BOWSER'S MATHEMATICS. ACADEMIC ALGEBRA. With numerous Examples. COLLEGE ALGEBRA. With numerous Examples. PLANE AND SOLID GEOMETRY. With numerous Exer- cises. AN ELEMENTARY TREATISE ON ANALYTIC GEOMETRY, embracing Plane Geometry, and au Introduction to Geometry of Three Dimensions. AN ELEMENTARY TREATISE ON THE DIFFERENTIAL AND INTEGRAL CALCULUS. With numerous Ex- amples. AN ELEMENTARY TREATISE ON ANALYTIC MECHANICS. With numerous Examples. AN ELEMENTARY TREATISE ON HYDROMECHANICS. With numerous Examples. COLLEGE ALGEBRA, FOR THE USE OF ACADEMIES, COLLEGES, AND SCIENTIFIC SCHOOLS. WITBE NUMEROUS EXAMPLES. BY EDWARD A. BOWSER, LL.D., PROFESSOR OF MATHEMATICS AND ENGINEERING IN RUTGERS COLLEGE, FOURTH EDITION. BOSTON, U.S.A.: D. C. HEATH & CO., PUBLISHERS. 1893. Copyright. 1888. Copyright, 1888, By E. A. BOWSER. STovtooofc }0rcBB: Berwick & Smith, Boston, U.S.A. t- _r — PREFACE. The present work is designed as a text-book for Acade- mies, Colleges, and Scientific Schools. The book is com- plete in itself ; it begins at the beginning of the subject ; and the full treatment of the earlier parts renders it unnecessary that students who use it shall have previously studied a more elementary Algebra. The aim has been to explain the principles in as concise and simple a manner as was consistent with clearness and careful accuracy, and to discuss all the elementary parts of the subject as completely as possible within the limits of a single volume. Copious illustrations have been given to make the work intelligible and interesting to young students ; and numerous explanatory notes have been all along inserted, to guard the pupil against the errors to which experience shows he is liable. Sturm's theorem has been omitted, because it has no application in the usual course of mathe- matical study, and because the mental power of the student can be spent more profitably on other branches of mathe- matics. In the earlier chapters, some of the most interesting practical applications of the subject have been introduced. Thus, a chapter on easy equations and problems precedes the chapters on Factoring and Fractions. By this course the beginner soon becomes acquainted with the ordinary Alge- braic processes without encountering too many difficulties ; and he is at the same time deriving the pleasure which a 797947 VI PREFACE. student always feels in using his knowledge to some practical end. Throughout the book are numerous examples fully worked out, to illustrate the most useful applications of important rules, and to exhibit the best methods of arranging the work. No principle is well learned by a pupil and thoroughly iixed in his mind till he can use it. For this purpose a very large number of examples is given at the ends of the chapters. These examples have been selected and arranged so as to illustrate and enforce every part of the subject. Each set has been carefully graded, commencing with some which are very easy, and proceeding to others which are more difficult. Complicated examples have been excluded, because they consume time and energy which may be otherwise spent more profitably. The numerous examples are given for the convenience of the teacher, that he may have, year by year, in using the book, a sufficient variety from which to select, and so avoid routine, rather than to require any one pupil to do them all. In preparing this work I have consulted the writings of some of the best authors. The chief sources from which I have derived assistance are the Treatises of Wood, De Morgan, Serret, Hind, Young, Todhunter, Coleuzo, Hall and Knight, Smith, Chrystal, and Whitworth. It remains for me to express my thanks to those of my friends who have kindly assisted me in reading and correct- ing the proof-sheets and verifying copy. My especial thanks are due to my old pupil, Prof. R. W. Prentiss of the Nautical Almanac Office, for reading the MS., and for valuable suggestions. E. A. B. Rutgers College, New Brunswick, N.J., May, 1888. TABLE OF CONTENTS. CHAPTER I. FIRST PRINCIPLES. ART. PAGE 1. Quantity and its Measure 1 2. Number 1 3. Mathematics 2 4. Algebra 2 5. Algebraic Symbols 2 6. Symbols of Quantity 2 7. Symbols of Operation 3 8. The Sign of Addition 3 9. The Sign of Subtraction 3 10. The Sign of Multiplication 4 11. The Sign of Division 5 12. The Exponential Sign 6 13. The Radical Sign 7 14. Symbols of Relation 7 15. Symbols of Abbreviation 8 16. Algebraic Expressions 9 17. Factor — Coefficient 11 18. A Term, its Dimensions, and Degree — Homogeneous ... 12 19. Simple and Compound Expressions 13 20. Positive and Negative Quantities 13 21. Additions and Multiplications Made in any Order .... 17 22. Suggestions for the Student in Solving Examples .... 18 Examples 18 CHAPTER II. ADDITION. 23. Addition — Algebraic Sum ■ 21 24. To Add Terms which are Like and have Like Signs ... 21 25. To Add Terms which are Like, but have Unlike Signs . . 22 26. To Add Terms which are not all Like Terms 22 27. Remarks on Addition 23 Examples 24 vii viii CONTENTS. CHAPTER III. SUBTRACTION. ART. PAGB 28. Subtraction — Algebraic Difference 26 29. Rule for Algebraic Subtraction 26 30. Remarks on Addition and Subtraction 29 31. The Use of Parentheses 30 32. Plus Sign before the Parenthesis 31 33. Minus Sign before the Parenthesis 31 34. Compound Parentheses 32 Examples 33 CHAPTER IV. MULTIPLICATION. 35. Multiplication 38 36. Rule of Signs 38 37. The Multiplication of Monomials 41 38. To Multiply a Polynomial by a Monomial 42 39. Multiplication of a Polynomial by a Polynomial 43 40. Multiplication by Inspection 48 41. Special Forms of Multiplication — Formulae 49 42. Important Results in Multiplication 53 43. Results of Multiplying Algebraic Expressions 54 Examples 55 CHAPTER V. DIVISION. 44. Division 68 4."). The Division of one Monomial by Another 58 46. Tbe Rule of Signs 60 47. To Divide a Polynomial by a Monomial 61 48. To Divide one Polynomial by Another 61 49. Division with the Aid of Parentheses 67 50. Where the Division cannot he Exactly Performed .... 67 51. Important Examples in Division 68 Examples 69 CONTENTS. CHAPTER VI. SIMPLE EQUATIONS OF ONE UNKNOWN QUANTITY. iRT. PAGE 52. Equations — Identical Equations 72 53. Equation of Condition — Unknown Quantity 72 54. Axioms 73 55. Clearing of Fractions 74 56. Transposition 75 57. Solution of Simple Equations with one Unknown Quantity . 76 58. Fractional Equations 79 59. To Solve Equations whose Coefficients are Decimals ... 81 60. Literal Equations 81 61. Problems Leading to Simple Equations 83 Examples 89 CHAPTER VII. FACTORING — GREATEST COMMON DIVISOR — LEAST COMMON MULTIPLE. 62. Definitions 96 63. When All the Terms Have one Common Factor 96 64. Expressions Containing Four Terms 97 65. To Factor a Trinomial of the Form x 2 + ax + b .... 99 66. To Factor a Trinomial of the Form ax' 1 + bx + c . . . .101 67. To Factor the Difference of Two Squares 104 68. When One or Both of the Squares is a Compound Expression 105 69. Compound Quantities as the Difference of Two Squares . . 106 70. To Factor the Sum or the Difference of Two Cubes . . . 107 71. Miscellaneous Cases of Resolution into Factors 107 Examples 108 GREATEST COMMON DIVISOR. 72. Definitions Ill 73. Monomials, and Polynomials which can be Easily Factored . Ill 74. Expressions which cannot be Easily Factored 113 Examples 115 CONTENTS. LEAST COMMON MULTIPLE. TAGE ART. 75. Definitions 120 76. Monomials, and Polynomials which can be Easily Factored . li'l 77. Expressions which cannot be Easily Factored 123 Examples 125 CHAPTER VIII. FRACTIONS. 78. A Fraction — Entire and Mixed Quantities 128 79. To Reduce a Fraction to its Lowest Terms 129 80. To Reduce a Mixed Quantity to the Form of a Fraction . . 132 81. To Reduce a Fraction to an Entire or Mixed Quantity . . . i:>5 82. To Reduce Fractions to their Least Common Denominator . 134 83. Rule of Signs in Fractions 136 84. Addition and Subtraction of Fractions 138 85. To Multiply a Fraction by an Integer 144 86. To Divide a Fraction by an Integer 144 87. To Multiply Fractions ]47 > 88. To Divide Fractions 147 89. Complex Fractions 148 90. A Single Fraction Expressed as a Group of Fractions . . .151 Examples lu2 CHAPTER IX. HARDER SIMPLE EQUATIONS OF ONE UNKNOWN QUANTITY. 91. Solution of Harder Equations 157 92. Harder Problems Leading to Simple Equations 161 Examples *^ CHAPTER X. SIMULTANEOUS SIMPLE EQUATIONS OF TWO OR MORE UNKNOWN QUANTITIES. 08. Simultaneous Equations of Two Unknown Quantities . . 179 L81 181 94. Elimination . . 95. Elimination by Addition or Subtraction CONTENTS. XI ART. PAGE 96. Elimination by Substitution 185 97. Elimination by Comparison , 186 98. Fractional Simultaneous Equations 188 99. Literal Simultaneous Equations 189 100. Simultaneous Equations witb Tbree Unknown Quantities . 191 101. Problems Leading to Simultaneous Equations 196 Examples 202 CHAPTER XI. INDETERMINATE PROBLEMS — DISCUSSION OP PROBLEMS INEQUALITIES. 102. Indeterminate Equations — Impossible Problems . . . .211 103. Discussion of Problems — Negative Results 213 104. Interpretation of the Forms, --, — , £ 216 oo 105. Problem of the Couriers 218 INEQUALITIES. 106. Inequalities 221 Examples 225 CHAPTER XII. INVOLUTION AND EVOLUTION. 107. Involution ' . 227 108. Involution of Powers of Monomials 227 109. Involution of Binomials 229 110. Involution of Polynomials 230 EVOLUTION. 111. Evolution — Evolution of Monomials 232 112. Square Root of a Polynomial 234 113. Square Root of Arithmetic Numbers 238 114. Square Root of a Decimal 240 115. Cube Root of a Polynomial 243 116. Cube Root of Aritbmetic Numbers 246 116a. Cube Root of a Decimal 24S Examples 250 Xii CONTENTS. CHAPTER XIII. THE THEORY OF EXPONENTS — SURDS. ABT. TAGS 117. Exponents that are Positive Integers 253 US. Fractional Exponents 25? 119. Negative Exponents 255 120. To Prove that (a m )» = a mn is True for All Values of m and n 257 121. To Prove that (ab) n = a n b n for Any Value of n 258 SURDS (RADICALS). 122. Surds — Definitions 2G1 123. To Reduce a Rational Quantity to a Surd Form 262 124. To Introduce the Coefficient Under the Radical Sign . . . 262 125. To Reduce an Entire to a Mixed Surd 262 126. Reduction of Surds to Equivalent Surds 263 127. Addition and Subtraction of Surds 2<>4 128. Multiplication of Surds 265 129. To Rationalize the Denominator of a Fraction 267 130. Division of Surds 268 131. Binomial Surds — Important Propositions 269 132. Square Root of a Binomial Surd 270 133. Equations Involving Surds 272 Examples 273 CHAPTER XIV. QUADRATIC EQUATIONS OP ONE UNKNOWN QUANTITY. 134. Quadratic Equations 280 135. Pure Quadratic Equations 280 136. Affected Quadratic Equations 282 137. Condition for Equal Roots 286 138. Hindoo Method of Completing the Square 2S8 139. Solving a Quadratic by Factoring 290 140. To Form a Quadratic when the Roots are Given i'!»2 141. Equations Having Imaginary Roots 284 142. Equations of Higher Degree than the Second 295 143. Solutions by Factoring 298 144. Problems Leading to Quadratic Equations 300 Examples 808 CONTENTS. CHAPTER XV. SIMULTANEOUS QUADRATIC EQUATIONS. ART 316 318 PAGE 145. Simultaneous Quadratic Equations 311 146. When One of the Equations is of the First Degree . . . .311 147. Equations of the Form x ± y = a, and xy = b 312 148. When the Equations Contain a Common Algebraic Factor . 314 149. Homogeneous Equations of the Second Degree . . . 150. When the Two Equations are Symmetrical 151. Special Methods 319 152. Quadratic Equations with Three Unknown Quantities . . 321 153. Problems Leading to Simultaneous Quadratic Equations . . 321 Examples 324 CHAPTER XVI. RATIO PROPORTION VARIATION. 154. Ratio — Definitions 331 155. Properties of Ratios 333 PROPORTION. 156. Definitions 336 157. Properties of Proportions 337 VARIATION. 158. Definition 342 159. Different Cases ot Variation 343 160. Propositions in Variation 344 Examples 348 CHAPTER XVII. ARITHMETIC, GEOMETRIC, AND HARMONIC PROGRESSIONS. ARITHMETIC PROGRESSION. 161. Definitions — Formulae 353 162. Arithmetic Mean 356 xiv , CONTENTS. GEOMETRIC PROGRESSION. ART. PAGE 163. Definition — Formulae 359 164. Geometric Mean 361 165. The Sum of an Infinite Number of Terms ,> 362 166. Value of a Repeating Decimal 364 HARMONIC PROGRESSION. 167. Definition 365 168. Harmonic Mean 366 169. Relation between the Different Means 368 Examples 369 CHAPTER XVIII. MATHEMATICAL INDUCTION — CONTINUED FRACTIONS — PERMUTATIONS AND COMBINATIONS. 170. Mathematical Induction — General Proofs of Theorems . . 374 CONTINUED FRACTIONS. 171. Definition 376 172. To Convert a Given Fraction into a Continued Fraction . . 376 173. The Convergents are Alternately too Great and too Small . 377 • 174. Law of Formation of the Successive Convergents .... 378 175. Difference between Two Consecutive Convergents .... 379 176. Limit of Error in Taking any Convergent 380 177. Approximate Value of a Quadratic Surd 383 178. Periodic Continued Fractions 384 PERMUTATIONS AND COMBINATIONS. 179. Definitions 385 180. The Number of Permutations 386 181. The Number of Combinations 388 182. To Divide m + n Things into Two Classes 800 183. Permutations of ?i Things not all Different 801 Examples 392 CONTENTS. CHAPTER XIX. INDETERMINATE COEFFICIENTS PARTIAL FRACTIONS - BINOMIAL THEOREM. ART. PAOK 184. Definitions oWj 185. Indeterminate Coefficients 396 186. Development by Indeterminate Coefficients 397 PARTIAL FRACTIONS. 187. Partial Fractions. Case 1 401 188. Partial Fractions. Case II .402 189. Partial Fractions. Case III 403 BINOMIAL THEOREM. 190. Positive Integral Exponent 405 191. The r th or General Term of the Expansion ....... 408 192. Any Exponent 409 Examples 413 CHAPTER XX. SUMMATION OF SERIES. RECURRING SERIES. 193. Definition 416 194. The Scale of Relation being Given, to Find Any Term . . 417 195. To Find the Sum of n Terms of a Recurring Series . . . .418 196. Series Summed by Means of Other Series 419 METHOD OF DIFFERENCES. 197. Definition — To Find Any Term of a Series 421 198. To Find the Sum of n Terms of Any Series . ... 422 199. Piles of Cannon-Bails .424 200. Interpolation 426 201. Reversion of Series 427 Examples 428 CONTENTS. CHAPTER XXI. LOGARITHMS EXPONENTIAL AND LOGARITHMIC SERIES INTEREST AND ANNUITIES. LOGARITHMS. ART. TAGE 202. Definitions 433 203. Properties of Logarithms 434 204. Comparison of Two Systems of Logarithms 430 205. Common System of Logarithms 437 206. Table of Logarithms 439 207. Exponential Equations 442 EXPONENTIAL AND LOGARITHMIC SERIES. 208. Exponential Series 443 209. Logarithmic Series 444 210. Computation of Logarithms 445 INTEREST AND ANNUITIES. 211. Simple Interest 447 212. Present Value and Discount . . 448 213. Compound Interest 449 214. Present Value and Discount 450 215. Annuities 452 216. The Amount of an Annuity 452 217. The Present Value of an Annuity 453 218. The Present Value of a Deferred Annuity 454 Examples 450 CHAPTER XXII. IMAGINARY QUANTITIES INDETERMINATE EQUATIONS THEORY OF NUMBERS. IMAGINARY QUANTITIES. 219. Definitions 460 220. Fundamental Principles 461 221. Geometric Meaning of the Imaginary Unit 462 INDETERMINATE EQUATIONS. 222. Indeterminate Equations t ,; ' CONTENTS. THEORY OF NUMBERS. ART. PAGE 223. Scales of Notation 468 224. To Express a Whole Number in Any Proposed Scale . . . 470 225. Properties of Numbers in the Common System 472 226. Theorems in Relation to Numbers 474 Examples 478 CHAPTER XXIII. PROBABILITY (CHANCE). 227. Definition 481 228. Simple Events 483 229. Compound Events 485 230. Mutually Exclusive Events 488 231. Expectation 492 232. General Problem 492 Examples 494 CHAPTER XXIV. THEORY OF EQUATIONS. 233. General Equation of the n th Degree 497 234. Divisibility of Equations 498 235. Number of Roots 499 236. Relations between Coefficients and Roots 500 237. Fractional Roots 502 238. Imaginary Roots 502 239. Descartes' Rule of Signs 503 240. Derived Functions 505 241. Equal Roots 507 TRANSFORMATION OF EQUATIONS. 242. Equation with Roots of Sign opposite those of f{x) = . . 509 243. Equation with Roots Multiples of those of f{x) =0 ... 510 244. Equation with Roots Reciprocals of those of f(x) - . . .511 245. Equation with Roots Squares of those of f{x) = . . . .511 246. Equation with Roots Less than those of f(x) = . . . . 512 247. Horner's Method of Synthetic Division 513 248. Removal of Any Given Term 515 xviii CONTENTS. LIMITS OF THE REAL HOOTS OF AN EQUATION. ART. PAGE 249. Definition 515 250. Superior Limit 51(5 251. Inferior Limit 517 252. Limits between which the Roots separately Lie 518 Examples 519 CHAPTER XXV. SOLUTION OF HIGHER NUMERICAL EQUATIONS. 253. Commensurable Roots 523 254. Reciprocal Equations 525 255. Binomial Equations 527 256. Cardan's Solution of a Cubic Equation 529 257. Incommensurable Roots 532 258. Horner's Method of Approximation 532 259. Newton's Method of Approximation 536 260. Approximation by Double Positiou 537 Examples 539 ALGEBRA CHAPTER I. FIRST PRINCIPLES. t 1. Quantity and its Measure. — Quantity is any thing that is capable of increase, diminution, and measure- ment ; as time, space, motion, weight, and area. To measure a quantity is to find how many times it con- tains another quantity of the same kind, taken as a standard of comparison. This standard is called a unit. For example, if we wish to determine the quantity of a weight, we must take a unit of weight, such as a pound, or an ounce, and observe how many times it is contained in the quantity to be measured. If we wish to measure area, we must take a unit of area, as a square foot, square yard, or acre, and see how many times it is contained in the area to be measured. So also, if we wish to measure the value of a sum of money, or any portion of time, we must take a unit of value, as a dollar or a sovereign, or a unit of time, as a day or a year, and see how many times it is contained in the quantity to be measured. 2. Number. — The relation between any quantity and its unit is always expressed by a number; a number therefore simply shows hoio many times any quantity to be measured contains another quantity, arbitrarily assumed as the unit. All quantities, therefore, can be expressed by numbers. All numbers are concrete or abstract. A Concrete Number is one in which the kind of quantity which it measures is expressed or understood ; as 6 books, 10 men, 4 days. 1 2 MATH EM A TICS. — ALGEBRA. An Abstract Number is one in which the kind of quantity which it measures is not expressed ; as 6, 10, 4. The word quantity is often used with the same meaning as number. Numbers may be either whole or fractional. The word integer is often used instead of whole number. 3. Mathematics. — Mathematics is the science which treats of the measurement and relations of quantities. It is divided into two parts, Pure Mathematics and Mixed Mathematics. Pure Mathematics consists of the four branches, Arithmetic, Algebra, Geometry, and Calculus. Mixed Mathematics is the application of Pure Mathematics to the Mechanic Arts. 4. Algebra. — Algebra is that branch of Mathematics in. which we reason about numbers by means of symbols. The different symbols used represent the numbers themselves, the manner in which they are related to one another, and the operations performed on them. In Arithmetic, numbers are represented by ten characters, called figures, which are variously combined according to certain rules, and which have but one single definite value. In Algebra, on the contrary, numbers are represented either by figures, as in Arithmetic, or by symbols which may have any value we choose to assign to them. 5. Algebraic Symbols. — The symbols employed in Algebra are of four kinds : symbols of quantity, symbols of operation, symbols of relation, and symbols of abbreviation. 6. Symbols of Quantity. — The symbols of quantity may be any characters whatever, but those- that are most commonly used are figures and the letters of the alphabet; and as in the simplest mathematical problems there aid certain quantities given, in order to determine other quail' tities which are unknown, it is usual to represent the known quantities by figures and by the first letters of the alphabet, a, b, c, etc. ; a', b' , c', etc., read a prime, b prime, c prime % etc. ; a v b v c v etc., read a one, b one, c one, etc. ; while the SYMBOLS OF OPERATION. 3 unknown quantities are represented by the final letters of the alphabet, v, x, y, z, i/, a/, y', z', etc. Known Quantities are those whose values are given. Unknown Quantities are those whose values are required. Since all quantities can be expressed by numbers (Art. 2), it is only these numbers with which we are concerned, and the symbols of quantity, whether figures or letters, always represent members. In Arithmetic a character has but one definite and invariable value, while in Algebra a symbol may stand for any quantity we choose to assign to it (Art. 4); but while there is no restriction as to the numerical values a symbol may represent, it is understood that in the same piece of work it keeps the same value throughout. Thus, when we say "let a — 2," we do not mean that a must have the value 2 always, but only in the particular example we are considering. Also, we may operate with symbols without assigning to them any particular value at all; and it is with such operations that Algebra is chiefly concerned. 7. Symbols of Operation. — The symbols of operation are the same in Algebra as in Arithmetic, or in any other branch of Mathematics, and are the following : 8. The Sign of Addition, +, is called plus. When placed before a number it denotes that the number is to be added. Thus, 6+3, read 6 phis 3, means that 3 is to be added to 6 ; a + b, read a pZws 6, denotes that the number represented by b is to be added to the number represented by a ; or, more briefly, it denotes that b is to be added to a. If a represent 8, and b represent 5, then a + b represents 13. Similarly a + b + c, read a plus b plus c, denotes that we are to add b to a, and then add c to the result. 9. The Sign of Subtraction, — , is called minus. When placed before a number it denotes that the number is to be subtracted. Thus, a — b, read a minus b, denotes that the number represented by b is to be subtracted from the number represented by a ; or, more briefly, that b is to be subtracted from a. If a represent 8, and b represent 5, then a — b represents 3. 4 THE SIGN OF MULTIPLICATION. Similarly a — b — c, read a minus b minus c, denotes that we are to subtract b from a, and then subtract c from the result. If neither + nor — stands before a quantity, -f- is always understood ; thus a means + a. Quantities which have the same sign, either + or — , are said to have like signs. Thus, + a and -f- b have like signs, also — a and — b ; but -f a and — b have unlike signs. Note. — Although there are many signs used in Algebra, when the sign of a quantity is spoken of, it means the + or — sign which is prefixed to it; and when we speak of changing the signs of an expres- sion, it means that we are to change + to — and — to + wherever they occur. The sign ~ is sometimes used to denote the difference of two numbers when it is not known which of them is the greater. Thus, a ~ b denotes the difference of the numbers represented by a and b; and is equal to a — b,orb — a, according as a is greater or less than b; but this symbol ~ is very rarely required. 10. The Sign of Multiplication, x , is read into, or times, or multiplied by. When placed between two numbers it denotes that they are to be multiplied together. Thus, a x b, read a into b, denotes that the number represented by a is to be multiplied by the number represented by b, or, more briefly, that a is to be multiplied by b, or that the two are to be multiplied together. The numbers to be multiplied together are called factors, and the result of the multiplica- tion is called a product. Thus 5, a, and b are the factors of the product 5 X a X 6. If a represent 8, and b repre- sent 4, then a X b represents 32 ; a and b are the factors of the product a X b, or 8 and 4 are the factors of 32. Simi- larly a X b X c denotes the product of the numbers a, 6, and c. If a represent G, b represent 8, and c represent 10, then a x b x c represents 480, and 5 x a x b x c repre- sents 2400. Sometimes a point is used instead of the sign x ; or. still mure commonly, one number is placed close after the other THE SIGN OF DIVISION. 5 without any sign between them. Thus, a x b, a-b, and ab all mean the same thing, viz., the product of a and b ; also, a X b X c, or a ■ b • c, or abc, denotes the product of the numbers a, 6, and c. If a, 6, and c represent 2, 5, and 10 respectively, then abc represents 100. If one factor of a product is equal to 0, the whole product must be equal to 0, whatever values the other factors may have. A factor is sometimes called a "zero factor." * The sign of multiplication must not be omitted when num- bers are expressed in the ordinary way by figures. Thus 23 cannot be used to represent the product of 2 and 3, because 23 is used to mean the number twenty-three. Nor can the product of 2 and 3 be represented by 2.3, because 2.3 is used to mean two and three-tenths. We must therefore represent the product of 2 and 3 by placing the sign of multiplication between them, as follows : 2 x 3. When the numbers to be multiplied together are represented by letters, or by letters and a figure, it is usual to omit the sign of multiplication for the sake of brevity, and write them in succession close to each other; thus, the product of the numbers 7, a, 6, c, and d would be written 7 abed, instead oi 7 x a x b x c x d, or 7 • a • b • c • d, and would have the same meaning. 11. The Sign of Division, -5-, is read divided by, or simply by. When placed between two numbers, it denotes that the number which precedes it is to be divided by the number which follows it. Thus, a -s- 6, read a divided by b, or a by b, denotes that the number represented by a is to be divided by the number represented by 6, or, more briefly, that a is to be divided by 6. If a represent 8, and b repre- sent 2, then a -s- 6 represents 4. Most frequently, to express division, the number to be divided is placed over the other * It is a common mistake of beginners to say that an Algebraic expression like a x or x a is equal to a, by supposing it to mean a not multiplied at all ; whereas a x 01 x a signifieB takeu a times, or a taken times, aud is therefore equal toO. 6 THE EXPONENTIAL SIGN. with a horizontal line between them, in the manner of ?i fraction in Arithmetic. Thus, - is used instead of a ■+- b, b and has the same meaning. Also, the sign of division may be replaced by a vertical line, straight or curved. Thus, a\b, or b)a is used instead of a -h b, and has the same meaning. Note. — It is important for the student to notice the order of the operations in such expressions as a -\- b x c and a — b -f c. The former means that b is first to be multiplied by c, and the result added to a. The latter means that b is first to be divided by c, and the result subtracted from a. 12. The Exponential Sign. — This sign is a small fig- ure or letter written at the right of and above a number to show how many times the number is taken as a factor, and is called an exponent. Thus, a 8 is used to denote a x a, or that a is taken twice as a factor ; a 3 is used to denote a x a X a, or that a is taken three times as a factor ; a 4 is used to denote a x a X « X a, or that a is taken four times as a factor; and a n is used to denote a x a x a x a, etc., to it factors, or that a is taken n times as a factor. Similarly a 2 6 4 cd 3 is used to denote aabbbbcddd, and 7a 3 cd 2 is used for laaacdd. If a factor be multiplied b} r itself any number of times the product is called a power of that factor. Thus, a x a is called the second power of a, and is written « 2 ; a x a x fl is called the third power of a, and is written « 3 ; a x a x a x u is called the fourth power of a, and is written a 4 ; and so on. Similarly aaabbc is called the product of the third power of a, the second power of b, and c, and is written u 3 b'~c. The second power of a, i.e., a 2 , is usually read a to the second power, or a square. The third power of a, i.e., a 8 , is usually read a to the third power, or a cube. There are no such words in use for the higher powers; the fourth power of «, i.e., a 4 , is usually read a to the fourth potcvr, TEE RADICAL SIGN — SYMBOLS OF RELATION. 7 or briefly, a fourth power ; and so on. When the exponent is unity it is omitted. Thus we do not write a 1 , but simply a, which is the same as a 1 , and means a to the first power. 13. The Radical Sign, V . — A root of a quantity is a factor, which, multiplied by itself a certain number of times, will produce the given quantity. The square root of a quantity is that quantity whose square or second power is equal to the given quantity. Thus the square root of 16 is 4, because 4 2 is equal to 16 ; the square root of a 2 is a, of 81 is 9. The square root of a is denoted by sa, or more simply vs. Similarly the cube, fourth, fifth, etc., root of any quantity is that quantity whose third, fourth, fifth, etc., power is equal to the given quantity. The roots are denoted by the symbols \/ , V , \/ , etc. ; thus, 'y/27a s denotes the cube root of 27a 3 , which is 3a, because 3a to the third power is 27a 3 . Similarly ^32 is 2. The small figure placed on the left side of the symbol is called the index of the root. Thus 2 is the index of the square root, 3 of the cube root, 4 of the fourth root, and so on ; the index, however, is generally omitted in denoting the square root ; thus \a is written instead of yja. The symbol V is sometimes called the radical sign. When this sign with the proper index on the left side of it is placed over a quantity it denotes that some root of the quantity is to be extracted. 14. Symbols of Relation. — The symbols of relation are the following : The sign of equality, =, is read equals, or is equal to. When placed between two numbers, it denotes that they are equal to each other. Thus a = 6, read a equals b, or a is equal to b, denotes that the number represented by a is equal to the number represented by b ; or, more briefly, that a equals b. And a + b = c denotes that the sum of the 8 SYMBOLS OF ABBREVIATION. numbers a and b is equal to the number c ; so that if a represent 8 and b represent 4, then c must represent 12. The signs of inequality, > and <, are read is greater than, and is less than, respectively. When either is placed between two numbers it denotes that they are unequal to each other, the opening of the angle in both cases being turned towards the greater number. Thus a > b, read a is greater than b, denotes that the number a is greater than the number b, and b < a, read b is less than a, denotes that the number b is less than the number a. The sign of ratio, : , is read is to or to. When placed between two numbers it denotes their ratio. Thus a : b, read a is to b, or the ratio of a to b, denotes the ratio of the number a to the number b. A proportion, or two equal ratios, is expressed by writing the sign = or the sign : : between two equal ratios. Thus a : b = c : d, or a : b : : c : d , read a is to b as c is to d, or the ratio of a to b equals the ratio of c to d. The sign of variation, cc , is read varies as. When placed between two numbers it denotes that they increase and decrease together, in the same ratio. Thus x a ?/, read x varies as y, denotes that x and y increase and decrease together. 15. Symbols of Abbreviation. — The symbols of abbre- viation are the following : The signs of deduction, .•. is read hence or therefore, and •.• is read .since or because. The signs of aggregation are the bar \ , the parenthesis ( ) , the bracket [ ] , the brace \ \ , and the vinculum . These are employed to connect two or more numbers which are to be treated as if they formed one number. Thus, suppose we have to denote that the sum of a and 6 is to be multiplied bye; we denote it thus (« + b) x c or \a -f b\ X c, or simply (a + b) c or {a 4- l>\ C, here we mean that the whole of a + b is to be multiplied by c. If we omit the ALGEBRAIC EXPRESSIONS. 9 parenthesis, or brace, we have a + be, and this denotes that b only is to be multiplied by c and the result added to a. Also (a + b + c) x (d + e) denotes that the result expressed by a -f- b + c is to be multiplied by the result ex- pressed by d + e. This may also be denoted simply thus (a -r- b + c) (d + e), just as a x b is shortened into ab. If we omit the parenthesis we have a + b + ccZ + e, and this denotes that c only is to be multiplied by d only, and the result added to a '+ b + e. Also V (a -f- & + c) denotes that we are to obtain the result expressed by a + & + c, and then take the square root of this result. Also (ab) 2 denotes ab X ab ; and (ab) 3 denotes ab x ab X a&. Also (a + & + c) -=- (d + e) denotes that the result ex- pressed by a + 6 + c is to be divided by the result expressed by d + e. This may also be expressed by the bracket thus [a + b + c] -7- [d + e] , or the brace ^a +'& + c\ -=- |d + ej, a + & + c or the vinculum a + 6 + c-r-d + e, or — 7f~_r~p — > where the line between the numerator and denominator acts as a vinculum. The signs of continuation are dots , or dashes , and are read and so on. 16. Algebraic Expressions. — The four kinds of sym- bols which have been explained are called Algebraic symbols (Art. 5). Any collection of Algebraic symbols is called an Algebraic expression, or briefly, an expression. Thus 4a + 5b — c + x is an expression ; 3& + 4c is the Alge- braic expression for 3 times the number b increased by 4 times the number c. The numerical value of an expression is the number obtained by giving a particular value to each letter, and then performing the operations indicated. We shall now give some examples in finding the numerical values of expressions, as an exercise in the use of the symbols which have been explained. 10 EXAMPLES. EXAMPLES. If a = 1, 6 = 2, c = 3, a" = 4, e = 5, find the numerical values of the following expressions : 1. 9a + 26 + 3c — 2d. Here we have 9a + 26 + 3c — 2d = 9x1+2x2 + 3x3-2x4 = 9 + 4 + 9-8 = 14 4n*. 2. 7ae + 36c + 9a\ Here we have lae + 36c + 9rf = 7x1x5 + 3x2x3 + 9x4 = 89 Ans. 3. a6ccZ + a6ce + abde + acde + 6c<2e. .Ans. 274. . 4ac . 86c bed r 4. 1 . G. 6 d e K cde , bbed 6ade Q , o. — - -j — . a-i. ab ae be If a = 1, 6 = 3, c = 5, and d = 0, find the numerical values of the following : 6. a 2 + 26 2 + 3c 2 + 4r/ 2 . 7. a 4 - 4a 8 6 + Ga 2 6 2 - 4a6 3 + 6 4 . g 12a 3 - 6 2 2c 2 _ a + lr + <•' 3a 2 a + 6 2 56 3 If a = 1, 6 = 2, c = 3, d = 5, and e = numerical values of the following : 9. 6 2 (« 2 + e 2 - c 2 ). 10. ^(26 + Ad + be). 11. (a 2 + 6 2 + c 2 )(e 2 - d 2 - c 2 ). 12. e -\ositive quantities and negative quantities, according as they are preceded by the sign -f- or the sign — ; and this is quite irrespective of any actual process of addition and subtraction. Illustration (1) Suppose a ship were to start from the equator and sail northward 100 miles and then southward 80 miles, the Algebraic statement would be 100 - SO = +20. Here the positive sign of the result indicates that the ship is 20 miles north of the equator. But if the ship first sailed 80 miles northward and then southward 100 miles, the Alge- braic statement would be 80 - 100 = -20. Here the negative sign of the result indicates that the ship is 20 miles south of the equator. (2) Suppose a man were to gain $40 and then lose $36, his total gain would be $4. But if he first gained $36 and then lost $40, he sustained a loss of $4. The corresponding Algebraic statements would be $40 - $36 = +$4, $36 - $40 = -$4. Here the negative quantity in the second case is interpreted as a debt, i.e., a sum of money opposite in character to the POSITIVE AND NEGATIVE QUANTITIES. 15 positive quantity or gain ill the first case. In Arithmetic we would call it a debt or loss of $4. In Algebra we make the equivalent statement that it is a gain of — $4. (3) Suppose a man starts at a certain point and walks 100 yards to the right in a straight line, and then walks back 70 yards, he will be 30 yards to the right of his starting point. If he first walks from the same point 70 yards to the right and then walks back 70 yards, he will be at the point from which he started. But if he first walks to the right 70 yards and then walks back 100 yards, he will be 30 yards to the left of his starting point. The corresponding Algebraic statements are 100 yards — 70 yards = 30 yards 70 " - 70 " = » 70 « - 100 " = -30 " . Here we see that the negative sign may be taken as indi- cating a reversal of direction. In Arithmetic we would say the man was 30 yards to the left of his starting point. In Algebra we say he was —30 yards to the right of his start- ing point. There are numerous instances like the preceding in which it is con- venient for us to be able to represent not only the magnitude but the nature or quality of the things about which we are reasoning. As in the preceding cases, in a question of position we may have to distin- guish a distance measured to the north of the equator from a distance measured to the south of it; or a distance measured to the right of a certain starting point from a distance measured to the left of it; or we may have to distinguish a sum of money gained from a sum of money lost ; and so on. These pairs of related quantities the Alge- braist distinguishes by means of the signs + and — . Thus if the things to be distinguished are gain and loss, he may denote by 4 or -f4 a gain, and then he will denote by —4 a loss of the same extent. In this way we can conceive the possibility of the independent existence of negative quantities. The signs -f and — , therefore, are used to indicate the nature of quantities as positive or negative, as well as to indicate addition and subtraction (Arts. 8 and 9). In Arithmetic we are concerned only with the numbers which begin at and are represented by the symbols 0, 1, 16 POSITIVE AND NEGATIVE QUANTITIES. 2, 3, 4, etc. without limit, and intermediate fractions. But the quantities which we usually measure by numbers in Alge* bra do not really begin at any point, but extend in opposite directions without limit. In order therefore to measure such quantities on a uniform system, the symbols of Algebra arc- considered as increasing from in two opposite directions ; i.e., besides the symbols used in Arithmetic, we cousidei another set —1, — 2, —3, —4, etc. without limit, and inter- mediate fractions. Symbols iu one direction are preceded by the sign +, and are called positive; and those iu the other direction are preceded by the sign — , and are called negative. Symbols without a sign prefixed are considered to have + prefixed. These two sets of symbols may be illustrated as follows : ... -8, -7, -6, -5, -4, -3, -2, -1, 0, +1, +2, +3, +4, +5, +6, +7, +8, . . . l l l l l !_J l I I l l I l i L_J I the positive being those in the right direction from zero, and the negative those in the left direction from the same point. Thus, if 4 represent a distance of 4 miles measured to the right of a certain point, —4 will represent a distance of 4 miles measured to the left of the same point. If +4 represent 4 degrees above zero, —4 will represent 4 degrees below zero. If +4 represent 4 years after Christ, —4 will represent 4 years before Christ. If +4 represent a fall of four feet, —4 will represent a rise of 4 feet. If +4 represent a (jain of $4, —4 will represent a loss of $4. In general, when we have to consider quantities the exact reverse of each other in their nature or quality, we may regard the quantities of either quality as positive, and those of the opposite quality as negative. It matters not which quality we take as the positive one so long as we take the opposite one as negative; but having assumed at the commencement of an investigation a certain quality as positive, the important point is to use it uniformly and consistently throughout. The absolute value of any quantity is the number repre- sented by this quantity taken independently of the sign which precedes the number. Thus, 3 and —3 have the same absolute value. MULTIPLICATIONS IN ANY ORDER. 17 Negative quantities are often spoken of as less than zero. For example, if a man's debts exceed his assets by $4, it is said that "he is worth $4 less than nothing." In the language of Algebra it would be said " he is worth — $4." A negative number is said to be Algebraically greater than another when it is numerically less, or when it has the smaller absolute value. Thus — 3 > — 6, since —3 is only 3 less than while — G is 6 less than 0, or as a person who owes $3 is better off than one who owes $6 ; or in the case of the thermometer, when the mercury is at 10° below (marked — 10°) at one hour, and at —5° at another hour, the temperature is said to be increasing; i.e., — 5° > — 10°. Also, in Algebra, zero is greater than any negative quantity, as a man who has no property or debt is considered better off than one who is in debt. Thus it is easy to see that in the series on page 16 each number is greater by unity than the one immediately to the left of it. 21. Additions and Multiplications may be Made in any Order. — (1) When a number of terms are con- nected by the signs + and — , the value of the result is the same in whatever order the terms are taken ; thus 6 + 5 and 5 + 6 give the same result viz., 11 ; and so also a + b and b + a give the same result, viz., the sum of the num- bers which are represented by a and b. We may express this fact Algebraically thus, a + 6 = 6 + a. Similarly a — b + c = a + c — b, for in the first of the two expres- sions b is taken from a, and c added to the result ; in the second c is added to a, and b taken from the result. Similar reasoning applies to all Algebraic expressions. Hence we may write the terms of an expression in any order we please, provided each has its proper sign. Thus it appears that a — b may be written in the equiva- lent form — b + a. As an illustration we may suppose that a represents a gain of a pounds, and —6a loss of b pounds ; it is clearly immaterial whether the gain precedes the loss, or the loss precedes the sain. 18 SUGGESTIONS FOR THE STUDENT. (2) When one number, whether integral or fractional, is multiplied by a second, the result is the same as when the second is multiplied by the first. The proof for whole numbers is as follows : Write down a rows of units, putting b units in each row, thus : I | | | | b in a row, I I I I I • I I I I I a rows. Then counting by rows there will be 6 units in a row repeated a times, i.e., b x a units. Counting by columns there will be a units in a column repeated b times, i.e., a X b units. .-. ba = ab. These two laws are together called the Commutative Laic, or Law of Commutation. 22. Suggestions for the Student in Solving Ex- amples. — In solving examples the student should clearly explain how each step follows from the one before it ; for this purpose short verbal explanations are often necessary. The sign " = " should never be used except to connect quantities which are equal. Beginners should be particularly careful not to employ the sign of equality in any vague and inexact sense. The signs of equality, in the several steps of the work, should be placed one under the other, unless the expressions are very short. In elementary work too much importance cannot be at- tached to neatness of style and arrangement. The beginner should remember that neatness is in itself conducive to accuracy. EXAMPLES. Find the numerical value of the following expressions, when a = 1, b = 2, e = 3, d = 4, and e = 5. 1. a 2 + b- + c 9 + d~ + <-. Ans. 55. 2. abc- + bed' 1 — dea". 94. 3. e* -f Mb 2 + 6 4 - 4e"6 - 4e6 3 . 81, EXAMPLES. 1^ 4 ^ de_32^ Am ^ 4a b 2 o 4 . 8a 2 + 36 2 4c 2 + 66 2 _ tf_±& 15 6. t 2 + &a c 2 - b 2 e a 4 + 4 a 3 & + 6a 2 6 2 + ±ab 3 + b* 3 a 3 + 3a 2 6 + 3ab 2 + & s 7 V + ( * c G. ' &2 + d 2 _ 6d 8. (a + b)(b + c)-(b + c)(cA-d) + (c + d)(d + e). 43. 9. ( a -2& + 3c) 2 -(o-2c + 3d) 2 -f(c-2a' + 3e) 2 . 72. 10. V^c 2 + 5d 2 + e). 11. 11. V^e 2 + «" 2 + c 2 - a 2 ). 7. If a = 8, b = 6, c = 1, a; = 9, y — 4, find the value of 14. Find the difference between abx and a + 6 + a,-, when a = 5, 5 = 7, and oj = 12. Ans. 396. 15. When a = 3, find the difference between a 2 and 2a, a 3 and 3a, a 4 and 4a, a 5 and 5a, a 6 and Ga. Ans. 3, 18, G9, 228, and 711. 16. Find the value of 3^ + 2as/(2a + b - x), when a = 6, 6 = 5, c = 4, x = 1. ^l»s. 54. 17. Find the value of (9 - y)(aj + l) + (® + 5) (y + 7) — 112, when a; = 3 and y = 5. -4ns. 0. 18. Find the value of x\Ra? - 8#) + y\R^ + %)i when a; = 5 and 2/ = 3. Ans. 20. 19. If a = 2, 6 = 3, x = 6, and ?/ = 5, find the value of y](a+byy\+f\(a + x)(y-2a)\+]/\(y-by-al Ans.9. Find the value of 20. a; 4 - 11a; 3 - 11a; 2 - 13a; + 11 for x = 12. — 1. 21. x 4 - x 3 — 4a; 2 — 3x — 5 for x = 3. 4. 22. a; 5 - 3a; 2 - 8 for x = 4. 968. 23. 3a; 4 - 60a; 3 + 54a? + 60a; + 58 for x = 19. 115. 20 EXAMPLES. 24. 3x 5 '- lG0.r 4 + 344a 3 + 700x 2 - 1910a; + 1200 for x = 51. Ans. 27. Express the following in Algebraic symbols: 25. Seven times a, plus the third power of b. la + b 3 . 26. Six times the cube of a multiplied by the square of 6, diminished by the square of c multiplied by the fourth power of d. Ans. Ga 3 b 2 - cH\ 27. 3 into x minus m times y, divided by m minus n. Ans. (3x — my) ~ (m — n) . 28. Four times the fourth power of a, diminished by six times the cube of a into the cube of b, and increased by four times the fourth power of b. Ans. 4a 4 — (Ja 3 b 3 + 4b 4 . 29. Six times the square of n, divided by a minus b, increased by six a into the expression m plus n minus b. Ans. -^~ + 6a (in + n - b). a — b 30. Three times the square root of x, diminished by y mto the expression, the square of x plus the product of a; and y plus the square of y, and increased by the square root of x into the expression x square plus y square. Ans. 3\x - y(x- + xy + y-) + \ f x(x- + y-). ADDITION — ALGEBRAIC SUM. 21 CHAPTER II. ADDITION. 23. Addition — Algebraic Sum. — Addition in Alge- bra is the process of finding the Algebraic sum of several quantities. The Algebraic sum of several quantities is their aggregate value, and it is usual to find the simplest equivalent expression for it. It is convenient to make three cases in Addition ; (1) when the terms to be added are like (Art. 18), and have like signs; (2) when they are like but have unlike signs; and (3) when they are unlike. 24. Case 1. To Add Terms which are Like and have Like Signs. — Let it be required to add 8x 2 y, <±x' 2 y, and lx 2 y. Here 8x 2 y is x 2 y taken 8 times, 4x 2 y is x~y taken 4 times, and lx 2 y is x 2 y taken 7 times ; therefore x~y is taken in all 8 + 4 + 7=19 times, and hence the sum is 19an/. The truth of this will be evident to the beginner when he remembers that the three quantities 8 lbs., 4 lbs., and 7 lbs., udded together, give 19 lbs. Similarly 12ab + Sab + hab + ab = 21ab. Let it be required to add — Sab, —lab. and — 9ab. Here — Sab is ab taken —3 times, — lab is ab taken —7 times, and — 9ab is ab taken —9 times ; therefore ab is taken in all —19 times, and hence the sum is — 19«&. The truth of this will be evident from the consideration that, if a sum of money be diminished successively by $3, $7, and $9, it is diminished altogether by $19. Therefore, to add like terms which have the same sign, add the numerical coefficients, prefix the common sign, and annex the common symbols. 22 TO ADD LIKE TERMS WITH UNLIKE SIGNS. For example, Ga + 3a -f a + la = 17a, and — 2a6 — lab -9ab = -18a&. 25. Case 2. To Add Terms which are Like, but have Unlike Signs. — Let it be required to add 9a and -4a. Here —4a destroys 4 of the 9 times a, and gives when added to it, 5a. This is usually expressed by saying —4a will cancel -f 4a in the term da, and leave +5a for the aggre- gate or sum of the two terms. # For if 9a denote $9 which a man has in his possession, and —4a denote a debt of $4, then the aggregate value of his money is $5. In like manner if it be required to add 8a, —9a, —a, 3a, 4a, —11a, a, we find the sum of the positive terms to be 16a, and the sum of the negative terms to be —21a; now + 16a will cancel —16a in the term —21a, which leaves —5a for the aggregate or sum of the terms. Therefore, to add like terms which have not all the same sign, add all the positive numerical coefficients into one sum, and all the negative numerical coefficients into another ; take the difference of these tivo sums, prefix the sign of the greater, and annex the common symbols. For example 7a — 3a+lla-f a — 5a — 2a — 19a — 10a = 9a, and 5a6-6a6+2ao — lab — 3a&+4a&=llao-16a& = -5a&. We need not, however, strictly adhere to this rule, for since terms may be added or subtracted in any order (Art. 21), we may choose the order we find most convenient. Thus, in the last example, we may say 5a& added to —Gab gives — ab ; adding — ab to +2a& gives +a&; adding +ab to — lab gives —Sab; adding — Gab to —3ab gives — 9a& ; adding —9db to + \ah gives — 5a&, for the sum, which is the same as was found by the rule.' 26. Case 3. To Add Terms which are not all Like Terms. — Let it be required to add la + bb — 1c + :\d, 3a — b + 2o + 5a*, 9a — 2b — c — d, and —a + 36 + 4c - 3d + e. TO ADD TERMS WHICH ARE NOT ALL LIKE TERMS. 23 It is convenient to arrange the terms in columns, so that like terms shall stand in the same column ; and then add each column, beginning with that on the left, as follows : 4a +56 -7c +3d* 3a — 6 + 2c + 5d 9a — 2b — c — d ~a +36 +4c -3a" +e 15a + 56 -2c +4d +e Here the terms 4a, 3a, 9a, and —a are all like terms ; the sum of the positive terms is 16a ; there is one negative term, viz., —a, so that the sum of the terms in the first column by Art. 25 is +15a; the sign + may be omitted by Art. 9. Similarly 5b — b — 2b + 3b — 56, — 7c + 2c - c + 4c = - 2c, and so on ; there being no term similar to e, it is connected to the other terms by its proper sign. Therefore, to add terms which are not all like terms, add together the terms which are like terms, by the ride in Case 2, and set doicn the other terms each preceded by its proper sign. In the two following examples the terms are arranged suitably in columns. x 3 +2x 2 - 3a; +1 a 2 + db + lr -c 4a; 3 +7x 2 + a; -9 3a 2 — 3a6 —7b 2 -2x 3 + x 2 - 9a; +8 4a 2 +5a6 +96 2 -3x 8 - x 2 +10aj -1 a 2 -3a6 -36 2 9a;2 _ x __i 9 ft 2 _c In the first example we have in the first column x 3 + 4te B — 2a; 3 — 3a; 3 = 5a; 3 — 5a; 3 = ; this is usually expressed by saying the terms which involve x s cancel each other. Similarly, in the second example, the terms which involve ab cancel each other ; and also those which involve 6 a cancel each other. 27. Remarks on Addition. — "We have seen that when two or more like terms are to be added together they may be collected into a single term (Arts. 24 and 25). If, how-. 24 REMARKS ON ADDITION — EXAMPLES. ever, the terras are unlike they cannot be collected. Thus we write the sum of a and b in the form a + b, and the sum of a and — b in the form a — b. From the foregoing examples it will be observed that in Algebra the word sum is used in a wider sense than in Arithmetic. Thus, in the language of Arithmetic, a — b signifies that b is to be subtracted from a, and has no other meaning ; but in Algebra it also means the sum of the two quantities a and —b without any regard to the relative magnitudes of a and 6. When quantities are connected by the signs -f- and — , the resulting expression is called their Algebraic sum. Thus the Algebraic sum of 12a — 29a + 14a is —3a. In Algebra, wherever the word sum is used without an adjective, the Algebraic sum is understood. EXAMPLES. 1. 2. 3. 4. 4aa; Gab 26a; 2 - 2a?b bax -lab Sbx 2 - a n -b — 3 ax -Sab — 9&a; 2 11 a 2 b 2ax bab Ibx 2 -ba*b — 7 ax -dab 2bx 2 4a 2 b ax 2ab —Abx- -da-b 2ax -6ab -56» 9 2a 2 b b. 7a; 2 -3xy + x 3a; 2 - f + 3a: - - V -2a 2 + 4.T?/ + b>f — x - -ty — "•<'.'/ - f + 9x - - :>// 4a 2 +4/ - 2x 12a; 2 — 6xy +ly 2 + 10a; —8y Add together the following expressions : 6. a+2b — 3c, —3a + b-\-2c, 2a— 36+c. Ans. 0. 7. 3a+26 — c, -a+36+2c, 2a-6+8c. 4o+46+4c EXAMPLES. 25 8. -3 x +2y+z, x-3y + 2z, 2x+y—3z. Ans. 0. 9. - X +2y+3z, Bx-y+2z, 2x+3y—z. 4x+±y+4z. 10. 4a+36+5c, — 2a+36— 8c, a — 6+c. 3a+56— 2c. $s 11. — 15a-196-18c, 14a + 14&+8c, a + hb +9c. -c. 12. 25a — 156+c, 13a — 106+4c, a+206-c. 39a— 56+4c. 13. _16a-106+5c, 10a+5&+c, 6a+5&-c. 5c. In adding together several expressions containing terms with different powers of the same letter, it will be found convenient to arrange all the expressions in ascending or descending powers of that letter (Art. 19). U. 3x 3 +7-5x 2 , 2a,- 2 - 8 -9a,-, 4x-2x 3 +3x\ Arranging the terms in the descending powers of x, we have 3a; 3 -5a; 2 +7 2ar -9a; -8 — 2a; 3 +3a/ 2 +4a; a; 8 —ox —1 Ans. 15. 3a& 2 -26 8 +a s , 5a 2 &-a£/ 2 -3a 3 , 8a 3 +5& 3 , 9a 2 Z>-2a 3 +ab\ Ans. 36 3 +3a& 2 +14a 2 & + 4« 3 . It will be observed that this answer is arranged according to descending powers of &, and ascending powers of a. 16. 2x*-2xy+3y*, ±y 2 +oxy-2x 2 , x 2 -2xy-Gy 2 . Ajis. x 2 +xy+y 2 . 17. a 8 — a 2 +3a, 3a 3 +4a 2 +8a, 5a 8 — 6a 2 — 11a. Ans. 9a 3 -3a 2 . 18. x*+Sx*y+3xy 2 , -3x 2 y-6xy 2 -x 3 , 3x 2 y+4:Xy 2 . Ans. 3x 2 y+xy 2 . 19. a; 8 — 2aa 2 +a 2 a;+a 8 , a; 3 +3aar, 2a 3 -ax 2 -2x 3 . Ans. a 2 a?-r-3a 8 . 20. 2a& - Sax 2 + 2a 2 a;, 1 2a& + 1 Oaa; 2 - Ga 2 .«, - 8a& + ax 3 — bct-x. Ans. Gab — 9a 2 x+7ax 2 +ax 3 . 21. x 2 +*f+z 3 , -4a; 2 -52 3 , 8ar-7?/ 4 +10z 3 , 6y 4 -6z 3 . Ans. ox 2 . 22. a; 4 -4a; 3 ?/-f6xV-4.^ 3 + ? y 4 , Ax 3 y- l2xY+12xy 3 -4y\ 6x 2 y 2 -12xy 3 +Gy\ -ixy 3 —iy\ y\ Ans. xK 26 SUBTRACTION — ALGEBRAIC DIFFERENCE. CHAPTER III. SUBTRACTION. 28. Subtraction — Algebraic Difference. — Subtrac- tion in Algebra is the process of finding the difference between two Algebraic quantities. The Algebraic Difference of two quantities is the number of units which must be added to one in order to produce the other. Thus, what is the difference between 2 and G means " how many units added to 2 will make G " ? The Difference is sometimes called the Remainder. The Subtrahend is the quantity to be subtracted ; or it is the one from which we measure. Thus, 2 is the subtrahend in the above example. The Minuend is the quantity from which the subtrahend is taken ; or it is the one to which we measure. Thus, G is the minuend in the above example. If the minuend is Algebraically greater than the subtra- hend, the difference is positive (Art. 20). If the minuend is Algebraically less than the subtrahend, the difference is negative. In Arithmetic we cannot subtract a greater number from a less one, because subtraction in Arithmetic means taking a less number from a greater. But in Algebra there is no such restriction, because Algebraic subtraction means finding a dfference. 29. Rule for Algebraic Subtraction. — Let distances to the right of the zero point be called positive, and those to the left of the same point be called negative (Figure, Art. 20). Also call measuring toward the right from any point positive, and measuring toward the left from any point negative. RULE FOR ALGEBRAIC SUBTRACTION. 27 Then the difference between 2 and 6 means either how many units must toe measure, and in what direction, in order to pass from 2 to 6 or to pass from 6 to 2. In the first case we begin at 2 and measure four units to the right and say 2 from 6 is +4. In the second case we begin at G and measure four units to the left and say 6 from 2 is —4. That is, if we subtract 2 from 6 the difference is 4 ; but if we subtract 6 from 2 the difference is —4. Also to find the difference between —1 and +1, we may begin at —1 and measure 2 units to the right and get +2, or we may begin at +1 and measure 2 units to the left and get —2 ; i.e., if we subtract — 1 from +1 the difference is +2, but if we subtract +1 from —1 the difference is —2. Similarly the difference between —2 and — 7 is — 5 or + 5, according as we measure from —2 toward the left to —7 or from —7 toward the right to —2; i.e., if we subtract —2 from —7 the remainder is —5, but if we sub- tract — 7 from —2 the remainder is +5. And also, the difference between —6 and +7 is +13 or —13 according as we measure from —6 to +7 or from +7 to — 6 ; i.e., if we subtract — 6 from +7 the difference is 13, but if we subtract 7 from — 6 the difference is —13. Hence we see that the remainder in each case is found by changing the Algebraic sign of the subtrahend, and then adding it Algebraically to the minuend. Otherwise thus. Suppose we have to take 9 + 5 from 16 ; the result is the same as if we first take 9 from 16, and then take 5 from the remainder ; that is, the result is denoted by 16-9-5. Thus 16 - (9 + 5) = 16 - 9 - 5. Here we enclose 9 + 5 in parenthesis in the first expres- sion, because we are to take the whole of 9 + 5 from 16 (Art. 15). Suppose we have to take 9 — 5 from 16. If we take 9 from 16, we obtain 16 — 9 ; but we have thus taken too much from 16, for we had to take, not 9, but 9 diminished 28 RULE FOR ALGEBRAIC SUBTRACTION. by 5. Hence we must increase the result by 5 ; and thus we obtain 16 - ( ( J - 5) = 1G - 9 + 5. Similarly, 1G — (G + 4 - 1) = 16 - 6 - 4 + 1. In like manner suppose we have to subtract b — c from a. If we subtract b from a, we obtain a — b ; but we have thus taken too much from a, for we are required to take, not b, but b diminished by c. Hence we must increase the result a — b by c ; and thus we obtain a — (b — c) = a — b + c for the true remainder. Similarly, a — (b + c — d) = a — b — c + d. Suppose we have to subtract b — c + d — e from a. This is the same thing as subtracting b + d — c — e from a (Art. 21). If we subtract b -\- d from a, we obtain a — b — d; but we have thus taken too much from o, for we were to take, not b + d, but b + d diminished by c and e. Hence we must increase the result by c + e, and thus obtain (6— (b — c + cl — e) = a — b — d+c+e = a— b+c — d+e. From considering each of these examples, it is evident that subtracting a positive number is the same thing as adding an equal negative number, and also that subtracting a negative number is the same thing as adding an equal positive number. Therefore, Algebraic subtraction is equivalent to the Alge- braic addition of a number with the opposite Algebraic sign. Hence for subtraction we have the following Rule. Change the signs of all the terms in the subtrahend, and then add the result to the minuend. EXAMPLES. 1 . Let it be required to subtract 3a;— y+z from 4aj— Sy+2z. Changing the signs of all the terms in the sulitrahend.it stands as follows: — 3a; + y — Z. Then collecting as in addition, we have 4a; — 3y + 2z — ox + y — z = x — 2y -f z. EXAMPLES. 29 2. From 3a; 4 + 5a; 3 — 6sc a — Ix + 5 take 2a; 4 — 2a; 3 + 5ar — 6aj — 7. Changing the signs of all the terms in the subtrahend, and proceeding as in addition, we have 3a; 4 + 5a; 3 - 6a; 2 - 7x + 5 -2x* + 2a; 3 - 5a; 2 + 6x + 7 a; 4 + 7a- 3 - lis 2 - x + 12 Kem. — The beginner may solve a few examples by actually changing the signs of the subtrahend and going through the operation as fully as we have done in these two examples; but he may gradually accustom himself to perform the subtraction without actually changing the signs, but merely changing them mentally, as in the following example. 3. From Sab + lac + 2c 2 take oab — 4ac + 3c 2 — d. Writing the subtrahend under the minuend so that similar terms shall fall in the same column, for convenience (Art. 26), we have 7 . „ . .-, '' Sab -f arenthesis is preceded by the sign -f , the parenthesis can be removed ivithout making any change in the expression within the parenthesis. This rule has already been illustrated in Arts. 25 and 26 ; it is in fact the rule for addition. 7 + (12 + 4) means that 12 and 4 are to be added and their sum added to 7. It is clear that 12 and 4 may be added separately or together without altering the result. Thus 7 + (12 + 4) = 7 + 12 + 4 = 23. Also a + (b + c) means that b and c are to be added together and their sum added to a. Thus a + (b + c) = a + b + c. 7 + (12 — 4) means that to 7 we are to add the excess of 12 over 4 ; now if we add 12 to 7, we have added 4 too much, and must therefore take 4 from the result. Thus 7 + (12 - 4) = 7 + 12 - 4 = 15. Similarly a + (b — c) means that to a we are to add b diminished by c. Thus a + (b — c) = a + b — c. Therefore Conversely : Any part of an expression may be enclosed within a parenthesis and the sign -f- placed before it, the sign of every term within the parenthesis remaining un- altered. Thus, the expression a — b + c — rf + e may be written in any of the following ways : a-6+c+(-d+e), a-6+(c-d+e), a+(— b+c— d+e), and so on. 33. Minus Sign before the Parenthesis.— When a parenthesis is preceded by the sign — , the parenthesis may be removed if the sign of every term within the parenthesis be changed. 32 COMPOUND PARENTHESES. This rule has already been illustrated in Art. 29 ; it is in fact the rule for subtraction. The rule is evident, because the sign — before a parenthesis shows that the whole ex- pression within the parenthesis is to be subtracted, and the subtraction is effected by changing the signs of all the terms of the expression to be subtracted. Thus a — (b + c) = a — b — c. Also a — (b — c) = a — b + c. Therefore Conversely : Any part of an expression may be enclosed within a p>arenthesis and the sign — placed before it, provided the sign of every term within the p>arenthesis be changed. The proof of this operation is to clear the paren- thesis introduced, and thus obtain the original expression. Thus a — b + c + d — e may be written in the following ways: a-b+c— (-cZ+e), a—b— (— c — d+e), a—(b-c— d+e), and so on. 34. Compound Parentheses. — Expressions may occur with more than one pair of parentheses ; these parentheses may be removed in succession by the preceding rules. We may either begin with the outside parenthesis and go inward, or begin with the inside parenthesis and go outward. It is usually best to begin with the inside parenthesis. The beginner is recommended always to remove first the inside pair, next the inside of all that remain, and so on. Thus for example ; a + \b + (c - d) \ = a + ]b + c - d\ = a + b + c - d. a + \b — (c — d) \ = a + $6 — c + d\ = a + 6 — c + d. a - \b + (c - d) \ = a - \b + c - d\ = a - b - c + d. a - \b - (c - d) \ = a - \b - c + d\ = a - b + c - d. It will be seen in these examples that, to prevent confusion between different pairs of parentheses, we employ those of different forms; and hence we use. besides the parenthesis, the brace, the bracket, and sometimes the vinculum (Art. 1 5 ) . COMPOUND PARENTHESES — EXAMPLES. 33 Thus, for example, a — [6 — \ c — (d — e~ = a — [6 — {c — (Z-f = a — & + c — d + ■«-/■ c + d - - e + /] Also -26 - [4a - 66 - ^3a — c + (5a — 26 - a - 3a — c : + 2&)fj = a - 26 - [4a - 66 - { 3a - c + (5a - 26 - 3a + c - 26) j ] = a _ 26 - [4a - 66 - \3a - c + 5a - 26 - 3a + c - 26 }] _ a _ 26 - [4a - 66 - 3a + c - 5a + 26 + 3a - c + 26] _ a - 26 - 4a + 66 + 3a — o + 5a — 26 — 3a + c — 26 = 2a, by collecting like terms. EXAMPLES. Simplify the following expressions by removing the paren- theses and collecting like terms. 1. a — (6 — c) + a + (6 — c)+6— (a+c). Ans. a+b—c. 2. a - [6 + \a - (6 + a)\]. a. J) a - [2a - ^36 - (4c - 2a) \]. a + 36 - 4c. (1t> \a.-(b-c)\ + \b-(c-a)\-\c-(a-b)\. 3a-6-c. 5. 2a-(56 + [3c-a])-(5a-[6+c]). -2a-46-2c. 6. -\a-\b-(c-a)X]-\b-\c-(a-b)X\. b-a. ~D -(-(-(-*)))-(-(- *))■ «- y. 8 . _[ 5a5 _ (iij, _ 3a;)] - [5y - (3a - 6?/)]. -5a;. 9. -[15a - \Uy - (15z + 12?/) - (10a; - 15s) J]- J.ws. —25a; + 2y. 10. 8a;— |16y — [3a;-(12j/ — x) — 8y'] + x\. 11a;- 36?/. 11. - [x - \z + (a; - s) — (s — a;) — sj - a?]. 2x - 2s. 12. —[a 4- \a - (« — ») — (« + ») - aj — «]• 2 «. 13. -[a — ja + (a - a) — (« — a) — a| - 2a]. a. 14. 2a - [2a - ^2a - (2a - 2a — a)}] . a. 15. 16 - x —[7x - ^8a; - (9jb - 3a; — 6a;) |]. 16 - 12a;. 16. 2a; - [3?/ - |4a> - (by - 6a; - 7t/)|]. 12a; - 15?/. 17. 2a-[36 + (25-c) -4c + j2a-(3&-c= 26)|]. 4c. 18. a- [56- {a- (5c-2c^6-46) + 2a - (a - 26 + c) \]. Ans. 3a — 2c. 34 EXAMPLES. 19. jb»_ [4. 1; 3_ s, 6aJ 2_ ( 4iC _ ! ) |] _ (. 1 .4 + 4r 3_ } _ fil . 2+47 . + j ^ .4?is. — -•<»- •>}-{*-? -(a -6)1. Ans. a - y-6 + V 1 EXAMPLES. 35 »■ HMHHHH')}-^)) ./Ins. 0. The terms of an expression can be placed in parentheses in various ways (Arts. 32 and 33). Thus, 33. ax — 6x + ex — ay + by — cy may be written (ax — bx) + (ex — a?/) + (&»/ - cy), or (ax — 6jb + ex) — (a?/ — by + cy), or (ax — a?/) — (6x — by) + (ex — cy). Whenever a factor is common to every term within a parenthesis, it may be placed outside of the parenthesis as a multiplier of the expression within. Thus, 34. ax 8 + 7 — ex — dx 2 — c + bx — dx 9 + bx 2 — 2x = (ax s - dx 3 ) + (bx 2 - dx 2 ) + (&x - ex - 2x) + (7 - c) = (a - d)x s + (b- d)x 2 + (b - c - 2)x + (7 - c). In this result, (a — d), (b — d), (b — c — 2) are regarded as the coefficients of x 3 , x 2 , and x, respectively (Art. 17). Hence we have here placed together in parentheses the coefficients of the different powers of x so as to have the sign + before each parenthesis. 35. — a 2 x — la + a 2 y + 3 — 2x — db _ -(a 2 x - a 2 y) - (7a 4- aft) - (2x - 3) = -( X _ y ) a 2 - (7 4- 6)a - (2x - 3). We have here placed together in parentheses the coeffi- cients of the different powers of a so as to have the sign — before each parenthesis. In the following four examples place together in paren- theses the coefficients of the different powers of x so that the sign 4- will be before all the parentheses. 36. ax 4 4- bx 2 + 5 + 2bx — ox 2 4- 2x 4 — 3x. Ans. (a + 2)x 4 + (b - 5)x 2 4- (26 - 3)x + 5. 37. 36x 2 — 7 - 2x 4- «6 4- 5ax 3 + ex — Ax 2 — 6x 3 . Ans. (5a - b)x 3 + (36 - 4)x 2 + (c - 2)x 4- ab - 7. 38. 2 — 7x 3 + 5ax 2 - 2cx 4- 9ax 3 + 7x — 3x 2 . Ans. (9a - 7)x 3 4- (5a - 3)x 2 4- (7 - 2c)x + 2. 36 EXAMPLES. 39. 2CX 5 - Sabx + idx - 3bx* - a*x* + x\ Ans. (2c - a?)x* + (1 - 36)aJ* + (4d - 3a6)a;. In the following four examples place together in paren- theses the coefficients of the different powers of x so that the sign — will be before all the parentheses. 40. ax 2 + 5.r 3 - a¥ — 2bx* - 3x 2 - bx\ Ans. -(a 2 + 6).t 4 - (26 - 5).r 3 - (3 - a)x 2 . 41. 7.i- 3 — 3c*x — abx 5 + 5aa; + 7.r 5 — (ibex*. Ans. — (ab — 7)x- 5 — (abc — l)x 3 — (3c 2 — 5a)x. 42. ax 2 + a 2 a: 3 — bx 2 — 5 a; 2 — ex 8 . Ans. — (c — a 2 )# 3 — (6 + 5 — «).r. 43. 36 2 a; 4 - 6a; - ax* - rac 4 - 5c 2 a; - 7x*. Ans. -(a + c + 7 - 36 2 )«* - (6 + 5c 2 )a;. Simplify the following expressions, and in each result place together in parentheses the coefficients of the different powers of oj. This is known as re-grouping the terms accord- ing to the powers of x. 44. ax 8 — 2cx— [6a; 2 — \cx—dx— (6a; 3 +3ca; 2 ) \ — (ex*— bx)~\. Ans. (a - b)x 3 - (6 + 2c) a 2 - (6 + c + d)x. 45. ax 2 - 3^-aa; 3 + 36a; - 4[£c.f 8 - §(ax - bx*)]\. Ans. (3a + 2c)a; 3 + (a + Sb)x 2 - (8a + 9b)x. 4G. x .5_4^4_ir i2ax _4 ( 35aj*_9/^_ 6aA-|oaJ*l |. Ans. (66 + l)x 6 - (a +2b)x* - (2a + 3c)a;. "We shall close this chapter with a few examples in Addi- tion and Subtraction. 47. To the sum of 2a — 36 — 2c and 26 — a + 7c add the sum of a — 4c + 76 and c — 66. Ans. 2a -f 2c. 48. Add the sum of 2y — Sy 2 and 1 — 5y s to the remainder left when 1 — 2y 2 + y is subtracted from b>f. Ans. —y 2 -f y. 49. Take a; 2 — y 2 from 3xy — 4y 2 , and add the remainder to the sum of Axy — x 2 — Sy 2 and 2x 2 -f- 6i/ a . ^w. 7.r/./. 50. Add together 3.t 2 — 7x + 5 and 2a; 3 + 5a; — 3, and diminish the result by fix* + 2. Ans. 2x* — 2&. 51. What expression must be subtracted from 3a — 56 -f c so as to leave 2a — 46 + c? Ans. a — b. «*£ EXAMPLES. 37 52. From what expression must 3a5 + o&c — Gca be sub- tracted so as to leave a remainder Gca — 56c? Ans. Sab. 53. Subtract 3a; 3 — 7x + 1 from 2a; 2 — ox — 3, then subtract the difference from zero, and add this last result to 2x 2 — 2x 3 - 4. Ans. x* — 2x. 54. Subtract 3ar — 5x + 1 from unity, and add 5a; 2 — 6a; to the result. Ans. 2x' 2 — x. 55. To what expression must Tar 5 — 6a; 2 — 5a; be added l so as to make \)x 3 — 6x — 7x 2 ? Ans. 2x 3 — a; 2 — x. 56. From 5a; 3 + 3a; — 1 take the sum of 2a; — 5 + 7 'x* and 3a; 2 + 4 — 2a; 3 + x. Ans. 7x 3 — 10a; 2 . &z). Subtract 3a — 7a 3 + 5a 2 from the sum of 2 + 8a 2 — a 3 and 2a 3 - 3a 2 + a - 2. Ans. 8a 3 — 2a. 58. If 3x 2 — 7a; + 2 be subtracted from zero, what will be the result? Ans. -3a; 2 + 7a; - 2. 59.' Subtract 5a; 2 + 3a; — 1 from 2a; 3 , and add the result to 3a; 2 + 3x - 1. Ans. 2x s - 2a; 2 . GO. What expression must be added to 5a; 2 — 7a; + 2 to produce 7a; 2 - 1? Ans. 2a; 2 + 7a; - 3. 61. What expression must be added to 4a; 3 — 3a; 2 + 2 to produce 4a; 3 + 7a; - 6 ? Ans. 3a; 2 + 7a; - 8. 62. What expression must be subtracted from 9a; 2 +lla;— 5 so as to leave 6a; 2 - 17a; + 3? Ans. 3a; 2 + 28a; — 8. 63. From what expression must 11a 2 — 5a& — Ibc be subtracted so as to give for remainder 5a 2 + 7a& + 76c ? Ans. 16a 2 + 2a6. 38 MULTIPLICATION — RULE OF SIGNS. CHAPTER IV. MULTIPLICA T I O N. 35. Multiplication in Algebra is the process of taking any given quantity as many times as there are units in any given number.* The Multiplicand is the quantity to be taken or multiplied. The Multiplier is the number by which it is multiplied. The Product is the result of the operation. The multiplicand and multiplier taken together are called Factors of the product. In Algebra as in Arithmetic, the product of any number of factors is the same in whatever order the factors may be taken (Art. 21). Thus, 2x3x5 = 2x5x3 = 3x5x2, and so on. In like manner abc = acb = bca, and so on. Also 2a x 36 = 2 x a x 3 x & = 2x3«XaX& = 6a&. 36. Rule of Signs. — The rule of signs, and especially the use of the negative multiplier, usually presents some difficulty to the beginner. (1) if +a is to be multiplied by + <\ this indicates that -\-a is to be taken as many times as there are units in c Now if +a be taken once, the result is -\-a ; if it be taken twice, the result is evidently +2a; if taken three times, the result is + 3«; and so on. Therefore if +a be taken c times, it is +ca or +a<-. That is +a x +c = +oc. (2) If —a is to be multiplied by -+-C, this indicates that —a is to be taken as many times as there arc units in c. * This •definition Is true only of whole numbers. RULE OF SIGNS. 39 Now if —a be taken once, the result is —a; if it be taken twice, the result is — 2a; if taken three times, the result is —3a; and so on. Therefore if — a be taken c times, it is —ca or —ac. That is, — a x +c = —ac. Similarly — 3 x +4 = —3 taken four times = -3 -3 -3 -3 = -12. (3) Suppose +« is to be multiplied by — c. We have illustrated the difference between +c and — c (Art. 20), by supposing that +c represents a line of c units measured in one direction, and — c a line of c units measured in the oppo- site direction. Hence if +a is to be multiplied by — c, this indicates that + a is to be taken as many times as there are units in +c, and further that the direction of the line which represents the product is to be reversed. Now +a taken +c times gives +ac ; and changing the sign, which corresponds to a reversal of direction, we get —ac. That is +a x — c = —ac. Similarly +3 X —4 indicates that 3 is to be taken 4 times, and the sign changed. The first operation gives +12, and the second —12. That is +3 x -4 = -12. (4) If —a is to be multiplied by — c, this indicates that — a is to be taken as many times as there are units in c, and then that the direction of the line which represents the product is to be reversed. Now —a taken c times gives — ac; and changing the sign, which corresponds to a reversal of direction, we get -\-ac. That is — a x — c = +ac. Similarly —3 x —4 indicates that —3 is to be taken 4 times, and the sign changed. The first operation gives —12, and the second +12. That is, -3 X -4 = +12. 40 EXAMPLES. (3) is sometimes expressed as follows : +« multiplied by — c iudicates that +a is to be taken as many times as there are units in c, and then the result subtracted. Now +fz. 2aj T yV. 4. a6 + 6c and a 3 6. a*6 s + a*b*c. 5. 5a + ."•// and 2a;-. 10.x 3 + 6afy. 0. 