64:16:4:l:l....^. 
 
PROGRESSION BY QUOTIENT. 019 
 
 64 
 
 ,,r u o « — ^y 65536*4 ^^ , 65525 
 WehaveS = 3— ^==_— 3; = 85 + — ^^g. 
 
 4 
 177. If in the formulas for S we substitute for I its value, 
 viz. Z = ag'''~\ we have 
 
 aq^' — a a — af 
 
 formulas, by means of which we obtain the sum of any number 
 of terms of a progression, when the number of terms, the first 
 term and ratio are given. 
 
 Thus to find the sum of the first 8 terms of the progression 
 rr 2 : 6 : 18 : 54 we have 
 
 S=-— P = -3--^=6560. 
 
 In the same manner, wc have for the sum of the 12 first terms 
 of the progression -rr 64 : 16 : 4 . . . 
 
 65535 
 
 a^aqn 64-64^)^^ 
 S = -.j-— = — j— = 85 + 
 
 q 1 ' 1966081 
 
 4 
 
 EXAMPLES. 
 
 1. The first term of a progression by quotient is 4, the ratio 
 3, and the last term 78372. What is the sum of all the terms ? 
 
 2. The first term of a progression by quotient is 8, the last term 
 
 ncxAci^ ^^^^ ^^® r^^io 9' What is the sum of the progression ? 
 
 3. The first term of a progression by quotient is 3, the ratio 
 
 7 
 
 -, and the number of terms 10. Required the sum of the pro- 
 gression. 
 
 4. The first term of a progression by quotient is -, the ratio 
 
 2 ' 
 
 s, and the number of terms 9. What is the sum of the pro- 
 
 gression ? 
 
220 ELEMENTS OF ALGEBRA. 
 
 INFINITE PROGRESSIONS BY QUOTIENT. 
 
 178. Let there be the decreasing progression 
 ir (t- b ' c: d . . , 
 consisting of an infinite number of terms. The formula for the 
 
 sum of any number of terms, viz. S = — may be put 
 
 under the form 
 
 But since the progression is decreasing, gf is a fraction ; ^ is 
 also a fraction ; hence as the number n becomes greater, or as 
 
 we take more terms, the expression -z . 9" becomes smaller, 
 
 and the value of S approaches nearer to -z . If then we 
 
 suppose n greater than any assignable quantity or infinite, 
 . g-" will be less than any assignable quantity or 0, and 
 
 will in this case represent the true value of the series. 
 
 \-q 
 
 a 
 
 l-q 
 
 We conclude, therefore, that the sum of the terms of a de- 
 creasing progression, in which the number of terms is infinite, 
 
 has for its expression S = , q being the ratio of the pro- 
 gression and a the first term. 
 
 179. Strictly speaking the quantity is the limit which 
 
 the sum of a decreasing progression can never surpass, but to 
 which it continually approximates as we take more terms. 
 Let there be, for example, the progression 
 
 we have a=\, q=\, whence 
 
 e__ O' a „_ 2 2 /iy_ 2 ]_ 
 
 — 1__^ l_g-^— SZTi— 2iriXV2/ ~2— 1 2"-»' 
 Here the greater the value of n or the more terms we take, 
 
 the less is the fraction ^^^zrv ^^^ the nearer the sum of the series 
 
PROGRESSION BY QUOTIENT. , 221 
 
 approaches to 2. If the number of terms be considered infinite, 
 the fraction ^^^—^ will be less than any assignable quantity or 
 
 zero, and the sum of the series will be equal to 2. 
 
 Strictly speaking, however, 2 is the limit, which the sum of 
 the proposed series can never surpass, but to which it constantly 
 approximates as we take more terms. 
 
 Thus let the nimiber of terms be 1, 2, 3, 4 . . . . succes- 
 sively, we have 
 
 1 
 
 = 2—1 
 
 1+i 
 
 = 2-4 
 
 1+i+i 
 
 = 2-J 
 
 l+J+i+4 
 
 = 2-4 
 
 i + 4 + J + 4 + tV=2-A- 
 
 Here the more terms we take, the nearer the sum of the pro- 
 gression will approach to 2, from which it may be made to differ 
 by a quantity as small as we please, though strictly speaking, it 
 can never become equal to 2. 
 
 180. When the series is increasing, that is, when q is greater 
 
 than unity, the expression S = cannot be considered 
 
 l — q 
 
 as the limit, which the s\im of the series can never surpass. 
 
 For the sum of a determinate number n of terms being 
 
 o « «?" . . ., , a(f M, . 
 
 o== , It IS evident, that :; — - — will mcrease 
 
 l_gr l — q l_gr 
 
 more and more numerically in proportion as n increases; by 
 consequence the more terms we take, the more will the sum of 
 
 the terms differ numerically from :; . In this case is 
 
 l — q l — q 
 
 merely the algebraic expression, which by its development gives 
 
 rise to the series 
 
 a-{'aq-\'aq^-{- ag^-\- . . . . 
 
 Indeed if we perform upon a the division indicated, we have 
 
 Yzr^=(^ + a9 + af + a^+ 
 
223 • ELEMENTS OF ALGEBRA. 
 
 181. In the above expression let a=l, g' = 2, we have 
 
 _1_ or- 1 = 1 + 2 + 4 + 8+16+... 
 
 an equation, in which the first member is negative, while the 
 second is positive, and greater in proportion as q is greater. 
 
 In order to interpret this result we observe, that if in the equa- 
 tion = a-\- aq-\- agf^ . . . . we stop the series at any par- 
 ticular term, it is necessary, in order to preserve the equality of 
 the two members, to complete the quotient by annexing to it the 
 fraction which remains. If, for example, we stop the series at 
 the fourth term a cf, we shall have by completing the quotient 
 
 ——=a + aq-\-aq^-}-a(f + Yir^' 
 an equation which is exact. Indeed if in this equation, we 
 make a=l, q = 2, we have 
 
 -l = l + 2 + 4 + 8 + i^; 
 
 from which we obtain — 1 = — 1. 
 
 EXAMPLES. 
 
 1. What is the sum of the infinite progression 
 
 1 : 1 : i : 2-V : . . . . 
 
 2. What is the sum of the progression 2 : f : f : con- 
 tinued to infinity ? 
 
 3. The first term of a geometrical progression is a, and the 
 
 ratio . What is the sum of this progression continued to 
 
 infinity ? 
 
 4. What vulgar fraction is equivalent to the repeating deci- 
 mal 3 ? 
 
 This decimal may be put under the form 
 
 ^ (tV + T^-^ + TTjW + TIT^XFTF + ) 
 
 5. What vulgar fraction is equal to the repeating decimal 25 ; 
 what to the decimal 375 ? 
 
 182. The equations lz=agr-\ S= ^ , contain all the 
 
 * ff_i 
 
J?BOGRESSION BY QUOTIENT. 
 
 tm 
 
 relations of the five quantities a, I, q, n, and S; we have then 
 the general problem, any three of the Jive quantities, a, 1, q, n and 
 S being given to find the remaining two. This general problem 
 gives rise to ten particular problems, the enunciations of which 
 will not differ from those relative to progressions by difference, 
 art. 171, with the exception, that the ratio is here expressed by 
 the letter q instead of d. 
 
 183. From the formula Z = ag'"~S we obtain 
 
 -v/^- 
 
 This expression for q enables us to resolve the following 
 problem, viz. to insert between two given numbers, b and c, m 
 mean proportionals, that is to say, a number m of quantities, 
 which comprised between b and c will form ivith them a progreS' 
 sion by quotient. 
 
 To resolve this problem it will be sufficient to determine the 
 ratio of the progression required ; for this we have given the first 
 term b, the last term c and the number of terms m -j- 2. 
 
 Substituting therefore b, c and m-\-2 ioT a, I and n in the 
 above expression for q, we have for the ratio of the required 
 progression 
 
 whence to find the ratio sought, ive divide the given numbers b 
 and c, one by the other, and extract the root of the quotient to the 
 degree marked by the number of terms to be i?iserted plus one. 
 
 Let it be required to insert six mean proportionals between the 
 numbers 3 and 384. Here ?/z = 6, we have therefore 
 
 . = ^f = ^Ils = . 
 
 The progression required is therefore 
 
 4f 3 : 6 : 12 : 24 : 48 : 96 : 192 : 384. 
 From what has been done, it will be easy to see, that if between 
 the terms of a progression by quotient taken two and two, we in' 
 
224 ELEMENTS OF ALGEBRA. 
 
 sert the same number of mean proportionals, the partial progres* 
 sions thus formed will together form a progression by quotient. 
 
 1S4. Of the ten particular problems, which may be proposed 
 upon progressions by quotient, four only can be resolved by 
 principles thus far laid down. Below we have the enunciation 
 of these problems with their answers. 
 
 1°. a, q, n being given to find I and S. 
 
 a{q^-\) 
 
 l-=.a^ 
 
 — 1 
 
 1 n-l 
 
 S— 1 
 
 2°. a,n,l being given to find q and S. 
 
 n_l 
 
 3**. q, n, I being given to find a and S. 
 
 a= -—7, o = 
 
 4®. q, n, S being given to find c and I. 
 
 S(?-l) ^ Sg-'(g-l) 
 
 Of the remaining problems, two, viz. those in which a and q, 
 I and q are the unknown quantities, depend upon the resolution 
 of equations of a degree superior to the second. The other four 
 depend upon the resolution of an equation of a nature altogether 
 different from any which we have yet seen, viz. upon an equa- 
 tion of the form c* = & in which the exponent is the unknown 
 quantity. 
 
 QUESTIONS PRODUCING PROGRESSIONS BY QUOTIENT. 
 
 1. There are three numbers in geometrical progression, the 
 greatest of which exceeds the least by 15. Also the difference of 
 the squares of the greatest and least is to the sum of the squares 
 of all the three numbers as 5 to 7. Eequired the numbers. 
 
 Let X, xy, xy^ be the numbers ; then by the question we 
 have xif — a: =15, 
 
 and 7 {x'y* ^a^) = 5 {x'f + a:V + :c»), 
 
 or by division 7{y* — 1) = 5 (i/* + ^ + 1). 
 
PROGRESSION BY QUOTIENT. 225 
 
 or performing the operations indicated, transposing and reducing 
 
 whence, resolving this last, we have 
 
 ^ = 4, and 2/ = 2. 
 Substituting next for y its value in the first equation, we ol 
 tain a: = 5. The numbers required are therefore 5, 10 and 2f •'. 
 
 2. The sum of three numbers in geometrical progression is 13, 
 and the product of the mean, and the sum of the extremes is 30. 
 Required the numbers. 
 
 X 
 
 Let the numbers be -, x and xy\ then by the question, we 
 have 
 
 J + ^ + ^y = 13. 
 
 and i--\-xy\x = ^0. 
 
 By transposition in the first equation, we have 
 — \-xy=.\^ — X ; 
 
 y 
 
 whence, by substitution in the second, we obtain 
 
 {n — x)x = '^0; 
 whence a? — 13 z = — 30, 
 
 from which we deduce 
 
 x=\0,x = 2. 
 Substituting the value a: = 3 in the first equation, we obtain 
 
 y = 3, or jr, and the numbers sought are 1, 3, 9. 
 
 3. The difference between the first and second of four num- 
 bers in geometrical progression is 36, and the difference between 
 the third and fourth is 4. What are the numbers ? 
 
 Ans. 54, 18, 6, and 2. 
 
 4. A gentleman divided £210 among three servants in geo- 
 metrical progression; the first had £90 more than the last. 
 How much had each ? 
 
 5. There are three numbers in geometrical progression, the 
 
 15 
 
226 ELEMENTS OF ALGEBRA. 
 
 sum of the first and second of which is 9, and the sum of the 
 first and third is 15. Required the numbers. 
 
 6. The sum of three numbers in geometrical progression is 
 35, and the mean term is to the difference of the extremes as 
 2 to 3. Required the numbers. Ans. 5, 10, 20. 
 
 7. The sum of £ 14 was divided between three persons, whose 
 shares were in geometrical progression ; the sum of the shares 
 of the first and second was to the sum of the shares of the 
 second and third as 1 to 2. Required the shares. 
 
 Ans. 2, 4, 8. 
 
 SECTION XXII.— Theory of Continued Fractions. 
 
 185. In order to form a more exact idea of a fraction, the 
 
 terms of which are large numbers and prime to each other, we 
 
 seek approximate values of this fraction, which are expressed in 
 
 more simple numbers. 
 
 159 
 Let there be, for example, the fraction j^. Dividing both 
 
 terms of this fraction by the numerator, an operation which will 
 
 not change its value, it becomes r^ 
 
 Ifi 
 
 If then we neglect, for the moment, the fraction r-^ in this ex- 
 
 lo9 
 
 pression, the result - will be greater than the proposed, since the 
 o 
 
 denominator has been diminished. 
 
 Ifi 
 On the other hand, if instead of neglecting the fraction r-^, 
 
 we substitute 1 for it, the result - will be less than the proposed, 
 since the denominator has been increased. 
 
 We conclude therefore, that the fraction ^qq is comprised be- 
 tween ^ and 2> we are thus enabled to form a very exact idea of 
 its value. 
 
CONTINUED FRACTIOT^S. 227 
 
 If a greater degree of approximation be required, we have 
 
 only to operate upon r-r^ m the same manner as we have al- 
 
 159 
 ready done upon ^^; we have thus 
 
 .16 1 
 
 159 ,. . 15 
 
 ^ + 16 
 
 and the proposed fraction becomes 
 1 
 
 3 + ^ 
 
 9-4-^. 
 ^+16 
 
 If we ne2:lect -^^^ ^ is greater than 7777 ; it follows therefore 
 lb y lo9 
 
 ^ 1 . , ^ 159 ^ 1 ^ 1 9 
 
 that 7 IS less than 77^; but becomes kf, or ^rp,; 
 
 ^ , 1 493' 1 28 28' 
 
 ^+9 ^ + 9 -9 
 
 1 9 
 
 thus the proposed is comprised between - and tr^. The differ- 
 
 ence between these two fractions is pr7 ; the error therefore com- 
 
 84 
 
 19 
 
 mitted in taking k or rr^ for the value of the proposed fraction is 
 
 less than — . 
 
 
 
 
 
 
 
 
 To attain to { 
 
 I still 
 
 [ greater degree 
 
 of J 
 
 ipproximation, 
 
 we 
 
 1 
 operate 
 
 in the same manner 
 
 15 
 
 upon -; 
 
 thus 
 
 we 
 
 have 
 
 
 
 * 
 
 
 15 
 16 ~ 
 
 1 
 1 + 
 
 T 
 
 15 
 
 
 
 
 and the proposed fraction becomes 
 
 1 
 
 « . 1 
 
 
228 ELEMENTS OF ALGEBRA. 
 
 11 15 
 
 Neglecting --p- the fraction - or 1 is greater than ~ ; hence 
 xOf 1 lb 
 
 7 or VFJ is less than —^ ; therefore ~ 
 
 , + ' "> »»» 3 + 1 
 
 or KT is greater than -r^; thus the proposed is comprised be» 
 
 9 10 
 
 tween ^r^ and -^, The difference between these two- fractions m 
 Jo ol 
 
 1 1. . 1 ...... ^ 9 10 
 
 5^3-; the error committed, therefore, m takmg either ^^ 01 -^ 
 
 for the value of the proposed is less than 3-3. 
 
 000 
 
 The expression is called a continiied frac' 
 
 =+; 
 
 '+A 
 
 tvon. We understand therefore, by a continued fraction a frac- 
 tion^ which has unity for its numerator, and for its denominator 
 an entire number plus a fraction, which fraction has also unity 
 for its nuTnerator and for its denoTmnutor an entire naimher plus 
 a fraction, and thus in order. 
 
 It sometimes happens, that the proposed fractional number is 
 greater than unity ; to generalize, therefore, the above definition,, 
 we say, that a continued fraction is an expression composed of an 
 entire number plus a fraction which has unity for its numerator t 
 and for its denominat&r an entire number plus a fraction, Sf&, 
 
 159 
 186. If we examine the above process for converting j^ into 
 
 a continu-ed fraction, it will be perceived, that we have divided 
 first 493 by 159, which gives three for a quotient and a remain- 
 der 16; we then divide 159 by 16, which gives 9 for a quotient 
 and a remainder 15 ; we next divide 16 by 15, which gives 1 for 
 a quotient and a remainder 1 ; from which we readily infer the 
 following rule for converting a fraction or fractional number into 
 a continued fraction, viz. 
 
CONTINUED FRACTIONS. 
 
 Apply to the two terms of the fraction the process of finding 
 fheir greatest comrmn divisor ; pursue the operation until a re- 
 mainder is obtained equal to ; the successive qicotients, thus ob- 
 tained, will he the denominators of the fractions, which constitute 
 the continued fraction. 
 
 If the proposed be greater than unity the first quotient will be 
 the entire part, which enters into the expression of the continued 
 
 fraction. 
 
 73 829 
 
 Examples. Let the fractions r-r-, jr^ be converted into con- 
 tinued fractions. 
 
 187. From what has been said a continued fraction may be 
 represented generally by the expression 
 
 c+^ 
 
 a, b, c, d , , , being entire and positive numbers. The fractional 
 number, to which this expression is equivalent, may moreover be 
 represented by x. 
 
 The fractions y, -,-..., the assemblage of which constitutes 
 oca 
 
 the continued fraction, are called integrant fractions. The de- 
 nominators b, c,'d . .. . are called incomplete quotients, since b, 
 for example, is only the entire part of the number expressed 
 
 hy b-\ r and c only the entire part of the number ex- 
 
 ^ + 1^7. 
 
 pressed by c -f- j-j- — and thus in order. Conversely the ex- 
 
 1 1 
 
 pressions b -] =■ c -}- -i — are called complete quotients. 
 
 + 1 a , . 
 
 n 
 
 The results obtained by converting successively into a single 
 fractional number each of the expressions 
 
 a -j- T) ^-\ T &c« are called reductions. 
 
 rp t: 
 
230 ELEMENTS OF ALGEBRA. 
 
 188. The formation of these reductions is subject to a very 
 simple law, which we shall now develop. 
 
 The first is a, which may be put under the form y, the second 
 
 is a 4- -T) or reducing the whole expression to a fraction, — y-—' 
 
 To form the third, represented by 
 
 . 1 
 
 '+i 
 
 c -|- -T for c in the third ; which gives 
 
 it will be sufficient to substitute h -\ — for h in the second ; 
 
 c 
 . * . 
 making this substitution, we have 
 
 ' c ' c 
 
 To form the fourth reduction, it will be sufficient to substitute 
 
 J third ; which gives 
 
 [{ab + \)c-\-a]d + ab+l 
 {bc + \)d + b 
 
 The first four reductions therefore will be 
 
 a ab-\-l {ab-\-\)c-\-a {{ab -\- \) c -[- a] d -{- ab -{- I 
 V b ' bc+\ ' {bc+l)d-\-b 
 
 Without proceeding further, it will be perceived, that the 
 numerator of the third reduction is formed by» multiplying the 
 numerator of the second by the third incomplete quotient c, 
 and adding to this product the numerator of the first reduction. 
 With respect to the denominator, it is formed in the same 
 manner by means of the denominators of the second and first 
 reductions. 
 
 In like manner, the numerator and denominator of the fourth 
 reduction is formed, it will be perceived, by multiplying re- 
 spectively the two terms of the third reduction by the fourth 
 
JO 
 
 form the reduction ^^, and let it be supposed, that we have 
 Jx 
 
 CONTINUED FEACTIONS. SSI 
 
 incomplete quotient d and adding to the two products respectively 
 the two terms of the second reduction. 
 
 From what has been done it will be readily inferred, that the 
 above law of formation for the third and fourth reductions should 
 be extended to those which follow. To demonstrate this law, 
 however, in a rigorous manner, we shall show that if it be true 
 in regard to any three successive reductions whatever, it will be 
 true for the reduction, which follows ; thus this law being already 
 found true for the first three reductions will be true for the 
 fourth, and being true for the second, third and fourth, it will 
 be true for the fifth, and thus in order; it will therefore be 
 general. 
 
 POT? 
 
 Let ^„ 7r-„ :^ be any three successive reductions whatever,* 
 
 let r be the incomplete quotient, at which we stop in order to 
 , and let it be 
 
 R _ Qr + P 
 R'~~ Q'r + P'' 
 
 Let - be the integrant fraction, which follows r, and let ~^, 
 
 S O 
 
 be the corresponding reduction. In order to form this reduc 
 tion, it is sufficient to substitute in the expression for -^, r -{-- 
 instead of r ; making this substitution, we have 
 
 S _ ^y+'^) + ^ _ {'Qr+ P)s+Q _ Rs+Q 
 
 S 
 We see, therefore, that — is formed from the two preceding 
 o 
 
 reductions according to the law enunciated above. This law is 
 therefore general ; whence, To form the numerator of any re- 
 duction whatever, we multiply the numerator of the preceding re- 
 duction hy the incomplete quotient, which corresponds to it, and 
 add to the product the numerator of the reduction, which precedes 
 hy tivo ranks the one which we wish to form ; the denominator is 
 
ELEMENTS OF ALGEBRA. 
 
 formed by the same law by means of th£ two 'preceding denomina- 
 tors. 
 
 189. When the number reduced to a continued fraction is 
 
 less than unity, we substitute y instead of a, in order to apply the 
 
 law, which supposes necessarily, that we have already the first 
 two reductions. 
 
 Let it be proposed to find the successive reductions of the 
 continued fraction 
 
 65_0 1 
 
 1 ~r 
 
 149 1 ' . 1 
 
 2 + i 
 
 1+i 
 
 The first two reductions being - , -, we have for those which 
 
 J. /i 
 
 follow 
 
 3 7 17 24 65 
 
 7' 16' 39' 55' 149* 
 
 In like manner we have for the several reductions of the 
 
 continued fraction arisinf? from ^r— -, 
 ° 347 
 
 2 5 7 12 43 829 
 
 I' 2' 3' T' 18' 347* 
 
 29 
 So also the fraction — being converted into a continued frac- 
 tion gives the following reductions, viz. ^ 
 1 1 2 3 29 
 1' 2' 3' 5' 8' 77* 
 190. The successive reductions, it will be perceived, are alter- 
 nately less and greater than the whole continued fraction, and 
 they approximate this fraction nearer and nearer. 
 
 The first reduction is always less thah the whole continued 
 fraction x. The reductions of an even rank are, therefore, greater 
 than the vjhole continued fraction, and those of an odd rank are 
 less. And since these reductions approach nearer and nearer 
 
CONTINUED FRACTIONS. 8iB 
 
 the value of a;, the reductions of an odd rank must go on increas- 
 ing, while those of an even rank decrease. Thus the reductions 
 form two series, the terms of which approach nearer and nearer 
 the value of the whole continued fraction. 
 
 191. The difference between any two consecutive reductions 
 whatever has unity for its numerator. The numerator of the 
 first difference is -j- 1, that of the second — 1, that of the third 
 -f- 1, and thus in order. In general, the numerator of any differ- 
 en£e whatever will he +1, if the second of the reductions under 
 consideration is of an even rank, but — 1 if it be of an odd rank. 
 
 From this property, it follows, that the two terms of any 
 
 72 
 
 reduction whatever ^7 are prime to each other. 
 
 Indeed let it be supposed, that R and R' have a common 
 factor h; by the preceding property, we have 
 
 whence dividing both terms by h, we have 
 
 RQ' QR'__l, 
 
 h h h 
 
 but the first member of this equation is an entire number since 
 by hypothesis R and K are divisible by h, while the second is 
 essentially a fraction; jR and R' cannot therefore have a com- 
 mon factor. 
 
 From this it follows, that if a fraction, the terms of which are 
 not prime to each other, be converted into a continued fraction, 
 and all the reductions be formed to the last inclusive, the last 
 reduction will not be the proposed fraction, but this fraction 
 reduced to its lowest terms. 
 
 348 
 Let there be, for example, the fraction ^^; converting this 
 
 into a continued fraction, we have for the successive reduc- 
 
 1 1 2 3 29 29 
 
 tions -, ^, 35, ^, ^, ==. The last reduction -- is the proposed 
 
 reduced to its lowest terms. 
 
 192. Since the value of the whole continued fraction x is 
 
ELEMENTS OF ALGEBRA. 
 
 always comprised between any two consecutive reductions 
 •p:rj, :g7, it follows that the error committed in taking —,. for z is 
 
 O TJ. 
 
 less than -z^ — ^^; but from what has been said we have 
 
 Q R 1 
 
 Q' R' Q'R' 
 and since Q'<R' gives Q"^< Q'R', we have 
 
 Q'R' ^Q' 
 
 The error therefore committed in taking any reduction what- 
 ever for the value of the whole continued fraction is less than 
 unity divided by the denominator of this reduction multiplied by 
 the dcTwrninator of the reduction ivhich folloios, or less exactly 
 but in terms more simple, less than unity divided by the square 
 of the denominator of the reduction, which is taken for the whole 
 continued fraction. 
 
 193. The ratio of the circumference to the diameter of a 
 
 circle being expressed by the fraction fH|. the terms of 
 
 which are prime to each other, let it be proposed to find a 
 
 fraction, the terms of which will be more simple, and which 
 
 will express the same ratio nearly. Converting the proposed 
 
 into a continued fraction, we have for the successive reductions 
 
 3 22 333 855 9208 9563 76149 314159 
 
 r y 106' 113' 2931' 3044' 24239' 100000* 
 
 The error committed in taking the second of these reductions 
 
 1 22 
 
 for the proposed fraction will not exceed -^; ■— is therefore fre- 
 quently employed to express the ratio of the circumference of a 
 circle to its diameter. This is the ratio given by Archimedes. 
 
 If a greater degree of approximation is required, we take 
 the fourth reduction, which it is easy to see, is but little more 
 complicated than the third. The error committed in taking this 
 
 1 855 
 
 reduction for the proposed will not exceed iio ^ t)Qoi 5 TTo will 
 
EXPONENTIAL EQUATIONS. 
 
 therefore approximate the proposed very nearly. This is the 
 ratio given by Adrian Metiiis. 
 
 We thus see the use, which may be made of continued frac- 
 tions in estimating approximatively the value of fractions, the 
 terms of which are large numbers and prime to each other. 
 
 EXAMPLES. 
 
 1. A's property is to that of B as 5743 to 80937. By what 
 smaller numbers ma^ the ratio of their property be expressed ? 
 
 2. The lunar month or the time in which the moon completes 
 its revolution, is found by calculation to be 27.321661 days. 
 Thus in 27321661 days it would perform 1000000 revolutions. 
 How may this relation be expressed in smaller numbers ? 
 
 SECTION XXIII. — Exponential Equations and Logarithms. 
 
 194. An equation of the form a' = 3, in which the exponent 
 z is the unknown quantity is called an exponential equation. 
 The solution of this equation consists in finding the power, to 
 which it is necessary to raise a given quantity a in order to pro- 
 duce another given quantity h. 
 
 Let there be, for example, the equation 2"" == 64 ; raising 2 to 
 its different powers, we soon find that 2^ = 64 ; a: = 6 answers 
 therefore the conditions of the equation. 
 
 Again, let there be the equation 3'' = 243; raising 3 to its 
 different powers we find 3^ = 243 ; whence a; = 5. In a word, 
 so long as the second member i is a perfect power of the given 
 number a, x will be an entire number and its value may be found 
 by raising a successively to its different powers, beginning with 
 that, the exponent of which is 0. 
 