6c + ca — a6 and a6c. a6V 4- «~b< : ~ — "-/'■'■. 7. 5xhf + a*/ 9 - 7. ( '-//-and S.ry\ 40«V + 8«Y - 56*Y. 8. 6a?bc - 7a& 2 c 2 and 6 . 2a'b 3 x 6 - .0. 8a 2 6 8 - §& 2 c 3 and §a& 4 . 12a 3 i THE MULTIPLICATION OF POLYNOMIALS. 43 - 7rt 3 6 4 C 2 . lOaW 5 . — a6 6 c 8 . 39. The Multiplication of a Polynomial by a Polynomial. — Suppose we have to multiply c + d by a + &. Here we are required to take c + d as many times as there are units in a + &, i.e., we are to take c + d as many times as there are units in a, and then add to this product c + d taken as many times as there are units in b. Hence (a+&) (c+d) = (c+d) taken a times together with (e+d) taken & times = (c+d)a+(c+d)& = ac+ad+&c+5d[(l)ofArt.38]. (1) Again (a— 6) (c+d) = (c+d) taken a times diminished by (c+d) taken & times = (c+d)a— (c+d)b = ftc+ad- (bc+bd) [(l)of Art. 38] = ac+ad-&c-6d (Art. 33) . . . (2) Also (a + 6 ) (c — d) = (c — d) taken a times together with (c— d) taken & times = (c— d)a+(c— d)6 = ac-ad+6c-6d[(2)ofArt.38]. (3) Lastly (« — 6) (c— d) = (c—d) taken a times diminished by ( c ~ d ) taken 6 times = (c— d)a— (c— d)& = ac'—ad- (bc—bd) [(2) of Art. 38] = oc-od-6c+6d (Art. 33). . . (4) Hence, to multiply one polynomial by another, we have the following Rule. Multiply each term of the multiplicand by each term of the multiplier; if the terms multiplied together have the same sign, prefix the sign + to the product, if unlike, prefix the sign — ; then add the partial products to form the complete product. 44 EXAMPLES. If we consider each term in the second member of (4), and the way it was produced, we find that +a x +c = +ae. -fa x — d — — ad. -b x -fc = -6c. _6 x -d = +bd. These results enable us again to state the rule of signs, and furnish us with another proof of that rule, in addition to the one given in Art. 36. This proof of the rule of signs is perhaps a little more satisfactory than the one given in Art. 36, though it is not quite so simple. EXAMPLES. 1. Multiply x + 7 by x -f- 5. The product = (x + 7) (x + 5) = x' 2 + Ix + 5x + 35 = x 2 + 12b + 35. Rem. — It is more convenient to write the multiplier under the multiplicand, and begin on the left and work to the right, placing like terms of the partial products in the same vertical column, as follows : x + 7 x + 5 addition x- + 7x + 5x + 35 by x* + 12x + 35. 2. Multiply 3x — iy by 2x — 3y. 3x — Ay 2x - Sy Gx 2 - 8xy - 9xy + 12?/ 2 by addition 6x* — 11 xy + 12jr. Here the first line under the multiplier is the product of the multiplicand by 2x ; the second line is the product of the EXAMPLES. 45 multiplicand by —3?/; like terms are placed in the same vertical column to facilitate addition. Find the product of the following : 3. a; — 7 and x — 10. Ans. x 2 — 17a; + 70. 4. x — 7 and x + 10. a; 2 + 3a; — 70. 5. aj — 12 and 05 — 1. x' - 13x + 12. 6. x — 15 and x -\- 15. a; 2 - 225. 7. — .c - 2 and —a; — 3. a; 2 + 5a; + 6. 8. —x + 5 aud — x — 5. x 2 - 25. 9. a; — 17 aud % + 18. a; 2 + x - 306. .0. —x — 16 and — flJ + 16. a; 2 - 256. .1. 2x — 3 and x + 8. 2a; 2 + 13a; - 24. .2. 3a; - 5 and 2aj + 7. 6a; 2 + 11a; - 35. .3. 4a a - 5a6 + 66 2 and 2a 2 - - Sab + 46 3 . 4a 2 - 5a6 + 6b 2 2« 2 - 3a6 + 4& 2 8a 4 - 10a 3 6 + V>a 2 b 2 - 12a 8 6 + 15a 2 6 2 - 18«& 3 + i6a 2 6 2 - 20a6 3 + 246 4 . 8a 4 - 22a 3 6 + 43a 2 6 2 - 38o6 3 + 246 4 . Here the first line under the multiplier is the product of the multiplicand by 2a 2 ; the second line is the product by — 3ab ; the third by 4& 2 ; like terms are set down in the same vertical column to facilitate the addition. The student will observe that both the multiplicand and multiplier are arranged according to the descending powers of a (Art. 19). Both factors might have been arranged according to the ascending powers of a. It is of no conse- quence which order we adopt, but we should take the same order for the multiplicand and multiplier. If the multiplier and multiplicand are not arranged accord- ing to the powers of some common letter, it will be convenient to rearrange them. Thus : 46 EXAMPLES. 14. Multiply 3.r -f 4 + 2a; 2 by 4 + 2x 2 — Bx. Arranging the factors according to the descending powers of x, the operation is as follows : 2x- + Bx -f 4 2a; 2 - 3a; + 4 4a; 4 + 6a; 3 + 8a; 2 - 6x 3 - 9x* - 12a; + 8.r + \2x + 16 4x 4 + 7.c 2 + 16. 15. Multiply a 2 4- b 2 + c 2 — a& — be — ca by a ■ + 6 + c. Arrange according to descending powers of a. a 2 — a& — ac 4- & 2 — be 4- c 2 a 4-6 + c a 3 — a 2 6 — a 2 c 4- o& 2 — a&c 4- ac 2 + «' 2 6 — ab 2 — abc 4- b 3 — b 2 c + 6c 2 4- a 2 c — abc — ac 2 + 6 2 c — 6c 2 4 c 3 a 3 - 3a&c + b z + e 8 Rem. — The student should notice that he can make two exercises in multiplication from every example in which the multiplicand and multiplier are different polynomials, by changing the original multi- plier into the multiplicand, and the original multiplicand into the multiplier. The result obtained should be the same in both opera- tions. The student can therefore test the correctness of his work by interchanging the multiplicand and multiplier. Multiply together 16. a 2 — ab + b 2 and a 2 4- ab + b 2 . Ans. a* 4 a 2 b 2 4 o 4 . 17. x 2 4- By 1 and x 4 4y. x 3 4 Ax 2 :/ 4- oxy 2 4 12// ? . 18. a; 4 — x 2 y 2 4 ?/ 4 and x 2 + y 2 . 19. a 2 — 2ax 4- Ax 2 and a 2 + 2ax + 4 x 2 . 20. 16a 2 + 12ab + 9& 2 and 4a - 36. 21. a 2 x — ax 2 4 x 3 — a 3 and & + Q» 22. 2a; 3 - 3a; 2 + 2sc and 2a; 2 + 3a; + - 23. -a 5 4- « 4 & - a 3 b 2 and -a - 6. a 6 + « 8 & 8 - 24. a 3 4 2a s 6 4 2a& 2 and a a - 2a& 4 26 s . a 5 4- 4a6 4 . a; 6 + y r '. a 4 4- 4a 2 a; 2 + 16a; 4 . 64a 3 - 276". a; 4 - a 4 . 4a; 5 - x 3 4 !•'•■ EXAMPLES. 47 When the coefficients are fractional we use the ordinary process of multiplication, combining the fractional coefficients by the rules of Arithmetic* 25. Multiply 4a 2 - 4a& + f& 2 by 4a + 4> 4a 2 - fa& + f& 2 4a + lb l a 3 _ l a 2ft _|_ l a &2 l fl 3 _ _5^ + l a& 2 + | &8< Multiply together 26. |a 2 + 4a + i and A/t — ^. J.?is. |a 8 + ^a — ^. 27. f« 2 + a^ + \tf and 4a - |#. -fa 3 - \tf. 28. 4s 2 - fe - f and 4s 2 4- f a - |. K - ffa; 2 + tV 29. fax + fa 2 + 4a 2 and fa 2 + fa; 2 - faa?. ±a 4 + a; 4 . It is sometimes desirable to indicate the product of poly- nomials, by enclosing each of the factors in a parenthesis, and writing them in succession. When the indicated multi- plication has been actually performed, the expression is said to be expanded, or developed. Expand the following : 30. (2a + 36) (a — 6). Ans. 2a 2 + ab — 3b 2 . 31. (a 2 + ax + a; 2 ) (a 2 — ax + x 2 ). x A + a 2 x 2 + a*. 32. (a 2 + 2a6 + 2& 2 ) (a 2 - 2a& + 26 s ). « 4 + 4& 4 . 33. [x - 3)(x + 4) (a; — 5)(a> + (5). -4ns. a; 4 + 2x 3 - 41a? - 42a; + 3G0. 34. (a 2 + ab + & 2 ) (a 3 - a 2 6 + & 3 ) (a -6). ^?is. a 6 - a 6 & + a 2 & 4 - & 6 . 35. (2a? + 4x 2 + 8a; 4- 16) (3a; - 6). 6a; 4 - 96. 36. (x* -{- x 2 + x - \){x - \). x i - 2x + 1. 37. (x + a) [(a; + b)(x + c) - (a + 6 4- c)(a; + 6) + (a 2 4- ab + ft 2 )]. -4ws. a; 3 + a 3 . * The student is supposed to be familiar with Arithmetic fractions, which are the only fractions that are used in this work previous to Chapter VIII. 48 MULTIPLICATION BY INSPECTION — EXAMPLES. 40. Multiplication by Inspection. — Although the result of multiplying together two binomial factors can always be obtained by the methods explained in Art. 3 ( J, yet it is very important that the student should learn to write down the product rapidly by inspection. This is done by observing in what way the coefficients of the terms in the product arise ; thus (x + 5) (a; + 3) = a 2 -f 5a + 3x + 15 = x 2 + 8x + 15. (x - 5) (x + 3) = x 2 - bx + 3a; - 15 _ x i _ 2x - 15. (x + 5)(aj - 3) = a- 2 + 5x - 3a; - 15 = x 2 + 2x — lf>. (x - 5)(cc - 3) = x 2 - 5x - Sx + 15 = x 2 - 8x + 15. It will be noticed in each of these results that : 1. The product consists of three terms. 2. The first term is a; 2 , and the last term is the product of the second terms of the two binomial factors. 3. The middle term has for its coefficient the Algebraic sum of the second terms of the two binomial factors. Hence the intermediate steps in the work may be omitted, and the product written down at once, as follows : (x + 2) (x + 3) = a; 2 + 5as + 6. (x - 3) (a; + 4) = x 2 + x - 12. (x + 6) (a; - 9) = x 2 - 3a; - 54. (x - 4y)(x - lOy) = x 2 - Uxy + 40y 2 . (as - by) \x + Gy) = x 2 + xy - 30t/ 2 . EXAMPLES. Write down the values of the following products, l. (a: + 8) (x - 5). Ana. x* + 3aj - 40. •_'. (./• - 3) (x + 10). x 2 + 7a? - 30. 3. (.c + 7) (a; - !)). x 2 - 2x - 63. I. (r - 4) (a; + 11). x 2 + 7as - 44. SPECIAL FORMS OF MULTIPLICATION. 49 5. (»+ 2) (a- 5). Ans . x 2 - Sx - 10. 6. (*+ 9)(tB- 5). x 2 + 4x — 45. 7. (» -8)(oj + 4). a-a _ 4x _ 32. 8. (x - G)(x + 13). a;' 2 + 7x — 78. 9. - 11) 0+ 12). x- 2 + x - 132. 10. (x — 3a) (x + 2a) . # 2 — a# — 6a 2 . 11. (x - 9b)(x + 86). x 2 - bx - 72& 2 . 12. (x - ly) (x - 8y) . X 2 — 15xy + 562/ 2 . 43. Special Forms of Multiplication — Formulae. — There are some examples iu multiplication which occur so often in Algebraic operations that they deserve especial notice. If we multiply a + b by a + b we get (a + b) (a + b) = a 2 + 2a6 + b 2 ; that is (a + b) 2 = a 2 + 2ab + b 2 . . . . (1) Thus £/*e square of the sum of two numbers is equal to the sum of the squares of the two numbers increased by twice their product. Similarly, if we multiply a — b by a — b we get (a - b) 2 = a 2 - 2ab + b 2 (2) Thus the square of the difference of two numbers is equal to the sum of the squares of the two numbers diminished by twice their product. Also, if we multiply a + b by a — b we get (a + b) (a - b) = a 2 - b 2 (3) Thus the product of the sum and difference of two numbers is equal to the difference of their squares. Because the product of two negative factors is positive (Art. 36), it follows that the square of a negative number is positive. For example, (-a) 2 =a* = (+a) 2 , and (b — a) 2 = a 2 — 2ab + b 2 = (a — b) 2 . 50 EXAMPLES. Hence ft 2 — 2ab -f b 2 is the square of both a — b and b — a. Rem. 1. — Equations (1), (2), and (3) furnish simple examples of one of the uses of Algebra, which is to prove general theorems respecting numbers, and also to express those theorems briefly. For example, the result (a -+- b)(a — b) = a 2 — b 2 is proved to be true, and is expressed thus by symbols more compactly than it could be by words. A general result thus expressed by symbols is called a formula; hence a formula is an Algebraic expression of a general rule. Rem. 2. — We may here indicate the meaning of the sign ± which is made by combining the signs + and — , and which is called the double sign. By using the double sign we may express (1) and (2) in one formula thus : (a ± b) 2 = a 2 ± 2ab + b 2 , . . . . (1) where ± , read plus or minus, indicates that we may take the sign + or — , keeping throughout the upper sign or the lower sign. Formulae (1), (2), and (8) arc true whatever may be the values of a and b. The following examples will illustrate the use that can be made of formulae (1), (2), and (.'5). The formulae will sometimes be of use in Arithmetic calculations. Thus EXAMPLES. 1. Required the difference of the squares of 127 and L23. By formula (3) we have (127) 2 - (128) 2 = (127 + 128) (127 - 128) = 2;")0 x 1 = 1(«)0. 2. Required the square of 29. By formula (2) (2!»)-= (30 - l) 9 = 900 - GO + 1 = 841. EXAMPLES. 51 3. Required the product of 53 by 47. By formula (3) 53 X 47 = (50 + 3) (50 - 3) = (50) 2 - (3) 2 = 2500 - 9 = 2491. 4. Required the square of 34. By formula (1) (34) 2 = (30 + 4) 2 = 900 + 240 + 1G = 1156. 5. Required the square of Ax + oy. We can of course obtain the square by multiplying Ax + 3y by itself in the ordinary way. But we can obtain it by formula (1) more easily, by putting Ax for a and 3y for b. Thus (4a; + 3y) 2 = (Ax)* + 2(4a%) + (3yY = 16x- 2 + 2Axy + 9t/ 2 . G. Required the square of x + y + %• Denote x + y by a ; then x-\-y + z = a-\-z; and by (1) we have (a + z) 2 = a 2 + 2az -f- z 2 = (a> + yY + 2(x + y)g + 2 2 = x 2 + 2xy + ^ + 2xz + 2tf2 + z 2 Thus (» + y + z) 2 = f 2 -f- r + 2 3 + 2an/ + 2yz + 2xz. That is, the square of the sum of three numbers is equal to the sum of the squares of the three numbers increased by twice the products of the three numbers taken two and two. 7. Required the square of p — q + r — s. Deuote p—q by a and r—s by b ; then p — q+r— s=a+b ; and by (1) we have (a+&) 2 =a a +2a&+& 2 = (p- q y+2(p-q) (r-s) + (r-s) 2 . Then by (2) we expand (p — qY and (/• — s)' 2 . Thus (p — q + r — s) 2 = p 2 — 2pq + g 2 + 2 (jpr — ps — qr + " + s 9 + 2pr + 2gs — 2pg — 2ps — 2qr — 2rs. 52 EXAMPLES. 8. Required the product of p — q+r—s and p— q— r+s. Let p — q — a and r—s = b ; then p — q + r — s = a + 6, and p — g — r + s = a — 6; and by (3) we have (a+6) (a-6) =a 2 -6 2 = (p-g) 2 - (r-s) 2 = p 2 -2pg+g 2 -(V 2 -2rs+s 2 ) by (2). Thus(p— tf+r— s)(p— q — r+s)=p 2 +g 2 — r 2 — s 2 — 2pq+2rs. From these examples we see that by using formula? (1), (2), and (3), the process of multiplication may be often simplified. The student is advised first to go through the work fully as we have done ; but when he becomes more familiar with this subject, he may dispense with some of the' work, and thereb}' simplify the multiplication still more. Thus in the last example he need not substitute a and &, but apply formula (3) at once, and then (2), as follows : [(P - - q) - (r - «)] = (p - y 2 ) 2 . 49a; 4 - 70aV + 25y 4 . 12. (a; 2 + 2x - 2) 2 . a; 4 + 4a; 3 - 8x + 4. 13. (a; 2 - bx + 7) 2 . a; 4 - 10a; 3 + 39a; 2 - 70a + 49. IMPORTANT RESULTS IN MULTIPLICATION. 53 14. (2x 2 -3x-4:) 2 . Ans. 4x 4 - 12a 3 - 7a; 2 + 24.K+ 16. 15. (x + 2y + 3z) 2 . x 2 + 4y 2 + 9z 2 + Axy + 6a-z -f 12yz. 16. (aj a + a^ + y 2 )(» 2 + a^-y 2 ). jb* + 2a?y + ary - &. 17. (x 2 + an/ + y 2 ) (a? - any + r) . a?* + x 2 y 2 + y\ 42. Important Results in Multiplication. — There are other results in multiplication which are important, although they are not so much so as the three formulae in Art. 41. We place them here in order that the student may be able to refer to them when they are wanted ; they can be easily verified by actual multiplication. (a + &)(a 2 - ab + b 2 ) = a 3 + b 3 . . . . (1) (a - b) (a 2 + «& + & 2 ) = « 3 - b 3 . . . . (2) (a + 6) 3 = (a+&) (a 2 + 2ab + & 2 ) = a 3 + 3a 2 & + 3a& 2 + & 3 . (3) (a- 6)8= (a- 6) (a 2 - 2ab + & 2 ) = a 3 - 3a 2 & + 3a& 2 - Z> 3 . (4) (a + b + c) 3 = a 3 + 3a 2 (& + c) + 3a(6 + c) 2 + (6 + c) 3 =a 8 +6 8 +c 8 +3a 2 (&+c) +3& 2 (a+c) +3c 2 (a+&) +6abc. (5) Kem. — It is a useful exercise iu multiplication for the student to show that two expressions agree in giving the same result. For example, show that (a — b)(b- c){c - a) = a 2 (c - 6) + b 2 (a - c) + c 2 (6 - a). Here we proceed as follows: Multiplying (a — b) by (6 — c) we obtain (a — 6)(6 — c) = ab — b 2 — ac + be, then multiplying this equation by c — a we obtain (a — b)(b — c)(c — a) = cab — c& 2 — ac 2 + be 2 — a 2 & + a& 2 + a 2 c — «6c - a 2 (c — 8) + & 2 (a — c) + c-(6 — a) . . . (0) Show that (a - 6) 2 + (b — c) 2 + (c - a) 2 = 2(c - b)(c - a) + 2(6 - a)(b - c) + 2(a - 6)(« - c). By (2) of Art. 41 we obtain (a - b) 2 + (b- c) 2 + (c - a) 2 = a 2 — 2ab + 6 2 + 6 2 — 2bc + c 2 + c 2 - 2ac + a 2 = 2(o 2 + b 2 + c 2 — a6 — 6c — co) (7) Now (c — 6)(e — a) = c 2 — ca — cb + a6, (6 — a)(b — c) — 6 2 — be — ab + ac, (a — 6)(a — c) = a 2 — ac — ab + 6c; 54 RESULTS OF MULTIPLYING ALGEBRAIC EXPRESSIONS. therefore, by adding these three equations, we obtain (c _ 6)(c - a) + (b - a)(b - c) + (a - b)(a - <■) — a 2 + b' 2 + c 2 — ab — ac — be; (8) therefore, from (7) and (S) we have (a - b) 2 + (b _ c) 2 + (c - a) 2 = 2(c - 6){c - a) + 2(6 - a)(6 - c) + 2(a - b)(a - c). (9) 43. Results of Multiplying Algebraic Expressions. — From an examination of the examples in multiplication, the student will recognize the truth of the following laws with respect to the result of multiplying Algebraic expres- sions. (1) In the multiplication of two polynomials, when the partial products do not contain like terms, the xohole number of terms in the final product will be equal to the product of the number of terms in the multiplicand by the number of terms in the multiplier, but will be less if the 2wrtial products contain like terms, owing to the simplification produced by collecting these like terms. Thus as we see in Ex. 17, Art. 39, there are two terms in the multiplicand and two in the multiplier, and four in the product, while in Ex. 13 there are three terms iu the multi- plicand and three in the multiplier, and only five in the product. (2) Among the terms of the product there are always two that are unlike any other terms; these are, tliat term which is the product of the tioo terms in the factors which contain the highest power of the same letter, and that term which is the product of the two terms in the factors which contain the loio- est poiver of the same letter. Thus in Ex. 13, Art. 39, there are the terms 8a 4 and 246 4 , and these are unlike any other terms ; in fact, the other terms contain a raised to some power less than the fourth, and thus they differ from 8a 4 ; and they also contain a to some power, and thus they differ from 246 4 . (3) Wlwn the multiplicand, and multiplier are both homo- geneous (Art. 18) the product is homogeneous, and the degree EXAMPLES. 55 of the product is the sum of the numbers which express the degrees of the multiplicand and multiplier. Thus in Ex. 13, Art. 39, the multiplicand and multiplier are each homogeneous and of the second degree, and the product is homogeneous and of the fourth degree. In Ex. 15, Art. 39, the multiplicand is homogeneous and of the second degree, and the multiplier is homogeneous and of the first degree ; the product is homogeneous and of the third degree. This law is of great importance, as it serves to test the accuracy of Algebraic work ; the student is therefore recommended to pay great attention to the degree of the terms in the results which he obtains. EXAMPLES. Multiply 1. 4cr - 36 by 3a6. Ans. 12a s b - 9a6 2 . 2. 8a 2 ' - 9a6 by 3a 2 . 24a 4 - 27a 8 6. 3. 3a> 2 - Ay 2 + 5z 2 by 2x 2 y. 6x 4 y - Sxhf + 10x 2 yz 2 . 4. x 2 y 3 — y 3 z A + z*x 2 by x 2 y 2 z 2 . x 4 y 5 z 2 — x 2 y 5 z 6 + x^yh*. 5. 2xy 2 z 3 + 3x 2 y 3 z — bx 3 yz 2 by 2xy 2 z. Ans. 4x 2 y 4 z 4 + Cx s y 5 z 2 — 10xYz s . 6. -2a 2 6 - 4a6 2 by -7a 2 6 2 . ' 14a 4 6 3 + 28a 3 6 4 . 7. 8xyz — 10x s yz 3 by —xyz. —8x 2 y 2 z 2 + 10# 4 ?/V. 8. abc — a 2 bc — ab 2 c by —abc. —a 2 b 2 c 2 + a 3 b 2 c 2 + a 2 b 3 c' i . 9. x + 7 by x — 10. or — 3# — 70. 10. x + 9 by x - 7. x 2 + 2x - 63. 11. 2x - 3 by x + 8. 2a; 2 + 13a; - 24. 12. 2a; + 3 by x - 8. 2a; 2 - 13x - 24. 13. x 3 — 7a; + 5 by a; 2 — 2x + 3. Ans. x 5 - 2a; 4 - 4a; 3 + 19a; 2 - 31a; + 15. 14. a 2 - 5a6 - b 2 by a 2 + hab + b 2 . Ans. a 4 - 25a 2 6 2 - 10a& 3 - b\ 15. a; 2 — xy + x + y 2 + y + 1 by x + y — 1. ^.ws. a; 3 + Sxy -\- y s — 1. 16. a 2 + Zr + c 2 — 6c — ca — a6 by a + 6 + c. ^ws. a 3 + 6 3 + c 3 — 3a6c. 56 EXAMPLES. 17. 3cr + 2a + 2a 3 + 1 + a 4 by a 2 - 2a + 1. ^?is. a 6 — 2a 3 + 1. 18. — a.r 2 + Baasy 3 — day 1 by —ax — Say 2 . Ans. a¥ + 27«y. 19. -2arty + ?/ 4 + 3ar# 2 + a 4 - 2a;# 3 by a 2 + 2«y + ?/ 2 . ^ws. x 6 + 2a; 2 ?/ 3 + y*. 20. 2aa;+a> 2 +a 2 bya 2 +2aa;— a? 2 . a 4 +4a 3 a;+4a 2 a; 2 -a; 4 . 21. 26 2 +3a6— a 2 by7a— 56. -106 8 — a6 2 +26a 2 6— 7a 8 . 22. a 2 -ab+b 2 by a 2 +a6-6 2 . a 4_ a 3 6 a +2a6 8 — 6*. 23. ±x 2 -3xy-y 2 by3x-2y. 12x»-17x 2 y+3xy 2 +2y 3 . 24. x 5 — x i y-\-xy i — ?/ 5 by x+y. x G —x i y 2 +x 2 y i —y 6 . 25. a; 4 +2afy-f4a; 2 ?/ 2 +8a;?/ 3 +16?/ 4 bya— 2y. x 5 -S2y 5 . 26. 9ajy+27afy+81»*+3a;y 8 +y 4 by 3a;-?/. 243a-- 5 -?/ 5 . 27. x+2y—Bzbyx-2y+Sz. x*—4y 2 +12yz—9z*. Write down tbe values of the following products by inspection. 28. (a; + 7) (a? + 1). Ans. x 2 + 8a; + 7. 29. (x — 7) (a? + 14). a; 2 + 7a; - 98. 30. (a + 36) (a - 26). a 2 + ab - 66 s . 31. (a - 6) (a + 13). a 2 + 7a - 78. 32. (a + 2) (a - 12). a 2 - 10a - 24. 33. (x - l)(x + 12). x 2 + 11a; - 12. 34. (2a; - 5) (a; - 2). 2a; 2 - 9a; + 10. 35. (3a; - l)(a? + 1). 3a; 2 + 2a; - 1. 36. (3a; + 7) (2a; - 3). 6a; 2 + ox - 21. 37. (3a; + 8) (3a; - 7). 9a; 2 + 3a; - 56. 38. (2a; + 7) (5a; + 4). lOa?" + 43a; + 28. Solve the following examples by formula 1 (1), (2), (3) in Art. 41. 39. (x 2 +xy-\-y 2 ) (x 2 — xy— y 2 ). Ans. x i —x 2 y 2 — 2xy 8 —y i 40. (x 2 +xy-y 2 )(x 2 -xy + >r). x i -x 2 y 2 +2xy*-y i 4 1 . (a; 3 + 2x 2 + 3a; + 1 ) (a; 3 - 2a' 2 + 3a- - 1 ) . a- 6 + 2a; 4 + 5a; 2 - 1 42. (a;-3) 2 (a; 2 +6a;+9). a; 4 -18a; 2 +81 43. (x+y) 2 (x 2 — 2xy — y 2 ). a; 4 — 4a; 2 ?/ 2 — 4a# 3 - EXAMPLES. 57 44. (2a? + Sy) 2 (ix 2 + 12xy - 9y 2 ). Ans. 16s 4 + 96x s y + 144s 2 ?/ 2 — 81y 4 . 45. (a» + by) (ax — by)(a 2 x 2 + b 2 y 2 ). a¥ — &V« 46. (as + 6y) 2 (aa; - 6?/) 2 . oV - 2a 2 b 2 x 2 y 2 + &V- Show that the following results are true : 47. (a s +6 2 ) (c 2 +d 2 ) = (ac+&d) 2 + (ad-&c) 2 . 48. (a+&+c) 2 +a 2 +& 2 +c 2 =(a+&) 2 +(&+c) 2 +(c+a) 2 . 49. (a— 6) (6— c) (c— a) =&c(c— 6) +ca(a— c)+a&(&— a). 50. (a-&) 3 +& 3 -a 3 =3a&(&-a). 51. (a 2 +a&+& 2 ) 2 -(a 2 -a&+& 2 ) 2 =4a&(a 2 +& 2 ). 52. (a-f&+c) 3 -a 3 -& 3 -c 3 =3(a+&) (&+c) (c+a). 53. (a+&) 2 +2(a 2 -& 2 ) + (a-6) 2 =4a 2 . 54. (a-i) 3 +(&-c) 3 +(c-a) 3 =3(a-&)(6-c)(c-a). 55. (a-&) 3 +(a+o) 3 + 3(a - &) 2 (a+&) + 3(a+6) 2 (a-6) = (2a) 3 . 56. (a + &) 2 (& + c - a) (c + «_&) + ( a - &) 2 (a + & + c) (a+6— c)=4a6c 2 . 57. [(as + by) 2 + (ay - bxy]\_(ax + by) 2 - (ay + to) 2 ] = (a 4 -& 4 )(s 4 -?/ 4 ). 58 THE DIVISION OF ONE MONOMIAL BY ANOTHER. CHAPTER V. DIVISION. 44. Division in Algebra is the process of finding, from a given product and one of its factors, the other factor ; or it is the process of finding how many times one quantity is contained in another. Division is therefore the converse of multiplication. The Dividend is the given product ; or it is the quantity to be divided. The Divisor is the given factor ; or it is the quantity by which we divide. The Quotient is the required factor; or it is the number which shows how many times the divisor is contained in the dividend. The above definitions may be briefly written quotient X divisor = dividend, or dividend -=- divisor = quotient. It is sometimes better to express this last result as a fraction ; thus dividend ^. , —-—, = quotient. divisor It is convenient to make three cases in Division, (1) the division of one monomial by another, (2) the division of a polynomial by a monomial, (.')) the division of one poly- nomial by another. 45. The Division of one Monomial by Another. — Since the product of 4 and z is 4x, it follows that when 4x is to be divided by .»■ the quotient is i. < >r otherwise 4x -7- x = 4. v\lso since the product of a and b is ah. the quotient of ah divided by a is b ; that is ab -i- a = b. THE DIVISION OF ONE MONOMIAL BY ANOTHER. 59 Similarly abc -f- a = be ; abc -r- b = ac ; abc -+- c "= a& ; abc -r- a& = c ; a&c -=- &c = a ; abc -h ca = b. These results may also be written abc a = be ; abc _ 6 ac ; a&c _ c rt& abc ab = c; a&c _ be ~ a ; a&c _ ca &. Also 36a 6 9a 4 = 36a 6 _ 36aaaaaa 4a = 4aa, by remov- 9a 4 daaaa ing from the divisor and dividend the factors common to both, just as in Arithmetic. Therefore 36a 6 -=- 9a 4 = 4a 2 . 45aaaabbbcc Similarly 45a 4 & 3 c 2 -=- 9a 2 6c 2 = 9aa6cc 5a 2 & 2 . Hence we have the following Rule. To divide one monomial by another, divide the coefficient of the dividend by that of the divisor, and subtract the exponent of any letter in the divisor from the exponent of that letter in the dividend. For example 12xhf -r- \2xhf = 6x 5 ~ 3 y 3 ~ 2 = Gx^y. Also Soa^Y -=- Ua 2 xif = 5a 2 xy. Rem. — If the numerical coefficient, or the literal part of the divisor he not found in the dividend, we can only indicate the division. Thus if la is to be divided by 2c, the quotient can only be indicated by la + 2c or by — . In some cases, however, we may simplify the expression for the quotient by a principle already used in Arithmetic. Thus if 16a 3 6 2 is to be divided by 12abc, the quotient is denoted by 16a8&2 . Here the dividend = 4a& x 4r/ 2 & and the divisor = 4ab x 3c; 12abc thus the factor 4o6, which occurs in both dividend and divisor, may be removed in the same way as in Arithmetic, and the quotient will be denoted by — . That is 3c Wa s h 2 _ 4ab x 4rt 2 ft _ 4a~b V2abc ~ 4ab x 3c 3c ' by removing the common factor 4o6. 60 THE RULE OF SIGNS — EXAMPLES. NoTBf. — If we apply the above rule to divide any power of a letter by the same power of the letter, we are led to a curious conclusion. Thus by the rule a 3 -f a 3 = a 8 ~ 3 = a ; but also o 3 -f a 3 =z — =1, a 3 by removing the common factor a 3 ; . •. cfi = 1 ; that is, any quantity whose exponent is is equal to 1. The true significance of this result will be explained in Art. 119. 46. The Rule of Signs for division may be obtained from an examination of the cases which occur in multiplica- tion, since the product of the divisor and quotient must be equal to the dividend. Thus we have +a x ( + &) = +ab, —a x ( + 6) = — aft, +a X ( — b) = — ab, —a x (-&) = +ab, Hence in division as well as in multiplication, like signs produce +, and unlike signs produce — . EXAMPLES. Divide 1. Sa% by 4a&. Ans. 2a. 2. — 15xy by Sx. —5?/. 3. -21« 2 6 3 by -la%\ 36. 4. 45a 6 o 2 .« 4 by -9a 3 bx 2 . —5a*bx*. 5. -36aWby — 24aW. $a& 9 . 6. 3a 3 bya 2 . " Sx. 7. -58aWby -2 = —a. -aft - -b = +a. +a& h- — b= —a. TO DIVIDE A POLYNOMIAL BY A MONOMIAL. 61 47. To Divide a Polynomial by a Monomial. Since (a — b)c = ac i — be ; ,, a ac — be 7 therefore = a — o. c Also since (a — fr) x (— c) = —ac + 6c; therefore ~ ac + 6c = a - b. — c Also since (a — 6 + c)a& = a 2 5 — ab 2 + a&c ; ,, fi a?b — ab 2 + abc , , therefore — = a — b + c. a& Hence we have the following rule : To divide a polynomial by a monomial, divide^ each term of the dividend sejmrately by the divisor. For example (8a 3 - Ga 2 b + 2a 2 c) -=- 2a 2 = 4a - 3b + c. (9a; - 12y + 3z) -s- — 3 = -3a: + 4y - z. (36a 3 6 2 - 24a 2 6 5 - 20a 4 Z> 2 ) -j- 4a 2 & = 9ab - 6b 4 - 5a 2 b. (2x 2 — 5xy — \x 2 y z ) -= \x = —Ax + 10?/ + 3a;?/ 3 . EXAMPLES. Divide 1. — 35a; 6 by 7a 3 . Ans. — 5a 8 . 2. a 3 ?/ 3 by arfy. a;?/ 2 . 3. 4aW by a&V. 4ac. 4. a 2 — 2xy by .r. a; — 2y. 5. a; 6 - 7a; 5 + 4a 4 by a; 2 . x* - 7x 3 -f- 4a; 2 . 6. 10a; 7 - 8a; 6 + 3a; 4 by a; 3 . 10a; 4 - 8a; 3 + 3a;. 7. 15a; 5 - 25a; 4 by - 5a; 3 . -3a; 2 + ox. 8. 27a; 6 - 36a; 5 by 9a; 5 . 3a; - 4. 9. -24a; 6 - 32a; 4 by -8a; 3 . 3a; 3 + 4a;. 10. a 2 — ab — ac by —a. —a + b + c. 48. To Divide one Polynomial by Another. — Let it be required to divide a 3 + 2a 2 — 3a by a 2 + 3a. Here we are to find a quantity which when multiplied by the divisor will produce the dividend. Hence the dividend 62 TO DIVIDE ONE POLYNOMIAL BY ANOTHER. is composed of all the partial products arising from the multiplication of the divisor by each term of the quotient (Art. 39). Arranging both the dividend and divisor accord- ing to descending powers of a, we see that the first term a 3 of the dividend is the product of the first term a 2 of the divisor by the first term of the quotient (Art. 43) ; therefore, dividing a 3 by a' 2 we obtain a for the first term of the quotient. Multiplying the whole divisor by a we obtain ft 3 _j_ 3 a 2 f or the partial product of the divisor by the first term of the quotient ; subtracting this product from the dividend we obtain the first remainder —a 2 — 3a, which is the product of the divisor by the remaining terms of the quotient, and consequently the first term —a 2 of this product is the product of the first term of the divisor by the second term of the quotient. Dividing therefore this first term —a 2 by the first term of the divisor a 2 we obtain —1 for the second term of the quotient. Multiplying the whole divisor by —1 we obtain —a 2 — 3a for the product of the divisor by the second term of the quotient ; subtracting this product there is no remainder. As all the terms in the dividend have been brought down, the operation is completed. Hence a — 1 is the exact quotient. The work may be arranged as follows : Divisor. Dividend. Quotient. a 2 + 3a) a 3 + 2a 2 - 3a (a - 1 a 3 + 3a 2 - a 2 - 3a - a 2 - 3a It will be observed that in getting each term of the quotient in this example, we divide that term of the divi- dend containing the highest power of a by the term of the divisor containing the highest power of the same letter , and therefore, when the dividend and divisor are arranged according to descending powers of a, any term of the quotient is found by dividing the first term of the divisor TO DIVIDE ONE POLYNOMIAL BY ANOTHER. G3 into the first term of the dividend, or into the first term of one of the remainders. Hence for the division of one polynomial by another, we have the following Rule. Arrange both dividend and divisor according to ascending or descending powers of some common letter. Divide the first term of the dividend by the first term of the divisor, and write the result for the first term of the quotient; multi-ply the whole divisor by this term, subtract the product from the dividend, and to the remainder join as many terms from the dividend, taken in order, as are required. Divide the first term of the remainder by the first term of the divisor, and write the result for the second term of the quotient ; multiply the whole divisor by this term, and subtract the product from the last remainder. Continue this operation until the remainder becomes zero, or until the first term of the remainder will not contain the first term of the divisor. This method of dividing is similar to long division in Arithmetic, i.e., we break up the dividend into parts, and find how often the divisor is contained in each part ; and then the sum of these partial quotients is the complete quotient. Thus, in the example just solved, a 3 + 2a 2 — 3a is divided by the above process into two parts, viz., a 3 + 3a 2 , and — a 2 — 3a, and each of these is divided by a 2 + 3a, giving for the partial quotients a and —1 ; thus we obtain the complete quotient a — 1. Note. — The divisor is often put on the right of the dividend and the quotient beneath the divisor as follows : Dividend. Divisor. a 2 + 2ab + b 2 \a + b a 2 + ab a + b Quotient. ab -f- b 2 ab + b 2 64 EXAMPLES. It is of great importance to arrange both dividend and divisor according to ascending or descending powers of some common letter ; and to attend to this order in every part of the operation. EXAMPLES. 1. Divide 24a* 2 - 6oxy + 2\y 2 by 8a; - By. The operation is conveniently* arranged as follows : 8x — 3?/) 24a* 2 — Goxy + 21y 2 (3x — ly. 24a* 2 - 9xy — 56a:?/ + 2\y 2 — 56a*?/ + 21y 2 Divide 2. x 2 + Sx + 2 by x + 1. Ans. x + 2. 3. x 2 — 7x + 12 by x — 3. x — 4. 4. x 2 — \\x -f 30 by x — 5. a — 6. 5. a; 2 - 49a: + 600 by x - 25. as - 24. 6. 3a* 2 + 10a* + 3 by x + 3. 3a* + 1. 7. 2a* 2 + 11a* + 5 by 2x + 1. a? + 5. 8. 5a* 2 + Ha; + 2 by a* + 2. 5a? + 1. 9. 2a* 2 + 17a* + 21 by 2a* + 3. a> + 7. 10. 5a* 2 + 16a: + 3 by a: + 3. 5a; + 1. 11. Divide 3a 4 - 10a 8 6 + 22a 2 6 2 - 22ab 3 + 15& 4 by a 2 _ 2ab + 3b 2 . The operation is written as follows. a a _ 2ab + 36 a )3a 4 - 10a 8 ft + 22a 2 & 2 — 22a& 3 + 156*(3a 9 - iab + 5b- 3a 4 - 6a 8 6 + 9a 2 & 2 - 4a 3 i + 13a 2 & 2 - 22r//, 3 - 4o*6 + 8«-V- a - 12a6 8 5a 2 6 2 - 10a& 8 + 15& 4 5 s%a_ I0a6 8 + 15/> 4 12. Divide a* 7 - 5a* 5 + 7a* 8 + 2a: 2 - 6a: - 2 by 1 + 2a* - 3.x* 2 + & . Arrange both dividend and divisor according to descending powers of x, and arrange the work as follows : X' EXAMPLES. 65 - 5a; 5 + 1.x? + 2x 2 - 6x - 2 la; 4 - 3a; 2 + 2x + 1 a; 7 — 3a 5 4- 2.c 4 + « 3 » 3 — 2a; - 2a; 5 - - 2X 5 - 2x* + Gx 3 + Gx 3 + 2x- - l.r- - Gx - 2x - 2x* - 2a; 4 + Gx 2 - Ax - - Ax - - 2 - 2 "We might have arranged the dividend and divisor accord- ing to ascending powers of x as follows : _2 - 6a> + 2a; 2 + 7a; 3 - 5a; 5 + a; 7 1 1 -f- 2a; - 3a; 2 + a; 4 _2 - Ax + 6a; 2 - 2a; 4 -2 - 2a; 4- a; 3 - 2a; - 2x - Ax 2 + Ix 3 + 2a; 4 - 5a; 5 - 4a; 2 + 6a; 3 - 2a; 5 x 3 + 2a; 4 - 3a; 5 + x 1 x 3 + 2a; 4 - 3a; 5 + x? We thus obtain the same quotient we had before, though the terms are in a different order. 13. Divide a 3 + b 3 + c 3 - Sabc by a + b + c. Arrange the dividend according to descending powers of a. Sabc 4- 6 3 + c 8 [a + & a 3 + a 2 b + « 2 c a 2 — ab — ac + & 2 — be + c 2 — a 2 6 — a 2 c — — a 2 b — a& 2 — 3a6c abc - o 2 c + — a?c ab 2 — 2a,bc — abc — ac 2 air — abc + ac 2 + ft 3 o6 2 + 6 3 4- &*c — a&c 4- 8C 2 — ?; 2 c — a&c — 6 2 c — 6c 2 ac 2 4- 6c 2 4- c 3 ac 2 + &C 2 4- c« 66 EXAMPLES. Iii this example we arrange the terms according to de- scending powers of a ; then when there are two terms, such :is d-b and a?c, which involve the same power of a, we select a new letter, as 6, and put the term which contains b before the term which does not ; and again of the terms air and a&c, we put the former first as involving the higher power of b. 14. Divide x 4 + 4a 4 by ar + 2ax + 2cr. x 4 + 4« 4 . | a 2 + 2a x + 2a 2 X* + 2ax 3 2a.x 3 2ax* + 2a 2 x 2 - 2« 2 x 2 — 4a 2 x 2 x 2 — 2ax -f- — 4a 3 x 2a 2 .2a 2 x 2 2a 2 x 2 + 4a 3 x + 4a 4 + 4a 3 x + 4a 4 Divide 15. x 5 - 5.x 4 + 9x 3 - 6x 2 - a + 2 by a 2 - Sx + 2. ^dris. x 3 — 2a; 2 + x + 1. 16. x 5 — 2x 4 - 4.x 3 + 19»j 2 - 31a + 15 by x 3 - 7x + 5. ^4?)s. a; 2 — 2x + 3. 17. u 3 + & 3 + 3a&c — c 3 by a + & — c. Ans. -' ; 7/ + tV'.V 2 •, 7 r + A»" W 2 + fay* DIVISION WITH THE AID OF PARENTHESES. 67 Divide 21. %a 3 - fa%s + 2£aa? - 27a* 8 by \a - Sx. Arts. \<& - Sax + 9a 2 . 22. fra* - T Vr + T y< - A b > T i a ~ i- K ~ i a + tV 49. Division with the Aid of Parentheses. — Some- times it is found convenieut to divide with the aid of paren- theses thus : 1. Divide x 3 — (a+b-\-c)x 2 + (ab-\-ac+bc)x—abc by x—c. x 3 — (a + b + c)ar+ (ab + ac -f- bc)x — abc \x — c x 3 — ex 2 a* 2 — (a+b)x+ab — (a + b)x 2 + (ab + ac + be) x — (a + b)x 2 + (a + 6) ex* abx — abc abx — abc Divide 2. a 2 a* 4 +(2ac— 6 2 )a* 2 +c 2 by ax 2 — bx+c. Ans. ax 2 +bx+c. 3. x 3 + (a + & + c)a: 2 + (a& 4- ac + 6c) a + abc by a + &• -4ws. a 2 + (a + c)a; -+- ac. 4. ax 3 — (a 2 + &)ar + b 2 by ckc — b. x 2 — aa* — &. 5. a 2 (6 + c) + 6 2 (a — c) + c 2 (a — 6) + a&c by a + & + c. .4ns. a (6 + c) — 6c. 50. "Where the Division cannot be Exactly Per- formed. — In the examples given thus far the divisor has been exactly contained in the dividend. It may happen, as in Arithmetic, that the division cannot be exactly performed. In such cases it is a good exercise for the student to deter- mine the accurate value of the remainder. Divide a 3 - 6a* 2 + lia* + 2 by a* — 2. x 3 — 6a: 2 + 11a* + 2 [as — 2 a; 3 - 2x 2 x 2 - 4a* + 3 — 4a* 2 + - 4a* 2 -f- llaj 8a* 3a* + 2 3a; - 6 68 IMPORTANT EXAMPLES IN DIVISION. The division can be carried no further without fractions, because x will not go into 8. We therefore express the result in the same way as in Arithmetic, that is, by adding to the quotient a fraction of which the numerator is the remainder and the denominator the divisor. Thus the result is * 3 -<^ + ll* + 2 = a* - Ax + 3 + 1A_. x — 2 x — 2 EXAMPLES. Find the remainder when x 3 - 6a; 2 + 12a; - 17 is divided by x - 3. .4ns. 3a; 3 — 7a; — 9 is divided by x + 1. 5a; 2 — 4a; — 7 is divided by a; + 2. 7ar — 3a; — 33 is divided b}' 4x — 5. 27a; 3 -f 9ar — 3a; — 5 is divided by 3x — 2. 1 6a; 3 - 19 + 39a; - 46a; 2 is divided by 8x - 3. 7. 8a; — 8a; 2 + 5a; 3 + 7 is divided by 5a; — 3. 2a; 8 + 4a; 3 -8. — 5. 5. -18. 5. -10. 10. 51. Important Examples in Division. — The follow- ing examples are very important ; they may be easily verified, and should be carefully noticed. = x + ?/, = x 2 + xy + y 2 , = x 3 + x'hj + xy 2 + y», x — y and so on ; the terms in the quotient all being jiositive. -x 2 - y 2 .1- — :'/- X - y X s — y 3 X - y X* — t 11. x 3 — cry + x>/' 2 x + y x* — y 4 x + y ~ -L = .t- 5 — a; 4 ^ -f x 8 y 2 — an/ 8 -j- xy* — ?/ 6 , .(• :i + ?/ 3 X X 6 + y + y h X x> + y IMPORTANT EXAMPLES IN DIVISION. 69 and so on ; the terms in the quotient being alternately posi- tive and negative. 3 1 - xy + y\ : a 4 - xhj + xY - vy* + y\ >-x 5 y + x 4 y 2 - xY 4- A 4 - xy 5 + y\ x + y and so on ; the terms in the quotient being alternately positive and negative. The student can verify these results in any particular case, and carry on these operations as far as he pleases, and he will thus gain confidence in the truth of the following state- ments for which we shall hereafter give a general proof. See Art. 170. These different cases may be conveniently arranged in the following concise statements : x n — y n is divisible by x — y if n be any whole number. (1) That is : the difference of any two equal poivers of two numbers is ahvays divisible by the difference of the two tumbers. x n — y n is divisible by x + y if n be any even whole number. (2) That is : the difference of any tivo equal even powers of tivo numbers is always divisible by the sum of the numbers. x n + y n is divisible by x -\- y if n be any odd whole number. (3) That is : the sum of any two equal odd powers of two numbers is always divisible by the sum of the numbers. x n -\- y n is never divisible by x + y or x — y, when n is an even whole number. EXAM PLES. Write the results in the following by these three state ments. 1. a 5 — b 5 h- a - 6. Ans. a* + a z b + a%- + ab* + b*. 2. X 3 - 1 -r- X - 1. JB 2 + X + 1. TO EXAMPLES. 3. ga* - b z -H 2a - h. Ans 4a 2 + 2a& + b\ 4. 8a; 3 - 27?/ 8 -*• 2x - :'.?/. 4a; 2 + 6xy + 9?/ 2 . 5. a; 4 — 16?/ 4 -s- a? -f 2y. a; 3 - 2a; 2 ?/ + 4a;?/ 2 — 8?/ 3 . 6. 16a; 4 — y 4 -j- 2x + y. 8a; 8 — 4a; 2 ?/ + 2xy~ — y z . 7. a; 3 + 1 -=- x + 1. a; 2 - a? + 1. 8. a 3 + G4 -s- a + 4. a 2 - 4a + 16; 9. 8a; 3 + 27 y 3 -=- 2a; + Sy. 4a; 2 — 6a;?/ + 9y-. Divide 10. 3a; 3 - 9a; 2 ?/ - \2xif by -3a;. -a; 2 + Sxy + 4?/ 2 . 1 1 . 4a; 4 ?/ 4 — 8a; 3 ?/ 2 + 6a;?/ 3 by - 2a;?/. — 2a; 3 ?/ 3 + 4a; 2 ?/ — Sy 2 . 12. x s y — 3a; V 4- 4a-?/ 3 by a;?/. a; 2 — 3xy -f 4?/ 2 . 13. -15a 8 6 8 - 3a 2 6 2 + 12a6 by -3o6. 5a 2 6 2 + a6 - 4. 14. 60a W - 48a W + 36aW by 4a6c 2 . Ans. 15a 2 6 2 - 12a& 8 + 9«6c 2 . 15. _3a 2 + %ab - 6ac by -fa. 2a - 36 + 4c. 1G. fa; 5 ?/ 2 - 3a- 3 ?/ 4 - Go- 4 ?/ 8 by -fa; 3 ?/ 2 . -3a- 2 + 2y 2 + 4a;?/. 17. £a 2 a; — -^abx — facaJ by faa;. |a — £6 — c. 18. a; 2 - 11a; + 30 by x - 5. x - 6. 19. x- - 7a- + 12 by x - 3. a- - 4. 20. 3a; 2 + a- - 14 by a: - 2. 3a; + 7. 21. 6a; 2 - 31a; + 35 by 2a; - 7. 3a; - 5. 22. 15a 2 + 17aa; - 4ar by 3a + Ax. ha — x. 23. 60a; 2 - 4a;?/ - 45?/ 2 by 10a; - 9y. 6x + by. 24. -4xy - 15y 2 + 96a; 2 by 12a- - by. 8a; + By. 25. 100a; 3 - 3a- - 13a- 2 by 3 + 25a;. 4a- 2 - x. 26. 7a; 8 + 96a- 2 - 28a- by 7x - 2. a- 2 + 14.r. 27. a; 5 - 4a; 4 + 3x 3 + 3ar - 3a; + 2 by a; 2 - a- - 2. Ans. X s — 3a- 2 4- 2a- — 1. 28. x 6 +x 4 y—x s y 2 -{-x s —2xy 2 +y s hyx 2 +xy—y-. x s +x—y. 29 . 2a; 8 - 8a- + a; 4 + 12- 7ar by x 2 + 2 - 3a-. ar+ 5a; + 6 . 30. 8a; 8 - 4a; 2 - \2*x + a; 4 - 192 by a- 2 - 16. a- 2 + 8x + 12. 3 1 . a; 9 — if by a; 2 + xy 4 y 2 . a; 7 — x 6 y + x*y* — x*y* + xy* — if' . 32. 2a 4 +27a& 8 — 816* bya+36. 2a 8 -6a 2 &+18a& 9 — 276 8 . 33 . a; 5 + x*y 4 xhf 4 xhf + xy* + y 6 by b 8 + y 8 . a- 2 + ay/ + >/-. 34. u 5 + 2 A + Safy 2 - a; 2 ?/ 8 - 2xy* - Sy" by r 1 - ?/ 8 . An 8. x- 4- 2ajf + 8y". EXAMPLES. 71 35. x 12 + x* - 2 by a; 4 + a; 2 + 1. .dn.s. x* - x 6 + 2a; 2 - 2. 36. a; 4 — x 3 y — xy z + y 4 by a; 2 + xy + y 2 . a? 2 — 2#y + y 2 . 37. 49a; 2 + 21a;?/+ 12?/z - 16z 2 by 7x + 3y—4z. 7x + 4z. 38. a 2 + 2a& + 6 2 - c 2 by a + 6 - c. a + 6 + c. 39. a 3 + 3a 2 6 + & 3 - 1 + Bab 2 by a + b - 1. J.ws. a 2 + 2a& + 6* + a + 6 + 1. 40. a; 8 — y s by a-- 3 + x 2 y + xy 2 -f y 3 . x 5 — x 4 y + xy* — y 5 . 41. a 12 + 2a 6 6 6 + b u by a 4 + 2a 2 b 2 + i 4 . ^ws. a 8 - 2« 6 6 2 + 3a 4 6 4 - 2a 2 & 6 + & 8 . 42. §a 2 c 8 + T f ^a 5 by £a 2 + \ac. fta* - fa 2 c + fac 2 . 43. T ^a 4 _|a 3 -|a 2 +|a+^byfa 2 -|-a. ^_i a -§. 44. 36a; 2 + ±y 2 + \ - 4xy - 6x + %y by 6» - \y - f .4ns. 6a; — -|y — £. 45. (a; + ?/) 2 — 2(z + ?/)z + 2 2 bya; + ?/ — z. ar + y — z. 46. aa; 2 — a& 2 -f- b 2 x — a; 3 by (x + 6) (a — a;). a; — 6. 47. (b-c)a 3 +(c-a)b z +(a-b)c s by a 2 —ab—ac+bc. Ans. a{b-c) + b 2 -c 2 . 48. a; 6 — 1 by a; - 1. (Art. 51.) a; G + a ,4 +a; 3 +a; 2 -|-a;+l. 49. a; 4 — 81y 4 by x — By. x 3 + 3a; 2 y + 9xy 2 + 27y 3 . 50. a; 5 — ?/ 5 by a; — y. a; 4 + x 3 y -+- a; 2 ?/ 2 + xy 3 -f- y*. 51. a 9 - 6 9 by a 3 - Z> 3 . a 6 + a 3 b 3 + 6 6 . 52. 27a; 3 + 8?/ 3 by 3a; + 2y. 9x 2 - 6xy + 4?/ 2 . 53. 64a; 6 — y 6 by 2a; — y. Ans. 32a; 5 + 16x*y + 8afy 2 + ix 2 y* + 2a;?/ 4 + y 5 . 54. 32a; 5 + y 5 by 2a; + y. 16a; 4 - 8x- 3 y + 4afy a — 2a;?/ 3 + ?/ 4 . 55. Divide the product of a 2 + ax -f- a; 2 and a 3 + a; 8 by a 4 + a 2 x 2 + a; 4 . J.«s. a + a\ 56. Divide (ax + %) 2 + (ay - &a;) 2 + c 2 a; 2 + c 2 y 2 by a; 2 + y 2 . Ans. a 2 + b 2 + c 2 . 57. Divide a?b — bx 2 + a?x — x 8 by (a;+&) (a— a;) . a+x. 58. Divide &(a; 8 + a 3 ) + ax{x 2 - a 2 ) + a 8 (a; + a) by (a + 6) (a; + a) . ./1ms. a; 2 — aa; + a 2 . 59. Divide (x + y) 7 - (a 7 + y 7 ) by (a; 2 + «y + y 2 ) 2 . .4ms. 7xy{x 4- y). 72 EQUATION OF CONDITION — UNKNOWN QUANTITY. CHAPTER VI. SIMPLE EQUATIONS OF ONE UNKNOWN QUANTITY. 52. Equations — Identical Equations. — An Equation is a statement in Algebraic language that two expressions are equal. Thus, 2a; + 4 = a? + 8 is an equation ; it states that the expression 2x -f 4 is equal to the expression x + 8. The two equal expressions thus connected are called sides or members of the equation. The expression to the left of the sign of equality is called the first side or member, and the expression to the right is called the second side or member. Every equation has two members. An Identical Equation, or briefly an Identity, is one in which the two members are equal whatever numbers the letters represent. Thus, the following are identical equa- tions : „ ' , _ x + 3 + x + 4 = 2x -f- 7, (a -f- x) (a — x) — a 2 + x z — ; that is, these Algebraic statements are necessarily true, whatever values we assign to x and a. All the equations used in the previous chapters to express the relations of Algebraic quantities are identical equations, because they are true for all values of these quantities. 53. Equation of Condition — Unknown Quantity. — An Equation of Condition is one. which is true only when the letters represent some particular value. For example, the equation, „ _ .., cannot be true unless x — 5, and is therefore an equation of condition. AXIOMS. 73 An equation of condition is called briefly an equation. The letter whose value, or values, it is required to find is called the unknown quantity. Thus x is the unknown quantity in the above equation. To solve an equation means to find the value, or values, of the unknown quantity for which the equation is true. These values of the uukuown quantity are said to satisfy the equation, and are called the roots of the equation. An equation which contains ouly one unknown quantity is called a simple equation, or an equation of the first degree, when the unknown quantity occurs only in the first power. It is usual to denote the unknown quantity b} T the letter x. The equation is said to be of the second degree or a quadratic equation when x 2 is the highest power of x which occurs, and so on.* Thus 2x + G = a; + 8, and ax -\- b — c are simple equatious. x 2 - 2x = 3 is a quadratic equation. 54. Axioms. — An Axiom is a self-evident truth. The operations employed in solving equations are founded upon the following axioms : 1. If equal quantities be added to equal quantities, the sums will be equal. 2. If equal quantities be taken from equal quantities, the remainders will be equal. 3. If equal quantities be multiplied by equal quantities, the products will be equal. 4. If equal quantities be divided by equal quantities, the quotients will be equal. f * The equation is supposed to be reduced to such a form that the unknown quantity is found only in the numerators of the terms, and that the exponents of its powers are expressed by positive integers. t If the divisors are different from zero. 74 CLEARING OF FRACTIONS. 5. Like powers and like roots of equal quantities are equal. These axioms ma} r be summed up in the following one : If the same operations be performed on equal quantities, the results will be equal. In the solution of equations there are two operations of frequent use. These are (1) clearing the equation of fractions, and (2) transposing the terms from one member to the other so that the unknown quantity shall finally stand alone as one member of the equation. 55. Clearing of Fractions. — Consider the equation X X x 2^36 Multiplying each term by 2 X 3 X 6 (Axiom 3), we get 3 X 6x + 2 X 6s + 2 X 3x* = 2 X 3 x G X 2, or 18a- + 12a: + 6as = 72; dividing each term by 6 (Axiom 4), we get 3a- 4- 2x + x = 12, or Gx =12. Instead of multiplying each term by 2 x 3 x G, we might multiply each term by the least common multiple of the denominators, which is 6, and get immediately 3x + 2a; + x = 12. Hence to clear an equation of fractions, we have the following Rule. Multiply each term of the equation by (he least common nulla pie of the denominators. Clear the following equation of fractions : X _|_ * _ X _ g 3 5 7 Here the least common multiple of the denominators is the product of the denominators, 3, 5, and 7. Multiplying each term by it, we get 35a; + 21as - 15a: = 315, or 41a; = 315. TRANSPOSITION. 75 Clear the following equation of fractions : 2 + 2 + 2 + ± = 4. 4 6 8 12 Here 24 is the least common multiple of the denominators. Multiplying each term by it, we get 6x + 4a> + 3a; + 2x = 96, or 15a? = 96. 56. Transposition. — To transpose a term is to change it from one member of an equation to the other without destroying the equality of the members. Suppose, for example, that x — a = b. Add a to each member (Axiom 1) ; then we have x — a -\- a = b + a; therefore, since — a and +a cancel each other, we have x = b + a. Again, suppose that x + b = a. Subtract b from each member (Axiom 2) ; then we have x + b — b = a — b ; therefore, since +6 and —b cancel each other, we have x = a — b. Here we see, in these two examples, that —a has been removed from one member of the equation, and appears as +a in the other ; and -j-b has been removed from one member and appears as —b in the other. It is evident that similar steps may be employed in all cases. Hence we have the following Rule. Any term may be transposed from one member of an equation to the other by changing its sign. It follows from this that the sign of every term of an equa- tion may be changed; for this is equivalent to transposing every term, and then making the first and second members change places. Thus, for example, suppose that 4x — 8 = 2x — 16. 76 SOLUTION OF SIMPLE EQUATIONS. Transposing every term, we have — 2x + 16 = — Ax + 8, or —4a,- + 8 = —2x + 16, which is the original equation with the sign of every term changed. This result can also be obtained by multiplying each term of the original equation by —1 (Axiom 3). 57. Solution of Simple Equations with One Un- known Quantity. — Find the value of x in the equation x 18 _ x x 2 5 ~ 4 5' The least common multiple of the denominators is 20. Multiplying each term by 20, we get 10a — 72 = hx — Ax. Transposing the unknown terms to the first member, and the known terms to the second, we have 10a — hx + Ax = 72. Collecting the terms, we have 9a; = 72. Dividing each member by 9 (Axiom 4), we have x = 8. We can now give a general rule for solving any simple equation with one unknown quantity. Rule. Clear the equation of fractions, if necessary; transpose all the terms containing the unknown quantity to the first member of the equation, and the knoivn quantities to the second, and collect the terms of each member. Divide both members by the coefficient of the xinknown quantity, and the second member is the value required. EXAMPLES. 77 EXAMPLES. 1 . Solve 9x + 35 = 75 + ox. Here there are no fractions ; transposing, we have 9x — 5x = 75 — 35. Collecting terms, Ax = 40. Dividing by 4, x = 10. It is very important for the student to acquire the habit of occasionally verifying, that is, proving the truth of his results. The habit of applying such proofs tends to con- vince the student, and to make him self-reliant and confident in his own accuracy. To verify the result, in the case of simple equations, we substitute the value of the unknown quantity in the original equation ; if the two members are equal the result is said to be verified, or the equation satisfied. Thus, in the last example, 10 is the root of the proposed equation (Art. 53). We may verify this, i.e., we may show that x = 10 satisfies the original equation by putting 10 for x in that equation. Thus 9 x 10 + 35 = 75 + 5 x 10, or 90 + 35 = 75 + 50, or 125 = 125, which is clearly true. Hence, since the two members are equal, x = 10 satisfies the equation. 2. Solve 5(x - 3) - 7(6 - a;) + 3 = 24 - 3(8 - a;). Removing parentheses, 5x _ 15 _ 42 + 7x + 3 = 24 - 24 + 3 - 21 + 5a;] = 4aj + 1, 8a; - 5a; + 105 - 25a; = 4a; + 1. .-. -26a; = -104. x = 4. Solve the following equations : 5. 2a; + 7 = Bx + 3. Ans. 4. 6. 24a; - 49 = 19a; - 14. 7. 7. 16a; - 11 = 7x + 70. 9. 8. 8(x - 1) + 17(a> - 3) = 4 (4a; - 9) + 4. 3. 9. 5.x - 6(.r - 5) = 2(.t + 5) + 5(a> - 4). 5. 10. 8 (a; - 3) - (6 - 2x) = 2(x + 2) - 5(5 - .r). 3. 11. 3(169 - aj) - (78 + aj) = 29aj. 13. 12. 7a; - 39 - 10a; + 15 = 100 - 33a; + 26. 5. FRACTIONAL EQUATIONS — EXAMPLES. 79 13. 118 - 65a; -123 = 15a; + 35 - 120a;. Ans. 1. 14. 157 - 21 (x + 3) = 163 - 15(2a; - 5). 16. 15. 97 - 5(.t + 20) = 111 - S(x + 3). 30. 16. x — [3 + \x — (3 + x)\~\ = 5. 5. 17. 14a; - (5a; - 9) - {4 - 3a; - (2x — 3)\ = 30. 2. 18. 5a; - (3a; - 7) - £4 - 2x - (6a; — 3) | = 10. 1. 58. Fractional Equations. — The following are some of the most useful methods of solving fractional equations. EXAMPLES. , , 5x -f- 4 7x + 5 r 3 x — 1 1. Solve ! ! — =5 . 2 10 5 2 q3 — m_ • the least common multiple of the denominators is 10 ; multiplying by 10, we have 5 (5a; + 4) - (7a; -f 5) = 56- — 5(a; — 1) ; removing parentheses, 25a; + 20 - 7a; - 5 = 56 - 5x + 5 ; transposing, 25a; — 7x -f- 5i» ; = 56 -+- 5 — 20 + 5 ; collecting terms, 23.« = 46 ; .-. x = 2. Note. — Mistakes with regard to the signs are often made in clearing an equation of fractions. In this equation the fraction — x ~ is regarded as a single term with the minus sign before It; it is equivalent to — \{x — 1). When multiplied by 10, it is well to put the result first in the form — 5(x — 1), and afterwards in the form — 5x + 5, in order to secure attention to the signs. 2. Solve 4 - °L^lA = SL - I. Ans. 33. 8 22 2 3. « i(5aj + 3) - ^(16 - 5a;) = 37 - 4a:. 6. . Gx + 15 8a; — 10 ±x — 7 11 3. Note. — In certain cases it is more advantageous not to clear the equation of fractions at once by multiplying it throughout by the least common multiple of the denominators (Art. 55), but to clear it of fractions partially, and then to effect some reductions, before we remove the remaining fractions. SO EXAMPLES. 1 2x - 3 _ 5a; — 32 a; + 9 3 35 9 28 First multiply by 9, ami we have 3a; - 12 4- I 8 *" 27 = 5x - 32 - »* + 81 ; T 35 28 18a; - 27 . 9a; + 81 ~ on transposing, — 1 — — = 2.x — 20. Now clear of fractions by multiplying by the least common multiple of the denominators, which is 5 x 7 X 4, or 140, and we get 72a- - 108 + 45a; + 405 = 280a - 2800 ; transposing, 72.x- + 45a; - 280a: = -2800 + 108 - 405 ; collecting terms, — 163a; = —3097; dividing by -163, x = 19. Solve the following equations : 6 . ^lA + *L+J° = 5 . Am. 8. 2 9 7. ■+!? = 3 + * 16. 5 17. 8. x — 1 8 i+2 + 1 18 9. 4(a; + 2) 5 = 7 6x 13 .0. sb + 20 9 3a; _ 7 = 6. 11 1) + 2x - 1. -2. 7. 2(x - 1) - 3 (a; - 2) + 4 (a - 3) + 2 = 0. 2. 8. 5a; + 6(3! + 1) - 7(x + 2) - 8 (a + 3) = 0. -8. 9. (a + i)(2as + 1) = (a? + 8)(2as + 3) - 14. 1. 10. (x+l) 2 -(a; 2 -l)=a;(2a;+l) -2(«+2) (x-4-1) +20. Ans. 2. 11. 6(« 2 -3x+2)-2(.i- 2 -l)=4(a;+l)(a;+2;)-24. 1. 12. 2a; - 5^3a; - 7 (4a; - 9)\ = 66. 3. 13. 3(5 - 6x) - 5[> - 5|1 - 3(aj - 5)|] = 23. 4. 14. (a; + l) 2 + 2(3! + 3) 2 = 3x(x + 2) + 35. 2. 15. 84+(s+4)(s-8)(a+5) = (aj+l).(*+2)(»+S). 1. 16. (a + 1) (a; + 2) (a; + 6) = x 3 + 9x 2 + 4(73! - 1) . 2. 17 * _ ?. = 1. -20. 5 4 18. 5^zi + *Zli = 3. 19. 1(3! 4- 1) - f(3! - 1) = 20. A(2 - 3!) - i(5.r + 21) = 21 - ~2~~ + 3 + 4 22. *^J? - x ~ 4 = g ~ 3 - (3! - 2). 2*. 2 3 2 V 5. -5. a; + 3. 912 8 = 0. -9A ±_i _ 2 - c - i + ii = 0. -16. 2 5 3x + 5 21 23 - "1 5~ 24. 25. Z* - T \(x - 11) = f(x- - 25) + 34. 25. 5 90 EXAMPLES. 26. i(., _ 8) + :L± ^ + --=-^- = 7 - 23 - * 4 7 27. — (•—^io 1 )- *<■—"> a. 28. 1^-1^ + 13 = 0. x. 3 G 29. ! L±I _ ^J^i + 4 X = 12 + 2a? - 1 . 3. 3 4 6 30. *±1 _ ^|J = b ^l_ZL£ J + i. 28. 31. §«LZLl _ 13 ~ * = 25 _ ¥ (., + 3). 2. o 2 o 32. IZL* + *-lJ> + 1^ + LlL? + f = 0. 4. o 4 o b 33. 5a? ~ 3 - "LzLi; = ^ _|_ i9 (it; _ 4 ). 2 . 7 3 2 ^ V ; 34. p 4- .25a; - .Sx = x - 3. 12. 35. J)x — .2a; = .3a; — 1.5. 27. 3G. 1.5 = ^ - - 09a? ~ - 18 . .2 .9 37. (a + &)» + (a - b)x = • 4- a) (x 4- &). J//.S-. — $(a 4- 6). 44. (a?-a) 8 +(a!-6)»+(aj-c) 8 =3(aj-a)(aj-6)(a;-c). Ans. l(a 4-6 4- c). 15. Twenty-three times a certain Dumber is as much above 11 as 16 is above seven times the number: liml it. J/**. 1. EXAMPLES. 91 46. Divide 105 into two parts, one of which diminished Irv 20 shall be equal to the other diminished by 15. Ans. 50, 55. 47. The sum of two numbers is 8, and one of them with 22 added to it is five times the other : find the numbers. l Ans. 3, 5. 48. A and B begin to play each with $60. If they play till A's money is double B's, what does A win? Ans. $20. 49. The difference between the squares of two consecutive numbers is 121 : find the numbers. Ans. 60, 61. 50. Divide $380 between A, B, and C, so that B may have $30 more than A, and C may have $20 more than B. Ans. A $100, B $130, C $150. 51. A father is four times as old as his son ; in 24 years f he will only be twice as old : find their ages. Ans. 48, 12. 52. A is 25 years older than B, and A's age is as much above 20 as B's is below 85 : find their ages. Ans. 65, 40. 53. The sum of the ages of A and B is 30 years, and five y years hence A will be three times as old as B : find their present ages. Ans. 25, 5. 54. The length of a room exceeds its breadth by 3 feet ; if the length had been increased by 3 feet, and the breadth diminished by 2 feet, the area would not have been altered : find the dimensions. Ans. 15 ft., 12 ft. 55. There is a certain fish, the head of which is 9 inches long ; the tail is as long as the head and half the body ; and the body is as long as the head and tail together : what is the length of the fish? Ans. 6 ft. 56. The sum of $76 was raised by A, B, and C together ; B contributed as much as A and $10 more, and C as much as A and B together : how much did each contribute ? Ans. $14, $24, $38. 57. After 34 gallons had been drawn out of one of two equal casks, and 80 gallons out of the other, there remained just three times as much in one cask as in the other : what did each cask coutain when full? Ans. 103. 92 EXAMPLES. 58. Divide the number 20 into two parts such that the sura of three times one port, and live times the other part, may be 84. Ans. 8, 12. 59. A person meeting a company of beggars gave 4 cents to each, and had 16 cents left; he found that he should have required 12 cents more to enable him to give the beggars 6 cents each : how many beggars were there ? Ans. 14. 60. Divide 100 into two parts such that if a third of one part be subtracted from a fourth of the other, the remainder may be 11. Ans. 24, 76. 61. Divide 60 into two parts such that the difference between the greater and 64 may be equal to twice the difference between the less and 38. Ans. 36, 24. 62. Find a number such that the sum of its fifth and its seventh shall exceed the sum of its eighth and its twelfth by 113. Ans. 840. 63. An army in defeat loses one-sixth of its number in killed and wounded, and 4000 prisoners ; it is re-enforced by 3000 men, but retreats, losing one-fourth of its number in doing so ; there remain 18000 men : what was the original force? Am. 30000. 64. One-half of a certain number of persons received 18 cents each, one-third received 24 cents each, and the rest received 30 cents each ; the whole sum distributed was $5.28 : how many persons were there? Ans. 24. 65. A father has six sons, each of whom is four years older than his next younger brother ; and the eldest is three times as old as the youngest : find their respective ages. Ans. 10, 14, 18, 22, 26, 30. 66. A man left his property to be divided between his three children in such a way that the share of the eldest was to be twice that of the second, and the share of the second twice that of the youngest ; it was found that the eldest received $3000 more than the youngest: how much did each receive? Ans. §4000, §2000, §1000. ' EXAMPLES. 93 67. A sum of money is divided among three persons ; the first receives $10 more than a third of the whole sum ; the second receives $15 more than a half of what remains ; and the third receives wha\ is over, which is $70 : find the original sum. Ans. $270. 68. In a cellar one-fifth of the wine is port and one-third claret ; besides this it contains 15 dozen of sherry and 30 bottles of spirits : how much port and claret does it contain ? Ans. 90 port, 150 claret. 69. Two-fifths of A's money is equal to B's, and seven- ninths of B's is equal to C's : in all they have $770, what have they each? Ans. A $450, B $180, C $140. 70. A, B, and C have $1285 between them ; A's share is greater than five-sixths of B's by $25, and C's is four- fifteenths of B's : find the share of each. Ans. A $525, B $600, C $160. 71. The width of a room is two-thirds of its length; if the width had been 3 feet more, and the length 3 feet less, the room would have been square ; find its dimensions. Ans. 12 ft., 18 ft. 72. A man left $2200 to be divided among his four children A, B, C, D ; of whom B was to have twice as much as A, C as much as A and B together, and D as much as C and B together : how much had each ? Ans. $200, $400, $600, $1000. 73. A sum of money is to be distributed among three persons, A, B, and C ; the shares of A and B together amount to $240 ; those of A and C to $320 ; and those of B and C to $368 : find the share of each person. Ans. $96, $144, $224. 74. Two persons A and B are travelling together ; A has $100, and B has $48 ; they are met by robbers who take twice as much from A as from B, and leave to A three times as much as to B : how much was taken from each? Ans. $88, $44. 75. A wine merchant has two sorts of wines, one sort 04 EXAMPLES. worth $2.40 a quart, and the other worth $1 a quart ; from these he wants to make a mixture of 100 quarts worth $2.80 a quart : how many quarts must he take from each sort ? Ans. 75, 25. 76. In a mixture of wine and water the wine composed 25 gallons more than half of the mixture, and the water 5 gallons less than a third of the mixture : how man}' gallons were there of each? Ans. 85, 35. 77. In a lottery consisting of 10000 tickets, half the number of prizes added to one-third the number of blanks was 3500 : how many prizes were there in the lottery? Ans. 1000. 78. In a certain weight of gunpowder the saltpetre com- posed 6 lbs. more than a half of the weight, the sulphur 5 lbs. less than a third, and the charcoal 3 lbs. less than a fourth : how many lbs. were there of each of the three ingredients? Ans. 18, 3, 3. 79. A general, after having lost a battle, found that he had left fit for action 3600 men more than half of his army ; 600 men more than one-eighth of his army were wounded ; and the remainder, forming one-fifth of the arm}', were slain, taken prisoners, or missing : what was the number of the army? Am. 24000. 80. How many sheep must a person buy at $28 each bo that after paying 20 cents a score for folding them at night he may gain $319.20 by selling them at $32 each ? Ans. 80. 81. A certain sum of money was shared among five persons, A, B, C, D, and E ; B received $10 less than A ; C received $16 more than B; D received $5 less than C; and E received $15 more than D ; and it was found that E received as much as A and B together : how much did each receive? Ans. $26, $16, $32, $27. $12. 82. A tradesman starts with a certain sum of money ; at the end of the first year he had doubled his original stock, all but $100; also at the end of the second year he had doublet! the stock at the beginning of the second year, all EXAMPLES. 9") but $100; also in like manner at the end of the third year; and at the end of the thfyd year he was three times as rich as at first : find his original stock. Ans. $140. 83. A person went to a tavern witli a certain sum of money ; there he borrowed as much as he had about him, and spent a dollar out of the whole ; with the remainder he went to a second tavern, where he borrowed as much as he had left, and also spent a dollar ; and he then went to a third tavern, borrowing and spending as before, after which he had nothing left : how much had he at first? Ans. 87£ cents. 84. A crew which can pull at the rate of nine miles an hour, finds that it takes twice as long to come up a river as to go down : at what number of miles an hour does the river flow? Ans. 3. 85. A and B play at a game, agreeing that the loser shall always pay to the winner $1 more than half the money the loser has ; they commence with equal sums of money, but after B has lost the first game and won the second, he has twice as much as A : how much had each at the commence- ment? Ans. $6. 86. A and B find a purse with dollars in it. A takes out $2 and one-sixth of what remains ; then B takes out $3 and one-sixth of what remains : and then they find that they have taken out equal shares. How many dollars were in the purse, and how many did each take? Ans. $20 in the purse ; $5 taken by each. 9G WHEN ALL TERMS HAVE ONE COMMON FACTOR. CHAPTER VII. FACTORING — GREATEST COMMON DIVISOR- LEAST COMMON MULTIPLE. 62. Definitions. — Factoring is the process of resolving a quantity into its factors. The Factors of a quantity are those quantities which multiplied together produce it. A factor of a quantity is therefore a divisor of the quantity, i.e., it wiU divide the quantity without a remainder. Thus, a is a factor or divisor of abc, and b is a factor or divisor of ab — b 2 . Note. —In Division (Chap. V.) we had given the product of two factors and one of the factors, and we showed how to find the other factor. In the present chapter we shall consider cases in which the factors of an expression can be found when none of the factors are given. A Prime Quantity is one which has no integral factor except itself and unity. Thus, a, b, and a + c are prime quantities ; while ab, and ac + be are not prime. Quantities are said to be prime to each other or relatively prime, when unity is the only integral factor common to both. Thus, ah and cd are prime to each other. A Composite Quantity is one which is the product of two or more integral factors, neither of which is unity or the quantity itself. Thus, ax + x 2 is a composite quantity, the factors of which are x and a + x. 63. When All the Terms have one Common Factor. — When each term of a polynomial is divisible by a common factor, the polynomial may be simplified by the following EXAMPLES. 97 Rule. Divide each term of the polynomial separately by the com- mon factor, and enclose the quotient within parentheses, the common factor being placed outside as a coefficient; then the divisor will be one factor and the quotient the other. EXAMPLES. 1 . Factor the expression 3a 2 — 6a6. Here we see that the terms have a common factor, da ; therefore, dividing the polynomial by 3a, we obtain for the quotient a — 26. Hence the two factors are 3a and a — 2b. .-. 3a 2 - Gab = 3a(a - 26). Similarly 2. 5a 2 bx* - lbab 2 x 3 - 20ab s x i = 5abx 3 (ax - 36 - 46 2 .r). Factor the following expressions : o 3. x 2 — ax. Ans. x{x — a). 4. x 3 — x 2 . x 2 (x - 1). 5. a 2 - ab 2 . a(a - b 2 ). (§ 8x - 2a; 2 . 2.t-(4 - x). .7. bax — ba 3 x 2 . bax (I — a 2 x). 8. x 3 — x 2 y. x 2 (x — y). 9. bx - 2bx 2 y. bx (1 — bxy). 10. 16a; 2 + 6ix 2 y. 16a; 2 (l + Ay). 11. 54 - 81a. 27(2 - 3a). 12. 3,<- 3 — Gar + 9x. 3.r(a; 2 - 2» + 3) . 13. 6a 2 bx 3 + 2ab 2 x i + 4a6ar\ 2a6« 8 (3a + 6a; + 2a; 2 ). l4\ 72x 2 y — 84:xy 2 + 60ar> 2 . 12a;?/ (6a; — ly + bxy). 64. Expressions containing Four Terms. — When a polynomial contains four terms which can be arranged in pairs that have a common binomial factor, the polynomial may be simplified by the following Rule. Divide the polynomial by the common binomial factor ; then the divisor will be one factor and the quotient the other. 98 EXAMPLES. EXAMPLES. 1. Resolve into factors x 2 — ax + bx — ab. Here we see that the first two terms contain a factor x, and the last two terms a factor b ; therefore we factor the first two and last two terms by Art. G3, and obtain x(x — a) and b (x — a). We now see that the two pairs have the common binomial factor x — a. Dividing by x — a we obtain the quotient x + b for the other factor. The work therefore will stand as follows : x 2 — ax + bx — ab = x(x — a) + b(x — a) = (x - a)(x + b). 2. Resolve into factors 6x 2 — 9ax + 46a; — Gab. 6x 2 - Sax + Abx - Gab = 3x(2x - 3a) + 2b(2x - 3a) = (2x - 3a) (Bx + 2b). 3. Resolve into factors 12a 2 — 4«6 — 3aar + &0 2 . 12a 2 - 4a& - 3aar + frr = 4a (3a - 6) - .r(3a - 6) = (3a - b) (4a - x 2 ). Note. — It is not necessary always to factor in the same way. In the first line of work it is usually sufficient to see that each pair contains some common factor; and any suitably chosen pairs will bring out the same result. Thus, in the last example, we may have a different arrangement, and enclose the first and third terms in one pair, and the second and fourth in another as follows : 12a 2 - 4ab - 3ax 2 + bx 2 = \'2.<•-) _ 3rt(4a — x 2 ) - 6(4a - x z ) - (4a _ x 2 ) (3a - b), which is the same result as before. Resolve into factors 4. a 2 + ab + ac + be. Ans. (a + b)(a + c) f>. a 2 - ac + ab - be. (a - c)(a + 6) <;. ( /-r- + oca* + abc + bd. (ac + d)(ac + b) 7. a 2 + 3a + ac + 3c. (a + 3) (a + c) 8. 2ax + a?/ + 2&aj + by. (2x + 2/) (a + 6) 9. 3ax - bx - 3a?/ + &//. ( 3a - & ) (•'* ~ SO 10. a.i; 2 + 6x 2 + 2a + 26. (a + b){x 2 + 2) 11. x 1 - 3x - a# + 8y. (a - 3) (x - y) TO FACTOR A TRINOMIAL OF THE FORM X 2 + CtX + b. 99 65. To Factor a Trinomial of the Form x 2 +ax-\-6. — Let x' 2 + ax + b be any trinomial in which the coefficient of a; 2 is +1 , and the signs of a and b either pins or minus. Before proceeding to explain this case of resolution into factors, the student is advised to refer to Art. 40, and examine the relation that exists between two binomial factors and their product. Attention was there called to the way in which, in forming the product of two binomials, the coefficients of the different terms combined so as to give a trinomial result. Therefore, in the converse problem, namely, the resolution of a trinomial expression into its component binomial factors, we see, by reversing the results of Art. 40, that any trino- mial may be resolved into two binomial factors, when the first term is a square, and the coefficient of the secoud term is the sum of two quantities whose product is the third term. Hence the following Rule. The first term of each factor is x, and the second terms are two numbers whose Algebraic sum is the coefficient of the second term, and whose product is the third term. The application of this rule will be easily understood from the following EXAMPLES. 1. Resolve into factors x" + llx + 24. Here the first term of each binomial factor is x, and the second terms of the two binomial factors must be two numbers whose sum is 11 and whose product is 24. It is clear therefore that they must be +8 and +3, since these are the only two numbers whose sum is 11 and whose product is 24. .-. x 2 + 11a + 24 = (x + 8)( + 3y). 4. Resolve into factors 26d — c 2 — a 2 + d 2 + 6 2 + 2ac Here we see that the expression is composed of two trinomials, each of which is the square of a binomial [(1) and (2) of Art. 41]. .-. 2bd — c 2 — a 2 +d 2 +b 2 + 2ac = b 2 + 2bd + d 2 -(a 2 -2a<- + r-) = (6+* 15. Resolve a 2 x 3 - — - a; 3 + 4 iuto four fa ctors. f y oW-^-a»+ i = *»(«S-l) --^(« 2 -l) ?/ 3 y z y = (« 2 -l) W =(a+1)( „_ 1} HX^ + ! + i). 16. Resolve a 9 — 64a 3 — a 6 + 64 into six factors. The expression = a 3 (a 6 -64)-(o, 6 -64) = (a 6 -64)(« 3 -l) = (a 8 +8)(a 8 -8)(a 8 -l) = (a + 2) (a 2 -2a+4) (a-2) (a 2 +2a+4) (a-1) (« 2 + a + 1 ) . Resolve into factors 17. a 4 + 16a; 2 + 256. ^, JS . (asF+4as+16) (a*-4as+16). 18. s 4 + ?/ 4 - 7xV. (a,- 2 + 3a^ + ?/) (a; 2 - 3zy + */ 2 ) . 19. 81a*+9a 2 & 2 +6 4 . (9a 2 + 3a& + 6 2 ) (9a 2 - 3a& + & 2 ) . 20. a; 4 - 19x 2 2/ 2 +25/. (x 2 + 3ay- 5y 2 ) (a 2 - 3xy- by 2 ) . 110 EXAMPLES. By a skilful use of factors, the actual processes of multi- plication and division can often be partially or wholly avoided. 21. Multiply 2a + 36 — c by 2a — 36 + c. The product = [2a + (36 - c)][2a - (36 - c)] = (2a) 2 - (36 - c) 2 [(3) of Art. 41] = 4a 2 - 9b 2 + 66c - c 2 . 22. Divide the product of x 2 — bxy + Gy 2 and x — 4y by x 2 — Ixy -+- I2y 2 . We might multiply the first two expressions together and then divide the result by the third. But by factoring the first and third expressions, and denoting the division by means of a fraction (see Art. 78), the work will be much shorter. Thus, the required quotient _ (x 2 — oxy + Gy 2 ) (x — 4y) x 2 — Ixy + I2y 2 _ (x — 3y) (x — 2?/) (x — 4y) (x - Ay) (x - 3y) = x - 2y. 23. Divide the product of 2x 2 + x — 6 and Gx 2 - 5x -f 1 by 3x 2 + bx — 2. Ans. (2x - 3)(2as - 1). Find the product of 24. 2x— 7y + 3z and 2x+7y— 3z. 4x 2 —49y 2 +42yz—9z 2 . 25. x 3 +2x 2 y+2xy 2 +y 3 and X s — 2x 2 y + 2xy 2 — y 3 . x 6 —y 6 . 26. x s -4x 2 + 8x - 8 and x 3 + 4a; 2 + 8a; + 8. x e - 64. Divide 27. 2a,-(a; 2 - 1 ) (x -f 2) by x 2 + x-2. 2x(x + 1 ) . 28. (x 2 + 7x + 10) (x + 3) by x 2 + 5x + 6. x + 5. 29. 5a?(as- 11) (a; 2 -a;- 156) by .r s +a: 2 -132a;. 5(«-18). 30. a, 9 - 6 9 by (a 2 + ab + 6 2 ) (a, 6 + a 3 b s + 6 6 ) . a - 6. 31. [a 3 + (a - 6)oj - aft] |V - (a - &)a - 06] by a 3 + (a + 6).« + ab. Ans. (as — a)(x - b). GREATEST COMMON DIVISOR — DEFINITIONS. Ill GREATEST COMMON DIVISOR. 72. Definitions. — A Common Divisor of two or more expressions is an expression that will divide each of them exactly. Hence, every factor common to two or more expressions is a common divisor of those expressions (Art. 62) . Thus, in 4a 2 6, 6a 3 b 2 , and a 4 b 3 , a 2 occurs as a factor of each quantity ; b also occurs as a factor of each quantity ; a 2 and b are therefore common divisors of these three quantities. The Greatest Common Divisor of two or more Algebraic expressions is the expression of highest degree (Art. 18) which will divide each of thern exactly. Note. — The term greatest common divisor, which has been adopted from Arithmetic, does not imply in Algebra that it is numerically the greatest, but that it is the factor of greatest degree. The student is cautioned against being misled by the analogy between the Algebraic and the Arithmetic greatest common divisor. He should notice that no mention is made of numerical magnitude in the definition of the Algebraic greatest common divisor. In Arithmetic, the greatest common divisor of two or more whole numbers is the greatest whole number which will exactly divide each of them. But in Algebra, the terms greater and less are seldom applicable to those expressions in which definite numerical values have not been assigned to the various letters which occur. Besides, it is not always true that the Arith- metic greatest common divisor of the values of two given expressions obtained by assigning any particular values to the letters of those expressions, is the numerical value of the Algebraic greatest common divisor when those same values of the letters are substituted therein, as will be shown later (Art. 74). For this reason, some writers have used the terms, highest common divisor, and highest common factor, instead of the term greatest common divisor. But to avoid employing a new phrase, and in conformity with well-established usage, we shall retain the old term greatest common divisor. The abbreviation G. C. D. will often be used for shortness instead of the words greatest common divisor. 