 Let it be proposed next to resolve the equation 2' = 6. 
 Putting successively a; = 2, a; = 3, we have 2^ = 4, 2^ = 8; 
 the value of x is, therefore, comprised between the numbers 2 
 and 3. 
 
33d elements of algebra. 
 
 Let us put therefore, x = 2-\-—, x' being greater than 1 ; 
 
 substituting this value in the proposed, we have 
 
 2 + ^ 1 1- Q 
 
 2 " = 6 or 2=^ X 2"' = 6, whence 2'' = 1 
 or raising both members to the power x\ we have 
 
 To determine the value of x' we make successively x' •=.\^ 
 
 (3\ 3 /3\* 9 
 
 - J or ^ less than 2, but I- j or j 
 
 greater than 2 ; a:' is, therefore, comprised between 1 and 2. 
 
 Let us put then a;'= 1 + -77, x" being greater than 1. Sub- 
 stituting this value, we have 
 
 (|)-i=,„|x(|)»=., 
 
 /4\'"_3 
 \?) "^2 
 
 To determine the value of x" ^ we make successively x" •=. 1, 
 
 ^ 16 
 
 whence 
 
 (4\ 4 3 /4\'' 
 
 jr I or - less than -, but 1 - 1 
 
 3 
 
 greater than - ; x" is, therefore, comprised between 1 and 2. 
 
 Let us put then a;"= 1 -| — —^ x'" being greater than unity; 
 
 X 
 
 we have by substitution 
 
 Uj =2 °' 3 X W =2' ^^^^^^ Vsj =3- 
 
 (Q\ 8 ft! 
 ft) ^^fil'^^^"^' 
 
 4 /9\^ 729 4 
 
 ber less than - but ( - I = — ^, a number greater than -^ \ thus 
 
 x'" is comprised between 2 and 3. 
 
 Let x'" = 2 -| — 7777, the equation in x'" becomes 
 
 /9\=+b^ 4 , /256\*"" 9 
 
 is; =3' ^^^^^H243J =8- 
 
KXFONENTIAL EQUATIONS. SP^ 
 
 Operating upon this last equation as upon the preceding, we 
 find two entire numbers k and ^+1, between which x"" will 
 
 be comprised. Putting x"" z=Jc-\-—, we determine a;*, in the 
 
 same manner as we have already done x"'\ and thus in order. 
 Bringing together the equations 
 
 we obtain the value of x under the form of a continued fraction, 
 thus 
 
 1+ 
 
 But we have seen that in a continued fraction the greater the 
 number of integrant fractions, which are taken, the nearer we 
 approach the value of the number reduced to a continued frac- 
 tion ; we shall, therefore, be able to determine the value of x in 
 the equation 2' :^ 6, if not exactly, at least with such degree of 
 approximation as we please. 
 
 Forming the first four reductions, for example, we have 
 
 2 3 5 13 , ^ , . 13 ,.^ ^ 
 
 T' T' 9' "T ' ^ reduction — diners from a: by a quantity 
 
 less than — . 
 
 To attain a greater degree of approximation, we determine 
 
 (256\*"" 9 
 oTq) =qJ we thus find 
 
 a;"" =— 2 4" -;• We shall have, then, for the fifth reduction 
 
 X 
 
 31 1 
 
 •r^. This differs from a; by a quantity less than r^. 
 
 195. From the preceding examples the course to be pursued 
 in the solution of equations of the form a* = 3 will be readily 
 
m" 
 
 ELEMENTS OF ALGEBRA. 
 
 inferred. In the application of this method to particular cases it 
 is necessary to remark, 1°. If the quantity b be less than a, the 
 value of X will be comprised between and 1 ; we put, therefore, 
 
 x = —. ^°. If & is a fraction and a greater than unity, the 
 value of X will be negative, we put, therefore, x = — y ; the 
 
 equation is then reduced to the form or'z^-; having found the 
 
 value of y in this equation according to the method explained 
 above, the value of x will be equal to that of y taken negatively. 
 
 i 
 
 EXAMPLES. 
 
 I*'. Given 3^= 15 to find the value of x. Ans. x = 2.46. 
 
 2?. . 10'==3 . . . . Ans. a; = 0.47. 
 
 2 
 3*>. . 5*=- to find the value of a;. Ans. a: = — 0.25. 
 
 ? . . . . Ans. a; = 0.53. 
 
 4 
 
 In the above examples the reductions furnished by the method 
 are converted into decimal fractions, and the value of x is deter- 
 mined to within .01. 
 
 THEORY OF LOGARITHMS. 
 
 196. If in the equation a" = y, we assign a constant value 
 different from unity to a, and suppose that of z to vary, as may 
 be required, we may obtain successively for y all possible num- 
 bers. 
 
 Let us suppose first a greater than 1. 
 
 If we make successively a; = 0, 1, 2, 3, 4, . . . . 
 we have y=lf a, a^, a^, a*, . . . . 
 
 Thus by means of the powers of a, the exponents of which are 
 positive, entire or fractional, we may produce all possible posi^ 
 tive numbers greater than 1. 
 
 Again, let a; = 0, — 1, — 2, — 3, — 4, 
 
 1. ,1111 
 
 wehav« j^«l, -,-,_, ^,.... 
 
THEORY OF LOGARITHMS. 
 
 m 
 
 Thus by means of the powers of a, the exponents of which are 
 negative entire or fractional, we may produce all possible positive 
 numbers less than 1. 
 
 If on the other hand we suppose a less than unity, still all 
 possible positive numbers may be produced by means of the 
 different powers of a, only the order in which they are produced 
 will be reversed. 
 
 We see therefore, that all possible positive iiumbers may be 
 produced by means of any positive number whatever a, different 
 from unity, by raising this number to the requisite powers. 
 
 It is necessary, that a should be different from unity, other- 
 wise the same number will be produced, whatever value we 
 assign to x. 
 
 197. Let it now be supposed that wfe have made a table con- 
 taining in one column all entire numbers, and by the side of 
 these in another column the exponents of the powers, to which it 
 is necessary to raise a constant number in order to produce these 
 numbers; this would be a table of logarithms. 
 
 The logarithm of a number, is, therefore, the exponent of the 
 potoer, to lohich it is necessary to raise a given or invariable 
 number, in order to produce the proposed number. 
 
 Thus in the equation a'-==y, x is the logarithm of y ; in like 
 manner in the equation 2^ = 64, 6 is the logarithm of 64. The 
 logarithm of a number is indicated by writing before it the first 
 three letters of the word logarithm, or more simply by placing 
 before it the letter L. 
 
 The invariable number, from which the others are formed JB 
 called the base of the table. It may be taken at pleasure either 
 greater or less than unity, but should remain the same for the 
 formation of all numbers, that belong to the same table. 
 
 Since a°= 1, and d^ = a, whatever number may be assumed 
 for the base of the table, the logarithm of the base will be unity 
 and the logarithm of unity will be 0. 
 
 198. We proceed to show the properties of logarithms in rela- 
 tion to numerical calculations. 
 
240 ELEMENTS OF ALGEBRA. 
 
 1. Let there be the series of numbers y, y\ y'\ .... to be 
 multiplied together. Let a represent the base of a system of 
 logarithms, which we suppose already calculated, and let z, a;', 
 x" . . be the logarithms of 2/, y\ y", . . . ; by the definition of a 
 logarithm we have 
 
 y = a%y'=a'\y" = a"'; 
 multiplying these equations member by member, we have 
 
 yy'y"=a'+''+'", 
 whence log yy'y" =^X'\- x' -\- x" = log y + log y' •\- log y'\ 
 
 That is, the logarithm of a 'prodwct is equal to the sum of the 
 logarithms of the factors of this product. 
 
 If then a multiplication be proposed, we take from a table of 
 logarithms the logarithms of the numbers to be multiplied ; the 
 sum of these logarithms will be the logarithm of the product 
 sought. Seeking therefore this logarithm in the table, the num- 
 ber corresponding to it will be the product sought. Thus by 
 means of a table of logarithms addition may he made to take the 
 place of multiplication, 
 
 2. Let it be required to divide the number y by the number 
 y' ; let a;, x' be the logarithms of these numbers, we have the 
 equations 
 
 dividing these equations member by member, we have 
 
 whence log ~, = x — x' =■ log y — log y'. 
 
 That is, the logarithm of a quotient is equal to the difference 
 between the logarithm of the divisor and that of the dividend. 
 
 If then it be proposed to divide one number by another, from 
 the logarithm of the dividend we subtract the logarithm of the 
 divisor, the result will be the logarithm of the quotient ; seeking 
 therefore this logarithm in the tables the number corresponding 
 will be the quotient sought. Thus, by means of a table of loga- 
 rithms, subtraction may be made to take the place of division. 
 
THEORY OF LOGARITHMS. 241 
 
 3. Let it next be required to raise the number y to the power 
 denoted by tw, we have the equation a* = y ; 
 
 raising both members to the mth power, we have 
 
 whence the logarithm oi y"^ = mx =■ m\og y. 
 
 That is, the logarithm of any power of a number is equal to the 
 product of the logarithm of this number by the exponent of the 
 power. 
 
 To form any power whatever of a number by means of a table 
 of logarithms, we multiply, therefore, the logarithm of the pro- 
 posed number by the exponent of the power, to which it is to be 
 raised; the number in thfe table corresponding to this product, 
 will be the power sought. 
 
 4. Again, let it be required to find the n\k root of y. We 
 have as before a* = y ; 
 
 whence taking the wth root of both members, we have 
 
 a" = y" ; whence log 2/" = - = — 2_^ 
 
 That is, the logarithm of the root of any degree whatever of a 
 number is equal to the logarithm of this number divided by the 
 index of the root. 
 
 Thus by the aid of a table of logarithms a number Tnay he 
 raised to a power by a simple multiplication, and its root Tnay he 
 extracted by a simple division. 
 
 FORMATION OF TABLES. 
 
 199. The properties of logarithms demonstrated above are 
 altogether independent of the number a or their base. We 
 may therefore form an infinite variety of tables of logarithms 
 by putting for a all possible numbers except unity. 
 
 If it be required to construct a table of logarithms the base of 
 which is 2, in the equation 2' = y, we make y equal successively 
 to the numbers 1, 2, 3 ... ., and determine by the methods 
 explained, art. 195, the values of x corresponding. 
 
 We thus obtain the values of x exactly, if y be a perfect power 
 of 2, or otherwise with such degree of approximation as we please 
 
242 
 
 ELEMENTS OF ALGEBRA. 
 
 To calculate the logarithm of 3, for example, we have the 
 
 equation 2* = 3, from which we deduce 
 1 
 
 x=l 
 
 l-f 
 
 1 
 
 1 
 
 Whence stopping at the fourth integrant fraction, and forming 
 
 the reduction corresponding, we have x= j^, or reducing this 
 
 last to a decimal we have z= 1.583 accurate to the third deci- 
 mal figure, 
 
 200. In the calculation of a table of logarithms, it will he 
 sufficient to calculate directly the logarithms of the prime num- 
 bers 1, 2, 3, 5 . . . , the logarithms of compound numbers may then 
 be obtained by adding the logarithms of the prime factors, which 
 enter into them. To find the logarithm of 35, for example, we 
 have 35 = 5 X 7 ; whence log 35 = log 5 -f- log 7 ; having al- 
 ready calculated the logarithms of 5 and 7, the logarithm of 35 
 will be found therefore by adding the logarithm of 5 to that of 7. 
 
 Since moreover the logarithm of a fraction will be equal to the 
 logarithm of the numerator minus the logarithm of the denomi- 
 nator, it will be sufficient to place in the tables the logarithms of 
 entire numbers. 
 
 201. Below we have a table of logarithms of numbers from 
 1 to 30 inclusive, the base of the system is 2, and the logarithms 
 are calculated to 4 places of decimals. 
 
 N. 
 
 Log. 
 
 N. 
 
 Log. |N. 
 
 Log. 
 
 1 
 
 0.0000 
 
 11 
 
 3.4594 
 
 21 
 
 4.3922 
 
 2 
 
 1.0000 
 
 12 
 
 3.5849 
 
 22 
 
 4.4594 
 
 3 
 
 1.5849 
 
 13 
 
 3.7000 
 
 23 
 
 4.5235 
 
 4 
 
 2.0000 
 
 14 
 
 3.8073 
 
 24 
 
 4.5849 
 
 5 
 
 2.3219 
 
 15 
 
 3.9065 
 
 25 
 
 4.6438 
 
 6 
 
 2.5849 
 
 16 
 
 4.0000 
 
 26 
 
 4.7000 
 
 7 
 
 2.8*073 
 
 17 
 
 4.0874 
 
 27 
 
 4.7548 
 
 8 
 
 3.0000 
 
 18 
 
 4.1699 
 
 28 
 
 4.8073 
 
 9 
 
 3.1699 
 
 19 
 
 4.2479 
 
 29 
 
 4.8577 
 
 10 
 
 3.3219 
 
 20 
 
 4.3219 
 
 30 
 
 4.9065 
 
THEORY OF LOGARITHMS. 243 
 
 202. The most convenient number for a base to a system of 
 logarithms, and the one employed in the construction of the 
 tables in common use is 10. 
 
 If in the equation 10* = y we make successively 
 
 
 a:=:0, 
 
 1, 2, 3, 4 . 
 
 we have 
 
 y=l, 
 
 10, 100, 1000, 10000 . 
 
 Again if 
 
 we make 
 
 
 
 x = 0, 
 
 -1, -2, -3, -4 . . 
 
 we have 
 
 y=i, 
 
 1111 
 10' 100' 1000' 10000' 
 
 Therefore in a table of logarithms, the base of which is 10, 
 1**. the logarithms of numbers greater than unity are positive 
 and go on increasing from to infinity. 2°. The logarithms of 
 numbers less than unity are negative, and their absolute values 
 are so much the greater as the fractions are smaller ; whence if 
 we take a fraction less than any assignable quantity, the loga- 
 rithm of this fraction will be negative, and its absolute value will 
 be greater than any assignable quantity. On this account we 
 say that the logarithm of is an infinite negative quantity. 
 3°. The logarithms of all numbers below 10 are fractions; the 
 logarithms of numbers between 10 and 100 are 1 and a fraction ; 
 the logarithms of numbers between 100 and 1000 are 2 and a 
 fraction ; those of numbers between 1000 and 10000 are 3 and a 
 fraction ; and in general, the whole number which precedes the 
 fraction in the logarithm is less by one than the number of 
 figures in the number corresponding to the logarithm. On this 
 account it is called the index or characteristic of the logarithm, 
 since it serves to indicate the order of units, to which the number 
 corresponding to the logarithm belongs; Thus in the logarithm 
 3.75527 the characteristic 3 shows that the number correspond- 
 ing to this logarithm consists of 4 figures or is comprised be- 
 tween 1000 and 10000. 
 
 203. The logarithm of a number being given, the logarithm 
 of a number 10, 100, . . . times greater is found by adding 
 1, 2, . . . units to the characteristic only ; indeed log 
 {y X 10") = log 2/ + log 10" = log 2/ + 7i; 
 
24^ ELEMENTS OF ALGEBRA. 
 
 whence it will be sufficient to add n units to the logarithm of y in 
 order to obtain the logarithm of a number 10" times as great ; an 
 addition which may be performed upon tbe characteristic only. 
 
 Conversely, log :^ = logy — log 10" = log 2/ — n; thus it is 
 
 sufficient to subtract n units from the logarithm of y^ in order to 
 find the logarithm of a number 10" times smaller than y. 
 
 204. The fractional parts of logarithms in the tables are ex- 
 pressed by decimals. From what has been said the decimal part 
 of the logarithm of a number will be the same for this number 
 multiplied or divided by 10, 100, ... On this account the sys- 
 tem of logarithms, the base of which is 10, is more convenient 
 than any other system, since we have frequent occasion to multi- 
 ply or divide by 10, 100, . . . operations reduced in this case to 
 tfie simple addition or subtraction of units. 
 
 205. Since the characteristic of the logarithm may be easily 
 determined by the number, and the number of figures in the 
 number by the characteristic of the logarithm, it is usual to 
 omit the characteristic in the tables to save the room. It is 
 also convenient to omit it ; because the same decimal part with 
 different characteristics forms the logarithms of several different 
 numbers. 
 
 206. Having already calculated a system of logarithms, it 
 will be easy from this to form as many other systems as we 
 please. 
 
 Indeed, let N designate any number whatever, log N its loga* 
 rithm in the system the base of which is a, X its logarithm in a 
 different system the base of which is 3, we have 
 
 i^ = N. 
 Taking the logarithms of both members of this equation in the 
 system, the base of which is c, we have 
 X . log * = log N J 
 
 whence X = ,^ , . 
 
TflEORY OF LOGARITHMS. 245 
 
 Having calculated therefore a set of tables for a particular 
 base, to find the logarithm of a number in a proposed system 
 with a different base, we take from the tables already calculated 
 the logarithm of the number , and also the logarithm of the base 
 of the proposed system; the former of these logarithms, divided by 
 the latter, will give the logarithm of the number in. the proposed 
 system. 
 
 The logarithm of 6, for example, in the system, the base of 
 which is 10, is .77815, and that of 3 is .47712 ; the logarithm 
 of 6, therefore, in the system the base of which is 3, will be 
 
 ^=1.63093. 
 47712 
 
 loffN 
 
 207, The expression X = may be put under the form 
 
 X = I — r log N. Thus having already formed a table of loga- 
 logd 
 
 rithms, the base of which is a, to construct from this a new table, 
 
 the base of which shall be ^, we multiply the logarithms of the 
 
 first table by the quantity -, — r. This quantity by means of 
 
 which we are enabled to pass from the old to the new table, is 
 called the modulus of the new table in relation to the old. 
 
 MODE OF USING THE TABLES. 
 
 208. As it is impossible to place in the tables the logarithms 
 of all numbers, it is usual to place in them the logarithms of 
 numbers from unity to within a certain limit In what follows it 
 is supposed, that the student has in his hands tables containing 
 the logarithms of entire numbers from 1 to 10000. 
 
 In order to use such a set of tables, we have the two following 
 questions to resolve, viz. 1°. Any number whatever being given, 
 to, find its logarithm. 2**. Any logarithm being given, to find 
 the number tohich corresponds to it. 
 
 The following examples will exhibit the method of resolving 
 these questions. 
 
 1. Let it now be proposed to find the logarithm of 9748 
 
246 ELEMENTS OF ALGEBRA. 
 
 Seeking the proposed in the column of numbers, against it in 
 the column of logarithms we find 98892 ; this will be the deci- 
 mal part of the logarithm ; or, as is the case with most tables, 
 if the column of numbers contain but three places of figures, 
 we look for 974, the first three figures of the proposed, in the 
 first column, and at the top of the table we look for the fourth 
 figure 8 ; directly under the 8 and in the same line with 974, 
 we find the decimal part 98892 as before ; then since the pro- 
 posed consists of four places, the characteristic will be 3, thus 
 log 9748 = 3.98892. 
 
 2. Let it be required to find the logarithm of 76.93. Re- 
 moving for the moment the decimal point, we find as above 
 log 7693 = 3.88610, whence, art. 203, subtracting 2 units from 
 the characteristic 3 of this logarithm, we shall have the logarithm 
 of the proposed ; thus log 76.93 = 1.88610. 
 
 3. To find the logarithm of .75. The logarithm of this num- 
 ber may be presented under two different forms. Writing it 
 
 75 
 
 in the form of a vulgar fraction, it becomes — rr-. The loga- 
 rithm of 75 is 1.87506, and that of 100 is 2.00000; whence 
 subtracting the logarithm of the denominator from that of the 
 numerator, art. 198, we have — 12494 = log .75. This loga- 
 rithm, being altogether negative, is inconvenient in practice; 
 
 it will be observed, however, that .75 = -rTrTrX75; whence 
 
 log .75 = log ^ + log 75 = - 2 + 1.87506, 
 
 = _ 1-^87506, 
 or placing the sign — over the 1 to show that the characteristic 
 only is negative, we have log .75 == 1.87506. 
 
 This last form of the logarithm of the proposed is derived, it 
 will be perceived, immediately from the continuation of the prin- 
 ciple, art. 203, according to which the logarithm of a number 
 10, 100 . . . times less than a proposed number is found by sub- 
 tracting 1, 2 . . . units from the characteristic of its logarithm. 
 
THEORY OF LOGARITHMS. 247 
 
 Thus since the logarithm of 750 = 2.87506, we have 
 
 log 75 = 1.87506 
 
 log 7.5 = 0.87506 
 
 log .75 =1.87506 
 
 log .075 = 2.87506 
 
 log .0075 = 3.87506 
 4 
 
 4. To find the logarithm of - ; we have log 4 = .60206, 
 
 og 5 = .69897; whence subtracting this last logarithm from 
 
 4 
 the former, we have log - = — .09691, in which the logarithm 
 o 
 4 
 is entirely negative. But - reduced to a decimal becomes .8, 
 
 the logarithm of which is 1.90309, the characteristic only being 
 negative. 
 
 493 
 
 5. To find the logarithm of 54|- ; we have 5^1 = -3- ; log 
 
 493 = 2.69285, log 9 = 0.95424 ; whence subtracting the latter 
 
 493 
 logarithm from the former, we have log —^ or 54^ = 1.73861. 
 
 6. To find the logarithm of 675437. This number exceeds 
 the limits of the table ; its logarithm, however, may be readily 
 found. The greatest number of places in a number, the loga- 
 rithm of which can be found in the tables, is 4 ; separating there- 
 fore the four left hand figures of the proposed from the rest by a 
 point, we consider for the moment those on the right as deci- 
 mals. The logarithm of 6754.37 is comprised between the loga- 
 rithm of 6754 and that of 6755; the diflference between these 
 
 37 
 
 two logarithms is .00007 ; -jrr of this difference therefore added 
 
 to the less logarithm will give the logarithm of 6754.37 nearly ; 
 thus log 6754.37 = 3.82959 ; whence adding 2 units to the 
 characteristic of this last to obtain the logarithm of the proposed, 
 we have log 675437 = 5.82959. 
 
 209. We proceed next to the second of the proposed questions, 
 viz. A logarithm being given, to find the number which corres- 
 ponds to it. 
 
 1. To find the number corresponding to the logarithm 2.10449. 
 The decimal part of this logarithm is contained in the tables ; in 
 the left hand column and on the same line with it according to 
 the arrangement of the tables, in which there are but three places 
 
His ELEMENTS OF ALGEBRA. 
 
 of figures in the column of numbers, we find 127, and at the top 
 of the table directly over it we find 2 ; the characteristic of the 
 logarithm being 2, we have therefore 127.2 for the number cor- 
 responding to the proposed. 
 
 2. To find the number corresponding to the logarithm 3.42674. 
 This logarithm is not found in the tables ; it is comprised however 
 between 3.42667 the logarithm of 2671, and 3.42684 that oi 
 2672 ; the diflference between these two logarithms is .00017, the 
 difference between the proposed and 3.42667 is .00007 ; we have 
 then the following proportion : 
 
 .00017 : 1 : : .00007 :.41 nearly. 
 The number corresponding to 3.42674 is, therefore, 2671.41. 
 
 3. To find the number corresponding to the logarithm 
 — 2.45379. The number corresponding to this logarithm will 
 be comprised, it is evident, between .01 and .001 ; to obtain this 
 number let us add to — 2.45379 a sufficient number of units to 
 make it positive, 5 for example, we have 5 — 2.45379 = 2.54621 ; 
 the number corresponding to this last is 351.73 ; but by adding 5 
 units to the proposed logarithm, we have multiplied the number, to 
 which it belongs, by 100000, whence, dividing 351.73 by 100000, 
 we have .0035173, the number corresponding to the proposed. 
 
 4. To find the number corresponding to the logarithm 3.86249. 
 Adding three units to the characteristic, the proposed becomes 
 0.86249, the number corresponding to which is 7.286 ; whence, as 
 it is easy to see, the number corresponding to 3.86249 is .007286 
 
 SECTION XXIV.— Application of the Theory of 
 Logarithms. 
 
 multiplication and division. 
 
 1. Let it be required to multiply 872 by .097. 
 
 log 872 = 2.94052 
 
 lo<T .097 = 2.98677 
 
 log 84.584 Ans. 1.92729 
 
APPLICATION OF LOGARITHMS. 249 
 
 2. Let it be required to multiply .857 by .0093. 
 
 log .857 = 1.93298 
 log .0093 = 3.96848 
 
 log .00797 Am. 3.90146 
 
 3. Let it be required to divide 5672 by .0037. 
 
 log 5672 = 3.75374 
 log .0037 = 3.56820 
 
 log 1533000 Ans. 6.18554 
 
 4. Let it be required to divide .053 by 797. 
 
 log .053 = 2.72428 = 3 + 1.72428 
 log 797= 2.90146 
 
 log .0000665 Ans. 5.82282 
 
 To render the subtraction required in this example possible, we 
 change the characteristic 2 into 3 + 1, which has the same value; 
 this furnishes a ten to be joined with 7 for the subtraction of 9, 
 the left hand figure of the decimal part. A similar preparation, 
 it is evident, must be made in all cases of the same kind. 
 
 FORMATION OF POWERS AND EXTRACTION OF ROOTS. 
 
 210. Let it be required to find the 5th power of .125. 
 log .125 = 1.09691 
 5 
 
 log .000030519 Ans. nearly 5.48455 
 2. To find the 7th power of .73. 
 
 log* .73 =1.86332 
 7 
 
 log .11047 Ans. nearly 7 + 6.04324=1.04324. 
 3. To find the third root of .01356. 
 The logarithm of .01356 is 2.13226. The negative charac- 
 teristic 2 of this logarithm is not divisible by 3, the index of the 
 root required, neither can it be joined to the positive part on 
 account of the different sign. If however we add — 1 + 1 to 
 
250 ELEMENTS OF ALGEBRA. 
 
 the characteristic, which will not alter its value, it becomes 
 3 -|- 1 ; the negative part is then divisible by 3, and the 1 being 
 positive may be joined to the fractional part, we have then 
 
 log .01356 = 2.13226 = 3 + 1.13226; 
 whence dividing by 3, we have 
 
 1.37742 = log .23846 Ans. nearly. 
 In all cases, if the negative characteristic is not divisible by 
 the index of the root required, it must be made so in a similar 
 manner. 
 
 ARITHMETICAL COMPLEMENT. 
 
 211. The arithmetical complement of a logarithm is the dif- 
 ference between this logarithm and 10; thus the arithmetical 
 complement of 3.472584 is 10 — 3.472584 = 6.527416. The 
 arithmetical complement of a logarithm is obtained by subtract- 
 ing the right hand figure^ if it be significant, from 10, and the 
 others from 9. 
 
 Let it be proposed to find the value of z in the expression 
 x = l — l'J^l" — V"—V"' 
 Z, V, I" . . being logarithms ; this expression, it is evident, may 
 be put under the form 
 
 :c = Z + (10 — Z') + r + (10 — V") + (10 — I"") — 30 ; 
 that is, to find the value of x, we take the sum of the logarithms 
 to be added and the complements of the logarithms to be sub' 
 tracted, from this sum subtract as many times 10, as there are 
 complements employed. 
 
 '-^ Thus when there are several multiplications and divisions to 
 be performed together, by using the complements of the loga- 
 rithms of the divisors the whole may be reduced to the addition 
 of logarithms. 
 
 EXAMPLES. 
 
 1. To find the value of x in the expression 
 
 ■ 75 X 73 X .056 \t 
 .7498 X 125.13 / 
 
 _/37 
 
APPLICATION OF LOGARITHMS. 251 
 
 log 3.75 0.57403 
 
 log 73 1.86332 
 
 log .056 2.74819 
 
 log 1.7498 Comp. 9.75701 
 
 log 125.13 Comp. 7.90264 
 
 18.84519 
 Subtracting next 20 from the characteristic, and taking f of 
 the remainder, we have 2.07532 = log .011803 Ans. . 
 2. To find the value of x in the expression 
 
 ^_/132x(7.356)Y Ans. 144.5972. 
 ^ (3.25)2- / 
 
 PROPORTIONS. 
 
 212. Let it be required to find the fourth term of the proportion, 
 of which the numbers 963, 1279, 8.7, are the first three terms. 
 
 log 1279 3.10687 
 
 log 8.7 0.93952 
 
 log 963 Comp. 7.01637 
 
 log 11.555 Am. nearly 1.06276 
 
 From the proportion a:b:: c:d, we have - = - ; 
 
 whence log a — log ^ = log c — log dj 
 
 therefore log a . log ^ : log c . log d, 
 
 that is, if four numbers form a proportion^ their logarithTns vnU 
 
 form an equidifference. 
 
 EXPONENTIAL EQUATIONS. 
 
 213. We have already explained a method for finding the 
 value of X in the equation a' = b, from which the theory of 
 logarithms is derived ; but a table of logarithms being once con- 
 structed, there is nothing to prevent its use in the solution of 
 equations of this kind. 
 
 Let it be required to find the value of x in the equation 
 3'= 15. 
 
 Taking the logarithms of both sides, we have 
 a; log 3 = log 15; 
 
252 Elements of algebra. 
 
 whence ^._log 15 _ 1-17609 _ 
 
 Whence a— j^^^_ 47712-" ^^ 
 
 The division required in this example may be performed, it is 
 easy to see, by subtracting the logarithm of .47712 from that of 
 1.17609, as in the case of any other numbers. 
 
 PROGRESSION BY QUOTIENT. 
 
 214. Logarithms are particularly useful in the solution of 
 questions in progression by quotient. 
 
 Let it be proposed to find the 20th term in the progression 
 3 9 27 
 • 2 '• 4 ' 8 * • 
 Putting u for the last term of a progression by quotient, we 
 have, art. 176, ** 
 
 u = aq"~^; whence, log u = log a-\- {n — 1) log q. 
 We have, therefore, for the 20th term in the progression pro- 
 posed 
 
 log t^ = log 1 + 19 (log 3 — log 2) = 19 (log 3 — log 2) 
 the term required will therefore be 2216.84 to within .01. 
 
 Let it be required next to insert between the numbers 2 and 
 15 fifty mean proportionals ; We have for the ratio, art. 184, 
 
 - , losri — logfl 
 whence locf q = -^ , , 
 
 
 in the question proposed, we have, therefore, 
 , W 15 — loff 2 
 
 '°gg= 51 ■ 
 
 or, performing the calculations, we obtain 
 ^=1.040286. 
 215. Let it be required to find the sum of the first ten terms in 
 the progression -rf 5 . 15 . 45 . . . ; we have, art. 178, 
 
 8 = —^^ — :; — ; whence 
 ^—1 
 
 log S = log a + log (g" — 1) — log {q — 1). 
 Applying this formula to the proposed question, we have 
 
 log S = log 5 + log (3^° — 1) — log (3 — 1). 
 Calculating 3^° by logarithms, we have 
 log 3^° = 10 X loe- 3, 
 
A/'^'i.iUATlON OF LOGABITHMS. 253 
 
 from which we obtain 3'° = 59048 ; 
 
 whence log S = log 5 + log (59048 — 1) — log 2, 
 
 or, performing the calculations, we obtain 147620 for the sum 
 
 required. 
 
 Let it be proposed next to find the^ number of terms in the 
 progression of which the first term is 3, the ratio 2, and the last 
 term 6144. 
 
 From the formula w = ag^""^ we have 
 
 log u = log a-{-{n — 1) log q ; 
 
 , , , \q^u — Wa 
 
 whence w = 1 H — 2_- 2-_, 
 
 \ogq 
 
 Applying this formula to the proposed question, we have 
 
 log 2 
 
 216. Let us take next the progression 
 
 rrr d'-^'^cd'. e:f: g . . 
 from the nature of the progression, we have 
 a b c d f 
 b^^ c d e g* 
 
 whence log - = log - = log ^ = log - . . . . 
 b c. a e 
 
 wherefore, log a — log i = log b — log c = log c — \ogd= , . 
 from this last we have 
 
 -7- log a . log i . log c . log d . . . , 
 If, therefore, the numbers a^ b, c, d , . form a progression by 
 quotient, their logarithms will form a progression by difference. 
 Logarithms may therefore be defined a series of numbers in 
 arithmetical progression corresponding term to term to aTwther 
 series of numbers in geometrical progression. This is the defini- 
 tion of logarithms given in arithmetic. 
 
 COMPOUND INTEREST. 
 
 217. One of the most important applications of logarithms ia 
 to questions upon the interest of money. 
 
 Interest is of two kinds, simple and compound. If interest be 
 paid upon the principal only, it is called simple interest; but if 
 
254 ELEMENTS OF ALGEBRA. 
 
 the interest, as it becomes due, be added to the principal, and in 
 terest be paid upon the whole, it is then called compound interest. 
 
 We have already investigated formulas for simple interest. 
 Let it now be proposed to determine what sum a given principal 
 p will amount to, in a number n of years, at a given rate r at 
 compound interest. 
 
 The amount of unity for one year will be 1 -[- r ; that of p 
 units will be therefore p{\-\-r). 
 
 For the second year p{^-\-r) will be the principal, and its 
 amount will he p{\-\- r) (1 -\-r) oi p{l-\- rf. 
 
 The original sum p therefore at the end of the second year 
 will amount to p{l-\-rY' In like manner at the end of the 
 third year it will amount to p(l +r)^; whence putting A for 
 the amount required, we have 
 
 A=p{l + r)\ 
 This is a general formula for compound interest; taking the 
 logarithms of both sides we have, 
 
 log A = log^ -f" 71 log (1 -f- r). 
 
 Let it be proposed to determine what sum $30000 will amount 
 to, in 30 years, at 5 per cent, compound interest. 
 
 We have log A = log 30000 + 30 log 1 .05, 
 whence, we obtain $ 1^9658.27, Ans. 
 
 218. The equation A=p{l-\-rY contains four quantities 
 A, p, r, and n^ any one of which may be determined, when the 
 others are known. It gives rise therefore to the four following 
 questions, viz. 
 
 1**. To determine A, when p, r, and n are given, or the princi' 
 palf rate, and number of years being given, to find the amount. 
 
 This question we have already solved. 
 
 2°. To determine p lohen A, r, and n are given, or to find 
 what principal put at compound interest will amount to a given 
 sum, in a certain number of years, at a given rate. 
 
 Resolving the general equation with reference to p, we have 
 _ A 
 
 or by logarithms log j9 = log A — w log ( 1 + ^)' 
 
APPLICATION OF LOGARITHMS. 255 
 
 3". To determine r, when A, p, and n are hnown^ that is, to 
 find at what rate a given sum must be put at compound interest^ 
 in order to amount to another given sum in a given time. 
 
 Resolving the general equation with reference to r, we have 
 
 or by logarithms log (1 -f- r) = — ^ —. 
 
 Having by means of this last determined the value of 1 -|- r, 
 that of r will be easily found. 
 
 4°. To determine n, when A, p, and r are given, that is, to 
 find for what time a given sum must he put at compound interest 
 at a certain rate in order to amount to a given sum. 
 
 Making n the unknown quantity in the general formula, we 
 
 obtain 
 
 log A — log^ 
 
 """ log(l + r)' , 
 If it be asked what must be the value of n in order that the 
 sum at interest may be doubled, tripled, &c. ; we put in the 
 general formula A. = kp, k denoting 1, 2, 3 . . . , we thus have 
 
 Tcp=p{l + rY; whence 71 = j^^^i-p^; 
 
 n is therefore independent of p, that is, whatever the sum put 
 out, it will be doubled, tripled, &c. in the same time. 
 
 EXAMPLES. 
 
 1. What the amount of $1000 for 25 years at 5 per cent, 
 compound interest ? Ans. $3386. 
 
 2. "What will $600 amount to in 6 years at 4J per cent, 
 compound interest, supposing the interest to be payable half 
 yearly? Ans. $783.63. 
 
 3. In a certain province there are at present 200000 inhab- 
 itants. If the population increases -^jj part yearly, what will it 
 be 100 years hence ? Ans. 1448927, nearly. 
 
 4. How much money must be placed out at compound inter- 
 est to amount to $1000 in 20 years, the interest being 5 per 
 cent.? Ans. $376.89. 
 
256 ELEMENTS OF ALGEBRA. 
 
 5. A sum of $201.22 is payable 12 years hence without inter- 
 est. What sum put out at 6 per cent, compound interest will be 
 sufficient to meet the payment at the end of that time ? 
 
 Ans. $100.00. 
 
 6. The sum of $ 500 put out at 5 per cent, compound inter- 
 est has amounted already to $900. How long has it been at 
 interest ? Ans. 12.04 years. 
 
 7. A capital of $3200 having been at compound interest for 
 80 years has amounted to $34050.84, at what rate per cent, was 
 it put out ? Ans. 3 per cent. 
 
 8. In what time will a principal be doubled at 5 per cent.? 
 In what time will it be tripled at 6 per cent. ? 
 
 ANNUITIES. 
 
 219. An annuity is a sum of money payable yearly for a cer 
 tain number of years or forever. 
 
 Let it be proposed to determine what sum must be put at 
 interest to pay an annuity of b dollars for n years, the interest 
 being reckoned at the rate r compound interest. 
 
 According to the rule for compound interest, the amount of 
 the first payment, at the expiration of the n years, will be 
 i(l + r)'*~\ the amount of the second payment will be 
 
 ^(1 _|_ rY~\ that of the third will be ^(1 +r)"-' the 
 
 last payment will be b. Putting A for the sum placed at interest 
 for the payment of the annuity, its amount at the end of the 
 n years will be A(l-(-r)"; we shall have therefore 
 A{l + rY = b{l + rY-' + b{l + rr-^-\-b{l + rY-K . . h, 
 but the second member of this equation forms, it is evident, a 
 
 progression by quotient the ratio of which is . , or, the ordei 
 
 of the series being reversed, 1 + r ; taking its sum we have 
 
 A(l + r)- = ^[<^+;>"-^; 
 
 whence • ^^ m+rfU 
 
 r (1 + r)* 
 
PRAXIS 257 
 
 This equation gives rise also to four different questions, ac- 
 cording as we make A, b^ r, or n the unknown quantity. The 
 following examples exhibit particular cases of these questions. 
 
 1. A man wishes to purchase an annuity which shall afford 
 him $ 1500 a year for 12 years. What sum must he deposit in 
 the annuity office to produce this sum, supposing he can be 
 allowed 7J per cent, interest? Ans. $11602.91. 
 
 2. A man purchased an annuity for 15 years for $100000. 
 How much can he draw annually, the interest being reckoned at 
 5 per cent.? Ans. $9634.22. 
 
 3. A man has property to the amount of $ 34580, which yields 
 him an income of 4 per cent. His annual expenses are $2000. 
 How long will his property last him ? Ans. 30 years nearly. 
 
 SECTION XXV.— Praxis. 
 
 I. — equations of the first degree. 
 
 1 1 
 
 1. Given {x + 40)^ = 10 — x^, to find the value of x. 
 
 Squaring both sides of the equation, we have 
 
 a:_j.40=100 — 20a:* + a; 
 whence a: = 9. 
 
 2. Given {x — 16)^ = 8 — a;^, to find the value of x. 
 
 Ans. a; = 25. 
 
 . ^. a:* + 28 a:* + 38 , . ,. , - 
 
 3. CjTiven —^ = —r-^ , to find the value of x. 
 
 a;^ + 4 a;5 _j_ 6 
 
 Freeing from denominators and reducing, we have 16 = 8a; , 
 whence a; = 4. 
 
 (9a:)*— 4 15 + (9a:)* ^ , ^ , 
 
 4. Given ^ — j = — r-i— ^ , to find the value of x. 
 
 a;* + 2 a:^+40 
 
 Ans. a: = 4. 
 11 4 
 
 5. Given (2 + a;)^ + a:^ = j, to find the value of x. 
 
 (2 + a:)* 
 17 Ans. a:=|. 
 
258 ELEMENTS OF ALGEBRA. 
 
 6. Given z^ A- {x — 9)^ = j-, to find the value of sc* 
 
 {x-9f 
 
 Ans. a; = 25. 
 
 n. — INCOMPLETE EQUATIONS OF THE SECOND DEGREE. 
 
 1. Given a:^ + ?/^= 189) , ^ ,, , . , 
 
 1 « , ' % 180 ( values ol x and y. 
 
 Adding 3 times the second equation to the first and extracting 
 the third root, we have a; -|- 2/ = 9, 
 
 but C(Py'{-X'i/' = xy{x-\-y), whence 9^:2/ =180, 
 
 and xy = 20; subtracting 4 times this last from the equation 
 a: -j- 2/ = 9 raised to the square, and extracting the square root 
 of both sides of the remainder, we obtain x — yz=:l; whence 
 2: = 5, 2/ = 4. 
 
 2- ^'21 ^'^ll^ll] to findthe values of x and y. 
 
 Ans. a; = dz 2, y = db 4. 
 ^^ 1 "^ 2ZI 18 [ to find the values of a; and y. 
 
 Ans, a; = zfc 9, 3/ = ^t: 3. 
 \ o ^ 2 IZ 1 A ( to find the values of x and y. 
 
 Ans. a: = 4 or — 2, y = 2 or — 4. 
 
 13 1 
 
 5. Given x^-\-y^ = 
 
 /> -^ V to find the values of x and y. 
 
 anda;v = 
 
 a; — 2/J 
 
 Ans. a; = 3 or — 2, y = 2 or — 3. 
 
 6. Given x +y^ = 13) ^^ ^^^ ^^^ ^^^^^^ ^^ ^ ^^^ ^^ 
 
 and x^ -^y^ = 5 ) 
 Squaring the second equation 
 
 a;^ + 2a:*2/*+2/^ = 25 
 
 but ^ +y^ = ]3 
 
 whence by subtraction 2 a; y = 12 
 
 subtracting this last from the first equation 
 
 x^-.2x^y^+y^=l 
 whence a: — y^ = ± 1 
 
PRAXIS. 
 
 from which compared with the second equation, we obtain 
 a: = 27 or 8, y = S or 27. 
 
 alTJ^f = 34 I *° ^"^ ^^® ^^^^^^ °^ ^ ^'^^ ^• 
 
 Ans. 2; = 5 or 3, y = 3 or 5. 
 
 IndJy' tf ^ 333 | ^° ^""^ ^® ^^^^®' °^ ^ ^"^ y* 
 
 Ans. a: = 2, y==3. 
 
 9. Given a;^ + 2/^ = 20 K i. i ,i, i r j 
 
 2 1 Mo in^d the values of x and ^. 
 
 anda;^ + 2/^= 6 ) 
 
 Ans. x = ±8 or ± V8, 2/ = 32 or 1024. 
 
 10. Given x-{'X y^ -\-y= 19 / to find the values 
 
 and ar^ + a:2/-j-^= 133 j of a; and 2/. 
 
 Dividing the second equation by the first, we have 
 
 x-'X^y^+y = 7; 
 adding this last to the first and dividing by 2, we obtain 
 a: -|- 2/ == 13 ; subtracting it from the first, dividing by 2 and 
 squaring both sides of the result, we have xy = 36; comparing 
 the equations thus obtained, we have a: = 9 or 4, y = 4 or 9. 
 
 11. Given ^^ ==48^ 
 
 4 
 x^y 
 
 and fi^ = 
 
 i2f = 24[ 
 x^ J 
 
 to find the values of x and y. 
 
 Ans. a; ^36, y = 4. 
 
 12. Given a:^ — v* = 369) , n a ,u ^ c a 
 
 lo ^ 9(^® ^^ *"® vames of x and y. 
 
 Ans. x = :^5i y =z= ^ 4. 
 
 and^a.-^Z-f J^= 12 | ^^ ^"^ ^^® ^^^^^^ °^ ^ ^"^ ^• 
 Ans. a; = 2 or 1, y = 1 ox 2. 
 
 III.— COMPLETE EQUATIONS OF THE SECOND DEGREE. 
 
 1. Given a; -|- ^ = ''^^j ^^ find the values of a;. 
 
 n 1 r .1. f f f I 1 3025 
 
 Completing the square, a:*4"^+T = —j- 1 
 
 extracting the root ^ a;^ -|- - = ± -q- ; 
 
 8 
 
 from which we obtain z = 243 or ( — 28) . 
 
260 ELEMENTS OF ALGEBRA. 
 
 2. Given 3^ — a;= 56, to find the values of ar. 
 
 Ans. a: = 4, or (— 7f. 
 
 3. Given 3 a;* + a;^ = 3104, to find the values of x. 
 
 Ans. a; = 64, or (— ^7-)*. 
 
 4. Given z^ -{-z^ = Qfx^,\.o find the values of x. 
 
 Ans. a: = 2, or — 3. 
 
 5. Given ar — a; = 2 a: , to find the values of x, 
 
 Ans. ar = 4 or 1. 
 
 6. Given a; -|- 5 — (a; -}- 5)^ = 6, to find the values of x. 
 Completing the square, iC + 5~(a:-{-5)^+7==X' 
 
 extracting the root {x-\-5)^ — - = 2t:oJ 
 
 from which we obtain a: = 4, or — 1. 
 
 7. Given (x + 12) ^ + (a; + 12)* == 6, to find the values of x, 
 
 Ans. x = ^y or 69. 
 
 8. Given a: + 16 — 7 (a: + 16)* = 10 — 4 (a; + 16)^ to find 
 the values of x, Ans. a; = 9, or — 12. 
 
 9. Given a;^ + {5x + a;*)^ = 42 — 5a:, to find the values of a:. 
 
 Ans. a: = 4, or — 9. 
 
 10. Given a:^ — 2a; + 6(ar^ — 2a: + 5)* = 11, to find the 
 values of x. 
 
 Adding 5 to each member 
 
 ar' — 2a: + 5 + 6 (a:* — 2a; + 5)* = 16, 
 completing the square 
 
 ar' — 2a: + 5 + 6(ar' — 2ar + 5)* + 9 = 25, 
 extracting the root and reducing, we obtain 
 a:=l, or=fc2yv/15. 
 
 11. Given 9a; — 4ar»+ (4ar' — 9a: + 11)* = 5, to find the 
 Talues of X, Ans. a; = 2, or \, 
 
 12. Given (a;^ + 5)*— 43?* = 160, to find the values of x. 
 
 Ans. a; = 3, or V — ^^' 
 
PRAXIS, 261 
 
 13. Given (ar' — 7 a;) + (ar^ — 7 a; + 18)^ = 24, to find the 
 values of z. Ans. a; = 9, or — 2. 
 
 14. Given 2ar^ + 3a; — 5(2a:' + 82; + 9)^ + 3 = 0, to find 
 the values of a;. Ans. a; = 3, or — 4J. 
 
 15. Given a; + (a; + 6)^ = 2 + 3 (a; -f- 6)^, to find the values 
 of z. Ans. z = 10, or — 2. 
 
 z I a?^ 3; — a; 
 
 16. Given "^ = — - — , to find the values of z. 
 
 ^ — ^^ Ans. a; = 4, or 1. 
 
 (Q\ 2 Q 
 a; -|- - I -j- a; = 42 , to find the values of z. 
 
 Ans. a; = 4, or 2. . 
 
 ^^* ^^^^^ (^ITlp + ?:Z4 "= 25?' *^ ^"^"^ ^^^ ^^^"^^^ °^ ^' 
 
 Ans. a; = ± 3. 
 19. Given [(a; — 2)'' — a;P— (a;--2)'' = 90 — a;, to find the 
 values of z. Ans. a; = 6, or — 1 . 
 
 ^' ^id " + ^ = f ~ ""^ 1 '» '^"'j t'^^ ^^1"^^ °f ^ ='"'' 2'- 
 
 From the first equation 7?y^ -\- ^zy = 96, completing the square 
 and extracting the root zy = S, or — 12. 
 
 Ans. a; = 4 or 6, ?/ = 2 or 4. 
 
 21. Givenar^y^ — 7a;2/^ — 945 = 765^) to find the values of 
 
 and zy — y = 12 J z and y. 
 
 Ans. z-==-5, y == 3. 
 
 22. Given ar'-[-a;-[-y=18 — 'f) to find the values of z 
 
 anda;2/=6 ( and y. 
 
 From the first equation 3? -\' 'f -\- z -\- y =\Q 
 
 from the second 2 zy =12 
 
 by addition • x^ -j" ^^V + 2/' + 2;-[-2/ = 30 
 
 or {x + yf+{z + y)=20 
 
 whence a; = 3 or 2, y=2 or 2. 
 
 23. Given z^-\-if — z — y==78) to find the values of x 
 
 and a;y-|"Z-|-y = 39 S and?/. 
 
 Ans. a; = 9 or 3, y = 3 or 9. 
 
 24. Given z^ 4-3 z-{-y ==73 — 2zy)to find the values of x 
 
 and y^ -^^ 3 y-\-z = 4:4: ) and y. 
 
 Ans. a; = 4 or 16, 2/ = 5 or — 7. 
 
/ 
 
 262 ELEMENTS OF ALGEBRA. 
 
 25. Given x — 2x^y^-{'y = x^—y^lto^nd the values of 
 
 ^ndxi+y^ = 5 .) ^^^^2/. 
 
 From the first equation we have {x^ — y^f — {x^ — 2/^) = 0. 
 
 Ans. x = 9, 7/ = 4. 
 
 26. Given r* + 2 z?/ + ?/^ + 2a: == 120 — 2 ?/ ) to find the val- 
 
 and a;?/ — y^ = S j ues of a; and y. 
 
 Ans. 2/ :^ 4 or 1, a; = 6 or 9. 
 
 27. Given a: + 42:^4-42/ = 21 4- 82/2 -i-4a:2 y2 ) to find a 
 
 , anda:i+2/i = 6 ' ) ^^^2/- 
 
 Ans. a; = 25, y=l. 
 
 IV. — MISCELLANEOUS QUESTIONS. 
 
 1. A farmer has a stack of hay, from which he sells a quan- 
 tity, which is to the quantity remaining in the proportion of 4 
 to 5. He then uses 15 loads and finds that he has a quantity 
 left, which is to the quantity sold as 1 to 2. How many loads 
 did the stack at first contain ? Ans. 45. 
 
 2. A person engaged to reap a field of 35 acres, consisting 
 partly of wheat and partly of rye. For every acre of rye he 
 received 5 shillings ; and what he received for an acre of wheat, 
 augmented by one shilling, is to what he received for an acre of 
 rye as 7 to 3. For his whole labor he received £ 13. Required 
 the number of acres of each sort. 
 
 Ans. 15 acres of wheat and 20 of rye. 
 
 3. A person put out a certain sum at interest for 6J years at 
 5 per cent, simple interest, and found that if he had put out the 
 same sum for 12 years and 9 months at 4 per cent, he would 
 have received $ 185 more. What was the sum put out ? 
 
 Ans. $1000. 
 
 4. Two persons, A and B, were partners. A's money re- 
 mained in the firm 6 years, and his gain was one-fourth of his 
 principal, and B's money, which was £50 less than A's, had 
 been in the firm 9 years, when they dissolved partnership, and it 
 appeared that if B had gained £ 6. 5s. less, his gain and princi- 
 
PRAXIS. 263 
 
 pal would have been to A's gain and principal as 4 to 5. What 
 was the principal of each ? Ans. £200 and £ 150. 
 
 5. The crew of a ship consisted of her complement of sailors 
 and a number of soldiers. Now there were 22 seamen to e very- 
 three guns and ten over. Also the whole number of hands was 
 5 times the number of soldiers and guns together. But after an 
 engagement, in which the slain were one-fourth of the survivors, 
 there wanted 5 to be 13 men to every 2 guns. Required the 
 number of guns, soldiers, and sailors. 
 
 Ans. 90 guns, 55 soldiers, and 670 sailors. 
 
 6. A shepherd in time of war was plundered by a party of 
 soldiers, who took J of his flock and J of a she6p ; another party 
 took from him \ of what he had left and J of a sheep ; then a 
 third party took ^ of v/hat now remained and J of a sheep. 
 After which he had but 25 sheep left. How many had he at 
 first? Ans. 103. 
 
 7. A trader maintained himself for 3 years at the expense of 
 $50 a year; and in each of those years augmented that part of 
 his stock, which was not so expended by one-third thereof. At 
 the end of the third year his original stock was doubled. What 
 was his stock? Ans. $740. 
 
 8. When wheat was 5 shillings a bushel and rye 3 shillings, 
 a man wanted to fill his sack with a mixture of rye and wheat 
 for the money he had in his purse. If he bought 7 bushels of 
 rye, and laid out the rest of his money in wheat, he would want 
 two bushels to fill his sack ; but if he bought 6 bushels of wheat, 
 and filled his sack with rye, he would have 6 shillings left. 
 How must he lay out his money and fill his sack ? 
 
 ^s. He must buy 9 bushels of wheat, and 12 bushels of rye. 
 
 9. In one of the corners of a garden there is a rectangular 
 fish-pond, whose area is one-ninth part of the area of the garden ; 
 the garden is rectangular and its periphery exceeds that of the 
 fish-pond by 200 yards. Also if the greater side be increased 
 by 3 yards and the other by 5 yards, the garden will be enlarged 
 
264 ELEMENTS OF ALGEBRA. 
 
 by 645 square yards. Required the periphery of the garden, and 
 the length of each side. 
 
 Ans. The periphery is 300 yards, and the 
 sides are 90 and 60 yards respectively. 
 
 10. A sets out express from C towards D, and three hours after- 
 wards B sets out from D towards C, travelling 2 miles an hour 
 more than A. When they meet it appears that the distances they 
 have travelled are in the proportion of 13 to 15 ; but had A trav- 
 elled 5 hours less and B gone 2 miles an hour more, they would 
 have been in the proportion of 2 to 5. How many miles did each 
 go per hour, and how many hours did they travel before they met? 
 
 Ans. A went 4, and B 6 miles an hour, and 
 they travelled 10 hours after B set out. 
 
 11. There is a number consisting of two digits, which being 
 multiplied by the digit on the left hand, the product is 46 ; but if 
 the sum of the digits be multiplied by the same digit, the product 
 is only 10. Required the number. Ans. 23. 
 
 12. A detachment of soldiers from a regiment being ordered to 
 march on a particular service, each company furnished four times 
 as many men as there were companies in the regiment; but 
 these being found to be insufficient, dach company furnished 3 
 more men; when their number was found to be increased in 
 the ratio of 17 to 16. How many companies were there in the 
 regiment? Ans. 12. 
 