73. The Greatest Common Divisor of Monomials, and of Polynomials which can be easily Factored. — Let it be required to find the greatest common divisor of 21a 4 z 3 y, 35a 2 afy, 28a s xY, and Ua 5 x 2 y\ 112 GREATEST COMMON DIVISOR OF MONOMIALS. By separating each expression into its prime factors, we have 2la*x 3 y = 7 X 3aaaaxxxy. 35a 2 x 4 y = 7 X baaxxxxy. 28a 3 x 2 y i = 7 x 2 x 2aaaxxyyyy. 14a b x 2 y 2 = 7 X 2aaaaaxxyy. By examining these expressions we find that 7, aa, xx, and y are the only factors common to all of them. Hence all the expressions can be exactly divided by either of these factors, or by their product, la 2 x 2 y, which is therefore their greatest common divisor. Find the G. C. D. of 4cx s and 2cx 3 + 4cV. Resolving each expression into its factors, we have Acx 3 — 2cx 2 x 2x. 2cx 3 + Ac 2 x 2 = 2cx 2 {x + 2c). Here it is clear that both expressions are divisible (1) by 2, which is the numerical greatest common divisor of the coefficients, (2) by c, and (3) by x 2 . .-. G. C. D. = 2cx\ Find the G-. C. D. of 3a 2 + 9a&, a 3 — dab 2 , a 3 + 6a 2 & + 9a& 2 . Resolving each expression into its factors, we have 3a 2 + 9a6 = 3a (a + 36). a 3 - 9a6 2 = a(a + 3b) (a - 3b). a 3 + Ga 2 b + dab 2 = a(a + 3b) (a + 3b). .-. G. C. D. = a(a + 36). Find the G. C. D. of x(a — a) 2 , ax (a — x) 3 , 2ax(a - x)\ Resolving into factors, we have x(a — x) 2 = x(a — x)(a — x). ax(a — x) s = aoj(a — x)(a — x)(a — x). 2ax(a — a-) 4 == 2a,x(a — x)(a — x){a — a;) (a — x). .-. G. C. D. = x{a - x) 2 . Hence the following Rule. Resolve each expression into its prime factors, and take the product of all the factors common to all the expressions, giving to each factor the highest power tohich is common to cdl the given expressions. GREATEST COMMON DIVISOR OF POLYNOMIALS. 113 EXAMPLES. Find the G. C. D. of 1. 4a& 2 , 2a 2 6, 6ab 3 . . Am. 2ab. 2. 3xY, x 3 y 2 , x 2 y 3 . x 2 y 2 . 3. 6xy 2 z, 8x 2 y 3 z 2 , 4xyz 2 . 2xyz. 4. 5a s b 3 , loabc 2 , Wa 2 b 2 c. bob. 5. 9x 2 y 2 z 2 , 12xy z z, 6x 3 y 2 z 3 . Sxy 2 z. 6. 8a 2 x, 6abxy, 10abx 3 y 2 . 2ax. 7. a 2 + ab, a 2 - b 2 . a + 6. 8. (x + y) 2 , x 2 - y 2 . x + y. 9. x 3 + x 2 y. x 3 -\- y 3 . x + y. 10. a 3 — a 2 x, a 3 — ax 2 , a 4 — ax 3 . a(a — x). 11. x 4 - 27a 3 x, (x - 3a) 2 . x - 3a. 12. xy - y, »V - xy. y(x - 1). 13. ax 2 + 2a 2 x + a 3 , 2ax 2 - 4a 2 x - 6a 3 , 3 (ax + a 2 ) 2 . -4ws. a(x + a). 74. The Greatest Common Divisor of Expressions that cannot be Readily Resolved into Factors. — To find the G. C. D. in such cases, we adopt a method analogous to that used in Arithmetic for finding the G. C. D. of two or more numbers. The method depends on two principles. 1. If an expression contain a certain factor, any multiple of that expression is divisible by that factor. Thus, if F divides A it will also divide mA. For let a denote the quotient when A is divided by F; then A = aF; therefore mA = maF\ and therefore F divides mA. 2. If two expressions have a common factor, it will divide their sum and their difference; and also the sum and the difference of any midtiple of them. Thus, if F divides A and B, it will divide mA ± nB. For since F divides A and B, we may suppose A=aF, and B—bF; therefore mA ± nB = maF ± nbF = F(ma ± nb). Therefore F divides mA ± nB, 114 GREATEST COMMON DIVISOR OF POLYNOMIALS. We can now prove the rule 'for finding the G. C. D. of any two compound Algebraic expressions. Let A and B denote the two expressions. Let them be arranged in ascending or descending powers of some common letter ; and let the highest power of that letter in B be either equal to or greater than the highest power in A. Divide B by A ; let p be the quotient and C the remainder. Suppose C to have a simple factor m. Remove this factor, and so obtain a new divisor D. Suppose further, that in order to make A divisible by D it is necessary to multiply A by a simple factor n. Divide nA by D ; let q be the next quotient, and E the remainder. Divide D by E ; let r be the quotient, and suppose that there is no remainder. Then E will be the G. C. D. required. The operation of division will stand thus : A)B(p pA mjc D)nA(q qD E)D(r rE First, to show that E is a common divisor of A and B. From the above division we have the following results : D = rE. nA = qD + E = qrE + E = {qr + \)E. B= pA + C = pA + mD = 1WE + pE + mrE fpqr -f p \ Therefore E is a common divisor of A and B. Second, to show that E is the greatest common divisor of A and B. EXAMPLES. 115 By (2) of this Art. every common factor of A and B divides also B — pA, that is O, and therefore D (since m is a simple factor) . Similarly as it divides A and D it divides nA — qD, that is E. But no expression of higher degree than E can divide E. Therefore E is the greatest common divisor of A and B. The greatest common divisor of three expressions, A, B, C, may be obtained as follows : First find D, the G. C. D. of any two of them, say of A and B ; next find F, the G. C. D. of D and C; then F will be the G. C. D. of A, B, C. For D contains every factor which is common to A and B (Art. 72) ; and as F is the G. C. D. of D and C, it contains every factor common to D and C, and therefore every factor common to A, B, and C. Hence F is the G. C. D. of A, B, C. EXAMPLES. 1. Find the G. C. D. of x 2 -ix -f- 3 and 4a; 3 - 9x 2 - 15a;-f-18. x 2 - 4a + 3)4a; 3 - 9a? - 15a + 18 (4a + 7 4a 3 - 16a 2 + 12a 7a 2 - 27a + 18 7a 2 - 28a + 21 a - 3) a 2 — 4a + 3(a - - 1 a 2 - 3a - a + 3 — a + 3 Therefore the G. C. D. is a - 3. Explanation. — First arrange the given expressions according to descending powers of x. Take for dividend that expression whose first term is of the higher degree; and continue each division until the first term of the remainder is of a lower degree than the first term of the divisor. When the first remainder, x — 3, is made the divisor, we put the first divisor to the right of it for a dividend, and after obtaining the new quotient, x — 1, we have nothing for a remainder. Hence, as in Arithmetic, the last divisor, x — 3, is the G. C. D. required. 116 EXAMPLES. 2. Find the G. C. D. of 8a; 8 4a; 3 - 3x 2 - 24a; - 9. x 4a; 3 4x 3 2x 2 53a; - 39 and 3a; 2 - - 24a; - 9 8a; 3 - 2a; 2 - 53a; - 39 5a; 2 - - 21a; 8a; 3 - Ga- 2 - 48a; - 18 2x- - - 3a; — 9 4a; 2 - 5a; - 21 2x 2 - - Gx 4a; 2 - 6a; - 18 3a - 9 x — 3 3a; - 9 2x Therefore the G. C. D. is x - 3. Explanation. — First arrange the given expressions according to descending powers of x. The expressions so arranged having their first terras of the same order, we take for divisor that whose highest power has the smaller coefficient, and arrange the work in parallel columns, as above. (1) At the first division we put the quotient 2 to the right of the dividend. (2) When the first remainder 4a- — 5.r — 21 is made the divisor we put the quotient x to the left of the dividend. (3) When the second remainder 2x 2 — 3x — 9 is made the divisor we put the quotient 2 to the right of the dividend. (4) When the third remainder x — 3 is made the divisor we put the quotients 2x and 3 to the left of the dividend, and so on. This method is used only to determine the compound factor of the G. C. D. Simple factors of the given expressions must first be separated from them, and the G. C. D. of these, if they have any, must be reserved and multiplied into the compound factor obtained by the rule. 3. Find the G. C. D. of 6a; 4 - 2Ga; 3 + 4()ar - 42a; and 18a; 4 + 3x 3 - 132a; 2 + 63a;. We have 6a; 4 -26x 3 + 46a; 2 and 18a; 4 + 3a; 8 - 132a; 2 -(-63a; 42a;=2a;(3a; 3 -13a; 2 +23a;-21). (1) 3a;(6a; 3 -r- a; 2 -44a;+21). . (2) The simple factor 2 is found in the first expression and not in the second ; therefore it forms no part of the G. C. D., and may be rejected. Likewise the simple factor 8, occurring in the second expression ami not in the first, may EXAMPLES. 117 be rejected as forming no part of the G. C. D. But the simple factor x is common to both expressions, and is therefore a factor of the G. C. D. and must be reserved. Rejecting therefore the simple factors 2 and 3 as forming no part of the G. C. D., and reserving the common factor x as forming a part of the G. C. D., and arranging in parallel columns, we have 3a; 3 - 13a; 2 + 23a; 21 6a; 3 6a; 3 + X 2 26a; 2 + 44a; 46a; + 21 - 42 27a; 2 - 90a; + G3 The first division ends here, since 27a;' 2 is of a lower degree than 3a; 3 . If we now make 27a;' 2 — 90a; + 63 a divisor we find that it is not contained in 3a; 3 — 13a;' 2 -+- 23a; — 21 with an integral quotient. But, noticing that 27a; 2 — 90a; + G3 may be written in the form 9 (3a; 2 — 10a; + 7), and remem- bering that the G. C. D. we are seeking is contained in the remainder 9 (3a; 2 — 10a; + 7), and that, since the two expres- sions 3a; 3 - 13a; 2 + 23a; - 21 and 6.c 3 + a;' 2 - 44a; + 21 have no simple factors, therefore their G. C. D. can have none, we conclude that the G. C. D. must be contained in the factor 3a; 2 — 10a; + 7, and that therefore we can reject the simple factor 9, and go on with the divisor 3a; 2 — 10a; + 7. Resuming the work, we have 3a; 3 - 13a; 2 + 23a- - 21 3a; 3 - 10a;' 2 + 7a; 3a; 2 — 10a; + 7 3a; 2 — 7a; — 3a;' 2 + 16a; — 21 — 3a; 2 + 10a; — 7 — 3a; + 7 - 3a; + 7 2) 6a; - 14 3a; — 7 Therefore the G. C. D is a;(3a; - 7). The factor 2 was removed for the same reason as the factor 9. 4. Find the G. C. D. of 2a; 3 + a; 2 - x - 2 (1) and 3a; 3 - 2a; 2 + » - 2 (2). 118 EXAMPLES. As the expressions stand neither can be divided by the other without obtaining a fractional quotient. This difficulty cannot be obviated by removing a simple factor, since neither expression contains a simple factor. We may however intro- duce a suitable factor into either expression, just as in Ex. 3 we removed a factor when we could no longer proceed with the division without a fractional quotient. The given expres- sions (1) and (2) have no common simple factor, therefore their G. C. D. can have no simple factor, and hence cannot be affected if we multiply either of them by any simple factor. Multiplying (2) by 2 and taking it for dividend, we have -2a! 17a; 11 2a; 3 + a; 2 - 7 x- 2 6a; 3 - 4a; 2 + 2x - 4 6x 3 + 3a; 2 — 3a;- 6 14a; 3 + 7a; 2 - 14a; 3 - 10a; 2 - 7x — 14 4a; — 7a; 2 + 5a; -f- 2 17 17a;' 2 - 17a; 2 — 3a;- 14 17a; — 119a; 2 + 85a; + 34 - 1 19a; 2 + 21a- + 98 14a; - 14 14a; - 14 64) 64a; -64 x- 1 Therefore the G. C. D. is x - 1. In this example, after the first division the factor 7 is introduced because the first remainder —7a;' 2 + 5a; + 2 will not divide the first divisor 2a; 3 + a; 2 — a; — 2. After the second division the factor 17 is introduced because the second remainder 17a; 2 — 3a; — 14 will not divide the second divisor — 7ar -f 5x + 2. Finally the factor 64 is removed as explained in Ex. 3. Note. —The difference between the Algebraic G. C. D. and the Arithmetic G. C. D. can be seen by an example. Factor (1) and (2) of last example as follows: 2a; 3 + x* - x - 2 = (x - l)(2a- 2 + 8a + 2), and 3a; 8 - 2a; 2 + x — 2 = (x — 1) (3a; 2 + x + 2). EXAMPLES. 119 Now since the G. C. D. of these expressions is x — 1, the factors 2x 2 -f 3.« + 2, and 3ar + x + 2, have no common factor. But if we put x — 4, then 2a; 3 + x 2 - x - 2 = 138, and 3x s - 2x 2 + x - 2 = 162, and the G. C. D. of 138 and 162 is 6, while 3 is the numerical value of the Algebraic G-. C. D., as — 1. Thus the numerical value of the Algebraic G. C. D. does not agree with the numerical value of the Arithmetic G. C. D. The reason may be explained as follows ; the expressions 2X 2 + 3a; + 2 and 3» 2 + x + 2 have no Algebraic common factor ; but when x = 4 they become equal to 46 and 54 respectively, and therefore have a common Arithmetic factor 2, which, multiplied into x — 1 or 3, gives 6 for the nu- merical value of the Arithmetic G. C. D., while 3 is the numerical value of the Algebraic G. C. D. In the same way it may be shown that if we give particular numerical values to the letters in any two expressions, and in their Algebraic G. C. D., the numerical value of the G. C. D. is by no means necessarily the Arithmetic G. C. D. of the values of the expressions. We may now enunciate the rule for finding the greatest common divisor of two compound Algebraic expressions. Rule. Arrange the given expressions according to the descending poivers of the same letter. Divide that expression which is of the higher degree by the other; or, if both are of the same degree, divide that whose first term has the larger coefficient by the other; and if there is no remainder the first divisor will be the required greatest common divisor. If there is a remainder divide the first divisor by it, and continue thus to divide the last divisor by the last remainder, until a divisor is obtained which leaves no remainder; the last divisor ivill be the greatest common divisor required. 120 LEAST COMMON MULTIPLE — DEFINITIONS. Note 1. — Before beginning the division, all simple factors of the given expressions must be removed from them, and the greatest common divisor of these must be reserved as a factor of the (Jr. C. D. required. (See Ex. 3.) Note 2. — Either of the given expressions or any of the remain- ders may be multiplied or divided by any factor which does not divide both of the given expressions. (See Ex. 4.) Note 3. — Each division must be continued until the remainder is of a lower degree than the divisor. Find the G. C. D. of 5. x 3 + 2x 2 — 13a; + 10, and x 8 + x°- — 10a; 4- 8. Ans. x 2 — 3a; + 2. 6. a; 3 -5a- 2 -99a;-f-40, and a; 3 - Ga; 2 - 86a; + 35. ^ 2 -13x'+5. 7. x s -x 2 —5x— 3, and a; 3 — 4ar— 11a: — 6. a; 2 + 2a; + 1. 8. a; 3 + 3a; 2 - 8a; - 24, and a- 3 + 3a; 2 - 3a; - 9. x + 3. 9. 2a; 3 4-4a 2 -7.f-14,andGa; 3 -10a- 2 -21a;+35. 2ar-7. LEAST COMMON MULTIPLE. 75. Definitions. — A Multiple of an expression is any expression that can be divided by it exactly. Hence, a multiple of an expression mast contain all the factors of that expression. Thus, Ga 2 b is a multiple of 3 or 2 or G or a or h. A Common Multiple of two or more expressions is an expression that can be divided by each of them exactly ; or, it is one of which all the given expressions are factors. Thus, the expression aire 3 is a common multiple of the expressions, a, 6, c, ah, abc, ab 2 , 6 a c 8 , etc., or of the expres- sion itself ; but it is not a multiple of a 2 , nor of b 3 , nor of any symbol which does not enter into it as a factor. The Least Common Multiple * of two or more Algebraic expressions is the expression of least degree which is divisi- ble by each of them exactly. * Called alHo lowest common multiple. The term, Uaat common multiple, i* objected to l>y some, for a reason similar to tho ouc for which they object to tho term greatest common dioisor. LEAST COMMON MULTIPLE OF MONOMIALS. 121 Hence, the least common multiple of two or more expres- sions is the product of all the factors of the expressions, each factor being taken the greatest number of times it occurs in any of the expressions. The abbreviation L. C. M. is often used instead of the words least common multiple. Note. — Two or more expressions can have only one least common multiple, while they have an indefinite number of common multiples. 76. The Least Common Multiple of Monomials, and of Polynomials which can be easily Factored. — Let it be required to find the least common multiple of 2la*x 3 y, 35aVy, 28a 3 x 2 y*, and lia 5 x 2 y 2 . By separating each expression into its prime factors, we have 2\a i xhj = 3 X 7a 4 arty, S5a 2 x*y = 5 x IdWy, 28a 3 x 2 y i = 2 2 x 7a 3 x 2 y 4 , 14a & x 2 y 2 = 2 x 7a 5 arfy 2 - Hence, the L. C. M. = 7 x 3 X 5 x 2 2 a 5 afy i = 42Ua 5 .i'Y; for 420 is the numerical L. C. M. of the coefficients ; a 5 is the lowest power of a that is divisible by each of the quan- tities a 4 , a 2 , a 3 , a 5 ; x* is the lowest power of x that is divisible by each of the quantities a,- 8 , a; 4 , x 2 ; and y i is the lowest power of y that is divisible by each of the quantities y, y*, y 2 . 2. Find the L. C. M. of Gx s (a - x) 2 , 8a 2 (a - x) 3 , and \2d 2 x 2 (a — a;) 4 . Resolviug into factors, we have 6x 3 (a - x) 2 = 3 X 2x 3 (a - x)\ 8a 2 (a - x) 3 = 2 X 2 X 2a 2 (a - x) 3 , 12a 2 x 2 (a - xY = 3 x 2 x 2a 2 x 2 (a - x)\ Hence, the L. C. M. = 3 X 2 3 a 2 x 3 (a - a;) 4 = 2ia 2 x 3 (a - a;) 4 . For it consists of the product of (1) the numerical L. C. M. of the coefficients, and (2) the lowest power of each factor 122 EXAMPLES. which is divisible by every power of that factor occurring in the given expressions. 3. Find the L. C. M. of 3a 2 + 9a6, 2a? - 18ab 2 , a 3 + 6a 2 6 + dab 2 , a 3 -f ba 2 b + 6a6 2 . Resolving into factors, we have 3a' 2 + 9a6 = 3a (a + 3b), 2a 3 - 18a6 2 = 2a(a + 36) (a - 36), a 3 + 6a 2 6 + 9a6 2 = a (a + 36) 2 , a 3 + oa 2 6 + Gab 2 = a (a + 36) (a -f 26). Hence the L. C. M. = Ga(a + 36) 2 (a - 36) (a + 26). Hence the following Rule. Resolve each expression into its prime factors, and take the product of cdl the factors, giving to each factor the highest exponent which it has in the given expressions. If the expressions are prime to each other, their product is the least common multiple. EXAMPLES. Find the least common multiple of 1. 5a a 6c 8 , 4a6 2 c. 2. 12a6, 8xy. 3. 2a6, 36c, 4ca. 4. a 2 6c, 6 2 ca, c 2 ab. 5. 5a 2 c, Gc6 2 , 36c 2 . 6. x 2 , x 2 - 3a. 7. 21a 3 , 7a- 2 (a +1). 8. a 2 + ab, ab + b 2 . 9. Ga 2 — 2x, 9a 2 — 3a. x 2 + 2x, x 2 + 3a + 2. x 2 + la + 4,x 2 + 10. 11. 12. + 6. Ans. 20aW 24abxy 12abc a 2 6 2 c 2 30aW a 2 + x — 20, x 2 x 2 (x - 3) 21a 3 (a + 1) ab(a + b) 6a*3a - 1) a(a + 2)(» + 1) (,• + 2) 2 (a + 3) 10a + 24, a' J - x - 30. Ans. (x -f- 5) (a — 4)(« — 6) LEAST COMMON MULTIPLE OF POLYNOMIALS. 123 X 2a; 4 + x 3 - 20a; 2 - Ix + 24 2a; 4 + 7a; 3 —9x 2a; 4 +3a; 3 -13a; 2 -7a;+15 2a; 4 + x 3 -20x 2 -lx+2± 3 -6x 3 -20x 2 +2x+24: -6x 3 -2lx 2 +27 2a; 3 + lx 2 - 9 2a; 3 + 4a; 2 -6a; x 2 +2x- 3 3a; 2 +6a;— 9 3a; 2 +6a;- 9 77. "When the Given Expressions cannot be Resolved into Factors by Inspection. — To find the least common multiple of two compound Algebraic expres- sions in such cases, the expressions must be resolved by findiug their G. C. D. 1. Find the L. C. M. of 2x* + x 3 - 20a; 2 - 7a; + 24 and 2a; 4 + 3a; 3 - 13a; 2 - lx + 15. We first find their G. C. D. 1 ■2x .-. G. C. D. = x 2 + 2a; - 3. Hence, by division, we obtain 2x*+ x 3 - 20a; 2 - 7a; + 24 = (a; 2 + 2a; - 3) (2x 2 - Sx - 8) , and 2 x i +Bx 3 - 13a; 2 - 7a; + 15 = (a; 2 + 2a; - 3) (2a; 2 - x - 5). Therefore the L. C. M. = (a; 2 + 2a; - 3) (2a; 2 - Sx - 8)(2a; 2 - x - 5). "We may now prove the rule for finding the least common multiple of any two compound Algebraic expressions. Let A and B denote the two expressions, F their greatest common divisor, and M their least common multiple. Sup- pose that a and b are the respective quotients when A and B are divided by F ; then A = aF, and B = bF. . . ... (1) Since F contains all the factors common to A and B, the quotients a and b have no common factor, and therefore their least common multiple is ab, and hence the least 124 LEAST COMMON MULTIPLE OF POLYNOMIALS. common multiple of aF and bF, or of A and JS, from (1) is abF, by inspection. That is M = abF (2) But from (1) we have AB = abFF = MF, from (2) (3) M = 4*. F Rule. The least common multiple of two expressions may be found by dividing their product by their G. C. D., or, by dividing either of the expressions by their G. C. D. and multiplying the quotient by the other. From (3) we see that the product of any two expressions is equal to the product of their G. C. D. and L. C. M. To find the least common multiple of three expressions, A, B, C. First find M, the L. C. M. of A and B. Next find N, the L. C. M. of M and C; then N will be the required L. C. M. of A, B, G. For N is the expression of least degree which is divisible by M and C, and M is the expression of least degree which is divisible by A and B. Therefore N is the expression of least degree which is divisible by all three. In a similar manner we may find the L. C. M. of four expressions. Note. — The theories of the greatest common divisor and of the least common multiple are not necessary for the subsequent chapters of the present work, and any difficulties which the student may find in them may be postponed till he has read the Theory of Equations. The examples however attached to this chapter should be carefully worked, on account of the exercise which they afford in all the fundamental processes of Algebra. EXAMPLES. 125 EXAMPLES. Resolve iuto factors 1. x 2 + xy. 2. x s — x 2 y. 3. 10a; 3 — 2ox i y. 4. x 8 — x 2 y + xy 2 . 5. 3a 4 - 3a 3 6 + 6a 2 6 2 . 6. 38a 3 x 5 + 57aV. 7. aa; — 6a; — az + 6«. 8. 2a« + ay + 26a; + 6?/. 9. 6x 2 -f- 3a;?/ — 2ax — ay. 10. 2x 4 — x 3 + 4x - 2. 11. 3a: 3 + 5a; 2 + 3a + 5. 12. a; 4 + a; 3 + 2a; + 2. 13. 2/ 8 - y 2 + y - 1. 14. 2aa; 2 + Baxy — 2bxy — Sby' 15. a 2 - 19a: + 84. 16. x 2 - 19a; + 78. 17. a 2 - 14a6 + 496 2 . 18. a 2 4- 5a6 + 6b 2 . 19. m 2 — 13mw + 40n 2 . 20. m 2 - 22mn + Won 2 . 21. x 2 - 23xy + 132?/ 2 . 22. 130 + Blxy + x 2 y 2 . 23. 132 - 23a; + x 2 . 24. 88 + 19a; + a; 2 . 25. 65 + Sxy - xhf. 26. a; 2 + 16a; - 260. 27. a; 2 - 11a; — 26. 28. a 2 b 2 - Babe - 10c 2 . 29. a; 4 - a 2 x 2 - 132a 4 . 30. 4a; 2 + 23a; + 15. 31. 12a; 2 --2Sxy + 10?/ 2 . 32. 8a; 2 - 38a; + 35. 33. 12a; 2 - 31a; - 15. Ans. x(x + y). x 2 (x - y). 5a; 3 (2 — Sxy). x(x 2 — xy + y 2 ). Ba 2 (a 2 - ab + 2b 2 ). 19a 8 a; 2 (2a; 3 + 3a). (a — b) (x — z) . (2x + y)(a + b). (2x + y)(Bx - a). (2x - l)(a; 3 + 2). (3a: + 5)(a; 2 + 1). (x+ l)(a; 3 + 2). (!/- l)0/ 2 +l). . (2a; + By) (ax - by). (*- 12) (x- 1). (x - 13) (a; - 6). (a — lb) (a — lb) . (a + 36) (a + 26). (m — 8m) (m — 5«) . (m — 15n)(m — In). (x- 12y)(x- Uy). (26 + asy)(5 + a?y). (12 - a;) (11 - a). (8 +aj)(ll + •»). (5 + a^)(13 - xy). (x + 26) (a; - 10). (x + 2) (as - 13). («6 + 2c) (ab — 5c). (a; 2 + 11a 2 ) (a; 2 - 12a 2 ). (as + 5)(4a> + 3). (3a; - 2y)(4x - by). {2x - 7) (4a; - 5). (12a; + 5) (a; - 3). 126 EXAMPLES. 34. 3 + 11a; - 4a; 2 . Ans. (3 - a;)(l + 4a;). 35. 6 + 5a; - 6x 2 . (2 + Sx) (3 - 2a;). 36. 4 - 5a; - 6a; 2 . (4 + 3a;) (1 - 2x). 37. 5 + 32a; - 21a; 2 . (1 + 7a;) (5 - 3a-). 38. 20 - 9a; - 20a; 2 . (5 + 4a;) (4 - 5a;). 39. 25 - 64a; 2 . (5 + 8a?) (5 - 8a;). 40. 81pV - 25Z> 2 . (9/rz 3 + ob)(0p 2 z s - 56). 41. (1811) 2 - (689) 2 . 2500 x 1122 = 2805000. 42. (8133) 2 - (8131) 2 . 16264 x 2 = 32528. 43. (24a; + y) 2 - (23a; - y) 2 . 47a;(.r -f 2y). 44. (5a; + 2y) 2 - (3a; - y) 2 . (8a; + y)(2x + By). 45. 9a; 2 - (3a; - by) -. by (6a- - by). 46. 16a; 2 - (3a; + l) 2 . (7a; + \){x - 1). 47. a 6 + 7296 3 . [a 2 + 9&) (a 4 - 9a 2 6 + 81& 3 ). 48. x 3 y s - 512. (a-.y - 8)(x 2 y 2 + 8xy + 64). 49. b00x 2 y - 20y 3 . 20y(bx + y)(bx - y). 50. (« + 6) 4 -l. [(« + 6)2 + i]( ft+ ^ + i)( a + ^_ i). 51. (c + cl) 3 + (c - cZ) 3 . 2c(c 2 + 3cp). 52. a; 4 ?/ — a; 2 ?/ 3 — xhf + a-?/ 4 . .r?/(.r + y) (x — y) (x - y) . 53. a; 4 — 6x 2 y 2 + y*. (a; 2 + 2xy — y' 2 ) (x 2 — 2xy - y 2 ). 54. ab(x 2 + 1) + x(a 2 + b 2 ). (ax + &)(6as + a). 55. a 3 + (a + &)u.e + bx 2 . (a 2 + &.r) (a + x). Find the greatest common divisor of 56. 66aW, 44a W, 24a 2 6 3 c 4 . 2a W. 57. a; 2 + x, (x + l) 2 , x 3 + 1. x + 1. 58. a; 8 + 8y 8 , a; 2 + xy — 2y 2 . x + 2y. 59. 12a; 2 + x - 1, 15a; 2 + 8a; + 1. 3a; + 1. 60. 2a; 2 + 9a; + 4, 2a; 2 -f 11a? + 5, 2x 2 - Bx - 2. 2x + 1 . 61. Bx*— 3a 8 — 2a? a — x— 1, 6a; 4 -3a; 3 -a- 2 -a;-l. ;ir-+l. 62. 2a^ - 9aa; 2 + 9a 2 a; - 7a 3 , 4a; 3 - 20• EXAMPLES. 127 Find the least common multiple of 65. 35cm,* 2 , 42a/, 30az 2 . Ans. 210ax 2 y 2 z 2 . 66. x 2 - 3a; + 2, x 2 - 1. (x + 1) (a - l)(x - 2). 67. or - ox + 4, a; 2 - Gx + 8. (a; — 4) (a? - 1) (x — 2) . 68. a; 2 - x - 6, x 2 + a; - 2, a; 2 - 4a; + 3. Ans. (x - 3) (a; + 2) (a; — 1). 69. 3a; 2 + 11a; + 6, 3a; 2 + 8a; + 4, a; 2 + 5a; + 6. Ans. (x + 2) (a; + 3) (3a; + 2). 70. 5a; 2 + 11a; + 2, 5a; 2 + 16a; + 3, a; 2 + 5a; + 6. Ans. (x + 2) (a; + 3) (5a; + 1). 71. 3a; 2 - x - 14, 3a; 2 - 13x + 14, x 2 - 4. Ans. (x + 2)(x - 2) (3a; - 7). 72. a; 2 - 1, x 3 + 1, a; 3 - 1. a; 6 - 1. 73. x 2 — 1, x 2 + 1, x i + 1, a; 8 - 1. x 8 - 1. 74. x 2 - 1, x s + 1, x s - 1, a; 6 + 1. a; 12 - 1. 75. a; 3 + 2a; 2 - 3a;, 2a; 3 + 5a; 2 -3a;. a;(a;-l) (a;+3) (2a;— 1). 76. a; 2 - 10a; + 24, a; 2 - 8a; + 12, a; 2 - 6a; + 8. Ans. (x - 4) (a; - 6) (a- - 2). 77. a; 3 -r y 3 , x 3 y - y\ y 2 {x — ?/) 2 , a; 2 + xy + y' 2 . Ans. y' 2 (x — y) 2 (x 2 + xy + ?/ 2 ). 78. a 2 - 6 2 , a 3 - 6 3 , a 3 - a 2 b - ah 2 - 2b 3 . Ans. (a + 6) (a — 6)(a — 2?>)(« 2 + aft + & 2 ). 79. 21a?(ajy - y 2 ) 2 , 35(x*y 2 - a; 2 ?/ 4 ), 15?/(a; 2 + xy) 2 . Ans. 105x 2 y 2 (x + y) 2 (x — y) 2 . 128 A FRACTION — ENTIRE AND MIXED QUANTITIES. CHAPTER VIII. FRACTIONS. Note. — In this chapter the student will find that the definitions, rules, and demonstrations closely resemble those with which he is already familiar in Arithmetic. 78. A Fraction — Entire and Mixed Quantities. — A Fraction is an expression of an indicated quotient by writing the divisor under the dividend with a horizontal line between them (Art. 11). In the operation of division the divisor sometimes may be greater than the dividend, or may not be contained in it an exact number of times ; in either case the quotient is expressed by means of a fraction. Thus, the expression - indicates cither that some unit is b divided into b equal parts, and that a of these are taken, or that a times the same unit is divided into b equal parts, and one of them taken. In any fraction the upper number, or the dividend, is called the numerator, and the lower number, or divisor, a 6' which is read a divided by b, a is called the numerator and b the denominator, and the two taken together are called the terms of the fraction. Thus the denominator indicates into how many equal parts the unit is to be divided, and the numerator indicates how many of those parts are to be taken. Every integer or integral expression may be considered as a fraction whose denominator is unity ; thus, a , , a -\- b a = -, a + b = — ^— . TO REDUCE A FRACTION TO ITS LOWEST TERMS. 129 An entire quantity or integral quantity is one which has no fractional part ; as ah or or — 2ab. A mixed quantity is one made up of an integer and a fraction ; as b + -. a A proper fraction is one whose numerator is less than its denominator ; as . a -f x An improper fraction is one whose numerator is equal to or greater than the denominator ; as - and -. a a The reciprocal of a fraction is another fraction having its numerator and denominator respectively equal to the denom- inator and numerator of the former. 79. To Reduce a Fraction to its Lowest Terms. — Let- denote any fraction, and — denote the same frac- b mb tion with its terms multiplied by m. Now - means that a unit is divided into b equal parts, and b that a of these are taken (1) And — means that the same unit is divided into mb equal mb parts, and that ma of these are taken (2) Hence b parts in (1) = mb parts in (2). 1 part in (1) = m parts in (2), and .'• a parts in (1) = am parts in (2), that is, Conversely, Therefore, the value of a fraction is not altered if the numerator and denominator are either both multiplied or both divided by the same quantity. a b = ma mb ma, mb = a b 130 EXAMPLES. When both numerator and denominator are divided by all the factors common to them, the fraction is said to be reduced to its lowest terms. Hence to reduce a fraction to its lowest terms we have the following Rule. Divide both numerator and denominator by their greatest common divisor. Dividing both terms of a fraction by a common factor is called canceling that factor. EXAMPLES. Reduce the following fractions to their lowest terms. 1 Qa 2 bc 2 = 2ac 9ab 2 c 36' The greatest common divisor Sabc of both terms is can- celed. 2> lx 2 yz _ J_ 28x 3 yz 2 Axz The factor 7x 2 yz, which is the greatest common divisor of both terms, is canceled. „ 24a s c 2 x 2 24a s c 2 x 2 Aac 2 18a s x 2 - 12a 2 * 3 ~ Ga 2 x 2 (3a - 2a) ~ 3a - 2x Here 6a 2 x 2 is canceled since it is the greatest common divisor of both terms. Note. — .In each of these examples, the resulting fractions have the same value as the given fractions, but they are expressed in a simpler form. The student should be careful not to begin canceling until he has expressed both terms of the fraction in the most con- venient form, by factoring when necessary. Thus, 4 6x 2 — 8xy = 2x(Sx — Ay) _ 2x [)xy - \2y 2 ~ 3?/ (3a - Ay) ~ Sy' Instead of reducing a fraction to its lowest terms by dividing the numerator and denominator by their G. C. D., EXAMPLES. 131 we may divide by any common factor, and repeat the process till the fraction is reduced to its lowest terms. Thus, 24a 3 & 2 c 3 _ 12a 2 6c 2 _ Gac __ 2a 36aW ~ 18a6 2 c 2 96c 3b Reduce the following fractions to their lowest terms. G 8 ^ c * Ans.^. 12ab 2 cd 3bd „ So? - Gab 3 9. 10. 2,r0 - Aab 2 2b Ax 2 - 9y 2 2x - 3y ^ 4x 2 -(- 6xy' 2x 20 (x 3 - f) ^ A(x-y). 5a; 2 + bxy + by 2 x 3 - 2a;?/ 2 x x * _ 4ajy _|- 4y*' x 2 — 2y 2 ' When the factors of the numerator and denominator cannot be found by inspection, their greatest common divisor may be found by the rule (Art. 74), and the fraction then reduced to its lowest terms. ' „ „ ., , w 3a; 3 - 13a: 2 + 23a; - 21 11. Reduce to its lowest terms — — — — — — — . 15a; 3 — 38a; 2 — 2x + 21 The G. C. D. of the numerator and denominator is 3x — 7. Dividing the numerator and denominator by 3x — 7, we obtain the respective quotients x 2 — 2x + 3 and 5ar — x — 3. Therefore 3a; 3 -13a; 2 +23a;-21 _ (3a; - 7) {x 2 - 2x + 3) = a; 2 - 2x + 3 15a; 3 - 38a; 2 - 2a; + 21 (3* -7) (5a; 2 - x- 3) 5a; 2 - x -3 This example may also be solved without finding the G. C. D. by the rule (Art. 74) as follows : By 2 of Art. 74, the G. C. D. of the numerator and denominator must divide their sum 18a; 3 — 51a; 2 -f- 21a;, that is, 3cc(3a5 — 7)(2» — 1). If there be a common divisor it must clearly be 3x — 7. Hence, arranging the numerator 132 TO REDUCE A MIXED QUANTITY TO A FRACTION. and denominator so as to show 3s — 7 as a factor, we have the fraction = x*(3x - 7) - 2s(3s - 7) + 3 (3s - 7) 5s 2 (3s - 7) - s(3s - 7) - 3 (3s - 7) _ (3s - 7) (a 8 - 2s + 3) = s' 2 - 2x + 3 (3s — 7) (5s 2 — s — 3) 5s 2 — s — 3 When either the numerator or denominator can readily be factored we may use the following method : 12. Reduce to its lowest terms — - 1 — — — : — — — ■ . 7s 3 — 18s 2 + 6s + 5 The numerator = x(x 2 + 3s — 4) = x(x + 4) (s — 1). The only one of these factors which can be a common divisor is s — 1, since the denominator does not contain s, and 5 the last term in the denominator does not contain 4. (See Art. GO.) Hence, arranging the denominator so as to show s — 1 as a factor, the fraction = *(«+*; (*-l) = *(*+*> . 7s'-(s— 1)— lls(s— 1)— b(x— 1) 7s'-— lis— 5 Reduce to lowest terms a s - a 2 b - ab 2 - 2b s a - 2b 13. 14. 15. + 3a 2 6 + 3a& 2 + 26 8 a + 25 - 5s 2 + 7s - 3 s - 3 x 3 — 3s + 2 x + 2 4o 8 + 12« 2 & - ab 2 - 15& 8 2a + 56 6a 8 + 13a 2 6 - 4a& 2 - 156 3 3« + bb 80. To Reduce a Mixed Quantity to the Form of a Fraction. — Let it be required to reduce a + b + c to the form of a fraction. The entire part a = - (Art. 78) = a ^ b + ^ (Art. 79). tt i &c a(b + r) . be Hence a -\ = v - — ' H 6 + c b + c 6 + o _ ab + ac, -j- be b + c ' TO REDUCE A FRACTION TO A MIXED QUANTITY. 133 Hence we have the following Rule. Multiply the entire part by the denominator, and to the product add the numerator with its proper sign; under this sum place the denominator, and the result ivill be the fraction required. EXAMPLES. Reduce the following to fractional forms : x 2 1. a — x + 2. a? + 1 + 3. ar — xy + y i 4. a + b a + x 4 x - 3' x + y a- + ft 2 Ans a 2 a -\- x x 2 - 2x + 1 X - 3 X s x + y 26 2 81. To Reduce a Fraction to an Entire or Mixed Quantity. — Let it be required to reduce — to a ^ ■ ■* a + x mixed quantity. Performing the division indicated, we have ax + 2x 2 _ x + x- a + X a + x Hence we have the following Rule. Divide the numerator by the denominator, as far as possible, for the entire part, and annex to the quotient a fraction having the remainder for numerator, and the divisor for denominator ; it will be the mixed quantity required. 134 TO REDUCE FRACTIONS TO THEIR L. C. D. EXAMPLES. Reduce to whole or mixed quantities the following : 1. — 24a 7 a* + 3a6 a + b 36ac + 4c 9 8a 2 + 3& 4a x 2 + 3x + 2 SB ■+ 3 2x 2 - Gx - - 1 &• — 3 x 4 + 1 J.WS. 3a + y. a 2a6 a + 6' 4ac + f. 2a + » 4a , 2 ' a? + 3" 2x 1 x — 3 4- 1 + - 2 a? - 1 a; - 1 82. To Reduce Fractions to their Least Common Denominator. — Let it be required to reduce — , — , — to yz zx xy equivalent fractions having the least common denominator. The least common multiple of the denominators is xyz. Dividing this L. C. M. by the denominators, yz, zx, and xy, we have the quotients x, y, and z, respectively. By Art. 79, both terms of a fraction may be multiplied by the same number without altering its value; therefore we may multiply both terms of the first fraction by x, both terms of the second fraction by ?/, and both terms of the third fraction by z, and the resulting fractious will be equivalent to the given ones. a_ _ cm b_ _ by_ c _ cz yz xyz zx xyz xy xyz Hence — That is, the resulting fractions (IX xyz' and — have the xyz xyz xyz same values respectively as the given fractious — , — , and — , yz zx xy and they have the least common denominator xyz. Hence, RULE OF SIGNS IN FRACTIONS. 135 for reducing fractions to their least common denominator, we have the following Rule. Find the least common multiple of the given denominators, and take it for the common denominator; divide it by the denominator of the first fraction, and multiply the numerator of this fraction by the quotient so obtained; and do the same toith all the other given fractions. Note 1. — It is not absolutely necessary to take the least common denominator. Any common denominator may be used. But in practice it will be found advisable to use the least common denomina- tor, as the work will thereby be shortened. Note 2. — It frequently happens that the denominators of the fractions to be reduced do not contain a common factor. Thus, tbe denominators of the fractions -, -, and - have no common factor: b d' f therefore the least common denominator of tbese fractions is bdf, the product of all their denominators. EXAMPLES. Reduce the following fractions to their L. C. D. 1 JL _± _JL_ Ans — -^- — " Ax 6b 2 ' 12a; 3 ' 12a; 3 ' 12a; 3 ' 12a; 8 ' 2 a x a 2 a(x + a) x(x + a) a 2 x — a x — a as 2 — a 2 ' a; 2 — a 2 ' a; 2 — a 2 ' a; 2 — a 2 83. Rule of Signs in Fractions. — The signs of the several terms of the numerator and denominator of a fraction relate only to those terms to which they are prefixed, while the sign prefixed to the dividing line relates to the fraction as a whole, and is the sign of the fraction. Thus, in the fraction — ^-^ — , the sign of a, the first term of the numer- a + b ator, is + understood, the sign of the second term b is — , and the sign of each term a and b of the denominator is +, while the sign of the fraction itself is — . 136 RULE OF SIGNS IN TRACTIONS. The symbol means the quotient resulting from the — b division of —a by — b; and this is obtained by dividing a by b, and prefixing + , by the rule of signs in division (Art. 46). Therefore — = +- = - (1) Also, ^— is the quotient of —a divided by b ; and this is obtained by dividing a by b, and prefixing — , by the rule of signs. Therefore — = -- (2) b b K ' In like manner, is the quotient of a divided by — b; and this is obtained by dividing a by b, and prefixing — , by the rule of signs. Therefore — = -^ (3) Hence, we have the following rule of signs : (1) If the signs of both numerator and denominator be changed, the sign of the whole fraction remains unchanged. (2) If the sign of the numerator alone be changed, the sign of the whole fraction will be changed. (3) If the sign of the denominator alone be changed, the sign of the whole fraction will be changed. . Or they may be stated as follows : (1) We may change the sign of every term in the numera- tor and denominator of a fraction without altering the value of the fraction. (2) We may change the sign of a fraction by changing the sign of every term either in the numerator or denominator. EXAMPLES. 137 EXAMPLES. a — b — a-\-b b — a m — n —m + n n — m b — a —b + a a — b —m + n -b + a 2x 3x 2x 2x 2x Sx 3x Sx x — x 2 —x + x 2 x 2 — x The intermediate steps may usually be omitted. From Art. 36 we have abmx _ (~a)(-b)(-m)x _ a(-b)(-m)x _ etc _ pqr (-p)qr (~I>)(-ff) r That is, if the terms of a fraction are composed of any number of factors, any even number of factors may have their signs changed without altering the value of the fraction; but if any odd number of factors have their signs changed, the sign of the fraction is changed. Thus (a - b )( b ~ °> _ ^ - q)(6 - c) = (6 - a)(c - b) ' (x-y)(y-z) (y-x)(y-z) (y-x)(z-y) When the numerator is a product, any one or more of its factors can be removed from the numerator and made the multiplier. Thus, -^ = ab-^-r = abed- * a + b a -f b a + b Change the signs of the following fractions so as to express them altogether in four different ways. 4. a — b x — y . b — a b — a a — b Ans. , , y — x x — y y — x X — X —X X y — z z - y y - z * - y b — a a — b a — b b — a a — b + c b — c — cl a — b + c b — c — a abed a(- -b)(-c)d, a(-b)cd abed xyz xyz ' xy(-z) ' x(-y)(-z) 138 ADDITION AND SUBTRACTION OF FRACTIONS. 84. Addition and Subtraction of Fractions. — Let it be required to add together - and -. c c Here the unit is divided into c equal parts, and we first take a of these parts, and then b of them; i.e., we take a + b of the c parts of the unit ; and this is expressed by the fraction — — — . c a j_b _ a + b c e c Similarly a _ b _ a — b c c c Let it be required to add together - and -. b d w , a ad , c be We have - = — , and = — . b bd d bd Here in each case we divide the unit into bd equal parts, and we first take ad of these parts, and then be of them ; i.e., we take ad + be of the bd parts of the unit; and this is expressed by the fraction — -. bd a , c _ ad + be Similarly b d bd a c ad — be b d bd Here the fractions have been reduced to a common denom- inator bd. But if b and d have a common factor, the product bd is not the least common denominator, and the fraction will not be in its lowest terms. To avoid working bd with fractions which are not in their lowest terms, it will be found advisable to take the least common denominator, which is the least common multiple of the denominators of the given fractions (Art. 82, Note 1). EXAMPLES. 139 Hence we have the following Rule. To add or subtract fractions, reduce them to the least common denominator ; add or subtract the numerators, and icrite the result over the least common denominator. EXAMPLES. 6 6 b Here the fractions have already a common denominator,* and therefore need no reducing. Hence we have a + c a — c a + d _ a + c + a — c + a + d _ 3a + d 6 6 6 b b 2. Add — — and — . 3a 9a Here the least common denominator is 9a. Hence we have (Art. 82) 2x + 3 a ^ ( + bx - 9a 4a _ 3 (2x 9 + «) a , 5x 9a 4a, Cu- + 3a + ").)• — 1, l 11.x- — a 9a 9a 3. From 4a - c 26 , . 3a take — - 36 c 4a - 2b 3a — 3 b 4a - 26 - - (3a c 36) c c 4a - 26 - - 3a + 36 a + 6 Note 1. — To insure accuracy, the beginner is recommended to put down the work in full; and when a fraction whose numerator is not a monomial is preceded by a — sign, he is recommended to enclose its numerator in a parenthesis as above before combining it with the other numerators. 140 4. Add Here EXAMPLES. a a 6 b — a b b - a a b -(Art. 83). a b b — a a b a a — b a — b a — a — b b = 1. X — 2y 3y — a and — Sx xy ay ax 5. Add The least common denominator is axy. x — 2y 3y — a 2a — 3.x- xy ay "~ ax _ a(x - 2ij) + a(3?/ - a) + ?/(2a - 3a>) a.*:y a# — 2ay + Sxy — ooj + 2a?/ — 3xy axy since the terms in the numerator destroy each other. n T . a + b . , a — b 6. Irom — ! — take . a — b a -\- b The least common denominator is a 2 — Zr. a + b _ a — b _ (a + ?>)' 2 _ (a — 6) 2 0, a — b a + b a' 2 — If a- — lf a 2 + 2ab + b 2 — (a 2 - - 2aft + & 2 ) a 2 - 6 2 4a6 a 2 - & 2 ' _ . , , 2as — 3a , 2x — a 7. Add - and . x — 2a x — a The least common denominator is (x — 2d) (x — a). EXAMPLES. 141 Hence we have 2x — 3a _ 2x - a _ (2x — 3a) (x — a) — (2x — a) (x - 2a) x — 2a x — a (x — 2a) {x — a) _ 2a; 2 — 5ax + 3ft 2 — (2a; 2 — oax -+- 2a 2 ) (x — 2a) (a; — a) 2X 2 - box + 3a 2 - 2a; 2 + 5ax - 2a 2 (x — 2a) (x — a) (x — 2a) (x — a) Note 2. — In finding the value of an expression like — (2x — a)[x — a), the beginner should first express the product in parentheses, and then after multiplication, remove the parentheses, as we have done. After a little practice he will be able to take both steps together. Note 3. — In practice, the foregoing general method may some- times be modified with advantage. When the sum of several fractions is to be found, it is often best, instead of reducing at once all the fractions to their L. C. D., to take two or more of them together, and combine the results. o c ,.-. ft -(- 3 « + 4 8 8. Simplify — — — . ft — 4 . ft — 3 ft' 2 — 16 Here, instead of reducing all the fractions to the least common denominator at once, we may take the first two fractions together, as follows : q + 3 _ 0+4 _ 8 _ ft 2 - 9 — (ft 2 — 16) _ a — 4 a- a 2 - 16 (ft - 4) (ft - 3) « 2 - 16 7 8 (ft -4) (« -3) (« + 4) (ft - -4) 52 — ft (a - 4)(« + 4)(ft- 3) 1 1 2x Ax* 9. Simplify ft — X ft + X ft" + x- ft 4 + X* Here we see that the first two denominators give L. C. D. ft 2 — a; 2 ; and this with « 2 -f- x- gives L. C. D. « 4 — x 4 ; and this with ft 4 + a; 4 gives L. C. D. a s — x s . Hence, instead 142 EXAMPLES. of reducing all the fractious to the least common denom- inator, we proceed as follows : The first two terms The first three terms The whole expression = a + x — (a — x) _ 2x a 2 - X' 2 a s - X 2 2., 2x d' ■ — X 2 a 2 -+- X 2 •2. :((/' + * 2 )- - 2.s(a 2 — X 2 ) a 4 - a 4 4x 3 a 4 - x i 4x 3 a 4 - a; 4 8a; 7 4.x- 3 t 4 + x* 10. Simplify (a-b)(a-c) (b-c)(b-a) (c-a)(c-6) The beginner is very liable, in this example, to take the product of the denominators for the least common denom- inator, and thus to render the operations very laborious. The denominator of the second fraction contains the factor b — a, aud this factor differs from. the factor a — b, which occurs in the denominator of the first fraction, only in the sign of each term. Also the denominator of the third fraction contains the factors c — a and c — b, which differ from the factors a — c and b — c iu the denominators of the first two fractions, only in the signs of each term. It is better to arrange these factors so that a precedes b or c, and that b precedes c. By Art. 83 we have b b and (b - c) (b - a) (b - c) (a - b) c c (c — a) (c — b) (a — c) (b — c) Hence the given expression may be put in the form a be (a - &)(a - c) (6 - c)(a - &) (a - c)(& - c) whose L. C. D. we see at once is (u — b) ( 46 9a 46 X 9a 36 146 EXAMPLES. 2. Multiply 2a2 by < a + b> >\ 2a 2 (a + &) 2 _ 2a 2 (a + 6) (a -f b) a?-b 2 4a 2 b (a + 6) (a - 6)4a 2 6 a + 6 26(a - 6)' by canceling those factors which are common to both numer- ator and denominator. q tit u- i 2a 2 4- 3a -. 4a 2 — 6a . ,, 3. Multiply 2- — and together. J 4a 3 12a + 18 fe 2a 2 + 3a 4a 2 - 6a _ a (2a + 3)2a(2a - 3) 4a 8 12a + 18 4a 3 (2a + 3)6 _ 2a - 3 12a 4. Multiply f + ^+ lby| + ^- 1. b a b a « + I + 1 = ft2 + *' + « & (Art. 84), 6 a a& and « + * - 1 = q ' + h2 ~ ab . b a ab a& a& a 2 & 2 J a 4 + fr 4 + a 2 6 2 a 8 & 8 Otherwise thus : ,2 G+s+'XH-O-G+y = «^ , 6_ 2 , j = a 4 + 6 4 + a g ft 8 & 2 + a 2 + a 2 b 2 Simplify 5. ^X^X .*-*, 4» * aj _ l x 1 - 1 (a; + 2) 2 (as - 1) (a + 2) 6. X j. x — a. x + a \a xj TO DIVIDE FRACTIONS — EXAMPLES. 147 Ans.-f U ' \ a - b)\ a + b) b 1 x) ax a 2 + 2ax + x' 1 " a 2 — 2ax v(a — x) a (a + a; ) ax — x — 88. To Divide Fractions. — Let it be required to divide - by « b J d Denote the quotient b} r x. Then, since the quotient mul- tiplied by the divisor gives the dividend (Art. 44), we have d b Multiplying by -, we have xx S x d = a x d d c b c Therefore, Art. 87, and canceling factors common to the numerator and denominator, we have _ ad X ~ ~bc That is, » + S. = 2* = x* — KS'. c(a — c) (c — 6) EXAMPLES. 155 40. £ r + £-_ 41. (a-6)(a-c) (6_a)(6_ c ) (c-a)(c-6) -4ns. 1. 1 + I + 1 . (a-b)(a-c)(x-a)^(b-a)(b-c)(x-b) ( C -a)(c-b)(x-c) 1 Ans. (x — a)(x — b)(x — c) Simplify the following examples in multiplication and division : 42> 2a- 2 4- 5a- + 2 a; 2 4- 4a; , ^ 2as a + 9a; + 4 x — 2 43 2a;' 2 - x - 1 4a; 2 + ar — 14 * — 1 2a; 2 + 5a; + 2 16a; 2 - 49 4a; + 7 44 ^ a; 2 +a;-2 ^ ^ + 5a; + 4 . / a; 2 + 3a- + 2 x x + 3\ ^ a; 2 — a;— 20 a; 2 — a; \a; 2 — 2a— 15 a- 2 / 45 a; 4 - 8a- a; 2 + 2a- + 1 ^ a; 2 + 2a; + 4 j x i — 4a- — 5 a; 3 — ar — 2x x — 5 46> (q + ?>) 2 -c 2 x a x (q-fe) 2 -^ 1 a 2 + ab — ac (a-j-c) 2 — b 2 ab — b' 2 — bc b 3x x - 1 47 I ! 1. ¥(*+i) "|- 2 * x - 1 + 6 x — 6 k — 4 4o. . . x - 2 + -1- K ~ 5 « - 6 49 1 ~ a * _,_ i/ j . _L_\ i 1 + aa-) 2 - (a + a?) 2 ' 2 \1 - » 1 + a;/' ( a- 2 + i/ 2 a; , n # a; 2 - w 2 50 ' i_i x ^T7«- as 156 EXAMPLES. 51. 1 7. Ans. i . + j £B 52. 1 + ^ 1 + x 1+X + -**-' 1 + + I — X 53. 1 x + 1. x 1_ i+i a; 54. 1 + 1 + a; 2 1 + x , 2x 1 + x + 1 - a; V« + 2/ a;-?/ x'-y-J \x + y x'-y) *+i 56. - ! 1 y&yz + x + z) X 57. ' " ^ v ' x \2/ •'•/\.'/" - -''7 a 2 + an/ ay/ - f x + v/ 58. x — 1 x - 1 x + 3 _ x + 3 3 x — 2 ^ 7 x + 4 28(s + 4) s + 2 + x + 2 ' a; — 2 + a: - 2 Q(x + 3) 4 a; — 3 3 x — 1 SOLUTION OF HARDER EQUATIONS — EXAMPLES. 157 CHAPTER IX. HARDER SIMPLE EQUATIONS OF ONE UNKNOWN QUANTITY. 91. Solution of Harder Equations. — We shall now give some simple equations, involving Algebraic fractions, which are a little more difficult than those in Chapter VI. These may be solved, by help of the preceding chapter on fractions, and by the same methods as the easier equations given in Chapter VI. The following examples worked in full will sufficiently illustrate the most useful methods. EXAMPLES. 1. Solve 6 ^^ = ^- 2 . 2x + 7 x + 5 The L. C. M. of the denominators is (2x + 7)(x -+■ 5). Clearing the equation of fractions by multiplying each term by (2x -f 7) (x + 5), we have* (6x - 3)0 + 5) = (3a - 2) (2a; + 7), or 6a; 2 -f- 27a; — 15 = 6.c 2 + 17a; - 14 ; .-. 10a; = 1 ; .'. x = T V We may verify this result by putting -^ for x in the original equation, as in Chapter VI. ; it will be found that each member then becomes — 1> Note 1. — When the denominators of the fractions involved con- tain both simple and compound factors, it is frequently best to multiply the equation by the simple factors first, and then to collect tbe integral terms; after tins tbe simplification is readily completed by "multi- plying across " by tbe compound factors. * This is called " multiplying across." 158 EXAMPLES. 2 Solve 8x + 23 _ bx + 2 = 2x + 3 _ i 20 3a; + 4 5 Multiplying by 20, the L. C. M. of the simple factors in the denominators, we have Sx -f 23 - 20(5a; + 2) = 8* + 12 - 20. 3x + 4 Transposing, 31 = 20{5x + 2) . 3a; + 4 Multiplying across by 3a; + 4, we have 93a; + 124 = 20(5a; + 2), or 84 = 7x; .-. x = 12. We may verify this result as before ; it will be found that each side becomes - 2 ^. Note 2. — The student will see that, even when the denominators of the fractions contain all simple factors, it is sometimes advanta- geous to clear of fractions partially, and then to effect some reduc- tions, before removing the remaining fractions. 3. Solve X -±^ _ 2x-lS 2 JL±1= fa + 4 11 3 4 3 12 Multiplying by 12, the L. C. M. of 3, 4, 12, 12 + 6 ) _ 4(2a; - 18) 4- B(2x + 3) = 16 x 4 + 3a; 4- 4 or 12(s + 6) _ 8x + 72 + Gx + 9 = 64 + Bx + 4> Transposing and reducing, we have 12 (* + 6 > = 5x - 13. 11 Multiplying by 11, we have 12(a; + 6) = 11 (5a; - 13), or 12a; + 72 = 55a: - 143 ; .-. 43a; = 215; .-. x = 5. "We may verify this result as before. Note 3. — When two or more fractions have the same denominator, they should be taken together and simplified. EXAMPLES. 159 4. Solve 13 ~ 2a; + 23a; + 8 i = 16 -^- x + 3 4a + 5 x + 3 Transposing, we have 23a; + 8^ _ 4 = 16 - *a; - 13 + 2a?. 4a; + 5 x + 3 then 7a? ~ ¥ = 3+fr 4a; + 5 a; + 3 Multiplying across, we have (x + 3) (7a; or 7a; 2 - -% 5 -x + 21a; - 35 = 12a; + 7a; 2 + 15 + -\ 5 -x ; ... -l&x = 50; .-. a; = -ffs. 5. Solve g - 8 + ^=^ = ^^ 4- ^^. 35—10 a? — 6 a; — 7 x — 9 Note 4. — This equation might be solved by clearing of fractions, by multiplying by the four denominators, but the work would be very laborious. The solution will be much simplified by transposing two of the fractions as follows : Transposing, we have x — 8 _ x — 5 _ x — 7 x — 4 x — 10 x — 7 ~~ a; — 9 a? — 6* Simplifying each side separately, we have (a?-8) (a;-7)-(a;-5) (a;-10) = (x-7) (x-(>)-(x-4) (x-9) (ar-10) (a;-7) (x-9) (aj-6) or a?- 15a; + 56 - (x 2 - 15a; + 50) = a: 2 -13a;+42-(a; 2 -13a;+36) (a;-10)(a.--7) (a;-9) (a;-6) 5 = 5 . (a?-10) (a;-7) (a?-9)(a;-6) Dividing by 6 and clearing of fractions, we have (a; - 9) (a; - 6) = (x - 10) (a; - 7), or a; 2 - 15a; 4- 54 = a; 2 - 17a; + 70 ; .-. x = 8. 160 EXAMPLES. Note 5. — This example may also be solved very neatly by writing the equation at first in the form x — 10 -f 2 x — 6 + 2 _ x — 7 + 2 x — + 2 x — 10 x — x — 7 x — U ' Reducing each fraction to a mixed number (Art. 81), we have 1 • 1 • 1,1 1,1 which gives — H = -\ 3 x — 10 x — 6 a; — 7 x — 9 Transposing, = — — -. 1 a x — 10 x — 7 x — 9 x — 6 3 3 (x - 10) (x - 7) (x - 9)(x - 6)' and the solution may be completed as before. „ c , 5a; — 64 2a; — 11 4a; - 55 a? — 6 6. Solve = . x — 13 x — 6 x — 14 x — 7 Proceeding as in the second method of Ex. 5, we have x - 13 \ x-6j x - 14 \ x-l) 1111 x — 13 x — 6 a; — 14 a: — Simplifying each side separately, we have 7 7 (a; - 13) (a; - 6) (a; - 14) (a: - 7) Clearing of fractions, or, since the numerators are equal, the denominators must be equal, we have (a; - 13) (a; - 6) = (x - 14) (x - 7); .-. x 2 - 19x + 78 = x 2 - 21a; + 98 ; .-. x = 10. Solve the following equations : Ans. 10. 6. 7. 8. 9. 12. 1 x 12a; ^ 4 * 45 57 2a; + 3 4a; — 5 3a; - 1 2a; - 5 x^-J _ x = x _ ? 4 6 HARDER PROBLEMS. 161 10. 6 -^-±- 8 - 2x + 38 = 1. Ans. 2. 2x +1 « + 12 11. i(2»-10)- T 1 r (3a;-40) = 15-i(57-.i-). (See Note 2). ^4hs. 17. 12. x— 1 a— 5 15 — 2a;_9 — a; 4 32 40 " 2 |5? aj + 4 a; + 5 (See Note 2). 5. 14. 3a; - 8 6x + 13 15 3x — 1 2x - 1 6a; + 7 3a; + 5 2a; 5a; - 25 ~~ 5 ' 17. 9x + 6 x - 12a; + 8 4 2 5 a;+3 a;+l 2a;+6 2a;+2 (See Note 1). 20. *• -6f. ( . (See Note 3). 1. 18. ^^2 _|_ ® £ = 5 * + 5 ^ (See Ex. 5) . 4. a; - 2 a; - 6 a; - 5 a; - 3 v ' , n a; — 1 a; — 2 a; — 5 a; — 6 .. iy. = . 4*. a; — 2 a; — 3 a; -6 a;— 7 J 20 6a; + 1 _ 2 -^ - 4 _ 2a; - 1 _ 2 15 7a; - 16 5 92. Harder Problems Leading to Simple Equations with One Unknown Quantity. — We shall now give some examples which lead to simple equations, but which differ from those of Art. 61 in being rather more difficult. The statement of the problem is rather more difficult than in the examples of that Article, and the equations often involve more complicated expressions. 162 EXAMPLES. EXAMPLES. 1. A alone can do a piece of work in 9 days, and B alone can do it in 12 days: in what time will they do it if they work together? Let x = the number of days required for both to do the work ; then - = the part that both can do in one day. Also i = the part that A can do in one day, and i 1 ^ = the part that B can do in one day.. Since the sum of the parts that A and B separately can do in one day is equal to the part that both together can do in one day, we have * + «4 Clearing of fractions by multiplying by 36a;, we have Ax + 3x = 36; .-. x = 5£, which is the number of days required. 2. A workman was employed for 60 days, on condition that for every day he worked he should receive $3, and for every day he was absent he should forfeit $1 ; at the end of the time he had $48 to receive : required the number of days he worked. Let x = the number of days he worked ; then 60 — x = the number of days he was absent. Also 3x = the number of dollars he received, and 60 — x = the number of dollars he forfeited. Hence, from the conditions of the problem, we have 3x - (60 - x) = 48. .-. Ax = 108. .-. x = 27. That is, he worked 27 days and was absent 33 days. 3. A starts from a certain place, and travels at the rate of 7 miles in 5 hours ; B starts from the same place 8 hours after A, and travels in the same direction at the rate of 5 EXAMPLES. 163 miles in 3 hours ; how far will A travel before he is over- taken by B? Let x = the number of hours A travels before he is overtaken ; then x — 8 = the number of hours B travels before he overtakes A. Also -| = the part of a mile which A travels in one hour, and -§ = the part of a mile which B travels in one hour, Therefore %x = the number of miles which A travels in x hours, and ^{x — 8) = the number of miles which B travels in x — 8 hours. Since, when B overtakes A, they have traveled the same number of miles, we have for the equation f* = f (x - 8). .-. 21oj = 25a; - 200. .-. x = 50. Therefore f# = f x 50 = 70 miles, the distance which A travels before he is overtaken by B. 4. A cistern could be filled with water by means of one pipe alone in 6 hours, and by means of another pipe alone in '8 hours ; and it could be emptied by a tap in 12 hours if the two pipes were closed : in what time will the cistern be filled if the pipes and the tap are all open? Let x = the required number of hours. Then £ = the part of the cistern the first pipe fills in one hour ; therefore - = the part of the cistern the first pipe fills in x hours. And | = the part of the cistern the second pipe fills in one hour ; therefore - = the part of the cistern the second pipe fills in x hours. 164 EXAMPLES. Also hour ; therefore — = the part of the cistern the tap empties in x hours. Since in x hours the whole cistern is filled, we have, repre- senting the whole by unity, X x x_ _ 1 6 8 12 ~ ' Multiplying by 24, we have 4* + Sx - 2x = 24. .-. x — 44- . 5. A smuggler had a quantity of brandy which he ex- pected would bring him $198 ; after he had sold 10 gallons a revenue officer seized one-third of the remainder, in con- sequence of which the smuggler gets only $162 : required the number of gallons he had at first, and the price per gallon. Let x = the number of gallons ; 198 then ■ — = price per gallon in dollars. x - 10 o = the number of gallons seized ; 10 ., 198 X — = the value of the quantity seized . in 3 it- dollars. Hence we have the equation «L=J° x 128 _ m _ 162 = 86 _ 3 x Clearing of fractions 66(3 — 10) = 36a?. .-. 30aj = 660. x = 22, the number of gallons ; 198 198 ftQ .. and = — - = $9, the price per gallon. x 22 EXAMPLES. 165 6. A colonel, on attempting to draw up his regiment in the form of a solid square, finds that he has 31 men over, and that he would require 24 men more in his regiment in order to increase the side of the square by one man : how many men were there in the regiment? Let x = the number of men in the side of the first square ; then or -f 31 = the number of men in the regiment. Also (x + l) 2 — 24 = the number of men in the regiment. Hence, we have the equation x 2 + 31 = (x + l) 2 - 24, or x 2 + 31 = x 2 + 2x - 23. .-. x = 27. Hence (27) 2 + 31 = 760 is the number of men in the regiment. Note 1. — In this example it was convenient to let x represent the number of men in the side of the first square instead of the number of men in the whole regiment. 7. At the same time that the up-train going at the rate of 33 miles an hour passes A, the down-train going at the rate of 21 miles an hour passes B ; they collide 18 miles beyond the midway station from A : how far is A* from B? Let x = the distance from A to B in miles ; then ® = half the distance. Also - + 18 = the number of miles the up-train goes, and 2 A t _ distance in miles ,, ,. . , JNow : = the time in hours. rate in miles per hour ^ + 18 Therefore = the time the up-train takes, DO - - 18 2 and — — — = the time the down-train takes. 21 166 EXAMPLES. Hence, since these times are equal, we have the equation | + 18 ~ - 18 33 21 Solving, we get x = 102, which is the distance from A to B in miles. 8. A cask, A, contains 12 gallons of wine and 18 gallons of water ; and another cask, B, contains 9 gallons of wine and 3 gallons of water : how many gallons must be drawn from each cask so as to produce by their mixture 7 gallons of wine and 7 gallons of water ? Let x = the number of gallons to be drawn from A ; then 14 — x = the number of gallons to be drawn from B, since the mixture is to contain 14 gallons. Now A -contains 30 gallons, of which 12 are wine ; that is, ^| of A is wine. Also B contains 12 gallons, of which 9 are wine ; that is, T 9 ^ of B is wine. Hence 3$x = the number of gallons of wine in the x gallons drawn from A ; and x%(14 — x) = the number of gallons of wine iu the 14 — x gallons drawn from B. Since the mixture is to contain seven gallons of wine, we have i2 >T + T 9_ (14 - X ) = 7; that is, f« -f f (14 - x) = 7. Solving, we get as = 10, the number of gallons to be drawn from A, and 14 — x = 4, the number of gallons to be drawn from B. 9. At what time between 4 and 5 o'clock is the minute- hand of a watch 13 minutes in advance of the hour-hand? Let x = the required number of minutes after 4 o'clock ; that is, the minute-hand will move over x minute divisions of the watch face in x minutes ; and as it moves 1 2 times as x \2 linute divisions in x minutes. At 4 o'clock the minute-hand EXAMPLES. 167 is 20 minute divisions behind the hour-hand, and finally the minute-hand is 13 minute divisions in advance ; therefore, in the x minutes, the minute-hand moves 20 -f- 13, or 33, divisions more than the hour-hand. Hence a,* = - 1 - + 33 ; therefore lis =12 x 33. .-. x = 36, or the time is 36 minutes past 4. If the question be asked, " At what times between 4 and o o'clock will there be 13 minutes between the two hands? " we must also take into consideration the case when the minute-hand is 13 divisions behind the hour-hand. In this case the minute-hand gains 20 — 13, or 7 divisions. Hence x = — + 7. .-. 11a; = 84. .-. x = 7 T V Therefore the times are 7^ minutes past 4, and 36 minutes past 4. Note 2. -«- The student is supposed to have obtained from Arith- metic some knowledge of ratio and proportion. When two or more unknown quantities, in any example, have to each other a given ratio, it is best to assume each of them a multiple of some other unknown quantity, so that they shall have to each other the given ratio. Thus, if two unknown numbers are to each other as 2 to 3, it is best to express the numbers by 2x and 3a 1 , since these two numbers are to each other as 2 to 3. This will be illustrated in the next two examples. 10. A number consists of two digits of which the left digit is to the right digit as 2 to 3 ; if 18 be added to the number the digits are reversed : what is the number? The student must remember that any number consisting of two places of figures is equal to ten times the figure in the ten's place plus the figure in the unit's place ; thus, 46 is equal to 10x4 + 6; likewise 358 is equal to 100 X 3 + 10 X 5 + 8. Let 2x = the left digit ; then 3x = the right digit, and 10 x 2x + 2>x = number. 168 EXAMPLES. Hcuce we have the equation (10 x 2x + 3a?) + 18 = (10 x 3x + 2x), or 20a; + 3a? + 18 = 30a: + 2a\ x = 2 ; .-. 2a; = 4, and 3a: = G; therefore the number is 4G. 11. A hare takes 4 leaps to a greyhound's 3, but 2 of the greyhound's leaps are equivalent to 3 of the hare's ; the hare has a start of 50 leaps : how many leaps must the greyhound take to catch the hare? Let 3.r = the number of leaps taken by the greyhound ; then 4a: = the number of leaps taken by the hare in the same time. Also, let a denote the number of feet in one leap of the hare ; then fa denotes the number of feet in one leap of the greyhound. Therefore 3x x fa = the distance in 3a; leaps of the grey- hound ; and (ix + 50) a = the distance in Ax + 50 leaps of the hare. Hence we have the equation fax- = (4a: + 50)a. Dividing by a and multiplying by 2, we have 9a; = 8a: + 100. .-. x = 100. Therefore the greyhound must take 300 leaps. Notk 3. — It is often convenient to introduce an auxiliary symbol, as a was introduced in the above example, to enable us to form the equation easily; this can be removed by division when the equation is formed. 12. A person bought a carriage, horse, and harness for $600 ; the horse cost twice as much as the harness, and the carriage half as much again as the horse and harness : what did he give for each? Ans. $360, $160, $80. 13. In a garrison of 2744 men, there are two cavalry soldiers to twenty-five infantry, and half as many artillery as cavalry: lind the numbers of each. Ans. 2150, 196, 98. EXAMPLES. 169 14. A and B play for a stake of $5 ; if A loses he will have as much as B, but if A wins he will have three times as much as B : how much has each? Ans. $25, $15. 15. A, B, and C have a certain sum between them ; A has one-half of the whole, B has one-third of the whole, and C has $50 ; how much have A and B? Ans. $150, $100. 16. A number of troops being formed into a solid square, it was found that there were 60 over j but when formed into a column with 5 men more in front than before and 3 less in depth, there was just one man wanting to complete it: find the number of men. Ans. 1504. 17. A and B began to pay their debts ; A's money was at first f of B's ; but after A had paid $5 less than f of his money, and B had paid $5 more than f of his, it was found that B had only half as much as A had left : what sum had each at first? Ans. $360, $540. 18. In a mixture of copper, lead, and tin, the copper was 5 lbs. less than half the whole quantity, and the lead and tin each 5 lbs. more than a third of the remainder : find the respective quantities. Ans. 20, 15, 15 lbs. 19. A and B have the same income ; A lays by a fifth of his ; but B, by spending annually $400 more than A, at the end of four years finds himself $1100 in debt: what was their income? Ans. $625. 20. There are two silver cups and one cover for both ; the first weighs 12 ozs., and with the cover weighs twice as much as the other cup without it ; but the second with the cover weighs a third as much again as the first without it : find the weight of the cover. Ans. 6| oz. 21. Two casks, A and B, contain mixtures of wine and water ; in A the quantity of wine is to the quantity of water as 4 to 3 ; in B the like proportion is that of 2 to 3. If A contain 84 gallons what must B contain, so that when the two are put together, the new mixture may be half wine and half water? Ans. 60. 170 "" EXAMPLES. EXAMPLES. Solve the following equations. 1. 2LzJ + 4 = x - 2x ~ 1 . Ans. -6. 4 3 2. 4 * + 1 7 to - 10 = ?j 2> a; + 3 cc — 4 3. ^-=-i + (a; - 1)'(* - 2) = x- - 2x - 4. 7. 3 4 3(7 + Gx) _ 35 + 4x 1 2 + 9a? 9 + 2x ' X + -i- = 1. 2. . x -f 2 a; + 6 „ 21-5 . a-3 4.x- - 3 , . 5 2x - 15 10 liy „ 4 (a; + 3) _ 8a; + 37 _ 7x - 29 9 18 5x - 12' 7 60 10* 4 5a; - 30 3a; — 12 x — 6 6. -10. Q 3a; 2 - 2a; - 8 (7a? - 2) (3a; - G) 5 35 10. *±1°- t (te-4)+ (»«-«)(»»-») = ».- A . 2. 3 G 11. — L_ + -1- = — ^— . |*. 2a; - 3 x - 2 8a? + 2 " J 12 ft — 4 _ x — 5 _ x — 7 _ x — 8 13. _£- + "-^ = *±^ + 2_ZL_ 8 . 4. a; — 2 x — 7 x — 1 a; — G - . 3 - 2a? 2a; - 5 - 4a; 2 - 1 14. — = 1 — ;. — 1. 1 — 2x 2x — 7 7 — IGa; + 4ar 15. "zJ+gMJ-^zl,^— +!+>. -23. EXAMPLES. 171 16. 3 30 3 5 ^lns. -4. 4 - 2x 8(1 - x) 2 - x 2 - 2a;' 17. 30 -f Gx 60 + 8a; _ u , 48 x + 1 x + 3 a; + l" 3. 18. x x + 1 a: - 8 a; — 9 4 as — 2 a; — 1 a; — 6 a; — 7 19. x + 5 a; — 6 _ a; — 4 a; — 15 a; + 4 a; — 7 a; — 5 a; — 16 6. 20. aj — 7 a; — 9 a; — 13 x — 15 a; — 9 a; — 11 x — 15 x — 17 13. 21. a; + 3 a; + 6 _ a; + 2 cc + 5 o; + 6 a; + 9 a; + 5 a; + 8 -7. 22. a; + 2 x — 7 a; + 3 _ a; — 6 a; a; — 5 a; + 1 a; — 4 2. 23. 4a; - 17 10a; - 13 _ 8a; - 30 5a; - x — 4 2a; — 3 2a; — 7 a; — 4 1 H. 24. 5aj _ 8 6a; - 44 10a: - 8 _ x - 8 a;— 2' a;— 7 a;— 1 a; — 6 4. 25. (a;+lXa;+2Xa;+3) = (a;-lXa!-2Xa;-3)+3(4a;-2Xa;+l). Ans. 3. 26. (a;-9) (a;-7) (a;-5)(a;-l) = (a;-2)(a;-4)(a;-6)(a;-10). Ans. 5i. 27. (8a; - 3) 2 (a; - 1) = (4a; - l) 2 (4a; - -5). *■ 28. x " ~ x + l + a ' 2 + x + 1 = 2a;. a; - 1 a; + 1 0. Q9 6a; + 7 2a; — 2 _ 2a; + 1 3. 15 7a; — 6 5 30. .ox — 2 = .25a; + .2a; — 1. 20. 31. .5a; + .6a; — .8 = .75a; + .25. 3. 32. .15., + - 135 * ~ " 225 = ^ - ' 09x ~ - 18 . 5. .6 .2 .9 33 2x - 3 = - 4a; ~ - 6 p .3a; - .4 .06a; - .07' 2 ' 172 EXAMPLES. 34. .3.t - 1 .5 + 1.2a; .bx - A 2x - .1 * 35. (.3.i-- 2)(.3.r- 1) .2x - 1 * ( - 3x 2) = ,4a: - 2. 20. 36. a?(x — a) + b 2 (x — 6) = o&aj. a + b. 37 2a; + 3a _ 2 (3a; + 2a) x + a 3x + a ss. f(| + i) = *(|-i> »•«— >-( t P) , -K-S- £ 40. x 2 + a(2a - a?) - — = (x - -Y -f- a 2 . a + 6. a)^*+| t ) = 4^-.rVKa-4a ; )(2a + 3x). 42. 1 1 a - b 2ab x — a x — b x- — ab a + b 43. x — a x + a 2ax a — b a + b a 2 — 6 s a 2 b — a 44. a! — a a; — a — 1 a - - b X -&- 1 a; — a — 1 jb — a — 2 a; — 6 — 1 a; — 6 — 2 Jws. $(a + & + 3). 45. (a-«) 3 (x+«-2&) = (.T-&) 3 (.i--2«+&). |(a+&). . P 3a&c , crlr (2a + 6)?/ 2 a; _ „ te afr a + & (a + 6) 3 a(a + 6) s « ' a + &' 47. A person wishing to sell a watch by lottery, charges $1.20 each for the tickets, by which he gains $1G ; whereas, if he had made a third as many tickets again and charged $1 each, he would have gained one-fifth as many dollars as he had sold tickets : what was the value of the watch? Ana. $128. EXAMPLES. 173 48. There is a number of two digits, whose difference is 2, and if it be diminished by half as much again as the sum of the digits, the digits will be reversed : find the number. Ans. 75. 49. Find a number of 3 digits, each greater by 1 than that which follows it, so that its excess above a fourth of the number formed by reversing the digits shall be 36 times the sum of the digits. Ans. 654. 50. A can do a piece of work in 10 days, which B can do in 8 ; after A has been at work upon it 3 days, B comes to help him : in how many days will they finish it? Ans. 3^ days. 51. A and B can reap a field together in 7 days, which A alone could reap in 10 days : in what time could B alone reap it? Ans. 23^ days. 52. A privateer, running at the rate of 10 miles an hour, discovers a ship 18 miles off, running at the rate of 8 miles an hour : how many miles can the ship run before it is over- taken^ Ans. 72. 53. The distance between London and Edinburgh is 360 miles ; one traveler starts from Edinburgh and travels at the rate of 30 miles an hour, while another starts at the same time from London and travels at the rate of 24 miles an hour : how far from Edinburgh will they meet? Ans. 200 miles. 54. Find two numbers whose difference is 4, and the dif- ference of their squares 112. Ans. 12, 16. 55. Divide the number 48 into two parts so that the excess of one part over 20 may be three times the excess of 20 over the other part. Ans. 32, 16. 56. A cistern could be filled in 12 minutes by two pipes which run into it, and it could be filled in 20 minutes by one alone : in what time would it be filled by the other alone ? Ans. 30 minutes. 57. Divide the number 90 into four parts so that the first increased by 2, the second diminished by 2, the third multi- plied by 2, and the fourth divided by 2, may all be equal. Ans. 18, 22, 10, 40. 174 EXAMPLES. 58. Divide the number 88 iuto four parts so that the first increased by 2, the second diminished by 3, the third multi- plied by 4, and the fourth divided by 5, may all be equal. Arts. 10, 15, 3, 60. 59. If 20 men, 40 women, and 50 children receive $500 among them for a week's work, and 2 men receive as much as 3 women or 5 children, what does each woman receive for a week's work? Ans. $5. GO. A cisterD can be filled in 15 minutes by two pipes, A ^ and B, ruuuing together ; after A has beeu running by itself N^or 5 minutes B is also turned on, and the cistern is filled in -^13 minutes more : in what time would it be filled by each pipe separately? Ans. 37|, and 25 minutes. 61. A man and his wife could drink a cask of beer in 20 days, the man drinking half as much again as his wife ; but -§§ of a gallon having leaked away, they found that it only lasted them together for 18 days, and the wife herself for two days longer : how much did the cask contain when full ? Ans. 12 gallons. Let x = the number of gallons the woman could drink in a day. 62. A man, woman, and child could reap a field in 30 hours, the man doing half as much again as the woman, and the woman two-thirds as much again as the child : how many hours would they each take to do it separately ? Ans. 62, 93, 155. Let 2x = the man's number of hours, ox = the woman's, and 5x = the child's. 63. A and B can reap a field together in 12 hours, A and C in 16 hours, and A by himself in 20 hours : in what time (1) could B and C together reap it, and (2) could A, B, and C together reap it? Ans. 21 r 9 T hours, 10^-!} hours. 64. A can do half as much work as B, B can do half as much work as (', and together they can complete a piece of work in 24 days : in what time could each alone complete the work? Ans. 168, 84, and 42 days. EXAMPLES. 175 65. There are two places 154 miles apart, from which two persons start at the same time with a design to meet ; one travels at the rate of 3 miles in two hours, and the other at the rate of 5 miles in four hours : when will they meet ? Ans. At the end of 56 hours. 66. Three persons, A, B, and C, can together complete a piece of work in 60 days ; and it is found that A does three- fourths of what B does, and B four-fifths of what C does : in what time could each one alone complete the work? Ans. 240, 180, 144 days. Let x = C's time of completing the work, in days. 67. A general, on attempting to draw up his army in the form of a solid square, finds that he has 60 men over, and that he would require 41 men more in his army in order to increase the side of the square by one man : how many men were there in the army? Ans. 2560. 68. A person bought a certain number of eggs, half of them at 2 for a cent, and half of them at 3 for a cent ; he sold them again at the rate of 5 for two cents, and lost a cent by the bargain : what was the number of eggs ? Ans. 60. 69. A and B are at present of the same age ; if A's age be increased by 36 years, and B's by 52 years, their ages will be as 3 to 4 ; what is the present age of each? Ans. 12. 70. A cistern has two supply pipes which will singly fill it in 4£ hours and 6 hours respectively ; and it has also a leak by which it would be emptied in 5 hours : in how many hours will it be filled when all are working together? Ans. 5^. 71. A person hired a laborer to do a certain work on the agreement that for every day he worked he should receive $2, but that for every day he was absent he should lose $0.75 ; he worked twice as many days as he was absent, and on the whole received $39 : how many days did he work ? Ans. 24. 72. A sum of money was divided between A and B, so that the share of A was to that of B as 5 to 3 ; also the share of A exceeded five-ninths of the whole sum by $200 : what was the share of each person? Ans. $1800, 1080 176 EXAMPLES. 73. A gentleman left his whole estate among his four sons. The share of the eldest was $4000 less than half of the estate ; the share of the second was $600 more than one- fourth of the estate ; the third had half as much as the eldest ; and the youngest had two-thirds of what the second had : how much did each son receive ? Ans. $11000, $8100, $5500, 5400. Let x = the number of dollars in the estate. 74. A draper bought a piece of cloth at 38 cents per yard ; he sold one-third of it at 48 cents per yard, one-fourth of it at 44 cents per yard, and the remainder at 40 cents per yard ; and his gain on the whole was $1.70 : how many yards did the piece contain? Ans. 30. 75. A and B shoot by turns at a target ; A puts 7 bullets out of 12 into the bull's eye, and B puts in 9 out of 12; between them they put in 32 bullets : how many shots did each fire? Ans. 24. 76. Two casks, A and B, are filled with two kinds of sherry, mixed in the cask A in the proportion of 2 to 7, and in the cask B in the proportion of 2 to 5 : what quantity must be taken from each to form a mixture which shall consist of 2 gallons of the first kind and 6 of the second kind ? Ans. 4£, 3|. 