 13. A farmer has- two cubical stacks of hay. The side of one 
 is three yards longer than the side of the other ; and the differ- 
 ence of their contents is 117 solid yards. Required the side of 
 each. Ans. 5 and 2 yards respectively. 
 
 14. A and B purchased a farm containing 900 acres of land, 
 at the rate of $2 an acre, which they paid equally between 
 them; but on dividing the same, A got that part of the farm, 
 which contained the best of the improvements, and agreed to pay 
 45 cents an acre more than B. How many acres had each, and 
 at what price ? Ans. A had 400 acres at $ 2,25 an acre, 
 
 and B 500 acres at $ 1,80 an acre. 
 
GENERAL THEORY OF EQUATIONS. S05 
 
 SECTION XXVI. — General Theory of Equations. 
 
 221. The equations thus far considered are of the first and 
 second degrees only. Those of the third degree come next in 
 order. We now proceed, however, to develop the general the- 
 ory of equations. 
 
 No general fonnulas can be given for the solution of equa- 
 tions of a degree higher than the fourth. And the attention of 
 mathematicians has been directed chiefly to the solution of 
 numerical equations, that is, to those which arise from the alge- 
 braic translation of a problem in which the given things are 
 particular numbers. We shall give an elementary view of the 
 principles by means of which this object has been successfully 
 accomplished. 
 
 222. In the numerical operations required in the solution of 
 equations of this kind, particularly that of division, certain sim- 
 plifications are of great utility. We will first explain them. 
 
 DETACHED COEFFICIENTS. 
 
 1. To multiply x^ _ Sa;^ + 3^; — 1 by x^ — 2a; + 1. 
 The operation may be abridged by first performing the 
 multiplication upon the coefficients detached from the letters, 
 and afterwards annexing the letters raised to the proper powers. 
 Commencing with the coefficients the work will stand thus : 
 1—3+ 3— 1 
 1 — 2+ 1 
 1 — 3+ 3— 1 
 _2+ 6— 6 + 2 
 
 1_ 3 + 3—1 
 
 1 — 5+10—10 + 5-1. 
 
 The product of a:^ by a;^ is a:* ; the highest power of x in the 
 product will be, therefore, ar* ; and since from the arrangement, 
 the powers of this letter go on decreasing by unity, we shall 
 have, it is evident, for the powers, 
 
 23 
 
266 ELEMENTS OF ALGEBRA. 
 
 Annexing these to the coefficients, the required product will 
 
 be 
 
 x^ — bx^-^lOa^—lOx'^ + bx — l. 
 
 2. If any powers are wanting in either of the factors, they 
 must be supplied by writing them with as a coefficient. Thus, 
 let it be required to multiply 
 
 S3^—l:i^y + 8xy-6y^ hj 2x^ — 3xy + f 
 Here a term xi^ in the multiplicand is wanting, which must 
 be supplied, thus, Oxy^. The operations upon the coefficients 
 will then be as follows : 
 
 3- 7+ 8+ 0- 5 
 
 2- 3+ 1 
 
 6—14 + 16+ — 10 
 — 9 + 21 — 24— + 15 
 
 3— 7+ 8+ — 5 
 6 — 23+40 — 31— 2 + 15 — 5. 
 
 The powers of x go on decreasing by unity, and those of y in- 
 creasing by unity. Supplying these, the product required will be 
 6x^ — 23x'y-\-A0xy — 3lxy — 2xY + 15xy^—5f 
 
 3. Multiply 6x^ — Sax^ + 5a^x — a^ hy a" -{-Sax -{- 6xK 
 In this case we reverse the order of the terms in the multiplier, 
 so that the arrangement may be the same as in the multiplicand. 
 The operation performed upon the coefficients, as above, will give 
 
 25 — + 21 + 7+2 — 1. 
 And the required product will be 
 
 25 a;« + 2 1 2; V + 7 a: V + 2 a: a* - a', 
 
 4. Multiply2a^ — 3fli2^53'^by2a2__5^2. 
 
 Ans. 4:a' -^160^5'+ 10 a'b^+ 15ab'-2bb'. 
 
 5. Multiply x"^ — aa^ -{- a^x^ — a^x -\- a"^ hy x -}- a. 
 
 Ans. x^ + a^. 
 
 The process above is called Multiplication by Detached Co- 
 efficients. The examples, art. 24, will serve as an additional 
 exercise. 
 
 223. The process of division may, in like manner, be 
 
GENERAL THEORY OF EQUATIONS. 267 
 
 abridged by first performing the operation upon the coefficients 
 detached from the letters, and then supplying the letters. 
 
 1. Let it be required, for example, to divide c^ — ha^x-\- 
 ]OaV— lOaV + 5aa;* — a;^ by a2 _ 2 aa; + x^. 
 The operation upon the coefficients will be as follows : 
 
 1-2+1 
 1_34.3-1 
 
 1 
 
 1 
 
 -5 + 
 -2 + 
 
 10- 
 1 
 
 10 + 5- 
 
 1 
 
 
 -3 + 
 
 -3 + 
 
 .9- 
 6- 
 
 10 
 3 
 
 
 
 
 3- 
 
 3- 
 
 7 + 5 
 6 + 3 
 
 
 
 — 
 
 1 + 2- 
 1 + 2- 
 
 -1 
 -1. 
 
 The coefficient of the quotient will be 1 — 3-}-3 — 1. And, 
 m order to supply the letters, we take the quotient of the letters 
 in the first term of the dividend by those of the divisor ; thus, 
 a* divided by c^ gives c^. The letters in the first term of the 
 quotient will then be a^, and in the succeeding terms they will 
 follow, it is evident, the law of the dividend. The quotient 
 required will then be c^— 3 a^a; + 3 a a;^ — a:^. 
 
 2. Divide 6 a%^ + 3 a%^ — 4 a^i* + ^« by 3 a^Zi - 2 a 3^ + h\ 
 The operations upon the coefficients will stand thus : 
 
 6+3-4+0+1 
 6+0-4+2 
 
 3+0-2+1 
 3 + 0-2 + 1. 
 
 3-1-0-2+1 
 2+1 
 
 Supplying the letters we shall have for the quotient, 2ah -^-W. 
 
 Before commencing the operations, the dividend and divisor 
 should, it is evident, be arranged both in reference to the same 
 letter. The process is called Division by Detached Coefficients. 
 
 SYNTHETIC DIVISION. 
 
 224. The operation for finding the coefficients of the quotient 
 may be still further abridged. 
 
268 ELEMENTS OF ALGEBRA. 
 
 In the ordinary process of division, we multiply the divisor 
 by each term of the quotient as it is found, and subtract suc- 
 cessively the partial products from the dividend. The effect, 
 it is evident, will be the same, if we change the signs of the 
 divisor and add the j)artial products to the dividend. Thus, in 
 the first example above, if we change the signs of the divisor, 
 and then find the terms of the quotient by the first term of the 
 divisor with its sign unchanged, the partial products may be 
 added, and the work will stand as follows : 
 
 -1+2-1 
 1_-3_|-3__1 
 
 1-5 + 
 
 -1 + 2- 
 
 10- 
 
 1 
 
 10 + 5-1 
 
 -3 + 
 3- 
 
 9- 
 6 + 
 
 10 
 3 
 
 
 3- 
 
 -3 + 
 
 7 + 5 
 6-3 
 
 
 
 
 1 + 2—1 
 1 — 2 + 1. 
 
 In this operation it is easy to see that the terms + 9 — 10 
 in the second partial dividend, — 7 + 5 in the third, and + 2 
 — 1 in the fourth, may be omitted ; and the first term in each 
 partial dividend found by adding all the terms in each column 
 as the Work proceeds. With this modification the work will 
 stand thus : 
 
 1-5 + 10-10 + 5-11-1 + 2-1 
 
 -1+2 - 1 I 1-3+3-1 
 
 -3 
 
 
 
 
 + 3 + 
 
 6 + 
 3 
 
 3 
 
 
 - 
 
 -3 + 
 
 6- 
 
 -3 
 
 
 — 
 
 1 
 
 
 
 + 
 
 1- 
 
 -2+1 
 
 0. 
 
 In this process there is liability to error in the signs of the 
 quotient, in consequence of the necessity of finding each term 
 
GENERAL THEORY OF EQUATIONS. 269 
 
 of the quotient by means of the first term of the divisor with its 
 sign unchanged. To avoid this liability, recollecting that the 
 first term in each successive dividend is always cancelled by 
 the product of the first term of the divisor by the corresponding 
 term of the quotient, we retain the first term of the divisor with 
 its sign unchanged, and change all the rest. The operation 
 will then stand thus : 
 
 1 — 5+10-10 + 5-1 11+2-1 
 
 2- 1 
 
 l__3_i_3_l 
 
 -3 
 
 -6+3 
 
 - 3 
 
 + 6-3 
 
 - 1 
 
 -2+1 
 
 
 
 The work may be written more concisely thus : 
 
 1 
 
 1_5_|_ 10 -10 + 5-1 
 
 2 
 
 2_ 6+ 6-2 
 
 -1 
 
 _ 1+ 3-3 + 1 
 
 First term of Dividends, 
 
 _3+ 3-100 
 
 Quotient, 1 — 3+3—1 
 
 The divisor is placed at the left of the dividend in a vertical 
 
 column. Beneath, in a horizontal line, are placed the fitst 
 
 terms of the successive partial dividends ; and under the whole 
 
 is written the quotient also in a horizontal line. The partial 
 
 products are written under the tenns of the dividend to which 
 
 they belong, in a diagonal line from the left downwards toward 
 
 the right. 
 
 23=* 
 
270 ELEMENTS OF ALGEBRA. 
 
 2. Divide 2d} - 6^* + 4^^ — 7^^ + 9 
 by2a3 + 6a2__io. 
 
 10 
 
 2 0-6+4-7 0+9 
 
 --6 + 18 -54+ 150 -372 
 
 
 
 10- 30+ 90-250+620 
 
 _ 6 + 18 — 50 + 124 - 289 - 250 + 629. 
 
 1__3+ 9-25+ 62. 
 
 The operation, it is evident, terminates when the partial 
 products have reached the right hand column. This is the 
 case, in the present example, when the term 62 of the quotient 
 is obtained. And since the columns to the right of this do not, 
 when added, severally reduce to 0, there will be a remainder, 
 of which the sums of these columns respectively will be the 
 coefficients. 
 
 Supplying the letters, we shall have, therefore, a^ — 3 a^ + 9 
 a^ — 25 a + 62 for the quotient, with a remainder — 289 c? — 
 250 a + 629. 
 
 3. Divide x^ — hs? -\-\hx'' -2^7? -\-'n x^ -\^x-\-h by 
 a:* _ 2 a:^ + 4 a;2 — 2 a; + 1 . kx\&,:^ — ^x-^-h. 
 
 4. Divide a:* + 2 a;*?/ + 3 a:^?/^ — a:y — 2 a: ^ — 3 ?/« by x^ 
 + 2 a: y + 3 2/^. Ans. o? — if. 
 
 The process with the modification above is called Synthetic 
 Division. The examples, art 39, will furnish an additional 
 exercise foiH;he learner. 
 
 General Properties of Equations. 
 
 225. Any expression which involves a quantity is called a 
 fimction of that quantity. 
 
 Thus, x^ -\-pXja3^-{'by {a-\-xf are all functions of x,. 
 
 In like manner, ax^ — by^, x^y + j/^ar, are functions of z 
 and y. 
 
 2. A function is usually indicated by some one of the letters, 
 /, F, &c., the quantity or quantities of which the expression is 
 
GENERAL THEORY OF EQUATIONS. 271 
 
 a function being inclosed in a parenthesis. Thus,/ (2:) indicates 
 a function of a;, / {x, y) a function of x and y. 
 
 3. If hyf{x) we denote a particular function of x, then f {a) 
 will denote the same function of a. Thus, if the first function 
 is x^ -{- 6 X -{- 6, the second will he a^--\-ba-\-6. 
 
 4. It will be recollected that by the root of an equation we 
 understand any quantity which, being substituted in the equa- 
 tion, will satisfy its conditions. . 
 
 5. An equation of the second degree is sometimes called a 
 quadratic equation, one of the third degree a cubic^ and one of 
 the fourth a bi-quadratic equation. 
 
 6. A complete equation of the nth. degree with one unknown 
 quantity, n being an entire and positive number, may be re- 
 duced to the form, 
 
 a:"+Aa:"-i4-Ba:'-2 + Ca;"-3-|- . . . Ta: + U = 0,in 
 
 which the coefficients A, B, C . . . T, U, are any num- 
 bers whatever, positive or negative, entire or fractional. 
 
 Every equation of this description, since it is supposed to be 
 derived from a problem with sufficient and properly limited 
 conditions, may be assumed to have at least one root. 
 
 We now proceed to investigate the general principles neces- 
 sary to the solution of numerical equations of any degree. 
 
 Divisibility of Equations. 
 
 226. 1. Resuming the general equation, 
 a:« + Aa;"-i + Ba;'-2+ . . . Ta: + U=0, (1) 
 if a is a root of the equation, then the jirst member is divisible 
 by X — a. 
 
 For if the division is not exact, let Q be the quotient, and R 
 the remainder arising from the division by x — a ; then we 
 have 
 a:'' + Aa;"-i+ .... T a: + U = Q (2: - c) + R. (2) 
 
 But the left hand member of this equation is equal to ; and 
 
272 ELEMENTS OF ALGEBRA. 
 
 since a is by hypothesis a root of the equation, we have a; = a, 
 or » — a = 0, and the equation (2) reduces to 
 
 0=:0 + R, orR = 0, 
 that is, there is no remainder, and the division is exact. 
 
 2. Conversely, if the first member of the equation (1) is divisi- 
 ble by a; — «, then a is a root of the equation. For Q being 
 the quotient arising from the division by a; — «, the equation 
 returns to Q (a: — a) = 0, 
 
 which is satisfied by the value x = a\ hence a is a root of the 
 equation. 
 
 In the solution of equations we have frequent occasion to 
 ascertain, by trial, whether a particular number is a root of the 
 equation. From the preceding principle it is obvious that this 
 may easily be done by division. 
 
 Ex. 1. To determine whether 4 is a root of the equation, 
 a:3_9^2_^26a:-24 = 0. 
 
 Dividing by z — 4, and performing the operation by synthetic 
 division we have 
 
 1-9 + 26 — 24 
 
 4_20 + 24 
 
 1-5+ 6 
 Ans. 4 is a root, and if the proposed be divided by a; ~ 4 
 the equation which results will be 
 
 Ex. 2. To determine whether 5 is a root of the same equa- 
 tion. 
 
 Ans. 5 is not a root, since the division by a; — 5 leaves a 
 remainder of 6. 
 
 Ex. 3. Is 2 a root of the equation a;^ — 7a; + 6 = 9? 
 
 Ex. 4. Is 3 a root of the equation s^-^Qx^ + Sx'-l^ 
 = 0? 
 
GENERAL THEOEY OF EQUATIONS. 273 
 
 Number of the Roots. 
 
 • 227. In order to the solution of an equation, we must first 
 determine the number of its roots. An equation of the second 
 degree with one unknown quantity has, we have seen, two 
 roots. We shall now show that every equation with one 
 unknown quantity has as many roots as there are units in the 
 highest power of the unknown quantity, and no more. 
 Let a be a root of the equation 
 
 x-J^Ax^-^ + Bx^-^^ . . . Tx-\-V = 0; 
 since by the last article this equation is divisible by a; — a, it 
 returns to 
 
 {x-a) (a;'-i + AV-'^+ . . . Tx-\-V\)=:0, 
 A', &c., being the new coefficients which arise from the division. 
 But this equation is satisfied by a: — a = 0, or by 
 
 a;"-i + AV-2+ .... Tx + V' = 0. 
 Let 3 be a root of this last equation, then we have 
 (a:-i)(a;"-2 + A'V-3+ . . . T"a: + U'0 = O, 
 which is satisfied by a: — 3 = 0, or by 
 
 x^^ + A"x^-i- .... T''a:+U' = 0. 
 Continuing the operation, it will be seen that for every new 
 factor obtained, the exponent of x is made one less, and that we 
 shall have finally a:" + A a;"-! 4- Ba;''-2+ , . . Tx + U 
 e={x — a){x — b) {x-^ c) . . . . {x — p)j in which the 
 number of binomial factors, x — a, x — i, &c., is equal to n or 
 to the number of units in the index of the highest power of the 
 unknown quantity. And since there are as many roots as 
 factors, there will be as many roots as units in the highest 
 power of Xf the unknown quantity. 
 
 An equation, moreover, cannot have a greater number of 
 roots than there are units in the highest power of x. 
 
 LetV=a;'' + Aa;'-i + Ba;'-2+ .... Ta: + U = 0, 
 the roots of which are a, i, c . . . . ^, respectively ; then 
 
 Y = (x-'a){x—b){z-.c) {x-ph 
 
 18 
 
274 ELEMENTS OF ALGEBRA. 
 
 If it be possible, let a! be another root differing frpm a, 3, c, 
 . .• . p\ then we shall have 
 
 \^{a' -a){a! -h){d — c) . , . . («'_j9)=0; 
 but this equation is impossible, since a' being different from a, 
 3, c, . . . . ^, no one of the factors of V can be equal to 0. 
 
 'Every equation^ therefore^ will have as manij roots as there are 
 units in the highest power of the unknxmn quantity, and nx) more. 
 
 These roots may not, however, be all different. In fact, any 
 number of them may be equal, as a and 3, or a, h, and c, &c. 
 
 If the equation has two roots, each equal to a, for example, it 
 will be divisible by {x — aY; if it has three roots, each equal to 
 fl, it will then be divisible by {x — of, and so on. 
 
 A part of the roots, moreover, may be imaginary. But, from 
 what has been said, every equation will have at least one real 
 root. 
 
 228. From what has been done, it' will be seen that if one or 
 more roots of an equation are known, the reduced equation con- 
 taining the other roots may easily be found by division. 
 
 Ex. 1. One root of the equation x^ — \bx'^-\-1bx — 125 
 = 0, is 5. What is the equation which contains the other 
 roots? 
 
 By Synthetic Division, 1 
 5 
 
 1^15 + 75_125 
 5_50-Ll25 
 
 1 _ 10 + 25. 
 
 Ex. 2. Two roots of the equation a:'* — ba^—\2x'^-\-lQx 
 — 80 = 0, are 2 and 5. What is the reduced equation which 
 contains the other roots 1 
 
 Operation. 
 
 1st Division, 1 
 2 
 
 2d Division, 1 
 5 
 
 1__5__ 12 + 76 — 80 
 2— 6 — 364-80 
 
 1 — 3 _ 18 + 40 
 5+10 — 40 
 
 1+2— 8 
 
 Ans. a;2 + 2a: — 8 = 0. 
 
GENERAL THEORY OF EQUATIONS. 
 
 275 
 
 If, as in the preceding example, the reduced equation is a 
 quadratic, the remaining roots may be found by the methods 
 already explained. 
 
 Ex. 3. One root of the equation a^ + 3x^— I6x+I2=z0 
 IS 1 ; what are the remaining roots ? Ans. 2, and — 6. 
 
 Ex. 4. Two roots of the equation x^ — 12a^ -\-48x^ — G8z 
 -|- 15 = 0, are 3 and 5. What are the remaining roots ? 
 
 Ans. 2 + V 3, and 2 — V 3. 
 
 Ex. 5. One root of the equation a^ — x^ — 7 a; -f- 15 = 0, is 
 — 3, What are the other two roots ? 
 
 Ans. 2 + a/ — 1, and 2 — yy/ — 1' 
 
 Coefficients. 
 
 229. The roots of an equation are obviously involved in the 
 coefficients. We proceed next to determine the law of the 
 coefficients, or the manner in which they are connected with 
 the roots. 
 
 Let it be proposed, then, to form the equation whose roots 
 shall he a, b, c . . . . respectively. 
 
 The left hand member will be equal, it is evident, to the con- 
 
 tinued product of X — a, X — bj x 
 the multiplication we have 
 
 {x — a) {x — b) -zzzT? — 
 
 {x — a) (a: — b) {x 
 
 x-\'ah 
 
 Performing 
 
 abc 
 
 c)=^a^ — a a^-^ab 
 — b ac 
 
 •— c be 
 
 and so on, as in art. 128. 
 
 From what has been done we have the following properties, 
 yviz. : 
 
 1^. The coefficient of the second term in the required equa- 
 tion will be the sum of all the roots with their signs changed. 
 
 2®. The coefficient of the third term will be the sum of the 
 products of every two roots with their signs changed. 
 
270 ELEMENTS OF ALGEBRA. 
 
 3*'. The coefficient of the fourth term will be the sum of the 
 products of every three roots with their signs changed, and so on, 
 
 4°. The last, or absolute terrri will be the jn'odvx^t of all the 
 roots with their signs changed. 
 
 That this law is general may be shown, as in art. 129. 
 
 From these principles, it follows, 
 
 1°. If the coefficient of the second term in any equation is 0> 
 that is, if the second term is wanting, the sum of the positive 
 roots is equal to the sum of the negative roots. 
 
 2°. If the signs of the terms of the equation are all positive, 
 the roots are all negative ; and if the signs are alternately posi- 
 tive and negative, the roots are all positive. 
 
 ,3°. Every root of an equation is a divisor of the last or 
 absolute term. 
 
 1. The following examples exhibit the manner in which the 
 coefficients are derived from the roots. 
 
 Ex. 1. Find the equation whose roots are 2, 3, 4 and — 5- 
 Indicating the equation it will be 
 
 {X — 2) (a; — 3) (a;— 4) (a; + 5) = 0. 
 
 The coefficients may be found by the principles just demon- 
 strated, or by actual multiplication as follows, 
 
 1 — 2 
 
 — 3 + 6 
 
 — 3 
 
 — 4 
 
 5+ a 
 
 4-f 20— 24 
 
 1_9_|_26— 24 
 
 5 — 45+130 — 120. 
 
 1_4_19_|.106_120. 
 Ans. a:* — 4 a:3_ 19^2^ 106 a;— 120 = 0. 
 Ex. 2. What is the equation whose roots are 1, 3, and — 4? 
 
 Ans. a:» — 13a;+12 = 0. 
 Ex. 3. What is the equation whose roots are — 1, — 2, — 3> 
 and— 5? 
 
 Ans. a:*+ll«» + 41x2 + 61a;+30 = a 
 
GENERAL THEORY OF EQUATIONS. 277 
 
 Ex. 4. Find the equation whose roots are 2, 3, 5, and 6. 
 Ans. ai'—16u^-\-91x' — 216x-\-l80 = 0, 
 
 2. By means of the first of the preceding principles one of 
 the roots of an equation may be found, when all the rest are 
 determined. 
 
 By means of the fourth, the integral roots may all b^ found. 
 In order to this, we seek among the divisors of the last term 
 those that will satisfy the equation. 
 
 Ex. 1 . Find the integral roots of the equation x^ — 8 a:^ -j- 19 a: 
 — 12 = 0. The divisors of the last term are 1, 2, 3, 4, and 6 ; 
 of these 1, 3, and 4, substituted respectively for x, satisfy the 
 equation, and are, therefore, roots. The equation being of the 
 third degree only, they are all the roots. 
 
 Ex. 2. Find the roots of the equation a^ — 2z^ — bx-\-6 
 = 0. Ans. 1, 3, and — 2. 
 
 Ex. 3. Find the roots of the equation a^ — x — 6 = 0. 
 
 Ans. 2 is the only integral root. Depressing the equation by 
 this root, the remaining roots found from the resulting equation 
 are — 1± V — 2. 
 
 FORM OF THE ROOTS, 
 
 230. The roots of an equation may be entire, fractional, surd 
 or imaginary. 
 
 Let there be the equation 
 
 a;--f Az^-i-f Ba:'-2+ Ta: + U = 0, 
 
 in which the coefficient of the first term is unity, and A, B, &;c., 
 entire numbers. To deteraiine whether this equation can have 
 
 a fractional root : 
 
 a ' 
 If it be possible, let the fraction-, the terms of which are 
 
 prime to each other, be a root of this equation. 
 
 Substituting - for a:, multiplying both members by 3"~S and 
 
 transposing we obtain 
 
 f=-Aa'-i-Ba'-2*- .... Tab'^' — Vb'^K 
 ^ 24 / 
 
278 ELEMENTS OF AtGfiBRA. 
 
 The right hand member of this equation is an entire quantity, 
 since it is composed of terms each of which is integral. The 
 left hand member must, therefore, be entire, or we shall have a 
 whole number equal to a fraction, which is absurd. Hence an 
 equation, whose coefficients are all integers and that of the highest 
 power of the unknown quantity equal to uniti/, cannot have a 
 fractional root. 
 
 It does not follow from this, however, that all the roots are 
 whole numbers. The equation may have other roots, which can- 
 not be expressed in whole numbers or definite fractions, such as 
 surds or imaginary quantities. 
 
 231. But surds and impossible roots enter equatimis by pairs ^ 
 so that if there be one, there will necessarily be two ; and if 
 three, there will necessarily be four, and so on. 
 
 Let a + V —- 5, for example, be one of the roots of an equa- 
 tion, the coefficients of which are real. 
 
 Suppose the equation reduced by division until two only of 
 its roots remain. It will then be a quadratic. And if one of 
 its roots \s a -{■ /s/ b, the other will necessarily he a — a/ — b. 
 In the same way it may be shown that if there are three surd 
 or imaginary roots, there will necessarily be four, and so on. 
 
 From this it follows, 
 
 1°. An equation of an even degree may have all its roots 
 imaginary; but if they are not all imaginary, two of them, at 
 least, will be real. 
 
 2^. The product of every pair of imaginary roots being of the 
 form, a^ -j- b, is positive ; hence the absolute term of an equation 
 whose roots are all imaginary must be positive. 
 
 3°. Every equation of an odd degree has at least one real 
 root ; and if there be but one, that root must necessarily have a 
 contrary sign to that of the last term. 
 
 4°. Every equation of an even degree whose last term is 
 negative has, at least, two real roots ; and if there be hoA two, 
 one of these will be positive and the other negative. 
 
 These principles are illustrated in the following examples. 
 
GENERAL THEORY OF EQUATIONS. 279 
 
 In forming the equations, the most convenient process will be to 
 multiply together, in pairs the factors containing the imaginary 
 roots, and then combine the factors thus obtained. 
 
 Ex. 1. Form the equation whose roots are 2 -}- V — 3, 2 
 
 — V —3, 3 -f- V — 1. and 3 — V — 1- 
 
 Ans. a;*— 10a;3 + 41a;2 — 82a: + 70 = 0. 
 
 Ex. 2. Form the equation whose roots are 3 -f- v^ — 5, 3 — 
 a/— 5,and5. Ans. 3^ -^ 12 x^ -\- bO x — 84. = 0. 
 
 Ex. 3. Form the equation whose roots are 5 -}- V — 1j and 
 5 ~ V - 1- Ans. x^ — 10 X + 26 = 0. 
 
 Ex. 4. Form the equation whose roots are 2, 3 -|- V — 4, 
 3 _ V - 4, and - 5. 
 
 Ans. a:*-32:3_ 15^2^99 2. _ 130^0. 
 
 2. An equation which has imaginary roots is divisible by 
 (x — a + b/sZ-^l) {x — a — bA/ — 1), or, {x — af + b^; a 
 -\-b /s/ —l^a — b/s/ — 1, representing any pair of the roots ; 
 hence 
 
 1°. Every equation may be resolved into rational factors, 
 simple or quadratic. 
 