77. The national debt of a country was increased by one- fourth in a time of war. During a long peace which followed, $250000000 was paid off, and at the end of that time the rate of interest was reduced from \\ to 4 per cent. It was then found that the amount of annual interest was the same as before the war. What was the amount of the debt before the war? Ans. $2000000000. 78. How many minutes does it want of 4 o'clock, if three- quarters of an hour ago it was twice as many minutes past 2 o'clock? Ans. 25. Let x = the number of minutes it wants of 4 o'clock. 7!). At whal time between 3 o'clock and 4 o'clock is one hand of a watch exactly in the direction of the other hand produced? Ans. 41)^ minutes past three. EXAMPLES. 177 80. The hands of a watch are at right angles to each other at 3 o'clock : when are they next at right angles? Ans. 32 T 8 T minutes past three 81. At what time between 3 and 4 o'clock is the minute hand one minute ahead of the hour-hand? 17^ minutes past three 82. A clock has two hands turning on the same centre ; the swifter makes a revolution every 12 hours, and the slower every 16 hours : in what time will the swifter gain just one complete revolution on the slower? Ans. 48 hours. 83. A watch gains as much as a clock loses ; and 1799 hours by the clock are equivalent to 1801 hours by the watch : find how much the watch gains and the clock loses per hour. Ans. 2 seconds. Let x ■= the number of seconds which the watch gains and the clock loses per hour. 84. A and B made a joint stock of $2000 by which they gained $640, of which A had for his share $128 more than B : what did each contribute to the stock ? Ans. $1200, $800. 85. A hare is 80 of her own leaps ahead of a greyhound ; she takes 3 leaps for every 2 that he takes ; but he covers as much ground in one leap as she does in 2 : how many leaps will the hare have taken before she is caught? Ans. 240. 86. A and B play at a game, agreeing that the loser shall always pay to the winner one dollar less than half the money the loser has ; they commence with equal quantities of money, and after B has lost the first game and won the second, he has $2 more than A : how much had each at the commencement? Ans. $6. 87. An officer can form his men into a hollow square 4 deep, and also into a hollow square 8 deep ; the front in the latter formation contains 16 men fewer than in the former formation : find the number of men. Ans. 640. 88. A person has just a hours at his disposal ; how far may he ride in a coach which travels b miles an hour, so as 178 EXAMPLES. to return home in time, walking back at the rate of c miles an hour? Ans. abc b + c 89. A man rides one-third of the distance from A to B at the rate of a miles an hour, aud the remainder at the rate of 26 miles an hour ; if he had traveled at a uniform rate of 3c miles an hour, he could have ridden from A to B and back again in the same time : prove that - = l + l- cab 90. A certain article of consumption is subject to a duty of $2.25 per cwt. ; in consequence of a reduction in the duty the consumption increases one-half, but the revenue falls one-third : find the duty per cwt. after the reduction. Ans. SI. 91. A vessel can be emptied by three taps; by the first alone it could be emptied in 80 minutes, by the second alone in 200 minutes, and by the third alone in 5 hours. In what time will the vessel be emptied if all the taps are opened ? Ans. 48 minutes. 92. A cask A contains 12 gallons of wine and 18 gallons of water ; and another cask B contains 9 gallons of wine and 3 gallons of water : how many gallons must be drawn from each cask so as to produce by their mixture 7 gallons of water and 7 gallons of wine ? Ans. 10 from A, 4 from B. 93. A ship sails with a supply of biscuit for 60 days, at a daily allowance of a pound a head ; after being at sea 20 daj's she encounters a storm in which 5 men are washed overboard, and damage sustained that will cause a delay of 24 days, and it is found that each man's daily allowance must be reduced to five sevenths of a pound. Find the orig- inal number of the crew. Ans. 40. SIMULTANEOUS EQUATIONS. 179 CHAPTER X. SIMULTANEOUS SIMPLE EQUATIONS OF TWO OR MORE UNKNOWN QUANTITIES. 93. Simultaneous Equations of Two Unknown Quantities. — If we have a single equation containing two unknown quantities x and y, we cannot determine any thing definite regarding the values of x and y, because whatever value we choose to give to either of them, there will be a corresponding value of the other. Thus, from the equation, 2x + Sy = 24, (1) we may deduce the equation, 24 - 2x v = ; u 3 but we cannot find the value of y from this equation unless we know the value of x. "We may give to x any value we choose, and there will be one corresponding value of y ; and thus we may find as many pairs of values as we please which will satisfy the given equation. For example, if x = 3, then y = (24 If x = 6, then y = (24 If x = 9, then y = (24 If x = 20, then y = (24 and so on. Any one of these pairs of values f ""' }, ( ), ( \ J \y=«J \y=y \y=i) etc., substituted for x and y in (1) will satisfy the equation. Hence a single equation containing two unknown quantities is not sufficient to determine the definite value of either. 6)- -3 = 6. 12)- -3 = 4. 18)- -3 = 2. 40) - - 3 = -51 180 SIMULTANEOUS EQUATIONS. Suppose we have a second equation of the same kind, expressing a different relation between the unknown quanti- ties, as for example, 3x + 2y = 26 ; (2) then we can find as many pairs of values as we please which will satisfy this equation also. Now suppose we wish to determine values of x and y which will satisfy both equations (1) and (2) ; we shall find that there is only one pair of values of x and y, i.e., only one value of x and one value of y that will satisfy both equations. For, multiply equation (1) by 2, and equation (2) by 3, and the equations become 4 X + Qy = 48, (3) and 9a; + Gy = 78 (4) The coefficients of y are now the same in (3) and (4) ; hence if we subtract each member of (.">) from the corre- sponding member of (4), we shall obtain an equation which does not contain y : the equation will be 5 - 10?/ = 17G (6) Subtracting (G) from (5), we have 31y = 124. .-. y = 4. In this solution we eliminated x by subtraction. Note 1. — The student will observe that we might have made the coefficients of x equal by multiplying (1) and ('2) by 12 and 8 respec- tively, instead of by 3 and 2; but it was more convenient to use the smaller multipliers, because it enabled us to work with smaller numbers. 2. Solve 2a? -f- Sy = 81 (1) 12.»; - 17y = -59 (2) Here it will be more convenient to eliminate x. ELIMINATION BY ADDITION OR SUBTRACTION. 183 Multiplying (1) by G, to make the coefficients of x in both equations equal, we have 12a; + I82/ =186; .... (3) and from (2) 12a; - Uy = -59 ..... (4) Subtracting (4) from (3), 35y = 245. ■■■ y = 7. Substituting this value of y in (1), we have 2x 4- 21 = 31. . • . x = 51 and y = 7j " Note 2. — When one of the unknown quantities has been found, it is immaterial which of the equations we use to complete the solution, though it is sometimes more convenient to use a particular equation on account of its being less involved than the other. Thus, in this example, we substituted the value of x in (1) rather than in (2), because it rendered the process simpler. In these two examples we have eliminated by addition and subtraction. Hence to eliminate an unknown quantity by addition or subtraction, we have the following Rule. Multiply the given equations, if necessary, by such numbers as will make the coefficients of this unknown quantity numeri- cally equal in the resulting equations. Then, if these equal coefficients have unlike signs, add the equations together; if they have the same sign, subtract one equation from the other. Rem. — It is generally best to eliminate tbat unknown quantity which has the smaller coefficients in the two equations, or which requires the smallest multipliers to make its coefficients equal. When the coefficients of the quantity to be eliminated are prime to each other, we may take each one as the multiplier of the other equation. When these coefficients are not prime to each other, find their least common multiple; and the smallest multiplier for each equation will be the quotient obtained by dividing this L. C. M. by the coefficient in that equation. Thus, in Ex. 1, first solution, 7 and 5 (the coefficients of y) are prime to each other. We multiplied (1) by 5 and (2) by 7. In the second solution of Ex. 1, the L. C. M. of S and 12 (the coeffi- cients of .r) is 24; and hence the smallest multipliers of (1) and (2) are 3 and 2 respectively, which we used in that solution. 184 ELIMINATION BY ADDITION OR SUBTRACTION. 3. Solve 171a; — 213?/ = 642 (1) 111.-' - 320?/ =244 (2) Here we see that 171 and 114 contain a common factor 57 ; so we shall make the coefficients of x in (1) and (2) equal to the least common multiple of 171 and 114 if we multiply (1) by 2 and (2) by 3. Thus, 342a - 426# = 1284 342a - 978y = 732 Subtracting, 552// = 552. .-. y= i,l therefore a- = 5. j Note 3. — The solution is sometimes easily effected by first adding the given equations, or by subtracting one from the other. Thus, 4. Solve 127a + 5%= 1928. . . . (1) 59a; + I27y = 171)2. . . . (2, By addition 186a -4- 186y = 3720. .-. x + y = 20 . . . . (3) Subtracting (2) from (1), 68a; - 68?/ = 136. ••• x-y =2 .... (4) Adding (3) and (4), 2a; = 22. .*. x = \\. Subtracting (4) from (3), 2// = 18. .-. y = 9. Note 4. — The student should look carefully for opportunities to effect such reductions as are made in this example. He will find as he proceeds that in all parts of Algebra, particular examples may be treated by methods which are shorter than the general rules; but such abbreviations can only he suggested by experience and practice. Solve the following equations by addition or subtraction: 5. 3a; + 4y = 10, 4a + ?/ = 9. Ans. x = 2,y = 1. 6. x + 2y = 13, 3a + ?/ = It. x = 3, y = 5. 7. 4a; + ly = 29, x + 3y =11. x = 2, y = 3. 8. 8a - ?/ = 34, a; + 8?/ = 53. x = 5, y = 6. 9. 14a; - 3y = 39, 6a + 17?/ = 35. x = 8, y = 1. 10. 35a + 17# = 86, 56a - Vdy =17. a = 1, y = 3. ELIMINATION BY SUBSTITUTION. 185 11. 15a; + lly = 92, 55a; — 33?/ = 22. Ans. x = 1, y = 1. 12. 3a? + 2y = 32, 20a; - 3y = 1. a: = 2, ?/ = 13. 13. 7x + 5y = 60, 13a; — lly t= 10. a- = 5, ?/ = 5. 14. 10s -f- 9y = 290, 12a; - lly = 130. a- = 20, y = 10. 96. Elimination by Substitution. — Find the values of x and y in the equations 4a; '.+ Sy = 22 (1) 5x — 7y = G (2) Transpose 3?/ in (1), 4a; = 22 — 3?/; divide by 4, a; = 22 ~ 3y ; substitute this value of a; in (2), and we obtain ,/22 - 3?/\ 5 (— 4 j- 7 ' /= G; multiply by 4, 5(22 - %) - 28y = 24. .-. ?/ = 2. Substitute this value of y in eitfier (1) or (2), thus in (1) 4a; + G = 22. .-. x = 4 and ?/ = 2 ' In this solution we eliminated a; by substitution. Otherwise thus: from (1) we have 3?/ = 22 - 4a; ; divide by 3, y = £lz^ ; o substitute this value of y in (2), multiply by 3, 15a; - 7(22 - Ax) = 18 ; that is, 15a; — 154 + 28a; = 18. .-. x = 4. 1SG ELIMINATION BY COMPARISON. Substitute this value of x in either (1) or (2), thus in (1) 16 + By = 22. .-. y = 2. Here we eliminated y by substitution. Hence, to eliminate an unknown quantity by sxibstitution, we have the following Rule. From either equation, find the value of the unknown quan- tity to be eliminated, in terms of the other; and substitute this value for that quantity in the other equation. Solve by substitution the following equations : 2. 3a; — \y = 2, Ix — 9y = 7. Ans. x = 10, y = 7. 3. 11a,* — ly = 37, 8a; -f 9y = 41. a; = 4, ?/ = 1. 4. Ga; — ly = 42, 7x — Gy = 75. a? == 21, ?/ = 12. 5. 3a; - 4?/ = 18, 3a; + 2y = 0. a; = 2, ?/ = -3. 6. 4a; -2=11, 2a; - By = 0. a> = 3, y = 2. 7. 2x — y = 9, 3x — ly = 19. x — 4, ?/ = — 1. 8. 15a; + ly = 29, 9a; + 15?/ = 39. a; = 1, y = 2. 9. 2aj + y = 10, 7a; -f 8?/ = 53. a; = 3, ?/ = 4. 97. Elimination by Comparison. — Find the values of x and y in the equations 2a? + By a 23 (1) 5a; - 2y = 10 . . ' . . . . (2) Finding the value of x in terms of y from both (1) and (2), we have, from(l), x = 23 ~ 3?/ , (3) and from (2), a; = 10 + 2y (1) 5 Placing these two values of x equal to each other, we have 10 + 2y 23 - 3?/ 5 2 ELIMINATION BY COMPARISON. 187 Clearing of fractious, by multiplying by 10, we have 20 + 4?/ = 115 — Voy. .-. 19?/ = 95. .-. y = 5. Substitute this value of y in either (3) or (4), thus in (4) 10 + 10 x = ! = 4. 5 In this solution we eliminated x by comparison. Otherwise thus: find the values of y in terras of x from (1) and (2). (5) Therefore 2/ = 23 — 2x 3 y = 5x — 10 2 2z 5a - 10 J3 _- 3 2 Clearing of fractions 46 - 4flj = 15a? - 30. .-. 19a; = 76. x — 4, the same as before. Substituting this value of x in either (5) or (6), we deduce y = 5, as before. In this solution we eliminated y by comparison. Hence, to eliminate an unknown quantity by comparison, we have the following Rule. From each equation find the value of the unknown quan- tity to be eliminated, in terms of the other; then place these values equal to each other. Note. — Either of these methods of elimination may be employed, according to circumstances, and we shall always obtain the same result, whichever one we use; each method has its advantages in particular cases. Generally, the last two methods introduce fractional expressions, while the first method does not, if the equations be first cleared of fractions. As a general rule, the method by addition or 188 FRACTIONAL SIMULTANEOUS EQUATIONS. subtraction is the most simple and elegant. When either of the un- known quantities has 1 for its coefficient, the method by substitution is advantageous. When there are more than two unknown quantities, it is often convenient to use several of the methods in the same example. Solve by comparison the following equations : 2. Ix — by = 24, 4a: — Sy = 11. Ans. x = 17, y = 19. 3. | + Sy = 7, ^L_pJ = 32/ - 4. x= 3, y = 2. 4. 6* - by = 1, 7« — 4?/ = 8|. a; = 3£, y = 4. 5. 1±J/ + a = 15, ?L=J? +y = 6. a; = 10, y = 5. o o 6. — + 5w = 13, 2x + 4 ~ ly = 33. a: = 19, y = 2. 19 J '2 ' * 7. 2a + t-? = 21,4?/ + ^i = 29. a; = 10, y = 7. 5 b 98. Fractional Simultaneous Equations of the Form 12 + 8 = x y *1 - H = 3 (2) x y If we cleared these equations of fractions they would involve the product xy of the unknown quantities ; and thus they would become quite complex. But they may be solved by the methods already given, as follows : Multiply (1) by 3, Multiply (2) by 2, 3G X 54 X + !* = 24. y -M- 6. y Add Divide by 30, ?2=30. X 5= i. x = 3. LITERAL SIMULTANEOUS EQUATIONS. 180 Substitute this ^ 'alue of x in (1) , Transpose, + § = 8. y * = 4. y ring equations : y = 2. Solve the follow 2. •_* x y = !, 18 + ?0 = 16. ^4ns. a; = 3, 7/ = 2. 3. §_» a; y = i,i5+ « = a; ?/ 7. x = 2, y = 3. 4. 5+6 a; ?/ = 3, 15 + ? _ * 2/ 4. x = 5, ?/ = 3. 5 . 5_ 2 a; ?/ = » 2/ 3. a; = 2, ?/ = 7. 6. 5+10 * 2/ = 79, 1« - 1 = x y 44. * = h y = i- 99. Literal Simultaneous Equations. — Let required to solve ax + by = c . . . a'# + &'?y = c' . . . Multiply (1) by a', aa'as -+- a'62/ = a'c • • • Multiply (2) by a, aa'a + ab'y = ac' . . . Subtract (3) from (4), (ab' — a'b)y — ac' — a'c. Divide by (ab' — a'b) , y = it be (1) (2) (3) (4) a&' — a'6 To find the value of x, eliminate y from (1) and (2) thus : Multiply (1) by &' and (2) by 6, ab'x + bb'y = b'c . . . . (5) a'&a; + bb'y = be' . . . . (G) Subtract (6) from (5), (ab' — a'&)aj = &'c — be'. b'c - 6c' Divide by (ab' — a'&), ab' - a'b 190 LITERAL SIMULTANEOUS EQUATIONS. Note. — As previously explained (Art. 95), this value of x might be found by substituting the value of y in either (1) or (2); but sometimes the value of the second unknown quantity is more easily found by elimination, as in this case, than by substituting the value, of the unknown quantity already found. The method by Undeter- mined Multipliers is often advantageous in the solution of literal equations. Method by Undetermined Multipliers. — Let it be required to solve by undetermined multipliers the same equation as in the last example. ax + by = c (1) a'x + b'y = c' (2) Multiply either (1) or (2) by any finite constant whatever; thus, multiply (2) by m, ma'x + mb'y = mc' (3) Add (1) and (3), ax + by + ma'x + mb'y = c + mc'. Factoring, (a + ma')x + (b + mb')y = c + mc' .... (4) Now since m is at our disposal, we may so choose it that the coefficient of y in (4) shall be zero; that is, b + mb'= 0; (5) and (4) will reduce to (a -f ma')x = c + mc' (6) From (5) wc have m = — -^, and substituting this value for m, we deduce from (6) _ c + mc' _ b' _ be — be' ~ a + ma' _ ba/ ab' —a'b b' The value of y may next be obtained by so choosing m that the coefficient of x in (4) shall be zero; that is, a + ma' = 0, • (7) (b + mb')y — c + mc'. ....... (8) From (7) we have m = — - \, a and using this value for m, we deduce from (8) _ c + mc' _ (£_ _ a'c — ac '. ~ b + mb' ~ , _ ab' a'b — o&'' «' or, changing signs in the terms so as to have the same denominator as the value of X, we have ac' — a'c ab' — a'b EQUATIONS WITH THREE UNKNOWN QUANTITIES. 191 Solve the following literal equations by either method of elimination : 2. x -f- y = a + b, bx + ay = 2t(b. Ans. x = a, y = b. 3. (a + c)x — by = be, x + y = a + b. x = b, y = a, a 7 / j \ ac be 4. a + y = c,aa; — by = c(a — b). x= -— ,y=_ __. a + 6 a + w - x , y . x , y , «£> (t6 a b b a a + b a + b a b b a a 2 + b~ J a 2 + 6 2 7. The sum of two numbers is a and their difference is 6 : find the numbers. Ans. Greater - + - ; less . When the known quantities in a problem are represented by letters, the answer furnishes a general result or Formula (Art. 41); and a formula expressed in ordinary language, furnishes a Rule. Thus, in the present example, we have the following Rule. The sum and difference of two numbers being given, to find the numbers : The greater number is equal to half the sum plus half the difference ; the less number is equal to half the sum minus half the difference. 100. Simultaneous Equations with Three or More Unknown Quantities. — In order to solve simultaneous equations which contain two unknown quantities, we have seen that we must have two equations (Art. 93). Similarly we find that in order to solve simultaneous equations which contain three unknown quantities, we must have three equa- tions. And generally, when the values of several unknown quantities are to be found, it is necessary to have as many simultaneous equations as there are unknown quantities. Simultaneous simple equations involving three or more unknown quantities, may be solved by either of the three methods of elimination explained in the preceding articles ; 192 EQUATIONS WITH Til RICE UNKNOWN QUANTITIES. but the most convenient method of elimination is generally that by addition or subtraction. The unknown quantities are to be eliminated one at a time by the following Rule. If there be three simple equations containing three unknown quantities, eliminate one of the unknoum quantities from any two of the equations, by the methods already explained (Arts. 95, 96, 97) ; then eliminate the same unknown quantity from the third given equation and either of the former two; two equations involving two unknoum quantities are thus obtained, and the values of these unknown quantities may be found by the rules given in the preceding Articles. The remaining unknown quantity may be found by substituting these values in any one of the given equations. If four equations are given involving four unknown quan- tities, one of the unknown quantities must be eliminated from three pairs of the equations. Three equations involv- ing three unknown quantities will thus be obtained, which may be solved according to the rule. If five or more equa- tions are given, they may be solved in a similar manner. Note 1. — Either of the unknown quantities may be selected, as the one to be first eliminated; but it is best to begin with the quantity which has the simplest coefficients; and when an unknown quan- tity is not contained in all the given equations, it is generally best to eliminate that quantity first. EXAMPLES. 1. Solve 6a; + 2y - 5« = 13, (1) 3a; + By - 2« = 13 (2) 7x + By - Zz = 26 (3) Choose y as the first quantity to be eliminated. Multiply (1) by 3, and (2) by 2, 18a? + Gy - 15z = 39, <;.,: + c,// _ lv = 26. subtracting, 12* — 11* = 13 (O EXAMPLES. 193 Multiply (1) by 5, and (3) by 2, 30a; + lOy - 252 = 65, Ux + 10// - 62 = 52. subtracting, 16a; — 19z = 13 (5) We have now to find the values of x and z from (4) and (5) . Multiply (4) by 4 and (5) by 3 (Art. 95, Kern.), 48a; - 442 = 52, 48a; — 572 = 39. subtracting, 132 =13. .-. 2 = 1. Substitute this value of z in (4) ; thus 12a: - 11 = 13. .-. x = 2. Substitute these values of x and z in (1) ; thus 12 + 2y — 5 = 13. .-. y= 3, 1 and 2 = 1, [ Note 2. — Although the method of elimination given by the rule is generally the best, yet in particular examples solutions may be obtained more easily and elegantly by other means, which the student in ust learn by experience. After a little practice he will find that the solution may often be considerably shortened by a suitable combination of the given equations. Thus, Ex. 1 may be solved as follows: Add (1) and (2) and subtract (3), 2x - 42 = 0, or x = 22 (6) Substitute this value of x in (1) and (2), and we get 2)j + 72 = 13, 3?/ + Az = 13. Subtracting, y — oz = 0. .-. y = 3z (7) Substitute these values of x and y in (1) ; thus 122 + 62 - 52 = 13. .-. 2 = 1; therefore from (6) and (7), a; = 2, y= 3. 194 EXAMPLES. i-f. (2) a; 3?/ !-i+*->ft (3) a; 5// z Clearing of fractional coefficients, we obtain from (1) § +^ - * = 3, (1) aj ?/ 2 from (2) - - i =0, (5) a; ?/ from (3) — - - + — = 32 (6) x if z Choose z as the first quantity to be eliminated (Note 1). Multiply (4) by 15 and add the result to (6) ; thus M + i? = 77. x y Divide by 7, — + - = 11 CO x y Multiply (5) by 6, — - - = 0. x y ... »_ii. x x = 3, ] from (5) n = 1, J from (4) z = 2. j ■ 3. .Solve 5a; - 3y - z = 6, (1) 13a; - ly + 3z = 14, (2) 7a; - 4?/ =8 (3) Multiply (1) by 3 and add the result to (2), 28a; - IGy = 32. Divide by 4, 7a; — Ay = 8. Thus we see that the combination of equations (1) and (2) leads to an equation which is identical with (3) ; and so to find x and //, we have but a single equation, 7x — -1// = 8, EXAMPLES. 195 with two unknown quantities, which is not sufficient to determine the deiiuite value of either (Art. 93). The anomaly here arises from the fact that one of these three equations is deducible from the others ; in other words, that the three equations are not independent (Art. 93) . Note 3. — Sometimes it is convenient to use the following rule : Express the values of two of the unknown quantities from two of the equations in terms of the third unknown quantity, and substitute these values in the third equation. From this, the third unknown quantity can be found, and then the other two; thus 4. Solve 3;c 4- Ay - I62 = 0, (1) 5x - 8y + IO2 = 0, (2) 2x + 6y + 72 = 52 (3) Multiply (1) by 2 and add to (2) ; thus 11a - 22z = 0. .-. x = 2z. Multiply (1) by 5, and (2) by 3, and subtract; thus My - 110a = 0. .-. y = — . Substitute these values of x and y in (3) ; thus Az + 15a + Iz = 52. .-. 2 = 2,1 and x = 4, | y = 5. J Note 4. — The rule in Note 3 is especially convenient when all of the unknown quantities occur in only one equation; thus 5. Solve x + y + z = a + b + c, . . . . (1) x — y = b — «, (2) x — z = c — a (3) From (2) we have y = x + a — b (4) From (3) we have 2 = x + a — c (5) 19G PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS. Substitute these values of y and z in (1), x + x + a — b + x + a — c = a + b + c. ... 33 = _ a + 26 + 2c. ... 3.= |( a + 6 + c ) _ from (4) from (5) Solve the following equations G. 7» + 3y - 2z = 1G, 2x + 5y + 3z = 39, 5a; — y + 5z = 31. 7. 2xc + 32/ + 4z = 16, 3a + 2y — 5z = 8, 5a — Gy + 3z = 6. ."' x + 2y + 2z = 11, 2x + y + g = 7, Bx + 4?/ + a = 14. a + 3?/ + \z = 14, x + 2// + z = 7, 2x + y + 2z = 2. 1 2 _ 3 _ x y z 1, x y z 24, 7 _ 8 9 _ 14. |(a +6 + c ) - - &, |(a + & .+ < ) " - c. J r. . 2, yl»s. = 4, >« = 5. .r = 3, 2/ = 2. 2 = 1. ' a; = 1. 2/ = 2, z = 3. a; = - -2, 2/ = 4, z = 1. \ y = 101. Problems Leading to Simultaneous Equa- tions. — AVe shall now give some examples of problems which lead to simultaneous equations of the first degree with two or more unknown quantities. Many of the problems given in Chapter IX. really contain ttvo or more unknown quantities, but the given relations are there of so simple a nature that it is easy to express all of the unknown quanta- EXAMPLES. 197 ties in terms of one unknown quantity, and thus to require but a single equation. In the problems of the present chapter the relations between the unknown quantities are not so simple, and the solution will give rise to simultaneous equations ; and in all cases the conditions of the problem must be sufficient to furnish as many independent equations as there are unknown quantities to be determined (Art. 100) . EXAMPLES. 1. Find two numbers such that the greater exceeds twice the less by 3, and that twice the greater exceeds the less by 27. Let x = the greater number, and y = the less number. Then from the conditions, x — 2y = 3, and 2x - y = 27. Solving these equations, we have x = 17, y = 7. If the numerator of a fraction be increased by 2 and the denominator by 1, it becomes equal to f ; and if the numerator and denominator are each diminished by 1, it becomes equal to £ : find the fraction. Let x = the numerator, and y = the denominator. Then from the conditions, X_±_2_ 5 ii — 8' y + i and ^^i = i. y - 1 Solving, we have x = 8, y = 15. Hence the fraction is ^. (^3y -A- man and a boy can do in 15 days a piece of work which would be done in 2 days by 7 men and 9 boys : how long would it take one man to do it? 198 EXAMPLES. Let x = the number of days in which one man would do the whole, and let y = the number of days in which one boy would do the whole. Then - = the part that one man can do in one day, x and - = the part that one boy can do in one day. y Then from the conditions of the question, a man and a boy together do T V^ n °f the wol 'k in one day ; hence we have i + i = A 0) x y Also, since 7 men and 9 boys do half the work in a day, we have „ - + - = \ (2) x y Multiplying (1) by 9, and subtracting (2) from it, we have 2 x Thus one man would do the work in 20 days. 4. A railway train after traveling an hour is detained 24 minutes, after which it proceeds at six-fifths of its former rate, and arrives 15 minutes late. If the detention had taken place 5 miles further on, the train would have arrived 2 minutes later than it did. Find the original rate of the train, and the distance traveled. Let x = the original rate of the train in miles per hour ; and y = the number of miles in the whole distance traveled. Then y — x = the number of miles to be traveled after the detention. ^ ~ ' = the number of hours in traveling y — x miles at the original rate, and — -~ — - = the number of hours in traveling y — x miles at the increased rate. &. .-. s=20. EXAMPLES. 199 Since the train is detained 24 minutes, and yet arrives only 15 minutes late, it follows that the remainder of the journey is performed in nine miuutes less than it would have been if the rate had not been increased ; hence we have > x 6x ~^° (1) If the detention had taken place 5 miles further on, there would have been y — x — 5 miles left to be traveled after the detention : hence we have Subtract (2) from (1), 5 (.7 - x — 5) 6x 5 _ 25 x Co 6x (2) .*. x = 25, an( * y = 47£. ^ v 'There is a number consisting of three digits ; the middle digit is zero, and the sum of the other digits is 11 ; if the digits be reversed, the number so formed exceeds the original number by 495 : find the number. Let x = the digit in the unit's place, and y = the digit in the hundred's place. Then, since the digit in the ten's place is 0, the number will be represented by 100?/ + x (Art. 92, Ex. 10) ; hence from the conditions, we have x + y = 11, and 100a + y — (100// + x) = 495. Solving, we get x = 8, y = 3 ; hence the number is 308. ^6j A sum of money was divided equally among a certain number of persons ; had there been three more, each would have received SI less, and had there been two less, each would have received $1 more than he did : find the number of persons, and what each received. 200 EXAMPLES. Let x = the number of persons, and y = the number of dollars which each received. Then xy = the number of dollars to be divided, and from the conditions, wc have (x + 3)0/ - 1) = xy, and {x- 2){y + 1) = xy. Solving, we get x = 12 and y = 5. 7. A train traveled a certain distance at a uniform rate ; had the speed been 6 miles an hour more, the journey would have occupied 4 hours less ; and had the speed been 6 miles an hour less, the jouruey would have occupied G hours more : find the distance. Let x = the rate of the train in miles per hour, and y — the time of running the journey in hours. Then xy = the distance traversed, and from the conditions, we have (x + 6) (y — 4) = xy, and (x - 6) (y + 6) = xy. Solving, we get x = 30, and y = 2-4. Hence the distance is 720 miles. 8. A, B, and C can together do a piece of work iu 30 days ; A and B can together do it in 32 days ; and B and C can together do it in 120 days : find the time in which each alone could do the work. Let x = the number of days in which A could do it, y = the number of days in which B could do it, and z = the number of days in which C could do it. Then we have from the conditions, 1 + 1 4- 1 - i so y z 1,1 — I — = SZi x y l.l-i y + -z-^~- Solving, we get x = 40, y = 160, z = 480. EXAMPLES. 201 Find the fraction which is equal to ^ when its ferator is increased by unity, and is equal to \ when its denominator is increased by unity. Ans. §. JlO>) A certain number of two digits is equal to five times the sum of its digits; and if nine be added to the number the digits are reversed : find the number. Ans. 45. 11. If 15 lbs. of tea and 17 lbs. of coffee together cost $7.86, and 25 lbs. of tea and 13 lbs. of coffee together cost $10.34, find the price per pound of each. Ans. The tea cost 32 cents, and the coffee cost 18 cents, alb. 12. If A's money were increased by $36 he would have three times as much as B ; and if B's money were diminished by $5 he would -have half as much as A : find the sum possessed by each. Ans. A has $42, B has $26. 13. Find two numbers such that half the first with a third of the second may make 32, and that a fourth of the first with a fifth of the second may make 18. Ans. 24, 60. 14. A farmer parting with his stock, sells to one 9 horses and 7 cows for $1200 ; and to another, at the same prices, 6 horses and 13 cows for the same sum : what was the price of each? Ans. $96, $48. 15. Having $45 to give away among a certain number of persons, I find that for a distribution of $3 to each man and $1 to each woman, I shall have $1 too little ; but that, by giving $2.50 to each man and $1.50 to each woman, I may distribute the sum exactly : how many were there of men and women? Ans. 12, 10. 16. Find three numbers, A, B, C, such that A with half of B, B with a third of C, and C with a fourth of A, may each be 1000. Ans. 640, 720, 840. 17. A person spent $1.82 in buying oranges at the rate of 3 for two cents, and apples at 5 cents a dozen ; if he had bought five times as many oranges and a quarter of the number of apples he would have spent $5.30 : how many of each did he buy? Ans. 153, 192. 202 EXAMPLES. EXAMPLES. Solve the following equations : 1. bx — ly = 0, Ix + by = 74. Ar.s. x = 7, y = 5. 2. 5as = 7?/ — 21, 21.r — 9?/ = 75. x = 7, y = 8. 3. 6?/ - 5aj = 18, 12a: - 9y = 0. a; = G, ?/ = 8. 4. 7.x + 4y = 1, 9.f + Ay = 3. a? = 1, y = -1|. 5. x - Uy — 1, Illy — 9x = 99. x = 100, ?/ = 9. G. 8x — 21y = 5, 6x + Uy = — 26. £c=— 2,2/= — 1. 7. 39a; - 8# = 99, 52a; - I5y = 80. x = 5, y = 12. 8. 3a; = 7?/, 12?/ = 5a? — 1. x = —7, y = —3. 9. 93a; + 15?/ = 123, 15a: + 93?/ = 201. x = 1, ?/ = 2. 10. ? + £ = i 2 _ ?y = 3 . x = A y = _ 3> 2 3 4 3 J 11. £_±J + x = 15, EJZJ + y = 6 . « = 10, 3, = 5. 3 5 12. ^ + ^ = 34,^ + ^ = ^+12. ^=12^ = 12. 6 3 84 8 10 l-3a; , 3v — 1 3a; + w Q „ 13. _ + * =2, —-p£ + y = 9. a?=o,y = 7. 7 5 11 14. |-i(y-2) -i(*-3) =0, .,_ t ( f . ir cc-2 ?/ + 2 n 2a;— 5 11—2?/ ,. - , 15. Z-I— = 0, • = 0. #=5, y = 2. 3 4 5 7 16. - + -^ = 3a:-7?/-37, 3a:-7?/=37. a;=3,?/ = -4. o 4 1 7. (a;+l) 0/+5) = (aj+5) (//+ 1 ), «y+a;+y= (a?+2) (*/+2). Ans. x = —2,y = —2. 18. xy - (y - l)(x - 1) = 6(.y - \),x-y= 1. .Aws. .i- = 2h y = l J. 19. l±* = 1+J = n +•» + * x = 8, y = L6 3 5 7 '* EXAMPLES. 203 20. 21. 22. 23. 24. 26. 26. 27. 28. 29. 30. 31. 32. 33. ,3x + .125?/ = x — 6, 3oj — .5*/ = 28 - .25*/. ^4»is. a; = 10, y = 8. .08z + 3*t — 2v = 2, 52 -f 4m + 2« - 2a; = 3, Ga — 00 + l.o — 'ly = 6. ( •>•= 4, M= ( 2 = 16, M= (M = 4, a;: 5,2: x= 3,y= ?<= i),z = 1 2: 2, // = 3, w= 12, 7. 1, 5. 4, 3, [v= 1. = — 1, V EXAMPLES. 205 45. What fraction is that, to the numerator of which if 7 be added, its value is f ; but if 7 be taken from the denom- inator its value is f ? Ans. ^. 46. 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 : find its area. Ans. 108 sq. ft. 47. A party was composed of a certain number of men and women, and, when four of the women were gone, it was observed that there were left just half as many men again as women ; they came back, however, with their husbands, and now there were only a third as many men again as women : what were the original numbers of each? Ans. 12, 12. 48i The sum of the two digits of a certain number is 6 times their difference, and the number itself exceeds 6 times their sum by 3 : find the number. Ans. 75. 49. Divide the numbers 80 and 90 each into two parts, so that the sum of one out of each pair may be 100, and the difference of the others 30. Ans. 30, 50, and 70, 20 ; or 60, 20, and 40, 50. 50. Four times B's age exceeds A's age by 20 years, and one-third of A's age is less than B's age by 2 years : find their ages. Ans. A 36 years, B 14 years. ^- §if. In 8 hours A walks 12 miles more than B does in 7 brours ; and in 13 hours B walks 7 miles more than A does in 9 hours : how many miles does each walk per hour? .-^ Ans. A 5 miles, B 4 miles. [52/ The sum and the difference of a number of two digits ana of the number formed by reversing the digits are 110 and 54 respectively: find the numbers. Ans. 28, 82. 53. In a bag containing black and white balls, half the number of white is equal to a third of the number of black ; and twice the whole number of balls exceeds three times the number of black balls by four : how many balls did the bag contain? Ans. 8 white, 12 black. 206 EXAMPLES. 54. Twenty-eight tons of goods arc to be carried in carts and wagons, and it is found that this will require 15 carts and 12 wagons, or else 24 carts and 8 wagons: how much can ..each cart and each wagon carry? Ans. § tons, f tons. •-*• 55. The first edition of a book had 600 pages, and was divided into two parts ; in the second edition one quarter of the second part was omitted and 30 pages added to the first ; the change made the two parts of the same length : what were they in the first edition? Ans. 240, 360. 56. If A were to receive $10 from B he would then have twice as much as B would have left ; but if B were to receive $10 from A, B would have three times as much as A would have left: how much has each? Ans. $22, $26. 57. A farmer sold 30 bushels of wheat and 50 bushels of barley for $75 ; he also sold at the same prices 50 bushels of wheat and 30 bushels of barley for $77: what was the price of the wheat per bushel? Ans. $1. 58. A certain fishing rod consists of two parts ; the length of the upper part is to the length of the lower as 5 to 7 ; and 9 times the upper part together with 13 times the lower part exceeds 11 times the whole rod by 36 inches : find the lengths of the two parts. Ans. 45, 63. 59. A certain company in a tavern found, when they came to pay their bill, that if there had been 3 more persons to pay the same bill, they would have paid $1 each less than they did ; and if there had been 2 fewer persons they would have paid $1 each more than they did : find the number of persons, and the number of dollars each paid. Ans. 12, 5. 60. There is a rectangular floor, such that if it had been 2 feet broader, and 3 feet longer, it would have been 61 square feet larger; but if it had beeo 3 feet broader, and 2 feet longer, it would have been 68 square feet larger : find the length and breadth of the floor. Ans. 14 ft., 10 ft. Let x = the length, and y — the breadth, of the floor in feet; then .nj — the surface of the floor In square feet. EXAMPLES. 207 61. "When a certain number of two digits is doubled, and increased by 3G, the result is the same as if the number had been reversed, and doubled, and then diminished by 36 ; also the number itself exceeds 4 times the sum of its digits by 3 : find the number. Ans. 59. 62. Two passengers have together 560 lbs. of luggage, and are charged for the excess above the weight allowed 62 cents and $1.18 respectively ; if the luggage had all belonged to one of them he would have been charged $2.30 : how much luggage is each passenger allowed without charge ? Ans. 100 lbs. 63. A farmer has 28 bushels of barley at 56 cents a bushel ; with these he wishes to mix rye at 72 cents a bushel, and wheat at 96 cents a bushel, so that the mixture may consist of 100 bushels, and be worth 80 cents a bushel : how many bushels of rye and wheat must he take? Ans. 20, 52. 64. A and B ran a race which lasted 5 minutes ; B had a MJtart of 20 yards ; but A ran 3 yards while B was running 2, and won by 30 yards : find the length of the course and the rate of each per minute. Ans. 150 yards, 30 yards, 20 yards. v;/o5. A and B can together do a certain work in 30 days ; at the end of 18 days however B is called off and A finishes it alone in 20 days more : find the time in which each could do the work alone. Ans. 50, 75. j\ G6. A, B, and C can together drink a cask of beer in 15 days ; A and B together drink four- thirds of what C does ; and C drinks twice as much as A : find the time in which eachalone could drink the cask of beer. Ans. 70, 42, 35. ,Y.//^7. A and B run a mile ; at the first heat A gives B a 'start of 20 yards, and beats him by 30 seconds; at the second heat A gives B a start of 32 seconds, and beats him by 9^ yards : find the rate per hour at which A runs. Ans. 12 miles. 68? A and B are two towns situated 24 miles apart, on the same bank of a river. A man goes from A to B in 7 208 EXAMPLES. hours, 1)) T vowing the first half of the distance, and walking the second half. In returning he walks the first half at three-fourths of his former rate, but the stream being with him he rows at double his rate in going ; and he accomplishes the whole distance in G hours. Find his rates of walking and rowing up stream. Ans. 4 miles walking, 3 miles rowing. /69. A railway train after traveling an hour is detained 15 minutes, after which it proceeds at three-fourths of its former rate, and arrives 24 minutes late. If the detention had taken place 5 miles further on, the train would have arrived 3 minutes sooner than it did. Find the original rate of the train and the distance traveled. Ans. 33^ miles per hour, 48^ distance. 70; The time which an express train takes to travel a journey of 120 miles is to that taken by an ordinary train as 9 to 14. The ordinary train loses as much time in stopping as it would take to travel 20 miles without stopping. The express train loses only half as much time in stoppiug as the ordinary train, and it also travels 15 miles au hour faster. Find the rate of each train. • Ans. 45, 30 miles per hour. 71. Two trains, 92 feet long and 84 feet long respectively, are moving with uniform velocities on parallel rails ; when they move in opposite directions they are observed to pass each other in one second and a half ; but when they move in the same direction the faster train is observed to pass the other in 6 seconds : find the rate at which each train moves. Ans. 30, 50 miles per hour. 72. A railroad runs from A to C. A freight train starts from A at 12 o'clock, and a passenger train at 1 o'clock. After going two-thirds of the distance the freight train breaks down, and can only travel at three-fourths of its former rate. At 40 minutes past 2 o'clock a collision occurs, 10 miles from C. The rate of the passenger train is double the diminished rate of the freight train. Find the distance from A to C, and the rates of the trains. Ans. GO miles, 30 and 20 miles per hour. EXAMPLES. 209 73. If there were no accidents it would take half as long to travel the distance from A to B by railroad as by coach ; but 3 hours beiug allowed for accidental stoppages by the former, the coach will travel the distance all but 15 miles in the same time ; if the distance were two-thirds as great as it is, and the same time allowed for railway stoppages, the coach would take exactly the same time : required the dis- , tance. . Ans. 90 miles. 74. A and B start together from the foot of a mountain to go to the summit. A would reach the summit half an hour before B, but missing his way goes a mile and back again needlessly, during which he walks at twice his former pace, and reaches the top 6 minutes before B. C starts 20 minutes after A and B and walking at the rate of two and one-seventh miles per hour, arrives at the summit 10 minutes after B. Find the rates of walking of A and B, and the distance from the foot to the summit of the mountain. Ans. 2±, 2 miles per hour ; distance 5 miles. 7."i. A railway train after traveling for one hour meets with an accident which delays it one hour, after which it proceeds at three-fifths of its former rate, and arrives at the terminus 3 hours behind time ; had the accident occurred 50 miles further on, the train would have arrived 1 hour 20 minutes sooner : find the length of the line, and the original rate of the train. Ans. 100 miles, rate 25 miles per hour. 76. The fore-wheel of a carriage makes 6 revolutions more than the hind- wheel in going 120 yards ; if the circum- ference of the fore-wheel be increased by one-fourth of its present size, and the circumference of the hind-wheel by one-fifth of its present size, the 6 revolutions will be changed to 4 : find the circumference of each wheel. Ans. 4 yards, 5 yards. 77. A and B can perform a piece of work together in 48 days ; A and C in 30 days ; and B and C in 26f days : find the time in which each could perform the work alone. ■Ans. 120, 80, 40 days. 210 EXAMPLES. 7.S. There is a certain number of three digits which is equal to 48 times the sum of its digits ; and if 198 be sub- tracted from the number the digits will be reversed ; also the sum of the extreme digits is equal to twice the middle digit : find the number. Ans. 432. 79. A man bought 10 horses, 120 oxen, and 46 cows. The price of 3 oxen is equal to that of 5 cows. A horse, an ox, and a cow together cost a number of dollars greater by 300 than the whole number of animals bought ; and the whole sum spent was $9366. Find the price of a horse, an ox, and a cow respectively. Ans. $420, $35, $21. 80. A farmer sold at a market 100 head of stock consist- ing of horses, oxen, and sheep, so that the whole realized $9.40 per head ; while a horse, an ox, and a sheep were sold for $88, $50, and $6 respectively. Had he sold one-fourth the number of oxen, and 25 more sheep than he did, the amount received would have been still the same. Find the number of horses, oxen, and sheep, respectively, which he sold. Ans. 2, 4, 94. 81. Five persons, A, B, C, D, E, play at cards; after A has won half of B's money, B one-third of C's, C one-fourth of D's, D one-sixth of E's, they have each $30 : find how much each had to begin with. Ans. A $11, B $38, C $33, D $32, E $36. 82. A offers to run three times round a course while B runs twice round, but A only gets 150 yards of his third round finished when B wins. A then offers to run four times round to B three times, and now quickens his pace so that he runs 4 yards in the time he formerly ran 3 yards. B also quickens his so that he runs 9 yards in the time he formerly ran 8 yards, but in the second round falls off to his original pace in the first race, and in the third round goes only 9 yards for 10 he went in the first race, and accordingly this time A wins by 180 yards. Determine the length of the course. Ans. 600 vanls- INDETERMINATE EQUATIONS. 211 CHAPTER XL INDETERMINATE PROBLEMS — DISCUSSION OF PROBLEMS — INEQUALITIES. 102. Indeterminate and Inconsistent Equations — • Impossible Problems. — An Equation is said to be Inde- terminate when it can be satisfied by an indefinite number of sets of values of the unknown quantities that enter it. Thus, an equation containing two unknown quantities may be satisfied by as many pairs of values as we please (Art. 93). Therefore, Every equation containing tivo unknown quantities is inde- terminate. If there are two equations containing three unknown quantities, we may eliminate one of them, and reduce the two equations to a single equation containing two unknown quantities, which we have already seen is indeterminate. Similarly, if there is any number of equations containing more unknown quantities than there are equations, we may by elimination reduce them to a single equation containing two or more unknown quantities, which we have already seen is inde- terminate. Hence we have the following general principle : Any collection of equations containing more unknown quantities than there are equations can be reduced to a single indeterminate equation. A Problem is said to be Indeterminate when it imposes fewer conditions than there are unknown quantities, that is, when it leads to a less number of independent equations than there are unknown quantities. A problem which involves only one unknown quantity may be indeterminate, the equation which it leads to being an identical one (Art. 52). Thus, 212 INDETERMINATE EQUATIONS. What number is that whose | increased by the \ is equal to the ^ 7 () increased by the ^ ? Let a; = the number. Then - + - = *V + — 5 5 G 20 GO' clearing of fractious, 12a; + IOj; = 21a; + x, or 22a; = 22a;, which can be satisfied by any value whatever of x. It is uot always the case that the values of two unknown quantities can be found from two independent equations. Thus, take the equations x + 3y = 4, (1) 2.x- + Gy = G (2) Multiply (1) by 2, 2x + 6y = 8, and from (2), 2a + 6?/ = G. .-. = 2, an impossible result, which shows that there are no values of x and y which will satisfy both (1) and (2). Such equations are called Inconsistent Equations. When a problem leads to a greater number of independent equations than there are unknown quantities, it is Impossible. Thus, suppose we have a problem furnishing the three inde- pendent equations x + y = 12, (l) 2x + y = 17, . . * . . • (2) S X + y=28 (3) The values of x and y from (1) and (2) are x = 5, y = 7 ; from (1) and (3), x = 8, y = -I ; from (2) and (3), x = 11, y = — 5. From this it is evident that only two of these equations can be true at the same time. Hence the problem is impossible. If (3) had not been independent, but deduced from (1) and (2) by combining them in any way, as, for example, by adding them, giving 8a + ty = 29, DISCUSSION OF PROBLEMS. 213 then the values of x and y from any two of the three equations would have been x = 5, y = 7, and the problem would have been possible. In this case the last equation would have been unnecessary. When a problem furnishes more conditions than are necessary for its solution, those that are unnecessary are termed redundant. A problem which involves only one unknown quantity is sometimes impossible. Thus, What number is that whose f| increased by 3 is equal to the difference between its | and ^ increased by 11 ? Let x = the number ; then we have the equation y^ + 3 = — - — + ii. 15 6 10 Clearing of fractions, 22z + 90 = 25a; - 3x + 330. .-. = 240, which shows that the problem is absurd. Hence it appears that to determine any number of unknown quantities, there must be given as many independent equa- tions as there are unknown quantities. If there be fewer, the problem is indeterminate, but if there be more, the problem in general is impossible. 103. Discussion of Problems — Interpretation of Negative Results. — The Discussion of a Problem consists in assigning different values to the arbitrary quantities which enter the equation, and interpreting the results. An arbitrary quantity is one to which any particular value may be assigned at pleasure. It sometimes happens in the solution of a problem, that the result obtained has the minus sign. Such a result is termed a Negative Result ; and from the nature of positive and negative quantities (Art. 20), a negative result must have an interpretation diametrically opposite to that which it would receive if it were preceded by the plus sign. 214 INTERPRETATION OF NEGATIVE RESULTS. This interpretation we will illustrate by one or two prob- lems. 1. What number must be added to a in order that the sum may be b? Let x = the required number ; then a + x = 6. .-. x = b — a, (1) which gives the value of x corresponding to any assigned values of a aud b. Thus, if a = 12 and b = 25, we have from (1), x = 25 - 12 = 13. Here x is +13, which means that 13 must be added to 12 in a strict Arithmetic sense in order to make the sum 25. But suppose that a = 30 and b — 2-4 ; then (1) becomes x = 24 - 30 = -6. What is the meaning of this negative result? The problem now reads : what number must be added to 30 in order that the sum may be 24 ? It is evident that, if the words added and siim are to retain their Arithmetic meanings, the problem is impossible. If however the problem be modified so as to read : what number must be taken from 30 in order that the difference may be 24, it may be solved, and 6, which is the absolute value of the negative result, is the answer to it. This second problem differs from the given problem only in this : the words added to and sum are replaced b}' the words taken from and difference. Now, if the words, number, added to, and sum, be used Algebraically, then the negative result, — G, will be the correct answer to the given problem. In this case the nega- tive result is interpreted as meaning that —6 must be added Algebraically to 30 to produce 2-1, or that G must be sub- tracted Arithmetically from 30 to produce 24. This agrees exactly with the principle of Art. 29, that subtracting a positive number is the same as adding an equal negative number. DISCUSSION OF PROBLEMS. 215 2. A's age is 35 years, and B's age is 20 years : when will A be twice as old as B ? Suppose the required epoch to be x years after the present time ; then we have 35 + x = 40 + 2x (2) What is the meaning of this negative result ? It is evident that if a strictly Arithmetic meaning is to be given to the symbols x and +, equation (2) is impossible, for 40 is greater than 35, and 2x is greater than x, so that the two members cannot be equal. If however the problem be modified so as to read : A's age is 35 years, and B's age is 20 years ; when was A twice as old as B, it may be solved, and 5, which is the absolute value of the negative result, is the answer to it. Thus, we may say that, in the given example, the negative result indicates that the problem in a strictly Arithmetic sense is impossible, but that a new problem can be formed by appropriate changes in the original enunciation, to which the absolute value of the negative result will be the correct answer. Suppose the problem had originally read thus : A's age is 35 years, and B's age is 20 years ; find the epoch at which A is twice as old as B. Here, if we suppose that the required epoch is after the present date, we obtain for the required number of years x = —5, which indicates that this suppo- sition, in an Arithmetic sense, is incorrect, since it leads to a negative result. If we suppose that the required epoch is before the present date, we obtain for the required number of years x = 5, which indicates that this second supposition is correct, since it leads to an Arithmetic value for x. Hence 1. We may say that a negative result indicates either an erroneous enunciation of a problem, or a wrong choice out of two p>ossible suppositions which the problem allowed. 216 INTERPRETATION OF THE FORMS, -, — , -. CO 2. When a negative result is obtained, the question to which it is the answer may be so modified as to make the result true in an Arithmetical sense. 104. Interpretation of the Forms, ?, — , -. —We c» shall now examine the meaning of certain Forms which occur in the course of mathematical operations, and Inter- pret them. The symbol 0, called zero, is used to represent either abso- lute zero, or an indefinitely small quantity, i.e., a quantity which is less than any assignable quantit} 7 . If we subtract any quantity from itself, the remainder is absolute zero. Thus, the cipher in the equation, a — a = 0, is the abso- lute zero of Arithmetic. The symbol oo, called infinity, is used to represent an infinite quantity, i.e., a quantity which is greater than any assignable quantity. (1) Let - represent a fraction. If the numerator a re- mains constant while the denominator b diminishes, the value of the fraction increases. If the numerator remains con- stant while the denominator is divided by any number, the value of the fraction is multiplied by that number (Art. 85). When the denominator becomes veiy small the value of the fraction becomes very large ; and finally, when the denomi- nator becomes less than any assignable quantity, or 0, the value of the fraction becomes greater than any assignable quantity, or infinity. That is, - = oo (1) V This may be otherwise expressed as follows : If a fraction has a fixed numerator, no matter how small, the denominator can be made so much smaller than the numerator, that the fraction shall be greater than any assignable quantity. (2) If the numerator a remains constant while the de- INTERPRETATION OF THE FORMS, -, — , -. 217 cc nominator b increases, the value of the fraction decreases (Art. 86). When the denominator becomes very large the value of the fraction becomes very small ; and finally, when the denominator becomes greater than any assignable quan- tity, or co, the value of the fraction becomes less than any assignable quantity, or 0. That is, 2l = (2) 00 This may be otherwise expressed as follows : If a fraction has a fixed numerator, no matter how large, the denominator can be made so much larger than the numerator, that the fraction shall be less than any assignable quantity. The zero in (1) and (2) is an abbreviation to denote an indefinitely small quantity, and does not mean absolute zero. If a = and b is not =0, then of course the value of the fraction =0. (3) If a = and 6 = 0, the value of the fraction - takes the form g. In this case, the fraction may be said to have any value whatever; that is, since the divisor zero, multiplied by any number whatever, gives the dividend zero, therefore zero, divided by zero, may have any value whatever. Hence the fraction J} is indeterminate, and is called the Symbol of Indetermination. * This symbol, however, does not always mean indetermi- nation; but is often the result of a particular supposition, which makes a factor become zero, which is common to both terms of a fraction. Thus : 1 Let it be required to find the value of the expression x 2 — a" 1 , when x = a. If we suppose x = a, this becomes §, the value of which is indeterminate. But if we cancel the common factor, * Called also a Vanishing Fraction. 218 PROBLEM OF THE COURIERS. x — a, before we put x = a, this becomes x + a. If we now suppose x = a, the expression becomes 2a, which is its true value. Here we see that the expression becomes °- in consequence of the factor x — a appearing in both numerator and de- nominator. Now we cannot divide by a zero factor, but as long as x is not absolutely equal to a the factor x — a may be removed. 2 Find the value of the fraction or — a 2 , , when x = a. (x - a y If we suppose x = «, this expression becomes J). But if we first cancel the common factor, x — a, and then put x = a, the expression becomes — = . 3 Find the value of the fraction (x — a)' 2 , ^- '—, when x = a. x- — R The discussion of this problem will serve to further illus- trate the principles of Arts. 103 and 104, and to show that the results of every correct solution correspond to the cir- cumstances of the problem. Let the point P be taken as the origin of distances, and suppose that distances measured to the right from P, and time counted after 12 o'clock, are regarded as jwsitive. Let t = the number of hours from 12 o'clock to the time of meeting, and x = the distance A travels in t hours ; then x — a = the distance B travels in t hours. Now, since the distance traveled by each equals the rate per hour multiplied by the number of hours, we have x = mt, and x — a = nt. Solving these equations for t and x, we get «--2- (i) m — n ma . , as * = ■ (2) m — n We shall now examine these values of t and x on different suppositions as to the arbitrary quantities, a, m, and n. 1. Suppose m > n, and a > 0. This hypothesis makes both terms of the fractions (1) and (2) positive ; hence the values of both t and x are positive. That is, the couriers are together after 12 o'clock, and to the right of P. This interpretation evidently corresponds with the supposi- tion made. For if m > n, A travels faster than B, and will therefore be continually gaining on him ; hence at some time after 12 o'clock they must be together and at the right of P. 2. Suppose m < n, and a > 0. This hypothesis makes the numerator in each of the fractions in (1) and (2) positive, and the denominators 220 PROBLEM OF THE COURIERS. negative ; hence the values of both t and x are negative. Now, since we supposed distances measured to the right of P, and time counted after 12 o'clock to be regarded as positive, we conclude, from what we have observed respect- ing negative quantities (Arts. 20, 29, 103), that these values of t and x indicate that, instead of the couriers meeting after 12 o'clock and to the right of P, they met before 12 o'clock and to the left of P. This interpretation evidently corresponds with the sup- position made. For, if m < n, A travels more slowly than B, and hence, after 12 o'clock, they continually separate. But as A travels more slowly than B, and was behind B at 12 o'clock, it follows that, at some time before 12 o'clock and at the left of P, they must have been together. 3. Suppose vi = 7i, and a > 0. This hypothesis makes the numerator of each fraction finite and the denominator zero ; hence, the values of both t and x become (Art. 104). That is, the couriers will be together at an infinite distance from P; in other words, they will never be together. This interpretation evidently corresponds with the sup- position made. For, if m = n, the couriers travel at the same rate ; and since they were a miles apart at 12 o'clock, they always have been, and always will be separated by the same distance, so that they will never meet. 4. Suppose m > n or < n, and a — 0. This hypothesis makes the numerator of each fraction zero, and the denominator finite ; hence the values of t and a; become (Art. 104). That is, the couriers are together at the point P at 12 o'clock, and at no other time and place. This interpretation evidently corresponds with the sup- position made. For, if a = 0, the couriers are together :it 12 o'clock at the point P; and as m and n arc unequal, they travel at different rates ; hence they can never be together after 12 o'clock, nor could they ever have been together be- fore that time. EXAMPLES. 221 5. Suppose m = n, and a = 0. This hypothesis makes both terms of each fraction ; hence the values of both t and x assume the form -g. But when an expression takes this form, it may have any value whatever (Art. 104(3)). That is, there is an infinite number of times when the couriers are together. This interpretation evidently corresponds with the sup- position made. For, if m = n and a = 0, the couriers were together at 12 o'clock at the point P, and were traveling at the same rate ; hence they are always together, both before and after 12 o'clock. EXAMPLES. 1. Given 2x — y = 2, ox — dy = 3, ox + 2y = 17, 4x + dy = 24 ; to find x and y, and to show how many equations are redundant (Art. 102). Ans. x = 3, y = 4. 2. The sum of two numbers is 9, and their difference is 25: find the numbers. Ans. 17, and —8. Interpret the negative result obtained, and modify the question accordingly (Art. 103). 3. A is 50 years old, and B 40 : find the time when A will be twice as old as B. Ans. In —30 years. Interpret the negative result, and modify the question accordingly. 4, T^iihI flip vnlup nf* * wlion t — 1 Ans. %. X ill 11 Lll^ > ttlLLO KJL „ VV 11^11 •*/ — x« x 2 + x — 2 5. Find the value of — when x = a. x- — a 2 la. 6. Find the value of — when x = a. x' 2 — ax 2. 7. Find the value of ■ — :r — when x = a. 3. 222 JNE Q UALITIES. INEQUALITIES. 106. Inequalities. — It is often desirable to compare expressions that are unequal, and to learn which of them is the greater. An Inequality is a statement in Algebraic language that two expressions are unequal. Thus, a > &, is an inequality, showing that a is greater than &, or that a — b is positive. Also, a < b, is an inequality, showing that a is less than b, or that a — b is negative. The quantity on the left of the sign is called the first member, and that on the right, the second member of the inequality. The form a> b > c, indicates that b is less than a but greater than c. The form a > b, indicates that a may be cither equal to or greater than />, but cannot be less than b. Two inequalities are said to subsist in the same sense when the first member is the greater or the less in both. Thus, the inequalities a > &, and 5 > 3 ; or a < b, and 3 < 8, are said to subsist in the same sense. Two inequalities are said to subsist in a contrary sense when the first member is the greater in one, and the less in the other. Thus, the inequalities a > b, and 3 < 8 ; or a < b, and 7 > 2, are said to subsist in a contrary sense. Properties of Inequalities. — (1) An inequality will still subsist in the same sense after the same quantity has been added to or subtracted from each member. INEQUALITIES. 223 For suppose a > b, then a — b is positive ; therefore, a + c — (6 + '') and « — c — (b — c) are positive, since each equals a — b. Therefore a -f- c > b + c, and a — c> 6 — c. Hence, a?i?/ ienw. may 6e transposed from one member of an inequality to the other, if its sign be changed. (2) If the signs of all the terms of an inequality be changed, the sign of inequality must be reversed. For, to change all the signs is equivalent to removing each term of the first member to the second, and each term of the second member to the first. (3) An inequality will still subsist in the same sense after each member has been multiplied or divided by the same positive number. For suppose a > b, then a — b is positive ; therefore, if m is positive m(a — b) or —(a — b) m is positive ; and therefore , a b ma > mb or — >— . m 7/1 It may be shown in like manner that if each member of an inequality be multiplied or divided by the same negative number, the sign of inequality must be reversed. (4) If tivo or more inequalities subsist in the same sense, the corresponding members may be added together, and the resulting inequality will subsist in the same sense. For, if a > b, a' > b' ', a" > &", ...... then a — b, a r — b' , a" — b" , are all positive ; therefore their sum a — & + a! — .&' + a" — b" + is positive ; hence a + a' + a" + . . . . > b + b' + b" + (5) If a>b, a' > b', a" > b" , and all the quantities are positive, then it is evident that aa'a" > bb'b" .... 224 EXAMPLES OF INEQUALITIES. (G) If tioo positive members of an inequality be raised to any power, the inequality will still subsist in the same sense. For suppose a > 6, and let a and b be positive ; then if n is any positive number, it follows directly from (5) that a" > b". EXAMPLES OF INEQUALITIES. 1. Find the limit of x in the inequality Clearing of fractions, GO + 4a; > 9(5 + 3.x-. x > 36. 2. If a > b, prove that a- + b" > 2ab. The square of (a — b) or (6 - a) must be positive (Art. 36), and therefore greater than zero. ... r ,2 _ 2db 4- & 2 >0. ' .-. a 2 4- & 2 > 2a&. a. 4- tt„ 4- cl 4- . . . . a„ ,, 3. Show that the fraction - , * , * ~ f > the \ + 6 2 4- 6 8 4- . . . . &„ least, and < the greatest of the fractions, % \ f' m '' ' '/','' all of the denominators having the same sign. Suppose the fractions % ^ 2 , f \ . . . . ','" to be in ascend- °1 b-l }) :\ h " ing magnitude, and that all the denominators are positive. Then ... a, = \ X "b «2 > & a x ^b 2a > "> .-. a, > l>.. x "b 6, and so on. EXAMPLES. 225 Adding, a 1 +a 2 +a s + a n > (& 1 +& 2 +6 8 + . . . . h n )j ( h + °2 + 5. x>2. 4. 7a: — 3 > 32. 5. 7jb - % 3 > fa; + 5. 6. ax ,, r ft 2 5 5 7. Show that - -+- - > 2, unless a = b. b a 8. Show that w 3 + 1 > ?r + w, unless EXAMPLES. 1. Show that the following example is impossible : What number is that whose T \ diminished by 5 is equal to the difference between its f and £ increased by 7 ? Interpret the negative results in the following two ex- amples, and modify the enunciation accordingly. 2. The triple of a certain number diminished by 100 is equal to 4 times the number increased by 200 : find the number. Ans. -300. 3. The sum of two numbers is 00, and their difference is 120: find the numbers. Ans. 105, —15. x 3 — 1 4. Find the value of — — , when a; = 1. a; 3 — 2x 2 + 2x — 1 Ans. 3. 5. Find the value of ~ , when x = 1. 5. 226 6. Find the value of 7. Find the value of 8. Find the value of (4a; 3 - 9)(1 + x) Ans. %. Find the limit of x in the following four inequalities : 9. 3x — 5 > 13. Ans. x > 6. 10. I^_^>*-3. 5 3 3 EXAMPLES. & _ 5^ + 7. v _ 3 i c = 3. ar — a- — 5a; — 3 Ans. \ r z 4- 1 * — — — , when x = — 1. ar — 1 _3 3 (3 + 2a*) (. -5) X > 0. p — m q — ri x > a. 11. m — nx > p — gas. 12. ax + 5&a; — 5a& > a 2 . In the following examples the letters are unequal and positive. b 1 , 1 ijt-,>- + T- b" cr a b 14. Show that a" + b' 2 + c 2 > o/> + oc + &c. 15. Show that (a 2 + & 2 )(a 4 + & 4 ) > (a 8 + & 3 ) 2 . 16. If x*= a 2 +& 2 , and y 2 = c 2 + d 2 , show that xy > ac + bd. 17. Show that abc > (a + b — c)(a + c — b) (6 + c — 0). 18. Show that «b(a + 6) + be {b+c) + w'('' + a) > Gabc. 19. Show that (a + &)(& -I- c){c + a) > 8a6c. 20. Show that x 2 — 8x + 22 is never less than 6, what- ever may be the value of x. 21. Show that a + c + e > the least, and < the greatest b + d + f -, -, -, each letter being positive. b d f m INVOLUTION OF POWERS OF MONOMIALS. 227 CHAPTER XII. INVOLUTION AND EVOLUTION. 107. Involution is the process of raising an expression to any required power. Involution is therefore only a particular case of multiplication, in which the factors are equal (Art. 12). It is convenient, however, to give some rules for writing clown the power at once. It is evident from the Rule of Signs (Art. 36) that, (1) No even power of any quantity can be negative. (2) Any odd power of a quantity will have the same sign as the quantity itself. Thus, ( — a) 2 = ( — «)( — a) = +a~, ( — ay = ( — a) ( — a) ( — a) = +a 2 ( — a) = —a 3 , {-ay = (-«)(-«) (-a) (-«) = (-a*) (-a) = +a 4 ; and so on. Note. — The square of every expression, whether positive or negative, is positive. 108. Involution of Powers of Monomials. — From the definition, we have, by the rules of multiplication, (a 2 ) 8 = (a 2 ) (a 2 ) (a 2 ) = o 2+2+2 = a 6 . (-a 8 ) 2 =(-a 8 )(-a 8 )=a 8+8 = a 6 . (_3a 3 ) 2 =(-3a 3 )(-3a 3 ) = (- 3) 2 (a 3 ) 2 = 9a 6 . Generally, («,"')"= « m • a m • a m • a m . . . . to n factors __ a m + m + m+m fa n t erm g = a"" 4 . (ab) m =ab • ab • ab to m factors = (aaaa .... to m factors) x (bbbb .... torn factors) =a m b m . Hence (ab)" 1 = a'"b"\ 228 EXAMPLES. and so on for any number of factors. Thus, the m th poiver of a product is equal to the product of the m' h powers of its factors. Hence (a/W )"< = {a x ) m (p») m (c u 8. /-JLV =_«!_, \ b-cy 6 12 c 18 11. / . H/V \- = .rV 27a; 16 125**' INVOLUTION OF BINOMIALS. 229 109. Involution of Binomials. — We have already proved by actual multiplication (Art. 41), the two following cases of the involution of binomial expressions : (a + b) 2 = a- + 2ab + & 2 . . . . (1) (a - b) 2 = a 2 — 2ab + 6 2 . . . . (2) If we multiply (1) and (2) by a -{- b and a — 6 respec- tively, we have (a + 6) 3 = a 3 -(- 3a 2 6 + 3a& 2 + b 3 . . . (3) (a - b) s = a 3 - 3a 2 6 + 3a& 2 - & 3 . . . (4) If we multiply (3) and (4) by a + b and a — b respec- tively, we shall have (a + 6) 4 = a 4 + 4a 3 & + Ga 2 b 2 + 4ab 3 + b\ (a - by = a 4 - 4a 8 6 + 6a 2 & 2 - 4a& 8 + b\ By multiplying these two results by a -f- & and a — b respectively we should obtain (a + &) 5 and (a — &) 5 ; and by continuing the process we could obtain any required power of (a + b) or (a — b). Hence the following Rule. Multiply the binomial by itself, until it has been taken as a factor as many times as there are units in the exponent of the required poiver. This rule, however, would be very laborious in finding any high power, for instance (a + 6) 20 . In Chapter XIX. we shall prove a theorem, called the Binomial Theorem, by the aid of which any power of a binomial expression can be obtained without the labor of actual multiplication. Since the above formulae are true for all values of a and b, we can write down the squares and the cubes of any binomial expressions. Thus, 1. (a* - V) 2 = (a 4 ) 2 + 2«K-//) + (-b*) 2 = a 8 - 2a%* + b s . 230 INVOLUTION OF POLYNOMIALS. Show that 2. (2x + 3y) 2 = 4a- 2 + 12.r?/ + 9?/ 2 . 3. (3a; + 5?/) 2 == 9a- 2 + SOxy + 25?/ 2 . 4. (a; - 2#) 3 = a; 3 - 6afy + 12av/ 2 - Sy s . 5. (2a6 - 3c) 3 = 8a 3 Z> 3 - 36a 2 6 2 c + Mabc 2 - 27c 3 . G. (5a 2 - 36 2 ) 3 = 125a° - 225a 4 6 2 + 135a?b 4 - 276 6 . 110. Involution of Polynomials. — We may now apply the formulae of Art. 109 to obtain the powers of any trinomial or polynomial. Thus from (1) (« + & + c)-= [(a + b) + c] 2 = (a + &) 2 + 2(a 4- &)c + c 2 = a 2 + & 2 + c 2 + 2ab + 2ac + 26c . (1) In the same way we may prove (a+b+c-\-d) 2 =a-+b 2 +c-+d 2 +2ab+2ac+2ad+2bc+2bd+2rd . (2) We observe in both (1) and (2) that the square consists of (1) the sum of the squares of the several terms of the given expressions ; (2) twice the sum of the products two and two of the several terms, taken with their proper signs. The same law holds whatever be the number of terms in. the expression to be squared. Hence the following Rule. To find the square of any polynomial, write the square of each term together with twice the product of each term by each of the terms following it. From (3) of Art. 109 we obtain the cube of a trinomial as follows : (a+b+cf = [0+(6+c)]« - a»+Sa a (&+c)+8a(&+c)*+(&+c)" — a 8 +6 8 +c 8 +3a a (6+c)+86 8 (o+c)+8c a (a+5)+6oBc . (8) INVOLUTION OF POLYNOMIALS. 231 Hence to find the cube of a trinomial we have the following Rule. Write the cube of each term, together with three times the product of the square of each term by the sum of the other two, and six times the product of the three terms. Formulae (1), (2), and (3) may be used for obtaining the squares and cubes of any polynomial expressions, as ex- plained in Art. 109. Thus, if we require (1 — 2x + 3a; 2 ) 2 , iu formula (1) we put 1 for a, —2x for b, and 3a: 2 for c, and obtain 1. (l-2x+3x 2 y = (l) 2 +(-2x-) 2 +(3^) 2 +2(l)(-2.i-)+2(l)(3.r)+2(-2a;)(3^) = 1 + 4a: 2 + 9a; 4 -4a;+ Gar- 1 2a: 3 = l-4a;+10a; 2 -12a; 3 +9a: 4 . Similarly by (3) we have 2. (1-2Z+3Z 2 ) 3 = (l)3 + (_2^) 3 +(3a; 2 ) 3 -f3(l) 2 (-2.T+3.« 2 ) + 3(-2a;) 2 (l+3a ; 2 ) +3(3.x- 2 ) 2 (l-2a;)+6(l)(-2a;)(3a; 2 ) = l_8x 3 +27a; 6 +3(-2.T+3x- 2 ) + 12a; 2 (l+3^) + 27a; 4 (1 -2a;) -36a; 3 = l_Ga;+21a; 2 -44a; 3 +63a; 4 -54a; 5 +27a; 6 . Show that 3. (i _ x _|_ x 2)2 „ : _ 2x + 3a; 2 - 2.x- 3 + %*• 4. (1 + 3a;+2a; 2 ) 2 = 1 + 6a; + 13a; 2 + 12a; 3 + 4a; 4 . 6. (fa- 2 - x + f) 2 = ^ - ^ + 3a; 2 - 3a; + f . 7. (1 + x + a; 2 ) 3 = 1 + 3a; + 6a: 2 + 7a- 3 + Car 4 + 3a- 5 + a 6 . 8. (1 + a; - a; 2 ) 3 = 1 + 3a - 5a; 3 +• 3a; 5 - a; 6 . 232 EVOLUTION OF MONOMIALS. EVOLUTION. 11L Evolution — Evolution of Monomials. — Evo- lution is the operation of finding any required root of a number or expression. A root of any quantity is a factor which being multiplied by itself a certain number of times produces the given quantity (Art. 13). Hence Evolution is the inverse of Involution (Art. 107). The symbol which denotes that a square root is to be extracted is \j~ ; and for other roots the same symbol is used, but with a figure called the index written above to indicate the root (Art. 13). By the Rule of Signs (Art. 36), we see that (1) any even root of a positive quantity may be either positive or negative; (2) every odd root of a quantity has the same sign as the quantity ; (3) there can be no even root of a negative quantity. Thus, (l)a x a = a 2 , and (— a)(— a)= a 2 ; therefore there are two roots of a 2 , namely, -j-a and —a. (2) ( — «)( — a) ( — a) = — a 3 ; therefore the cube root of — a 3 is —a. (3) There can be no square root of —a' 2 ; for if any quantity be multiplied by itself, the result is a positive quantity. There can be no even root of a negative quantity, because no quantity raised to an even power can produce a negative result. Even roots are called impossible roots or imaginary roots. Since the n"' power of a'" is a m " (Art. 108), it follows that the /'"' root of a'"" is a" 1 . Also, the m ih power of a product is the product of the m' h powers of its factors (Art. 108) ; hence, conversely, the m th root of a product is the product of the m"' roots of its factors. Thus, yJaTc = \Ja sjb V 7 ' 7 ; "ylab = 7« V&. EVOLUTION OF MONOMIALS. 233 Again, we have (Art. 108) (a x b y cr ....)' therefore, conversely, m Ja xm V /m 7. V^y 1 * 7 = .«yz ; 8. V-^y 112. Square Root of a Polynomial. — Since the square of a + & is a 2 + 2«6 + & 2 , the square root of a 2 -f- 2a,b -f- 6 2 is a -f- 6. "We may deduce the general rule for the extraction of the square root of a polynomial by observing in what manner a + b may be derived Iron: a 2 + 2ab + b 2 . Arrange the terms of the square according to the descend- ing powers of a ; then the first term is a 2 , and its square root is a + 2«& + b 2 |a + b a, which is the first term of the &_ required root. Subtract its square, 2a + b)2ab + 6 a a 2 , from the given expression, and 2ab + & 2 bring down the remainder, 2ab -+- b 2 . Thus, b, the second term of the root, will be the quotient when 2ab, the first term of the remainder, is divided by 2a, i.e., by double the first term of the root. This second term, 6, added to 2a, twice the first term, completes the divisor, 2a + b ; multiply this complete divisor by b, the second term, and subtract the product, i.e., 2ab + b 2 , from the remainder, and the operation is completed. If there were more terras we should proceed with a + b as we have done with a ; its square, a 2 -f 2ab + & 2 » has already been subtracted from the given expression, so we Bhould divide the remainder by twice the first term, i.e., by 2(a + b), for a new term of the root. Then for a new subtrahend we should multiply the sura of 2(« + b) and the EXAMPLES. 235 new term by the new term. The process must be continued till the required root is found. Hence to extract the square root of a polynomial, we have the following Rule. Arrange the terms according to the powers of some letter ; find the square root of the first term for the first term of the square root; place this on the right, and subtract its square from the given polynomial. Double the root already found for a trial divisor; divide the first term of the remainder by this trial divisor for the second term of the root, and annex this second term to the root and cdso to the trial divisor for the complete divisor. Multiply the complete divisor by the second term of the root, and subtract the product from the remainder. If there are other terms remaining, repeat the process until there is no remainder, or until all the terms of the root have been obtained. EXAMPLES. 1 . Find the square root of 4a; 4 - 20a; 3 + 37a: 2 - 30a; + 9 12a; 2 - bx + 3. 4a; 4 Ax' 2 — ox 20a; 3 + 37a; 2 20a; 3 4- 25a; 2 4ar - 10a; + 3 12a; 2 - 30a; + 9 12a; 2 - 30a; + 9 The expression is arranged according to the descending powers of x. The square root of 4a; 4 is 2a; 2 , and this is placed at the right of the given expression for the first term of the root. By doubling this we obtain 4a; 2 , which is the trial divisor. The second term of the root, —5a;, is obtained by dividing — 20x 3 , the first term of the remainder, by 4a; 2 , and this new term has to be annexed both to the root and divisor. Next 236 EXAMPLES. multiply the complete divisor by — 5x and subtract the product from the first remainder. We then double the root already found and obtain 4x 2 — lOx* for a new trial divisor. Dividing 12ar, the first term of the remainder, by 4a 2 , the first term of the divisor, we get 3, which we annex both to the root and divisor. We now multiply the complete divisor by 3 and subtract. There is no remainder, and the root is found. 2. Find the square root of 15a 2 x 4 - Gaz 5 + x G - 20a s x 3 + a 6 + lhah? - 6a 5 x. Arrange in descending powers of x. x 6 -6ax 5 +l5a 2 x i -20a 3 x 3 +loa 4 x 2 -6a 5 x+a e | a; 3 -3ff.r 2 +3a 2 .r-a 3 x« 2x 3 -Mx" — 6ax 5 +locrx* — Gax 5 + 9a 2 x 4 2x z -6ax 2 +3a?x 2x 8 — Qax 2 +6crx 6a 2 a 4 -20a 3 a: 3 +15a 4 ar 6a 2 x 4 -18a 3 .c 8 + 9« 4 ar - 2a 3 x- 3 + 6aV-6a 5 .?+a 6 - 2a 3 .r 3 + 6a 4 a: 2 -6a 8 a;+a 6 Note. —All even roots admit of a double sign (Art. 111). Thus the square root of a 2 + 2ab + & a is either a + b or — a — b, as may be verified. In the process of extracting the square root of a 2 + 2ab + & 2 , we begin by extracting the square root of a 2 , and this may be either a or —a. If we take the latter, and continue the operation by the ride as before, we shall obtain — a — b. A similar remark holds in every other case. Thus, in Ex. 2 the square root of the first term x 6 is either x 8 or — x z . If we take the latter, and continue the operation as before, we shall obtain — x 3 + 3a.r 2 — Za*X + a*. Since the fourth power is the square of the square, the fourth root of an expression may be found by extracting the square root of the square root. Similarly the oiijhth root may be found by three successive extractions of the square root; and so on. For example, required the fourth root of 16a: 4 - 9 Gar 5 // + 216x-y 2 - 2\(Sxif + M//'. EXAMPLES. 237 By the rule we find that the square root is Ax 2 — 12a?/ 4- 9.?/ 2 ; and the square root of this is 2x — %, which is therefore the fourth root of the given expression. 3. Find the square root of 24 16?/ 2 8x x* 32y x 1 y y 2 x ' Arranging in descending powers of y, we have I6j/ 2 ,.2 l "// 32?/ , g^ 8x , x* y y 1 - 4 x 2 X - 4 _32y + 24 X X 8 c y 8 - 8 - 8x x 2 y f 8x x 2 y y 1 Here the second term of the root, —4, is found by the rule as usual, i.e term, -, is found by dividing 8 by -^ y x by dividing ^ by — , and the third xx EXAMPLES. Find the square root of each of the following 4. 25a; 2 - 30a?/ + dif. 5. 4a; 4 - 12a 3 + 29a; 2 - 30a- + 25. 6. 1 - 10a + 27a- 2 - 10a 3 + a 4 . 7. 4a 2 + 9y 2 + 25z 2 + 12ay — SOyz — 20az. 8. 34a 3 -22a 4 +a 6 + 121a 2 - 374a + 289. Ans. 5a — 3?/. 2a 2 - 3a + 5. 1 - 5a + a 2 . 2a + 3y — 5z. a 3 -lla-f 17. * The reason for this arrangement will appear in Chap. XIII. JLns. h 2. 32/ r t 4 . x' 2 , a 3 .a 2 4 a 2 05 or - 2. a 2 a a; 2 x a 238 SQUARE ROOT OF ARITHMETIC NUMBERS. 10. 113. Square Root of Arithmetic Numbers. — The rule which is given in Arithmetic for extracting the square root of a number is based upon the method explained in Art. 112. Since 1 = l 2 , 100 = 10 2 , 10000 = 100 2 , 1000000 = 1000 2 , and so on, it follows that the square root of a number between 1 and 100 is between 1 and 10; the square root of a number between 100 and 10000 is between 10 and 100 ; the square root of a number between 10000 and 1000000 is between 100 and 1000, and so on. That is, the square root of a number of one or two figures consists of only one figure ; the square root of a number of three or four figures consists of two figures ; the square root of a number of five or six figures consists of three figures ; and so on. Hence the Rule. If a point is placed over every second figure in any number, beginning with the units' place, the number of points toill show the number of figures in the square root. Find the square root of 5329. Point the number according to the rule. Thus, it appears that the root consists of two places of figures, i.e., of tens and units. Let a denote the value of the figure in the tens' 5321) (70 + 3 = 73. place of the root, and b that 4900 in the units' place. Then a must be the greatest multiple of 10 whose square is less than 5300; this we find to be 70. Subtract a 2 , i.e., the square of 70, from the given number, and the remainder is 429, which must equal (2a + b)b. Divide tins remainder by the 140 + 429 429 SQUARE ROOT OF ARITHMETIC NUMBERS. 239 trial divisor, 2a, i.e., by 140, and the quotient is 3, which is the value of 6. Then the complete divisor, 2a + 6, is 140 + 3 = 143, and (2a + 6)6, that is, 143 x 3 or 429 is the number to be subtracted ; and as there is now no remainder, we conclude that 70 + 3 or 73 is the required square root. In squaring the tens, and also in doubling them, the ciphers are omitted for the sake of brevity, though they are understood. Also the units' figure is added to the double of the tens by merely writing it in the units' place. The actual operation is usually performed as follows : If the root consists of three places of figures, let a repre- sent the hundreds and 6 the tens ; then hav- ing obtained a and 6 as before, let a represent 5329(73 the hundreds and tens as a new value ; and 49 find a new value of 6 for the units ; and in 143 \ 429 general, let a represent the part of the root 429 already found. Hence for the extraction of the square root of a number, we have the following Rule. Separate the given number into periods of two figures each, by pointing every second figure, beginning at the units' place. Find the greatest number ivhose square is contained in the left period, and place it on the right; this is the first figure of the root ; subtract its square from the first period, and to the remainder bring down the next period for a dividend. Double the root already found for a trial divisor, and see how many times it is contained in the dividend, omitting the last figure, and annex the result to the root and cdso to the tried divisor. Multiply the divisor thus completed by the figure of the root last obtained, and subtract the product from the dividend. If there are more periods to be brought down, continue the operation as before, regarding the root already found as one term. 240 SQUARE ROOT OF A DECIMAL. Extract the square root of 132496, and 1024G401. 1. 132496(364 2. 10246401(3201 9 9 66)424 62)124 396 124 724)2896 6401)6401 2896 6401 As the trial divisor is an incomplete divisor, it is sometimes found that the product of the complete divisor by the corre- sponding figure of the root exceeds the dividend. In such a case the last root figure must be diminished. Thus, in Ex. 1, after finding the first figure of the root, we are re- quired by the rule to divide 42 by 6 for the next figure of the root, so that apparently 7 is the next figure. On multi- plying however 67 by 7 we obtain a product which is greater than the dividend 424, which shows that 7 is too large, and we accordingly try 6, which is found to be correct. The student will observe in Ex. 2 that, in consequence of the dividend, exclusive of the right hand figure, not contain- ing the trial divisor, 64, we place a cipher in the root and also at the right of the trial divisor 64, making it 640 ; we then bring down the next period and proceed as before. 114. Square Root of a Decimal. — The rule for extracting the square root of a decimal follows from the rule of Art. 113. If any decimal be squared there will be an even number of decimal places in the result; thus (.2o) 2 = .0625, and (.111)- = .012321. Therefore there cannot be an exact square root of any decimal which has an odd number of decimal places. The square root of 32.49 is one-tenth of the square root of 3249. Also the square root of .0361 is one-hundredth of that of 361. Hence, for the extraction of the square root of a decimal, we have the following SQUARE ROOT OF A DECIMAL. 241 Rule. Separate the given number into periods of two figures each, by putting a point over every second figure, beginning at the units' place and continuing both to the right and to the left of it; then proceed as in the extraction of the square root of in- tegers, and point off as many decimal places in the result as there are periods in the decimal part of the proposed number. If there be a final remainder in extracting the square root of an integer, it indicates that the given number has not an exact square root. We may in this case place a decimal point at the end of the given number, and annex any even number of ciphers, and continue the operation to any desired extent. We thus obtain a decimal part to be added to the integral part already found. Also, if a decimal number has no exact square root, we may annex ciphers and obtain decimal figures in the root to any desired extent. Find the square root of 12 ; and also of .4 to three deci- mal places. 12.000000(3, 9 64)300 256 ,464 .400000(.632 36 123)400 369 686)4400 4116 1262)3100 2524 6924)28400 27696 Note. — We see here in what sense we can be said to approximate Vo the square root of a number. The square of 3.464 is less than 12, and the square of 3.465 is greater than 12. Also the square of .032 is less than .4, and the square of .633 is greater than .4. No fraction can have a square root unless the numerator and denominator are both square numbers when the fraction is in its low- est terms. But we may approximate to the square root of a fraction to any desired extent. Thus, 242 SQUARE ROOT OF A DECIMAL. Let it be required to find the square root of -f . Here (Art. Ill) i/| = zjL Therefore we find the square root of 5 and also of 7, ap- proximately, aud divide the former by the latter. Or, we may reduce the fraction f to a decimal to any required degree of approximation, and obtain the square root of this decimal. Otherwise thus : ~yTl V^5 x 7 055 . vl-vf X 7 yj-t x 7 7 then find the square root of 35 approximately, and divide the result by 7. Either of these last methods is preferable to the first. If the square root of a number consists of 2n + 1 figures, when the first n + 1 of these have been obtained by the ordinary method, the remaining n may be obtained by division. Let N represent the given number; a the part of the square root already found, i.e., the first n + 1 figures found by the rule, with n ciphers annexed; and x the part of the root which remains to be found. Then y/jv = a + x; N = a' 2 + 2ax + x 2 ; .-. £=* = .+ * (i) 2a 2« Now N — a- is the remainder after n + 1 figures of the root, rep- resented by a, have been found; and 2a is the corresponding trial divisor. We see from (1) that N — a 2 divided by 2a gives x, the rest x 2 of the square root required, increased by — . Now — is a proper fraction, so that by neglecting the remainder 2« arising from the. division, we obtain x, the rest of the root. For, X contains n figures by supposition, so that x 2 cannot contain more than _'// figures ; but a contains 2n + 1 figures (the last n of which are ciphers) and thus 2a contains 2n + I figures at least; therefore ^- is a proper fraction. Prom ibis investigation, by putting n = 1, we Bee thai al leasl two of the figures of ;i square root must have been obtained in order that CUBE ROOT OF A POLYNOMIAL. 243 the method of division may give the next figure of the square root correctly. We will apply this method to finding the square root of 290 to five places of decimals. We must obtain the first four figures in the square root by the ordinary method ; and then the remaining three may be found by division. 290 (17.02 1 27) 190 3402) 10000 6804 3196 We now divide the remainder 3196, which is N — a 2 , by twice the root already found, 3404, which is 2a, and obtain the next three figures. Thus, 3404) 31960 (938 13240 10212 30280 27232 3048 Therefore to five places of decimals, y/090 = 17.02938. In extracting the square root, the student will observe that each remainder brought down is the given expression minus the square of the root already found, and is therefore in the form 2V — a 2 , EXAMPLES. Find the square roots of the following numbers : 5. .835396. Am. .914. 6. 1522756. 1234. 7. 29376400. 5420. 8. 384524.01. 620.1. 115. Cube Root of a Polynomial. — Since the cube of a + b is « 3 -f Serb + oab 2 + ft 8 , the cube root of a 3 + Serb + Sab' 2 4- b 3 is a + b. We may deduce a general rule for the extraction of the cube root of a poly- 1. 15129. Ans. 123. 2. 103041. 321. 3. 3080.25. 55.5. 4. 41.2164. G.42. 244 CUBE ROOT OF A POLYNOMIAL. nomial by observing in what manner a -f- b may be derived from a 8 + Serb + Sab 2 + b 3 . Arrange the terms of the cube according to the descending- powers of a\ then the first term is a 3 , and its cube root is a, which is the first term of the required root. Subtract its cube, a 8 , from the given expression, and bring down the remainder 3a 2 & + Sab' 2 -+- b 3 . Thus, o, the second term of the root, will be the quotient when 3a 2 o, the first term of the remainder, is divided by 3a' 2 , i.e., by three times the square of the first term of the root. Also, since Sa 2 b + 3a& 2 + b 3 = (3a 2 + 3a?> + & 2 )o, we add to the trial divisor Sab -f & 2 , i.e., three times the product of the first term of the root by the second, plus the square of the second, and we have the complete divisor 3a 2 + Sab -f- b' 2 ; multiply this complete divisor by 6, and subtract the product, Sarb -j- Sab' 2 + b 3 , from the remainder, and the operation is completed. The work may be arranged as follows : a 3 + 3a 2 6 + oab 2 + b 3 \a + b a 3 3a 2 + Sab + V 2 S,rb + Sab 2 + b 3 :\a 2 b + Sab 2 + b 3 If there were more terms, we should proceed with a + b as we have done with a ; its cube, a 3 + :), CUBE ROOT OF ARITHMETIC NUMBERS. 247 which is the value of b. Then adding Sab, or 1200, and 6 2 , or 25, to the trial divisor 3a 2 , or 19200, we obtain the com- plete divisor 20425 ; and multiplying the complete divisor by 5 and subtracting the product 102125, there is no remain- der. Therefore 85 is the required cube root. 614125 1 80 + 5 512000 ~~ 3a 2 = 3(80) 2 = 19200 3ab = 3(80) (5) = 1200 b* = (5) 2 = 25 20425 102125 102125 In cubing the tens the ciphers are omitted for the sake of brevity, though they are understood. If the root consists of three places of figures, let a repre- sent the hundreds and b the tens, and proceed as before. See Art. 113. Hence for the extraction of the cube root of a number, we have the following Rule. Separate the given number into periods of three figures each by pointing every third figure, beginning at the units' place. Find the greatest number whose cube is contained in the left period, and place it on the right; this is the first figure of the root; subtract its cube from the first period, and to the re- mainder bring down the next period for a dividend. Take three times the square of the root already found for a trial divisor, and see hoiv many times it is contained in the dividend, omitting the last tivo figures, and annex the residt to the root. Add together, the trial divisor ivith two ciphers annexed; three times the product of the last figure of the root by the rest, with one cipher annexed; and the square of the last figure of the root. 248 CUBE ROOT OF A DECIMAL. ■ Multiply the divisor thus completed by the figure of the root last obtained, and subtract the product from the dividend. If there are more periods to be brought cloivn, the operation must be repeated, regarding the root already found as one term. Extract the cube root of 109215352. 109215352(478 64 3(4) 2 = 4800 3(4) (7) = 840 ( 10*»+ 1 : x 2 a . 10 2 » < Tu- Also X s 3a 2 IO 3 " or < 1 < 3 x 10 4n + 2 3.x 10" + U X 2 a X 8 + 3^<^ + 1 -a' 3 x 10^ and is therefore a proper fraction. EXAMPLES. Show that 1. (-2a 5 ) 4 = 16a 20 . 2. (-a 4 ) 5 = -a 20 . 3. (-3aW) 3 = -27 = \x- ; .r;r a 2 b s x~hj- a ~ ~b ~ 3 xy 27-1 =JL = 1 8 T- = i = i( b y( 8 )° fArt - 118 )- 27* V27 2 V3 6 3" Otherwise thus 27* (V27) 2 3 2 (3) To prove that a" 1 -~ a" = «'" _n /or a?Z values of m and n. a m -7- a n = — = a'" X a - " = a m ~", by the fundamental law. Examples, a 3 -=- a 5 = a 3-5 = a - a -=- a - J = a 1+ 5 = a" 3 ' . 2a* X a* x6a _ 5 9a~§ X a* V^ 3 x vV 2 | a i+l-5+f-i _ | ( 3a a* x ^ Vy~- X Vx 6 y •'* x a$ x$ $yi+-s =z x°y = y. Note. — It appears that it is not absolutely necessary to introduce fractional and negative exponents into Algebra, since they merely supply us with a new notation in addition to one we already had. It is simply a convenient notation, which the student will learn to appreciate as he proceeds. EXAMPLES. Express with positive exponents : i 3 z 1. 2x hi ■'. Ans. ~t~^"- 2x* X 3.):- 1 V* 8 3. ^X^X^ ina>g y 4 . vsr^ + VS. -, TO PROVE THAT (a m ) n = a mn IS UNIVERSALLY TRUE. 2o7 Express with radical signs : 5. bcftx $b %. 6. a - * x 2a' A, Wa_ \J.vb* _2_ Va" 5 " 7. a; - ' -r- 2a" 8. 7a~8 x 3a -1 . Ans. 21 120. To Prove that (a»)" = a""' is Universally True for All Values of m and n. 1. iei n be a positive integer, and m have any value. Then from the definition of a positive integral exponent (a'")" = a"' x a" 1 x a m x to n factors qJU + m +vi + to n terms = a mn (1) 2. Let n be a positive fraction *-, where p and q are positive integers, and m unrestricted as before. Then (a'") n = (a m y = V («"')" (Art. 118) = VaT p by (1) = aJ (Art. 118) = «"■" (2) 3. Let n be negative, and equal to —p, where p is a positive integer, and m unrestricted as before. Then («'»)" = («»)-i> = 7-^( Art ' 119 ) = — by (1) and (2) a mp J V t V * = a-™* = «•»» (3) Hence (a m ) n = (a") m = a"" 1 for all values of m and n. 4. Let n = -. Then we have (a m )" = (a") m = a" (4) 258 TO PROVE THAT ((lb)" = a"b" FOR ANY VALUE OF n. That is, the n' h root of the m' k power of a is equal to the m th power of the n th root of a. 5. Let m = — and n = -. m n 11 11 _L_ (a"') n = (a")™ = a mn (5) That is, the n' h root of the m' h root, or the m' h root of the n tA root of a is equal to the mn th root of a. Examples. (b*)$ = fi3 x ? = &'. [>- 2 ) 3 ]-3 = (x- 6 )-5 = a- 6 (-j) = x\ (fry = (3*) 3 = 3* = ^3. Vv/27^ 3 = [(27a; 3 )*]* = [(27a 8 ) -3]* = ^3x. 121. To Prove that (ab)" = a"b" for Any Value ositive fraction £_, lohere p and q a/e Q positive integers. Then (ab) n = (a&)«. Now [(«&)*]' = («&)' (Art. 120) = a p b" (Art. 108) p p = (a«&«)«. P 21 1! .-. (ab) q = aW () 2. Let n have any negative value, say —r, where i >s a positive integer. Then (ab)" = (ab)- r = — !— K > K (aby -JL-o-&-%. ... (2) a'b' which proves the proposition generally. EXAMPLES. 259 Iu this proof the quantities a and b are wholly unrestricted^ and may themselves have exponents. P _ 1 Let — — -. Then from ( 1 ) we have q n v ' i 11 (ab) n = a*b\ r. \ab = \a • n yjb (3) That is, the n th root of the product is equal to the product of the n' h roots. EXAMPLES. 1. («%-*)* -^ {xhj- 1 )-* = x$y~% -f- x~^y-> = xhj- 1 . Express with positive exponents Ans, 8. yab-h-* x (a-^-^j-*) -*. 9. Va^ X (a%- a )" ft . 1 4 9a 2 a; 2 ' 16ac 4 . 5. (aty-»)«(»V)- Ans. 6. y/x^x-K 7. (4a 9a 2 )- 1 .2a + 36* xK 3a.r 2 ' :,;'' 10. (a~*Va;)- 8 x ^x-*Jar\ 11. ^(a + 6) 5 X (a + &)-t. a + b. Rem. — Since the laws of the exponents * just proved are universally true, all the ordinary operations of multiplication, division, involution, and evolution are applicable to any expressions which contain frac- tional and negative exponents. (Jailed the index Itnvs. 2G0 EXAMPLES. The reason for the arrangement in Ex. 3, Art. 112, may now be seen. Thus the descending powers of x are .3 .2 ,111 X ■, X . X. 1, , — -, — , as may be seen (Art. 119) by writing the terms as follows : 12. Multiply 3x-? + x + 2xi by xs — 2. Arrange in descending powers of x. x -f 2xi + 3a; ~* a;* - 2 xs + 2x + 3 — 2x — 4xt — 6a; - * *t - 4o;t + 3 - 6a;-*. 13. Divide 3a;*2r*+B*— 3a?*2T*— 2T* by aj*+2T*— 2a;*#-*. Arrange in descending powers of x. x \ — 2xhj~l + y~*[ a;* — 3o5%-* + 3a;*?/-* — y~^ \ a* — y~s »i — 2aj%~6 + xhj-'t — x$y-$ + 2#*y~* — y* — aj%-i + 2x$y~i — y* Multiply 14. x§ + y\ by a$ — ?/*. j4ns. a$ — yt. 15. :« 4 + x 1 + 1 by a;- 4 - x~ 2 + 1. x 4 +l+ a~ 4 . 16. a -3 + a"* + 1 by a - * — 1. a" 1 — 1. Divide 17. 21a; + a$ + a* + 1 by 3a;* + 1. 7a;* - 2a;* + 1. 18. 15a— 3a*— 2a-*+8a _1 by 5a*+4. 3a*— Sarb+Var 1 . Find the square root of 19. 9a; - 12a;* -f 10 — 4a;-* + ar*. 3aJ* - 2 + .<■-'. 20. 4aJ" -f 9a;-" + 28 — 2Ax~* — Km:". 2.r" - 4 + ^ _ ^- SURDS — DEFINITIONS. 261 SURDS (RADICALS). 122. Surds. Definitions. — When the indicated root of a quantity cannot be exactly obtained, it is called an irra- tional quantity or a Surd. Thus, ^2, \^i, Va 3 , sja 2 + b 2 , a$, are surds. When the indicated root can be exactly obtained, it is called a rational quantity. Thus Vx*, ^9, Va 4 , are rational quantities, though in the form of surds. The order * of a surd is indicated by the index of the root. Thus 3 y/x, y/a are respectively surds of the third and fifth orders. The surds of the most common occurrence are those of the second order ; they are sometimes called quadratic surds. Thus \fs, Va, \lx + y are quadratic surds. Surds of the third and fourth orders are called cubic and biquadratic surds respectively. When the same root is required to be taken, the surds are said to be of the same order. Thus, Va, Va + &, and 5? are all surds of the third order or cubic surds. Surds are said to be like or similar when they are of the same order, or can be reduced to the same order, with the same quantity under the radical sign. Thus, 4^7 and 5^7 are like or similar surds ; also 5 3 /2 and 3Vl6 are like surds ; 2^3 and 3\/2 are unlike surds. A mixed surd is the product of a rational factor and a surd factor. Thus 4\/5, and 3^7 are mixed surds. When there is no rational factor outside of the radical sign, the surd is said to be entire. Thus \f% and V^ are entire surds. The rules for operating with surds follow from the propo- sitions of the preceding Articles of this Chapter. * Sometimes culled degree. 262 TO REDUCE AN ENTIRE TO A MIXED SURD. 123. To Reduce a Rational Quantity to a Surd Form. — It is often desirable to write a rational quantity in the form of a surd. Thus a =y^ = V«? = Vc?! 3 = yft = y/27. In the same way the form of any surd may be altered. Hence the Rule. Any rational quantity may be expressed in the form of a surd of any required order by raising it to the power corre- sponding to the root indicated by the surd, and prejixing the radical sign. Examples. 5 = y/25 = yl25 = y5 n . (a + b) = \l(a + by 3 = V(« + b) 3 . 124. To Introduce the Coefficient of a Surd under the Radical Sign. — We have 3^2 = fi X fe (Art. 123) = y/'J x 2 [(3) of Art, 121] = \Zl8. 2V5 = yip X VIF- V2 8 X 5 = VIo. ccy^a — x = sj'laxr — a- 3 . Rule. Reduce the coefficient to the form of the surd and then multiply the surds together. EXAMPLES. Express as entire surds. 1. 11^2. Aus. y/242. 3. 14\':>. Ans. y/'J.sO. 2. 5 ye. yiEo. 4. cyi ysci. 125. To Reduce an Entire to a Mixed Surd. We have y/32 = y/i6 x 2 = \/T<; x\/2= iy/2 ; also yaJb = V^" X yi = ir\b. REDUCTION OF SURDS TO EQUIVALENT SURDS. 263 Rule. Resolve the quantity under the radical sign into two factors, one of which is the greatest perfect power correspond- ing to the root indicated; extract the required root of this factor, and prefix the result as a coefficient to the indicated root of the other. When a surd is so reduced that the smallest piossible integer is left under the radical sign, it is said to be in its simplest form. Thus, The simplest form of \Jl28 = ^64 x 2 = 8^2, EXAMPLES. Express in the simplest form : 1. \/288. Ans. 12^2. 3. y/36o~ 3 . Ans. 6a^a. 2. y/1029. 7y/3. 4. \/27a 3 b 5 . 3ab°-\j3ab. 126. Reduction of Surds to Equivalent Surds.— To reduce surds of different orders to equivalent surds of the same order. For example, take y/5 and y/ll. y/5 = 5^ = 5* = y~5~ 3 = Vl25. yn = n$ = lit = yn 5 = ym. In general, let n \Ja m and q yjb p be two surds of different orders. Then we have to change these into equivalent surds whose fractional exponents have the same denominator. We can reduce both surds to the order nq as follows : y a m = a? = a"* — '"y/cr", p p? and \b p = b q = b qn = %/&*». Thus the equivalent surds of the same order are "\/«" !9 and Rule. Represent the surds with fractional exponents; re- duce these fractions to their least common denominator ; then express the resulting fractional exponents with radical signs, 2G4 ADDITION AND SUBTRACTION OF SURDS. and reduce the expressions under the radical signs to their simplest forms. Note. — In this way, surds of different orders may be compared. Thus, if we wish to know which is the greater, V^ or |*Kll, we have only to reduce them to the same order, as above; we see that the former is greater because 125 is greater than 121. EXAMPLES. Express as surds of the twelfth order, with positive ex- ponents : Ans. l y/a; 4 . 1 v S[af Express as surds of the same lowest order : 1. xK 2. a- 1 -:- ar 1 3. — • a * 4. xk Ans. \a?. 5. yx^r. i s/x^i. G. v&x V«" ljr " 2 - Va 7. y/a,V^ Ans. \/a 9 , 'y/a™. 8. v^v^- W,W- d.^x^y/x 10 . Ans. v^, 1 ^ . 10. y/cW,yab. ^M^y/aW. 11. Which is the greater \/l4 or \/G? *6. 127. Addition and Subtraction of Surds. — Let it be required to find the sum of ^12, ^7o, -\J±8, and fit). Here we have (Art. 125) \!\2 + \jTb - y/te + \J5Q = 2\/S + 5\/3 - 4\/3 + 5^2 = (2 + 5 - 4)v/3 + 5^2 = 3\/3 + 5y/2. Rule. .Reduce the surds to their simplest form; then add or subtract the coefficients of similar surds and prefix the result to the common surd, and indicate the addition or subtraction of unlike surds. Thus, 3y/20 + 4y/5 + fa + y/75 = Gy/5 + 40i + l\ ! -i + 5^3 - 10^' + 5^3. MULTIPLICATION OF SURDS. 2G5 EXAMPLES. Find the value of the following : 1. 3^45 + 7^5 - y/20. Ans. 14^5. 2. 4033 + 5\Il - 8^28. V^- 3. y/44 - 5\/l76 + 2^99. -12y/ll. 4. 2^303 - 5^243 + y/l92. -15^3. 5. 2VI + 8VS- 3 V 2 - g. y4o - iV 320 + V 135 - 3 V 5 - 128. Multiplication of Surds. — (1) When the surds are of the same order. To multiply a n y/x by b n \Jy. Here a"y/x x* b n y/y = ax n x &#" [Art. 118, (5)] ii i = abx n y n = ab(xy) n (Art. 121) = ab'yjxy. (2) T7/ie» the surds are of different orders. To multiply a\x by by/y. _ _ i 2. Here ayas X by/y = ax n x 6?/ m = abx""'y""> (Art. 126) = a6(«"y ( )'"" (Art. 121) = ab m ylx^Y. Rule. When the surds are of the same order, multiply sep- arately the rational factors and the irrational factors. When the surds are of different orders, reduce them to equivalent surds of the same order, and proceed as before. Thus, 3\/2 x 7v/6 = 21^12 = 42^3. 5^2 x 2^5 = 5V2*X 2^ = loysoo. A compound surd is an expression involving two or more 266 MULTIPLICATION OF SURDS — EXAMPLES. simple surds. Thus 203 — 3\Jb, and 3 0t + V& are compound surds. The multiplication of compound surds is performed like the multiplication of compound Algebraic expressions. Multiply 2Va; — 5 by 30c. The product = 3\/^(20c - o) = Gx - 150k. Note. — To multiply a surd of the second order by itself is simply to remove the radical sign ; therefore \x x \Jx = x. Multiply 603 - 5 0> by 205 -|- 302. The product = (603 - 5y/2) (203 + 302) = 36 + 18 ft - 1003 - 30 = 6 + Sy/S. The following case of the multiplication of compound surds deserves careful attention. The product of the sum and difference of any two quadratic surds is a rational quan- tity. Thus (303 + 403) (303 - 403) = (3 ft? - (4y/3) 2 = 45 - 48 = -3. Also (\Ja + 00(0*" - 00 = (ft) 2 ~ (ft) 2 = a - b. A binomial in which one or both of the terms are irra- tional, is called a binomial surd. When two binomial quadratic surds differ only in the sign which connects their terms, they are said to be conjugate. Thus 30* + 4y/3 is conjugate to 3\/5 — 403. Similarly, a — 0t 2 — b' 2 is conjugate to a + 0r — b 2 . The product of two conjugate surds is always rational. EXAMPLES. Find the value of 1. 2014 x 0>T. Ans. 1405. 2. 80$ X 03. 1203. 3. 2)JTr> x 80S 3003. 4. yi68 x 0147. Ans. 14*0). 5. 0* X \ 2. V 10 *- 6. V^2xV8xVixVI« V 2 - TO RATIONALIZE THE DENOMINATOR OF A FRACTION. 267 7. (30c — 5) X 2y/a\ Ans. 6x — 100c. 8. (V® — sja) x 20c. 2a; — 20X8. 9. (07 + 503) (2^7 - 403). 60*1 - 46. 10. (30i - 400(203 + 302). 6 + 010. 11. (5 + 802) (5 - 302). 7. 12. (30t + 0b — 9a)(30» — 0c — 9a). 18a — x. 129. To Rationalize the Denominator of a Frac- tion. — The process by which surds are removed from the denominator of any fraction is known as rationalizing the denominator. (1) When the denominator is a monomial. _2_ _ 203 _ 203 03 03 x 03 3 '2 X 9 l3 /l8 0L8 *W 3x9 V 27 3 Rule. Multiply both terms of the fraction by any factor which will render the denominator rational. (2) When the denominator is a binomial quadratic surd. b 2 Rationalize the denominator of > „ ■ y/a 2 + b 2 + a „, . b 2 y/a 2 + b 2 - a The expression = -== X , y/ a 2 + & 2 + a y/ a 2 + & 2 _ a = yfy/a" + fr 2 - a] = ^r^i _ a< (a 2 + &' 2 ) - a 2 Rule. Multiply both numerator and denominator of the fraction by the surd which is conjugate to the denominator. When the denominator of the fraction is rationalized, its numerical value can be more easily found, Thus, the numeri- - 2 cal value of f 03 can be found more easily than that of "7=* 268 DIVISION OF SURDS. Given ^5 = 2.2360G8, find the value of 29 2^/5 It might seem at first sight that we must subtract twice the square root of 5 from 7, and divide 29 by the remainder — a troublesome process, as the divisor would have 7 figures. We may avoid much of this labor by rationalizing the de- nominator. Thus, 29 = 29(7 + 2y/5) 7 - 2v/5 49 - 20 = 11.472136. = 7 + 2^5 EXAMPLES. Rationalize the denominators of V3 1. V» 2. \/f- 3. \Ti- 4. VI- 5 2 + y/5 V/5- - 1 Ans. W 15 - We- 7 + 3y/5 7^3 — 5^/2 iOy/6 - 2y/7 3\/6 + 2y 7 ' 8 - y/42. y/7- y/2 3 y/7 + y/_2_ _ 9 + 2^14 5 Since a n yjx x &>A/ = ab^Jxy 130. Division of Surds. (Art. 128), therefore ab n yjxy ■+■ a n \]x = b\y. Rule. When the surds are of the same order, divide separately the rational factors and the irrational factors. When the surds are of different orders, reduce them to equivalent surds of the same order, and proceed as before. Then the denominator may be rationalized (Art. 129). 4^75 = 4 X 5y/3 25^56 25 X 2^14 Thus, 4^75 -=- 25y/56 = = «Vft = ^l X 14 X 14 y/42 35' BINOMIAL SURDS — IMPORTANT PROPOSITIONS. 269 The only case of division of a compound surd which we shall consider is that in which the divisor is a binomial quadratic surd. We may express the division by means of a fraction, and then practically effect the division by ration- alizing the denominator. Thus, Divide y/3 + y/2 by 203 - 02. y/3_+ v/2_ = (y/3 + y/2) (203 + y/2) 203 - y/2 (203 - y/2) (2^3 + y/2) 8 + 30 3 ___ 8 + 303 12 — 2 10 The quotient EXAMPLES. Find the value of 1. 210384 -*- 80)8. J.ns. 3 y/3. 3. -I3y/l 25 -*- 5y/65. -4ns. — y/13 2. 5y/27-3y/24. ^. 4. 6014 -=- - 2021. y/(5. 5. 29 -j- (11 + 307). 11 - 307 2 6. (3^2 - 1) - (302 + 1 )• 19 - 602 17 7. (2y/3 + 700 -=- (5y/3 - 8. (2* - 0^) - (20^ - - 4y/2). y)- 2 + 03 131. Binomial Surds. Important Propositions. (1) The square root of a rational quantity cannot be partly rational and partly a quadratic surd. If possible, let y/a = b + y/c. Squaring, we have a = & 2 -f- 260; + c. V 26 that is, a surd is equal to a rational quantity, which is impossible. 270 SQUARE ROOT OF A BINOMIAL SURD. (2) In any equation consisting of rational quantities and quadratic surds, the rational parts on cadi side are equal, and als<> the irrational parts. If x + V y = « + \b, then will x = a, and y = b. For if x is not equal to a, suppose x = a + m ; then a + m + y/?/ = a + y'ft ; that is, m + \Jy = \Jb, which is impossible by (1). Hence x = a, and therefore s/y = sjb. Note. — When x + \fy = a + sjb, we can conclude that x = a and sjy = \l> only when \Jy and Sjb are really irrational. We cannot, for example, from the relation 6 + V'-A = 5 + V^, conclude that 6 = 5 and s/l = sl9. (3) If >Ja + s/b = sjx + sjy, then \a - sfb = Va: - \ly. For by squaring the first equation we have a + sjb = a; + y + 2^. a = x + ?/, and yo = 2\xy. Subtracting, a — sjb = x — 2sjxy -+- # ; .-. V a — \lb = \Jx - \ly. 132. Square Root of a Binomial Surd. — The square root of a binomial surd, one of whose terms is rational, may sometimes be expressed by a binomial, one or each of whose terms is a quadratic surd. Let a + y b be the given binomial surd. Assume Va + \b = yx + sjy. By (3) of Art. 131, Va - \lb = sfx - sjy. . Multiplying (1) by (2), ya 2 - b = x - y. . . Squaring (1) a + sjb = x + 2\'.ry + y. Therefore by (2) of Art, 181, a = x + y. . . , (1) (2) (3) (4) (5) SQUARE HOOT OF A BINOMIAL SUED. 271 Hence, from (3) and (5), b}' addition and subtraction, we have _ a + y a 2 2~~ (6) a — va 2 — b ,„. y = — 2 — ' (7) which substituted for x and y in (1) and (2) will give the values of \a + s/b and \ a — ^b. Find the square root of 16 + 2^55. Here a = 16, and ^b = 2^55. Then a 2 — b = 256 - 220 = 36, which in (6) and (7) gives x = ±(16 + 6) = 11. y = |(16 - 6) = 5. Hence Vl6 + 2055 = *Jli + 05. From the values of x and y in (6) and (7), it is clear that each of them is itself a complex surd unless \cr — b is rational ; and the exp ression \jx -f \Jy will be more compli- cated than \a + \lb itself. Hence the above method for finding the square root of a + ^b fails entirely unless a 2 — b is a square number ; and as this condition is not often satis- fied, the process has no great practical utility. The square root of a binomial surd may often be found by inspection. For we see from (4) and (5) that we have to find two numbers whose sum is a and whose product is 6 ; and if two rational numbers satisfy these conditions, they can generally be found at once by inspection. Thus 1. Find the square root of 11 + 2030. We have only to find two numbers whose sum is 11, and whose product is 30_; and these are, ev idently 6 and 5. Hence 11 -f 2030 = 6 + 2^6 X 5 + 5 _ = (^6 + fo)\ .'. y/6 + V 5 = the square root of 11 + 2030. 272 EQUATIONS INVOLVING SURDS. 2. Find the square root of 53 - 12^10. We must write the binomial so that the coefficient of the surd is 2. Thus 53 - 12v/l0 = 53 - 2\/360. The two numbers whose sum is 53 and whose product is 3G0 are 45 and 8. Hence 53 - 20560 = 45 - 2^45 X 8 + 8 __ = (\/45 - >Jzy. .-. ^53 - 12V/10 = ^45 - \J% = 3\/5 - 2V^2. EXAMPLES. Find the square root of 1. 7 + 2v/l0. Am. s/J + >J2. 2. 13 + 2y/30. )/l0 + y/3. 3. 5 + 2^ \Jj + l/2. 4. 47 - 4^33. 2^11 - 0*- 5. 15 + 2^56. Y^ + \ft> The c*i&e root of a binomial surd may sometimes be found by a method similar to the one just given for obtaining the square root. But the method is very imperfect, and is of no practical importance. 133. Equations Involving Surds. — Equations some- times occur in which the unknown quantity appears under the radical sign. In the solution of such equatious, special artifices arc often required. We shall here consider only a few of the easier cases, which reduce to simple equations. These can generally be solved by the following Rule. Transpose to one member of the equation a single radical term so it will stand by itself; then on raising each member to a power of the same degree as the radical, it will disappear. If there are still radical terms remaining, repeat the process till all are removed. EXAMPLES. 273 EXAMPLES. 1. Solve 2\/x — ^Ax - 11 = 1. Transposing, 2\Jx — 1 = ^Ax — 11. Squaring, Ax — A\Jx + 1 = Ax — 11. Transposing and dividing by — 4, ^x = 3. .*. a; = 2. Solve V^a? — ^1 — x + \/x = Transposing, \x — ^1 — a = Squaring, \Jx. 2)Jx + ». 4y/cc + 4a;. a; — y 1 — x = Canceling x and squaring, 1 — x = Transposing and squaring, 25a- 2 = 16a;. Dividing by 25a;, x = ^§. When radicals appear in a fractional form, the equation should be first cleared of fractions in the usual way before performing the involution. 3. Solve 6Vfc-ll = 2 V^ + 1. 3\/x \jx + 6 Clearing of fractions 6a; + 25y/a; - 66 = 6x + S)Jx. .-. 22y/a; = 66. .-. x = 9. Solve the following equations. 4. \jx - 5 == 3. Ans. 14. 5. SjAx - 7 = 5. 33. 6. y/5a; - 1 = 2\jx + 3. 13. 7. 13 - y/5a; 4 = 7, 203 - 7a; - 3^8x - 12 VI + ^3 + ^6x = 2. 44. f 6. 274 EXAMPLES. EXAMPLES. 1. Multiply a~% by a. EXAMPLES. 277 62. ;~>yi28 x 2y/432. ^ws. 240 \/4. 63. 3 y 6 x 4 y/«. 12 'y/aW. 64. y/2 x y§ x v*- J y/648000. 65. 4y/5 x 2'Vii. 8V15125. 66. 2^*V£ 2y/2 a 67. (2y/3 + 3y/2) 2 . a; — 30 + 1203. 68. (\/a? + V- c - 1)V*- !• 1 + y/a; 2 - a;. 69. (y/a; + a — y^a; — a) y/a; + a. x + < % — y/a; 2 — a 2 . 70. (y/2 + y/3 - y/5) (y/2 + 0* + y/5). 2y/6. 71. (Vl2 + v/l9)(Vl2 - Vl9)- 5. 72. (a,2 + a;V /2 + l)(a»- ^2 + : 0- a; 4 4- 1. Rationalize the denominator of 73. 203 + 3y/2 5 + 2y/6 ' ^4.Wi ?. 30> - 203. 74. 2 y/5 V/5 + 03* 5 - y/T5. 7."). 3+03 3 - y/5' i(7 + 303). 76. ■ ' x — 0a: 2 - if. x + y/^ _ ^2 77. ,■ • va; 2 + a 2 — a. y/a? + a 2 + a v T 78. y/l + x 2 - y/1 - a^ i _ y/F~ 4 y/l 4- a: 2 4- y/l - a; 2 ' ^ Find the numerical value of the following to five places of decimals : 79. It, i£. Ans. 9.89957, 3.77964. y/2 y/7 80. t?, — L. , .81649, .28867. ^3 203 278 EXAMPLES. gl y/5 + Vj^ Ans ^ _ ^ = .50402. 4 + y/l5 82. V* ~ 2 _. y/5 + 2 = 4.23607. 9 - 4y/5 Find the value of 83 . sy/n + * {ft. 2y/98 7\/22 84. l^^Gj/JM ^ 5y/ll2 y/392 85. (3 + y/5) (y/5 - 2) -=- (5 - ^5) 86 5 y/« _^_ \/a + y/g y/«a y/a — y/sc y/x a — x 87> 2y/l5 +_8 _^ 8y/3 - 6y/5 . 4 + ^ 5 + y/l5 5y/3 — 3y/5 1 + L=. 2a-. a; + y/a 2 — 1 x — y/ar 89 ^ + * + ^ - _1 + V** + 1 ~ Vglzii . 2a 2 yV + i _ y/a-a _ i y/^ + 1 + y/a; 2 - 1 90. _ » + ^ y/3. 5y/3 - 2y/l2 - y/32 + y/50 Find the square root of 91. 41 - 24y/2. -dns. 402 - 8. 92. 83 + 12035. 2y/5 + 3^7. 93. 101 - 28013. 2yi3 - 7. 94. 117 - 36y/io. 6^2 - 30i. 95. 280 + 56^21. 14 + 2y/21. 96. 8 + 40}. 05 + y/2. 97. 4 - y/15. y/f - y/f- 98. 75 - 12021. 3y/7 - 20J. EXAMPLES. 279 99. 16 + 5y/7. V^¥ + V^i- 100. 6 + 2y/2 + 20* + 205. 1 + 0> + 0*- 101. 5 + y^lO - ^6 - ^15. 1 + Vf - Vt- Solve the following equations : 102. S - 20b = 4. Ans. 4. 103. G + 0b = 2^/12 + x. 4. 104. \Jx - 3 - 0c + 12 = -3. 4. 105. y/3.K + 10 - \j3x + 25 + 3 = 0. -3. 91 10G. y/3a: -f ^3x + 13 = 12 y/3a; -f- 13 107. 2 +ya; - 5 = 13. 1336 108. 0J + 2x - \j2x y/9 + 2x 109. 0» - 1 + V^ = -7=- f no. 05 - 0> - 8 = _ 2 . 9 0c - 8 111. 0» + 11 - 0c = 1- 25 112. 03a; + 4 + y/3a; - 5 = 9. 7 113. «0 - a- = &0i - x. Ans. -~^— v v a + b 114. 0» + 13 - 0b - 11 = 2. 36 115. ^~ 1 = 1 + V^- \ 3 0jaj + 1 2 a(6 4- c) 16. V a + a V a — a V a 2 — x 2 b - c 117. vVa; + 3 - */\/x - 3 = V^20b. 9 118. y/(l+a) 2 +(l— a)a;+y/(l— a) 3 +(l+a)«=2a. 8 no ./ — I — , ./ — TT ./" a 2 +6 3 +c 2 — 2a&— 26c— 2ca 119. Va+a + v a; +& = V^'- — ! ! 4c 280 PURE QUADRATIC EQUATIONS. CHAPTER XIV. QUADRATIC EQUATIONS OF ONE UNKNOWN QUANTITY. 134. Quadratic Equations. — An equation which con- tains the square of the unknown quantity, but no higher power, is called a quadratic equation, or an equation of the second degree. A Pure quadratic equation is one which contains only the square of the unknown quantity ; it is sometimes called an incomplete quadratic equation. An Adfected, or Affected,* quadratic equation is one which contains both the square and the first power of the unknown quantity ; it is also called a complete quadratic equation. Thus, 2x 2 = 50, and ax 2 + 6 = 0, are pure quadratic equations ; and 2x 2 — hx — 4, and ax 2 + bx + c = 0, are affected quadratic equations. 135. Pure Quadratic Equations. — A pure quadratic may be solved for the square of the unknown quantity by the method of solving a simple equation. Let it be required to solve x 2 — 13 . a; 2 — 5 n 4- = b. 3 10 Clear of fractions, 10.tr - 130 + 3a; 2 - 15 = 180. ... iSx 1 = 325. , .-. aj a = 2'). Extracting the square root x = ±5. • The term adfected, or affectt . 11 + 19 — 4 m . Q . . X — —10 ^ TO — IT' °' «*• Hence, for solving affected quadratic equations, we have the Rule I. Reduce the equation so that the terms involving the unknoion quantity are alone in one member, and the coefficient of x 2 is +1 ; complete the square by adding to each member of the equation the square of half the coefficient of x; extract the square root of both members, and solve the resit/Zing simple equation. Note 4. — There are other ways of completing the square of an affected quadratic, which arc convenient in special cases, and some of which will be given as we proceed; hut the method just explained is the most important, and will solve every case. AFFECTED QUADRATIC EQUATIONS. 285 Instead of going through the process of completing the square in every particular example, it is more convenient to apply the following rule deduced from formula (1) of this Article : Rule II. Reduce the equation to the general form, x 2 + px = q. Then the value of x is half the coefficient of the first power of x ivith a contrary sign, plus or minus the square root of the second member increased by the square of half the same coefficient. Note 5. —The student should use this method in practice, and become familiar with it, but at the same time be careful that he does not lose sight of the complete method. 5. Solve 36a; - 3.x- 2 = 105. Transposing, changing signs, and dividing by 3, x 2 - 12a; = -35. Therefore by Rule II, x = 6 ± \Z-35 + 36 = 1. .-. x = 6 ± 1 = 7, or 5. 6. Solve Sx ~ 2 = — 2. 2x - 3 x + 4 c ,. ,., . 3a; — 2 3a; — 8 Simplifying, = . 1 J ° 2x - 3 x + 4 Clearing of fractions, 3x 2 -\- lOx — 8 = Gx 2 — 25.v + 24. Reducing, x 2 — ^-x — — - 3 g 2 -. Therefore, Rule II, x = % 5 - ± V 7 - .-. x = 5/ ± 2.9-= lOf, or 7. Solve x 2 — 4x = 1. Rule II, x = 2 ± ^1+4 = 5. .-. x = 2 ± 2.236 = 4.236, or -0.236. These values of x are correct only to three places of decimals, and neither of them will be found to satisfy the equation exactly. 286 CONDITION FOR EQUAL ROOTS. If the numerical values of the unknown quantity are not required, it is usual to leave the roots in the form 2 + Vo, and 2 - V5. 8. Solve x 2 - 10.x = -32. Rule II, x = 5 ± V-32 + 25 = -7. .-. x = 5 ± ^7. But — 7 has no square root, either exact or approximate (Art. Ill) ; so that no real value of x can be found to satisfy the given equation. In such a case the quadratic equation has no real roots ; the roots are said to be imagin- ary or imjjossible. In the examples hitherto considered, the quadratic equa- tions have had two different roots. Sometimes however, there is only one root. Take, for example, the equation, x 2 — 10a; + 25 = ; by extracting the square root we have x — 5 = ; therefore x = 5. It is found convenient how- ever in this and similar cases to say that the quadratic has two equal roots. EXAMPLES. Solve 9. x 2 = x + 72. 10. 9x - x 2 + 220 = 0. 11. x 2 - fee = 32. 12. l£x = £ - x 2 . 13. 5a; 2 = 8* + 21. 14. ^L±J = 3a; 4- 2. a; — 1 1K Sx - 8 5a; - 2 15. — = -. x — 2 x + 5 137. Condition for Equal Roots. — To find the rela- tion that must exist between the knoum quantities of a quadratic equation in order thai the two roots may be equal. Take the general equation ax 2 + bx + c = 0. Ans. 9, - 8. 20, -11. 6, -¥• b - 4. 3, - b 3, - 1. 4, V- CONDITION FOR EQUAL ROOTS. 287 Transpose c x 2 + .c a and divide by ( c a -6 ± v^ 2 - X, Rule II, x X a - 4ac ft 2 4a 2 _ & 2 — 4ac 4a- Denoting the root corresponding to the positive surd by x v and that corresponding to the negative surd by x 2 , we have -b + w - - 4ac 2a -6 - y/b* - - 4ac 2a Now we see that if b 2 — 4ac = 0, these two roots are equal, and each of them is . 2a Hence the relation, b 2 — iac = 0, is the condition that the two roots of the equation ax 2 + bx + c = may be equal. The two roots are real and unequal if b 2 — 4ac is positive, i.e., if b 2 is Algebraically greater than 4ac. The two roots are imaginary if b 2 — 4ac is negative, i.e., if b 2 is Algebraically less than 4ac. Hence the two roots of this equation are real and unequal, equal, or imaginary, according as b' 2 is greater than, equal to, or less than 4ac. Note 1. — If either of the roots of a quadratic equation is imaginary, they are both imaginary. By applying these tests, the nature of the roots of any quadratic may be determined without solving the equation. 1. Show that the equation 2x 2 — 6x + 7 = cannot be satisfied by any real values of x. Here a = 2, b = — 6, c = 7. .-. b 2 - Aac = 36 - 4 . 2 • 7 = -20. Hence the roots are imaginary. 288 HINDOO METHOD OF COMPLETING THE SQUARE. Determine the nature of the roots of 2. .c- + Bx +1 = 0. Ans. Real and surds. 3. 3.r - 4a; -4 = 0. Rational. 4. If the equation /- -\- 2(k -f- 2)x + [)/,• = has equal roots, find k. Ans. h = 4, or 1. When an equation is in the general form o.r- -j- bx + c = 0, instead of solving it by either of the rules in Art. lo6, we may make use of formula (1) above as follows : 5. Solve bx 2 + 11a- = 12. Here a = 5, b = 11, c = —12; substituting these values in(l) x _ -11 ± V'(ll) 2 - 4-5(- 12) 10 -li ± v^sTTi -11 ± 19 _ 4 If' or —3, 10 10 lieh agrees with the solution of Ex 4, (Ai t. 136). Solve by this method the following G. 3a: 2 = 15 - 4a;. Ans. f , -3. 7. 2.r- 4- 7x =15. 1. — 5. .s. 5a; 2 4- 4 4- 21a; = 0. —i, 1 l> * !). 8a; 2 = x + 7. 1, _7 8 - Id. 35 4- 9a; - 2x* = 0. 7, 2 ' Note 2. — Though we can always find the roots of a given quad- ratic equation by substituting in formula (1 ), yet tbe student is advised to solve each separate equation either by the method given in Art. 136, and embodied in Rule II, or by one of the two following. 138. Hindoo Method of Completing the Square. — When a quadratic equation appears in the general form ax 2 4- bx 4- c = 0, the first member may be made a complete square, without dividing by the coefficient of •'"'. thus avoid- ing fractions, by another method (called the Hindoo method), as follows : Transpose c, and multiply by 4a, 4a-.r 4- lubx — — 4ac. HINDOO METHOD OF COMPLETING THE SQUARE. 289 Now since the middle term of :my trinomial square is twice the product of the square roots of the other two (Art. 41), the square root of the third term must be equal to the second term divided by twice the square root of the first term. Hence, dividing Aabx by twice the square root of 4a 2 ic 2 , i.e., by 4acc, and adding the square of the quotient, & 2 , to both members, the first becomes a perfect square. Thus, 4a' 2 .e 2 + iabx + b 2 = b 2 — 4ac. Extracting the square root, 2ax + b = ± vV — 4ac. — b ± vV — 4ac .°. x = , 2a which are the same values we obtained in (1) of Art. 137. Rule. Reduce the equation to the form ax 2 -f bx + c = 0. Mul- tiply it by four times the coefficient of x 2 ; add to each member the square of the coefficient of x in the given equation; extract the square root of both members, and solve the resulting simple equation. Note. — This method may be used to advantage when Ave wish to avoid fractions in completing the square, and it is often preferred in solving literal equations. (See Note 4 of Art. 13G.) 1. Solve 2x 2 — ox = 3. Multiply by four times 2, or 8, 16a 2 - 40a = 24. Add to' each side 5 2 , or 25, 16.« 2 - 40.» + 25 = 49. Extract the square root, Ax - 5 = ±7. ... x = L±J = 3, or -a 290 SOLVING A QUADRATIC BY FACTORING. Solve by the Hindoo method the following: 2. 3a? 2 + 5a; = 2. Am. £, -2. 3. 63J 2 - 12 = x. H, -1|. 4. 3ar -f 2a; = 85. 5, -of. 5. ar.r — bcx + odte = 6rf. -, — • a c 139. Solving a Quadratic by Factoring. — There is still one method of solving a quadratic which is often shorter than either of the methods already given. 1. Consider the equation x 1 — 2x — .15 = 0. Resolving this into factors (Art. 65), we have (x - 5) (x + 3) = 0. Now it is clear that a product is zero when any one of its factors is zero ; and it is also clear that no product can be zero unless one of the factors is zero. Thus ab is zero if a is zero, or if b is zero ; and, if we know that ab is zero, we know that either a or b must be zero ; and so on for any number of factors. Similarly the product (a- - 5) (x + 3) is zero, when either of the factors, x — 5, x -f 3, is zero, and in no other case. Hence the equation (x - 5) (x + 3)\= 0, is satisfied if x -5 = 0, or if a- + 3 = ; i.e., if x = 5, or if x = —3, and in no other case. Therefore the roots of the equation are 5, and —3. 2. Solve x 2 — 5x + 6 = 0. Resolving this into factors, we have (x - 2) (x - 3) = 0. The first member is zero either when x — 2 = 0, or when x — 3 = 0; and in no other case. Hence the equation is satisfied by x = 2, or 3 ; and by no other values ; thus, 2 and 3 are the roots of the equation. From these examples it appears that when a quadratic equation has been simplified, and lias all its terms in the SOLVING A QUADRATIC BY FACTORING. 291 first member, its solution can alwaj's be readily obtained if the expression can be resolved into factors. Hence for the solution of such an equation, we have the following Rule. Reduce the equation to its simplest form, with all its terms in the first member ; then resolve the ivhole expression into factors, and the values obtained by equating each of these factors separately to zero will be the required roots. 3. Solve x 2 — Ax = 0. Factoring, we have x(x — 4) = 0. The equation is satisfied if x = 0, or if x — 4 = 0, and in no other case. Hence we must have either x = 0, or x — 4 = 0. .-. x = 0, or 4. Note 1. — In this example we might have divided the given equation by x and obtained the simple equation x — 4 = 0, whence x = 4, which is one of the solutions. But the student must be partic- ularly careful to notice that whenever we divide every term of an equation by x, it must not be neglected, since the equation is satisfied by x = 0, which is therefore one of the roots. Note 2. — When the factors can be written down by inspection, the student should always solve the example in this way, as he will thus save himself a great deal of unnecessary work. Solve the following by resolution into factors: 4. (3a; - 1)(3» + 1) = 0. Ans. ±a. 5. x 2 - lice = 0. 0, 11. 6. x 2 - 3a; + 2 = 0. 1, 2. 7. x 2 - 2x = 8. 4, -2. ' 8. x 2 - 2ax + 4a6 = 26a?. 2a, 26. 9. x 2 - 2ax + 8x = 16a. 2a, -8. Note 3. — When the student cannot factor the equation readily by inspection, he should solve it by Rule II, Art. 136, or by Art. 138. 292 TO FORM A QUADRATIC WHEN THE ROOTS ARE GIVEN. 140. To Form a Quadratic when the Roots are Given. — We have seen (Art. 139) that if x 2 + px + q = (as — a) (as — b) , then a and b are the roots of the equation rf+px + q = (1) Conversely, if a and b are roots of (1), then x — a and a; — b are factors of the expression as 2 -f pa: + (< r >) a a TO FORM A QUADRATIC WHEN THE ROOTS ARE GIVEN. 293 Adding together the two roots of (1), Art. 137, we have x. -J- x„ = ; a and by multiplication we have _ ^ _ (b a _ 4 qc) _ c 4a- a which confirms the proposition. Hence any quadratic may be expressed in the form x 1 — (sum of roots) x + product of roots = . (6) By this principle we may easily form a quadratic with given roots. Although we cannot in all cases find the roots of a given equation, it is easy to solve the converse problem, namely, the problem of finding an equation which has given rooW. These relations are useful in verifying the solution of a quadratic equation. If the roots obtained do not satisfy these relations, we know there is some error in the work. Relations analogous to those above hold good for equations of the third and of higher degrees. But we defer the proof to a subsequent chapter. When we know one root of an equation, we may by division lower the degree of the equation. Thus if a is one root of an equation, we may divide it by the factor x — a. Note 1. — In any equation the term which does not contain the unknown quantity is frequently called the absolute term. A quadratic equation cannot have more than two roots. For no other value of x besides a and b can make (x — a)(x — b) in (3) equal to 0, since the product of two factors cannot equal if neither factor is equal to 0. It may therefore be inferred that a cubic equation has three and only three roots ; and that in any equation the number of roots is equal to the degree of the equation. Note 2. — The student must carefully distinguish between a quad- ratic equation and a quadratic expression. Thus, in the quadratic equation x' 1 + px -f q = 0, we must suppose x to have one of two 294 EQUATIONS HAVING IMAGINARY ROOTS. definite values, or roots, but when we speak of the quadratic expres- sion x 2 + px + q, without saying that it is to he equal to zero, we may suppose x to have any value we please. EXAMPLES. Form the quadratic equation whose roots are 1. 2 aud 8. Ans. x 2 — 5x + 6 = 0. 2. 3 and -2. a? - x - 6 = 0. & 2 + V.3 and 2 - \]'d. x 2 - 4x +1 = 0. 4. 6 and 8. x? - Ux + 48 = 0. 5. 4 and 5. x 2 — 9x + 20 = 0. 141. Equations Having Imaginary Roots. — It was shown (Art. 137) that when b 2 is less than 4ac, i.e., when b 2 . c —— is less than -, the two roots are imaginary. Hence, from 4a 2 a * (5) and (G) of Art. 140, the roots are imaginary when the square of half their sum is less than their product. Now it is impossible to have two numbers such that the square of half their sum is less than their product, which may be shown as follows : Let a represent any number ; and suppose it to be divided into tw r o parts — f- x and " — x. Then the product is a 2 o which is evidently the greatest when x 2 is the least. But when a;' 2 or x = the parts are each - ; thus the product of two unequal numbers is less than the square of half their sum. Hence, The square of half the sum of two numbers can never be less than their product. If then any problem furnishes an equation of the general quadratic form x 1 + px + 7 = 0, in which q is positive and greater than the square of ^-, we infer that the conditions of EQUATIONS OF HIGHER DEGREE THAN THE SECOND. 295 the problem are incompatible with each other, and hence the problem is impossible. Thus, Let it be required to divide 6 into two' parts whose product shall be 10. Let x = one of the parts, then 6 — x — the other. .-. x(6 - x) = 10, whence x = 3 ± V — 1 . Thus, the roots are imaginary. Now we know from the preceding proposition, that the number 6 cannot be divided into any two parts whose product will be greater than 9. Hence when we are required to divide 6 into two parts whose product is 10, we are required to solve an impossible problem. Thus, the imaginary root shows that the problem is impossible. 142. Equations of Higher Degree than the Second. — There are many equations which are not really quadratic, but which may be reduced to the quadratic form, and solved by the methods explained in this Chapter. An equation is in the quadratic form when the unknown quantity is found in two terms, and its exponent in one term is twice as great as in the other. Thus, x* - 9x 2 = -20, (x 2 + x) 2 + 4(.« 2 + x) = 12, ax 2 " + bx n + c = 0, etc., are in the quadratic form, and may be solved by either of the preceding rules ; care however should be taken to use the one best adapted to the example considered. 1. Solve x i - 9x 2 = -20. Here we may complete the square and solve by Rule I, Art. 136, or we may write out the value of x 2 at once by Rule II, Art. 136 as follows : x 2 = f ± V- 20 + -V- = = f ± i = 5, or 4. x = ±VE, or ±2. Thus there are four roots, ±sJ-'>, ±2. 296 EQUATIONS OF HIGHER DEGREE THAN THE SECOND. Otherwise thus. Transposing and factoring the first member, (a? - 5)(jb 3 - 4) = 0. .-. as 2 — 5 = 0, giving x- = ±05, or a; 2 — 4 = 0, giving x — ±2. 2. Solve oas 2 " + 6a;" + c = 0. Transpose and divide by a, B* + ^ = _ C . Art. 136, Rule II, a- = -A ±v /_c + _&1 = V_-4ac 2a V a 4a 2 4a 2 - ~ b ± ^ - 4ac 2a from which x may be found by taking the n ,h root of both members. Note 1. — If the student prefers, he may let x n = y; then x" M = y'-. Substituting, the equation becomes «>J 2 + by + c = 0. After solving for y, he may replace tbe value of y. 3. Solve a: 6 - Gx s = 16. Art. 136, Rule II, x 3 = 3 ±V / 25 = 3 ± 5 = 8, or -1 .-. x = 2, or -V2. 4. Solve a? - * + -c -4 = 6. Solving fora; - *, a.- - ' = — $ x V^j + * = 2, or -3. .-. a;" 1 = 16, or 81. •'• x = TS> or ST- 5. Solve y/x* + 12 + VaJ 2 + 12 = 6. Solving for Vxr -j- 12, t/ x * + 12 = _i ± yV, + | = — i ± § = 2, or -3. ar + 12 = 16, or .Si. a? = I, or 69. x = ±2, or ±V69. EQUATIONS OF HIGHER DEGREE Til AS THE SECOND. 297 6. Solve x + \bx + 10 = 8. By transposing a;, and squaring, 5a + 10 = 64 - 16a; + x 2 . .-. x 2 - 2\x = -54. Solving, we get x = - 2 /- ± x £ = 18, or 3. If we proceed to verify these values of x by substituting them in the given equation, we shall find that 3 satisfies the equation, but that 18 does not, while it does satisfy the equation x - \jbx + 10 = 8. Now the reason is this : the equation x 2 — 21a; = —54, which we obtained from the given equation by transposing and squaring, might have been obtained as well from x — \Jbx + 10 = 8, since the square root of a quantity may have either the sign + or — prefixed to it; i.e., the resulting equation x 2 — 21a; = —54, of which 18 and 3 are the roots, would be obtained, whether the sign of the radical be — or — . Hence we see that when an equation has been reduced to a rational, form by squaring, we cannot be certain without trial whether the values which are finally obtained for the unknown quantity are roots of the given equation. 7. Solve x 2 - 7x + sjx 2 - 7a; + 18 = 24. Acid 18 to both members in order that the equation may be in the quadratic form. x* _ 7aj + 18 + sjx 2 - 7x + 18 = 42. Solving, Sfx 1 -7a,-+18=-f±V / 42 + i = — \ ± - 1 / = 6, or -7 .-. x 2 — 7x + 18 = 36, or 49. Solving the first quadratic, we obtain x — 9, or —2. Solving the second quadratic, we obtain x = \{1 ± VI 73), Only the first two values are roots of the given equation the other two are roots of the eq nation ar - 7x - V'aJ 2 — 1x + 18 = 24. 298 SOLUTIONS BY FACTORING. 8. Solve x* - 4x» - 2x* + 12as - 16 = 0. We proceed to form a perfect trinomial square with the first two terms and a part of the third. The square root of this square is evidently (x 2 — 2a;), the square of which is a 4 — Ax 3 + Ax 2 ; having added Ax 2 to the equation we must now subtract Ax 1 . Hence the equation becomes x 4 - Ax 3 + Ax- - Gx 2 4- 12a; - 16 = 0, or (a? - 2x) 2 - G(x 2 - 2x) = 16, which is in the quadratic form, and may be solved as those above. Hence x 2 - 2x = 3 ± ^10 + 9 = 8, or -2. .-. x = 1 ± ^8 + 1, or 1 ± V-2 + 1. .-. x = 4, or -2, or 1 ± f^l. Solve the following : 9. a; 4 - 13a? + 36 = 0. Ans. ±2, ±3. 10. x 2 + yV + 9 = 21. ±4. 11. 9v/;f 2 - 9x + 28" + 9a; = x 2 + 36. 12, -3. 12. x§ + x% = 1056. 64, (-33)*. 13. (x 2 - 9) 2 - 11 (x 2 — 2) = 3. ±5, ±2. 14. a,- 4 - 8x s + 10.r + 24a; + 5 = 0. 5, -1, 2 ± fa 143. Solutions by Factoring. — By the principles of Art. 139, many equations of a higher degree than the second ma}' be solved, which cannot be reduced to the quadratic form. If an equation can be reduced to the form (x - a)X = 0, in which X represents an expression involving aj, we have either «-a-0, orX=0; therefore x = a, is one value of x ; and if we solve the equation X = 0, we shall have the other values of x. Hence whenever we have one factor of an equation, we have at Least one root, and by division we may lower the degree of the equation by one (Art. 1 10). Thus SOLUTIONS BY FACTORING. 2dd 1. Solve (x - 5) (x 2 - 3a; + 2) = 0. Here the first member is zero either when x — 5 = 0, or when x 2 — ox + 2 = ; and in no other case. Hence we have x — 5 = 0, or x 1 - Sx + 2 = 0. From the first we have x = 5 ; and the other roots of the equation are those given by x" - 3x + 2 = 0, that is, (x - 2) (a; - 1) = 0. Thus the cubic equation (as — 5) (x 2 — 3x + 2) = has the three roots 5, 2, and 1. The difficulty to be overcome in this method consists in resolving the equation iuto factors ; and facility in separat- ing expressions into factors can be acquired only by experi- ence. 2. Solve x s - 1 = 0. Since x 3 - 1 = (a; - l)(x 2 + x + 1), we have (x - l)(x* + x + 1) = 0. .-. x — 1 = 0, or ar + a; + 1 = 0, the roots of which are 1, or — \ ± V— f. Hence there are three roots of the equation x 3 = 1, one being real and the other two imaginary. Thus there are three numbers whose cubes are equal to 1 ; that is, there are three cube roots of 1. 2 3. S( jive x — 1 = 2 + -p. yjx Ans. 4. 4. ' : ' 2x 3 - a; 2 — 6a; = 0. 0, 2, -f. 5. ■ " x 3 -\- x 2 — 4x = 4. -1, 2, -2. 6. " a 3 - 3x = 2. -1, 2. 7. « a; 2 -A = If. ' 3x 9 2 3' Hi ± Vio). 300 PROBLEMS LEADING TO QUADRATIC EQUATIONS. 144. Problems Leading to Quadratic Equations of One Unknown Quantity. — We shall now give some examples of problems wliieh lead to pure or affected quad- ratic equations of one unknown quantity. In the solution of such problems, the equations are found on the same principles as in problems producing simple equations (Art. 61). EXAMPLES. 1. Find two numbers such that their sum is 15, and their product is 54. Let x = one of the numbers, then 15 — x = the other number. Hence from the conditions, we have a(15 — x) = 54, j»r x" — 15a; = —54. Solving, we get x = 9, or 6. If we take x = 9 we have 15 — x = 6 ; and if we take x = 6 we have 15 — x = 9. Thus, whichever value of x we take, we get for the two numbers G and 9. Hence, although' the equation gives two values of x, yet there is really only one solution of the problem. 2. A man buys a horse which he sells agaiu for $96 ; he finds that he thus loses one-fourth as much per cent as the horse cost him : find the price of the horse. Let x = the price of the horse in dollars, x 2 then — = the man's loss in dollars. 100 Hence from the conditions, we have 400 9G. Solving, we get x = 240 or 1G0. That is, the price was either, $210 or $1G0, for each of these values satisfies the conditions of the problem. EXAMPLES. 301 3. A train travels 300 miles at a uniform rate ; if the rate had been 5 miles an hour more, the journey would have taken two hours less : find the rate of the train. Let x = the rate the train runs in miles per hour ; then 300 -J- x = the time of running on the first sup- position ; and 300 -f- (re -f- 5) = the time of running on the second supposition. 300 = 300 2 x -f 5 x Solving, we get x = 25, or —30. Only the positive value of x is admissible, and thus the train runs 25 miles per hour. Note. — In the solutions of problems it often happens that the roots of the equation, which is the Algebraic statement of the relation between the magnitudes of the known and unknown quantities, do not all satisfy the conditions of the problem. The reason of this is that the Algebraic statement is more general than ordinary language; and the equation, which is a proper representation of the conditions, will also express other conditions. Thus, the roots of the equation are the numbers, whether positive, negative, integral, or fractional, which satisfy that equation ; but in the problem there may be restric- tions on the numbers, expressed or implied, which cannot be retained in the equation. If for instance, one of the roots of an equation is a fraction, it cannot be a solution of a problem which refers to a number of men, for such a number must be integral. Thus 4. Eleven times the number of men in a group is greater by twelve than twice the square of the number : find the number of men in the group. Let x = the number of men ; then we have llaj = 2x- + 12, or 2a; 2 - 11a; = -12. Solving, we get x = 4, or 1^. Thus, there are 4 men ; the value 1| is plainly inadmissi- ble. 302 EXAMPLES. 5. Eleven times the number of feet in the length of a rod is greater by twelve than twice the square of the number of feet: how long is the rod? This question leads to the same equation as Ex. 4, only here we cannot reject the fractional result, since the rod may be either 4 feet long or 1^ feet long. 6. The square of the number of dollars a man possesses is greater by 600 than ten times the number : how much has the man? Let x = the number of dollars the man has. Then x 1 = 10a? + GOO. Solving, we get x = 30, or —20. Both these values are admissible, since a negative posses- sion is a debt (Art. 20) . 7. The sum of the ages of a father and his son is 80 years ; also one-fourth of the product of their ages, in years, exceeds the father's age by 240 : how old are they? Lqt x = the father's age in years ; then 80 — x = the son's age in years. Hence Ja?(80 - x) = x + 240, or x 2 — IGx = —960. .-. x = 60, or 16. Thus the father is 60 and the son 20 years old. The second solution 16 is not admissible, since it would make the father younger than his son. Note. — The student should examine each root of every equation to see if it satisfies the conditions of the problem, and reject those which do not. 5. A cistern can be filled by two pipes in 33^- minutes ; if the larger pipe takes 15 minutes less than the smaller to fill the cistern, find in what time it will be filled by each pipe singly. Ans. 75 and 60 minutes. 9. A person selling a horse for $72, finds that his loss per cent is one-eighth of the number of dollars that he paid for the horse : what was the cost price? Ans. $80, or $720. EXAMPLES OF QUADRATICS. 303 10. Divide the number 10 into two parts such that their product added to the sum of their squares may make 76. Ans. 4, 6. 11. Find the number which added to its square root will make 210. Ans. 196. /T2? A and B together can do a piece of work in 14f days ; and A alone can do it in 12 days less than B alone :* find the time in which A alone can do the work. Ans. 24 days. 13. A company dining together at an inn, find their bill amounts to $35 ; two of them were not allowed to pay, and the rest found that their shares amounted to $2 a man more than if all had paid : find the number of men in the company. Ans. 7. 14. The side of a square is 110 inches long : find the length, and breadth of a rectangle which shall have its perimeter 4 inches longer than that of the square, and its area 4 square inches less than that of the square. Ans. 126, 96. Note. — We will conclude this chapter with the following examples. In solving them care must he taken to select the method best adapted to the example considered. Many of them may be solved by special methods (Arts. 137, 138, 139); but the methods of Art. 136 are the most important, and will solve every example. EXAMPLES OF QUADRATICS. Ans. ±h. ±2. ±y/3. 2, 3. 1, 9. \, -13. 1. 0:^6 + a; 2 = 1 + x 2 . 2. lx 2 + 8 x 2 + 4 x 2 21- 8x 2 - 11 3 o n •>. « + vi + ^-= v/i + a;3 - 4. 1 , i x + ^2 — ar « — \/2 — x 2 _ X ~ 2 5. x 2 — ox = —6. 6. x 2 — 10a; = —9. 7. x 2 + 12a; = 13. 304 EXAMPLES. 8. x- + 9a = 22. Am. 2, -11. 9. x 2 + 2x = 143. 11, -13. 10. 15.x 2 - 2a.r = a 2 . 11. 21x 2 = 2ax + 3a 2 12. x + 3 2x - 1 2a - 7 x - 3 13. 1 _ 1 1+05 3 — a 11. 4 5 = &. a + 1 x — 1 19. a; -f 1 a — 2 _ „ a - 1 a + 2 ~ *' 20. 1 2 _ 3 a; - 2 a; -f 2 & ' 21. 3 1 2(ar - 1) 4(a; + 1) 22. 5 3 _ 14 x + 2 a; a? + 4 23. 4 5 12 a: + 1 x + 2 a; + 3 24. a; + 1 a; - 1 _ 2x - 1 x + 2 a _a 3' 5" ** ~3 M- 3, -£• x — 1 a; + 2 a) 15. (a; + l)(2a; + 3) = 4a; 2 - 22. - 5, -2£. 16. 2 + ^ - ^LzUf = l _ aj + x 2 . 1, 2. 17. 8s + 11 + - = — • 7, -£,. a; 7 18< 3(s - 1) _ 2(a- + 1) = 5 f _ 3 3, -I- 3, -4f. 3, -5. 3, -If 3, -If 4,0. j + 2 a — S \x-Sj EXAMPLES. 305 26. x — 1 x - 2 _ 2x + 13 •Hi x + 2 ~" a; + 16 ' ^4ns. 5, — 1^. 27. a; + 1 as + 2 _ 2a? 4- 13 a; — 1 sc — 2 a; + 1 5, 1*. 28. 2x - 1 Sx — 1 _ 5a; - 11 a: + 1 ' a; + 2 a:— 1 5, -1£. 29. 14a; _9 a; 2 - 3 8a; — 3 x + 1 ■ 1 1,1 a; + - 1 + - 2f,0. 30. —= + —?=* a; 1 a; a; 3, -If- 31. V/Ia; + 2 4 - y/a; 4 + V^ 85 V^ 4. 32. a 2 ar - 2a 3 x + a 4 - 1 = 0. a 33. 4a 2 a; = (a 2 - b 2 + a;) 2 . (a ± o) 2 . 34. a; . a _ a; , & a x b x ±V^&. 35. (3a 2 + 6 2 )(x- 2 - x + 1) = (36 2 + a 8 ) (» 2 + aJ+ 1). Ans. a — b a + b 36. 1 1 , o a; — a, —b. a + b + x a 37 a + c(a + x) a a + c(a — x) + a; a a, °( 1 + c ). c(2c + 3) a; a — 2ca; » + 2 _ I = o. a a; a 38. 1 ± y/l -a 2 - 39 * (7/1 7) *N~ — 1 . a, —b. ya + V^ & (o6 2 ) -2 + (a 8 6) "i 306 EXAMPLES. Form the equation whose roots are 40. 