 From what has been done, it is also evident that, 
 
 2°. An algebraic equation which has real coefficients is 
 alw^ays composed of as many real factors of the first degree as 
 it has real roots, and of as many real factors of the second 
 degree as it has pairs of imaginary roots. 
 
 Ex, Form the equation whose roots are 3, — 5, 1 -}- V — 3, 
 1 - V —3. Ans. x"^ - 15a;2 + 38a; - 60 = 0. 
 
 1. What are the factors corresponding to the real roots of 
 this equation ? Ans. a: — 3, and a: -J- 5. 
 
 2. What is the factor corresponding to the pair of imaginary 
 roots ? Ans. a:^ — 2 a; -f- 4. 
 
 Relation of the Signs to the Roots. 
 
 232. In the preceding article we have seen the important 
 relation between the sign of the absolute term of an equation 
 and the form and number of the roots. Let us now examine 
 
280 ELEMENTS OF ALGEBRA. 
 
 the relation of the signs of the terms generally to the roots. 
 Resuming the general equation, 
 
 a;'' + Aa;"-i + Ba;''-2+ .... Ta; + U = a, (1) 
 and changing the signs of tke alternate terms it becomes 
 
 a:" — Aa:"-i + Ba:"-2— .... dbTa;:q=U = 0; (2) or 
 changing all the signs in this last, which will leave the equation 
 identically the same, 
 
 -a;" + Aa:"-i — Ba:"-2_|_ .... zp T a: ± U = 0, (3). 
 
 Now if a be substituted for x in equation (1), and — « be sub- 
 stituted for X in (2) when n is an even number, or in (3) when 
 n is an odd number, the equations which result will be identi- 
 cally the same. If then a is a root of equation (1) this equation 
 will be verified by this substitution. Hence the equation (2), 
 or (3) as the case may be, will be verified by the substitution of 
 — a for a;, and, therefore, — a is a root of the equations (2) 
 and (3). 
 
 If J therefore^ the signs of the alternate terms in an equation 
 are changed, the signs of all the roots will be changed, 
 
 Ex. 1. Form the equations whose roots are 1, 2, 3 ; and — 1, 
 -2,-3. 
 
 Ans. The equations are x^ — 6a:2 -j- 11 a: — 6 = 0, ^^^^ ^ 
 '^Qx'-^- lla; + 6 = 0. 
 
 Ex. 2. The roots of the equation x'^ -\- 7? — \^ x"^ -\- l\ x -\- 
 30 = 0, are — 1, 2, 3, and — 5. What will be the roots if the 
 signs of the alternate terms are changed ? 
 
 Since the negative roots may be changed into positive by 
 simply changing the signs of the alternate terms, the finding the 
 real roots of an equation is reduced, by the preceding principle, 
 to finding positive roots only. 
 
 233. When in an equation the signs continue the same from 
 one term to the next following, there is said to be a permanence 
 of signs ; and when the signs change from one term to the next 
 following, a variation of signs. Thus, in the equation, x'^ — 3 
 r^ — 15 o:^ -f- 49 a: — 12 = 0, there is one permanence and 
 three variations of signs. 
 
• GENERAL THEORY OF EQUATIONS. 281 
 
 Let -) — I 1 1 — |- -f- be the order of signs in any 
 
 equation, and let us introduce into this equation a new positive 
 root a. In order to this, we multiply the equation by a; — a. 
 The operation, so far as the signs are concerned, will be as 
 follows : 
 
 + - 
 
 — +-+ 
 
 +±-+-+±±- 
 
 in which the ambiguous sign db indicates that the sign may be 
 + or — , according to the relative magnitude of the partial 
 products with contrary signs united in the terms to which it 
 corresponds. 
 
 If now this result be examined with attention, it will be seen 
 that each permanence is changed into an ambiguity by the in- 
 troduction of the new positive root + ^' It follows, therefore, 
 that the permanences, take the ambiguous sign as we may, 
 cannot be increased in the final product by the introduction of 
 the new positive root; but, as the number of signs is increased 
 by owe, the number of variations must be increased by one. 
 
 In the equation x — a = 0, there is one positive root, and 
 one variation. And since, from the reasoning above, the intro- 
 duction of each new positive root in this equation will produce 
 at least one variation, it follows that the number of positive 
 roots in any equation can never be greater than the number of 
 variations of sign. 
 
 By a process altogether similar, it may be shown that the 
 introduction of a new negative root produces at least one new 
 permanence, and that the number of negative roots can never 
 be greater than the number of permanences. Hence, generally, 
 in a complete equation of any degree, the number of 'positive 
 roots cannjot he greater than the number of variations of sign^ 
 nor the number of negative roots greater than the number of 
 permanences. 
 
 24* 
 
282 ELEMENTS OF ALGEBRA. 
 
 Ex. 1. How many permanences and variations of sign are 
 there in the equation v/hose roots are 2, 3, and — 4 ? 
 
 The equation x — 2 = gives one variation corresponding 
 to the positive root 2. If we now multiply this by a; — 3, the 
 result, x^ — 5 X -j- 6 = 0, gives an additional variation corre- 
 sponding to the new positive root -\- 3. Multiplying again by 
 2 + 4, the result, a:^ — a;^ — 14 a; -|- 24 = 0, gives, as before, 
 two variations corresponding to the positive roots 2, and 3, and 
 one permanence corresponding to the negative root — 4. 
 
 Ex. 2. How many permanences and variations in the equa- 
 tion whose roots are 2, — 3, and — 5 ? 
 
 Ans. The equation is a:^ -[- 6 a;^ — a; — 30 = 0, exhibiting 
 one variation and two permanences. 
 
 The whole number of- permanences and variations taken 
 together, it is evident, will be equal to the degree of the equa- 
 tion. If all the roots, therefore, are real, the number of positive 
 roots will be equal to the number of variations, and the number 
 of negative roots will be equal to the number of permanences. 
 
 234. In what precedes, the equation is supposed to be com- 
 plete. If there are missing terms their place must be supplied 
 with the coefficient 0. Any sign may be given to this coeffi- 
 cient without affecting the roots of the equation. 
 
 Let there be the equation a;* — 5 a:^ -|- 8 a: — 6 c= 0. In its 
 present form this equation exhibits variations only. It would 
 appear, therefore, that it can have no negative roots. But if 
 we supply the missing term it becomes 
 
 a;4 _j_ a:3 _ 5^_|_ 3^. _ g ^ 0. 
 
 Now which ever sign we take with the coefficient, the 
 equation will present one permanence. It may have, therefore, 
 one negative root. 
 
 The preceding principle enables us to determine the number 
 of positive and negative roots a proposed equation may have, 
 an object of importance in the research for the roots. 
 
 Ex. 1. The equation a;*^ — 3a:4 — 5a;^-f 15a:2^4a._12 
 = 0, has five real roots. How many of them are positive ? 
 
GENERAL THEORY OF EQUATIONS. 283 
 
 Ex. 2. The equation, a:^ — 3 a:' — 15 a;^ -j- 49 a; — 12 = 0, 
 has four real roots. How many of them are negative ? 
 
 Ex. 3. The equation a:* — 3a;' — 15a;2+ 99a:— 130=0, 
 has (art. 231) two real roots. What are their signs ? 
 
 235. By means of the above rule we are sometimes also 
 enabled, when there are missing terms, to detect the existence 
 of imaginary roots in an equation. 
 
 Thus, let there be the equation a:^ _|_ 30 = 0. 
 
 Supplying the missing term, it becomes 
 a:2±0a; + 30 = 0. 
 - If the upper sign be taken with the coefficient 0, there will be 
 no variations ; hence there can be no positive roots. And if the 
 lower sign be taken, there will be no permanences ; hence there 
 can be no negative roots. The two roots of the equation will 
 therefore be imaginary. 
 
 Let there be next the equation a:^ -)- 10 a; -f- 5 = 0. 
 
 Supplying the missing term, it becomes 
 
 a:3_j_0a;2_f-l0a;-|:5 = 0. 
 Here, if the upper sign of the coefficient be taken, the equation 
 exhibits only permanences ; it can have, therefore, no positive 
 root. If the lower sign be taken there will be but one per- 
 manence, and, therefore, there can be but one negative root. 
 Thus it appears that the equation must have two imaginary 
 roots. 
 
 Transformation of Equations. 
 
 236. When the solution of an equation is difficult, the work 
 may sometimes be accomplished with more facility by aid of 
 another equation, the roots of which shall bear to those of the 
 proposed some given relation. The roots of the latter being 
 found, those of the former will then be readily deduced from 
 them. 
 
 It may, therefore, be required to change a proposed equation 
 into another, the roots of which shall bear to those of the former 
 a given relation. One of the most simple, as well as useful, of 
 = 0? 
 
284 ELEMENTS OF ALGEBRA. 
 
 these transformations is to change the equation into another, 
 the roots of which shall be greater or less than those of the pro- 
 posed by a given quantity. 
 
 To effect the transformation required, it will be sufficient, 
 it is evident, to substitute for x in the proposed x diminished or 
 increased by the desired quantity. The resulting equation will 
 be of the same form with the proposed, the roots of which will 
 be of the required dimensions. 
 
 Let it be required, for example, to find an equation whose 
 roots shall be less by 2 than those of the equation 
 
 a:2_8a;-|-7 = 0. (1) 
 
 Substituting a; -|- 2 for x^ the equation becomes 
 
 (2: + 2)2- 8 (a; + 2) + 7 = 0; 
 
 or developing and reducing, 
 
 a;2_4a: — 5 = 0. (2) 
 
 Resolving equation (1) its roots are 7 and 1. Resolving 
 equation (2) its roots are 5 and — 1, or less by 2 than those of 
 the proposed, as was required. 
 
 Ex. 2. Let it be required to find next an equation whose 
 roots shall be greater by 2 than those of the equation 
 
 Ans. a:^— 12a:2 + 47a: — 60 = 0. 
 237. The process above is laborious, especially in the higher 
 equations. Let us see if one more simple can be found. 
 Resuming the general equation 
 
 a:" + Aa;'^i+Ba;"-2-l- . . . . Ta: + U=0, 
 let us make xz=.u-\-x\u being a new unknown, and x' an 
 indeterminate, to which any value, positive or negative, may be 
 assigned at pleasure. 
 
 Substituting u -f x\ or, which is the same thing, x! •\'uiox x 
 in the proposed, it becomes 
 
 (a;' + 2^)"-f A(a:' + w)'-i-|-B(a;'+t^)'-? . . . .T(a;'+M) 
 + U==0,(1) 
 
GENERAL THEORY OP EQUATIONS. 
 
 285 
 
 or developing the several terms by the binominal formula, and 
 arranging with. reference to the ascending powers of m, . 
 
 a:" 
 U 
 
 
 u + 
 
 1 
 
 1.2 
 
 u^ + 
 
 .... M'' = 0. (2) 
 
 Observing the manner in which the different coefficients of u 
 are formed, the following remarkable law will be discovered : 
 
 P. The coefficient of u° is simply what the first equation 
 becomes when x' is substituted for x. 
 
 Let us designate this coefficient by X. 
 
 2°. The coefficient of u is formed by means of the preced- 
 ing or X, by multiplying each of the terms of X by the exponent 
 of x' in that term, and diminishing this exponent by unity. 
 
 Denote this coefficient by Y. 
 
 3°. The coefficient of w^ is formed from Y, by multiplying 
 each term of Y by the exponent of x' in that term, dividing the 
 product by 2, and diminishing each exponent by unity. 
 
 Z 
 
 Let — be this coefficient. It is evident that Z is formed in 
 di 
 
 the same manner from Y that Y is formed from X. In general 
 
 the coefficient of any power of u is formed from the preceding 
 
 coefficient in the following manner, viz. : 
 
 By taking each term of the •preceding coefficient in succession, 
 multiplying it by the exponent of x\ dividing by the number 
 which maj'ks the place of the coefficient, and diminishing the ex- 
 ponent of x' by unity. 
 
 The preceding development will be represented generally by 
 
 X + Yu + ^u'+^u' + ko. 
 
 The polynomials, Y, Z, &c., are called derived polynomials of 
 X, since Z is derived in the same manner from Y, that Y is 
 
286 ELEMENTS OF ALGEBRA. 
 
 from X. Y is called the firsts Z the second, V the third, 
 derived polynomial of X, and so on. 
 
 Ex. Let X = a;3 _ 6 x^ -j- H a: — 6 = 0. To find the 
 derived polynomials. 
 
 Ans. Y = 32;2- 12a; +11 
 Z=6a; -12 
 V = 6. 
 
 238. Returning to our purpose, let it now be proposed to find 
 an equation, the roots of which shall be greater by unity than 
 those of the equation 
 
 4a;3_5a;2_|-7a: — 9 = 0. 
 Let us put x=:u — 1, from which we have u=.x-\-\. 
 From the preceding development the transformed equation will 
 be Z V 
 
 To find the coefficients, we have therefore 
 5(_l)2-l-7(— 1) _9 = — 25 
 10 (— 1) 4- 7 = 29 
 
 5 == — 17 
 
 = 4 
 
 The transformed required will be 
 
 4w« — 17 ^2 _(_ 29 w - 25 = a 
 Ex. 2. Transform the equation a:* — 4a;^ — 8a:4-32 = 
 into another, whose roots shall be 2 less. 
 
 Ans. a:*-[-4a;^ — 24 a; = 0. 
 
 239. The process above being still laborious, another more 
 simple is to be sought. 
 
 In order to this, we resume the general equation, and also 
 equation (2) art. 237, denoting the coefficients of this last by 
 AVB' .... TMJ', respectively, thus, 
 
 af + Aa;'^^ + B2:'-2+ T^_l_U^O, (1) 
 
 M- + A'M"-^ + B'w"-2_j_ ^ .... T'w + U' = 0. (2) 
 Substituting in this last for u its value x — x\ we obtain 
 
 in 
 
 which 
 
 2;' = - 
 
 -L T 
 
 
 X = 
 
 = 4(- 
 
 -\f- 
 
 
 Y = 
 
 :12(- 
 
 -\f- 
 
 
 Z 
 
 2 "" 
 
 12 (- 
 
 -If- 
 
 
 V 
 2.3 = 
 
 = 4 
 
 
GENERAL THEORY OF EQUATIONS. 287 
 
 ^x — afy-\-A'{x-xy-^ + B'{x-xy-^-\- .... T' (a; -a;') 
 + F = 0, (3) 
 
 which when developed must, it is evident, be identical with 
 equation (1) ; since the second is formed from the first by the 
 substitution of u-\-x' for x, and the third is formed from the 
 second by substituting x — x' for u, by which we necessarily 
 return to (1), the original equation. We have, therefore, 
 ^x — x'y + A'ix — xT-^-^ . . . T{x — x') + V' = x'' 
 + Aa;'^i+ . . . Ta: + U. (4) 
 
 Now, if we divide the first member of this equation by a; — x^, 
 ever)'- term will be divisible by it, except the last, which will be 
 the remainder after the division. And since the two members 
 of this equation are identically the same, we shall obtain the 
 same quotient and the same remainder by dividing the second 
 member by x — x'. But U' is the last or absolute term of the 
 transformed equation. It follows, therefore, that if we divide 
 the proposed equation by x — x', the remainder will be equal to 
 the last or absolute term of the transformed equation. 
 
 Again, let the quotient arising from the division of the first 
 member of (4) by a: — a;' be represented by 
 (a; __ a:')"-' + A' (a: - a;')"-2 _^ .... R' (a: — a:') -f T'. 
 If we now divide this quotient by a; — x', every term will be 
 divisible by it, except the last, or T', the coefficient of the last 
 term but one of the transformed. And, since the remainder 
 must be the same when the quotient of the second member of 
 (4) by a: — of is also divided by a:— •«', we shall obtain, it is 
 evident, the coefficient of the last term but one of the trans- 
 formed equation by dividing this last quotient also by a; — x'. 
 And it is easy to see that thus, by successive divisions, all the 
 coefficients of the transformed equation may be obtained. 
 
 Thus, resuming the equation already transformed, viz. : 4 x^ — 
 5a:^ + 7a; — 9 = 0, and dividing by a: -j- 1, we obtain a quo- 
 tient 4 a;^ — 9 a; -|- 16, and a remainder — 25. Dividing next 
 this quotient by a; -(- !» we obtain a new quotient 4 a; — 13, and 
 
28S 
 
 ELEMENTS OF ALGEBRA. 
 
 a remainder 29. Dividing next this last quotient by a; -(- 1, the 
 operation terminates, and we have a remainder — 17. The 
 transformed will be, therefore, as before, 
 
 4^3 — 17^2 ^29 w — 25 = 0. 
 
 We have, then, the following rule by which to obtain the 
 transformed equation : 
 
 Let x' be the quantity by which the roots of the equation are 
 required to be increased or diminished. Divide the equation 
 by z — x\ or x-\-x', as the case may require, and the quotient 
 thus obtained by the same quantity, and so on until the division 
 terminates. The coefficient of the first term^ or that containing 
 the highest poiver of the transformed^ will he the same with the 
 coefficient of the first term of the proposed equation ; and the 
 successive coefficients will be the remainders arising from the 
 successive divisioTis taken in a reverse order, the first remainder 
 being the last or absolute term of the transformed equation. 
 
 The successive divisions, which are tedious by the common 
 method, are performed in a very concise and elegant manner by 
 synthetic division. 
 
 Thus, in the last example. 
 
 1 
 1 
 
 4— 5H 
 — 4- 
 
 f- 7— 9 
 -9-16 
 
 
 — 9-\ 
 
 - 4- 
 
 P 16 - 25 
 -13 
 
 
 — 13H 
 
 — 4 
 
 [-29 
 
 
 — 17 
 
 
 In which the remainders, as before, are — 25, 29 and — 17. 
 
 Examples. 
 
 1. Transform the equation b x* -\- 28 a^ -j- 61 a^ -\- 22 x -- 1 
 =0 into another, having its roots greater by 2 than those of 
 the proposed equation. 
 
GENERAL THEORY OF EQUATIONS. 
 
 289 
 
 1 
 
 -2 
 
 Operation. 
 
 54-28 + 51 + 32-1 
 __10 — 36 — 30 — 4 
 
 5+18+15+ 2 
 — 10—16+ 2 
 
 5+ 8 — 
 -10 + 
 
 1 + 
 4 
 
 4 
 
 5— 2 + 
 — 10 
 
 12 
 
 Ans. b2^ — 123^ + Sx' + 4:X--b = 0. 
 2. Find an equation whose roots shall be less by ^ than those 
 of the equation 3x* — 13a^~\-lx^ — 8x-—9 = 0, 
 
 Placing the divisor at the right, as in ordinary division, and 
 omitting all the figures in the first column except the first, the 
 calculations may be more conveniently disposed, thus : 
 3 —13 +7 —8 — 9 I 1 
 1 - 4 1 - 2^ U 
 
 -11* 
 
 -12 
 
 1 
 
 3 
 
 -3f 
 
 — 7 
 
 - f 
 
 — 11 
 1 
 
 -3^ 
 
 -n 
 
 — 10 
 
 1 
 
 — 4 
 
 
 12 o; — 28 = into another, 
 
 - 9 
 
 Ans. 3 a:* 
 
 3. Transform the equation a^ 
 whose roots shall be less by 4. 
 
 In the use of synthetic division the missing term in this 
 example must be supplied by coefficient. 
 
 Ans. 3^-{-12a^-{-S6x^l2 = 0. 
 
 4. Transform the equation 3a;* — 4a;^ + 7a:^ + 8a: — 12 
 = into another, whose roots shall be less by 3. 
 
 Ans. 3a;* + 32a:3_|. i33^^266a: + 210=0. 
 19 
 
290 ELEMENTS OF ALGEBRA. 
 
 5. Find the equation whose roots shall be greater by 2 than 
 those of the equation x^ -\- 10 x^ + A2 x^ -\- 86 x"^ -{- 10 x -j- 
 12 = 0. Ans. x^^2x^ — 6x^—10x-\-S=0. 
 
 6. Find the equation whose roots shall be less by ^ than those 
 of the equation 2x^^6x^-\-^x^ — 2x-\-l^=:0. 
 
 Ans. 2x* — 2a^ — 2x^ — ^x + ^=:0, 
 
 240. The transformation above is the only one for which we 
 have immediate occasion. There are others, which are some- 
 times useful, which we will here explain. 
 
 1. From what has been done we may readily transform an 
 equation into another, deprived of its second term. 
 
 Resuming the general equation x"" -\- K a;"~^ -|- ... T a: -}- 
 U = 0, and putting a; = w -j- :r', developing and arranging with 
 reference to the descending powers of u^ we have 
 
 w''-(-(7ia;' + A) w"-i =0. 
 
 In order that the second term of this last may disappear, its 
 coefficient must be equal to 0. We shall have then the relation 
 
 ?za;'-|-A = 0; whence a:' = . 
 
 n 
 
 Hence, to make the second term of an equation disappear, 
 
 P. Divide the coefficient of the second term by the highest 
 
 power of the unknown quantity. 2°. Transform the equation 
 
 into another whose roots shall be less or greater by the quotient 
 
 thus obtai7iedi acco?'ding as the sign of the second term is nega- 
 
 tive or positive. 
 
 Examples. 
 
 Deprive the following equations of their second terms : 
 
 1. s^—6x^4.4:X — l = 0. Ans. a^ — Sx — 15==0. 
 
 2. a;*--8a:3+16a;2+7a;— 12 = 0. 
 
 Ans. x' — 8x'' + lx+18=:0, 
 
 3. a^^6x^-j-8x — 2. Ans. a:^— 4a: — 2 = 0. 
 
 4. ar'+15a;*+ 12a;3_20a:2_|_ i42._25^ 0. 
 
 Ans. a:^— 78a:« + 412a:2_757a; + 401 = 0. 
 
GENERAL THEORY OF EQUATIONS. 291 
 
 2. An equation may be transformed into another, the roots 
 of which shall be any multiple of those of the proposed equa- 
 tion. 
 
 In the general equation a;" -}- A a:""^ -f- . . . . T a; -f- U = 0, 
 
 X X 
 
 let a: = — . Substituting - for a;, and clearing of fractions, we 
 
 obtain 
 
 ar-'^- A77ta;'^i + Bm2af-2 T mP-'^ x -{-V 7nr = 0, 
 
 the roots of which are m times greater than those of the pro- 
 posed. The required equation is, therefore, found by multiply- 
 ing the second coefficient of the proposed by w, the third by 
 m^, and so on, ?» Representing the number of times the required 
 roots are to exceed those of the proposed. 
 
 Ex. Find an equation whose roots shall be three times 
 larger than those of the equation 3? — Q x'^ -{-Sx — 9 = 0. 
 Ans. a:^— 18a:2 + 72a: — 243 = 0. 
 
 By means of this principle we may transform an equation 
 with fractional coefficients into another, whose coefficients shall 
 be integral. In order to this, we have merely to transform the 
 proposed equation into another, the roots of which shall be 
 equal to those of the proposed multiplied by the least common 
 multiple of the denominators of the fractions. 
 
 Ex. 1. Transform the equation a:^-}-K^^"— t^ + 2 = in- 
 to another, the coefficients of which shall be integral. 
 
 Ans. a:»-|-4a:2 — 36a;-f-3456 = 0. 
 
 In this example the number 6, when raised to the square, will, 
 
 it is evident, be divisible by 4. The fractional coefficients may, 
 
 therefore, be removed, and a more simple result obtained, if 
 
 we multiply by the successive powers of 6. It will be easy to 
 
 apply the like simplification to other cases. 
 
 7 11 25 
 
 Ex. 2. Transform the equation j? — -x'^ -\- ^-x — — =0, 
 
 into another, the coefficients of which shall be integral. 
 
 Ans. a:^ — .Ua;2-|-lla; — 75 = 0. 
 
292 
 
 ELEMENTS OF ALGEBRA. 
 
 3. In like manner an equation may be transformed into 
 another, the roots of which shall be the reciprocals of the pro- 
 posed equation. 
 
 In order to this, substituting - for x in the general equation, 
 
 clearing of fractions, and reversing the order of the terms, it 
 becomes 
 
 U af' 4- T a:"-! 4- S a;"-2 ^ A a: 4- 1 == 0, 
 
 the roots of which are reciprocals of those of the proposed equa- 
 tion. 
 
 The transformation is therefore effected by simply changing 
 the order of the coefficients of the proposed equ&tion. 
 
 If the coefficients of the proposed equation are the same, 
 whether taken in the reverse or direct order, the transformed 
 equation, it is evident, will be identical with tlie given equation, 
 and will furnish the same series of roots. 
 
 Equations of this description are called recurring equations, 
 or, from the form of the roots, reciprocal equations. 
 
 4. From what has been done, an equation, it is evident, may 
 be transformed into another, whose roots shall be greater or 
 less than the reciprocals of the proposed equation. In order to 
 this, we reverse the order of the coefficients, and then apply the 
 process of art. 237. 
 
 Ex. 1. Transform the equation oi? — 1 x -\- 7 into another, 
 the roots of which shall be less by 1 than the reciprocals of 
 those of the proposed. Ans. 1 a? -{-l^a? -\-l x -\-\ = 0. 
 
 Ex. 2. Find the equation whose roots shall be the reciprocals 
 of those of 3 2:*— 13 2:5 + 7 2:^ — 8 2; — 9 = 0, increased by 2. 
 Ans. — 92:^ + 642:3 __i6ia.2^151^_23 = 0. 
 
 LDIITS OF THE ROOTS. 
 
 241. The methods for resolving numerical equations of any 
 degree consist, in general, in substituting particular numbers 
 for X in the equations, in order to see if these numbers will 
 verify them ; or for the purpose of determining the initial figures 
 
GENERAL THEORY OF EQUATIONS. 293 
 
 of the roots. It is important, therefore, as a preliminary step, 
 to determine the limits between which the roots of an equation 
 are to be sought. 
 
 1. A number numerically greater than the greatest positive 
 root of an equation is called a Superior Limit to the positive 
 roots. A number numerically greater than the greatest nega- 
 tive root, abstraction being made of the sign, is called a 
 Superior Limit to the negative roots. 
 
 The extreme limits to the positive roots, it is evident, will 
 be and + oo , and those of the negative roots and — oo , the 
 sign OD, being used to denote infinity. In practice, much nar- 
 rower limits than these will be required. 
 
 Since the lai^est of the positive roots when substituted for ar, 
 reduces the equation to 0, it follows that a number greater than 
 this, when substituted in like manner for x, will give a positive 
 result. If, therefore, a number substituted for x in an equation 
 gives a positive result ^ then this number will be a superior limit 
 to the positive roots. 
 
 1. It may be shown that in any equation the greatest Tiega- 
 tive coefficient ii^reaseA by unity is a superior limit to the posi- 
 tive roots. A much nearer limit may, however, be found. 
 