1 ± 5. Ans. x 2 — 2x -24 = 0. 41. -$, f. 35a; 2 + 13a: - 12 = 0. 42 - £ir| -frrf (^ 2 ~ *>* + 4 ^ - P a + ff 3 = o. 43. 7 ± 205. a 2 - 14a> 4- 29 = 0. 44. ±2y/3 - 5. x 2 + 10a: + 13 = 0. 45. -p ± 2y/2g. a; 2 + 2pa; + p 2 - 8g = 0. Show that the roots of the following equations are real : 46. x 2 - 2ax + a? - b 2 - c 2 = 0. 47. (a - 6 + c)x 2 + 4 (a - &)a + (a - 6 - c) = 0. For what values of m will the following equations have equal roots? 48. a; 2 - 15 - m(2x - 8) = 0. Ans. 3, 5 49. a; 2 - 2sc(l + 3m) + 7(3 + 2m) = 0. 2, — V "- EXAMPLES OF EQUATIONS REDUCIBLE TO QUADRATICS. 50. a; 4 - 5a; 2 + 4 = 0. Ans. ±1, ±2. 51. (a? 2 - a;) 2 - S(x 2 - x) + 12 = 0. 3, -1, ±2. _1 4- i/_'>7 52. (a; 2 + a?)(a*» + a; + 1) = 42. 2, -3, g — 53. (a; + ^Y+ 4^ + ~\ = 12. i, _ 3 ± 2y/2. 54. 55. x 2 + 3 - \/2x 2 - 3a; + 2 = §(a> + 1). 1, J. 35* _|_ 2a; 3 - liar + 4.« + 4 = 0. 1, 2. 56. 57. . , « j_ aV2\/2 - 1 x* + 4a 8 a; = a 4 . — F ± *t= • y/2 y/2 1+.^ _ , (1 + »)< 4«8. 1 4- ^3 ± ^3 + 2^3. 1 - V 3 ± ^3 - 2\<3. EXAMPLES. 307 58. 27a; 2 - ~ + y = — - — „ -f 5. 3a; 2 3 3a; 3a; 2 Ans. 2, -If, A(-2 ± V/-2GG). Solve as explained in Art. 143 : Ans. 1, |-(-l ± v/^7) 3, k-3 ± y/£3) hH-1 ± \/-35) Find a number whose square diminished by 119 is equal to ten times the excess of the number over 8. Ans. 13. 67. A man is five times as old as his son, and the sum of the squares of their ages is equal to 210G : find their ages. Ans. 45, 9. 68. If a train traveled 5 miles an hour faster it would take one hour less to travel 210 miles: what time does it take? Ans. 7 hours. 69. The sum of the reciprocals of two consecutive numbers is £f : find them. Ans. 7, 8. 70. The perimeter of a rectangular field is 500 yards, and its area is 14400 square yards : find the length of the sides. Ans. 90 and 160 yards. 71. A number of two digits is equal to twice the product of the digits, and the digit in the ten's place is less by 3 than the digit in the unit's place : what is the number? Ans. 36. 72. The sum of a certain number and its square root is 42 : what is the number? Ans. 36. 308 EXAMPLES. 73. A rectangular court is ten yards longer than it i3 broad; its area is 1131 square yards: find its length and breadth. Ans. 39 and 29. 74. One hundred and ten bushels of coal were divided among a certain number of persons ; if each person had received one bushel more he would have received as many bushels as there were persons : find the number of persons. Ans. 11. 75. A cistern can be supplied with water by two pipes ; by one of them it would be filled G hours sooner than by the other, and by both together in 4 hours : find the time in which each pipe alone would fill it. Ans. G, 12. (76J Two messengers A and B were sent at the same time to a place 90 miles distant ; the former by riding one mile per hour more than the latter arrived at the end of his journey one hour before him : find at what rate per hour each traveled. Ans. 10, 9 miles. 77. A person rents a certain number of acres of land for $280 ; he keeps 8 acres in his own possession, and sublets the remainder at $1 per acre more than he gave, and thus he covers his rent and has $8 over : find the number of acres. Ans. 56. 78. From two places 320 miles apart, two persons A and B set out in order to meet each other. A traveled 8 miles a day more than B ; and the number of days in which they met was equal to half the number of miles B went in a day : find how far each traveled before they met. Ans. 192, 128. 79. A certain number is formed by the product of three con- secutive numbers, and if it be divided by each of them in turn, the sum of the quotients is -17 : find the number. Ans. GO. ■so. A boat's crew row 3 J miles down a river ami back again in 1 hour and 40 minutes : supposing the river to have a cm-rent of 2 miles per hour, find the rate at which the crew would row in still water. Ans. 5 miles per hour. .si. A person rents :i certain number of acres of land for$33G ; he cultivates 1 acres himself, and letting the rest for EXAMPLES. 309 $2 an acre more than he pays for it, receives for this portion the whole rent, $336 : find the number of acres. Ans. 28 acres. 82. A person bought a certain number of sheep for $140 ; after losing two of them he sold the rest at $2 a head more than he paid for them, and by so doing gained $4 by the transaction : find the number of sheep he bought. Ans. 14. 83. A man sends a lad to the market to buy 12 cents' worth of oranges ; the lad having eaten a couple, the man pays at the rate of one cent for 15 more than the market price : how many did the man get for his 12 cents? Arts. 18. 84. What are eggs a dozen when two more in a shilling's worth lowers the price one penny per dozen ? Ans. Ninepence. 85. Two men were employed at different wages, and paid at the end of a certain time ; the first received £4. 16s., and the second, who had worked for 6 days less, received £2. 14s. If the second had worked all the time and the first had omitted 6 days, they would have received the same sum : how many days did each work, and what were the wages of each? Ans. 24 days at 4s., and 18 days at 3s. per day. /86^) A man bought a certain quantity of meat for $2.16; if meat were to rise in price one cent per lb. he would get 3 lbs. less for the same sum : how much meat did he buy ? Ans. 27 lbs. 87. The price of one kind of sugar per stone of 14 lbs. is Is. 9d. more than that of another kind; and 8 lbs. less of the first kind can be got for £1 than of the second : find the price of each kind per stone. Ans. 8s. 9d., 7s. 88. A person spent a certain sum of money in goods, which he sold again for £24, and gained as much per cent as the goods cost him : find what the goods cost. Ans. £20. 89. A person drew a quantity of wine from a full vessel which held 81 gallons, and then filled up the vessel with water ; he then drew from the mixture as much as he before drew of pure wine ; and it was found that 64 gallons of pure wine remained : find how much he drew each time. Aits. 9 gallons. 310 EXAMPLES. 90. A certain company of soldiers can be formed into a solid square ; a battalion consisting of seven such equal companies can be formed into a hollow square, the nun being four deep. The hollow square formed by the battalion is sixteen times as large as the solid square formed by one company : find the number of men in the compan}-. Ans. 64. 91. Find that number whose square added to its cube is nine times the next higher number. Ans. 3. 92. A courier proceeds from one place P to another place Q in 14 hours ; a second courier starts at the same time as the first from a place 10 miles behind P, and arrives at Q at the same time as the first courier. The second courier finds that he takes half an hour less than the first to go 20 miles : find the distance from P to Q. Ans. 70 mile.^. 93. A vessel can be filled with water by two pipes ; by one of these pipes alone the vessel would be filled 2 hours sooner than by the other; also the vessel can be filled by both pipes together in 1| hours: find the time which each pipe alone would take to fill the vessel. Ans. 5 and 3 hours. 94. There are three equal vessels, A, B, and C ; the first contains water, the second brandy, and the third brandy and water. If the contents of B and C be put together, it is found that the fraction obtained b}- dividing the quantity of brandy by the quantity of water is nine times as great as if the contents of A and C had been treated in like manner : find the proportion of brandy to water in the vessel C. Ans. Equal. 95. A person lends $5000 at a certain rate of interest ; at the end of a year he receives his interest, spends $25 of it, and adds the remainder to his capital ; he then lends his capital at the same rate of interest as before, and at the end of another } T ear finds that he has altogether 85382 : find the rate of interest. Ans. 1 per cent. SIMULTANEOUS QUADRATIC EQUATIONS. 311 CHAPTER XV. SIMULTANEOUS QUADRATIC EQUATIONS. 145. Simultaneous Quadratic Equations.— We shall now consider some of the most useful methods of solving simultaneous equations, one or more of which may be of a degree higher than the first. It should be remarked that it is, in general, impossible to solve a pair of simultaneous quadratic equations ; for, if we eliminate one of the unknown quantities, the resulting equation will be of the fourth degree with respect to the other unknown quantity, and we cannot, in general, solve an equation of a higher degree than the second. There are several cases however in which a solution of two equations may be effected, when one or both of them are of the second or some higher degree. Various artifices are employed for the solution of such equations, the proper application of which must be learned by experience. 146. Case I. When One of the Equations is of the First Degree. — This case may be solved by the following Rule. From the simple equation find the value of one of the unknown quantities in terms of the other, and substitute this value in the other equation. 1. Solve Zx + 4?y = 18 (1) 5x 2 - 3xy = 2 (2) From (1) we have y = - — - ; («*) and substituting in (2), 5a° - 3 »< 18 ~ 3 - c > = 2. 4 .-. 20a 8 - 54.x- + 9a: 2 = 8, or 29a; 2 - 54.x = 8. 312 SIMULTANEOUS QUADRATIC EQUATIONS. Solving, we get x = 2, or 4 . and by substituting in (3), y = 3, or 267 "58 '• Solve the following : 2. 3a> - 4ty = 5, Ans, , a; = 3, or -H^ 3 X * _ xy _ stf = 21. y = i, or 3. ox — y = 17, x = 4, or 8 1j ' SB? = 12. y = 3, or - 20. 4. 2a; - 5y = 3, a? =4, or -¥> x 2 + xy = 20. 2/= 1, or -ft- 5. Ax — by = 1, a? = 4, or -if, 2x 2 - xy + 3?/ 2 + 3aj - -42/ 17. 2/ = 3, or _ 4JL 147. Case II. Equations of the Form x ± y = a, and xy = b ; or x- + y 2 = a, and jt/ = 6. 1. Solve a? + y == 15. . . . (1) a# = 36 . . . . (2) Square (1), x 2 + 2xy + .7- = 225; . . . (3) multiply (2) by 4, 4xy =144; . . . (4) subtract (4) from (3), x 2 — 2xy + y 2 = 81 ; extract the square root, ® — y = ±9. ... (5) Combining (0) with (1), we have the two cases x + y = 15, ^ a; + y = 15, x - y = 9.j x - y = -!). from which we have a; = 12, j .r = 3, 2/ = 3. 1 ?/ = 12. 2. Solve a- 2 + ?/ 2 = 25 (1) •'// =12 (2) Multiply (2) by 2 ; then by addition and subtraction we have x 2 + 2xy + y 2 = 49, x 2 - 2. x y 1 + 1-5 ,.)\ Square (1), 1 - A + 1 = x (3 ) x- xy y Subtract (3) from (2), A = | ....... (4) Add (2) and (4) and take the square root, -+- = ±1 (5) x y From (1) and (5), - = f, or —\. 1 x 1 i - = *, or — £. y ^ .«. x = |, or —3, and y = 3, or — |. Solve the following : 5. x — y =12, Ans. x = 17, or — 5, xy = 85. y = 5, or —17. 6. x + ?/ = 74, x = 53, or 21, xy =1113. y = 21, or 53. 7. a; -f- ?/ = 84, x = 71, or 13, xy = 923. ?/ =13, or 71. 8. of + if = 1)7, aj = i», or 4, a; -f- y =13. y =, 4, or 9. 9. a; + y = 9, a; = 5, or 4, x- + .ry + if = 61. ?/ = I, or 5. 148. Case III. When the Two Equations Con- tain a Common Algebraic Factor. 1 v i le. Divide <>n<' equation by the other, ami caned the cum mint factor. SIMULTANEOUS QUADRATIC EQUATIONS. 315 1. Solve xy + if = 133 (1) * a - f = 95 (2) Divide (2) by (1) and cancel the common factor x + y, and substituting in (1) W + 2/ 2 = 133 - Solving, we get ?/ = ± 7 ; and by substituting in (3), x = ±12. Note. — This includes the case where one equation is exactly divisible by the other. 2. Solve cc 8 + y z = ISxy (1) x + y = 12 (2) Divide (1) by (2), x 2 - xy + y 2 = %xy. ... x * _ |., y + y 2 = (3) Subtract (3) from the square of (2), %xy = 144. .-. \xy= 16 (4) Add (3) and (4) and take the square root, x — y = ±4 . . . . » . . (5) From (2) and (5) x = 8, or 4, ?/ = 4, or 8. 3. Solve x* + ®y + 2/ 4 = 2613 (1) a 2 + xy + y 2 = 07 .... . (2) Divide (1) by (2), x 2 - xy + ?/ 2 = 39 (3) Add (2) and (3), x 2 + y 2 = 53. Subtract (3) from (2) , a^ = 14. .-. 05 = ± 7, or ±2, 2/ = ± 2, or ±7. 31G SIMULTANEOUS QUADRATIC EQUATIONS. Solve the following : 4. x 2 + xy = 84, ^4?is. a* 2 — 2/ 2 = 24. 5. x* + X Y + y = 133, x* + a^ + y 9 = 19. 6. a; 3 + f = 407, a; + y = li. 7. £C — y = 4, a; 3 - ?/ 3 = 988. a; = ±7, y = ±5. X = ±3, or ± 2, y = ±2, or ± 3. a; = 7, or 4, 2/ = 4, or 7. x = 11, or — 7, y = 7, or -11. 149. Case IV. When the Two Equations are of the Second Degree and Homogeneous. First method. 1. Solve 2a; 2 + 3xy + v/ 2 = 70 . . . . (1) Gar + xy — if — 50 . . . . (2) Divide (1) by (2), ^ + 3 ** + ^ h 6x 2 + xy - y 2 Hence 10a; 2 + loxy -f by" = 42a; 2 + Ixy — ly- ; or 32a: 2 - 8a# - 12/ = 0. This is a quadratic equation which we may solve, and find the value of one unknown quantity in terms of the other. Thus, solving for x, x 2 - \xy = |?/ 2 . By Art. 13G, Rule II, x « | ± 0// 2 + g = Mi/ 2 , or x = y H y = fy, or — $y. Take a; = |y, and substitute in either (1) or (2), and we have ?/ 2 = 1G. .-. y = ±4, and x = ±3. If we substitute the second value of a*, which is — ■?,//, we find an inadmissible result. Second method. Examples of this class are conveniently solved by substituting for one uf the unknown quantities the SIMULTANEOUS QUADRATIC EQUATIONS. 317 product of the other and a third unknown quantity, called an auxiliary quantity. 2. Solve 2y 2 - Axy + 3x 2 = 17 . ' . . . . (1) f - * 2 = 16 (2) Put y = vx, and substitute in (1) and (2). Thus x 2 {2v 2 - 4v + 3) = 17 (3) x 2 ( v 2 - 1) = 16 (4) _ ,. . . 2^ 2 -4^ + 3 17 By division, 2 _ = ti- .-. 32v 2 - 64v + 48 = 17v 2 -17; or 15u 2 - Mv = -65. Solving, we obtain v = f , or - 1 /. Take v = |, and substitute in either (3) or (4). From (4) o; 2 ( 2 ¥ 5 - - 1) = 16; .-. x 2 = 9. .-. x = ±3, and y = vx = %x = ±5. Again, take v = \ 3 -, x 2 ( W - 1) = 16; .'. z 2 = ^. .-. x = ±f, and y — vx = ±±£. Any pair of equations which are o/ £/te second degree and homogeneous, can be solved by either of these methods, though the second is usually preferred. Solve the following : 3. x 2 + Sxy = 28, Ans. x = ±4, or T 14, xy + Ay 2 = 8. y = ±1, or ± 4. 4. x 2 + a# + 2?/ 2 = 74, x = ±8, or ± 3, 2z 2 + 2xy + ?/ 2 =73. 2/ = q:5, or ± 5. 5. a;' 2 + a$ - 6y 2 = 24, SB- ±6, a? + 3^ _ 1( y _ 32. y = ±2. 6. a; 2 + xy - 6/ =21, x = ±9, a^ — 2?/ 2 =4. 2/ = ±4. Note. — Many examples of homogeneous equations of the second degree are easily solved by Case II or III. Only those examples of this class are to be solved by Case IV that rannot be solved by either Case II or III. 318 SYMMETRICAL EQUATIONS. 150. Case V. When the Two Equations are Symmetrical with respect to x and /. — An expres- sion is said to be symmetrical zvith respect to two letters ivhen these letters are similarly involved, i.e., when they can be interchanged loithout altering the expression. Thus, the expression a 3 + a 2 x + ax 2 + x 3 is symmetrical with respect to a and x, since if we write x for a, and a for x, we get the same expression. Also x* + 3x 2 y + 3xy 2 + y* is symmet- rical with respect to x and y. Examples of this class may frequently be solved by substituting for the unknown quantities, the sum and differ- ence of two others. 1. Solve x* + ?/ 4 = 82 (1) x -y = 2 (2) Put x = u + v, and y = u — v ; (2) becomes (u + v) — {u — v) = 2 ; .-. v = 1. (1) becomes (u + l) 4 -f (u - l) 4 = 82. .-. 2(?t 4 + Git 2 + 1) = 82; or u* + Gu 2 - 40 = 0. Hence, Art. 139, (a 2 + 10) (w 8 - 4) = 0. .-. u 2 = 4, or -10. .-. u = ±2, or ±\J^To. Thus x = 3, -1, 1 ± \/-10, 2/ = i, -3, -1 ± y/^To. 2. Solve (a 2 + r) (.r 3 + /) = 280 . . . . (1) a; + y = 4 .... (2) Put x = w + v, and y = w — v J (2) becomes (tt + v) + (w — v) = 4 ; .*. u = 2. Also /- + //- = (2 + v) 2 + (2 - v) 2 = 8 + 2v a , and a 3 + y A = (2 + v) a + (2 - u) 3 = 16 + 12v 2 . SPECIAL METHODS. 319 Hence (1) becomes (8 + 2v 2 )(16 + 12v 2 ) = 280, or (4 -(- v 2 ) (4 + 3v 2 ) = 35. ... v * + 3$.tf = Ig-. .-. v 2 = -f ± li = ^ or-y. .-. v = ±1, or ±y / -- 1 3 9 -- .-. a; = 3, 1, 2 ± ^-¥> ?/ = 1, 3, 2 T \J^K°. Solve the following : 0. x — y — 2, and x 5 — ?/ 5 = 242. ylws. a; = 3, or — 1 ; ?/ = 1, or —3. 4. x — y = 1, and a 5 — y 5 = 781. .4ns. a; = 4, —3 ; y = 3, —4. 5. x + y = 3, and x 5 + ?/ 5 = 33. .<4.7is. x = 1, 2; ?/ = 2, 1. 151. Special Methods. — The preceding cases will be sufficient as a general explanation of the methods to be employed ; but in some cases special artifices are necessary. One that is often used with advantage consists in consider- ing the sum, difference, product, or quotient, of the two unknown quantities as a single quantity, and first finding its value. Other artifices may also be used with advantage, but familiarity with them can be obtained only by experience. 1. Solve x 2 + 4xy + 3a: = 40 - 6y - 4y 2 . . . (1) 2xy - x 2 = 3 (2) From (1) we have x 2 + 4xy + 4y 2 -{- ox -\- Gy = 40 ; or (x + 2y) 2 + 3 (a; + 2y) - 40 = 0. Consider x -f 2y as a single unknown quantity, and find its value from this quadratic. Thus, (Art. 139), [(as + 2y) + 8] [(as + 2y) - 5] = 0. .-. x + 2y = -8, (3) or x + 2y = 5 (4) 320 SPECIAL METHODS. From (2) and (4) we obtain x = 1, orf ; y = 2, or J. From (2) and (3) we obtain _ -4 ± y/To 2 ' v = -UTt/lO * 2 2. Solve ay - 6fB = 34 - 3# n) 3xy + y = 18 + 2x (2) Multiply (2) by 3 and subtract the result from (1), xhf - dxy + 20 = 0. .-. (xy - 5)(xy -4) = 0. .'. xy = 5 . . . . (3) xy = 4 . . . . (4) From (2) and (3) we obtain x = 1, or -f, y = 5, or -2. From (2) and (4) we obtain x = ~ 3 ± V /l7 , and 2/ = 3 ± ^V7. Solve the following : 3. x 1 + y = 73 - 3a - 2xy, Ans. x = 4, 10, - 12 ± ^58, y 1 + a = 44 - 3y. * ?/=5, -7, -1 ^ y/58. 4. ^ + ^ = 12, x= 0, ^, y~ y x - y = 3. ?/ = 3,-?. 5. a 2 + 3.t?/ = 54, x = ±3, ±36, xy + 4y' 2 =115. y = ±5, T'-V 1 - G. x 4 - x 1 + .V 4 - ?/ 2 = S4, .r = ±3, ±2, .c- + x-y' 1 + y 1 = 49. y = ±2, ±3. PROBLEMS LEADING TO QUADRATIC EQUATIONS. 321 152. Simultaneous Quadratic Equations with Three Unknown Quantities. 1. Solve xy + xz = 27 (1) yz + yx = 32 (2) zx + zy = 35 (3) Add (1) and (2) and subtract (3) from the sum, 2xy = 24; .-. xy = 12 . . . . (4) Subtract (4) from (1), xz = 15 (5) Subtract (4) from (2), yz = 20 (6) Multiply (4) and (5), x'hjz =180 (7) Divide (7) by (G), x 2 = 9 .-..« = ±3. Heuce from (4), y = 12 h- (±3) = ±4. And from (5), z = 15 -f- (±3) = ±5. Thus x = ±3, ?/ = ±4, 2 = ±5, all the upper signs being taken together. Solve the following : 2. 3yz + 22KB — 4o^ =16, -4ws. x = ±1, 2y« — 3z.c + xy = 5, ?/ = ±2, 4*/z — Z£ — 3.t?/ = 15. 2 = ±3. 3. G(x- + 2/ 2 + « 2 ) = 13(0! + y + z) = ±£l, a? = f , f, a;?/ = z 2 . 2/ = f , |, z = ±2. 153. Problems Leading to Simultaneous Quad- ratic Equations. 1. The small wheel of a bicycle makes 135 revolutions more than the large wheel in a distance of 2G0 yards ; if the circumference of each were one foot more, the small wheel would make 27 revolutions more than the large wheel in a distance of 70 yards : find the circumference of each wheel. Let x = the circumference of the small wheel in feet, and y = the circumference of the large wheel in feet. Then the two wheels make — and - — revolutions respec- x y tively in a distance of 2G0 yards. 322 PROBLEMS LEADING TO QUADRATIC EQUATIONS. Hence 780 _ X 780 _ y 135; X y A • Similai •iy from the second condition, 210 210 = 27; x + i y + i i i (i) (2) x + 1 y + 1 52w , -, 61w + 52 From (1), x- = "' ; ••• x + 1 = „• , - • 9y + 52 9y + o2 Substituting in (2), ^__ _ __ = ^. .-. 9/ -USy = 52. Solving, we obtain y = 13, or — |. Substituting y = 13 in (1), we find a; = 4. The negative value of y is inadmissible. Hence the small wheel is 4 feet, and the large wheel is 13 feet in circumference. 2. A man starts from the foot of a mountain to walk to its summit ; his rate of walking during the second half of the distance is half a mile per hour less than his rate dining the first half, and he reaches the summit in 5£ hours. He descends in 3| hours by walking at a uniform rate, which is one mile per hour more than his rate during the first half of the ascent : find the distance to the summit, and the rates of walking. Let 2fl3 = the number of miles to the summit, and y = the rate of walking, in miles per hour, during the first half of the ascent. Then - = the time in hours for the first half of the ascent ; y an( j — fl J — = the time in hours for the second half of the y - \ ascent. PROBLEMS LEADING TO QUADRATIC EQUATIONS. 323 Hence - + -^-r = H 0) Similarly -^ = 3j (2) From (2) x = -^ (>/ + 1) (3) From (1) x(2y - \) = h}y{y - \) (4) Substituting (3) in (4), V-0/ + i)(2y-i)= V*(y-i). .-. 28/ - 89# = -15. Solving, we obtain y = 3, or 2 \. Substituting y = 3 in (3), we find a; = ^ 5 . The other value of y is inapplicable, because by supposition y is greater than ^. Hence the whole distance to the summit is 15 miles, and the rates of walking are 3, 2|, and 4 miles per hour. 3. The sum of the squares of two numbers is 170, and the difference of their squares is 72 : find the numbers. Ans. 11 ; 7. 4. The product of two numbers is 108, and their sum is twice their difference : find the numbers. Ans. 6 ; 18. 5. The product of two numbers is 6 times their sum, and the sum of their squares is 325 : find the numbers. Ans. 10 ; 15. 6. A certain rectangle contains 300 square feet ; a second rectangle is 8 feet shorter, and 10 feet broader, and also contains 300 square feet : find the length and breadth of the first rectangle. Ans. 20 ; 15. 7. Find two numbers such that their sum may be 39, and the sum of their cubes 17199. Ans. 15 and 24. 8. The product of two numbers is 750, and the quotient of one divided by the other is 3* : find the numbers. Ans. 50 and 15. 324 EXAMPLES. EXAMPLES OF SIMULTANEOUS QUADRATICS. Note. — In the great variety of simultaneous quadratic equations, it is impossible to give rules for every solution. The artifices employed in Algebraic work are very numerous. The student is cautioned not to go to work upon a pair of equations at random, but to study them until he sees how they can be reduced to a simpler equation by addition, multiplication, factoring, or by some other process, and then to perform the operations thus suggested. Solve the following : 1. x + Ay = 14, f + Ax = 2y + 11. 2. fix + 2y = 16, xy = 10. 3. x + 2y = 9, 3y 2 — 5x 2 = 43. 4. 3x — y = 11, 3a; 2 - y' 2 = 47. 5. Ax + 9y = 12, 2a; 2 + xy = 6y\ 6. Sx + 2y = 5xy, 15a; — 4?y = 4xy. 7. a; + ?/ = 51, xy = 518. 8. x — y = 18, a;?/ = 1075. 9. a; 2 + y 2 = 89, xy =40. 10. a; 2 + ?/ 2 = 178, x + 2/ = 1G. 11. a; 2 + ?/ 2 = 185, x - y = 3. L»s. a? = 2, -4G, Z/ = 3, 15. x = 2, V. y = 5, 3. X = 1, -ft. y = 4, w- X = 4, 7, y = 1, 10. a; = -24, f y = 12, t- a; = h o, 2/ = h 0. a; = 37, 14, ?/ = 14, 37. a; = 43, -25, y = 25, -43. x = ± 8, ± 5, y = ± 5, ± 8. X = 18, 3, y = 3, 13. X = U, - 8, y = - s , -11. EXAMPLES. 325 12. x 1 — xy + y* = 7G, ^Ins. a; = 10, 4, x + y = 14. y = 4, 10. 13. x - y = 3, » = 7, - 4, a 2 — 3^2/ + y 1 = -19. y = 4,-7. 14. 4 + - 2 = ttt, or ?/- a = f> 3 8' - + - = u- x y 2/ = 8' #• 15. -ji H r, = /(iff' a;?/ = 30. a; = ± G, ± 5, 2/ = ± 5, ± 6. 16. x 1 + y 2 + xy = 208, OJ — 12, 4, a + y = 16. y = 4, 12. 17. ^ _ yl _ K]^ a; = 5, a — y = 2. 2/ = 3. 18. a; 3 — y 3 = 7xy, a,* = 4, - 2, x — y = 2. 2/ = 2, - 4. 19. a; + y = 23, a; = 14, 9, a; 3 + y 3 = 3473. y = 9, 14. 20. aj + y = 35, x = 8, 27, aj^ + y$ = 5. 2/ = 27, 8. 21. 33 - y = \/x + V^i a; = 16, 9, a;i — y' = 37. 2/ = 9, 16. 22. a;4 _j_ ^2 _j_ ^4 _ 2923, a; = ± 7, ± 3, a; 2 — xy -f ?/ 2 = 37. 2/ = ± 3, ± 7. 23. a; 4 + kV + 2/ 4 = 9211, as = ± 9, ± 5, a; 2 — xy + ?/ 2 = 61. 2/ = ± 5, ± 9- 24. X s - y 3 = 56, x = 4, - 2, ar + xy + 2/ 2 = 28. y = 2, - 4. 25. x 3 + y 3 = 126, X = 5, 1, ar - xy + y* = 21. 2/ = 1, 5. 326 EXAMPLES. 26. — + — = 1 t ^tt, Ans. x = 5, 1, x 3 y 3 - + - = if y = i, s. a; y 27. a; 2 + a:?/ = 45, y 2 + a-v/ = 36. x= ± 5 ; y= ±4. 28. 2ar -xy = 56, 2xy — y 2 = 48. .<•= ± 7 ; y= ± 6. 29. x 2 — 2xy = 15, a*?/ — 2y- = 7. ;/■= ±15 ; y= ± 7. 30. x*y(x+y) =80, x*y(2x— 3y)=80. a-= ± 4;y=±l. 31. a; 3 + 1 = %, a; = 2, $, -1, ar 1 -f a; = %y. y = 1, a, 0. Note. — It will be seen that Examples 17 to 31 can be solved by Case III. 32. ar + 3xy = 54, Ans. x = ±3, ±36, xy + Ay 2 =115. y = ±5, T% 3 -. 33. a,- 2 + a;i/ = 24, x = ±4, ±6V^2, 2/ + 3xy = 32. 2/ = ±2, =F8V^2. 34. a; 2 - 3xy = 10, x = ±5, ±4, 4?/ 2 - xy = -1. ?/ = ±1, ±£. 35. a; 2 + xy - 2y 2 = -44, x = ±14, ±1, xy + 3?/ 2 = 80. y = ^ 8, ±5. 36. a; 2 + 3a;?/ = 54, x = ±3, ±36, xy + 4y 2 = 115. y = ±5, Tllf 37. a?(a; ± ? y) = 40, x = ±5, ±4V / 2, y(* - y) = 6. y = ±3, ± V2. 38. a;(a; + ?/) + ?/(.r - ?/) = 158, a- = ± !), ±8^2, 7x(x + y)= 72y(x-y). ?/=±7, ± V2. 39. a- + y = 4, a.- 4 + y 4 = *•_>. a; = 3, 1 ; y = 1, 3. 40. x — y = 3, a; 5 — ?/' = 3093. x = 5, — 2 ; y = 2, —5. 41. afy 3 + 13.tv/ -f 12 = 0, x = 4,-3, L=_V5. aj + y = 1. y= _S, 4,LT^. EXAMPLES. 327 42. x + y = 5, Ans. x = 6, — 1, 5 ± * 17 , 4a$ = 12 - a; 2 ?/ 2 , y = -1, G, 5 T * 17 . 43. xY + 5a;?/ = 84, a- = 7, 1, 4 ± \^8, a; + y = 8. y = 1, 7, 4 T V§8. 44. ar + 4?/ 2 + 80 = 15a + 30?/, a; = 4, 3, 6, 2, xy = 6. 7/ = |, 2, 1, 3. 45. 9a; 2 + ?/ 2 - 63a; - 21?/ + 128 = 0, x = 2, |, 4, -|, an/ = 4. y = 2, 6, 1, 12. 46. a" + 2/ 4 = 14a,- 2 /, a; = 5(1 ± Vs), |/l ± -±=\ - +f-«- ^2 (1=FV/3) ^( 1T ^) a^ - a; — y = 54. y = 12, -9. 48. x 3 + y(xy - 1) = 0, aj= V*(^§ - 1), y* - x(xy + 1) = 0. 1 V2(\/2 - 1) 49. x 5 + y° = an/(aj + ?/) 3 , x = fVlO ± ^V5, a-?/ 4 = (a; + y) 3 . (x«+l)y = (y*+l)x*, y = fV2 ± |V5. . 50. a ; =i[V / | 3 /3 + 3 + V / r3-l], (2/ 6 +l)a; = 9(ar+l)?/ 3 . ylns. ?/ = i [if 3 • V^3 + 3 ± V^s/9- 1], 51. a; + y + « = 13, x = 5, 2, 3 ± V^l, a-" 2 + ?/ a + z~ = 65, ?/ = 2, 5, 3 T V^T, xy =10. 2 = 6, 7. 52. (x + ?/)(a- -f- z) = 30, as = 4, -4, (2/ + 2)0/ + a,") = 15, y = 1, -1, ( 2 +a?)(2 4-y) = 18. z = 2, -2. 328 EXAMPLES. 53. if + yz + z 1 = 49, Ans. z' 2 + zx + a;'- = 19, x 2 + xy + if = 39. X = = ±2, V y = = ±5, *» 2! = = ±3, ** 54. x{y + z) = = / (g + ft-c)(a + c-fr) V 2(6 + c - a) x 7 ., /(« + 6 — c) (6 + c — a) / , x . /(& + c — a) (a + c — 6) 55. The product of two numbers is 60 times their differ- ence, and the sum of their squares is 244 : find the numbers. Ans. 10, 12. 56. Find two numbers whose difference added to the difference of their squares is 14, and whose sum added to the sum of their squares is 26. Ans. 4, 2. 57. Find two numbers such that twice the first with three times the second may make 60, and twice the square of the first with three times the square of the second may make 840. Ans. 18 and 8, or 6 and 16. 58. Find two numbers whose sum is nine times their difference, and whose product diminished by the greater number is equal to twelve times the greater number divided by the less. Ans. 5, 4. 59. Find two numbers whose difference multiplied by the difference of their squares is 32, and whose sum multiplied by the sum of their squares is 272. Ans. 5, 3. 60. Find two numbers whose product is equal to their sum, and whose sum added to the sum of their squares is 12. Ans. 2, 2. 61. Find two numbers whose sum added to their product is 34, and the sum of whose squares diminished by their sum is 42. Ana. 1, 6. EXAMPLES. 329 G2. A number consisting of two digits has one decimal place ; the difference of the squares of the digits is 20, and if the digits be reversed, the sum of the two numbers is 11 : find the number. Arts. G. 4, or 4.6. G3. A man has to travel a certain distance ; and when he has traveled 40 miles he increases his speed 2 miles per hour. If he had traveled with his increased speed during the whole journey he would have arrived 40 minutes earlier ; but if he had continued at his original speed he would have arrived 20 minutes later. Find the whole distance he had to travel, and his original speed. Ans. GO, 10. 64. A and B are two towns situated 18 miles apart on the same bank of a river. A man goes from A to B in 4 hours, by rowing the first half of the distance and walking the second half. In returning he walks the first half at the same rate as before, but the stream being with him, he rows 1£ miles per hour more than in going, and accomplishes the whole distance in 3h hours. Find his rates of walking and rowing. Ans. 4| walking, 4| rowing at first. 65. A and B run a race round a two mile course. In the first heat B reaches the winning post 2 minutes before A. In the second heat A increases his speed 2 miles per hour, and B diminishes his as much ; and A then arrives at the winning post 2 minutes before B. Find at what rate each man ran in the first heat. Ans. 10, 12 miles per hour. 66. Find two numbers whose product is equal to the dif- ference of their squares, and the sum of their squares equal to the difference. of their cubes. Ans. £05, -*-(5 + \/5). 67. The fore- wheel of a coach makes 6 revolutions more than the hind-wheel in going 120 yards ; but if the circum- ference of each wheel be increased 1 yard, the fore-wheel will make only 4 revolutions more than the hind-wheel in the same distance : find the circumference of each wheel. Ans 4 and 5 yards. 68. A vessel is to be filled with water by two pipes. The first pipe is kept open during three-fifths of the time which 330 EXAMPLES. the second would take to fill the vessel ; then the first pipe is closed and the second is opened. If the two pipes had both been kept open together the vessel would have been filled 6 hours sooner, and the first pipe would have brought in two-thirds of the quantity of water which the second pipe really brought in. How long would each pipe alone take to fill the vessel? Ans. 15 and 10 hours. 69. The number of men in both fronts of two columns of troops, A and B, where each consisted of as many ranks as it had men in front, was 84 ; but when the columns changed ground, and A was drawn up with the front that B had, and B with the front that A had, then the number of ranks in both columns was 91 : find the number of men in each column. Ans. 2304, 1296. 70. Two trains start at the same time from two towns, and each proceeds at a uniform rate towards the other town. When they meet it is found that one train has run 108 miles more than the other, and that if they continue to run at the same rate they will finish the journey in 9 and 16 hours respectively. Find the distance between the towns and the rates of the trains. Ans. 756, 36, 27. 71. Two travelers, A and B, set out from two places, P and Q, at the same time ; A starts from P with the design to pass through Q, and B starts -from Q and travels in the same direction as A. When A overtook B it was found that they had together traveled 30 miles, that A had passu 1 through Q 4 hours before, and that B, at his rate of travel- ing, was 9 hours' journey distant from P. Find the dis- tance between P and Q. Ans. 6 miles. 72. Two travelers, A and B, set out at the same time from two places, P and (,), respectively, and travel so as to meet. When they meet it is found that A has traveled SO miles more than 15, and that A will reach Q in 4 days, and I> will reach P in 9 days, after they meet, Find the distance between P and Q. Ana. 150 miles. RATIO — DEFINITIONS. 331 CHAPTER XVI. RATIO — PROPOR T ION — VARIATION. 154. Ratio — Definitions. — The relative magnitude of two quantities, measured by the number of times which the first contains the second, is called their Ratio. The ratio of a to b is usually written a : b ; a is called the first term, and b the second term of the ratio. The first term is often called the antecedent, and the second term the consequent. Magnitudes must always be expressed by means of num- bers; and the number of times which one number contains the other is found by dividing the one by the other. Hence the ratio a : b may be measured by the fraction -. Thus, the ratio a : b is equal to -, or is -. b b Concrete quantities of different kinds can have no ratio to one another ; thus, we cannot compare pounds with yards, or dollars with days. To compare two quantities, they must be expressed in terms of the same unit. For example, the ratio of 4 yards to 15 inches is measured by the fraction 4 x 3 x 12 _ 48 15" ~ " "*■ A ratio is called a ratio of greater inequality, of less in- equality, or of equality, according as the antecedent is greater than, less than, or equal to the consequent. Eatios are compounded by multiplying together the ante- cedents of the given ratios for a new antecedent, and the 332 RATIO — DEFINITIONS. consequents for a new consequent. Thus, the ratio com- pounded of the three ratios, 3a : 2b, 4ab : 5c 2 , c : a, is 3a X tab x c : 2b x 5c 2 x a, or 6a : 5c. When the ratio a : b is compounded with itself, the result- ing ratio is a 2 : b 2 , and is called the duplicate ratio of a : b. Similarly, the ratio a 3 : b 8 is called the triplicate ratio of a : b. Also the ratio a* : b* is called the swbduplicate ratio of a : b. If we interchange the terms of a ratio, the result is called the inverse ratio. Thus b : a is the inverse of a : b. The inverse ratio is the reciprocal of the direct ratio. When the ratio of two quantities can be exactly expressed by the ratio of two integers, the quantities are said to be commensurable; when the ratio cannot be exactly expressed by the ratio of two integers, they are said to be incommen- surable. Although we cannot find two integers which will exactly measure the ratio of two incommensurable quantities, yet we can always find two integers whose ratio differs from the required ratio by as small a quantity as we please. For example, the ratio of a diagonal to a side of a square cannot be exactly expressed b} T the ratio of two whole numbers, for this ratio is \J2, and we cannot find any fraction which is exactly equal to \J2 ; but by taking a suf- ficient number of decimals, we may find \J2 to any required degree of approximation. Thus \]2 = 1.4142135 and therefore \J2 > -\* ft •} J •> and < \ g ' ■;', jj J < . That is, the ratio of a diagonal to a side of a a square lies between f$ mx, but < (m + l)x, then a mx , (m 4- l)x - > — and < ^ ! — ; . b nx nx that is, - lies between — and — — — ; so that - differs from b n n h — by a quantity less than -. And since we cau choose x (our unit of measure) as small as we please, n can be made as great as we please, and therefore - can be made as small n as we please. Hence two integers, m and n, can be found whose ratio will express the ratio a : b to any required degree of accuracy. Note. — The student should observe that the Algebraic definition of ratio deals with numbers, or with magnitudes represented by numbers, while the Geometric definition of ratio deals with concrete magnitudes, such as lines or areas represented Geometrically, but not referred to any common unit of measure. 155. Properties of Ratios. — (1) If the terms of o. ratio be multiplied or divided by the same number the value of the ratio is unaltered. -c, a ma , . , „ nx For - = — (Art. 79). b mb Thus the ratios 2:3, 6:9, and 2m : 3m, are all equal to each other. Two or more ratios are compared by reducing the fractions which measure them to a common denominator. Thus, sup- pose a, : b and c : d are two ratios. Then - = — , - = — ; 1 b bd d bd hence the ratio a : b >, = , or < the ratio c : d, according as ad >, = , or < be. 334 PROPERTIES OF RATIOS. The ratio of two fractions can be expressed as a ratio of two integers. a Thus the ratio -:- is measured by the fraction - or — ; b d J c be and is therefore equivalent to the ratio ad : be. (2) A ratio of greater inequality is diminished, and a ratio of less inequality is increased, by adding the' same quantity to each <>f its terms; that is, the ratio is made more nearly equal to unity. Let a : b be the ratio, and let a + x : b + x be the new ratio formed by adding x to each of its terms. Then * - ±±* = »(° ~ & > ; b b + x b{b + x) and a — b is positive or negative according as a is greater or less than b. TT a + x a ,. 7 ,, . . Hence <, or > - according, as a >, or < b ; that is, b + x b the resulting ratio is brought nearer to unity. For example, if to each term of the ratio 3 : 2 we add 12, the new ratio 15 : 14 is less than the former, because || = l^ is clearly less than § = 1^. Also, if to each term of the ratio 2 : 3 we add 12, the new ratio 14:1;") is greater than the former, since \4 is clearly greater than f. (3) Similarly, it can be proved that a ratio of greater in- equality is increased, and a ratio of less inequality is dimin- ished, by taking the same quantity from both its terms. (4) The following is a very important proposition concern- ing equal ratios. If r = - = — = , then each of these ratios b d f _ / pa" + qc n + re n + . . . A " \/>t>" + qd n + rf" + / ' where p, q, r, n are any quantities whatever. EXAMPLES. 335 = Je ; 6 d / then a = &ft, c = dk, e = fk ; therefore pa* +qe?t « sencs of fractions are equal, each of them is equal to the .sum of all the numerators divided by the sum of all the denominators. EXAMPLES. 1. If - = i, find the value of b f ~ 3 J f . y 7x + -ly OX q hx — Sy _ y _ *£ — 3 7a; + -ly 7x g V" + 2 2/ 2. If a : & be in the duplicate ratio of a + a: : b + as, prove that :« 2 = o&. From the condition / a + a \ 2 _ « a-b + 2abx + &.e a = a& 2 + 2a&sc + ax 2 . ,-. x- = ab. 336 PROPORTION — DEFINITIONS. Find the ratio compounded of 3. The ratios i : 15 and 25 : 30. Aus. 5 : 27. 4. The ratio 27 : 8, and the duplicate ratio of 4 : 3. 6:1. 5. The ratio 10'J : 200, and the duplicate ratio of 15 : 2G. Arts. 9 : 32. 6. If Ax" + if = 4x77, fiud the ratio & : y. 1:2. 7. What is the ratio x : ?/, if the ratio Ax ■+■ by : 3x — y is equal to 2? Ans. 7:2. 8. If lx — Ay : 3x + y = b : \o, find the ratio x : >/. Ans. 3:4. PROPORTION. 156. Definitions. — Four quantities are said to be in proportion when the ratio of the first to the second is equal to the ratio of the third to the fourth ; and the terms of the ratios are said to be projiortionals. Thus, if - = -, then a, b, c, d, are called proportionals, b d or are said to be in proportion. The proportion is written a : b — c : d, or a: b :: c : d, which is read " a is to & as c is to d." The Algebraic test of a proportion is that the two fractions which represent the ratios shall be equal. The four terms of the two equal ratios are called the terms of the proportion. The first and fourth terms are called the extremes, and the second and third, the means. Thus, in the above proportion, a and d are the extremes and b and c the means. Quantities are said to be in continued proportion when the first is to the second, as the second is to the third, as the third to the fourth, and so on. Thus a, b, c, d, e, /, . . . are in continued proportion when a : b = b : c = c : d = d : e = e : / = PROPERTIES OF PROPORTIONS. 337 If a, b, c, be in continued proportion, b is said to be a mean proportional between a and c ; and c is said to be a tMrd proportional to a and b. If a, b, c, d be in continued proportion, b and c are said to be two mean jJroportionals between a and d ; and so on. 157. Properties of Proportions. — (1) If four quan- tities are in proportion, the product of the extremes is equal to the product of the means. Let the proportion be a : b = c: d. Then by definition (Art. 156), - = -. Multiplying by bd, ad = be (1) Hence if any three terms of a proportion are given, the fourth may be found from the relation ad = be. Note. — This proposition furnishes a more convenient test of a proportion than the one in Art. 156. Thus, to ascertain whether 2 : 5 : : 6 : 16, it is only necessary to compare the product of the means and extremes; and since 5 x 6 is not equal to 2 x 16, we see that 2, 5, 6, 16, are not in proportion. If b = c, we have from (1), ad — 6 2 ; .*. b = \ac. That is, the mean proportional between tioo given quanti* ties is equal to the square root of their product. (2) Conversely, If the product of two quantities be equal to the product of tioo others, two of them may be made the extremes, and the other two the means, of a proportion. For let ad = be. Dividing by bd, - = - ; b d that is, a : b : : c : d. In a similar manner it may be shown that the proportions arc: : b:d, b : a : : d: c, b:d : : a: c, c: d : : a:b, etc.. are all true provided that ad = be. 338 PROPERTIES OF PROPORTIONS. If four quantities are in proportion they icill be in propor- tion by (3) Inversion. — If a : b : : c : d, then h : a : : d : c. For - = - ; therefore 1 + ? = 1 -*- C - \ b d b d that is, - = - ; or & : a : : d : c. a c (4) Alternation. — If a : b : : c:d, then a : c :: bid. For ad = be; therefore — = — - ; cd cd that is - = - ; or a : c : : b : cl. c d (5) Composition. — If a : & : : c: d, then a+6 : 6 : : c+d: d. For - = - ; therefore j + 1 = C - + 1 ; 6 d b d that is ^±J> = 2J±J ;ora + &:&::c + d:d. & d (G) Division. — If a : & : : c : d, then a — & : & : : c — d:d. For - = - ; therefore ~ - 1 = 3 - 1 ; & d b d that is a ~ 6 = ^^ ; or a - 6 : 6' >•: c - d : d. 6 d In a similar manner it may be shown that the sum (or the difference) of the first and second of two quantities is to the first as the sum (or the difference) of the third and fourth is to the third. (7) Composition and Division. — If a:b : : c: d, then a + 6 : a — b : : c + d : e — d. For by (5) and (6), a + b c + rZ a — & _ c — i'oportion, the like powers, or roots, of these quantities will be in proportion. For if a : 6 : : c : d, then - = - ; b d a" c" therefore — = — : .-. a" : b n : : c" : d n . 6" d" fin r~n X 1 X L Also — i = —; .-. a*:b n ::c n :d*. b« d» (14) If any number of quantities are in proportion, any antecedent is to its consequent, as the sum of all the antecedents is to the sum of all the consequents. For if a : b : : c : d : : e : f, then by (1), ad = 6c, and af = be ; also ab = ba. Adding a(b + d ■+- /) = b(a + c + e); therefore by (2),a:6::a + c + e:6 + d + /. This also follows directly from (3) of Art. 155. (15) When -,-,—, are unequal, it follows from b d f Ex. 3 of Art. 10G, that a + c + e + g + b+d+f+h+ ■ is greater than the least, and less than the greatest, of the .. .. a c e q fractions -,-,—,, b d f h It is obvious from the preceding propositions that if four quantities are in proportion, many other proportions may be derived from them. The propositions just proved are often useful in solving problems. In particular, the solution of certain equations is greatly facilitated by a skilful use of the operations of composition and division. EXAMPLES. 341 EXAMPLES. 1. If a : b :: c : d, show that a 2 + ab : c 2 + cd : : 6 2 — 2ab : d a — 2cd. Let - = - = x ; then a = 6#, and c = d.t\ b d a 2 + a& _. 6 2 t- 2 + 6 2 o; = & •'* c a + C d rfV 2 + d 2 aj d 2 ' 6 2 - 2a6 _ b 2 - 2b 2 x _ 6 2 d a - 2cd d 2 - 2d 2 a; d 2 ' Therefore by (9), a 2 + ab : c 2 + cd : : 6 2 - 2ab : d 2 - 2cd. If 3a + 6b + c + 2d _ 3a + 66 — c — 2d 3a — 66 + c — 2d 3a — 66 — c + 2d' prove that a : & : : c : d. Bv m 2 (3a + c) = 2 (3a - c) * k ; 2(66 + 2d) 2(66 - 2d)' r r / 4 x 3a + c _ 66 + 2d y ^ ' 3a - c ~ 66 - 2d' Again by (7), 6a = 126 ^ . 6 . . c . rf . & J v 7 2c 4d 3. Find a fourth proportional to x 3 , xy, bx'hj. Ans. by 2 . 4. Find a mean proportional between \2cix 2 and 3a 3 . Ans. Ga 2 x. • 5. Find a third proportional to x 3 and ?« 2 . 4.x. 6. If a : 6 : : c : d, show that (1) ac : 6d :: c 2 : d 2 . (2) a6 : cd :: a 2 : c 2 . • (3) a 2 : c 2 : : a 2 - b 2 : c? - d*. 7. If a : 6 : : c : d, prove that ( 1 ) a6 + cd : «6 - cd : : a' 2 + c 2 f a 2 - c 2 . (2) a 2 + ac + c' 2 : a 2 - ac + c 2 : : 6 2 + &d + d 2 : 6' 2 - 6d + d 2 (3) a : 6 : : s/Sa 2 + 5c 2 : v'36 2 + 5d 2 . (4) a + 6 : c + d : : Va 2 + 6 2 : V^ 2 + d 2 . 342 VARIATION — DEFINITION. 8. If at : b : : c : d : : e : /, prove that 2a 9 + .V - 56 s : 2& 2 + 3d 2 - 5/ 2 : : ae : &/. .r 2 + x — 2 \.r + bx — 9. Solve the equation - 2 5a; - 6 Ans. x = 0, —2. 10. Find a; in terms of y from the proportions x: y : : a 3 : b 3 , and a : b : : \Jc + x : Vd + y. ^ fC = £ 11. If a, fr, c, d are in continued proportion, prove that a : d : : a 3 + & 3 + c 3 : & 3 + c 3 + d 3 . VARIATION. 158. Definition. — One quantity is said to vary directly as another when the two quantities depend upon each other in such a manner that if one be changed the other is changed in the .same proportion. Thus, if a train moving uniformly, travels 40 miles in an hour, it will travel 80 miles in 2 hours, 120 miles in :i hours. and so on ; the distance in each case being increased or diminished in the same ratio as the time. This is expressed by saying that when the velocity is uniform, the distance is proportional to the time, or more briefly, the distance varies as the time. We may express this result with Algebraic symbols thus: let A and a be the numbers which represent the distances traveled by the train in the times represented by the numbers B and b ; that is, when A is changed to any other value a, B must be changed to another value 6, so that A : a '.'. B : &; then A is said to vary directly as B, or more briefly, to vary as B. Another phrase,* which is also in use, is '■'■A is proportional to B." * Strictly Bpeaklog, this phrase is better than the one "varies as," which is somewhat antiquated ; but in deference to usage we retain it. The student must not suppose that the variation hrVr considered is the only kind. We are not here concerned with variation in gent rat, but merely « i 1 1 ■. the simplest of all the possible kimi» of variation. DIFFERENT CASES OF VARIATION. 343 This relation is sometimes expressed by the symbol oc, so that A oc B is read "A varies as B." It will thus be seen that variation is merely an abridged method of expressing proportion, and that four quantities are understood though only two are expressed. If A varies as B, then A is equal to B multiplied by some constant quantity. For suppose that or, a v « 2 , . . . . , b, b v & 2 , are corresponding values of A and B. Let a and b denote one pair of these values, so that when A has the value a, B has the value b ; then we have by the definition, A : a : : B : b. Hence A = -B = mB, b where m is equal to the constant ratio a : b. 159. Different Cases of Variation. — There are four different kinds of variation. (1) One quantity is said to vary Directly as another when the two increase or decrease together in the same ratio. Thus. A oc B, or A = mB (Art. 158). For example, If a man works for a certain sum per hour, the amount of his wages varies as the number of hours during which he works. (2) One quantity is said to vary Inversely as another when the first varies as the reciprocal of the other. Thus A varies inversely as B is written A oc — , or A ■= — , where m is a constant. B B For example, If a man has to perform a certain journey, the time in which he will perform it varies inversely as his speed. If he doubles his speed, he will go in half the time ; and so on. 344 PROPOSITIONS IN VARIATION. (3) One quantity is said to vary as two others Joint};/, when the first varies as the product of the other two. Thus A varies as B and C jointly is written A oc BC, or A = mBC, where m is a constant. For example, The wages to be received by a workman will vary as the number of days he has worked and the tvecges per clay jointly. (4) One quantity is said to vary Directly as a second and Inversely as a third, when it varies jointly as the second and the reciprocal of the third. Thus ^1 varies directly as B and inversely as C is written A B , B , A oc — , or A = m — , where m is a constant. C C For example, The base of a triangle varies directly as the area and inversely as the altitude. In the different cases of variation just defined, to deter- mine the constant m it will only be necessary to have given one set of corresponding values. Example 1. If Ace B, and A = 3 when B = 12, we have A = mB ; .-. 3 = m x 12; or m = £ ; . • . A = \ B. 2. If A varies as B and inversely as 0, and A — 6 when B = 2 and C = 9, we have A = in r ; .-. 6 = m x in = 27 ; A 0~ B ,. .1 = 2,-, 160. Propositions in Variation. — The simplest method of treating variations is to convert them into equations. (1) If A cc B, and 5 cc C, then A cc C. For let A = mil, and B = wC (Art. 158), where m and ft are constants. PROPOSITIONS IN VARIATION. 345 Then A = mnC ; .•. A , then AC cc Si). For let A = m23, and C = w2). Then AC = mnBD. .-. AC oz BD. (5) If ^1 cc J3, then u4" cc 23". For let .4 = m23 ; then A n = m n B". .-. yi n cc23\ (6) If A oz B when C is constant, and A cc C when B is constant, then A -l: :2, 2 5 or 0, or IX -1): H (2) (3) 350 EXAMPLES. Find the value of a; in each of the proportions : 20. 3x - 1 : Gx - 7 : : 7x — 10 : 9a + 10. Ans. 8 or 21. x 2 -2x + B:2x-3::x 2 - 22. 2./' - 1 :.r + 4::ar + 2a; 23. (V / ^TT + V / .^I):(V / a7+l 24. If a : 6 : : c : d : : e :/, prove that a 3 + c 3 + e 3 : & 3 + d 3 + / 3 : : ace : bdf. 25. If a : b : : c : d, prove that (1) «(c + d) = c(a + 6). (« + c)(« 2 + c-) = (h + rQ(ft 2 + d 2 ) (a - c)(a 2 - c 2 ) (6 - d)(& a - d 2 )' pa 2 + f the vessels. Ans. In the first the wine is J of the whole; in the second 5. EXAMPLES. 351 31. If the increase in the number of male and female criminals be 1.8 per cent., while the decrease in the number of males alone is 4.G per cent., and the increase in the number of females is 9.8 per cent., compare the number of male and female criminals respectively. Ans. Female criminals four-fifths of the male. 32. If xozy, and y = 7 when a; = 18, find x when y = 21. Ans. 54. 33. If x oc -, and y = 4 when x = 15, find y when x = 6. y Ans. 10. 34. A varies jointly as 5 and G; and A = G when 5 = 3 and (7=2: find A when B = 5 and C = 7. Ans. 35. 35. A varies jointly as B and C ; and A = 9 when 5 = 5 and C = 7 : find I? when ^1 = 54 and = 10. J.MS. 21. 36. A varies directly as B and inversely as C ; and A = 10 when B = 15 and C = 6 : find .1 when £ = 8 and C=2. Ans. 16. 37. If 3a + 7fr / = 3, find the equation between x and y. Ans. 27a 2 = \y 3 . 46. If y varies as the sum of two quantities, one of which varies as x directly, the other as x inversely, and if y = 4 when x = 1, and y = 5 when x = 2, find the equation between x and y. Ans. y = 2x H — x 47. If one quantity vary directly as another, and the former be f when the latter is f , find what the latter will be when the former is 9. Ans. 16. 48. If one quantity vary as the sum of two others when their difference is constant, and also vary as their difference when their sum is constant, show that when these two quan- tities vary independently, the first quantity will vary as the difference of their squares. 49. If y = the sum of two quantities, one of which varies directly as x, and the other inversely as ar ; and if y = 19 when x = 2, or 3 ; find y in terms of x. Ans. y = 5a; -\ — -. X' 50. If y varies as the sum of three quantities of which the first is constant, the second varies as aJ, and the third as or 2 ; and if y = when x = 1, y = 1 when x = 2, and y = 4 when x = 3 ; find y when x = 7. Ans. 36. 51. If y = the sum of three quantities, of which the first is constant, the second varies as x, and the third varies as a? ; also when x = 3, 5, 7, y = 0, —12, —32, respectively; find y in terms of x. Ans. y — 3 + 2x — x' 1 . 52. If y varies as the sum of three quantities, of which the first is constant, the second varies as x, and the third as x 2 ; also when x = a, 2a, 3a, y = 0, a, 4a, respectively ; show that when x = na, y = (n — l) 2 a. DEFINITIONS — FORM UL^E. 353 CHAPTER XVII. ARITHMETIC, GEOMETRIC, AND HARMONIC PROGRESSIONS. ARITHMETIC PROGRESSION. 161. Definitions — Formulae. — A number of terms formed according to some law is called a series. Quantities are said to be in Arithmetic Progression* when they increase or decrease by a constant difference, called the common difference. Thus, the following series are each in Arithmetic Progres- sion : 2, 5, 8, 11, 14, 17, 9, 7, 5, 3, 1, -1, —3, —5, a, a 4- d, a + 2(Z, a + 3d, a + 4(Z, The letters A. P. are often used for shortness instead of the term Arithmetic Progression. The common difference is found by subtracting any term of the series from that which immediately follows it. In the first series above the common difference is 3 ; in the second it is — 2 ; in the third it is d. The series is said to be increasing or decreasing, according as the common difference is positive or negative. Thus, the first series above is increasing, and the second is decreasing. If we examine the third series above, we see that the coef- ficient of d in any term is less by one than the number of the term in the series. Thus the 2d term is a + d, 3d term is a + 2d, 4th term is a + 3d, * Called also Arithmetic Series. 354 DEFINITIONS — FORM ULJE. and so on. Hence if n be the number of terras, and if I denote the last, or n th term, we have I = a + (n - l)d (1) Let S denote the sum of n terms of this series ; then we have 8 = a + (« + d) + (a + 2d) +...+(*- 2d) + (l- d) +1 : and, by writing the series in the reverse order, we have 8 = I + (I - d) + (I - 2rf) + . . . + (a + 2d) + (a + d) + a. Adding together these two equations, we have 2S = (a + V) + (a + I) + (a + + to n terms = w(a + ?). .'. iSf = ^(a + /) (2) By (1) and (2) we have S = ~[2« + (n - l)d] . . (3) We have here three useful formulae, (1), (2), (3), which should be remembered ; in each of these any one of the letters may denote the unknown quantity when the other three are known. For example, in (1), we can write down any term of an A. P. when the first term, the common dif- ference, and the number of the term are given. Thus, if the first term of an A. P. is 5 and the common difference is 3, the 10th term = 5 + (10 - 1)3 = 32, and the 20th term = 5 + (20 - 1)3 = 62. Also in (2), if we substitute given values for S, n, 1, we obtain an equation for finding a; and similarly in (3). Tims, 1. Find the sum of 20 terms of the series 1, 3, 5, 7, . . . Here a = 1, d = 2, n = 20; therefore by (3) S = %°[2 + 11) X 2] = 10 x 40 = 400. 2. The first term of a scries is 5, the last 4;".. and the sum 400 ; find the number of terms, and the common di 1'1'e re nee. DEFINITIONS — FORM UL^E. 355 Here a = 5, I = 45, S = 400 ; therefore by (2) 400 = -(5 + 45) = 25m. .-. n = 16. By (1) 45 = 5 + 15& .'. d = 2f. When any taco terms of an A. P. are given, the series can be completely determined ; for the data furnish two simul- taneous equations, with two unknown quantities, which may be solved by -methods previously given. 3. The 10th and 15th terms of an A. P. are 25 and 5 respectively ; find the series. Here 25 = a + 9d ; and 5 = a + 14tf. By subtraction, 20 = — 5d. .*. d = —4. Then a = 5 — 14cZ = 61. Hence the series is 61, 57, 53, 4. Find the sum of the first n odd integers. Here a . = 1, and d = 2 ; therefore by (3) S = "[2 + (« - 1)2] = ;i X 2n = n 2 . 2 2 Thus the sum of any number of consecutive odd integers beginning with unity, is the square of their number.* Find the last term and sum of the following series : 5. 14, 64, 414, to 20 terms. Ans. 964, 9780. 6. 9, 5, 1, to 100 terms. -387, -18900. l, -f , to 21 terms. -9f, -99f. 4' 4' Find the sum of the following series : 8. 5, 9, 13, to 19 terms. 779. 9. 12. 9, 6, to 23 terms. -483. Find the series in which 10. The 27th term is 186, and the 45th is 312. Ans. 4, 11, 18 * This proposition was knowu to the Greek geometers. 356 ARITHMETIC MEAN. 11. The 9th term is Am. 1, -£, -2, 12. The 16th term is 214, and the 51st is 739. Am. -11, 4, 19, 162. Arithmetic Mean. — When three quantities are in Arithmetic Progression, the middle one is called the Arith- metic Mean of the other two. Thus if a, b, c are in A. P., & is the arithmetic mean of a and c ; and by the definition of A. P. we have b — a = c — 6 ; ... i = l( ft + C ). Tims the arithmetic mean of any two quantities is half their sum. Between any two given quantities any number of terms may be inserted so that the whole series thus formed shall be in A. P. ; the terms thus inserted are called the arithmetic means. For example, to insert four arithmetic means between 10 and 25. Here we have to find an A. P. with 4 terms between 10 and 25, so that 10 is the first and 25 is the sixth term. By (1) of Art. 161, 25 = 10 + 5cZ; .-. d = 3. Thus the series is 10, 13, 16, 19, 22, 25 ; and the required arithmetic means between 10 and 25 are 13, 16, 19, 22. In general. To insert n arithmetic means between a and b. Here we have to find an A. P. with n terms between a and &, so that a is the first and b is the (w -f- 2) Ul term. By (1) of Art. 161, b = a + (u + •> - \)d = a -f (w + l)d .-. d = b ^^. n + 1 ARITHMETIC MEAN. 357 Thus the required means are , b — a . -.6 — a b — a a H , a + 2 , a + n n + 1 n + 1 ?i+l 1. Find the sum of the first £> terms of the series whose w th term is 3n — 1 . By putting ft = 1, and n = p respectively, we obtain first term = 2, last term = 3p — 1 . Hence by (2), Art. 161, S = 2(2 + 3j> - 1) = £(3p + l). In an Arithmetic Progression when a, S, and d are given, n is to be found by solving the quadratic (3), Art. 161. When both roots are positive and integral, there is no diffi- culty in interpreting the result corresponding to each. 2. How many terms of the series 24, 20, 16, must be taken that the sum may be 72 ? Here a = 24, d = -4, S = 72. Then from (3), Art. 161, we have 72 = 5[2 X 24 + (n - l)(-4)] = 24n - 2n(n - 1). .-. n 2 - 13n + 36 = 0, or (n - 4) (n - 9) = 0. .-. ?i = 4, or 9. Both of these values satisfy the conditions of the ques- tion ; for if we take the first 4 terms, we get 24, 20, 16, 12 ; and if we take the first 9 terms, we get 24, 20, 16, 12, 8, 4, 0, —4, — 8, in either of which the sum is 72; the last 5 terms of the last series destroy each other, so that the sum of the first 4 terms is the same as the sum of the first 9 terms. When one of the roots is negative or fractional, it is inap- plicable, for a negative or a fractional number of terms is, strictly speaking, without meaning. In some cases however a suitable interpretation can be given for a negative value of n. 358 ARITHMETIC MEAN. 3. How many terms of the series —9,-6,-3, must be taken that the sum may he G6 ? Here 66 = "[-18 + (n - 1)8]. .-. ?i 2 -7w-44 = 0; or (n - ll)(n + 4) = 0. .-. n = 11, or -4. If we take 11 terms of the series, we have -9, -6, -3, 0, 3, 6, 9, 12, 15, 18, 21 ; the sum of which is 66. If we begin at the last term and count backwards four terms, the sum is also G6. From this we see that, although the negative solution does not directly answer the question proposed, we are enabled to give it an intelligible meaning as follows : begin at the last term of the series which is furnished by the positive value of ??,, and count backwards for as many terms as the negative value indicates ; then the result will be the given sum. AVe thus see that the negative value for n answers a question closely connected with that to which the positive value applies. 4. How many terms of the series 26, 21, 16, must be taken that the sum may be 74? Here 74 = -[52 + (n - l)(-5)]. Solving, we get n = 4, or 7-|. Thus, the only applicable value of n is 4. Wo infer that of the two numbers 7 and 8, one corresponds to a sum greater, and the other to a sum less than 71. 5. Insert 3 arithmetic means between 12 and -JO. A ns. II. 16, 18. 6. Insert 5 arithmetic means between 14 and 16. -'"*• II?,. US 7. Insert 17 arithmetic means between 98 and 69. Arts. 91$, «»oi, 7,|i. GEOMETRIC PROGRESSION. 359 How many terms must be taken of the series 8. 42, 30, 3G, to make 315? Ans. 14, or 15. 9. _i6, -15,-14, to make -100? 8, or 25. 10. 20, 18f, 17|, to make 162£? 13, or 20. 11. The sum of three numbers in A. P. is 39, and their product is 2184 ; find them. Ans. 12, 13, 14. 12. The sum of 10 terms of an A. P., whose first term is 2, is 155 ; what is the common difference? Ans. 3. GEOMETRIC PROGRESSION. 163. Definition — Formulae. — Quantities are said to be in Geometric Progression when they increase or decrease by a constant factor, called the common ratio. Thus, the following series are each in Geometric Progres- sion (G. P.): 3, 6, 12, 24, 48, Q 1 lit °1 1 1 "SP "9' 2ll a, ar, ar 2 , ar 3 , ai A , The common ratio is found by dividing any term of the series by that which immediately precedes it. In the first series above the common ratio is 2 ; in the second it is ^ ; in the third it is r. The series is said to be increasing or decreasing, according as the common ratio is greater than 1, or less than 1. Thus, the first series above is increasing, and the second is decreasing. Note 1. — An Arithmetic Progression is formed by repeated addition or subtraction; a Geometric Progression by repeated multiplication or division. If we examine the third series above, we see that the exponent of r in any term is less by one than the number of the term in the series. Thus, the 2d term is ar, 3d term is ar 2 , 4th term is ar 3 , 360 GEOMETRIC PROGRESSION. and so on. Hence if n he the Dumber of terms, and if I denote the last, or u tu term, we have 1= <"- 1 (1) Let S denote the sum of n terms of this series ; then we have S = a + ar + ar + + ar" -3 + a?-" -1 ; multiplying by r, we have SV = ar + ar 2 + + ar"- 2 + a/-"- 1 + ar*. Hence by subtraction, we have Sr - S = ar* - a ; or (r - 1)S = a{r* - 1). s = a(f - 1) or «(1 - r") r - 1 1 - r w Multiplying- (1) by r, and substituting in (2), we get c rl — a a — rl , ON /S = , or , .... (3) r - 1 1 - r w a form which is sometimes useful. Note 2. — It will be found convenient to remember both forms given in (2) for S, and to use the first form in all cases when r is positive and > 1, and the second when r is negative or < 1. 1. Find the 8th term of the series — ■§■, -|, — f , Here a = — i, n = 8, r = | h- ( — ^) = — § : therefore ' J y(l) /_ lf_3\7__l/_2187\ 1 — ~ t\ 2) — tv. — rsrJ = fit = the 8th term - 2. Sum the series 1, 3, 9, to G terms. Here a = 1, n = G, r = 3; therefore by the first form ° f(2) ' ^ = 3^-j = 729 - 1 = 3G4j 3-1 2 3. Sum the series 81, 54, 36, to 9 terms. Here a = 81, v = 9, r = 54 ■— 81 = § ; therefore by the second form of (2), S = gl l 1 ~ (J Q!] = 248 [1 - (|)»] 1 — r! 213- -r =236|f. GEOMETRIC MEAN. 361 4. Sum the series 2, —3, |, — to 7 terms. Here a = 2, n = 7, r = — § ; therefore by the second form ° f (2) ' S = 2fl - (-I) 7 ] = 2fl + -¥M 1 - (-1) I = 3 x W^ = utf. 5. Find the Oth term of each of the following series : (1) 9, 3, 1, etc. ; (2) 2, -3, f, etc. ; (3) a 2 , aft, £/ 2 , etc. Jm. (l) ^i (2) -W; (3)|- Sum the following series : 6. 1, 4, 16, to 6 terms. Ans. 1365. 7. 25, 10, 4, to 4 terms. 40f . 8. |, -1, |, to 7 terms. - 4 gV- 9. 3,-1, $, to 6 terms. 2|^. 164. Geometric Mean. — When three quantities are in Geometric Progression the middle one is called the Geometric Mean between the other two. Thus if a, b, c are in G. P., b is the geometric mean between a and c ; and by the definition of G. P., we have 6 = c. a b ' b 2 = ac; .*. b = Sac. Thus, the geometric mean between any two quantities is the square root of their product. Quantities which are in G. P. are in continued proportion, and the geometric mean between .two quantities is the same as their mean proportional (Art. 156). Between any two given quantities any number of terms may be inserted so that the whole series thus formed shall be in G. P. ; the terms thus inserted are called the geometric means. For example, to insert three geometric means between 2 and 32. Here we have to find a G. P. with 3 terms between 2 and 32, so that 2 is the first and 32 is the fifth term. By (1) of Art. 163, 32 = 2r 4 ; .-. r = 2. 3G2 THE SUM OF AN INFINITE NUMBER OF TERMS. Thus the series is 2, 4, 8, 1G, 32, and the required geometric means between 2 and 32 are 4, 8, l(i. In general. To insert n geometric means between a and b. Here we have to find a G. P. with n terms between a and &, so that a is the first and b is the (u + 2)" 1 term. By (1) of Art. 1G3, b = ar n+1 : .-. V i+1 = - "VI- a (1) Thus the required means are ar, ar 2 , ar", where r has the value found in (1). 1. Insert 4 geometric means between 1G0 and 5. Ans. 80, 40, 20, 10. 2. Insert 6 geometric means between 5G and — l \. Ans. -28, 14, -7, |, -J, |. 3. Insert 4 geometric means between 5£ and 40^. Ans. 8, 12, 18, 27. 165. The Sum of an Infinite Number of Terms. — From (2) of Art. 1G3, we have s _ «(1 — r") _ __a «r^_ _ ^ 1 - r 1 - r 1 - r Now suppose r is a proper fraction, positive or negative ; then the greater the value of n the smaller is the absolute value of r", and consequently of ; and by taking n sufficiently large r" can be made as small as we phase. Hence, by taking n large enough, the sum of n terms of the series can be made to differ from — — by as small a quan- tity as we please. Thus, the sum of an infinite number of terms of a decreas- ing Geometric Progression is ; or more briefly, the sum to infinity is '■ — . THE SUM OF AN INFINITE NUMBER OF TERMS. 363 This quantity, , which we call the sum of the scries, 1 — r is the limit to which the sum approaches, but never actually attains; that is, although no definite number of terms will amount to , yet by taking a sufficient number, the sum 1 — r will reach it as near as we please. 1. Sum the series |, |, -J, For n terms we have by (2) of Art. 163, _.*-*) 1-1. 2" From this result it appears that however many terms be taken, the sum of this series is always less than 1. Also we see that by taking n large enough, the fraction — can be made as small as we please. Hence by taking a sufficient number of terms, the sum can be made to differ from 1 by as little as we please ; and when n is made infinitely great we have S = 1. This may be illustrated geometrically as follows: A\ 1 1 1 1 \B Let AB be a line of unit length. Bisect AB in P, ; bisect PJ> in P 2 , P 2 B in P 3 , P 3 B in P 4 , and so on indefinitely, always bisecting the remaining distance. It is evident that by a series of such bisections we can never reach B, because we shall always have a distance left eqnal to half the preceding distance ; but by a sufficient number of these bisections we can come nearer to B than any assigned distance, however small, because every bisection carries us over half the remain- ing distance. That is, if we take a sufficient number of terms of the we shall have a result differing from AB, i.e., from unity, by as little as we please. This is simply a geometric way of saying that ^ + i + ^- + 2 2* 2* 364 VALUE OF A REPEATING DECIMAL. Sum the following series to infinity : 2. 1, £,1 Arts. 2. 3. 9, -G, 4, - 5f. 4. 1, -*»*, |. 5- i,i,A, |. 166. Value of a Repeating* Decimal. — Repeating decimals furnish a good illustration of infinite Geometric Progressions. 1. Find the value of .423. .423 = .4232323 = ± 23 23_ 10 10 3 10 5 _ j4_ 23 10 10 j s V ^ io 2 io 4 ^ / ^ + ^{77 L l) [,,y(1)ofArt - 1G5] \ 10 2/ = ± 23 , _ 4 23 100 = 4 23 = 419 10 IO 3 99 10 990 ~ 990' which agrees with the value found by the usual rule in Arithmetic. The value of any repeating decimal may be found by the method employed in the last example; but in practice it maybe found more easily by a general rule, which may be proved as follows: Let P denote the figures which do not repeat, and suppose them p in number; let Q denote the repeating period consisting of (/ figures. Let 8 denote the value of the repeating decimal ; then 8 = .PQQQ ; .-. LOpS = P.QQQ ; and 10p+« tV 3V 3. Insert 4 harmonic means between 1 and 6. 1 1, 1^, 2, 3. 169. Relation between Arithmetic, Geometric, and Harmonic Means. (1) If A, G, H be the arithmetic, geometric, and har- monic means between a and b, then (Arts. 1G2, 164, 168), A = ^A (1) G = Va6 (2) H=™i- (3) a + b K ' Therefore AH = ?LdL& x -^- = a& = £ 2 ; 2 a + & that is, 6r is the geometric mean between A and //. Hence the geometric mean between any two real positive quantities, a and b, is also the geometric mean between the arithmetic and the harmonic means between a and b. (2) From (1) and (2) we have A - G = ±±± - s/al, = Wa ~ fb)*i and from (2) and (3), o-b=^-J^- = -%(.^- VS)'. a + b a -}- b Now if a and b arc both positive, \ G > II. EXAMPLES. 369 That is, the arithmetic, geometric, and harmonic means between any two real jjositive quantities are in descending order of magnitude.* Three quantities, a, b, c, are in A. I\, G. P., or H. P., according as °^± = ( \ % or % respectively. b — c a b c The first follows from the definition of A. P. (Art. 161). In the second, 6(a — b) = a(b — c) ; .-. b 2 = ac. See Art. 164. The third follows from (1) of Art. 167. Harmonic properties are interesting chiefly because of their im- portance in Geometry and in the Theory of Sound. If there be a series of strings of the same substance, the lengths of which are proportional to 1, £, j, \, z, h ail( l if these strings are stretched tight with equal force, and any two of them are sounded together, the effect is found to be harmonious to the ear. Notwithstanding the comparative simplicity of the law of its formation, there is no general formula for the sum of any number of terms in harmonic progression. EXAMPLES. Find the last term and sum of the following series : 1. 1, 1.2, 1.4, to 12 terms. Ans. 3.2, 25.2. 2. 3£, 1, -If, to 19 terms. —41 J, -361. 3. 64, 96, 128, to 16 terms. 544, 4864. Sum the following series : 4. 4, 5±, 6f, to 37 terms. 980f 5. -3, 1,5, to 17 terms. 493. 6. 3a, a, —a, to a terms. a 2 (4 — a). 7. 31, 2£, If, to n terms. \o * * These two propositions were known to the Greek geometers. 370 EXAMPLES. 8. 1|, HJ-, 2|$, to n terms. Arts. " (17 + 7 "\ 9. ■— - , V^, — — ^ — , .... to 7 terms. 7(^2 + 2). \ft + 1 ^2-1 Find the scries in which 10. The 15 th term is 25, and the 29 th term 46. Arts. 4, ;H, 7, 11. The 15 th term is -25, and the 23 d term -41. Ans. 3, 1, -1, 12. Insert 14 arithmetic means between — 7£ and — 2i. Ans. -OH, -6&, -2 T V 13. Insert 36 arithmetic means between 8| and 2£. Ans. &|, 8$, 2|. How many terms must be taken of the series 14. 15|, 15±, 15, to make 129? .4ns. 9, or 86. 15. -10£, -9, -lh, ... to make -42? 7, or 8. 16. -6f, -6f, -6, .... to make -52$ ? 11, or 24. 17. The sum of three numbers in A. P. is 33, and their product is 792 : find them. Ans. 4, 11, 18. 18. An A. P. consists of 21 terms ; the sum of the three terms in the middle is 129, and of the last three is 237 : find the series. Ans. 3, 7, 11, 83. 19. The first term of an A. P. is 5, and the fifth term is 11 : find the sum of 8 terms. Ans. 82. 20. The sum of four terms in A. P. is H, and the last term is 17: find the terms. Ans. 5, 9, 13, 17. 21. The seventh term of an A. P. is 12, and the twelfth term is 7 ; the sum of the series is 171 : find the number of terms. Ans. bs, or L9. 22. A sets ont from a place and travels 2.1 miles an hour. B sets out 3 hours after A, and travels in the same direction, 3 miles the first hour, .">j,- miles the second. 1 miles the third. and so on. In how many hours will B overtake A? Ans. •"». 23. In the series, 1, 8, 5, etc., the sum of 2n terms : the sum of n terms : : x : 1 : find the value of SB. Ans. 4. EXAMPLES. 371 24. Find an A. P. such that the sum of the first five terms is one-fourth the sum of the following five terms, the first term being unity. Ans. 1, — 2, —5, — 2G. 25. If the sum of m terms of an A. P. be always to the sum of n terms in the ratio of m 2 to n\ and the first term be unity, find the n ih term. Ans. 2n — 1. 26. If 2n + 1 terms of the series 1, 3, 5, 7, 9, ... . be taken, show that the sum of the alternate terms, 1, 5, 9, will be to the sum of the remaining terms 3, 7, 11, as n + 1 to n. 27. On the ground are placed n stones ; the distance between the first and second is one yard, between the second and third three yards, between the third and fourth five yards, and so on : how far will a person have to travel who shall bring them one by one to a basket placed at the first stone? * An ^ n^ _ l){2n _ 1} y . mls> o 28. Find a series of arithmetic means between 1 and 21, so that their sum has to the sum of the two greatest of them the ratio of 11 to 4. Ans. 9 means, 3, 5, 7, 19. 29. Find the number of arithmetic means between 1 and 19 when the second mean is to the last as 1 to G. Ans. 17. 30. If the second term of an A. P. be a mean proportional between the first and the fourth, show that the sixth term will be a mean proportional between the fourth and the ninth. Find the last term of each of the following geometric series : 31. 2, —6, 18, to 8 terms. Ans. —4374. 32. 2, 3, 4§ to G terms. Vr- 33. 3, —3'-, 3 3 , to 2n terms. —3-". Sum the following series : 34. 1, -.V. \, to 12 terms. Ans. i|||. 35. cj, _c, 4, to 7 terms. 5ff. * See Art. 170. 372 EXAMPLES. 36. 2, —4, 8, to 2;> terras. Ans. |(1 — 2 2p ). 37. ^2, Vg, 3^2, to 12 terms. 364(^6 + ^2). 38. Insert 3 geometric means between 486 and G. Ans. 162, 54, 18. 39. Insert 4 geometric means between | and 128. Ans. £, 2, 8, 32. 40. Insert 3 geometric means between 1 and 256. Ans. 4, 16, 64. 41. Insert 4 geometric means between 3 and — 729. Ans. -9, 27, -81, 243. Sum the following series : 42. §, J, £, to 6 terms. Ans. fff . 43. 1, — ^, £, to infinity. f. 44. 6, —2, f, to infinity. 4£. 45. }, |, ^\, to infinity. 1. 46. |,-1, f, to infinity. §f 47. .9, .03, .001, to infinity. §£. Find the value of the following repeating decimals : 48. .4282828 . . . Ans. §£§. I 50. .16. Ans. £. 49. .28131313... ^ r . I 51. .378. |f. 52. The sum of three terms in G. P. is 63, and the difference of the first and third terms is 45 : find the terms. Ans. 3, 12, 48 ; or 36, -54, SI. Let a, ar, «r 2 denote the numbers. 53. The sum of the first four terms of a G. P. is 40, and the sum of the first eight terras is 3280 : find the series. Ans. 1, 3, 9, 54. The sum of three terms in G. P. is 21, and the sura of their squares is 189 : find the terms. Ans. 3, 6, 12. 55. A person who saved every year half as much again as he saved the previous year had in seven years saved $102.95 : how much did he save the first year? Ans, $3.20. EXAMPLES. 373 5G. In a G. P. show (1) that the product of any two terms equidistant from a given term is always the same, and (2) that if each term be subtracted from the succeeding, the successive differences are also in G. P. 57. Show that the square of the arithmetic mean of two quantities is equal to the arithmetic mean of the arithmetic and geometric means of the squares of the same two quanti- ties. 58. There are four numbers, the first three of which are in G. P., and the last three in A. P. ; the sum of the first and last is 14, and the sum of the second and third is 12 : find the numbers. Ans. 2, 4, 8, 12. 59. Three numbers whose sum is 15 are in A. P. ; if 1, 4, and 19 be added to them respectively, the results are in G. P. : find the numbers. Ans. 2, 5, 8. Find the last term of the following harmonic series : 60. 6, 3, 2, to G terms. Ans. 1. Gl. 8, 2, li to 6 terms. J. 62. Insert three harmonic means between 2| and 12. Ans. 3, 4, G. G3. The arithmetic mean of two numbers is 9, and the harmonic mean is 8 : find the numbers. Ans. 6, 12. G4. Find two numbers such that the sum of their arith- metic, geometric, and harmonic means is 9|, and the product of these means is 27. Ans. 1, 9. 65. The arithmetic mean of two numbers is 3, and the harmonic mean is -| : find the numbers. Ans. 2, 4. 66. There are three numbers in II. P., such that the greatest is the product of the other two, and if one be added to each, the greatest becomes the sum of the other two : find the numbers. Ans. 2, 3, 6. 67. The sum of two consecutive terms in H. P. is T 2 n 9 T , 374 MATHEMATICAL INDUCTION. CHAPTER XVIII. MATHEMATICAL INDUCTION — CONTINUED FRACTIONS — PERMUTATIONS AND COMBI- NATIONS. 170. Mathematical Induction — General Proofs of Theorems Stated in Art. 51. — Many mathematical form- ula? are not easily demonstrated by a direct mode of proof ; in such cases it is often found convenient to employ a method of proof which is known as MatJiematiccU Induction. This method may be illustrated as follows : 1. Suppose it is required to establish the truth of the following formula : p + 2 » + 3 2 + + n 2 = »(tt + l)(g»+l). G We can easily see hy trial that this formula is true in simple cases, such as when n = 1, or 2, or 3 ; and from this we might be led to conjecture that the formula was true in all cases. Assume that it is true for n terms. Add (n + l) a to both members ; then !2 +2a+3a+ . . . +wa+(yi+1) 2 = n(n+l)(2n+l) + (w+1)a G = (n + l ^±J) +u + 1 j = (ft + l)(u + 2)('2 n + 3) . G which is the same formula for the sum of n + 1 terms of Ihe series that we assume to be true for n terms, n + 1 taking the place of n. In other words, if the formula is true when MATHEMATICAL INDUCTION. 375 we take a certain number of terms, whatever that number may be, it is true when we increase that number by one. But we see that it is true when 3 terms are taken ; therefore it is true when 4 terms are taken ; and therefore it is true when 5 terms are taken ; and so on. Hence the formula is true universally. 2. The first theorem stated in Art. 51 may be proved by this method as follows : By division a" - t __ x n-i + y(af- 1 -y B - 1 ) . x — y x — y hence if x"~ l — y n ~ l is divisible by x — y, then x n — y n is divisible by x — y. But we know that x 2 — y 2 is divisible by a; — y ; therefore X s — y 3 is divisible by x — y ; and therefore x* — y i is divisible by x — y ; and so on indefi- nitely. Hence x n — y n is always divisible by x — y, when n is any positive integer. Note. — The two theorems which we have here j)roved by the method of induction may be established by other methods. There are many other theorems which are capable of easy proof by the method of induction; and in the subsequent part of this book we shall some- times have occasion to use this method. Tbe student of Natural Philosophy will find the word induction used in a different sense in that subject from what it has here. There we cannot be sure that the law holds in any cases except those which we have examined. In fact, induction, as used in Natural Philosophy, is never absolutely demonstrative, and often far from it; whereas the method of mathe- matical induction is as rigid as any other process in mathematics. If we change y into —y,x—y becomes x — (—y) = x + y; also x" — y n becomes x n — ( — y) n , which is x n — y* or .<;" + y", according as n is even or odd. Hence, when n is even, x n — y n is divisible b}^ x -\- y ; and when n is odd, x 11 + y" is divisible by x + y. Since af* -f- y n = x n — y n + 2y n , it follows that when x n + y" is divided by x — ?/, the remainder is 2y" ; therefore x n + y n is never divisible by x — y. 376 CONTINUED FRACTIONS. CONTINUED FRACTIONS. 171. Definition. — A Continued Fraction is one whose denominator is a whole number plus a fraction ; the denomi- nator of this last fraction a whole number plus a fraction, and so on. Thus, a + b - a + l - c + =— , b + f — J —— c + g + etc. d + etc. are continued fractions; but the term is commonly restricted to the latter form, where all the numerators of the partial fractions are 1, and all the denominators, a, b, c, are positive integers. For economy of space the continued fraction is often written in the more compact form a + b+ c+ d -f etc. A continued fraction may either terminate with one of the denominators, or it may extend indefinitely. When the number of terms, a, b, c, is finite, the fraction is said to be terminating ; if the number of terms is unlimited, the fraction is called an infinite continued fraction. A ter- minating continued fraction may be reduced to an ordinary fraction by simplifying the fractions in succession, beginning with the lowest. 172. To Convert a Given Fraction into a Continued Fraction. Let — be the oiven fraction ; divide m by n. let a be the n J quotient and p the remainder ; thus m = a + P _ a + 1 n n n CONTINUED FRACTIONS. 377 Divide n byj9, let 6 be the quotient and q the remainder; thus 5 = & + 2 = 6 + l. 2> V P f l Divide p by q, let c be the quotient and r the remainder ; and so on. Thus m , 1 ,11 — = « -i = a ^ — ■ -. n 6 4- 1 6+ c + etc. c + etc. If m < n., the first quotient a is zero. It will be seen that the above process is the same as that of finding the greatest common divisor of m and n. Hence we shall at last arrive at a point where the remainder is zero, and the process terminates. Thus every fraction whose numerator and denominator are positive integers can be converted into a terminating continued fraction. The fractions formed by taking one, two, three, etc., of the quotients of a continued fraction are called the first, second, third, etc., converging fractions, or convergents, because, as will be shown in Art. 176, each successive con- vergent is a nearer approximation to the true value of the fraction than any of the preceding convergents. Thus, the first convergent is a ; the second convergent is formed from « H — , which is therefore — — ; the third convergent is b b formed from ft ^ , which is therefore a + 6+i bc + l c abc + « + c t — ^— ; and so on. 6c + 1 173. The Successive Convergents are Alternately Less and Greater than the Continued Fraction. — The first convergent, ft, is too small because the part 378 SUCCESSIVE CONVERGENTS. is omitted. The second convergent, a -\ — , is too b + etc. 6' great, because the denominator b is too small. The third convergent, a + , is too small because 6 + - is too c great ; and so on. When the given fraction is a proper fraction, a = ; if in this case zero be considered as the first convergent, the above results may be enunciated as follows : The convergents of an odd order are all less, and the con- vergents of an even order are all greater, than the continued fraction. 174. To Prove the Law of Formation of the Suc- cessive Convergents. — The first three convergents are (Art. 172) a ab + 1 c(ab + 1) + a . l' b ' be + 1 we see that the numerator of the third convergent may bo formed by multiplying the numerator of the second conver- gent by the third quotient, and adding the numerator of the first convergent ; also that the denominator of the third convergent may be formed in a similar manner by multiply- ing the denominator of the second by the third quotient, and adding the denominator of the first. We shall now prove by induction that this law holds universally. Let the numerators of the successive convergents be denoted by p v p 2 , jp 8 , etc., and the denominators by q v q.„ q s , etc.; o a , a 2 , a 3 , etc., the corresponding quotients; and assume that the law of formation above stated holds for the u"' convergent, so that Since the (n + 1)"' convergent differs from the » tt only DIFFERENCE BETWEEN CONVERGENTS. 379 iii having the quotient a„ -] instead of a„, it follows that this convergent, or <2n + l + JV <» + — YJn-i + q n ->. + l(«»P.-l +Pn-2) +P,-1 +iK&-i + g„_ 2 ) + n-1 «»+ig» + g„_i Hence if we put p B+ i=a» + ii\+P»-n g» + i = «» + i7»+f7»-n we see that the (n + l) th convergent follows the same law that was supposed to hold for the n th convergent. But the law does hold for the third convergent ; therefore it holds for the fourth, and so on ; therefore it holds universally. Reduce ff| to a continued fraction, and find the successive convergents. By the process of finding the greatest common divisor of 323 and 117, the successive quotients are found to be 2, 1, 3, 5, 1, 1, 2. Thus (Art. 172) m = 2 + — - — T 3 + 1 + i+i or 2 H . 1+ 3+ 5+ 1+ 1+2 The successive quotients are 2, 1, 3, 5, 1, 1, 2. The successive convergents are f, f , J^ 1 -, ff, ff, -^y, fff . 175. Difference Between Two Consecutive Con- vergents. — The difference between any two consecutive con- vergents is a fraction whose numerator is unity, and whose 380 LIMIT OF ERROR IN TAKING ANY CONVERGENT. denominator is the product of the denominators of the con- venient*. This is evident with respect to the first two convergents ; ~ A , . „ , ab + 1 a 1 for, Art. 1/4, ±— = -. b 16 Assume the law to hold for any two consecutive conver- gents, 2- , i-± ; so that Pi _ P __ PiQ ~ PQi * = J_ . ( h ~ 3 ^i 99j ' m ' ' ^ > then, a, a x , o 2 , etc., being the corresponding quotients, we have (Art. 174) Pz _ Pi _ P 2 9l ~ Plg 2 ?2 ~ ft ^1^2 _ (" 2 fli + P)ffi ~ K9i + g)p t = ^T- = ^' by(1) ' Hence if the law holds for one pair of consecutive conver- gents, it holds for the next pair. But the law does hold for the first pair, therefore it holds for the second pair, and so on. Hence it is universally true. Note. — In computing the numerical value of the successive convergents, this theorem furnishes an easy test of the accuracy of the work. All convergents are in their lowest terms; for if p and q had a common divisor it would divide }) x q ~ pq v or unity, which is impossible. 176. Limit of Error in Taking any Convergent. — (1) Every convergent is nearer to the continued fraction than an y of the preceding convergents. Let x denote the continued fraction, and — ■> — i — , three 9 Qi & • The sign ^ means " the difference between." LIMIT OF ERROR IN TAKING ANY CONVERGENT. 381 p., consecutive convergents; then x differs from — only in 7 2 J taking the complete quotient «., + instead of a 2 . a 3 + etc. Denote this complete quotient, which is always greater than unity, by k ; thus _ fc?>i + P . kq 1 + q ' ... x „P = Hf>~P^) = * .[Art. 175,(1)] (1) q q{kq x + q) q(kq x + q)' ' L WJ W andB^ a;= frg~.Wi = I . 9i 1, and q < q x ; hence for both these reasons Pi P . ' x < x ~ - ) 9i q Pi . p that is, — is nearer to x than — is. Hence every convergent is nearer to the continued fraction than the next preceding convergent is, and therefore than any preceding convergent. Comparing this result with that of Art. 173, we see that The convergents of an odd order continually increase, but are always less than the continued fraction. The convergents of an even order continually decrease, but are always greater than the continued fraction. (2) To find limits to the error made in taking any conver- gent for the continued fraction. Since k > 1, we see from (1) that the difference between x and | 1 (2) ^ Wi 7(7! + 7) K J p Also, since q l > q, the error in taking — instead of x < - - , and > — . q- 2ft 2 382 LIMIT OF ERROR IN TAKING ANY CONVERGENT. Therefore, as the error is less than — , it follows that in r order to find a convergent which will differ from the con- tinued fraction by less than a given quantity -, we have a only to form the consecutive convergeuts until we reach one P — , where cr is not less than a. (3) It appears from (2) that the error in taking any Vn convergent, — , instead of the continued fraction, x < , or — (Art. 174) ; that is, < QnQn + l Qni'tn + Sln + ^-j) «„ + ^ Pn Hence the larger a n + 1 is, the nearer does — approximate On to the continued fraction ; therefore Any convergent which immediately precedes a large quotient is a near approximation to the continued fraction. (4) Any convergent is nearer to the continued /ruction titan any other fraction whose denominator is less than that of the convergent. p p. Let — , — be two consecutive convergeuts, x the continued Q Q\ fraction, and - a fraction whose denominator s is less than q v r p>\ , r If possible, let - be nearer to x than — : then - must be 1 * v ( 1 ) ; and .since x lies between and V\ r V V\ — , it follows that - must lie between and — . Hence 7i s g 7l r p p, p 1 . ~ £<* -.that is <- (Ait. 17;")). rq ~. sp APPROXIMATE VALUE OE A QUADRATIC SURD. 383 that is, an integer less than a proper fraction, which is Pi impossible. Therefore — must be nearer to the continued Sri fraction than - is. 6' Find a series of fractions approximating to 3.14159. By the process of finding the greatest common divisor of 314159 and 100000, the successive quotients are 3, 7, 15, 1, 25, 1, 7, 4. Thus 3.14159 = 3 +JL_J_J_J_J_J_1. 7+ 15+ 1+ 25+ 1+ 7+ 4 The successive convergents are f , - 2 T 3 -, f$f, fff , ; this last convergent which precedes the large quotient 25 is a very near approximation, the error being less than - , by (3), and therefore less than .000004. 25(113)* J V " The method of continued fractions thus enables us to find two small integers whose ratio closely approximates to that of two quantities whose exact ratio can only be expressed by large integers. 177. Approximate Value of a Quadratic Surd. — By the properties of continued fractions we may approximate to the value of a quadratic siird. Thus, Express Vll as a contiuued fraction, and find a series of fractious approximating to its value. vTT = 3 + (vTl - 3) = 3 + 2 \/TT V41 + 3 - Vll - 3 „ 1 o — '-> + o — — — o -\ — — , 2 2 V/ll + 3 V/TT + 3 = G+ ^11-3 =6 + v/ ^^5 after this the quotients 3, 6, will be repeated indefinitely hence 1111 Vll = 3 + ' 3+ 6+ 3+ 6 + etc. 384 PERIODIC CONTINUED FRACTIONS. Explanation. —In each line of the above process we perform the same series of operations. Thus, in the first line we find the greatest integer in ^11, which is 3, and the remainder is . -J97. EXAMPLES. 385 EXAMPLES. Reduce each of the following fractions to a continued fraction, and find the fourth convergent to each. 1. 253. ^ S .l+J_J_J_JL_LJ_JL ; L7. 179 2+2+2+1+1+2+2+12 2 1189 11 111 1 1. 33 3927' 3+ 3+ 3+ 3+ 3+3+3' 109* 3 37 J_ JL J_ J_ _L I • _Z_ 2+ 1+ 2+ 2+1+3' 19' 1 84 1 1 1 1 1 11.7 '227' 2+ 1+ 2+ 2+ 1+ 3+ 2' 19' 5. Express y/l8 as a continued fraction, and find the limits of the error made by taking the fourth convergent. Ans. 4 -\ ; — , and the error lies 4+ 8+ 4+ 8+ etc. 136 between — and (Art. 176, (2)). 136x1121 136 x 1257 " PERMUTATIONS AND COMBINATIONS. 179. Definitions. — The different orders in which a number of things can be arranged, either by taking some or all of them, are called their Permutations. Thus, the permutations of the letters a, b, c, taken one at a time are three, viz., a, b, c; taken two a time, are six, viz., ab, ba, ac, ca, be, cb ; and taken three at a time, are also six, viz., abc, acb, bca, bac, cab, cba. The Combinations of things are the different groups or collections which can be made, either by taking a part or all of them, without reference to the order in which the things are placed. Thus, the combinations of the letters a, b, c, taken two at a time are three, viz., ab, ac, be; ab and ba, though differ- ent permutations, form the same combination, both consisting simply of a and b grouped together. 386 THE NUMBER OF PERMUTATIONS. It appears from this that in forming combinations we are concerned only with the number of things eacli group con- tains ; while in forming permutations we have also to consider the order of the things which make up each group ; thus the above six permutations of the letters a, b, c, taken three at a time, form but one combination. 180. The Number of Permutations. — To find the number of permutations of n different things, taken r at a time. Let the different things be represented by n letters, a, b, c, Set a aside ; write down the other n — 1 letters in a line ; put a before each of them in succession ; we thus obtain ab, ac, ad, etc., or n — 1 permutations, each of two letters in which a stands first. In the same manner there are n — 1 permutations, each of two letters in which b stands first. Similarly there are n — 1 permutations, each of two letters in which c stands first ; and so on for each of the other letters ; and as there are n of them, the whole number of permutations of the n letters, two together, is n(n — 1). Again, set a aside, and group the other n — 1 letters, two and two ; as has just been shown, there are (n — 1) (u — 2) such groups. Put a before each of them, and we have (n — l)(n — 2) permutations, each of three letters in which a stands first. Similarly there are (n — l)(?i — 2) per- mutations, each of three letters in which b stands first ; and so on for each of the other letters. Therefore the whole number of permutations of n letters taken three at a time is n(n - l)(w - 2). Proceeding thus, and noticing that at any stage, the number of factors is the same as the number of letters in each permutation, and that the negative number in the Inst factor is one less than the number of letters in each permu- tation, we shall have the number of permutations of n things, r together equal to n(n — 1 ) (n — 2) to r factors ; and the r th factor is n — (r — 1) or n — r + 1. THE NUMBER OF PERMUTATIONS. 387 Hence, the whole number of permutations of n things taken r at a time is n(n - 1) (n - 2) (n - r + 1) . . (1) If all the letters are taken together, r = n, and (1) becomes n(n - l)(n - 2) 3-2-1 . . . (2) Hence, the member of permutations of n things taken all at a time is equal to the product of the natural numbers from 1 up to n. It is usual to denote this product by the symbol [n, which is read " factorial n." * Thus, Factorial 6, or [6, means 6 • 5 • 4 • 3 • 2 • 1, or 720 ; factorial 5, or 15, means 5 • 4 • 3 • 2 • 1, or 120. From the law of formation it is clear that 17 = 716. More generally, \n -\- 1 = (n + l)[w. Thus \ n + 1 contains all the factors of [n, and one factor, n + 1? additional. Denoting the number of permutations of n things taken r at a time by the symbol „P r , we have from (1) and (2) H P r =n(n— l)(w— 2). . . .(»— r+1); and n P n =\n. Thus, n P 4 =w(n-l)0-2)0-3). Also „P 5 = n P 4 (?i-4) =n(w-l) (n-2) (»-3) (n-4) ; and so on. 1. Four persons enter a railway carriage in which there are six seats ; in how many ways can they take their places? Here n = 6, and r = 4 ; then by (1) we have 6 P 4 = 6 • 5 • 4 • 3 = 360. 2. Required the number of changes which can be rung, (1) with 5 bells out of 8, and (2) with the whole peal. Ans. (1) 6720; (2) 40320. * It in also sometimes denoted by n\. 388 THE NUMBER OF COMBINATIONS. 3. Required the number of different ways in which 6 persons can be seated at a dinner table. Ans. 720. 181. The Number of Combinations. — To find the number of combinations of n different tilings taken r at a time. The number of permutations of n things taken r at a time is n{n - l)(n - 2) (?i — r + 1). (Art. 180). But each combination of r things taken r at a time will make I r permutations, by (2) of Art. 180 ; therefore there are | r times as many permutations as combinations. Hence, calling „C r the required number of combinations, we have n W (w- 1)( W _ 2) {n-r + 1) This formula for n C r may also be written in a different form ; for if we multiply the numerator and the denominator by the product of the natural numbers from 1 up to n — r, it becomes c _ n(n-l)(n-2) (n-r+1) (n-r) 2-1 [r- (n-r)(n-r-l) 2-1 The numerator is now the product of the natural numbers from n to 1, or is In (Art. 180); the denominator is the product of the natural numbers from /• to 1, and from n — r to 1. Hence we have . \n • C -^h-r- • •' • ' - (2) It will be convenient to use (1) for „C r in all cases where a numerical result is required, and (2) when it is sufficient to leave it in an Algebraic shape. Note 1. — If in (2) we put r = n, we have C I— = 1 • " " |n [0 |0' but „C n = 1, so that if the Formula is to be true for r — », the symbol [0 must be considered as equivalent io 1. THE NUMBER OF COMBINATIONS. 389 The number of combinations of n things taken r at a time is the same as the number of them taken n — r at a time. For the number taken n — r ut a time is, from (2), \n \n „C»-r = j r J - 7 r = | L | , • (3) \n — r \n — (n — r) w — r \r which = „C y ,., from (2). The truth of this proposition is also evident from the consideration that for every different group of r things taken out of n things there is always left a different group of n — r things. Hence the number of groups of r things out of n must be the same as the number of groups of n — r things. Such combinations are called complementary. Note 2. — Put r = n; then from (2) and (3), »C„ = „C = 1. The proposition just proved is useful in enabling us to abridge Arithmetic work. Thus, 1. Required the number of combinations of 20 things taken 18 together. The required number is the same as the number taken 2 together. 90 v 1<> ''• » c s = T#T = 190 - If we had used the formula o Ci 8 i we should have had to reduce an expression whose numerator and denominator each contained 18 factors. 2. From 12 books, in how many ways can a selection of 5 be made when one specified book is always excluded? Since the specified book is always to be excluded, we have to select the 5 books out of the remaining 11. Hence = 11.10. 9-8-7 = 462< 11 5 1-2-3-4.5 3. How many combinations may be made of 10 letters taken 6 at a time? Ans. 210. 4. From 11 books, in how many ways can a selection of 4 be made? Ans. 330. 390 TO DIVIDE m + n THINGS INTO TWO CLASSES. 182. To Divide m + n Things into Two Classes. — To find the number of xcays in which m -f- n different things can be divided into two classes, so that one may contain m and the other n things. This is equivalent to finding the number of combinations of m + n things m at a time, for every time we select one group of m things, we leave a group of n things behind. Hence by (2) of Art. 181, Ihe required number = . [m \n In a similar manner it may be shown that the number of ways in which m + n + p different things can be divided into three classes containing m, n, p things respectively is \m + n + p [m In \p 1. There are three bookshelves capable of containing 14, 22, and 24 books ; in how many ways can 60 books be allotted to the shelves? Here we have to divide GO things into groups of 14, 22, and 24 things. 160 Hence the required number = == . 1 [14 [22 1 24 2. From 7 Englishmen and 4 Americans a committee of 6 is to be formed, containing 2 Americans ; in how many ways can this be done? Here we have to choose 2 Americans out of 4, and 4 Englishmen out of 7. The number of ways in which the Americans can be chosen is 4 C 2 ; and the number of ways in which the Englishmen can be chosen is 7 C 4 . Each of the first groups can be associated with each of the second Hence the required number of ways is (C . XiCl = l X I = J ! =210. EB l*E BB15 PERMUT. ■ITIONS OF n THINGS NOT ALL DIFFERENT. 39] 3. In how many ways can the 52 cards in a pack be ^2 divided among 4 players, each to have 13? Ans. . 183. Permutations of n Things not all Different. — To find the number of permutations of n things taken all at a time, when they are not all different. Let there be n letters ; and suppose p of them to be a, q of them to be b, r of them to be c, and the rest to be unlike. Let P be the required number of permutations. If in any one of the actual permutations, the p letters a were all changed into p letters different from each other and from all the rest, then from this single permutation, without alter- ing the position of any of the remaining letters, we could form \p new permutations. Hence if this change were made in each of the P permutations, there would be P X \p per- mutations. Similarly, if in any one of these new permutations, the q letters b were changed into q letters different from each other and from all the rest, then from this single permutation we could form \q new permutations. Hence the whole number of permutations would now be P X \p X \q. In like manner, if the r letters c were also changed so that no two were alike, the total number of permutations would be P X \p X \q X |r. But this number must be equal to the number of permutations of n different things taken all together, which is \n. Hence P X [p X \q X [r = \n. And similarly any other case may be treated. 1. How many different permutations can be formed out of the letters of the word Mississippi taken all together? 392 EXAMPLES. Here we have 11 letters, of which 4 are i, 4 are s, and 2 are p. |H .-. P= H— - = ILIO.9.7.5 = 34650. mm 2. TTow many different permutations can be made out of the letters of the word assassination taken all together? Arts. 10810800. EXAMPLES. Reduce each of the following fractions to a continued fraction, and find the fourth convergent to each. .832 .-,11111 157 1. . Ans. 5 H — — ; . 159 4+ 3+ 2+ 1+3 30 2 729 _L_LJ__L J_ _L_L L 11 2318' 3+5+1+1+3+2+1+5' 35* 3 3029 J_J_J_J_J_J_J_J_. ™. 3+ 3+ 3+ G+ 1+ 2+ 1+ 10' 208' 4. A metre is 39.37079 inches ; show by the theory of continued fractions that 32 metres is nearly equal to 35 yards. 5. A kilometre is very nearly equal to .62138 miles ; show by the theory that the fractions §, |f, §f, i ';, i . ! are successive approximations to the ratio of a kilometre to a mile. G. Two bells begin to ring together; the one rings 12 times in 7 minutes, and the other 17 times in 9 minutes; what strokes most nearly coincide in the first half-hour? Ans. The 49th of the first with the 54th of the second. 7. Express 27.321061 days, the average period of the moon's revolution, by the nearest equivalent fraction whose denominator is less than 1000. Ans. 27$$f days. 8. Find a series of fractions converging to .2422642, the excess in days of the true tropical year over 365 days. AriQ X 1 JL 39 47 321 r>+« EXAMPLES. 393 9. The relative sidereal periods of Venus and the earth are 224700 days and 3G525G days : find a series of fractions converging to fffHt- Ans ' *> h h h fs, Iff' etc - The 5th convergent, which precedes the large quotient 29, is a very near approximation, the error being less than [Art. 17(5, (3)]. In consequence of this, a transit occurs after a period of S years, and then again not till after 235 years have been completed. 10. Express \/l9 as a continued fraction, and find a series of fractions approximating to its value, and show that the 7th convergent will give the true value to at least 4 places of decimals. Ans. ^19 = 4+ — t^-t^-^-^-TT- > con " v ^2+1+ 3+ 1+ 2+ 8 + vergents are f, f , V. **, ft. W, W» • • j_- • • 11. Show that the 9th convergent to \J33 will give the true value to at least 6 places of decimals. 12. Find the limits of the error when %y is taken for ^23- Ans. and , ?— -. 44 x 49 44 x 93 13. Find the value of - — ■ - — - — 1+ 2+ 1+ 2 + Ans. Positive root of ar + 2x = 2. 14. Find the value of ±. ±. -L _L J- _L Ans. Positive root of Ix 1 + 8x = 3. 15. How many different numbers can be formed by using six out of the nine digits, 1, 2, 3, 9? Ans. G0480. 16. Required the number of changes which can be rung upon 12 bells taken all together. Ans. 479001G00. 17. Required the number of combinations of 24 different letters taken 4 at a time. Ans. 10G26. 18. Out of 14 men, in how many ways can 11 be chosen? Ans. 364. 394 EXAMPLES. 19. How man}' different products can be formed with any three of the figures 1, 3, 5, 7, 9? Ans. 10. 20. In how many ways can 6 copies of Horace, 4 of Virgil, and 3 of Homer be given to 13 boys, so that each boy may receive a book? Ans. G0060. 21. Out of 7 consonants and 4 vowels, how many words can be made each containing 3 consonants and 2 vowels? Ans. 25200. 22. How many parties of 12 men each can be formed from a company of GO men? ]60 Ans. II? II? 23. Out of 12 Republicans and 1G Democrats, how many different committees could be formed, each consisting of 3 Republicans and 4 Democrats? * llj H^ ' |3 [9 |4 J12/ 24. Out of 10 consonants and 4 vowels, how many words can be formed, each containing 3 consonants and 2 vowels? Ans. 86400. 25. There are 10 candidates for G vacancies in a commit- tee : in how many ways can a person vote for 6 of the can- didates? Ans. 210. 26. In how many ways can a cricket eleven be chosen out of fourteen players? Ans. 364. 27. In how many ways could 2 ladies and 2 gentlemen be chosen to make a set at tennis from a party of 4 ladies and G gentlemen? Ans. 90. 28. In how many ways could 2 ladies and 2 gentlemen be chosen to make a set at tennis from a party of G ladies and 8 gentlemen ? Ans. 420. 29. From 6 ladies and 5 gentlemen, in how many ways could you arrange sides for a game of croquet, so that there would be 2 ladies and one gentleman on each side' ^'• s '-(#7r or1800 - 30. Out of 6 ladies and < + E = 0, which gives E = (7 - D = 12. -7J + # + F = 0, which gives F = 2> - E - -19. * In these applications the Beriea are alwaya regarded as convergent DEVELOPMENT BY INDETERMINATE COEFFICIENTS. 399 Substituting these values of the coefficients in (1), it becomes — = 2 - 2a; + 5a 2 - 7x 3 + 12a 4 - 19a 5 + etc. 1 + x - x 1 In applying the method of indeterminate coefficients, the form of the given expression will determine what powers of x must be assumed. In the example just solved, we assumed the series to be arranged according to the ascending powers of x beginning with x° ; but in many series this is not the case, and the error in the assumed series will then appear, either by an absurd result, or by the coefficients of those terms which do not exist in the actual series reducing to zero. By dividing the term of the numerator having the lowest power of x by the term of the denominator having the lowest power of a;, we may always determine the expo- nent of x in the first term of the series. We may then assume a series beginning with this exponent of x, and arranged according to the ascending powers of x as usual. Thus, 2. Expand into a series. 1 3x - x 2 If we should assume the series to be A + Bx-\- Ox 2 + etc., on clearing of fractions, and proceeding to equate the coeffi- cients of like powers of a?, we would have no term on the right corresponding to 1 on the left ; so that we would have 1 = 0, an absurd result. Hence we infer that the expres- sion cannot be developed according to the ascending powers of x beginning with x°. r -i But, dividing 1 by 3x, we get ■ — for the first term of the o quotient ; hence we may assume a series beginning with a> -1 , and the exponents increasing by unity. Thus, ■ — = Ax' 1 + B + Cx + Dx 1 + Ex 3 + etc. 3x — x' 2 400 DEVELOPMENT BY INDETERMINATE COEFFICIENTS. Clearing this of f motions, and collecting the terms with the same powers of x, we have 1 = BA + (BB - A)x + (3(7 - B)x 2 + (3D - C)x 3 + etc. Equating the coefficients of the like powers of x, we find A = |, 5 = i, C = aV, ^ = J T , e tc. Therefore = — + - + — + — + etc. 3a; - x 2 3 T 9 T 27 T 81 T 3. Expand \l+x into s a series. Assume \l+x=A + Bx+ Cx 2 + Da 8 + etc. Squaring, l+a;= A 2 + 2ABx + (B 2 + 2AC)x*+ (2BC + 2AD)x*+ etc. Equating the coefficients of like powers of x, A 2 = 1, .-. A = 1. 2AB = 1, .-. 5 = f £ 2 + 2^1(7 = 0, .-. C = -J. 25(7 + 2.4Z) = 0, .-. D = T V, and so on. Note. — Since A 2 = 1, A = ±1 ; we might take the value .4 = — 1 . in which case we would find B — —\, C = $, D — — r b -, and so on, so that \J\ + x = _l_? + ? 2 _£ 3 _ etc. (See Note on Art. 112.) 2 T S 16 Expand the following to four terms : 4. L±_?? ^ ns . j + r jr + 15aj i + 45r s # 1 — 3a* 5. _ J + 2x 1 + 3.*: + 4.r + 7.v 8 . 6. vT^. 1 - •'' - £ - 2 a 16 PARTIAL FRACTIONS.' 401 PARTIAL FRACTIONS. 187. Partial Fractions. Case I. — When the factors of the denominator are all unequal. When the denominator of a fraction can be resolved into factors, we may by the principle of indeterminate coefficients decompose the fraction into the sum of two or more simpler fractions, called partial frac- tions, the denominators of which are the factors of the given denominator. We shall give a few examples illustrating the decomposition of a rational fraction into partial fractions. For a general discussion of the subject, the student is referred to Treatises on the Theory of Equations, and on the Integral Calculus.* K r 11 Resolve — — into partial fractions. 2x 2 + x — 6 Since the denominator 2x' 2 -f x — 6 = (« + 2) (2a — 3), we assume 5a; _ n A B (1) 2x 2 + a;-6 x + 2 2x - 3' where A and B are independent of x, and at present unknown. We have then to find such values for A and B as will make (1) an identity, that is, true whatever may be the value of x. Clearing (1) of fractions, it becomes 5x - 11 = A(2x - 3) + B(x + 2) . . (2) Since (2) is identically true, we may equate the coefficients of like powers of x ; thus, 2A + B = 5, -SA + 2B = -11 ; whence .4 = 3, and B = — 1, which in (1) gives 5x - 1 1 _ 3 1 2jt + x — 6 x + 2 2aj — 3* Second Method. In practice, in this case, there is a simpler method of finding the values of A, B, etc. than by equating * See Todhunter's Theory of Equations, Chap. XXIV., Integral Calculus, Chap. II., also Serret's Cours d'Algebre Superieure. 402 CASE II. the coefficients. Since A and B are independent of x, we may give to x any value we please. If in (2) we put x 4- 2 = 0, or a: = —2, then we have immediately -10 -11 = -1A, or ^1 = 3. In the same way, putting 2.»; — 3 = 0, or X = |, we get ¥ - 11 = (I + 2)-B, or B = -1, as before. 188. Case II. — When the factors of the denominator are equal. 2 3^2 1. Resolve — - into partial fractions. [x + 2) 3 Assume 2 ~ 3x * = -A- + * + O () (a + 2) 3 x + 2 (as + 2)* T (a + 2) 3 V Rem. — All the factors of the given denominator must be taken as denominators of the assumed fractions. Thus, in this example, the denominators of the partial fractions may he (,c + 2), (x + 2) 2 , (x + 2) s . It is therefore necessary to assume fractions with each of these denominators, so as to include every possible case which can produce the given fraction, although A or B may reduce to zero. Clearing (1) of fractions, it becomes 2 - 3z 2 = A(x + 2) 2 + B(x + 2) + C. Equating the coefficients of like powers of x, ±A + 2B + C = 2, 4A + B = 0, and .1 = —3. Solving these three equations, we get A = -3, B = 12, C = -10, which in (1) gives 2 - 3.-C 2 _ 3 12 10 (sb + 2) 3 x + 2 (.»• + 2) a (a: + 2) 3 ' NOTE. — In this case we cannot iind the values of .1, />', C, etc. by the second method of Case I., but have to employ the first. When both equal and unequal factors, however, occur in the denominator, both methods may be combined to advantage, as in the following example. CASE III. 403 2. Resolve — — into partial fractions. x'\l + x)* Assume A + ? + _^_ + *> (See Rem.) (1) x\l + X) 2 X x 2 1+x (1 + x) 2 Clearing (1) of fractions, it becomes Bx-l = Ax(l+xy + B(l+x) 2 + Cx 2 (l+x) + Dx* . (2) Here we may use the second method of Case I. as follows : Putting x = 0, we find B = — 1. Putting x = — 1, we find/) = —4. To find A and C, equate the coefficients of x 2 and a 3 ; thus, equating coefficients of x 2 , = -2 A + £ + + #, or 2^1 + C = 5 (3) Equating coefficients of a: 8 , yl + (7 = (4) Solving (3) and (4), we find A = 5, C = -5, which in (1) gives 3x - 1 = 5 _ J_ _ 5 _ 4 ar(l + x) 2 x x 2 1 + x (1 + a-)' 2 ' 189. Case III. — When some of . the factors of the denominator are of the second degree. 42 19a; 1. Resolve into partial fractions. (oF + l) 3a; 2 - 2a: - 1 3a; + 1 x — 1 Resolve into partial fractions : Q 8a; - 4 .5,3 3. —„ . Ans. x 2 — 4 x + 2 x - 2 23a; - 11a; 2 1 1 1 (2x - 1)(9 - a; 2 ) 2a; - 1 3 + * 3 - a; 33,2 + x _ 2 1 5 I («_ 2 )«(l -2a;)" 3(1 -2a;) 3(a--2) (a--2) 2 ' 2G.r 2 + 208.C (a; 2 + !)(» + 5)' \x + 5 .r + 1 / BINOMIAL THEOREM. 405 INOM1AL THEOREM. 190. Positive Integral Exponent. — The method of raising a binomial to any power by repeated multiplication has been explained in Art. 109. We shall now prove a formula known as the Binomial Theorem,* by which any binomial can be raised to any power without the labor of actual multiplication. To prove the Binomial Theorem for a positive integral exponent. By actual multiplication we obtain (x + a) (x + b) = x 2 + (a + b) x + ab, (x+a) (x+b) (x+ c) — x 3 + (a+b + c)x 2 + (ab + ac + bc)x + abc. In these results we see that the following laws hold : 1. The number of terms on the right side is one more than the number of the binomial factors on tJie left side. 2. The exponent of x in the first term is the same as the number of binomial factors, and decreases by one in each successive term. 3. The coefficient of the first term is unity ; of the second term, the sum of the letters a, b, c; of the third term, the sum of the products of the letters a, b, c, taken tioo at a time; and the fourth term is the product of all the letters. We shall now prove by induction (Art. 170) that these laws always hold whatever be the number of binomial factors. Suppose these laws to hold for n — 1 binomial factors, so that {x+a) (x+b) . . (x+Jc) =x n - 1 +Ax n -' 2 +Bx n - 3 +Cx n - i + . . K, (1) where A = a-\-b-\-c-\-. ...+&, the sum of the second terms, B = ab -\- ac + be + , the sum of the products of these terras taken two at a time. C = abc + abd +■ , the sum of the products of these terms taken three at a time. K = abed k, the product of all these terms. * This theorem was discovered by Newtou. 406 POSITIVE INTEGRAL EXPONENT. Multiply both sides of (1) by another factor (x + J) ; thus, (x+a) (x+b) .... (x+k)(x+-l)=x"+(A + l)x a - 1 + (B+Jl)x*- S +(C+Bl)ar*+ +KI . . • (2) Now A + l=a + b + c + -f k + I = the sum of all the terras a, 6, c, ?• 5 + Al = a&+ac+6c+ . . . +al+bl+ cl + . . . + W. = the sum of the products taken two at a time. C + Bl = abc + dbd + . . . + aW + ad + 6d + . . . = the sum of the products taken three at a time. Kl = <(bcd . . . kl = the product of all the terms o, &, c, . . . . I. Also the exponent of x in the first term is the same as the number of binomial factors, and decreases by 1 in each suc- cessive term. Hence if the laws hold when n — 1 factors are multiplied together, they hold when n factors are multiplied together ; but they have been proved to hold for 3 factors, therefore they hold for 4 factors, and therefore for 5 factors, and so on, generally, for any number whatever. Now let 6, c, d, Z, each = a ; then the binomial factors are all equal, and the first member of (2) becomes (x+a) (x+a) . . . = (x+a) taken n times as a factor = (x+a) n ; and the second member becomes A + I = a + a + a +-.... = a taken n times = na. B + Al = an + aa + . . . = a' 2 taken as many times as there are combinations of n letters taken 2 at a time = n ^ n ~ — '- I 2 (Art. 181). C + Bl = aaa + ami + = a? taken as many times as there are combinations of n letters taken ;i at a time = n(n-l)(n-2 ) Li POSITIVE INTEGRAL EXPONENT. 407 Kl = aaaa . . . . = a taken n times as a factor = a". Substituting in (2), we obtain (a, + a y = «■ + noa;"- 1 + '- <" ~ 1] «V"» + n(n - l)(n - 2) (tV _ 3 + § _ a n . . ( 3 ) I? This formula is called the Binomial Theorem; the series in the second member is called the expansion of (x + a) n . In this expansion we observe the following Rule. (1) The exponent of x in the first term is the same as the exponent of the power, and decreases by unity in each succeed- ing term; the exponent of a begins with one in the second term, and increases by unity in each succeeding term. (2) The coefficient of the first term is 1, that of the second is the exponent of the power, and if the coefficient of any term be multiplied by the exponent of x in that term, and the product be divided by the number of the term, the quotient will be the coefficient of the next term. By changing x to a and a to x, we have (a + x) n = a" + na n ~ y x + -*— '- a n W If + n(n- l)fo-2) a „- 8a;8+ i<<(f , .( 4) If we write —a for a in (3), we obtain (as - a) n = x" - noaf" 1 + " (w ~ 1} aV- a (5) Thus the odd powers of a are negative and the even powers positive, and the last term is positive or negative according as n is even or odd. Suppose a — 1, then (4) becomes (l+a0^1+na- + M( "r 1) .^+ ? l(n ~ 1 | ^~ 2) ^+ . . . *, (6) |2 [3 which is the simplest form of the binomial theorem. 408 GENERAL TERM OF THE EXPANSION. 1. Expand (as + a) 5 . By the rule, we have (x + a) 5 = x 5 + 5x 4 a + 10x 3 « 2 ■+- lOara 3 + 5xa* + a 5 . Similarly 2. (a-2o;) 7 =a 7 -7« 6 (2a;)+21a 5 (2a;) 2 -3oa 4 (2x) 3 +35a 3 (2.T) 4 -2kr(2x) 5 +7a(2x) 6 -(2x) 7 . =a 7 -14a 6 x+84a 5 x 2 -280a 4 x 3 +560a 3 x 4 -672a 2 x 5 +448ax 6 -128x 7 . Expand the following by the Binomial Theorem : 3. (as - 3) 5 . Ans. x 5 - lox 4 + 9 Ox 3 - 270x 2 + 405.x - 243. 4. (3x + 2?/) 4 . 81x 4 + 21Gx 3 ,y + 21Gx'-y 2 + 9Gxy 3 + 1G?/ 4 . 5. (x 2 + x) 5 . x 10 -f5x 9 + 10x 8 -|- 10x 7 + 5x 9 -|-x 5 . G. (2-fx 2 ) 4 . lG-48x 2 + 54x 4 -27x 6 -hfix 8 . The sum of the coefficients in the expansion of (1 + a:)" is 2*. For put x = 1 ; then (1 + .r)» = (1 + 1)» = 2» = 1 + n + " fr- 1 ) + e tc. = sum of the coefficients. Also, hy putting x = —1, we have (l-l)» = l-H + "(»- 1 >-ctc.; If .". = sum of the odd coefficients — the sum of the even ones; i.e., the sums of the odd and even coefficients are equal, and therefore each =|x2" = 2 n ~\ 191. The r th or General Term of the Expansion. — In the expansion of (x + a)", we see that the second term is naJ* -1 a ; the third term is n ' 71 ~ — 'x"-' 2 a-; and so on ; the last factor in the denominator of each term being one less than the number of the term to which it applies, one greater than the negative number in the hist factor of the numerator, and the same as the exponent of a ; and also ANY EXPONENT. 409 that the exponent of x is found by subtracting the exponent of a from n. Hence the n(n- l)(n- 2) (n - r + 2) x n ~ r + 1 a r ~ 1 1** term = |r — 1 This is called the general term, because by giving to r different numerical values, any assigned term may be ob- tained. The coefficient of the r th term from the beginning is equal to the coefficient of the r th term from the end. The coefficient of the v th term from the beginning is n(n - 1) (n - 2) O - r + 2) \IE± By multiplying both terms by \n — r + 1 , this becomes ^ . See (1) and (2) of Art. 181. \r — 1 \n — r + 1 The r th term from the end is the (n — r + 2) th term from the beginning, and its coefficient is *("" J > r , which also = 1 . \n — r + 1 \r — 1 \n — r + 1 Therefore the coefficients of the latter half of an expansion may be taken from the first half. 1. Find the fifth term of (a + 2x 3 ) 17 . Here n = 17, r = 5 ; therefore the 5th term = 17 ' 16 ' W ' U a ls X 16a; 12 = 38080a 13 ^ 12 . 1-2- 3-4 2. Find the 14th term of (3 - a) 1G . Ans. -945a 13 . 3. Find the 7th term of (a 3 + 3a6) 9 . 61236a 15 6 6 . 4. Find the 5th term of (a 2 - b 2 ) 12 . 495a 16 6 8 . 5. Find the 5th term of (3.^ - 4^) 9 . 12G x 3 6 ^4V- 192. Any Exponent. — To prove the Binomial Theorem when the exponent is fractional or negative. 410 ANY EXPONENT. Since, by Art. 190, every binomial may be reduced to one common type, (1 + %)", it will be sufficient to confine our attention to binomials of this form. We bave seen that when n is a positive integer, (i + xy -. We shall now prove that this formula is true when n is frac- tional or negative. In this proof, which has for its object the expansion of (1 + x) n in the form P + Qx -\- Ax- + Bx z + etc. , we shall first find P and Q, and then determine the othei coefficients, A, B, etc., in terms of P and Q. (1) To find P and Q. Since (1 + x)" = P + Qx + Ax 2 + for every value of x, the equation must be true when x — ; therefore P = 1. Hence v l + »)■ = 1 + Qx + Ax 2 + V Let n be a positive fraction = -, where p an d q are positive integers. v Assume (1 + ») 5 = 1 + Qx + etc. .-. (1 + x)*= (1 + Qx + etc.)', or 1 + px + etc. = 1 + qQx + etc. (Art. 11)0). .-. Q = |(Art. 185). Let n be a negative integer or fraction = — m ; then (i + .)■ = (i + .)- = _ 2_ = = 1 — iux + etc., by division. 1 + mx + t,,( '- Hence, generally, the numerical coefficient of a; in the ANY KXrONF.NT. 411 second term is the same as the exponent, and the form of the expansion is (1 + a;)" = 1 + nx + Ax 2 + Bx 3 + Cx* + (1) (2) To find the other coefficients. To lind A, B, etc., we put x + z for x in (1), and then expand 1 + x + z in two different ways ; first regarding x + z as one term ; and second regarding 1 -+- 2 as one term. Thus, [1 + (as + 2)]" = 1 + n(as + z) + A(x + z)*+B(x + 2) 3 + etc. = 1 + nx + Ac 2 + Bx 8 + etc. + nz + 2^1x2 + o.Ba; 2 z + etc. . . . (2) + etc. + etc. [(l+2)+.T]» = (l+2)'/l+~Y _(l + ,)^ + _ + __ +rt8 .j = (l+2)"+n(l+2)"- 1 .« + .l(l+2)"- 2 a; 2 +etc. = 1 +?io;+^lx 2 +etc. +w2 + w(n — l)2x-+yl(?t — 2)2ar+etc. . (3) + etc. + etc. + etc. Now as (2) and (3) are the expansions of the same expression in the same form, the coefficients of 2, zx, zx 2 , etc., must be the same in both. Equating them, we get n = n, 2A = n(n — 1), oB = ^4(n - 2), etc. . 1 _ n(n-l) J _ ^l(n-2) _ n(n-l)(n-2) ' [2 ' 3 ~ [3 ' ' where the law is obvious. Substituting these values in (1), we have /1 ■ \. 1 i n(n—l) , n(n — l)(n — 2) , , , ... (l+a,-)»=l+waH — v ; s 2 -j — * ^ ioi 3 + etc., (4) I 2 12 which proves the Binomial Theorem for all values of w, 412 ANY EXPONENT. whether fractional or negative. Hence the theorem is com- pletely established. Note, — If n is a positive integer in the expansion of (1 + x)» when we form the successive factors, n — 1, n — 2, etc., the u th factor will be n — n, or 0, and therefore all the coefficients after that of x n will vanish, for they will all have this factor. Hence, the series will end with the (n -f l) th term. But if n is negative or fractional, none of the factors, n — 1, n — 2, etc., can become 0, and therefore the series will be infinite. 1. Expand (1 — a?)« to four terms. 1 • / 1 • Z • O = l-fz+iarV^BM- 2. Expand (2 + o.r)" 4 to four terms. (2 + 3x)-* = 2-*(l + %*)-* = £,1 _«, + **_>*,.+ ,. The Binomial Theorem may also be applied to expand expressions which contain more than two terms. 3. Expand (a-- 2 + 2x - l) 3 . Regarding 2x — 1 as a single term, the expansion = (x- 2 ) 3 + 3(x- 2 ) 2 (2.x- - 1) + 3x-(2x - l) 2 + (2x - l) 3 = x 6 + 6a: 5 + 9a; 4 - 4.x 3 - 9a- 2 + 6a - l,.by reduction. A root may often be extracted by means of the Binomial Theorem. 4. Eind the cube root of 126 to 5 places of decimals. 126^ = (5 8 + 1)^ = 5(1 + V sfi + I.I + Mi^il 1 + S(i-n.(j^)i + V 3 5 8 1-2 5° 1-2-3 5 9 ( 1+ l)». EXAMPLES. 413 Expand to 1 terras : 5. (1 + x 2 )- 2 . Am. 1 - 2x 2 + 3a; 4 - 4x 6 . i + x + £ _ «!. 6 54 7. Find 0J8 to 5 decimal places. 9.89949. 8. Find 77= to 5 decimal places. 0.19842. EXAMPLES. Expand by the Method of Indeterminate Coefficients to 4 terms. 1. 1 + 2x o . ,4ns. 1 + 3a; + 4a; 2 + 7a; 3 . i + ix - fa; 2 + T V 3 - 3 + fa . + ¥aj2 + 2^3 -. 1 - ckk + a(a + 1 )x 2 - (a 3 + 2a 2 - 1 )x s . — o.»r— x'' 5. s/i + Resolve into partial fractions : 7x — 1 1 - X — X" 1 + X 2 + X + x- 3 + X 2 — X - X 2 ' 1 1+aa;- -ax 2 — x 8 1 _ 5 X + 6a; 2 1 — 3x 1 - 2x - 1 + 3x- + 2a 2 4 3_ (1 - 2a;)(l - a 2 )" 1 - 2x 1 - x g x 2 - 10a; + 13 2 3 4 (a; — l)(a 2 — 5a; + 6)' a; — 1 a; — 2 a; — 3 9 9 _J 1 §_ (a; - l)(aj + 2) 2 x — 1 a; + 2 (a; + 2) 2 in 3a; 3 -8a; 2 +1 5 7_ 1 , _3_ (a;-l) 4 ' (x-iy (a?-!) 8 («-l) a x-1 414 EXAMPLES. , . 7 -\- x . 3 . 4 — 3x 11. -. Ans. . (1 + »)(1 + x 1 ) 1 + x 1 + x- 12 2a;' 2 — lias + 5 3x 1 (x — 3) (a; 2 + 2x — 5) a; 2 + 2x — 5 a; — 3 13 a- 4 - 3a; 3 - 3a;' 2 + 10 ( ., + i )2(a; _ 3 ) 17 11 17 ^l^s. a; — 2 + 16 (a; +1) 4(a? + l)' 2 16 (a; - 3) Expand the following by the Binomial Theorem. 14. (2a; - ?/) 5 . Ans. 32a; 5 - 80afy + 80afy 2 - lO.r?/ 3 + lOajy* - f. 15. (3a -i-) 6 . Ans. 729a 6 - 972a 6 + 540a 4 - IGOa 3 + — - — + — . "•c-=y A 64a; 6 32a- 4 , 20a; 2 OA . 135 243 . 729 729 27 3 4a; 2 Xx* 64a! 6 17. (1 + 2a; - a; 2 ) 4 . Ans. 1 +8a;+20a-' 2 + 8a; 3 -26x 4 -8a; 5 +20.t- 6 -8.i- 7 -f .r\ 18. (3» 2 -2aa;+3a 2 ) 8 . Ans. 27a; 6 -5 loa^+ll 7aV-l 16«V+117(tV--5 la 6 a;+27o 6 . Write down and simplify : 19. The 4th term of (a; - 5) 13 . Ans. -35750a? 10 . 20. The 10th term of (1 - 2a') 1 ' 2 . -112640a?. 22. The 7th term oil- + 96 J. 21. The 4th term of ( f 9&) . 40a 7 6 3 . 10500 23. The 5th term of (aAr"< - r6"<) 8 . 70a/y""~^~' ; - EXAMPLES. 415 Expand to 4 terms 24. (1 - x)i. 25. (1 - 3®)i 26. (1 - 3a,-)~i 28. (1 + fa)" 4 . 29. (8 + 12a)i 30. (9 - 6a)-*. 81. (4a - 8x)-K flw.S'. 1 1 — x — a; 2 — fa; 3 . + x + 2x 2 + lAe 8 . 1 - x + fa; 2 ■J 7 XT'. 1 _ 2a + fa' 2 - f« 3 - 4(1 + a - £a 2 + |a»). AO + x + f* 2 + fK>- 2rt iV « 2 a 2 «/ Write down and simplify : 32. The 8th term of (1 + 2x)~$. 33. The 5th term of (3a — 26)" 1 . 34. The 14th term of (2 10 - 2V)^. 35. The 7th term of (3 8 + 6 4 e)V-. Find to 5 places of decimals the value of Ans. 16& 4 243a 5 ' -1848a; 13 . ^ 6 . 39. (630)-". Ans. .00795. 40. (3128)*. 5.00096. 36. V 998 - Ans - 9 -"332. 37. V 1003 - 10.00999. 38- (l^)i 1.00133. Expand to 5 terms : 41. (1 + 3a.- 2 - 6a; 3 )- 1 Ans. 1 - 2a, 2 + 4a; 3 + 5a; 4 - 20a; 5 . 42. (8 - 9a; 3 + 18a; 4 ) i . 16(1- fa; 3 + 3a; 4 + &x* - -p 1 ) . 416 RECURRING SERIES. CHAPTER XX. SUMMATION OF SERIES. RECURRING SERIES. 193. Definition. — The Summation of a Series is the process of finding an expression for the sum of all its terms. Examples of summation of Arithmetic and Geometric series have been given in Chapter XVII. We now proceed to give methods for summing other series. A Recurring Series is one in which, after a certain term, each term is equal to the sum of a fixed number of the pre- ceding terms multiplied respectively by certain constants. A geometric progression is a simple example of a recurring series ; for each term after the first is equal to the preceding term multiplied by the ratio. If u n _ 1 and v n denote respec- tively the (n — l) th term and the n th term, then u a — r*/„_,= ; the sura of the coefficients of u n and u n _ Y with their proper signs, i.e., 1 — r, is called the scale of relation. In the series 1 + 2x + 3a 2 + 4x 3 -f 5.x 4 + , each term after the second is equal to the sum of the two preceding terms multiplied respectively by the constants 2x and —a; 2 ; these quantities being called constants because they remain the same whatever term of the series we consider. Thus, 5x* = 2a-4•<•"; that is, w n = 2a; u„_! — x 2 u„_», or u n — 2xu„_i + x"u„_ 2 = 0. This law holds for all values of n greater than 1, so that SCALE OF RELATION GIVEN TO FIND ANY TERM. 417 every term after the second can be obtained from the two terms immediately preceding. Thus the series 1 + 2x + ox 2 + 4x 3 + 5x* + is a recurring series in which the scale of relation is 1 — 2x+x 2 . 194. The Scale of Relation being Given to Finfl Any Term. — If the scale of relation of a recurring series is given, any term can be found when a sufficient number of the preceding terms are known. As the method of proceed- ing is the same however many terms the scale of relation may have, the following illustration will suffice. If 1 — px — qx 2 — rx s is the scale of relation of the series a + a x x + a 2 x 2 -+- a 3 x 3 + , we have a n x n = px • a^x" -1 + qx 2 ■ a n _ 2 x n ~ 2 + ra? • a^x"' 3 , or a„ = 2>a _i + q^ n -i + ra n-z'i (1) thus any coefficient can be found when the coefficients of the three preceding terms are known. Conversely. To find the scale of relation when we have given a sufficient number of the terms of a series. Find the scale of relation of the recurring series 2 + 4a; + 14a- 2 + 46a; 3 + Let the scale of relation be 1 — })x — qx 2 ; then to find j? and q we have from (1) the equations, 14 - Ap — 2q = 0, ami 46 — 14p - 4q = ; whence p = 3, and q = 1 ; thus the scale of relation is 1 - 3aj - x 2 . If the scale of relation consists of 3 terms, it involves 2 constants, p and q ; and therefore we must have 2 equations. To obtain p we must have given at least 3 terms of the series, and to obtain q we must have one more term given. Thus, to obtain a scale of relation involv- ing 2 constants, we must have given at least 4 terms of the series. Similarly, if the scale of relation be 1 — px, — qx 2 — rx 3 , we must 418 TERMS OF A RECURRING SERIES. have X equations to find the '-> constants, and therefore at least (5 terms of the series must he given. In general, if the scale of relation involve m constants, we must have given at least 2m, consecutive terms. Conversely, if 2m consecutive terms are given, we may assume for the scale of relation 1 — p x x — p 2 x 2 — — p m -t m . 195. To Find the Sum of n Terms of a Recurring Series. — The method of finding the sum is the same what- ever be the scale of relation ; for simplicity we shall suppose the scale to contain only two constants. Let the series be a -f- a^ + a.,x' 2 -f- a s x z + , and let the scale of relation be 1 — px — qx 2 , so that for every value of n greater than 1 we have a n — pa n _ x — qa n -2 = 0. Denote the sum by S ; then S =s a + a t x + a»x 2 + + a n _ l x n ~ l . — pi— pa^x— [(pa„- l + qa„_ 2 )x n +qa n _ l x n+1 ~\ ... 1 — px — qx 2 Thus, the sum of a recurring series is a fraction whose denominator is the scale of relation. If the term \_{pa n _ x + qa n -'>)x n -f ga m _iaf +1 ] decreases indefinitely as n increases indefinitely, we have for the sum of an infinite number of terms, s _ a„ + («.,- pa )x . g v 1 — px — qx* If this fraction be developed into a series according to the ascending powers of .<•. we shall obtain the given recurring series (Art. 180) ; for this reason the fraction is called the SERIES SUMMED BY MEANS OF OTHER SERIES. 419 generating fraction* of the series. Thus, when the series is convergent, the summation reproduces the generating fraction. Find the generating fraction of the following series : 1. 4 + 9 6 3. 1 • 2, 2 • 3, 3 • 4, 4 • 5, 5 • G, tow terms. A w(n + !)(» + 2) 3 4. 3, 11, 31, G9, 131, to 20 terms. . 44330. 424 P1LE8 OF CAX.XOX-BALLS. 199. Piles of Cannon-Bails. — Cannon-balls are usually piled ou a level surface in horizontal courses, in piles of three different forms, triangular pyramids, square pyramids, and rectangular piles. TRIANGULAR PYRAMID. SQUARE PYRAMID (1) In a triangular pyramid the summit consists of 1 ball, the second course consists of 3 balls, the third course con- sists of 6 balls, the fourth of 10 balls, and so on. Hence the number of balls in a triangular pile of n courses is equal id the sum of n terms of the series, 1, 3, G, 10, 15, But from Ex. 2 of Art. 198, the sum of n terms of this ^ n(n + l)(n + 2) 6 (1) in which n is the number of balls in the side of the bottom course, as well as the number of courses. % (2) In a square pyramid the summit consists of 1 ball, the second course consists of 4 balls, the third course consists of 9 balls, and so on. Hence the number of balls in a square pile of n courses is equal to the sum of n terms of the series l 2 , 2 2 , 3 2 , 4 2 , n\ Here a — 1, d\ = 3, d a = 2, cl s , . . . . = 0. Hence by (3) of Art. 198, S=n + Sn(n 1) , n(n- l)(n-2) _»(H + l)(2n + l) (2) 2 ' 3 6 Here also n, is the number of 'mils in the side of the bottom eourse v as well :is the cumber of courses. PILES OF CANNON-BALLS. 425 (8) In a rectangular pile the summit consists of 1 row, the second course consists of 2 rows, the third of 3 rows, and so on. Let m + 1 denote the number of balls in the top row ; then the length of the second row will be m + 2, of the third m + 3, and so on. Hence the summit will consist of m + 1 balls, the second course will consist of 2(m -+■ 2) balls, the third course of 3(m + 3) balls ; and the n th course of n(m + n) balls. Therefore the number of balls in the rectangular pile of n courses is equal to the sum of n terms of the series m + 1, 2(ra + 2), 3(m + 3), n(m + n). Here a = m + 1, c7 x = m + 3, d 2 = 2, d 3 , = 0. Hence by (3) of Art. 198, S = n(m + 1) + n(n ~ 1} (m + 3) + "(" - l)(n - 2) & o _ n(n + l )(3m + 2n + 1) 6 If I = the number of balls in the longest row, then I = m + n, and therefore 3m + 2n = ol — w, and the formula becomes - g , .... (3) in which Z denotes the number in the length, and n the num- ber in the width, of the bottom course. To find the number of balls in an incomplete pile, we must find the number in the pile supposed complete, then the number in the pile which is wanting, and subtract the latter from the former. 1. Find the number of balls in a square pile of 15 courses. Ans. 1240. 2. Find the number of balls in a rectangular pile, the length and breadth of the bottom course containing GO and 30 balls respectively. Ans. 23405. 426 INTERPOLATION. 3. How many balls in an incomplete triangular pile, the number of balls in each side of the lower course being 44, and in each side of the upper 22 ? Ana. 13409. 200. Interpolation. — The process by which we intro- duce between terms of a series, intermediate terms which conform to the law of the series, is called Interpolation. Its most extensive application is in Astronomy, though it is often used for inserting intermediate terms between those given in Logarithmic and other Mathematical tables. To insert or interpolate a term in a series, is practically to find the n th term of the series by the method of differences (Art. 197). Let p represent the distance of the required term from a, the first term of the series, this distance being measured in terms of the distance between any two consecutive terms. Hence p will be a fraction ; and since it is the distance from a to the n tb term of the series, it is the distance to the (n — l)" 1 term from a; therefore^ = n — 1. Substituting in (1) of Art. 197, we obtain for the term to be interpolated a + M + S&^d, + pCp ~ 'Hp - * >rf, + . (1) 1. The cube roots of GO, Gl, 62 are respectively 3.91487, 3.93650, 3.95789 : find the cube root of 60* 1st order of differences = 0.02163, 0.02139, 2d order of differences = -0.00024, .... .-. a = 3.91487, d, = 0.02163, d, = -0.00024, j> = \. Substituting in (1), we have for the required term 3.91487 + K.02163) + j (-f)/- - " 024 ) = 3.91487 + .00541 + .00002 = 8.92030. REVERSION OF SERIES. 427 2. The logarithm of radius vector of Mars 1888, July 7 is 0.1782987. " 11 is 0.1767934. « " 15 is 0.1752860. " 19 is 0.1737790. What is it (1) for July 8th ; and (2) for July 10th? Ans. (1) 0.1779227; (2) 0.1771700. 3. Given V^ = 2.23607, \[H = 2.44949, V^ = 2.64575, vAs = 2.82843 : find ^5725, V^, and s/~5J). Ans. 2.29129, 2.32379, 2.34521. 4. Given V49 = 3.65931, y/51 = 3.70843, V53 = 3.75629, V55 = 3.80295 : find V50, V^ and V 5 ^- Ans. 3.68403, 3.73251, 3.77976. 201. Reversion of Series. — To Revert a Series is to express the value of the unknown quantity contained in the series by means of another series involving the powers of some other quantity. Thus, given the series 1. y = ax + bx 2 + ex 3 -f- (1) then to revert this series is to express the value of x in another series involving the powers of y. Assume x = Ay + By 2 + Cy 3 + (2) Substituting in (2) the value of y from (1), we have x = Aax + Abx 2 + Acx 3 + + Ba 2 x 2 + 2Babx 3 ■+- . . . . + Ca 3 x 3 + Therefore (Art. 185) , Aa — 1 ; ... 4 = 1. a Ab + Ba 2 = ; a 3 Ac + 2Bab + Co 3 = ; „ 26 2 - ac . . u = ■ '. a 5 which in (2) gives a a 3 a «c).v 3 _^ 428 EXAMPLES. When the given series contains only the odd powers of x, we may shorten the process by assuming for as another series containing only the odd powers of y. Thus, 2. Revert the series y = ox + bx* + ex 5 + Here we assume x = Ay + By 3 + Cy 5 + If we assumed as we did in Ex. 1, the alternate coefficients would each reduce to zero. Solving this example, we obtain y bf (3b 2 - ctc)y b x = a ~ a 4 + a 7 If the first term of the series to be reverted in any case be independent of x, thus if the series be of the form y = a' + ax + bx 2 + etc., we must put y — a' = z, and express x iu a series involving the powers of z, and /hen replace the value of z. Revert the following series : 3. y = X + X 2 + x* 4- Ans. x = y — y 2 + y 3 — y* + 4. V = x + - + |- + ?/ 2 . ?/ 3 ?/ 4 . ^•* = y 2 + JA~Ll + 5. y = x - 2x 2 + 3.x 3 - Ans. x = y + 2y 2 + 5?/ 3 + Uy* + EXAMPLES. Sum the following ten recurring series to infinity : 1 — 2aj 1.1+ 3a; + 11a;- + 43x 3 + . . . . -4ns. . . 2. 1 + 2k + 3z 2 + 4z 3 + 5a* 1 + _ • 3 . ! + 2 , + 8* 2 + 28., 3 + r^l^^^ 4 . 2 - x + 5ar» - 7* 3 + 1 + aT-V EXAMPLES. 429 7 20x 5. 7 - 6x + 9x' + 27z« + Am. t _ 2a . _ ^ 6. 1 + 3s + 5* + 1* + T=£t*- 7. 1 + Sx + 4s« + '*■ + 8. 2 + 3X + W + M+ f _ 1 + 2x — x - 2 - 3x 9. 3 _j_ 6a; + 14a; 2 4- 36a; 3 + 98a; 4 + 276a; 5 + .....'. A 3 - 12a; + H« a * ' 1 — 6a; + 11a; 2 — 6a; 3 ' 10. 2 - a; 4- 2a; 2 - 5a; 3 4- 10a; 4 - Ux 5 4- 2 + 5a; + 5x\ (1 + *)■ Sum the following six series : 11. i+ J- + J- +7^ + ... to infinity. Am. f f. l-o 2 • b 3-7 4 • « 12 _i I ■ H 1- (1) to n terms, and ' 8-8^6.12 9-16 (2) to infinity. [Multiply by 3 • 4.] 13. _1 1 1 1 to infinity. J. 1.3 2-4 3-5 14. — 1 1 h (1) to n terms, and (2) 3-4 4-5 5-6 to infinity. Ans. (1) g n + ; (2) f 15. — + — + — + (1) to n terms, and (2) 1 . 3 2-4 3-5 to infinity. Ans. (1) f(f - n J+ + + 2 ) 5 ( 2 ) *■ 16. — — I — I 1 + ... to infinity. 1. 1.5^5.9^9.13^13.17 T J 430 EXAMPLES. Find the n th term and the sum of n terms of the series : 17. 1, 4, 10, 20, 35, Ans n(n+ l)(w + 2) , 71(71 + l)(n + 2)(n + 3) [8 [± 18. 4,14,30,52,80,114, 3w 9 + »; n(n + l) 2 . 10. 8, 26, 54, 92, 140, 198, Aits. 5n 2 -f 3n ; £w(n + 1)(5» + 7). 20. 2, 12, 36, 80, 150, 252, Ans. n\n + 1) ; T V*0 + l)(w + 2)(3n + 1). 21. 8, 16, 0, -64, -200, -432, - Aris. — 4n 2 (?i — 3) ; — n(n + l)(n 2 — 3?i — 2). 22. 30, 144, 420, 960, 1890, 3360, Ans. n(n+l) (n+2) (w+4) ; ^-(w+l)(w+2)(n+3)(4w+21). Find the number of balls in 23. A triangular pile of 18 courses. Ans. 1140. 24. A rectangular pile, the length and breadth of the base containing 50 and 28 balls respectively. A)is. 16646. 25. An incomplete triangular pile, a side of the base having 25 balls, and a side of the top 14. A71S. 2470. 26. An incomplete square pile of 27 courses, having 40 balls in each side of the base. Ans. 21321. 27. An incomplete square pile having 169 balls in the top course, and 1089 in the lowest course. Ans. 11879. 28. A complete rectangular pile of 15 courses, having 20 balls in the longer side of its base. Ans. 1840. 29. An incomplete rectangular pile, the number of balls in the sides of its upper course being 11 and 18, and the number in the shorter side of its lowest course being 30. Ans. 11940. 30. An incomplete square pile of 16 courses when the number of balls in the upper course is 1005 less than in (he lowest course. Ans. L8296. EXAMPLES. 431 31. The number of balls in a complete rectangular pile is 24395 ; if there are 34 balls in the breadth of the base, how many are there in its length? A>is. 52. 32. The number of balls in a triangular pile is greater by 150 than half the number of balls in a square pile, the number of courses in each being the same : find the number of balls in the lowest course of the triangular pile. Arts. 300. 33. Show that the number of balls in a square pile is one- fourth the number of balls in a triangular pile of double the number of courses. 34. If the number of balls in a triangular pile is to the number of balls in a square pile of double the number of courses as 13 to 175, find the number of balls in each pile. Ans. Triangular 364 ; Square 4900. 35. The square roots of 30, 31, 32, 33 are respectively 5.47723, 5.56776, 5.65685, 5.74456 : interpolate the square roots of 30.3, 30.5, 30.8. Ans. 5.50454, 5.52268, 5.54978. 36. The cube roots of 9, 11, 13, 15 are respectively 2.08008, 2.22398, 2.35133, 2.46621 : interpolate the cube roots of 10, 12, 14. Ans. 2.15443, 2.28943, 2.41014. 37. The reciprocals of 74, 75, 76, 77 are respectively .01351, .01333, .01316, .01299: interpolate the reciprocals of 74.2, 74.4, 74.7. Ans. .01348, .01344, .01339. 38. The logarithms of 232, 233, 234, 235 are respectively 2.365488, 2.367356, 2.369216, 2.371068: interpolate the logarithms of 233.4, 233.6, 233.8. Ans. 2.368101, 2.368473, 2.368845. 39. The right ascension of Jupiter on a certain Monday at noon was 10 h. 5 m. 38.6 s. ; on Tuesday at noon it was 10 h. 6 m. 18.86 s. ; on Wednesday at noon 10 h. 6 m. 59.41 s. ; on Thursday at noon 10 h. 7 m. 40.24 s. Find its right ascension on Tuesday at midnight. Ans. 10 h. 6 m. 39.1 s. 40. The latitude of the moon 1888, March 26, Monday, at noon was 3° 9' 34."4 ; at midnight 3° 38' 42. "5 ; on Tuesday 432 EXAMPLES. at noon 4° 4' 23."5 ; at midnight 4° 20' l/'l ; on Wednesday at noon 4° 43' 4. "4. Find its latitude at !) p.m. on Tuesday. Ans. 4° 21/ l."3. Revert the followiug series : 41. y = x + 3-«' 2 + oar 5 + 7x 4 + Ans. x = y - 3/ + 13// 3 - b^f + 42., = *-f + f-...* = , + |° + £ + | + ... *, — f + ?-*+-..... Ans. x = y + £+ ^f + ^if + 44. y = 2x + Sx 3 + 4.r 5 + 5x 7 + Ans. x = \y - T %?/ 3 + T V 9 g.V 5 - 45. y = x + f + f+... x = y-£ + y~-£ + 2 6 2 3 4 46. ?/ = 1 — 2x + 3ar — Ans. x=- i0/ - 1)+ 10/ - I) 2 - MV ~ O 8 + 47. y = 1 + ^ + t + f- + 2 b ^ws. a? = (y - 1) - K^ - !) 2 + 1(2/ ~ !) 8 ~ • • 48. Given y + a?/' 2 + ^ 3 + . . . = px + go: 2 + rx 3 + , to express x in terms of y. Ans. x=V-+ («P 2 -?).V 2 + R>j> 4 -pr-2g(ap 2 -r 7 )] ? / 3 | _ _p p« p- LOGARITHMS — DEFINITIONS. 433 CHAPTER XXI. LOGARITHMS— EXPONENTIAL AND LOGARITHMIC SERIES — INTEREST AND ANNUITIES. LOGARITHMS. 202. Definitions. — The Logarithm of a number is the exponent of the power to which another number, called the base, must be raised to equal the given number. Thus, if a x = N, x is called the logarithm of iVto the base a. This is usually written log a iV, so that the same meaning is expressed by the two equations a x = N; x = log a JV. Examples. (1) Since 3 4 = 81, 4 is the logarithm of 81 to the base 3. (2) Since 10 1 = 10, 10 2 = 100, 10 3 = 1000, the natural numbers 1,2, 3, are respectively the loga- rithms of 10, 100, 1000, to the base 10. A System of Logarithms means the logarithms of all positive numbers to a given base. Thus, if we suppose a to remain fixed while N assumes in succession every value from to oo, the corresponding values of x will constitute a system of logarithms to the base a. The base is then called the base of the system. Any number might be taken as the base of the system, and corresponding to any such base a system of logarithms of all numbers could be found. Thus, suppose a = 9 ; then Since 9 1 = 9, 1 = log 9 9. Since 9'= 81, 2 = log 9 81. Since 9 3 = 729, 3 = log 9 729. That is, the natural numbers 1, 2, 3, . . . . are respectively 434 PROPERTIES OF LOGARITHMS. the logarithms of 9, 81, 729, .... in the system whose base is 9. Note. — When it is understood that a particular system of loga- rithms is in use, the suffix denoting the base is omitted. Thus in Arithmetic calculations in which 10 is the base, we usually write log 2, log 3, instead of log )0 2, log, 3, 203. Properties of Logarithms. — Before discussing the logarithmic systems in common use, we shall prove some general Properties which are true for all logarithms, whatever the base may be. (1) The logarithm of 1 is 0. For a = 1 for all values of a ; therefore log 1=0, what- ever the base may be. (2) The logarithm of the base itself is 1. For a 1 = a ; therefore log a « = 1. (3) The logarithm of in any system whose base is greater than 1 is minus infinity. For a~°° = — = — = 0; therefore logO = — a>. a 00 go Also, the logarithm of -+- is + co. (4) The logarithm of a product is equal to the s»m of the logarithms of its factors. For let mn be the product ; let a be the base of the sys- tem, and suppose x = log?>i, y = logn. " Then m = a 1 , n = a' J . Therefore the product, mn = a* X a" = a x + v ; whence, by definition (Art. 202), log?>iH = x + y = logm + logx. Similarly, \ogmnp = logm -f log n + logp; and so on for any number of factors. Thus, log 30 = log (2 x 3 X 5) = log 2 + logo + log 5. PROPERTIES OF LOGARITHMS. 435 (5) The logarithm of a fraction is equal to the logarithm of the numerator minus the logarithm of the denominator. For let — be the fraction, and suppose n x = log?u, y = logn. Then m = a x , n = a". Therefore the fraction, — = — = af~ y ; n a" whence by definition, log — = x — y = logm — logn. n Thus, log - 4 / = log 42 — log 5 = log 2 -f log 3 + log 7 — log 5. (G) The logarithm of any power of a number is equal to the logarithm of the number multiplied by the exponent of the power. For let x = log ra ; then m = a x . Therefore m ?> = (« x ) p = a px ; whence by definition, log m p = px = p log m. (7) The logarithm of any root of a number is equal to the logarithm of the number divided by the index of the root. For let x = log m ; then m = a x . ] 1 X Therefore m r = (a x ) r = a r ; whence by definition, - x 1 logfwi'-) = - = -logm. r r It follows from these results that by the use of logarithms, the operations of multiplication aud division may be replaced by those of addition and subtraction ; and the operations of involution and evolution by those ot multiplication and division. 436 COMPARISON OF TWO SYSTEMS OF LOGARITHMS. EXAMPLES. 1 . Express the logarithm of 5L_ m terms of log a, log b x and logc. loo; & c 6 6 2 = log — = °c 5 6 2 log a* - log(c 5 6 2 ) = f log a - (logc 5 + log& 2 ) = flog a — 5 log c - 2 log 6. Express the following logarithms k i terms of loga, logfr, and logc log Ans. G log « 2. y/(a 2 & 8 ) 6 + 9 log b. 3. log (V«' 2 x \/b*).__ flog a + f log &, 4. log- {\Ja~ 2 b x V a ^ -3 )- — flog a - £log&. 5. log \Jab~ ] L c~ 2 £ log a. 6. log (V«-v &• + v'6 V«) — •Hog a. 204. Comparison of Two Systems of Logarithms. Given the logarithm of a number to base a, to find the loga- rithm of the same number to base b. Let m be any number whose logarithm to base b is required. Let x = log b m ; then b x = m ; .-. \og a (b x ) = \og a m ; ora;log & = log a ?u. • •. x = x log a m, log a 6 , log-m , , , or Io". logZ) = — (n + 1) + a fraction. Therefore the characteristic of the common logarithm of any number may be written down by the following Rule. % (1) Tlie characteristic of the logarithm of a number greatei than unity is posit ire, and less by one than the number of digits in its integral part. (2) The characteristic of the logarithm of a decimal /rue/inn is negative, and greater by one than tin' number of ciphers immediately after the decimal point. Thus, the characteristic of log 2347.3 is 8 j characteristic of log .000459 is -4. TABLE OF LOGARITHMS. 