 2. Let. us take the general equation, 
 
 a:" 4- il a:"-i -f . . . — D 2;"-'^ ™ Pa;"-* . . . . U = 0, 
 
 and let — D a;"-*" be the first negative term, and — ' P the great- 
 est negative coefficient; then, if all the terms after the first 
 negative term are negative, the sum of these negative terms, 
 it is evident, must be equal to the preceding positive terms. 
 And any value, which, substituted for x, will make the sum of 
 the positive terms greater than the sum of the negative terms, 
 will be a superior limit. And, P being the greatest negative 
 coefficient, for a still stronger reason, any number will be a 
 superior limit, which substituted for x gives 
 
 ^n^p(^n_.^^.-r-l^ OT + 1), (1) 
 
 or, since the right hand factor of this inequality is a progression 
 by quotient, art. 176, 
 
 25* 
 
294 ELEMENTS OF ALGEBRA. 
 
 But this inequality will be satisfied if we have 
 
 ->KS) 
 
 or, reducing (^ — 1) a;'-^^ P; (3) 
 
 but X — 1 is less than 2:, and by consequence {x — 1) {x — 1)''"^ 
 or, (a; — l)'"is less than {x — l)a;'-^; the inequality (3) will, 
 therefore, be satisfied if we have 
 
 2 
 {x-^\Y =, or > P, or a; =, or > ?'•+ 1. 
 
 Thus, to obtain a superior limit of the positive roots, we in- 
 crease by unity the root of the greatest negative coefficiejit, whose 
 index is the number of terms which precede the first negative 
 term. 
 
 EXAMPLES. 
 
 Find superior limits to the roots of the following equations : 
 
 1. ic^ + 7a:^—12a;3 — 49^:24-520; — 13 = 0. 
 
 i- 1 
 
 Ans. P'-.f 1 = (49)2+1=8. 
 
 2. a:* — 5a;3 + 37a;2_3x -1-39 = 0. Ans. 6. 
 
 3. 3 a;3 — 2 a:2 — 1 1 a; + 4 = 0. Divide the equation first 
 by 3, then applying the rule the limit will be 5. 
 
 4. x^-\-\\x'^ — 25 a; — 67 = 0. The greatest negative co- 
 efficient, is that of tP^ or 67, and the missing term being counted, 
 
 j_ \_ 
 
 which is necessary, r = 3 ; whence P*"-]- 1 = (67) ^ -)- 1 = 6- 
 
 5. x^ + 11 0:2 - 25 a:— 61 = 0. Ans. 5. 
 242. A number numerically less than the least positive root 
 
 of an equation is called an inferior limit to thd positive roots. 
 
 In the preceding examples we have regarded as the inferior 
 limit of the positive roots. Thus the positive roots of the first 
 example are all comprised between and 8. A nearer inferior 
 limit may, however, be found. 
 
 Let X = be any equation, and in this equation let us put 
 
GENERAL THEORY OF EQUATIONS. 295 
 
 x = -. We shall obtain, it is evident, a derived equation Y = 
 
 y 
 
 in which the greatest value of y will correspond, to the least 
 value of X. If then we find the superior limit L to the roots of 
 
 the equation Y = 0, the reciprocal of this, ory-, will be the in- 
 ferior limit to the roots of the proposed equation, or X = 0. 
 
 Thus, let it be proposed to find the inferior limit of the posi- 
 tive roots of the equation 7? — 42 x"^ -{- 4:^1 x — 49 = 0. 
 
 Putting a: = - , we have, for the transformed equation, 
 
 which gives, by the rule, 10 for the superior limit of its posi- 
 tive roots. Thus the inferior limit to the positive roots of the 
 
 proposed, will be j^. 
 
 243. The particular form of the equation may sometimes 
 suggest artifices, by means of which closer limits may be ob- 
 tained than those given by the preceding rules. 
 
 Thus, the equation re* -f 11 a:^ — 25 a; — 61 = 0, may be put 
 under the form 
 
 in which it is evident that a: = 3, or any number greater than 
 3, will give a positive result. We shall have 3, therefore, for 
 the superior limit to the positive roots, which is much nearer 
 than 5, the limit obtained by the rule. 
 
 The equation x^-^ba^ + Sl x^ — S x + Sd = 0, maybe put 
 under the form 
 
 7^{x-5) + 31x(^x-~^+29 = 0; 
 
 and the equation 2^+lx'-l22^ — 49a^-\-b2x^l3 = 0, 
 under the form 
 
296 ELEMENTS OF ALGEBRA. 
 
 It is evident that 5, and any number greater than 5, sub- 
 otituted for x in the first of these equations, and 4, or any 
 number greater than 4, substituted for x in the second, will give 
 positive results. We have, therefore, 5 and 4 respectively for 
 the limits, instead of 6 and 8 obtained by the rule. 
 
 The artifice consists in decomposing the equation into parts, 
 each of which is composed of two factors, the first a positive 
 monomial, and the other a binomial in a;, the second term of 
 which is negative, and then determining x in such a manner, 
 that all the factors within parentheses shall be positive. 
 
 244. It remains to find the superior and inferior limits to the 
 negative roots. In order to this, we transform the proposed 
 equation into another whose roots shall be the same as the pro- 
 posed with contrary signs. The limits of the positive roots of 
 this equation, taken witn a contrary sign, will be the limits to 
 the negative roots of the proposed equation. 
 
 Thus, let it be proposed to find the limits to the negative root 
 of the equation :j^ — 7a;-[-7 = 0. Putting x-=. — x^ or, which 
 is the same thing, changing the signs of the alternate terms, 
 the missing term being supplied, we have 
 
 the limits to which are 4 and 3 respectively. The negative 
 root of the proposed will, therefore, lie between — 4 and — 3. 
 
 245. In the preceding numbers we have found the limits 
 between which the roots cHl lie. When the roots of the pro- 
 posed are incommensurable we shall wish to find limits between 
 which the iiidividual roots are situated, in order to determme 
 more readily the initial figures of the roots. 
 
 Let a, b, c, &c., be the real roots of an equation in the order 
 of their magnitude, so that we have a^b, b^ c, &c. ; and 
 let a', b\ c', be a series of numbers, such that a! is greater than 
 a, b' 2l number comprised between a and b, so that we have 
 b' <^a, b'"^ b, and so on. 
 
 The original equation, it is evident, will be 
 
 {x — a){x — b){x — c) =0. 
 
GENERAL THEORY OF EQUATIONS. 297 
 
 If now, in this equation, we substitute a' for a;, it becomes 
 (a' — a) (a' — i) (a'-r-c) . . . .=0; 
 
 and since o! is greater than a, 3, c, &c., the factors will each be 
 positive, and hence their product will be 'positive. 
 
 Again, let V be substituted for 2;, the equation becomes 
 ^}/^a){b' — b) {b' — c) =0; 
 
 here, since, b' is less than a, but greater than i, c, &c., the first 
 factor will be negative and the rest positive ; the product, there- 
 fore, will be negative. 
 
 If, again, c' be substituted for x, the equation becomes 
 
 {c'—a){c' — b){c'-'c) [ . . . . =0; 
 
 here, since c' is less than a and b, but greater than c, &c., the 
 first two factors will be negative and the rest positive; the 
 product will, therefore, be positive. 
 
 Hence, 1°, If a quantity, greater than t/ie greatest real root, 
 be substituted for x, the result will be positive. 
 
 2°. If quantities intermediate between the roots, beghining 
 with the greatest, be substituted, the results will be alternately 
 negative and positive. 
 
 Ex. 1. The roots of the equation a^ — 13 x -|- 12 = 0, are 
 3, 1, and — 4. Substitute 4, 2, 0, and — 5 for x, and observe 
 the signs of the results. 
 
 Ex. 2. Make a like substitution in the equation a^ — da^ -{- 
 2 a; + 8 = 0, the roots of which are 4, *2, and — 1. 
 
 From the preceding principles it follows, 
 
 1°. If two numbers be successively substituted for x in any 
 equation, and give results with different signs, then between 
 these numbers there must be otic, three. Jive , or some odd number 
 of roots. 
 
 2°. But if the numbers substituted for x give the same sign, 
 then between these numbers there will be either no root, or 
 there will be two, four, or some even number of roots. 
 
 3®. If any quantity q and every quantity greater than q 
 
298 ELEMENTS OF ALGEBRA. 
 
 renders the result positive, then q is greater than the greatest 
 root of the equation, and will be a superior limit of the roots. 
 
 4°. Hence, if the signs of the alternate terms are changed, 
 and if p and every quantity greater than p renders the result 
 positive, then — ^ is less than the least root, and will be an 
 inferior limit. 
 
 5°. If the degree of the equation be even, the substitution of 
 a number less than the least root will give a positive result. 
 But if the degree be odd, the result will be negative. 
 
 246. The substitution of the natural series, 0, 1, 2, 3, &c., 
 taken negatively as well as positively, will enable us to discover 
 the position, and determine, in general, the initial figure of 
 the real roots. 
 
 Ex. 1. Let it be required to find one of the roots of the 
 equation oi? — ^x^ — Qx-\-Q — 0. 
 
 Substituting for x, we have 8 fpr the result. Substituting 
 next 1, the result will be — 1. There will be a root, therefore, 
 between and 1, and very near 1. Try next .9. Putting .9 
 for X, the result is .089. There is, therefore, a root between 
 .9 and 1. We shall have then .9 for the initial figure of the 
 root. 
 
 Ex. 2. Find the first figure of one of the roots of the equation 
 a;4-|- 3 a;'^ + 2a:2 -f-6 a; — 143 ^ q. 
 
 Ans. The substitution of 2 gives a negative, and of 3 a posi- 
 tive result. There is, therefore, a root between 2 and 3. 
 Hence 2 is the first figure of the root. 
 
 Ex. 3. Find the first figure of one of the roots of the equation 
 x' + x'J^x-im^O, i Ans 4. 
 
 Ex. 4. Find the first figure of one of the roots of the equation 
 a:3_)_ 1,53.2^ .32: — 46=0. Ans. 3. 
 
 Ex. 5. Find the first figure of one of the roots of the equation 
 a:^ — 12a: + 8 = 0. Ans. .6. 
 
GENERAL THEORY OF EQUATIONS. 299 
 
 LIMITING EQUATION. EQUAL ROOTS. 
 
 247. An equation, the roots of which are intermediate be- 
 tween those of a proposed equation, will have for its roots limits 
 to those of the proposed equation, and is, therefore, called the 
 separating or limiting equation. 
 
 Let a, h, c, &c., taken in the order of their magnitude, be the 
 roots of the -equation 
 
 a-n _ A 3,n_i _|_ B ^n-2 ^ T 2: + U = ; (1) 
 
 it is required to find an equation the roots of which shall lie 
 between or separate those of the proposed equation. 
 
 Diminishing the roots of the proposed by x\ we put x = u 
 -j- x', and developing, as in art. 237, it becomes 
 
 w"+A'm"-i+ T'2^-f-U'=0, 
 
 in which the coefficient of the last term, U', will be 
 
 x'- -f A a:'"-! + P x'"-^ + . . . T aT + U ; 
 and the coefficient T' of the term before the last will be . 
 7^a:"-l-|-(7^ — 1) Aa;"'-2 + (7i — 2) Ba;'"-3+ . . . . T. (2) 
 
 But a, b, c, &c., being roots of equation (1), the roots of the 
 transformed will* be {a — xf), {b — x'), {c — x'), &cc. Hence, 
 by art. 229, the coefficient of the last term but one of the trans- 
 formed will be the sum of the products of every n — 1 of 
 these roots with their signs changed. Thus this coefficient 
 will be 
 
 {x' — b) {ctf — c) {x' — d) to n—1 factors, "j 
 
 --{:xf —a){x' — c){x' —d) " " | 
 
 -^{3f -a) {xf — b) {x' — d) " " I (3) 
 
 --l3f — a){x' — b){x'—c) " « 
 
 --&C. J 
 
 all the factors save one occurring, of course, in each term. 
 
 But the expressions (2) and (3) are equal, since they are but 
 different expressions for the same thing, viz., the last coefficient 
 but one of the transformed equation. Hence, whatever changes 
 are produced by substitution in one of these expressions, the 
 
. . . = + 
 
 • + 
 
 • += 
 
 + 
 
 . . . = — 
 
 + 
 
 + = 
 
 I — 
 
 . . . = — 
 
 . — 
 
 += 
 
 =+ 
 
 300 ELEMENTS OF ALGEBRA. 
 
 same changes will be produced by a like substitution in the 
 other. 
 
 If we now substitute a for x' in each term of (3), all the 
 terms, except the first, will vanish, and the coefficient will be 
 reduced to {a — h) [a — c) {a — d) . . . in which the signs of 
 the factors are each positive, since a is greater than b, c, d, &c. 
 Substituting b, in like manner, all the terms vanish except the 
 second, in which the sign of the first factor will be Jiegative, and 
 those of the rest positive. For the substitution of a, b, c, &c., 
 successively, we shall have the following results. 
 1st, a, {a — b) {a — c) {a — d) 
 2d,b, {b — a) {b — c) {b — d) 
 3d, c, {c — a) {c — b) {c — d) 
 ^th,d, (d — a) {d — b) {d--c) 
 
 Sec., &c. 
 and by consequence the same changes of sign will result from 
 a like substitution in (2). 
 
 If then we put (2) = 0, and write x for x', it becomes 
 
 wx"-i+(?z — 1) Aa:"-2 + B(?i — 2)a:"-3+ T = 0. (4) 
 
 -And since, from what has been done, the substitution of a, b, c, 
 &c., in this last gives results alternately positive and negative, 
 its roots will be intermediate between these quantities, that is, 
 between the roots of equation (1). 
 
 Equation (4) will, therefore, be the limiting equation to equa- 
 tion (1) ; and since the former is the derivative of the latter, 
 in general, the derivative of an equation will be its limiting 
 equation. 
 
 Ex. 1. What is the limiting equation to a;^ — 1 x^ -\-bx^ -\- 
 31a:-30 = 0? Ans. 4a:3 — 21 a:2+ 10a:+ 31 = 0. 
 
 248. The roots of equation (1) in the preceding article, being 
 
 represented by a, b, c, &c., respectively, if b', c', d', &c., in like 
 
 manner represent the roots of equation (4), the roots of the two 
 
 equations arranged in the order of their magnitude will stand 
 
 thus : 
 
 a, b', b, cfj c, d', &c. 
 
GENEEAL THEORY OF EQUATIONS. 301 
 
 Here, if the difference between a. and b is zero, the difference 
 between a and b' must also be zero ; that is, if a is equal to i, it 
 must also be equal to b' ; hence the factor x — a will be found 
 both in the proposed and its limiting equation. The two equa- 
 tions will, therefore, have a common measure x — a. 
 
 Again, if b and c are each equal to a, then b', c', will also be 
 each equal to a, and the two equations will have a common 
 measure {x — of, and so on. 
 
 Conversely, if a proposed equation and its derivative have a 
 common measure x — «, the proposed will have two roots, each 
 equal to a. If they have a common measure, {x — af, the 
 proposed will have three roots, each equal to a, and so on. 
 
 To determine, therefore, whether an equation has equal roots, 
 we form the derivative or limiting equation^ and then seek the 
 greatest commxm divisor of the two equations. 
 
 The factors of this common divisor being determined, it is 
 evident they must enter each once more into the proposed, and 
 thus the number and value of the equal roots will be deter- 
 mined. 
 
 Suppose the greatest common divisor of the proposed and its 
 derivative to be {x — af {x — bf {x — c). The proposed will 
 have 4 roots equal to a, 3 equal to b, and 2 equal to c. 
 
 Ex. 1. Find the equal roots of the equation 3 a:* — 10 a:' 
 + 15 a; + 8 = 0. 
 
 The derivative is 15 a:* — 30 a:^ -j- 15 = 0. And the greatest 
 common divisor is x^ -\- 2 x -{- \^ ox^ {x '\- \f. The proposed, 
 therefore, has 2 roots equal each to — 1. 
 
 Ex. 2. Find the equal roots of the equation x* — 14 a:^ -|-61 
 a:2_84a; + 36 = 0. 
 
 The greatest common divisor between the proposed and its 
 derivative is ar' — 7 a; + 6 = (a; — 6) {x — 1). 
 
 Ans. Two roots equal to 6 and two equal to 1. 
 
 Ex. 3. Find the equal roots of the equation a:* — 2 a:* -|- 1 
 = 0. Ans. Two equal to 1, and two equal to — 1. 
 
 26 , 
 
302 ELEMENTS OF ALGEBRA. 
 
 Ex. 4. Find the equal roots of the equation a:^ — 2x^ — 4 a: 
 + 8 = 0. Ans. Two equal to 2. 
 
 Ex. 5. Find the equal roots of the equation a:* — 6 a:' + 8 
 x^ -}- 6x — 9 = 0. Ans. Two equal to 3. 
 
 IMAGINARY AND REAL ROOTS. STURM's THEOREM. 
 
 249. In the search for the roots of an equation, it. is of the 
 first importance to determine, at the outset, the precise number 
 of real roots the proposed equation contains, m order that the 
 substitutions required in the solution may be restricted within 
 the narrowest possible limits. To separate the real and im- 
 aginary roots of a proposed equation so as to determine the 
 exact number of each, is a problem of great and acknowledged 
 difficulty. It entirely baffled the skill of mathematicians, until 
 in 1829 it was completely solved by the beautiful Theorem of 
 Sturm, which w^e shall now explain. 
 
 LetV = a:" + Aa:"-i + Ba:"-2+ . . . Ta: + U = 0, 
 be an equation of the nth. degree with no equal roots, and Vj = 
 be its derivative or limiting equation. 
 
 We now apply to V and V^ the process for finding their greatest 
 common divisor until a remainder is obtained independent of a?, 
 observing, however, to change the signs of each remainder as 
 we proceed. 
 
 Let the series of remainders, with their signs changed, be 
 represented by Vg, Vg, V4 . . . V„, V„ being the last remainder, 
 or that which is independent of x. 
 
 Let p and q be any numbers taken at pleasure, of which p is 
 the greater. Let p be substituted in the place of x in the func- 
 tions V, Vi, V2 V„, and write in order in one line 
 
 the signs of the results. Substitute next q in the same manner, 
 and write in order the signs of the results. The difference in 
 the number of variations between the two rows of signs resulting 
 from these substitutions, will be equal to the number of real roots 
 of the equation V=;= 0, comprised between the nurribers p and q. 
 
GENERAL THEORY OF EQUATIONS. 303 
 
 This is the Theorem of Sturm, which we now proceed to 
 demonstrate. 
 
 In order to this, let Q, Qi, Q2, &c., represent the quotients ob- 
 tained in the successive divisions of V by Vj, Vj by Vg, and so 
 on. From the manner in which these functions are derived, 
 we shall have this series of equations : 
 V=Q V,-V/ 
 
 * n-2 = Qn-2 * n-1 » n 
 
 This being premised, we remark • 
 
 1°. No two coTisecutive functions can become 0, or vanish^ far 
 the same value ofx. 
 
 For if two consecutive functions Vg, Vg, for example, can at 
 the same time become equal to 0, then we shall have ¥4 = 0; 
 and Vg, V4 being each equal to 0, then V5 will be equal to 0, 
 and so on, until finally V„ or the last remainder will be equal to 
 0, which is impossible, since this remainder is independent of 2:, 
 and cannot be affected by any change in the value of x. 
 
 2°. If one of the functions, Vg, for example, becomes for a 
 particular value of x, then the adjacent functions between which 
 it is placed, have for that value contrary signs. 
 
 For we have V2==Q2 V3 — V^; 
 hence, if V3 = 0, Vg = — V4. 
 
 Thus the adjacent functions have in this case contrary signs. 
 
 Let now p be greater than the greatest root, and negative 
 roots being regarded as less than corresponding positive ones, 
 let q be less than the least root of the equations 
 
 V = 0,Vi = 0,V2 = v_2 = o, 
 
 and let q increasing by insensible degrees until it reaches the 
 value of p, be substituted successively for x in these equations. 
 We remark again, 
 
 1°. So long as q remains less than the least root of the equa- 
 
304 ELEMENTS OF ALGEBRA. 
 
 tions, no change will be produced in the signs by its substitu- 
 tion. 
 
 2°. But when q in the process of its increase becomes equal 
 to the least root of the equations, in Vg, for example, then for 
 this value V3 vanishes. But there will be no change in the 
 number of variations of the signs produced by this circumstance. 
 
 Indeed, q being still less than the least root of Vg and V4, the 
 signs of these will not change by the substitution which causes 
 Vg to vanish. And since when Vg vanishes, Vg and V4 have 
 necessarily opposite signs, the signs of the three consecutive 
 functions must before have been, either 
 
 + =F - 1 " } - ± + 
 which give each, whichever of the double signs is the true one, 
 one permanence and one variation. 
 
 Now, when Vg vanishes, the signs become 
 
 V2, V 3, V 4 ) ( V2, Vg, V4 
 
 + - 1 " I - + 
 
 in each of which there is still one variation. 
 
 If q now becomes greater than the least root of Vg, but is still 
 less than the roots of Vg, V4, the signs of Vg, V4 will remain as 
 they were before ; but the sign of Vg will change. The three 
 consecutive signs will then be 
 
 + db— ,or — =F+, 
 still exhibiting^one permanence and one variation. 
 
 The same reasoning obviously applies to any of the inter- 
 mediate functions between V and V„. 
 
 The function V„, being the last remainder and independent 
 of a;, can undergo no change of sign, whatever value is sub- 
 stituted for X. It follows, therefore, that there can he no change 
 in the number of variations of the signs, unless it arise from a 
 change of sign in the primitive function. 
 
 2r°. Of the two equations V = 0, Vi = 0, one will necessarily 
 be of an even degree and the other odd. If, therefore, a number 
 less than their least roots be substituted in them, the results 
 
GENERAL THEORY OF EQUATIONS. 305 
 
 (art. 245) will be of different signs. Let us then suppose next 
 that the value of q in its increase has now become greater than 
 the least root of V = 0, but is still less than the least root of 
 Vi = 0, the roots of the latter (art. 247) being necessarily- 
 greater than the least root of the former. When q passes 
 the least root of V = 0, there will be a change of sign in this 
 equation ; thus the results of the substitution in the equations 
 V =i 0, Vi t= will now have the same sign. That is, the 
 results of the substitution in these two equations, which, before 
 exhibited a variation, will, in its stead, exhibit a permanence. 
 And by consequence the whole number of variations is in this 
 case diminished by unity. 
 
 If q goes on to increase until it has passed the least root of 
 Vi = 0, this function will change sign, so that V and Vi will 
 give different signs, but will again have the same sign when the 
 second root of V = is passed. Thus there will be no change 
 in the number of variations until the second root of V = is 
 passed, when the number of variations will again be diminished 
 by unity. 
 
 In like manner it may be shown that when q in its progress 
 towards p passes successively each of the remaining roots of the 
 primitive function V = 0, the number of variations in each case 
 will be diminished by unity. 
 
 It follows, therefore, since no change in the number of varia- 
 tions is made when any of the functions except the primitive 
 are reduced to zero, that the difference in the number of varia- 
 tions, when p and q are successively substituted in the functions 
 V, Vj, Vg, (f*c., will always be equal to the number of real roots 
 comprised between p and q. 
 
 This is the proposition which was to be demonstrated. We 
 will illustrate by an example. 
 
 For this purpose let us form the equation whose roots shall be 
 1, 2, 3, and 4. It will be a:* — 10 3:^ + 35 a;^ - 50 a; + 24 =0. 
 
 The functions formed according to the rule, and their roots, 
 
 when the functions are made equal to 0, will be as follows : 
 20 
 
306 ELEMENTS OF ALGEBRA. 
 
 Functions. ' Roots. 
 
 V := 2;4_10a:«-f-35a:'^ — 50a; + 24 1, 2, 3, 4 
 
 Vi = 4 a:^ — 30 a:^ + 70 a: — 50 1.3, 2.5, 3.6 nearly 
 
 V2 = 5a;2 — 25a; +29 1.8,3.1 
 
 V8 = 2a: - 5 2.5 
 V, = 9 
 
 Since the least root of the functions is 1, we will begin the 
 
 substitutions with a number less than this .8, for example. 
 
 
 
 V Vi 
 
 V, 
 
 Va 
 
 V4 
 
 
 
 = .8 i 
 
 gives -\- — 
 
 + 
 
 — 
 
 + 
 
 4 variations. 
 
 .9 
 
 (( 
 
 + - 
 
 + 
 
 — 
 
 + 
 
 4 
 
 (( 
 
 1 
 
 <( 
 
 — 
 
 + 
 
 — 
 
 + 
 
 
 
 1.1 
 
 (( 
 
 — — 
 
 + 
 
 — 
 
 + 
 
 3 
 
 (( 
 
 1.9 
 
 (( 
 
 - + 
 
 — 
 
 — 
 
 + 
 
 3 
 
 (( 
 
 2 
 
 (( 
 
 + 
 
 — 
 
 — 
 
 + 
 
 
 
 2.1 
 
 (( 
 
 + + 
 
 — 
 
 — 
 
 + 
 
 2 
 
 (( 
 
 2.5 
 
 (( 
 
 + 
 
 — 
 
 
 
 + 
 
 2 
 
 it 
 
 2.9 
 
 (( 
 
 + - 
 
 — 
 
 + 
 
 + 
 
 2 
 
 « 
 
 3 
 
 (( 
 
 — 
 
 — 
 
 + 
 
 + 
 
 
 
 3.1 
 
 (( 
 
 — — 
 
 — 
 
 + 
 
 + 
 
 1 variation. 
 
 3.9 
 
 (( 
 
 - + 
 
 + 
 
 + 
 
 + 
 
 1 
 
 (( 
 
 4 
 
 (( 
 
 + 
 
 + 
 
 + 
 
 + 
 
 
 
 4.1 
 
 (( 
 
 + + 
 
 + 
 
 + 
 
 + 
 
 no 
 
 variation. 
 
 From inspection of this table, we see 
 
 1*. So long as the value substituted for x is less than the 
 least root of the functions there is no change in the signs. 
 
 2°. When in the process of the substitution Vj, V3 vanish, 
 the adjacent functions are of opposite signs, and no change 
 takes place in the number of the variations. 
 
 3®. That whatever changes of sign take place in the secondary 
 functions Vi, &c., the number of variations is not affected by 
 these changes. 
 
 4°. That in every case when the primitive function changes 
 sign, there is a loss of one variation and of one only. 
 
 5°. The roots of the equation being comprised within the 
 
GENERAL THEORY OF EQUATIONS. 307 
 
 limits of the values assigned to :c, the number of variations 
 lost is precisely equal to the number of the real roots of die 
 equation. 
 
 250. If the number only of the real roots is required, it will 
 be sufficient to substitute -|- oo, and — QO, the extreme limits, 
 instead of x, 
 
 Ex. 1. Required the number of real roots in the equation 
 a:3_92,2_|_23a;_i5==o. 
 We shall have for the functions 
 
 V== a?- 9a;2 + 23a;— 15' 
 Vi = 3a:2— 18a:+23 
 V2= a; - 3 
 
 V3=:4. . 
 
 Substituting in these functions -j- oo and — oo successively, 
 we have the following results : 
 
 a:= 4- 00 gives -f- -|- -j- -}- No variations. 
 
 xz=. — 00 " 1 1- 3 variations. 
 
 Here the difference of the variations is 3. There will be, 
 therefore, three real roots in the proposed equation. And as 
 the equation contains no permanence, the three roots will all be 
 positive. 
 
 If the situation as well as number of the roots is required, we 
 proceed as follows. Beginning with 0, we substitute the series 
 of natural numbers 1, 2, 3, 4, &c., for x in the functions, thus, 
 V Vi V^ V3 
 
 a: = gives 1 \- 
 
 3 variations. 
 
 
 x = l " 
 
 0+- + 
 
 2 
 
 
 a: = 2 " 
 
 + + 
 
 2 
 
 
 x = S " 
 
 — + 
 
 1 variation. 
 
 
 a: = 4 " 
 
 — + + 
 
 1 
 
 
 x = b " 
 
 0+ + + 
 
 no variation. 
 
 
 Since the numbers 
 
 1, 3, 5, reduce 
 
 the proposed 
 
 to 0, these 
 
 are the roots of the et 
 
 luation, which - 
 
 are thus completely deter- 
 
 mined by the process. 
 
 
 
 
308 
 
 ELEMEJiTS OF ALGEBRA. 
 
 Ex. 2. To find the number and initial figures of the roots of 
 the equation a;^ — 4:x'^ — 6x-\-8=0, « 
 
 The functions are V = x^ — 4^x^ — 6 x-\-& 
 
 Vi= 3a:2_8a:— 6 
 
 V2=17a: —12 
 
 The sign only of Vg is written, since this is all that is neces- 
 sary, and this may be determined without actually performing 
 the division. 
 
 Substituting + c», and — oo, 
 
 x = -\-<X) gives -f- 4" + ~f" ^^ variations. 
 x = — 00 " 1 [-3 variations. 
 
 There will be, therefore, three real roots ; and as there is one 
 permanence in the proposed, one of these will be negative. 
 
 To determine the situation of the roots, we substitute the 
 series of natural numbers 0, 1, 2, &c., positive and negative, 
 thus: 
 
 X = gives -\ 
 
 '2Vs 
 
 7ar. 
 2 
 
 1 
 
 a: = 2 " 
 
 
 
 1 
 
 
 
 
 1 
 
 1 
 
 
 xz= gives -j f- 
 
 x = ^l « + + - + 
 
 X=:—2 » 1 }- 
 
 Van 
 2 
 2 
 3 
 
 From the column of variations, it is evident that the roots lie 
 between and 1, 4 and 5,-1 and — 2. The initial figures 
 of the roots are, therefore, 0, 4, and — 1. To obtain the first 
 decimal figure of the root between and 1, we substitute in 
 the order of tenths. And since 1 is very near the root we begin 
 with .9. Thus 
 
 x=l gives 1- 4" 1 variation. 
 
 xsss.9 " -\ h+^ variations. 
 
 The initial figures of the roots are, therefore, .9, 4 and — 1. 
 
 Ex. 3. How many real roots has the equation a^ — 6x^-j- 
 11 X — 6 = » Ans. Three, viz., 1, 2, and 3, 
 
GENERAL THEORY OF EQUATIONS. 309 
 
 Ex, 4. How many real roots has the equation a:^ — 5 a:^ -|" 
 6a;— 1=0? Ans. 1. 
 
 Ex. 5. How many real roots has the equation 3^ — 7 a: + 
 7 = 0? 
 Ans. 3. Two between 1 and 2, and one between — 3 and — 4. 
 
 SECTION XXVII. — Solution of numerical equations of 
 
 ANY DECREE. 
 
 The principles now obtained are sufficient for the solution of 
 numerical equations of any degree with one unknown quantity. 
 We proceed to apply them. 
 
 commensurable roots. 
 
 251. A commensurable root, it will be recollected, is one 
 which has a common measure with unity, h may, therefore, 
 be an integer, or a definite fraction, such as, \, ^, ^, &c. We 
 commence with the research for integral roots. 
 
 Since every root of an equation (art. 229) is a divisor of the 
 last or absolute term, the integral roots of equations must, as we 
 have seen, be sought among the entire divisors of this term. 
 They will be comprised, moreover, between the limits of the 
 roots. Every equation, the coefficients of which are entire 
 numbers and that of the first term equal to unity must have 
 (art. 230) entire numbers for its commensurable roots. We 
 begin with equations of this description. 
 
 Ex. 1. Let it then be proposed to find the integral roots of 
 the equation ar^ — 5 a;*+ ar^ -f- 16 a:^ — 20 a: -)- 16 = 0. 
 
 The limits of the roots are 5 and — 4. The integral roots 
 of the equation must, therefore, be found among the entire divi- 
 sors of 16 comprised between 5 and — 4. These are 4, 2, 1, 
 — 1, — 2, — 4. But 1 and — 1, it will be seen at once, are 
 
310 
 
 ELEMENTS OF ALGEBRA. 
 
 not roots, since they will not satisfy the equation. The trial 
 
 divisors, are, therefore, reduced to 4, 2, — 2, — 4. These we 
 
 try in sucpession. 
 
 1-5 + 1 + 16 — 20+16 
 + 4-4—12+16—16 
 
 _1_3+ 4— 4 
 
 2 + 2— 2+ 4 
 
 — 2 
 
 1_1+ 2 
 
 2 + 2— 2 
 
 1-1 + 1, 
 
 From the operation, 4, it is evident, is a root, since the divis- 
 ion by a: — 4 leaves no remainder. Dividing the quotient of 
 this division by a: — 2, there is no remainder ; 2 is, therefore, a 
 root. Dividing next this last quotient by x-\-2, there is no 
 remainder ; hence — 2 is a root. We proceed no further with 
 the negative divisors, since the equation exhibiting but one per- 
 manence can have but one negative root, and that is now found. 
 The coefficients of the remainder after these successive divisions 
 are 1 — 1 + 1.^ Hence we shall have fpr the depressed equa- 
 tion containing the remaining roots of the proposed, x^ — a: + 1 
 == 0, the roots of which are imaginary. 
 
 Ex. 2. What are the integral roots of the equation 
 a;3__i0a;2 + 31ar — 30 = 0? 
 
 Since the equation exhibits only variations of sign there can 
 be no negative roots. It will be necessary, therefore, to find 
 only the superior limit of the positive roots, and to employ the 
 divisors of 30 between and this limit. The limit is 10 , and 
 since 1 is obviously not a root, the trial divisors will be 2, 3, 5, 
 and 6, by means of which we obtain 2, 3, and 5, for the roots. 
 And the equation being of the third degree only, these are all 
 the roots, and the equation is completely solved. 
 
 Ex. 3. What are the integral roots of the equation x^ — 10 
 a,-5 + 37a;2 — 60a; + 36==0? 
 
 The equation has no negative roots. Applying the process 
 we obtain 2 and 3 for the positive roots. These, it is evident^ 
 
GENERAL THEORY OF EQUATIONS. 311 
 
 may be equal roots. To discover whether this is the case, we 
 continue the division by 2 and 3 upon the depressed equation 
 after taking out the roots 2 and 3. Or we substitute 2 and 3 
 successively in the derivative of the proposed. And since they 
 satisfy this equation also, they are (art. 247) equal roots. Thus 
 the integral roots of the proposed are 2, 2, 3, 3 ; and since it is 
 of the fourth degree only, these are all its roots. 
 
 252. When the number of trial divisors is large, the process 
 is laborious. It will admit of some simplification in the manner 
 we will now explain. 
 
 Let it be proposed, for example, to find the integral roots of 
 the equation a^ — I6x^ + 14:X — 120 = 0. 
 
 The equation has no negative roots. The superior limit to 
 the positive roots is 15 ; and since 1 is obviously not a root, the 
 trial divisors will be 12, 10, 8, 6, 5, 4, 3, 2. 
 
 In making trial of the divisors it will be found most conven- 
 ient to invert the order of the coefficients, ffiat is, arrange with 
 reference to the ascending powers of x, and then change the 
 signs of the divisor, the effect of which will be merely to change 
 the signs of the quotient. 
 
 Thus, in the present example, if we begin with 3, the work 
 will be 
 
 _ 120 + 74— 15+1 
 — 40 
 
 34. 
 
 3 is not a root, since 34 is not divisible by 3, and the total divis- 
 ion by 3 is, therefore, impossible. Try next 4. 
 
 -120 + 74-15 + 1 
 _30 + ll_l 
 
 44—4 0. 
 
 4 is a root, since there is no remainder. Proceeding with the 
 process we find 5 and 6 for the remaining roots. 
 
 If the operations performed above be examined with attention, 
 the following principle, which may easily be proved to be 
 general, will be discovered, viz. : In order that a number may 
 be a root, the first coefficient or absolute term, must be divisible 
 
312 ELEMENTS OF ALGEBRA. 
 
 by it, so must also the sum of the quotient and the next co- 
 dfficient, the sum of this last quotient and the next coefficient, 
 and so on throughout, the last quotient being always — 1. 
 Any number which will not sustain these tests in succession is 
 not a root. 
 
 Taking advantage of what has now been said, the entire 
 work in the example proposed may be conveniently performed 
 as follows : 
 
 -120 
 
 12, 
 
 10, 
 
 8, 6, 5, 4, 
 
 3, 
 
 2, 
 
 + 74 
 
 -10,- 
 64, 
 
 -12,- 
 62, 
 
 . 15, _ 20, — 24, — 30,- 
 59, 54, 50, 44, 
 
 -40,- 
 34, 
 
 -60, 
 14, 
 
 — 15 
 
 ^ 
 
 # 
 
 ^ 9, 10, 11, 
 _6,_ 5,- 4, 
 
 # 
 
 7 
 — 8 
 
 + 1 
 
 
 
 -1,- 1,- 1, 
 
 • 
 
 — 4 
 
 — 3 
 
 The coefficient* of the proposed are placed in a vertical 
 column with their proper signs. On the same horizontal line 
 with the first coefficient, or absolute term, are arranged the trial 
 divisors. Beneath these, in the same horizontal line with the 
 second coefficient, are placed the quotients arising from the 
 division of the absolute term by the trial divisors. Beneath 
 these, in the next line, are placed the sums of the first quotient 
 and second coefficients. These are the dividends to be divided 
 next by the trial divisors ; and the quotients of this division are 
 placed beneath them in the next line below, and in the same 
 line with the third coefficient, and so on in order. 
 
 On the second division the divisors 12, 10, 8, and 3 are re- 
 jected, and the divisor 2 on the fourth. The only divisors which 
 sustain the required tests are 6, 5, and 4, and these are, there- 
 fore, roots of the equation. 
 
 The following examples will serve as an additional exercise. 
 
 Ex. 1. What are the integral roots of the equation a^ -\-3 
 a:2_8a;4_iO = 0? Ans. — 5. 
 
 Ex. 2. What are the integral roots of the equation x^ -\- 4: 
 a^^x'—\Qx — l2 = 0'i Ans. 2, — 1, — 2, and —3. 
 
GENERAL THEORY OF EQUATIONS. 313 
 
 Ex. 3. What are the integral roots of the equation x^ — a:* 
 — 13a:2_|_i6a._4Q^0? Ans. 4,and— 4. 
 
 Ex. 4. What are the integral roots of the equation x^ — 7 
 d?J^nx^-^Y\x-\-^^^% Ans. 1, l,2,and3. 
 
 253. The method above is applicable also to equations, the 
 coefficients of which are entire numbers and that of the first 
 term difierent from unity. In this case the last quotient will 
 be — 1 multiplied by the coefficient of the first term. 
 
 Ex. 1. Find the integral roots of the equation 2 a:^ — 22 a;'' 
 4-62a: — 42 = 0. Ans. 1, 3, and 7. 
 
 Ex. 2. What are the integral roots of the equation 3 a;^ — 23 
 a:2_|_44a,_2o==0? Ans. 2 and 5. 
 
 254. We proceed next to equations which contain fractional 
 commensurable roots. 
 
 Let it be proposed, for example, to find the commensurable 
 roots of the equation 
 
 To find the integral roots, we free the equation from denomi- 
 nators, which will not alter the value of the roots. This gives 
 24a;*_ 74:c3 + 61 a;2 — 19 a: + 2 = 0. 
 
 Applying next the process above to this equation we obtain 
 2 for the integral root. " Removing this from the proposed, the* 
 reduced equation will be 
 
 Since 2 is the only integral root of the proposed, the commen- 
 surable roots of this last, if it have any, must be fractional. 
 To determine these we transform the equation into another 
 whose roots shall be integral. In order to this (art. 240, no. 2) 
 
 X . 
 
 we put a; = — , 24 being the least common multiple of the de- 
 nominators, and we have for the transformed 
 
 a:s _ 26 a:2 + 2 16 a: — 576 = 0. (3) 
 Applying the process for integral roots to this equation the 
 27 
 
314 ELEMENTS OF ALGEBRA. 
 
 roots are found to be 12, 8, and 6. But these, it is evident, 
 are 24 times larger than those of equation (2). Hence the 
 roots of (2) are ^, ^, and ^. 
 
 The roots of the proposed equation will be, therefore, 2, ^, ^, 
 and ^. 
 
 Ex. 2. Find the commensurable roots of the equation a? — 
 31 1 1 
 
 3Q^' + 3^^ - 30=^^- ^"s- h h and f 
 
 Ex. 3. Find the commensurable roots of the equation 3 a:^ — 
 14 a:2 + 21 a; - 10 = 0. Ans. 1, |, and 2. 
 
 Ex. 4. What are the commensurable roots of the equation 
 12a;*4-20a;3- lla:2-5a; + 2 = 0? 
 
 Ans. ^, ^, — I, and — 2. 
 
 INCOMMENSURABLE ROOTS. 
 
 255. Incommensurable Eoots are those which have no com- 
 mon measure with unity, such as surds and interminable deci- 
 mals. 
 
 We proceed next to equations having incommensurable roots. 
 These may be found by the principles explained above to any 
 degree of approximation we please. 
 
 Let it be proposed, for example, to find the roots of the equa- 
 tion a:^ _|. ^ ^2 _ io2 z + 181 = 0. 
 
 The substitution of + oo and — 00 shows that the equation 
 has three real roots. And since there is but one permanence, 
 two of the roots will be positive. The functions are 
 
 V= a:3_[_ii^_io2a:+181 
 Vi== 3a:2-t-22a: — 102 
 ¥2= 122 a; — 393 
 
 Putting a; = gives -| \- 2 variations. 
 
 a:=l " + + 
 
 a; = 2 " -| (- 
 
 a; = 3 « -j [- 2 « - 
 
 2: = 4 " -f- -|- _|- ._j_ no variation. 
 
GENERAL THEORY OF EQUATIONS. 315 
 
 The two positive roots are, therefore, comprised between 3 
 and 4. To approach them more nearly, we transform the pro- 
 posed equation into another whose roots shall be less by 3. 
 The functions will be, 
 
 V= a^^20x^ -Qx+l 
 
 Vi= Sz'-\-4S)x-9 
 
 V2=122a; — 27 
 
 V3 = +. ■ . 
 
 Putting in these 
 
 a; = gives -\ \- 2 variations. 
 
 x = .l « + + 
 
 x = .2 « -| 1- 2 « 
 
 x = .3 " + -h "t" -f* ^° variation. 
 The two positive roots of the reduced equation are comprised, 
 therefore, between .2 and .3. Hence those of the proposed are 
 comprised between 3.2 and 3.3. 
 
 To approach the roots still nearer we transform the last 
 transformed equation, a^ -\-20 x^ — 9 a; -f- 1> into another, the 
 Toots of which shall be less by .2. The functions will then be 
 V = u^ -\-20.6 x^ — .S8x -}- .008 
 \\= 3x' + 4.l.2x^,Q8 
 
 Va=122a: — 2.6 
 
 
 v,=+. 
 
 
 In these ,x= gives -| \- 
 
 2 variations. 
 
 x = m " + f- 
 
 2 
 
 a; = .02 « 1- 
 
 1 variation. 
 
 ^ = .03 " + + + + 
 
 no variation. 
 
 The two positive roots of the last transformed are comprised, 
 therefore, hetween .01, .02 and .02, .03. By consequence the 
 first three figures in those of the proposed are 3.21, and 3.22 ; 
 and since the sum of the three roots is — 11, the negative root 
 will be — 17.43. We shall have, therefore, for the approximate 
 roots of the proposed 3.21, 3.22, and — 17.43. 
 
 By the process above we may approach the roots of an equa- 
 tion as nearly as we please. The process is, however, laborious, 
 
316 ELEMENTS OF ALGEBRA. * 
 
 and may be much abridged. Th^ following method of accom- 
 plishing this object was first published by W. G. Horner, of 
 Bath, England, in 1819. 
 
 Horner's method of approximation. 
 
 256. This method is founded upon the following principles. 
 
 Let there be the equation 
 
 V==a;'^ + Aa;"-i+ Ta: + U = 0. 
 
 Let x' be the part of the root of this equation already found, 
 and y the remaining part, y being very small compared with x'. 
 Then, transforming the proposed into another, the roots of wjiich 
 shall be less by x\ we obtain 
 
 V' = 2/" + A'2/"-^+ . .*. . r2/ + U' = 0; 
 then, since 2/ is a very small quantity, all the terms of this equa- 
 tion, in which the power of y is above the first, may be neglected, 
 and the equation T' ?/ -(- U' = 0, will give the value of y very 
 nearly. Resolving this equation we have 
 
 U' 
 
 2/ = — ^• 
 
 Thus, in the equation a:^ — 5 o;^ -)- 8 a: — 1 = 0, the value of 
 a:, it is easy to see, lies between and 1. Neglecting the terms 
 which involve x above the first power, we have for a trial equa- 
 tion 8 a; — 1 = 0, from which we obtain x = .125. The first 
 figure of this decimal is true, since by substitution it will be 
 found that the value of x lies between .1 and .2. Thus the 
 first figure of the approximate root is easily determined. In 
 order to proceed to the next, we transform the equation V = 
 into another whose roots shall be less by the part last found. 
 This will give a new trial equation, by which to find the second 
 figure of the root, and so on. The rule may be thus stated : 
 
 1°. Find by trial, or by Sturm's Theorem, the situation and 
 first figure of the real roots. 
 
 2°. Transform the proposed equation into another, whose 
 roots shall be less than those of the given equation by the part 
 of the root already found. 
 
GENERAL THEORY OF EQUATIONS. 317 
 
 3°. With the absolute term in this transformed equation for 
 a dividend, and the coefficient of x for a divisor, find the next 
 figure of the root and verify it in the transformed equation. 
 
 4°. Diminish the roots of the transformed equation by the 
 value of the figure last obtained, divide as before for the next 
 figure, and so on. 
 
 Ex. 1. Find the roots of the equation y? -\- \(i y? -\- h x — 
 260 = 0. The functions will be 
 
 Vi = 3a;2-j-20a; +5 
 ¥3= 17a; + 239 
 
 V3 = -. 
 
 Substituting in these functions as above, the equation, it will 
 be found, has but one real root, the first figure of which is 4. 
 Transforming it into another, the roots of which shall be less 
 by 4, the operation will be 
 
 1 I 1 +10 4-5 —260 
 4| 4 56 244 
 
 14 61 -16 
 
 _4 J72 
 
 18 133 
 
 22 
 
 and the transformed will be a:^ + 22 a:^ -f- 133 a; — 16 = 0. (2) 
 We have, therefore, for the first trial equation 133 a; — 16 
 = 0, which gives .1 for the second figure of the root. Thus 
 the root of the proposed will be 4.1 nearly. Transforming 
 next the equation (2) into another, whose roots shall be less by 
 .1, the operation will be 
 
 1|1 _i.22. +133 —16 
 
 .1 I 0.1 2.21 13.521 
 
 22.1 135.21 -2.479 
 
 .1 2.22 
 
 22.2 137.43 
 
 .1 
 
 22.3 
 27* 
 
318 ELEMENTS OF ALGEBRA. 
 
 and the transformed will be a:^ + 22.3 a;2_|- 137.43^ — 2.479 
 = 0. (3) 
 
 We have, therefore, for the next trial equation, 137.43 a: — 
 2.479 = 0, which gives .01 for the next figure of the root, 
 The root will be, therefore, 4.11 nearly. Transforming, next, 
 equation (3) into another, whose roots shall be less by .01, the 
 operation will be 
 
 1 
 
 .01 
 
 1 
 
 + 22.3 
 
 .01 
 
 + 137.43 
 .2231 
 
 — 2.479 
 1.376531 
 
 
 22.31 
 .01 
 
 137.6531 
 .2232 
 
 — 1.102469 
 
 
 
 22.32 
 .01 
 
 137.8763 
 
 
 22.33 
 
 and the transformed will be 
 
 2^ + 22.33 x^ + 137.8763 x — 1.102469 = 0. (4) 
 And we have for the next trial equation 137.8763 x — 
 1.102469 = 0, by which we obtain .007 for the next figure of 
 the root. Transforming (4) into another, whose roots shall be 
 less by .007, the operation will be 
 
 1 
 
 .007 
 
 1 
 
 + 22.33 
 .007 
 
 H 
 
 • 137.8763 
 .156359 
 
 - 1.102469 
 .966228613 
 
 
 22.337 
 .007 
 
 
 138.032659 
 .156408 
 
 - .13624038^ 
 
 
 
 22.344 
 .007 
 
 
 138.189067 
 
 
 22.351 
 
 and the transformed will be 
 
 a^ + 22.351 a:^ + 138.189067 x - .136240387 = 0, (5) 
 and the next trial equation is 138.189067 a; — .136240387 = 0, 
 from which we obtain .0009 for the next figure of the root. 
 The root of the proposed will be, therefore, 4.1179 nearly. In 
 like manner the approximation may be pushed as far as we 
 please. 
 
GENERAL THEORY OF EQUATIONS. 
 
 31» 
 
 The calculations may be performed more concisely as in the 
 following table : 
 
 1 
 
 4 
 
 1 +10 
 4 
 
 56 
 
 - 260 1 4.11 
 244 
 
 
 14 
 4 
 
 61 
 
 72 
 
 ^-16 
 13.521 
 
 
 18 
 4 
 
 «=133 
 2.21 
 
 #_ 2.479 
 1.376531 
 
 1. 
 .1 
 
 =^22.1 
 .1 
 
 135.21 
 2.22 
 
 =^ — 1.102469 
 
 .966228613 
 
 
 22.2 
 .1 
 
 ^ 137.43 
 .2231 
 
 «= — .136240387 
 
 1 
 .01 
 
 =^22.31 
 .01 
 
 137.6531 
 .2232 
 
 
 
 22.32 
 .01 
 
 ^ 137.8763 
 .156359 
 
 
 1 
 .007 
 
 ^ 22.337 
 
 7 
 
 138.032659 
 .156408 
 
 
 
 22.344 
 
 7 
 
 =^ 138.189067 
 
 
 =^22.351 
 
 The process, when compared with the previous work, will be 
 easily understood. The coefficients of the successive trans- 
 formed equations are marked with a star. The .1 placed at 
 the right of =^22 in the first column is the .1 added in the first 
 step of the process for obtaining the second transformed equa- 
 tion, the addition being more conveniently made in this manner. 
 A similar remark applies to the right hand figure of the other 
 coefficients in the same column. 
 
 The successive figures may be verified as they are obtained 
 in the transformed equation. The process, it is evident, be- 
 comes more accurate as we proceed. After four or more 
 decimals have been obtained, two or three more may in general 
 be found by simple division. 
 
 The sign of the last term will sometimes change in the 
 course of the operation. Unless there is in this case a change 
 
320 
 
 ELEMENTS OF ALGEBRA. 
 
 of sign in the preceding column also, the figure which has 
 given rise to the change must he incorrect. This change may 
 not, however, always occur at the same figure of the root. 
 
 257. In the preceding example a greater number of decimal 
 places has been employed than is necessary to obtain the root 
 to the degree of approximation attained, and the work might 
 have been abridged by the omission of some of them. 
 
 Let it be required, as a second example, to find the roots of 
 the equation 7^ — \1 x^ -\-b^z — 350 = 0, to three places of 
 decimals. 
 
 The equation has but one real root, the first two figure's of 
 which, found by trial, or Sturm's Theorem, are 14. The re- 
 mainder of the work, according to the rule, will be as follows : 
 
 1 
 
 14 
 
 1 —17 
 14 
 
 + 54 
 — 42 
 
 
 — 350 1 14.954 
 168 
 
 
 - 3 
 14 
 
 12 =^- 
 154 
 
 -182 
 170.379 
 
 
 11 
 14 
 
 =^25.9 
 .9 
 
 26.8 
 .9 
 
 =^166 ^. 
 23.31 
 
 189.31 
 24.12 
 
 =^ 213.413 
 
 1.3|875 
 
 - 11.621 
 10.740 
 
 875 
 
 1 
 .9 
 
 =^ — 880 
 865 
 
 125 
 275664 
 
 
 — 14 1 849336 
 
 1 
 
 .05 
 
 ^217.75 
 5 
 
 214.8 175 
 1.3 900 
 
 
 
 27.80 
 5 
 
 ^216. 
 
 2075 
 111416 
 
 
 1. 
 
 .00^ 
 
 ^27.854 
 \ 4 
 
 27.858 
 4 
 
 216. 
 
 318916 
 111432 
 
 
 
 216. 
 
 430348 
 
 
 27.862 
 
 If this work is examined with attention, it will be seen that 
 the result will still remain the same, if all the figures to the 
 right of the vertical lines, and those below the vertical line in 
 
GENERAL THEORY OF EQUATIONS. 
 
 821 
 
 the left hand column are omitted, by which the labor will be 
 much abridged. 
 
 The object is to retain no more decimal places in the last 
 column than are necessary ; and these, in general, will not be 
 greater than the number of places required in the root. Thus, 
 in the present example, three places only being required in the 
 root, we cut off in the last column, by a vertical line, all the 
 remaining figures after three places have been obtained. This 
 occurs after the operations with the figure 9 of the root have 
 been completed. Then, since the multiplication by each new 
 figure of the root produces one new decimal place in each 
 column as we proceed from left to right, in order that no new 
 decimal places may occur in the right hand column we must, 
 it is evident, cut off one figure in the column next preceding, 
 two figures in the column next preceding that, and so on. The 
 operation with the figure 5 of the root being completed, we 
 again cut off one figure in the column next preceding the last, 
 and two in the column next preceding that, and so on. The 
 work will then stand thus : 
 
 1 
 
 14 
 
 -17 
 14 
 
 54 
 
 -42 
 
 
 — 350 14. 
 168 
 
 
 -3 
 14 
 
 12 
 154 
 
 ^—182 
 170.379 
 
 
 11 
 14 
 
 =^^166 
 23.31 
 
 *— 11.621 
 10.741 
 
 1 
 .9 
 
 ^«=25.9 
 9 
 
 189.31 
 24.12 
 
 — .880 
 865 
 
 
 26.8 
 9 . 
 
 * 213.4 
 1.3 
 
 3 
 
 8 
 
 -15 
 15 
 
 1 
 
 .05 
 1 
 .004 
 
 #2 
 
 7.7 
 
 1 
 
 214.8 
 1.3 
 
 1 
 9 
 
 
 1-1.21 
 
 7.8 
 
 2|1|6|.2 
 
 
 
 Care, it is evident, must be taken that, in the result of each 
 operation, the figure immediately preceding the one cut off 
 remain the same as if the contraction were not made. Thus 
 = 0? 21 
 
322 ELEMENTS OF ALGEBRA. 
 
 in the operation with the figure 5 of the root, we continue the 
 use of the 77, cut off in the left hand column, until the operation 
 with the 5 is completed. We retain also, in the column next 
 following, one of the decimal places cut off. This gives, when 
 the operation with the 5 is completed, 216.2 in this column, 
 precisely as if the contraction were not made. Cutting off next 
 one figure, the 2, in the column before the last, and two in the 
 column before that, there will be none remaining in this column, 
 and the multiplication by 4, the next figure of the root, will 
 produce no effect upon the following columns. 
 
 Having obtained the 4, two additional figures may be 'found, 
 by simple division. In order to this, cutting off the 6 in the 
 column before the last, we have 21 for a divisor and 15 for a 
 dividend, which gives for the next figure of the root. Again 
 cutting off the 1 in the column before the last, we have 2 for a 
 divisor and 15 for a dividend, which gives 7 for the next figure 
 of the root; and the operation, true to 5 places of decimals, is 
 now terminated. 
 
 There is room for the exercise of judgment in the use of the 
 figures cut off. The object being to keep the right hand figure 
 in the result of each operation what it would be if the contrac- 
 tion were not made, the learner must judge, in each case, what 
 is necessary for this purpose. What has been done will serve 
 as a general direction. After a little practice the operations 
 will be easily executed, and the root obtained, to any degree of 
 approximation required, with extreme facility. 
 
 Ex. 3. To find the roots of the equation oi? — 7 a; -[- 7 = 0. 
 There will be one negative and two positive roots. To find the 
 negative root, we change the sign of the alternate terms, and 
 proceed as for a positive root. The result, with its sign changed, 
 will be the negative root sought. 
 
 The yots are 1.356895, 1.692021, and -- 3.048917, true to 
 six places of decimals. 
 
 Two of the roots being obtained, the other may be found by 
 
GENERAL THEORY OF EQUATIONS. 323 
 
 the principle, art. 229. Thus, to find the negative root, we take 
 the sum of the two positive roots with the contrary sign. 
 
 Ex. 4. Find one root of the equation a^-^-Sx^-^bx— 178 
 = 0, true to 6 places of decimals. Ans. 4.538825. 
 
 Ex. 5. Find the roots of the equation 3^ — 5x — 3 = 0, 
 trae to four places of decimals. 
 
 Ans. 2.4908, - 0.6566, - 1.8342. 
 
 Ex. 6. Find a root of the equation 3^-\-2 x^ — 23 a; — 70 = 
 0, true to four places of decimals. " Ans. 5.1345u 
 
 Ex. 7. Find the root of the equation x^-\-Sx* '}-2x^ — 3 x^ 
 — 2x — 2 = 0, to four places of decimals. Ans. 1.0591. 
 
 Ex. 8. Find the roots of the equation x* — x^ -\-2x^ -\- x — 4 
 = 0. Ans. 1.14699459, and — 1.0905935. 
 
 258. The preceding process may he applied to the extraction 
 of the roots of numbers. 
 
 Ex. 1. Let it be required to extract the third root of 9. 
 
 In this case, we have to solve the equation a:^ — 9 = 0, 
 which, by the preceding process, gives 2.0800838 for the 
 answer. 
 
 Ex. 2. To find the roots of the equation x"^ — 2=0. 
 
 Ans. ± 1.414213. 
 
 Ex. 3. Find the fifth root of 2. Ans. 1.148699. 
 
 SECTION XXVIII. Elimination. — Solution of equations 
 
 WITH two or more UNKNOWN QUANTITIES. 
 
 259. The equations of any degree, thus far solved, contain 
 one unknown quantity only. We proceed next to equations 
 with more than one unknown quantity. 
 
 Let there be two equations with two unknown quantities. It 
 is proposed to find the systems of values for the unknown 
 quantities x and y, that will satisfy these equations. In 
 order to this, we must first eliminate one of the unknown quan- 
 tities. 
 
 Let the equations, arranged in reference to a:, be represented 
 by A = 0,B = 0. 
 
324 ELEMENTS OF ALGEBRA. 
 
 Applying to these the principle of the greatest common divisor, 
 
 let Q = the quotient of A by B, and R = the remainder ; then 
 
 A = BQ + R. 
 
 It follows from this equality, that all the values of x and y, 
 which give A = 0, B = 0, must also give R = 0. The sys- 
 tem of equations A = 0, B = 0, may, therefore, be replaced 
 by the more simple system, B = 0, R = 0. 
 
 Dividing next B by R, let a new remainder R^ be reached. 
 We may, in like manner, substitute for B = 0, R = 0, the sys- 
 tem R = 0, R' = 0, in which R' is of a lower degree in respect 
 to X than R. And we may thus continue until a remainder, R", 
 for example, is obtained independent of x. The original sys- 
 tem, A = 0, B = 0, may then be replaced by the system R' = 
 0, R" = 0, in which R" contains y only, and R' is generally of 
 the first degree in x. The equation R" = 0, from which x is 
 eliminated, is called the JiTicd equation. 
 
 The values of y being obtained from the final equation, those 
 of X will be easily found ; and thus the systems of values for x 
 and y^,proper to satisfy the proposed equations, be determined. 
 
 Ex. 1. To find the values of 2: and y in the equations 
 
 X -^y — 8 = 0. 
 Applying the process of the greatest common divisor, 
 
 x'+{y^8)x 
 
 x + y-8 
 
 x — y + 8 
 
 ^{y — 8)x—f — M 
 
 — (y — 8)0: — y^-|-16y~64 
 
 2 2/2 _ 16 2^ _|_ 30 = Remainder. 
 Putting this remainder equal to 0, we have for the final equa- 
 tion, 2^^ — 8^+15 = 0. 
 
 Resolving this equation we obtain 2/ = 3or5; and, substi- 
 tuting in the divisor, we obtain for the corresponding values 
 of ar, a: = 5 or 3. Thus the system of values for x and y, 
 which satisfy the proposed equations, are y = 3, 2: = 5, and 
 y = 5,a: = 3. 
 
CENERAL THEOEY OF EQUATIONS. 325 
 
 Ex. 2. Find the values of x and y in the equations, 
 ^2/* + 2/" — 333 = 0, 
 xf^y— 21 = 0. 
 Ans. y = 3, 2; = 2, or y = 18, a: = j^,. 
 Ex. 3. Find the values of x and y in the equations 
 4a; — 22^+ 2/^—11 = 0, 
 a; + 42^ —14 = 0. 
 Ans. y = 3, a; = 2,*or y = 15, a: = — 46. 
 
 260. In the process for finding the greatest common divisor, 
 it may be necessary, in order that the division may be exactly 
 performed, to multiply one of the quantities by a factor contain- 
 ing y. In this way roots may be introduced into the final 
 equation foreign to the proposed equations, and which must be 
 rejected. 
 
 In like manner, for convenience, factors containing y may be 
 suppressed in the course of the operations, which, when put 
 equal to 0, may give values for y proper to satisfy the equations, 
 but which, from the suppression of the factors, will not be found 
 in the final equation. We must, therefore, in order to a com- 
 plete solution, make the factors introduced or suppressed equal 
 0. We shall thus establish relations which, with the final 
 equation, will enable us readily to detect the foreign solutions 
 and to determine all the values of the unknown quantities 
 proper to satisfy the proposed equations. 
 
 261. Various simplifications may be introduced into the opera- 
 tions, and the process improved so as to avoid the foreign solu- 
 tions. The general idea of the process we have given is all 
 our limits admit. We subjoin a few additional examples. 
 
 Ex. 1. To find the values of x and y in the equations, 
 
 x^^^xy^r 2/^-1 = 0. 
 Ans. 2/ = — 1, a: = 2, and y =: — 2, a: == 1. 
 Ex. 2. To find the values of x and y in the equations, 
 :^^'iyx^J^{^f-y+\)x-f-\-f-2y = ^, 
 a^—.2yx-\-f — y = 0, Ans, a; = 2, 7=L 
 
 28 
 
326 ELEMENTS OF ALGEBRA. 
 
 Ex. 3. To find the values of x and y in the equations, 
 
 Ans. The equations are incompatible. 
 It will be easy to see how we are to proceed with three equa- 
 tions with three unknown quantities, and so on. 
 
 SECTION. XXIX. Infinite Series. 
 
 262. An infinite series is one in which the number of terms 
 is UTilimited ; the law of the series being generally discoverable 
 by an examination of a few of the terms. 
 
 A converging series is one whose successive terms decrease, 
 or become less and less. 
 
 A diverging series is one whose successive terms increase, or 
 become greater and greater. 
 
 An ascending series is one in which the exponents of the 
 unknown quantity continually increase ; and a descending series 
 is one in which the exponents continually decrease. 
 
 When the value of an algebraic expression cannot be exactly 
 determined, we expand the expression into a series, and thus 
 endeavor to obtain an approximate value. We have, therefore, 
 two questions to solve in respect to series. 1°. To expand 
 algebraic expressions into series. 2°. To find any term of a 
 series, and the sum of all the terms. 
 
 undetermined coefficients. 
 
 263. In the development of algebraic expressions in series, 
 the method of undetermined coefficients is found of great 
 utility. 
 
 We will give a brief exposition of this method. 
 
 Let there be the equation, 
 
 = A + Ba;-t.Ca:2-f-Da:*+... 
 in which the coefficients A, B, C . . are independent of x; it is 
 required to determine these coefficients so that the equation 
 
UNDETERMINED COEFFICIENTS. * 327 
 
 may be true whatever value is assigned to x. Since the coeffi- 
 cients A, B, C . . are to be determined, they are on this account 
 called undetermined coefficie7its. 
 
 Since by hypothesis the proposed equation must be true, 
 whatever value is assigned to x, it must be true for the particu- 
 lar value a: = 0. Putting a: = 0, the equation is reduced to 
 = A ; we have, therefore, A = 0. Substituting this value 
 of A in the equation, and dividing both sides by x, it becomes 
 
 = B + Ca: + Da:2+ .. 
 but since this equation must also be true whatever the value of 
 ic, putting a; = 0, we obtain B =: 0. By the same course of 
 reasoning it may be shown also that C = 0, D = 0. 
 
 If, then, we have an equation of the form 
 
 = A + Ba;4-Ca:2 + Da:^+&c., 
 in which the coefficients A, B, C, &c., are independent of a:, in 
 order that the equation may be true whatever value is assigned 
 to X, each separate coefficient must necessarily be equal to zero. 
 
 This is the principle upon which the method of undetermined 
 coefficients is founded. We pass to some applications of the 
 method. 
 
 Ex. 1. Let there be a dividend x^ — px-{-p', let the divisor 
 be a; — a, and the quotient x — q; to determine the conditions 
 necessary in order that the division may be exact. 
 
 Since, when there is no remainder, the divisor multiplied by 
 the quotient should produce anew the dividend, we must have 
 
 {x — a){x—q)=zx^—px-{- p\ 
 or, performing the multiplication indicated, transposing and re- 
 ducing, 
 
 ^ = {a-\-q—p)x-\'{j^ — aq)\ 
 but since this equation is true whatever the value of a:, we must 
 have 
 
 a -|- ^ -r— p = 0, and p' — a g = ; 
 eliminating q from these last, we obtain 
 (i^=.ap—pf. 
 
 In order, therefore, that the division may be exact, we must 
 
328 ELEMENTS OF ALGEBRA. 
 
 have the relation c^z=ap — p\ or, which is the same thing, 
 a2 — ap-{-p' = (i. 
 
 Ex. 2. Let it be proposed to decompose -:; — —- into 
 
 X" O X -j- o 
 
 fractions, whose sum is the given fraction, and whose denomi- 
 nators are the factors of the given denominator. 
 
 The factors of the denominator are x — 3, a: — 2; we assume, 
 
 therefore, 
 
 3 + 5a; _ A B 
 
 (a;— 3) (a; — 2) a; — 3 ' x — 2 
 
 Freeing from denominators, transposing and reducing 
 
 = (A + B— 5)a: — (2A + 3B + 3; 
 
 whence A-i-B = 5, and2A4-3B = --3; 
 
 from which we obtain A = 18, B = — 13, and we have 
 
 3 + 5a: _ 18 13 
 
 a:^ — 5a;-|-6 a; — 3 "~ a; — 2* 
 
 51x. 3. Find for A and B values, such that we may have 
 
 7 + 9a: A B 
 
 (a; — 5) (a; — 3) ""a; — 5"^ a; — 3' 
 
 Ans. A = 26, B=-17. 
 
 Ex. 4. Find for A and B values such that we may have 
 
 3a; --5 A B 
 
 (a: — 4)(a:— 2)~^a; — 4 a: — 2' 
 
 Ans. A = |-, B = J-. 
 
 264. We proceed to the application of the method to the 
 
 development of algebraic expressions in series. 
 
 z * 
 Ex. 1. Let it be proposed, to develop the expression — r— in 
 
 X — T~ z 
 
 series, according to the ascending powers of x. In order to this, 
 we assume 
 
 -4— = A + Ba; + Ca:2 , D3^ , E^4 j_ ^ ^ ^ 
 z-^x ' ' ' ' ' 
 
 the coefficients A, B, C, being independent of x. 
 
 Freeing the first member from its denominator, transposing 
 
 and arranging with reference to ar, we obtain 
 
UNDETERMINED COEFFICIENTS. 
 
 329 
 
 — 2\ A B C| 
 
 we have, therefore, the series of equations 
 Az — z = 0, Bz-}-A = 0, Cz+B = 0, Dz + C=0, &c., 
 from which to deduce the values of A, B, C, &c. Performing 
 the operations and substituting in the assumed expression, we 
 obtain for the development required 
 
 Ex. 2. Expand the fraction — ^ into an infinitis 
 
 X ~~~ <fe X — r" X 
 
 series. 
 
 Assume 
 fee. 
 
 Z X 
 
 Z -\- X z 
 
 1 
 
 l^2x + s^ 
 
 =:A4-Ba: + Ca:2 + D^3 + Ea:* + 
 
 Freeing from denominator, transposing and reducing 
 
 1 = A+ B x-\- Q. 2:2+ D ^_j. E a;* + &c. 
 — 2A 
 
 xJr C 
 
 2:2+ D 
 
 2:^+ E 
 
 — 2B 
 
 -2C 
 
 — 2D 
 
 A 
 
 B 
 
 G 
 
 from which we obtain A = 1, B 
 1 
 
 whence 
 
 \^2x-\-7? 
 
 2, C==3, D = 4, &c., 
 = l+2a; + 3rc2 + 42:» + 52r5+&c. 
 
 1 +22: 
 Ex. 3. Expand the fraction - — -5- -^ into an infinite series. 
 
 Ans. l+32: + 42:2 + 72:^+ll2;^ + 182:« + 292;«+,&c. 
 
 1 + 22; 
 Ex. 4. Expand ^ into an infinite series. 
 
 1 O X 
 
 Ans. l + 52: + 152:2+452.-3 + 1352:^+, &c. 
 • What is the law of the coefficients in the three preceding 
 series ? 
 
 Ex. 5. Develop (a — xY in series. 
 
 Assume (a — 2:)*=A + Ba: + C2r» + Da:» + &c. 
 
 Squaring, (a— 2;)= A^ + 2 A B 2; + B^ 
 
 2AC 
 
 2 2:2 + 2BC|2^+&c.. 
 2AD 
 
 whence A = ai , B 
 
 1 
 2ai 
 
 C = 
 
 2.4 aJ 
 
 &c. 
 
 from which the development sought will be easily obtained* 
 
330 ELEMENTS OF ALGEBRA. 
 
 265w Let it be required next to expand ^ —^ into an infi- 
 
 nite series. * 
 
 If we assume ^ = A-f-Ba; + Ca;2 4-Dr'-j-> &c., and 
 
 determine the coefficients accordingly, we shall have — 1 = 0, 
 3 A=0, 3 B — A = 0, &c., the first of which is absurd. 
 
 The proposed cannot, therefore, be developed in this form. 
 Indeed, if we put in the proposed fraction a; = 0, the fraction 
 takes the form of infinity. The development, therefore, for this 
 hypothesis, should take the same form. But in order to this it 
 must contain, it is obvious, a term in x with a negative exponent. 
 "We assume, therefore, 
 
 from which we obtain successively 
 
 It is usual to assume the development so that it shall proceed 
 according to the ascending powers of x, beginning with a:". But 
 this form will not always, apply. And in any case in which it 
 is not applicable the fact will become evident, as in the preced- 
 ing example, by the appearance of some absurdity in the result 
 of the operation. 
 
 The form which the development should take may, in 
 general, be discovered at the outset, by putting a; = in the 
 function to be developed, and observing the result. If the func- 
 tion, on the hypothesis a; = 0, is finite, the development should 
 be taken according to the ascending powers of x, beginning with 
 Qp. If the function on this hypothesis becomes 0, the first term 
 of the development should contain x. If it takes the form of 
 infinity, the first term of the development should contain x with 
 a negative exponent. 
 
 266. The following miscellaneous examples will serve as an 
 additional exercise. 
 
THE DIFFERENTIAL METHOD. 331 
 
 Ex. 1. To develop y-* — in series. 
 
 Ans. l + 2a: + 2a:2 + 2a:3+, &c. 
 
 Ex. 2. Develop -^ — ; — r^ in series. 
 (l+a:)2 
 
 Ans. 1 — 2a; + 3a;2_42;3 + 5a:*— , &c. 
 
 Ex. 3. Expand (1 — x)^ into an infinite series. 
 
 ^^'- ^-2"24~2A6-2A6:8^'^'- 
 
 3 ^2 J 
 
 Ex. 4. Decompose — - — — into partial fractions. 
 
 X [x ~\- i) [x — 1) 
 
 Ans. 
 
 a; ' a:-j- 1 x — 1 
 The Differential Method. 
 
 267. Let there be any series represented by a, b, c, d, &c. ; 
 if we subtract the first term from the second, the second from 
 the third, and so on, the differences thus obtained will form a 
 new series called the Jirst order of differences. If we subtract 
 again, the first term from the second, the second from the third, 
 &c., in this last series, the differences thus obtained will form a 
 third series, called the second order of differences, and so on. 
 
 Thus, the series of square numbers, with the several orders 
 of differences, is as follows : 
 
 1 4 9 16 25 36 49 
 3 5 7 9 11 13 1st DifTerence. 
 
 2 2 2 2 2 2d " 
 
 3d 
 
 If in a proposed series the first differences are all the same, 
 or constant, the series is called a difference series of the first 
 order. If the second differences are constant, it is called a 
 difference series of the second order, and so on. Thus, the 
 series of square numbers above is a difference series of the 
 second order. 
 
 Ex. Of what order is the series 1, 4, 10, 20, 35, 56 1 
 
332 ELEMENTS OF ALGEBRA. 
 
 268. From wiiat has been done we shall have the following 
 questions to solve in respect to series : 1°. To find the succes- 
 sive differences of the terms of a series. 2°. By means of 
 these to find any intermediate term, and the sum of all the 
 terms. 
 
 1. Resuming the general series, we have 
 
 a, b, c, df e, &c. 
 
 3— a, c — b, d — c, e — d 1st DifT. 
 
 c — 2b-{-a,d — 2c-\- b, e — 2d-{-c 2d " 
 
 ^__3c-}-3i — a, &c., 3d " 
 
 <( 
 
 If we now represent the first terms of the successive orders 
 of differences by Di, D2, &c., we shall have, reversing the order 
 of the terms, 
 
 Di = — «+ b 
 
 D2= a — 2b-^ c 
 
 D3 = — a + 33 — 3c + d: 
 
 D4== &c. 
 In which it will be perceived that the coefficients of the several 
 terms correspond with those in the expansion of a binomiaL 
 And we shall have, generally, 
 
 D. = ±a^^7ib±n-j^ c^^n ^ ^ ^(^zt&c., 
 
 the upper sign corresponding to the case in which the difference 
 is even, and the lower to the case in which it is odd. 
 
 By means of this formula, we readily find the first term in 
 &ny order of differences. 
 
 Ex. 1. What is the first term of the third order of differences 
 in the series of cubes 1, 8, 27, 64, 125, &c. ? 
 
 In this case, n being odd, we use the lower signs of the for- 
 mula, and we have 
 
 D3 = — 1+3.8 — 3.27 + 64 = 6. 
 , Ex. 2. What is * the first term of the fourth order of differ- 
 ences in the series 7, 12, 21, 36, 62 ? . Ans. 3. 
 
THE DIFFERENTIAL METHOD. 3^ 
 
 Px. 3. What is the first term of the fifth order of differences 
 in the series 1, 6, 21, 56, 126, 252, &c. ? Ans. 1. 
 
 2. Let it be required next to find any intermediate term of 
 the series. 
 
 From the expressions Dj, D2, D3, &c., we obtain 
 
 c = a-f2Di+ D2 
 
 ef = a -f- 3 Di + 3 D2 + D3, 
 
 in which it will be perceived that the coefficients of the succes- 
 sive terms are the coefficients of the power of a binomial one 
 degree less than the number of the term ; whence, putting L 
 for the 72th term of the series 
 
 The series will terminate if the diflferences become after a 
 certain order. Otherwise it will be infinite, and L can be deter- 
 mined only approximately. 
 
 Ex. 1. What is the 50th term of the series 1, 4, 8, 13, &c. ? 
 
 Here a = 1, Di == 3, D2 = 1, D3 = 0. Ans. 1324. 
 
 Ex. 2. Required the tenth term of the series 1, 4, 8, 13, 19, 
 &c. Ans. 64. 
 
 Ex. 3. Required the twentieth term of the series 1, 5, 15, 35, 
 70, 126, &c. Ans. 8855. 
 
 Ex. 4. What is the twelfth term of the order of cubes, or, in 
 other words, whafls the cube of 12? Ans. 1728. 
 
 3. Let it be required next to find the sum of any number n 
 terms of the series 
 
 a, ^, c, dt e, &c. 
 Assume the series 
 
 ^, a^ a -\- bt a -\- b -\' Ct a -\- h -\- c '\- d^ &c. 
 Subtracting each term from the next succeeding, we obtain a, 
 b, c, <?, &c., the proposed series. From this it follows that the 
 sum of n terms of the proposed series is the (?t -|- l)th term of 
 the assumed series, and that the wth order of differences in the 
 first series is the same as the (n-(- l)th order in the second. 
 
334 ELEMENTS OF ALGEBRA. 
 
 We shall have, therefore, by substituting in the formula for L 
 
 above, 
 
 for a,n-\-\ for n, a for Di, Dj for D2, &c., . . . 
 
 71^1 {n—\){n — 2) 
 S = 7za + 7i-^— Di + ?i^^ -^A iD2 + ... 
 
 which is the expression for the sum of any number n tearms of 
 the series. 
 
 Ex. 1. Required the sum of 12 terms of the series 
 1, 4, 8, 13, 19 
 
 Here a = 1, Di = 3, D2 = 1, D3 = 0, % = 12. Ans. 430. 
 
 Ex. 2. Required the sum of n terms of the series 1, 2, 3, 4, 
 
 5, &c. . 11? -\-n 
 
 Ans. -^. 
 
 Ex. 3. Required the sum of ten terms of the series 1, 5, 15, 
 35, 70, 126, &c. Ans. 2002. 
 
 Interpolation. 
 
 269. The formula for L may be applied to the process of 
 interpolation, or that of finding numbers intermediate between 
 those contained in tables. 
 
 Suppose that we have a table of the square roots of numbers, 
 from 1 to 100, and that we wish, for example, to insert the 
 square roots of the intermediate numbers to every ^ of a unit. 
 If the first differences of the roots of the numbers already in the 
 tables were constant, we should have merely to find proportional 
 parts of the difference between any two consecutive roots, which 
 we should add to the roots respectively. This would be 
 the method by first differences. But the first differences not 
 being constant, the second must be taken into consideration, 
 and so on, until the diflferences become constant, or the requisite 
 degree of approximation has been obtained. For this purpose 
 we employ the formula L above. Putting, for convenience, in 
 this formula w, to represent the distance of any term from the 
 first, we have x = n — 1, and the formula becomes 
 
SUMMATION OF INFINITE SERIES. 
 
 335 
 
 L = a + a: Di -f- 
 
 x{x—l) 
 
 ^(^-l)(^-2) 
 
 Let three contiguous numbers of the table, with the square 
 roots and differences to D2, be as follows ; it is required to find 
 the square root of 94^, or 94.25. 
 
 No. 
 
 Square root. 
 
 D: 
 
 D. 
 
 94 
 95 
 76 
 
 9.69536 
 9.74679 
 9.79796 
 
 .05143 
 .05117 
 
 - .00026 
 
 In the present case we shall have a = 9.69536, Dj = .05143, 
 D2 = — .00026, anda; = :^; whence 
 
 L== 9.69536 + ^ (.05143) + ^3_ (.00026) 
 
 = 9.70824, Ans. 
 Ex. 2. Given the logarithms of 6, 7, 8, 9, and 10, to find 
 that of 6.5. 
 
 No. 
 
 Logs. 
 
 Di 1 D, 
 
 D3 
 
 D4 
 
 6 
 
 7 
 
 8 
 
 9 
 
 10 
 
 0.778151 
 0.845098 
 0.903090 
 0.954243 
 1.000000 
 
 .066947 
 .057992 
 .051153 
 .045757 
 
 — .008955 
 
 — .006839 
 
 — .005396 
 
 .002116 
 .001443 
 
 - 000673 
 
 L = a 4- i Di — ^ D2 + -iV D3 — xf ^ D4 = 0.812901 Ans. 
 
 Ex. 3. Given the natural sine of 37° equal to .60182 
 
 " " 38° " .61566 
 
 " " 39° " ' .62932 
 
 " " 40° " .64279 
 
 to find that of 37° 30'. Ans. .60876. 
 
 Summation of Infinite Series. 
 
 270. 'The summation of an infinite series consists in finding 
 a finite expression equivalent to the series. 
 
 Since the sum of a series must evidently depend upon the 
 law of the series, no formula can be given for the summation of 
 series which will apply universally. A great variety of useful 
 series may be summed by the following principles. 
 
336 
 
 ELEMENTS OF ALGEBRA. 
 
 Let there be a series, the terms of which shall be represented 
 by the fraction, 
 
 pq q q 
 
 whence 
 
 n{n-{-p) '71 n-\-p 
 
 p \n n-^-pJ 
 
 n{n-{-p) p 
 If, then, a series be represented by 
 
 , its sum, it IS 
 
 n {7i-\-p) 
 evident, will be equal to a pth part of the difference of the two 
 
 series represented by - and — ^ — . 
 ^ *' w n-{-p 
 
 Ex. 1. What is the sum of n terms of the series 
 
 1_ J_ i_ i_ 2 ' 
 
 1.2' 2.3' 3.4' 4.5 * * ' * 
 Here, in the general term g'= 1, ^ = 1, ?i5= 1, 2, 3, &c., 
 whence we have for the sum S, 
 
 S = l+i + i+...i 
 
 2 
 — 1 
 
 2 
 
 « + l' 
 
 Ans. 
 
 n 71+ 1 
 If 71 = 00, the sum, it is evident, will be 1. 
 Ex. 2. What is the sum of n terms of the series 
 
 1.3 
 Here q==:l,p = 2, and 7i= 1, 3, 5, 7 . . . ; whence 
 
 3:5' 577' 7:9' 9li- -to infinity. 
 
 S = 
 
 ^+1+1+1 
 
 1 
 
 2n—l 
 1 
 
 271-1 271+1 
 
 -, Ans. 
 
 271+r 
 
 If w = 00, the sum, it is evident, will be J. 
 
UNIVERSITY OF CALIFORNIA LIBRARY 
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