439 206. Table of Logarithms. — The common logarithms of all integers from 1 to 200000 have been found and tabulated. In most tables they are given to six places of decimals ; and tables of logarithms to seven places of deci- mals are in common use for Astronomical and Mathematical calculations. This is the system in practical use, and it luis two great advantages : (1) From the rule (Art. 205) the characteristics can be written down at once, so that only the mantissa} have to be given in the tables. (2) The mantissa? are the same for the logarithms of all numbers which have the same significant digits ; so that it is sufficient to tabulate the mantissse of the logarithms of integers. For, since altering the position of the decimal point without changing the sequence of figures merely multiplies or divides the number by an integral power of 10, it follows that its logarithm will be increased or diminished by an integer, i.e., that the mantissa of the logarithm remains unaltered. In general. If N be any number, and p and q any in- tegers, it follows that N X 10 p and N -=- 10 17 are numbers whose significant digits are the same as those of N. Then log (N x 10") = \ogN + p log 10 = log N + p . (1 ) Also log (N + 10?) = log A 7 " - q log 10 = log N - q . (2) In (1) the logarithm of A 7 " is increased by an integer, and in (2) it is diminished by an integer ; that is, the mantissa, or decimal portion of the logarithm, remains unchanged, a property which is true only in the common system. Thus, from a table of logarithms we find the mantissa of the logarithm of 2583 to be 412124 ; therefore, prefixing the characteristic with its appropriate sign according to the rule, we have log 2583 = 3.412124, log 258.3 = 2.412124, log 25.83 = 1.412124, log 2.583 = 0.412124. 440 EXAMPLES. The mantissa is always kept positive. In the case of a negative logarithm the minus sign is written over the charac- teristic, and not before it, to indicate that the characteristic alone is negative, the mantissa being always positive. Thus, 3. 30103, the logarithm of .002, is equivalent to — 3 + .30103, and must be distinguished from —3.30103, an expression in which both the integer and the decimal are negative. Sometimes in working with negative logarithms an Arithmetic artifice will be necessary to make the mantissa positive. For example, a result such as —2.69897, in which the whole expression is negative, may be transformed by subtracting 1 from the characteristic, and adding 1 to the mantissa. Thus, -2.69897 = -3 + (1 - .69897) = 3.30103. Note. — When the characteristic of a logarithm is negative, it is often, for convenience, made positive by the addition of 10; which can lead to no error, if we are careful to subtract 10. Thus, instead of the logarithms 3.603582 and 4.026011, we may write 7.6035S2 — 10 and 6.026011 - 10. EXAMPLES. The following examples will illustrate the utility of loga- rithms in facilitating Arithmetic calculations. For infor- mation as to the use of Logarithmic Tables the student is referred to works on Trigonometry and to the introductions to Tables of Logarithms. 1. Required the logarithm of .0002432. In the tables of logarithms we find that 3859636 is the mantissa of log 2432 (the decimal point, as well as the char- acteristic, being omitted); and by the rule (Art. 205). the characteristic of the logarithm of the given number is —4. .-. log .0002432 = 4.3859636. 2. Find the value of \/.00000165, given log 165 = 2.2174889, ami log 697424 = 5.8434968. EXAMPLES. 441 Let x = the value required ; then log a- = log (.00000165)* = \ log (.00000165) [Art. 203, (7)] = 1(6.2174839) = £(10 + 4.2174839) = 2.8434968, and .8434968 is the mantissa of log 697424; hence k is a number consisting of these same digits, but with one cipher after the decimal point, by the rule (Art. 205). .-. x = .0697424. It will be seen that we subtracted 4 from the characteristic — 6 to make it exactly divisible by the index 5, and there- fore we added the same number 4 to the mantissa, to keep the value of the logarithm unchanged. 3. Given log 3 = . 4771213; findlog [(2.7) 8 x(.81)*-=-(90)l]. The required value = 3 logfj + f ^StVo ~~ I log 90 = 3(log3 3 - 1) + f(log3 4 - 2) - £(log3 2 + 1) = (9 + ¥ -f)log3 -(3 + f + |) = ?■£ log 3 - 5i£ = 4.6280766 - 5.85 = 2.7780766. Note. — The logarithm of 5 and its powers can always be obtained from log 2 ; thus log5 = log^ = log 10 — log2 = 1 - log2. 4. Find the number of digits in 875 16 , given log 2 = .3010300, and log 7 = .8450980. log(875 16 ) = 16 log (7 x 125) = 16(log7 + 3 log 5) = 16(log7 + 3-3 log 2) = 47.072128; hence the number of digits is 48 (Art. 205). Given log 2 = .3010300, log 3 =.4771213, log 7 =.8450980 ; find the value of 8. log 4f. Ans. .6690067. 9. log V 12. .3597271. 10. log V- 01 05. 1.5052973. 11. Given log 2 and log 3 ; find Log V/ ~. 1.1998692. 5. log 64. Ans. 1.8061800. 6. log. 128. 1.1072100. 7. W.0125. 2.0909100. 442 EXPONENTIAL EQUATIONS — EXAMPLES. 207. Exponential Equations. — An Exponential Equo Hon is one in which the exponent is the unknown quantity. Such equations are easily solved by logarithms. Thus, 1. Given a x = b ; find the value of x. Taking the logarithms of both members, we have x log a = log b log& X = — 2 — . log a 2. Find the value of x in 2 X = 128. Here a; log 2 = log 128 = log2 7 = 7 log 2. " X ~ log2 ~ 7 ' 3. Find the logarithm of 32y/4 to base 2y/2. Let x be the required logarithm ; then by definition (Art. 202), (2^2)* = 32y/4 = 2 5 ■ 2*. Taking logarithms of both sides, we have x log 2* = log 2 5 ° , or fa: log 2 = % 7 log 2. ••• « = ¥ + f = ¥ = 3.6. 4. Given log 2 and log 3 (Art. 206, Ex. 5) ; find to two places of decimals the value of x from the equation Taking logarithms of both sides, we have (3 - 4a;) log G + (a; + 5) log 4 = log 8 ; .-. (3 - Ax) (log 2 + log 3) + (a + 5)2 log 2 = 3 log 2; .-. x{— 4 log2 -4 log 3 + 21og2)=: ?, log2 - 31og2 - 3 log3 - 101og2; x = 10 log 2 + 3 log 3 = 4.4410639 _ j 7? 2 log 2 -t- 41og3 2.5105452 Solve the following equations : 5. a x = cb x . Ans. a - los& EXPONENTIAL SERIES. 6. a 2 ^ 3 * = c 6 . Ans. 7. a' 443 5 lo p; c 2 log a + 3 log 6 )og(logc) — log (log a) log b b x Note. — In this example, by a is meant a raised to the power c. expressed by b x , and not a b raised to the X th power. 8. C" ^ + ... (4, This result is called the Exponential Theorem. If we put x = 1, we have from (3) + L1 LI Li LI L^L From this series we may readily compute the approximate value of e to any required degree of accuracy. This con- stant value of e is called the Naperian base (Art. 205). To ten places of decimals it is found to be 2.718281.8284. 209. Logarithmic Series. — To expand % e (l + a;) in a series of ascending powers of x. By the Binomial Theorem we have a * = (1 + o - 1)* = 1 + x{a - 1) + X ( X ~ X \ a- - \y + <—iK«-i) (g _ 1)1 + _ Bl+a;[a-l-|(o-i)»+Ko-l)*-Ka- !)*+•••] + terms involving x~, sc", etc. COMPUTATION OF LOGARITHMS. 445 Comparing this value of a x with that given in (4) of Art. 208, and equating the coefficients of x, we have log e a = a - 1 - |(a - 1)' J + £(« ~ 1 ) 3 - £(« - l) 4 + • • • Put a = 1 ■+- x ; then log e (l + a) = * - | 2 + | 3 - -| + (1) This is the Logarithmic Series; but unless x be very small, the terms diminish so slowly that a large number of them will have to be taken ; and hence the series is of very little use for numerical calculations. If x = 1, the series is altogether unsuitable. We may, however, deduce from it other series by the aid of which Tables of Logarithms may be constructed. Changing x into — jc, (1) becomes log e (l - x) = -* - |V | - t - (2) Subtracting (2) from (1), we have log e (l + c) - log c (l - x) = 2x + y +~ + , 1 4- x J . a; 3 . x 5 . x 1 , \ (3) Put LilJL ; = ?L±i, and .-. aj = ; (3) thus 1 ~ x n 2n -f- 1 v ' becomes gfi w " [_2w + l 3(2% + l) 3 5(2w + l) s J or 1<)g .( B+ l) = l C)g . n+ 2[^ +3li ^ +675 i I?+ ...](4) 210. Computation of Logarithms. — By means of series (4) of Art. 209, the logarithm of either of two con- secutive numbers may be computed to any extent when the logarithm of the other number is known, and thus a table of logarithms can be constructed. Logarithms to the base e 446 NAPERIAN LOGARITHMS. are called Naperian Logarithms (Art. 205). They are also called natural logarithms, because they are the first log- arithms which occur in the investigation of a method of calculating logarithms. When logarithms are used in theo- retical investigations, the base e is always understood, just as in practical calculations the base 10 is invariably employed. It is only necessary to compute the logarithms of pHrru numbers from the series, since the logarithm of a composite number may be obtained by adding together the logarithms of its component factors. The logarithm of 1 is 0. Making u = 1, 2, 4, 6, etc. successively iu (4) of Art. 209, we obtain the followiug NAPERIAN LOGARITHMS. log, 2 = log, 1 + 2 (- + — + — - + -L- + — + . . . V or, since log e 1 = 0, log e 2 = 2(.33333333 + .01234568 + .00082305 + .00006532 + . 00000565 +. 00000051 -f .00000005+. . .) =±2(0.34657359) =0.69314718. log e 3==log e 2 + 2A+^i-4--V r + ^i-+. . \ = 1.09861228. \.> 3 • o 3 5 • 5 d 7 • o 7 / log e 4 = 21og e 2 =1.38029436. log c 5 = log e 4 + 2/'-+— !-H — L .-J — L_ + . A= 1.60943790. oe T \9 8-9* 5-9 5 7-9 7 / log e 6 = log, 3+log e 2 = 1 . 79 1 75946. log e 7 = log e 6 + 2^-- + -1—+-1— + . . .\ =1.94590996. log e 8 = 31og e 2 =2.07944154. log c 9 = 21og,,3 =2.19722456. log c I0=log e 5+log c 2 =2.80258509. In this way the Naperian logarithms of all numbers may be computed. Where lln v numbers are large, their logarithms are computed mure easily thau in the case of small numbers. INTEREST AND ANNUITIES — SIMPLE INTEREST. 447 Thus, in computing the logarithm of 101, the first term of the series gives the result true to seven places of decimals. By changing b to 10 and a to.e in (1) of Art. 204, we have log M m = kg^L = log.™ = .43429448 log.m, 010 ]og e 10 2.30258509 oe ' or common log m = Naperian log m x .43429448. The number .43429448 is called the modulus of the common system. Hence, the common logarithm of any number is equal to the Naperian logarithm of the same number multiplied by the modulus of the common system, .43429448. Conversely. Having given the common logarithm of a number, to find its Naperian logarithm; divide the common logarithm by .43429448. Note. — The base e, the modulus of the common system, and the Naperian logarithms of 2, 3, and 5 have all been calculated to more than 260 places of decimals. See the Proceedings of the Royal Society of London, Vol. XXVII, p. 88. Denoting the modulus by M, and multiplying (4) of Art. 209 by it, we obtain a series by which common logarithms may be computed ; thus, 1. Find by means of (1) the common logarithms of 65 and 131 ; given log 2 = .3010300, and M = .43429448. Ans. 1.8129134, 2.1172713. 2. Find the Naperian logarithm of 1325.07. Ans. 7.18923. INTEREST AND ANNUITIES. 211. Simple Interest. — To find the interest and amount of a given sum in a given time at simple interest. Interest is money paid for the use of money. The Princi- pal is the sum lent. The Amount is the sum of the Principal and Interest at the end of the time. When interest is paid on the principal alone, it is called Simple Interest. 448 PRESENT VALUE AND DISCOUNT. Let P be the principal in dollars, r the interest of $1 for one year, n the number of years, I the interest, and M the amount. Since the interest of %\ for one year is r, the interest of P for one year is Pr, and therefore for n years it is Pur; that is. I = Pnr (1) And for the amount, M = P + I, or M = P(l + nr), (2) From (1) and (2), if any three of the quantities P, n, r, 7, or P, n, r, M, are given, the fourth may be found. 212. Present Value and Discount. — To find the present value and discount of a given sum due at the end of a given time, at simple interest. The Present Value of a sum due at the end of a given time, is the principal which with its interest for the given time will amount to the sum. The Discount is the difference between the sum and its present value. Let M be the given sum or amount, r the interest of $1 for one year, n the number of years, P the present value, and D the discount. .Since P is the sum which will in n years amount to il/, we have from (2) of Art. 211, M = P(l + nr) ; .-. P = M . . (1) 1 -f- nr And Z>= M-P= Mnr . . (2) 1 + nr Note. — This value of 7) is called the true discount. In practice, when a sum of money is paid before it is due, it is customary to deduct the interest on the debt instead of the true discount, and the money so deducted is called the banker's discount; tlms, Banker's Discount = Mnr. True Discount = — . (3) 1 + nr K } 1. Required the rate when $300 in 2 years at simple interest amounts to $318. COMPOUND INTEREST. 449 Here M - 318, P = 300, n = 2. Solving (2) of Art. 211 for r and substituting, we have 318 - 300 nQ . . m r = = .03. .•. rate per cent* =3. 300 x 2 2. The difference between the true discount and the bank discount on $1900 paid 4 months before it is due is 33^ cents : find the rate per cent. Here M = $1900, and n = *; therefore from (3) we have 1900r 1900r h 3 1 + Solving, we get r = .04. .•. rate per cent = 4. 213, Compound Interest. — To find the interest and Amount of a given sum in a given time at compound interest. When interest as soon as it becomes due is added to the principal, and interest charged upon the whole, it is called Compound Interest. Let P be the principal, 11 the amount of $1 in one year, so that II = 1 + r, n the number of years, / the interest, and M t4ie amount. Since the amount of $1 for one year is E, the amount of P at the end of the first year is Pli ; and since this is the principal for the second year, the amount at the end of the second year is PR x R or PR 2 . Similarly the amount of PR 2 at the end of the third year is PR S , and so on ; hence the amount of P in n years is PR" ; that is, M=PR\ .-. I = P(R" - 1) . . . (1) If the interest is pa}*able more frequently than once a * It must be borne iu mind that r is not the rate per cent, but only the hundredth part of it. 450 PRESENT VALUE AND DISCOUNT. year ; for example, if it is payable semi-annually, the interest for $1 for half a year will be -. .-. The amount of Pin half a year = P(l + -)• The amount of P in one year =P\ 1 + - ) • The amount of P in n years =P[ 1 + -) • Similarly if the interest is payable quarterly. The amount of P in n years =P(l+-\ . And generally, if the interest is payable q times a year at equal intervals, the interest of $1 for each interval will be -. 1-1 — ) • If the interest is due, i.e., converted into principal, every moment, then q becomes infinitely great, i.e., the intervals between the payments are infinitely small. To find the value of the amount, put - = -, so that q = rx ; thus the amount q x of P in n years -^7-^-*+ ST = Pe" r [(2) and (3) of Art. 208], since x is infinite when q is infinite. 214. Present Value and Discount. — To find the present value and discount of a given sum due at the end of a given time, at compound interest. Let M be the given sum or amount, P the present value. 7) the discount, A' the amount of §1 for one year, n the number of years. PRESENT VALUE AND DISCOUNT. 451 Since P is the sum which in n years will amount to 31, we have (Art. 213) M = PR ... P=K. D= M -P= M ^ n -'\ . (1) 1. Find the amount of $100 in one year at 5 percent, when the interest is converted into principal semi-annually. Here P = 100, /• = .05, n = 1, and 1 + - = 1.025. .-. Amount = 100(1.025) 2 = $105.0025. 2. The present value of $G72 due in a certain time is $126 ; if compound interest at 4£ per cent be allowed, find the time, having given log 2 = .30103, and log 3 — .47712. Here r = .04 J = fo and R = 1 -f r == §£. Let n = the number of years ; then by (1) of Art. 213, we have 672 = 126(|5)«; ... n log f| = log 1 ^. .-. n (log 100 - log DO) = log 16 - log 3. 4 lose 2 - log3 .72700 „ . • • n ~ \ ^ ^ > : — / ^„ F ., = 41, very nearly; 2 - 5 log 2 - log 3 .01773 J J thus the time is very nearly 41 years. 3. At simple interest, the interest on a certain sum of money is $180, and the discount on the same sum for the same time and at the same rate is $150 : find the sum. Ans. $900. 4. If a sum of money doubles itself in 40 years at simple interest, find the rate of interest. Ans. 2|. 5. Find in how many years $100 will become $1050 at 5 per cent compound interest ; having given log 1 1 = 1.14613, log 15 = 1. 17G0D, log 16 = 1.20412. Ans. Between 48 and li). 452 Till: AMOUNT OF AN ANNUITY. 215. Annuities. — An annuity is a fixed sum payable at equal intervals of time under certain stated conditions. An annuity certain is one which is payable for a fixed number of years independent of any contingency. A life annuity is one which is payable during the lifetime of a person, or of the survivor of a number of persons. A deferred annuity, or reversion, is an annuity which does not begin until after the lapse of a certain number of years ; and when the annuity is deferred for p years, it is said to com- mence after p years, and the first payment is made at the end of p + 1 years. A perpetuity is one which is to continue forever ; if it does not commence at once it is called a deferred perpetuity. An annuity left unpaid for a certain number of years is said to be forborne for that number of years. 216. The Amount of an Annuity. — To find the xmount of an annuity left unpaid for a given number of years. ( 1 ) At simple interest. Let ^4 be the annuity, n the number of years, r the interest of $1 for one year, M the amount. At the end of the first year A becomes due, and the amount of this sum in the remaining n — 1 years is A + (n — \)Ar ; at the end of the second year another A is due, and the amount of this sum in the remaining n — 2 years is A + (n — 2)rA; and so on to the end of the ?i ,h year. Hence, calling M the sum of all these amounts, we have M = A+(n-l)rA+A+(n-2)rA+ +A+rA+A = nA + (1 + 2 + 3 + + n - l)rA = nA + w(M ~ }) rA. [(:;) of Art. L61] ... (1) (2) At compound interest. Let 11 be the amount of $1 for one year. At the end of the firsl year .1 is due and the amount of this sum in the remainingn — 1 years is AU"" i (Art. 218) ; at the end of THE PRESENT VALVE OF AN ANNUITY. 453 the second year another A is clue, and the amount of this sum in the remaining n — 2 years is AR"~ 2 ; and so on to the last annuity payable at the end of the u th year. Hence the entire amount due at the end of n years is the sum of these amounts. •. M = AH"- 1 + AR'- 2 + .... + AR 2 + AR + A = (1 + R -f- B 2 -f . . . . to n terms)J. = f-^-y-i. [(2) of Art. 163] .... (2) 217. The Present Value of an Annuity. — To find the present value of an annuity to continue for a given number of years, at compound interest. Note. — In finding the present value of annuities, it is always customary to reckon compound interest. Let P be the required present value, A the annuity, R the amount of $1 in one year, n the number of years. Then the amount of P in n years must equal the amount of the annuity in the same time ; that is (Arts. 213, 210), J?' 1 — 1 1 — 7?-" PRn = K l A. .-. P= l ^ A. . (1) R - 1 R - 1 v ' If we make n infinite, the annuity becomes a perpetuity (Art. 215). Hence, making n infinite in (1), we have for the present value of a perpetuity, A A R - 1 r C-?) Rem. — If the present value of an annuity A for any number of years be mA, the auuuity is said to be worth m gears' purchase. Iu the case of a perpetuity, mA = — , by (2). 1 ioo .-. m = - = ; . . . . (3) r rate per cent that is, the number of years' purchase of a perpetual annu- ity is obtained by dividing 100 by the rate per cent. 454 THE PRESENT VALUE OF A DEFERRED ANNUITY. 218. The Present Value of a Deferred Annuity. — To find the 'present value of a deferred annuity, to commence at the end of p years and to continue for n years, at com- pound interest. Let A be the annuity, R the amount of $1 in one year, P the present value. The present value of an annuity to com- mence at the end of p years, and then to continue n years, is found by subtracting the present value of the annuity for p years from the present value of the annuity for p + n years. By (1) of Art. 217, we have i j?- p the present value of the annuity for p years = A ; the present value of the annuity for p + n years = A. R — 1 .-. P = ^ A - ^— A R-\ R - 1 AR-p AR~»-' R - 1 R - 1 (1) The present value of a deferred perpetuity (Art. 215) to commence after p years is P=^Z1 (2 ) R - 1 K } since n becomes infinite. A F 'reeh ■old Estate is an estate which yields a perpetual annuity called the rent; and thus the value of the estate is equal to the present value of a perpetuity equal to the rent. If, therefore, we know the number of years' purchase that a tenant pays to buy his farm, we obtain the rate per cent at which interest is reckoned by dividing 100 by the number of years' purchase [Art. 217, (•'>)]• Given the following logarithms for use when required : log 1.04 = .017033; log 1.05 = .021 189; log 1.8009 = .255495; log2.19U = .340660; logl.980 = .296646; log2.7859 = .444969 THE PRESENT VALUE OE A DEFERRED ANNUITY. 455 1. An annuity of $500 was unpaid for 15 years. What was the amount due, at 4 per cent compound interest? By (2) of Art. 216, M = (^Q 4 ) 15 ~ 1 x 50 = $9983. J v ' .04 2. Find the present value of an annuity of $500 for 20 years, at 4 per cent interest. By (1) of Art. 217, P = O- 04 ) 20 - 1 X fi0 ° 7n = $6787. J v ' .04 (1.04)' 20 3. Find the present value of an annuity of $1000 to com- mence at the end of 7 years and to continue 14 years, at 5 per cent interest. By (1) of Art. 218, P = A l R " ~ ^ J K J B - 1\ R p + n J = 1000 (1.05)" - 1 = S7030> .05 (1.05) 21 4. The reversion after 6 years of a freehold estate is bought for $20000 ; what rent ought the purchaser to receive, at 5 per cent compound imterest? Given log 1.05 = . 0211893, log 1.340096 = .1271358. The rent is equal to the annual value of the perpetuity, deferred for 6 years, which may be purchased for $20000. Let A be the value of the annuity in dollars. Then by (2) of Art. 218, 20000 = ^M)-\ .05 .-. .4(1.05)-° = 1000; log.1 - 6 log 1.05 = 3. log^l = 3.1271358 = log 1340.096. .-. A = $1340.090, which is the rent. 456 EXAMPLES. EXAMPLES. 1. Express log y/a -4 ^ 3 in terms of log a and log&. Ans. — | log a -f- -jlogfr. 2- Simplify log [(£!)- + (^ 3 ) S ]. -5 logo. 3. Simplify log V 729 V9- 1 x 27~i log 3. 4. Simplify logff - 21ogf + log^. log 2. 5. Show that log .V 5 ,' V 2 _ = 1 W 5 - — log 2 - 2 log 3. V18.V2 4 15 3 Given log 2 =. 3010300, log3=. 4771213, log 7 = .8450980: find the value of 6. log 84. Ans. 1.9242793. 7. logy/ff. .0563520. 8. log 14.4. 1.1583626. 9. log [V 48 (108)* -j- ^6~]. 1.0039238. 10. Given log 2, log 3, log 7; also log 44092388 =7.6 143636, and log 194.8445 = 2.2896883: find (1) the 7th root of .00324; and (2) the 11th root of (39.2) 2 . Ans. (1) .44092388, (2) 1.948445. 11. Given log 2, log3, log 7; alsolog9076. 226 = 3.9579053 : find to 6 decimal places the value of V[(294 x 'l25)-s-(42 x 32)] a . - Ans. 9.076226. 12. Find the product of 37.20:;, 3.7203, .0037203, 37203, having givdh log 37.203 = 1.5705780, and log 1915631 = 6.2823120. Ans. 19156.31. 13. Find the number of digits in 3 U x 2 8 . 9. 14. Determine the number of ciphers there arc between the decimal point and the first significant digit in (.\) 1000 . Ans. 301, 15. Show that (21 ~ 20) 100 > 100. EXAMPLES. 457 Solve the following equations : a log a -f- log 6 16. a x + 1 -=- b x ~ x = c 2 *. Ans. — s__T — &_ — 21ogc — log a + log?; log (^2 + 1) 17 . 2ato + ^ = a - 21oga 18. a? = y* , a; 3 = 2/ 2 . K = 2i, y = 3| 19. a 8 *- ft 89 = m 5 , x _ 4 log m __ log m a 3 * ■ ft 2 " = m 10 . log a ' ' log b ».«•-•.*■-«■+■.*■. i og6 10 -" g« ■21. <«e-W+i«)~»-(.-»)-(« + »)-. ||g=g. Find the logarithms of 22. .0625 to base 2, and 1000 to base .01. ^4ws. -4, -f. 23. .0001 to base .001, and .1 to base 90$. |, — |. 24. Va 1 ', 1, Vtf-V to base a. |, -|, -f. a* 25. Find the value of log^^ and log 343 49. — -|, f. Solve the following equations, having given log 2, log 3, and log 7 : 26. 2 X - 6* -2 = 5 21 • V~ x . Ans. 4.562. 27. 2*+» =6% ^ = log 3 . = log 2 3* = 3-2 y + 1 . log 3 — log 2' " log 3 — log 2' 28. 3 1 - T -"=4- y , 3 log 3 - 2 log 2 log 3 4(log3 - log 2)' J 4(log3 - log 2) 29. Given log 10 2 = .30103; find log 26 200. 1.6465. 30. Given log 10 2 = .30103, log 10 7 = .84509 ; find log,^ and log V2 7. Ans _log2. = 1?81 2]og_7 = ^ 6U , 2 log 7 log 2 458 EXAMPLES. 81. Given M = .43429448, and log 2 = .3010800; find the common logarithms of 7, 11, 17, and 31 by (1) of Art. 210. Arts. .8450980; 1.0413927; 1.2304489; 1.4913017. 32. Given M and log 10 2 : find log 10 1025. Ans. 3.0107239. When required, the following logarithms may be used : log 2, log 3, log 7 (See Ex. 6) ; also log 11 = 1.0113927. 33. At simple interest, the interest on a certain sum is $90, and the discount on the same sum for the same time and at the same rate is $80 : find the sum. Ans. 720. 34. A tradesman marks his goods with two prices, one for ready money, and the other for a credit of G months : find what ratio the two prices ought to bear to each other, at 5 per cent simple interest. Ans. 40 : 41. 35. Find the amount of $100 in 50 years at 5 per cent compound interest ; given log 114.074 = 2.0594G50. Ans. $1140.74. 36. In how many years will a sum of money double itself at 5 per cent compound interest? Ans. 14.2. 37. Find the present value of $10000 due 8 years heuce at 5 percent compound interest ; given log67683.94 = 4.83048,")6. Ans. $6768. 39|. 38. In how many years will $1000 become $2500 at 10 per cent compound interest? Ans. 9.6 years. 39. Show that money will increase more than a hundred- fold in a century, at 5 per cent compound interest. 40. What sum of money at <"> per cent compound interest will amount to $1000 in 12 years? Given log 106=2.0253059, log 49697 = 4.6963292. " Ans. $496.97. 41. In how many years will a sum of mone}' treble itself at 3£ per cent compound interest? Given log 10850 = 4.01494. Ana. Nearly 82. 42. A person borrows $672 to be repaid in 5 years by annual instalments of $120: find the rate of interest at simple interest. ^ins. 6 per cent. EXAMPLES. 459 43. A person borrows $600.25 : find how much he must pay annually that the whole debt may be discharged in 35 years, allowing simple interest at 4 per cent. Arts. $24.50. 44. Find the amount of an annuity of $100 in 20 years, at 4| per cent compound interest. Ans. $3137.11. 45. A freehold estate worth $100 a year is sold for $2500 : find the rate of interest.* Ans. 4 per cent. 46. The reversion, after 2 years, of a freehold worth $168.10 a year is to be sold : find its present value at 1\ per cent. Ans. $6400. 47. If 20 years' purchase must be paid for an annuity to continue a certain number of years, and 26 years' purchase for an annuity to continue twice as long ; find the rate per cent. Ans. 3^. 48. When 4 per cent is the rate of interest, find what sum must be paid now to receive a freehold estate of $400 a year 10 years hence; having given log 104 = 2.0170333, log 6.75565 = .8296670. Ans. $6755.65. 49. If a perpetual annuity is worth 20 years' purchase, find the annuity to continue for 3 years which can be pur- chased for $2522. Ans. $800. 50. If a perpetual annuity is worth 25 years' purchase, find the amount of an annuity of $625 to continue for 2 years. Ans. $1275. * In these examples the interest is compound unless otherwise stated. 460 IMAGINARY QUANTITIES —DEFINITIONS. CHAPTER XXII. IMAGINARY QUANTITIES. — INDETERMINATE EQUATIONS.— THEORY OF NUMBERS. IMAGINARY QUANTITIES. 219. Definitions. — An Imaginary Quantity is an indi- cated even root of a negative quantity.* Thus, yda, V-«*> 2 + V^"^ 3 ~ V"^ etc - are imaginary quantities. They are also called impossible or unreal quantities. Since no number whatever can have a negative square, it follows that if we have to extract the square root of a nega- tive quantity, no answer is possible, in ordinary positive or negative numbers (Art. 107). But although the square root of a negative quantity is thus the symbol of an impossible operation, yet these square roots frequently occur in mathe- matical investigations, and it becomes important to explain in what sense such roots are to be regarded, and how to deal with them. Every imaginary quantity may be reduced to one common form, which is a real quantity multiplied by the imaginary quantity V^— T- Thus, Imaginary quantities may be added, subtracted, and di- vided, the same as other surds; but when we attempt to multiply them together there occurs one point of importance which deserves notice. Thus, by the rale for multiplying surds together (Art. 128), we have s/~i x v^ = n/( -«■)(-«) = tf* 5 = ±«- All other quantities, whether rational or irrational, are ealled real quantities. FUNDAMENTAL PRINCIPLES. 461 But since the square root of any number multiplied by itself must, by the definition, give the original number; therefore y/— a x V— a = —a. .Similarly, y/-a • >f-b = \/a • \^-l ■ \Jb . V^l = ^d)(sj~iy = -\fab~. The surd V — 1 is called the imaginary unit. The imaginary unit is often expressed by the letter i ; but until the student has had a little practice in the use of imaginary quantities, he will find it easier to retain the symbol V— 1. It is useful to notice the successive powers of V'— 1 or i. Thus, since the imaginary unit is such a quantity as when squared, the result is —1, therefore i = V^l, i»= (V^l) 2 = -1, r» = (V/-1) 3 = (\^l) 2 .\d = -\^i, ? 4 = (v^) 4 = (v 7 -!) 2 - (v^T) 2 = (-i)(-i) = i, and so on, repeating the results, \f— 1, — 1, — \] — 1, 1. The sum of a real and an imaginary quantity is called a Complex Quantity. It is usual to apply the term imaginary to all expressions which arc not wholly real. Thus, a + b\J —\ may be taken as the general type of all imaginary expressions. Here a and b are real quantities, but not necessarily rational. 220. Fundamental Principles. — ( 1 ) If a+b\J~\ = 0, then a = 0, and b = 0. For, if a + ?V— 1 = 0, then ZV— 1 = -a; .-. -b 2 = a 2 ; .-. a 2 + b 2 = 0. Now a 2 and b 2 are both positive ; therefore their sum cannot be zero unless each of them is separately zero ; i.e., a = 5 and b = 0. 462 FUNDAMENTAL PRINCIPLES. (2) If a + b^l = c + dtf—L, then a = c, and b = d. For, by transposition, a — c + (6 — d)\J — 1 = ; therefore by (1), a — c = 0, and b — d = 0; i.e., a = c, and & = d. Thus, in order that tivo imaginary expressions may be equal, it is necessary and sufficient that the real parts should be equal, and the imaginary parts should be equal. When two imaginary expressions differ only in the sign of the imaginary part, they are said to be conjugate. Thus, a — &V— 1 is conjugate to a + b\l — 1. The sum and the product of tivo conjugate imaginary expressions are both reed. For, a + b\l~\ + a - b>T^i = 2a . . . . (1) Also, (a + b)J~l)(a - Vf^X) = a" + b 2 . . (2) The use of imaginary units enables us to factor expressions which are prime when only real factors are admitted. Thus, (2) shows that the sum of two squares can always be expressed as the product of two complex factors. The positive value of the square root of a- + & 2 is called the modulus of each of the conjugate expressions, a + 6^1 and a - &V^— 1. Divide v'^a 2 by V^^. TT ./ :. j — To \l— a 2 rV-1 a Here V— « 2 -*■ V — b- — , = , =s -. \J-b 2 by/ -i h Hence imaginary expressions are divided the same as other surds. 221. Geometric Meaning of the Imaginary Unit. — It is clear that V— 1> or i, cannot represent a number, or any real quantity such as the symbols a, b, c, etc. represent; it must signify something which a real quantity does not take account of. As it occurs only as a multiplier of one of these other symbols, it must denote sonic operation performed on the real quantity denoted by these symbols; and the result of operating with it must be entirely different from that of operating with a real quantity. EXAMPLES. 463 All real numbers may be represented in magnitude and direction by a horizontal line, the positive numbers increasing in one direction from zero, and the negative ones in the opposite direction from the same point (Art. 20). Now ia must denote the result of some operation performed on a, and i 2 a must similarly denote the result of performing this operation twice. But i 2 a = —a (Art. 219). Hence, to multiply the line +a twice by the imaginary unit i reverses the sign of a. We have seen that if a denote a line measured in one direction, — o denotes a line of equal length measured in the opposite direction, a line which could be obtained by rotating the first line in any plane through two right angles. Hence, Multiplying by the imaginary unit i may be taken to denote the operation of turning the line a through one right angle, in any plane which contains Us original iwsition. Thus, let BB' be drawn perpendicular to AA' at its middle point 0. Denote OA by the symbol a ; then multiplying a by the imaginary unit i turns the line OA through the right angle into the position OB. Multiplying again by i turns the line through the second right angle into the position OA', opposite OA. Multiplying a third time by i brings the line into the position OB 1 , opposite OB. Multiplying by i a fourth time restores the line to its original position OA. That is, OA = a; OB = ia; OA' = i 2 a = -a; OB' = i 3 a = —ia; OA = i 4 a = a; which corresponds to the law governing the powers of i (Art. 219). Note. — This is but a brief sketch of Imaginary Quantities. The subject might be extended to great length, but this would clearly be beyond the limits of this work. The student who wishes to pur- sue the subject further is referred to Peacock's Algebra, Warren's Treatise on the square root of negative quantities, Price's Calculus, Vol. I. etc. EXAMPLES. Multiply the following : 1. 2V^9 and S^l. 2. 24 + V-49 and 24 - ^-49. 3. 2V^ - 6V^5 and 4\/3 - ^/^~5. 4. Divide 6V^3 by 2V / ^4. 40 1 INDETERMINATE /CQUATJONS. Express with rational denominator: 1 11 3 - Sl-'l 3V/-2 - >l\P-b -19 - 6^10. 7 3 + 5i -9 + 19* 13 INDETERMINATE EQUATIONS. 222. Indeterminate Equations. — We have seen (Art. 93), that the definite determination of the values of two unknown quantities requires two independent equations, and that when only one equation is given involving two unknown quantities, there will generally be an infinite number of solu- tions ; also, if there be any number of independent equa- tions involving more than the same number of unknown quantities, there will in general be an unlimited number of solutions. Such equations are called indeterminate equations. If however the conditions of the problem be such as to exclude all solutions except positive integers, the number of solutions may be limited. We shall discuss only the simplest kinds of indeterminate equations, confining our attention to 'positive integral values of the unknown quantities ; the others being more tedious than elegant, and of not much practical use. Any equation of the first degree involving two unknowns, x and y, can be reduced to the form ax ± by = ±c, where a, b, c are positive integers. It is clear that the equation ax + by = — c has no positive integral solution ; and that the equation ax — by = — c is equivalent to by — ax = c ; hence it will be sufficient to consider the equations ax±by=c. Neither of the equations ax ± by = c can be satisfied by integral values of x and y if a and b have a factor which i Iocs not divide c. INDETERMINATE EQUATIONS. 465 For, if either equation has such a solution, then, dividing both members by the common factor, the first member is integral and the second fractional, which is impossible. Thus, neither of the equations 6a; ± 9y = 11 can be solved in integers, because the sum or difference of 6a; and 9y must be divisible by 3 whatever values x and y have. If a, 6, c have a common factor, it can be removed by division, so that in the examples of this Article we shall suppose that a and b are prime to each other. 1. Solve 7a; + 12?/ = 220 in positive integers. . . (1) Divide by 7, the smaller coefficient ; thus, x + v + |' = 31 + f ; • •• x + y + y = PI. Since x and y are to be integers, ^zl = integer (2) 7 Multiplying (2) by 3 in order to make the coefficient of y differ by unity from a multiple of 7, we have 15 y - 9 ^ 2y 1 + V - 2 - integer. 7 J 7 .*. ^ — = integer — p, suppose. .-. y — 2 = 7p; or y = 7j> + 2. . . (3) Substituting in (1), a; = 28 - 12^. ' • (4) This is called the general solution of the equation ; and by giving to p any integral value, we obtain from (3) and (4) corresponding integral values of x and y ; but if p > 2, we see from (4) that x is negative ; and if p is a negative inte- ger, we see from (3) that y is negative. Thus, the only 466 INDETERMINATE EQUATIONS. positive integral values of x and y are obtained by putting p = 0, 1, 2. Thus we have p = 0, 1, 2 a; = 28, 16, 4 y = 2, 1), 16 Note. — An artifice similar to the one used with (2) should always be employed hefore introducing a symbol to denote the integer. 2. Solve 5x — ly = 29 in positive integers. . . . (1) Divide by 5, the smaller coefficient ; thus, z - 2/ - f.'/ = 5 + £ k 2y + 4 . , x — y — 5 = -^ — ! — = integer. 5 Multiplying by 3, 6v + 12 . „ . y + 2 - = '/ + 2 + •' ^ = integer. 5 5 V + 2 . 4 • \ « — — = integer = p, suppose, 5 y/ = 4, x is negative. Thus the only positive integral values of x and y are obtained by putting p = 0, —1, —2, —3 ; therefore the number of ways is 4. If, however, the sum may be paid either in guineas or five-pound notes, p may also have the value —4. If p = — 4. then x = and y = 100, aud the sum is paid entirely in five- pound notes. Thus, if the zero value of x is admissible, the number of ways is 5. 4. The expenses of a party numbering 43 were $114.50 ; if each man paid $5, each woman $2.50, and each child $1, how many were there of each? Let », y, z denote the number of men, women, and children, respectively ; then we have x + y + 2 = 43 (1) 5x + 2.5// + z = 114.5. Eliminating z, we obtain 8x -\- 3y = 143. The general solution of this equation is x = Sp + 1 ; y = 45 - 8p. Substituting in (1), we obtain z = 5p — 3. Hence p cannot be negative or zero, but may have positive integral values from 1 to 5. Thus we have p= 1, 2, 3, 4, 5; x = 4, 7, 10, 13, 16; y = 37, 29, 21, 13, 5; z = 2, 7, 12, 1% 22. 468 TI1E0RY OF NUMBERS. Solve in positive integers : 5. 3s+8#=103. Ans. 3=29, 21, 13, 5; y=2, 5, 8, 11. 6. 5a; +2?/= 53. sal, 3, 5, 7, 9 ; # = 24, 19, 14, 9, 4. 7. 8a? + 65y = 81. # x = 2; y = 1. 8. A farmer spends $3760 in buying horses and cows ; if each horse costs $185 and each cow $115, how many of each does he buy? Ans. 11 horses, 15 cows. 9. In how many ways can $100 be paid in $1 pieces and half-dollar pieces, including zero solutions? Ans. 101. THEORY OF NUMBERS. 223. Scales of Notation. — In the ordinary method of expressing numbers * 1)3' figures, the number represented by each figure is always a multiple of some power of ten. For instance, in 256 the 2 represents 2 hundreds, i.e., 2 times 10'-; the 5 represents 5 tens, i.e., 5 times 10 1 ; and the 6 represents 6 units, i.e., 6 times 10°. Thus, 4639 2« an abbreviated way of writing 4 x 10 3 + 6 X 10 2 + 3 X 10 1 + 9. This method of representing numbers is called the common scale of notation, and 10 is said to be the base or the radix of the scale. The figures by means of which a number is expressed are called digits. The symbols employed in this system of notation are the nine digits and zero. In like manner any other number than ten may be taken as the radix of a scale of notation ; thus if 7 is the radix, a number expressed by 3256 represents 3x7 3 + 2x7 2 + 5x7 + 6; and in this scale no digit higher than 6 can occur. In a scale whose radix is denoted by r the above number 3256 represents 3r 8 + 2r + 5/- + 6. And generally, if in a scale whose radix is /• the digits, beginning with that in the * The word number in thto subject is used ;is an abbreviation tor positive <"/<r + dm 3 + + a multiple of p = p -f a multiple of /> ; thus the expression p' is divisible by p, and is therefore not a prime number. (2) The number of primes is infinite. For, if not, let p be the greatest prime number ; then the product 2 • 3 • 5 • 7 • 11 • • ]> of all the prime numbers up to p, is divisible by each of these prime factors. It follows that the number formed by adding unity to this product is not divisible by any of these factors; hence this number is either itself a prime number, or it is divisible by some prime number greater than p ; thus in either case p is not the greatest prime number. Therefore the number of primes is infinite. (3) A number can be resolved into prime factors in only one icay. Let N denote the number; suppose X = abed . . . . , where a, &, c, + 1)(8 + !*)('■+ 1) 476 THEOREMS IN RELATION TO NUMBERS. But there are two divisors of N corresponding to each way in which N can be resolved into two factors ; hence the required number is Up + 0(2 + !)(*•+ i) Second ; suppose N a perfect square, so that all the expo- nents p, q, r, . . . . are even. In this case, one way of resolution into factors is ViV x V 7 -^ and to this way there corresponds only one divisor, y/jSf. If we exclude this, the number of ways of resolution is *[(!»+ !)(? + !)(*■+ "l]i if to this we add the one way of resolution, ViV X v'iV, we obtain for the required number, *[(P+ 1)(9+ !)(*•+ 1) .... + !]• In this proposition JV X 1 is counted as one of the ways of resolving iVinto two factors. (G) To find the sum of the divisors of a number. Let the number be denoted by a v b q c r as before. Then each terra of the product (l+a+cr+ . . +a p )(l+b+b*+ . . +6«)(l+c+c 2 + . . +c r ) . . . is a divisor, and therefore the sum of the divisors is equal to this product; that is (Art. 163), the sum required - ((P + 1 ~ l ° q+1 ~ l . r ' H 1 - 1 a - 1 6-1 c — 1 (7) T<> find the number of ways in which a composite num- ber can be resolved into two factors which arc prime to each other. THEOREMS IN RELATION TO NUMBERS. All Let the number N = a p b n c r as before. Since the two factors are to be prime to each other, we cannot have some power of a in one factor, and some power of a in the other factor, but a p must occur in one of the factors. Simi- larly b q must occur in one of the factors only ; and so on. Hence the required number is equal to the number of ways in which the product abc can be resolved into two factors ; and is therefore, by (5), 1(1 + 1)(1 + 1)0 + 1) ,or2- 1 , where n is the number of different prime factors in N. 1. Given the number 21 GOO, find (1) the number of divisors, (2) the sum of the divisors, and (3) the number of ways in which it can be resolved into two factors prime to eacli other. Smce 21600 = 6 3 - 10" 2 = 2 5 • 3 3 • 5' 2 , therefore by (4) , the number of divisors = (5 + 1 ) (3 + 1 ) (2 4- 1 ) = 72 ; 2 6 _ l o 4 — 1 5 8 — 1 6v (6) , the sum of the divisors = • • - JV/ 2 — 13 — lo — 1 = 63 x 40 x 31 = 78120; I y (7) , the number of ways = 2 3 " 1 = 4. 2. If n is odd show that n(n" — 1) is divisible by 24. We have w(n 2 - 1) = n(n - l)(n + .1). Here n — 1, n, n + 1 are three consecutive numbers; hence one of them is divisible by 3 ; and since n is odd, n — 1 and n + 1 are two consecutive even numbers ; hence one of them is divisible by 2 and the other by 4. Thus the given expression is divisible by the product of 2, 3, and 4, i.e., by 24. 3. Find the least multiplier of 3675 which will make the product a perfect square. Ans. 3. 4. Find the least multiplier of 1845 which will make the product a perfect cube. Ans. 3 • 5 a • 41". 5. Find the number of divisors of 140. Ans. 12. 478 EXAMPLES. EXAMPLES. Multiply !. 2\^3 + 3V^2 by 4\^3 - 5^2. 4ns. 6 - 2^0. 2 . 3y/~7 - 5\^2 by 2>\l~l + 5^-1. -13. 3. f~ l + e"^ 1 by e^" 1 + V-3 byx _L^B. 2 Express with rational denominator 5. 3 + 2V^1 j 3 - 2^-1 2 _ 5^-1 2 + 5\/^T ^ 8 vlns. — — . 29 6. X - V-l » + V-l 2 (3a r - 1)V^1 x- + 1 (a + V^T) 3 - (a - V~T) 3 3«- - 1. (a + S/-1) 2 - (a-s/-l>" 2a 8. (i + o 2 3 - i -1 -f 3t 5 9. \fe - /\/2 8 - ?V6 2\ll - iV2~" 14 10. 1 + i i. 1 - j 11. Find the value of (— V 7 — l) 4 ' 1 " 13 , when ?i is a positive integer. -Ih.s. V — 1. 12. Find the square of ^9 +40^^1 + ^9-40^^!. 100. Solve in positive integers : 13. 7x + 12y = 152. Ans. x = 20, 8 ; y = 1, 8. 14. 13a; + 11// = 414. x = 9, 20, 31 ; y = 27, 14, 1. 15. 23a + 25y = 915. x = 30, 5 ; // = 9, 32. 16. 41a; + 47// = 2191. a; = 50, 3 ; y = 3, 1 I. 17. 17a; + 23?/ = 183. x = 4 ; // = 5. 18. 19.r + 5// = 119. a; = 1, 6; y = 20, 1. EXAMPLES. 479 19. In how many ways can $2000 be paid in $21 bills and $5 bills, supposing there to be sueh bills? Ans. 19, or 20. 20. In how many ways can $4000 be paid in $21 and $40 bills? Ans. 4, or 5. 21. In how many ways can $39 be paid in $4 and $5 bills? Ans. 2. 22. Find how many dollars and half-dollars, whose diam- eters are respectively .81 and .666 of an inch, may be placed in a row together, so as to make a yard in length. Ans. 28, 20. 23. A farmer buys oxen, sheep, and ducks. The whole number bought is 100, and the whole sum paid is £100. Supposing the oxen to cost £5, the sheep £1, and the ducks Is. per head, find the number he bought of each. Ans. 19 oxen, 1 sheep, 80 ducks. 24. I buy 40 animals consisting of sheep at $20, pigs at $10, and oxen at $85 ; if I spend $1505, how many of each do I buy? Ans. 28 sheep, 1 pig, 11 oxen ; or 13 sheep, 14 pigs, 13 oxen. 25. Add 303478, 150732, 264305 in the scale of nine. Ans. 728626. 26. Find the product of 4685 and 3483 in the scale of nine. Ans. 17832126. 27. Divide 14332216 by 6541 in the scale of seven. Ans. 1456. 28. Divide the difference between 121012 and 11022201 by 1201 in the scale of three. Ans. 2012. 29. Divide 102432 by 36 in the scale of seven. Ans. 1625. 30. Divide 17832126 by 4685 in the scale of nine. Ans. 3483. 31. Extract the square root of 33224 in the scale of six. Ans. 152. 32. Extract the square root of 11000000100001 in the scale of two. Ans. 1101111. 33. Extract the square root of 67556^21 in the scale of twelve. Ans. 8te7. 480 EXAMPLES. 34. Extract the square root of e.etOOX in the scale of twelve. Ans. eee. 35. Express 624 in the scale of five. 114 1. 36. Express 1458 in the scale of three. 2000000. Transform 37. 213014 from scale six to scale nine. Arts. 2G011. 38. 400803 from scale nine to scale five. 30034342. 39. 123456 from scale ten to scale seven. 1022634. 40. 1357531 from scale ten to scale live. 321420111. 41. 15951 from scale eleven to scale ten. 22441. 42. Find the least multiplier of 3234 which will make the product a perfect square. Ans. 66. 43. Find the least multiplier of 6480 which will make the product a perfect cube. Ans. 2' 2 • 3'- • 5 2 . 44. Find the least multiplier of 13168 which will make the product a perfect cube. 2- • (.S2:3) J . 45. Find the least multipliers of the numbers 4374, 18375, 74088, respectively, which will make the products perfect squares. Ans. 6, 15. 12. 46. Find the least multipliers of the numbers 7623, 109350, 539539, respectively, which will make the products perfect cubes. Am. 1617, 180, 18.".:). 47. Find the number of divisors of 8064. Ans. 48. 48. In how many ways can the number 7056 be resolved into two factors? Ans. 22. 49. Find the number of divisor,s of 1845"; and the num- ber of ways iu which it can be resolved into two factors. Am. 12; 6. 50. Find the number of divisors there are in |_9_, and the sum of these divisors. Ans. 160; 1181040. 51. Find the number of divisors of 1000, 3600, and 14553. Ans. 16, 45, 24. 52. In how many ways can a line 100R00 inches long lie divided into equal parts, each some multiple of an inch? Ans. 126. PROBABILITY {CHANCE) — DEFINITION. 481 CHAPTER XXIII. PROBABILITY (CHANCE). 227. Definition. — If an event can happen in a ways and fail in b ways, and if each of these ways is equally likely, the probability or chance of its happening is ; and the probability or chance of its failing is -. a -f b Otherwise, thus : If an event can happen in a ways and fail in b ways, and each of these ways is equally likely to happen, we can assert that the chance of its happening is to the chance of its failing as a to b ; therefore the chance of its happening is to the sum of the chances of its happen- ing and failiug as a to a + b. But the event must either happen or fail, hence the sum of the chances of its happen- ing and failing must represent certainty. Therefore the chance of its happening is to certainty as a to a + b. If then we represent certainty by unity, we have the chance that the event will happen is ; a + b and the chance that the event will not happen is -. a + b Hence we have the following definition : The probability of an event happening is the ratio of the number of favorable cases to the whole number of cases both favorable and unfavorable. * Thus, if a bag contain 5 white and 12 black balls, and one of them be drawn at random from the bag, the chance of this being a white ball is ^, and of its being a black ball IS TTT« 482 PROBABILITY (CHANCE) — DEFINITION. If p is the chance of the happening of an event, the chance of its not happening is 1 — p. Note. — When the chance of the happening of an event is to the chance of its failing as a to b, it is sometimes expressed in popular language thus: the odds are a to b in favor of the event, or b to a against the event. If the chances for and against an event are even, the chance of the event is %. 1. From a bag containing 3 white and black balls a man draws one at random : what is the chance that this will be a white ball? There are here 3 ways of drawing a white ball and 12 ways of drawing either a white or a black ball. Hence the required chance = T % = J. 2. From a bag containing 4 white and 5 black balls a man draws 3 at random : what is the chance of these being all black? The total number of ways in which 3 balls can lie drawn from 9 balls is 9 <7 8 (Art. 181), and the number of ways of drawing 3 black balls from 5 is 5 C 8 ; therefore the chance of drawing 3 black balls = & = t±il 1 A. 9 C 3 9.8-7 42 3. Find the chance of throwing at least one ace in a single throw with two dice. As a common die has 6 symmetric faces, it may fall in any one of 6 ways ; and each of these 6 ways may unite with each of the six ways of the second die, giving 36 different cases in all. An ace on one die may lie associated with any of the G numbers on the other die, and the remaining 5 numbers on the first die may each be associated with the ace on the second die ; thus the number of favorable eases is 11. Hence the required chance = ££. SIMPLE EVENTS. 483 Otherwise thus : There are 5 ways in which each die can be thrown so as not to give an ace ; hence 25 throws of the two dice will exclude aces, so that the chance of not throwing one or more aces is §§. Hence the chance of throwing at least one ace = 1 — |-| 4. Find the chance of throwing more than 15 in one throw with 3 dice. A throw amounting to 18 must be made up of Go, 6, 6, and this can occur in 1 way ; 17 can be made up of G, G, 5, and this can occur in 3 ways ; 16 can be made up of G, 6, 4, and G, 5, 5, and these can each occur in 3 waj's. There- fore the number of favorable cases is 1 + 3 + 3 + 3, or 10. And the whole number of cases is 6 3 , or 21G. Hence the required chance = ^ = T § ¥ . 228. Simple Events. — Suppose there to be a number of events, A, B, C, , of which one must, and only one can, occur ; and suppose a, b, c, to be the numbers of ways respectively in which these events can occur, and that each of these ways is equally likely to happen. Then, since A can occur in a ways, out of a _|_ _|- c + ways, therefore by Art. 227, the chance that A will happen is . a + b + c . . . Similarly, the chance that B will happen is — ■ ; J a + 6 + c . . . and so on. 1. From a bag containing 3 white, 4 black, and 5 red balls a man draws 2 at random : what are the probabilities of the different cases ? The number of pairs that can be formed out of 12 things 484 SIMPLE EVENTS. is V2 C 2 (Art. 181), or 66. The number of pairs that can be formed out of the three white balls is 8 C 2 or 3 ; hence the chance of drawing two white balls = -fa. Similarly, the chance of drawing two black balls == & ; and the chance of drawing two red balls = £§. Also, as each of the 3 white balls might be associated with each of the 4 black balls, the number of pairs consisting of one white and one black ball is 3 X 4, or 12 ; hence the chance of drawing a white and a black ball = £f . Similarly, the chance of drawing a black and a red ball = §#; and the chance of drawing a red and a white ball = $f . The sum of these six chances is unity, as it should be (Art. 227). 2. A has 3 shares in a lottery in which there are 3 prizes and 6 blanks ; B has 1 share in a lottery in which there are 1 prize and 2 blanks : show that A's chance of success is to B's as 16 to 7. A may draw three prizes in one way ; he may draw 2 prizes and 1 blank in 3 C 2 X G, or 18 ways ; he may draw 1 prize and 2 blanks in 3 x 6 C 2 , or 45 ways : the sum of these numbers is 64, which is the number of ways in which A can draw a prize. Also he can draw 3 tickets in 9 6 3 , or 84 ways ; .-. A's chance of drawing a prize = §£ = -|f ; and clearly B's chance of drawing a prize = £ ; therefore A's chance : B's chance : : 16 : 7. Or ive might have reasoned thus : A may draw 3 blanks in 6 C 8 , or 20 ways. .-. A's chance of drawing a blank = §£ = ^ ; .-. A's chance of drawing a prize = 1 — 7j 5 r = £-$- (Art. 227) . 3. In a single throw with two dice find the chance (1) of throwing live, and (2) of throwing six. Ans. (1) J ; (2) /' . COMPOUND EVENTS. 485 4. From a pack of 52 cards two are drawn at random : find the chance that one is a knave and the other a queen. Ans. ^. 5. Find the chance of drawing two black halls and one red hall from a hag containing 5 black, 3 red, and 2 white halls. Ans. J. 229. Compound Events. — Hitherto we have consid- ered only those oecurrences which in the language of Proba- bility are called Simple events. When two or more of these simple events occur in connection with each other, the joint occurrence is called a Compound event. Events are said to be dependent or independent according as the occurrence of one does or does not affect the occurrence of the other. Dependent events are sometimes said to be contingent. (1) If there are two independent events of which the re- spective probabilities are Jcnoivn, to find the probability that both will happen. Suppose that the first event may happen in a ways and fail in b ways, all these ways being equally likely to occur; and suppose that the second event may happen in a' ways and fail in V ways, all these ways being equally likely to occur. Each of the a + b cases may be associated with each of the a' + b' cases, thus forming (« + b) (a' + b') compound cases all equally likely to occur. In aa' of these compound cases both events happen, in bb' of them both events fail, in ab' of them the first event hap- pens and the second fails, and in alb of them the first event fails and the second happens. Thus, . — is the chance that both events happen ; (a + &) («' + U) is the chance that both events fail ; ( a + 6) ( a ' + i') is th» chance that the first happens and (a + 6) (a' + &') the second fails. 486 COMPOUND EVENTS. — is the chance that the first fails and the (a + b) (a' + b') second happens. Hence, If tioo events be independent of each other, the probability that they will both happen is equal to the product of their separate probabilities. Thus, if p and p' are the respective chances of two in- dependent events, pp' is the chance that both events will happen. Similar reasoning will apply in the case of any number of independent events. For example, suppose there are three independent events, and that p, p\ p" are the re- spective chances of their happening separately. From the above rule, the chance that the first two will happen is pp' ; then in the same way the chance that the first three will happen is pp' x p", or pp'p" ; and so on for any number. Similarly, the chance that all the events will fail is (1 — p)(l — j/)(l — p")- The chance that the first two events will happen and the other fail is pp'{l — p") ; and so on. Hence, If there be any number of independent events, the probability that they will all happen is equal to the product of their separate probabilities. If p is the chance that an event will happen in one trial, the chance that it will happen in any assigned succession of r trials is p r ; this follows at once by supposing that p, p' , p" ', are each equal top. 1 . Find the chance of throwing an ace in the first only of two successive throws with a single die. Here we require the ace to appear at the first throw, and at the second throw it is not to appear. The chance of throwing the ace at the first trial = I ; and the chance of not throwing the ace at the second trial = |. .-. the chance of the compound event = $ K f = - ; [; ; . 2. Suppose 5 white and 8 black balls to be placed in a bag; find the chance that in two successive drawings of 3 COMPOUND EVENTS. 487 balls each, the first will give 3 white and the second 3 black balls, the balls first drawn, being replaced before the second trial. The number of ways in which 3 balls may be drawn is The number of ways in which 3 white may be drawn is 6 C 8 ; The number of ways in which 3 black may be drawn is 8 C 3 . .-. the chance of 3 white at the first trial = 5 C 3 -^ 13 C 3 = T *£^ ; and the chance of 3 black at the second trial = 8 C 3 -r- 13 C 3 = ^fe. .-. the chance of the compound event = T f 3 X T 2 j3 = ?hi%9- Note. — To find the chance that some one at least of the events will happen we proceed thus: the chance that all the events fail is (1 —p)(l — 2/)(l — P") , and except in this case some one of the events must happen; hence the required chance is 1_(1 - p)(l - p')U - P") 3. Suppose that it is 9 to 7 against a person A who is now 35 years of age living till he is 65, and 3 to 2 against a person B now 45 living till he is 75 : find the chance that one at least of these persons will be alive 30 years hence. The chance that A will die within 30 years = T % ; and the chance that B will die within 30 years = f ; . •. the chance that both will die within 30 years = | X fo = §$■ Therefore the chance that both will not be dead, that is, that one at least will be alive, is 1 — f£ = §§. (2) The probability of the concurrence of two dependent events is the product of the probability of the first into the probability that when that has happened the second will follow. This is only a slight modification of the principle estab- lished in (1), and is proved in the same way. Thus, suppose that when the first event has happened, d denotes the number of ways in which the second event will follow, and b' the number of ways in which it will not follow ; then the number 488 MUTUALLY EXCLUSIVE EVENTS. of ways in which the two events can happen together is aa', and the probability of their concurrence is (a + b) (a' + b') Thus, if p is the probability of the first event, and p' the contingent probability that the second will follow, the prob- ability of the concurrence of the two events is pp f . 4. In a hand at whist find the chance that a specified player holds both the king and queen of trumps. Denote the player by A ; then the chance that A has the king is clearly 13 -e- 52 ; for this particular card can be dealt in 52 different ways, 13 of which fall to A. The chance that, when A has the king, he can also hold the queen is then 12 -r- 51 ; for the queen can be dealt in 51 ways, 12 of which fall to A. Therefore the chance required = if X Jf = T V 5. Suppose 5 white and 8 black balls to be placed in a bag: find the chance that in two successive drawings <>f :) balls each, the first will give 3 white and the second 3 black balls, the balls first drawn not being replaced before the seconk trial. At the first trial, 3 balls may be drawn in 13 C 3 ways ; and 3 white balls may be drawn in 5 C 3 ways. At the second trial, 3 balls may be drawn in l0 C 3 ways ; and 3 black balls may lie drawn in 8 C 3 ways. .-. the chance of 3 white at first trial = 5 C 3 -=- i 3 C 3 = T |^ ; and the chance of 3 black at second trial = 8 C 3 -5- 13 C 3 = ^-. the chance of the compound event = T iu X it = ila- The student should compare this solution with that in Ex. 2. 230. Mutually Exclusive Events. — If an event can happen in two or more different ways which are mutually exclusive, tin' chance that it will happen is the sum of the chances of its happening in these different ivays. MUTUALLY EXCLUSIVE EVENTS. 489 If the different ways of happening are all equally probable, this proposition is merely a repetition of the definition of probability (Art. 227) ; and if they are not equally probable, the proposition seems to follow so naturally from that defi- nition that it is often assumed as self-evident. It may, however, be proved as follows : Suppose the event can happen in two ways which cannot concur; and let -, — be the chances of the happening of b b the event in these two ways, respectively. Then out of bb' cases there are ab' in which the event may happen in the first way, and a'b in which the event may happen in the second way ; and these ways cannot concur. Therefore, out of the whole bb' cases there are ab' -f- a'b cases favorable to the event ; hence the chance that the event will happen in one or the other of the two ways is ab' + a'b _ a , a[ bb' ~ b b'' Similarly for any number of exclusive ways in which the event can happen. Hence if an event can happen in n ways which are mutu- ally exclusive, and if p v p 2 , p 8 , p n are the proba- bilities that the event will happen in these different ways, respectively, the probability that it will happen in some one of these ways is Pi + Pi + P 3 + + P- 1. Suppose two urns, A and B; let A contain 2 white and 3 black balls, and let B contain 3 white and 4 black balls ; find the chance of obtaining a white ball by a single drawing from one of the urns taken at random. Since each urn is equally likely to be taken, the chance of selecting the urn A is ^ ; and the chance then of drawing a white ball from it is § ; hence the chance of obtaining a white ball so far as it depends on A is \ X f , or \. Simi- 490 MUTUALLY EXCLUSIVE EVENTS. larly, the chance of obtaining a white ball so far as it depends on B is | x -2, or T 3 T . .-. the required chance = \ + T \ = f$. 2. Find the chance of throwing 9 at least in a single throw with two dice. There are 36 cases in all ; 9 can be made up in 4 ways, 10 in 3 ways, 11 in 2 ways, and 12 in 1 way. .-. the chance of throwing 9 = 4 -5- 36 ; the chance of throwing 10 = 3 -f- 36 ; the chance of throwing 11 = 2 -=- 36 ; and the chance of throwing 12 = 1 -f- 36. Now the chance of throwing a number not less than 9 is the sum of these separate chances. .. . , , 4 + 3 + 2 + 1 5 .-. the required chance = — ! ! — ■ = — . 36 18 Note 1. — In each of these two examples the probability of the event is the sum of the probabilities of the separate events. It is important, however, to notice that the probability of one or other of a series of events is the sum of the probabilities of the separate events, only when the events are mutually exclusive, i.e., when the occurrence of ar.y one is incompatible with the occurrence of any of the others. 3. From 20 tickets marked with the first 20 numerals, one is drawn at random : find the chance that it is a multiple of 3 or of 7. The chance that the number is a multiple of 3 is ._,';,. and the chance that it is a multiple of 7 is .f v ; and these events are mutually exclusive. .-. the required chance = .{'„ + L ,- = -?.. But if the question had been : find the chance that the number is a multiple of 3 or of 7. it would have been incorrect to reason as follows : Because the chance that the number is a multiple of 8 is ,,';,. and the chance thai the number is a multiple of 1 is .,";,. therefore the chance that it is a multiple of <"> <>r 1 is ._,';, + ./;,, MUTUALLY EXCLUSIVE EVENTS. 191 or £J. For the number on the ticket might be a multiple both of 3 and of 4, so that the two events considered arc not mutually exclusive Note 2. — The distinction between simple and compound events is, in many cases, purely an artificial one; it often amounts to nothing more than a distinction between two different ways of viewing the same occurrence. 4. A bag contains 3 white and 4 black balls; if 2 balls are drawn what is the chance that one is white and the other black? (1) Regarding the occurrence as a simple event, the chance = 3 x 4 -=- 7 C 2 = f. (2) Regarding the occurrence as the happening of one or other of the two following compound events : (a) drawing a white and then a black ball, the chance of which is 8 y 4- 4. 7 A "6 — 14 ' (b) drawing a black and then a white ball, the chance of which is 4 3 _ 4 7 x T; — T4- And since these events are mutually exclusive, the required chance = T 4 X + tj = 7- 5. What is the chance of throwing an ace in the first only of two successive throws with an ordinary die ? Ans. ■£$. 6. Three cards are drawn at random from an ordinary pack : find the chance that they will consist of a knave, a queen, and a king. Ans. g I !;,-,. 7. Find the chance of throwing an ace and only one in two successive trials with one die. Ans. £#. 8. In one throw with a pair of dice find the chance that there is neither an ace nor doublets. Ans. ,;. 9. When 6 coins are tossed what is the chance that one, and one only, will fall with the head up ? Ans. ^. 10. The odds against a certain event are 5 to 2, and the odds in favor of another event independent of the former are 6 to 5 : find the chance that one at least of the events will happen. • Ans. - t i. 192 EXPECTATION — GENERAL PROBLEM. 231. Expectation. — The value of a given chance of obtaining a given .sum of money is called the expectation. Suppose that there are a + b tickets in a lottery for a prize .1/; then since each ticket is equally likely to win the prize, and a person who possesses all the tickets must win. the money value of each ticket is M -=- (a -f 1>) ; in other words this would be a fair sum to pay for each ticket; hence a person who held a tickets might reasonably expect aM -r- (a + b) as the price to be paid for his tickets ; that is, he would estimate this amount as the value of his chance, or expectation. Therefore, if E be the expectation, we have £ _ _ a a Thus, the expectation is the sum which may be toon multiplied by the chance of winning it. 1. A bag contains 3 white and 7 black balls. Find the expectation of a man who is allowed to draw a ball from the bag, and who is to receive $1 if he draws a black ball, and $5 if he draws a white one. The chance of drawing a black ball is T 7 ^ ; and therefore the expectation from drawing a black ball is $0.70. The chance of drawing a white ball is ^ V) ; and therefore the expectation from drawing a white ball is $1.00. Hence, as these events are exclusive, the whole expectation is $2.20. 2. A purse contains 2 sovereigns, 3 half-crowns, and 7 shillings. What should be paid for permission to draw (1) one coin, and (2) two coins? Ans. (1) As. 6]<1. ; (2) 9s. Id. 3. Two persons toss a dollar alternately on condition that the first who gets "heads" win the dollar: find their ex- pectations. Ans. 0.66| ; 0.3af 232. General Problem. — The 2^'obability of the hap- pening of an event in one trial being known, required the probability of Us happening once twice, three times, etc., exactly in n trials. Letjj be tin' probability <>!' the happening <>t' the event in one trial, and q the probability of its failing, so that q=l—j>. GENERAL PROBLEM. 493 Then the probability that in n trials the event will happen in one particular tried, and fail in the other n — 1 trials is pq"~ l (Art. 229) ; and since there are n trials, the proba- bility of its happening in some one of these and failing in the rest is npq n ~ 1 . The probability that in n trials the event will happen in two particular trials, and fail in the other n — 2 trials, is p' 2 q"~- ; and there are „C 2 ways in which the event may happen twice and fail n — 2 times in n trials ; therefore the probability that it will happen exactly twice in n trials is „0 2 p 2 g" -2 . In general, the probability that in n trials the event will happen in any particular set of r trials and fail in the other n — r trials is p r q n ~ r ; and as a set of r trials can be selected in n C r ways, all of which are equally probable and are mutually exclusive, the required probability of the event happening exactly r times in n trials is n C r p'q n ~ r . Thus, if (p + q) n be expanded by the binomial theorem, the successive terms will be the probability of the happening of the event exactly n times, n — 1 times, n — 2 times, etc., in n trials. If the event happens n times, or fails only once, twice, . . . . (n — r) times, it happens r times or more ; therefore the chance that it happens at least r times in n trials is V" + nC^-'q + n C 2 p n -V + • • • + n C n -rP r q n - r , (1) or the sum of the first n — r + 1 terms of the expansion of (2> + q) n - 1. In four throws with a pair of dice, what is the chance of throwing doublets twice at least? In a single throw the chance of doublets is ^, or f. ; and the chance of failing to throw doublets is f ; therefore P = h ? = h n = 4 > r = 2 - The required event follows if doublets are thrown four times, three times, or twice ; hence the required chance is the sum of the first three terms of (1). Thus the chance = L(l + 4 • 5 + G • 5 2 ) = f Vx- 6 4 494 CI SI R l L PROBLEM — EXAMPLES. 2. In five throws with .1 single die, what is the chance (1) of throwing exactly throe aces? and (2) of throwing at least three aces? Herep = £, q = f, n = 5, r = 3. Hence (1) the chance = .C^tf-fl)" = f^|(*) 8 (f) 3 = *¥&■ (2) the chance = (i) 5 + 6 (*)*(§) + ~(£) 3 (t) 2 = >>¥&■ 3. Find the chance that a person with two dice will throw aces at least four times in six trials. Ans. — '*' -'. (3G) 6 4. Find the chance of throwing an ace with a single die 4.1 4. ■ • 4. • 1 a 31031 once at least in six trials. Ans. — - — . 6 6 Notp:. — The subject of Probability is so extensive that, in an elementary work on Algebra, it is impossible to give more than a very brief sketch. The student who wishes to investigate this subject further, is referred to Todhunter's Algebra, and Hall and Knight's Higher Algebra; also to Whitworth's Choice and Chance, an admira- ble English work; and when he becomes acquainted with the Integral Calculus he may consult Professor Crofton's article Probability, in the Encyclopaedia Britannica. For a complete account of the origin and development of the subject, see Todhunter's History of the Theory of Probability from the time of Pascal to that of Laplace; in this work the reader is introduced to almost every process and every species of problem which the literature of the subject can furnish. For the practical applications of the theory of Probability to commercial transactions, we may refer to the articles, Annuities and Insurance in the Encyclopaedia Britannica. EXAMPLES. 1. Find the chance of drawing a red ball from a bag which contains 5 white and 7 red balls. Ans. , 7 ,. 2. Two balls are to be drawn from a bag containing ."» red and 7 white balls: find the chance that they will both be white. Ans. ./.,. 3. Show that the odds are 7 to 8 against drawing 2 red balls from a bag containing '■) red and 2 white balls. EXAMPLES. 495 4. A bag contains 5 white, 7 black, and 4 red balls : find the chance that 3 balls drawn at random are all white. Ans. -5V 5. If 4 coins are tossed, find the chance that there will be 2 heads and 2 tails. Ans. |. 6. If from a pack 4 cards are drawn, find the chance that they will be the 4 honors of the same suit. Ans. ^^ 7T5 . 7. In shuffling a pack of cards, 4 are accidentally dropped : find the chance that the missing cards should be one from each suit. Ans. ^VsVo- 8. A has 3 shares in a lottery containing 3 prizes and 9 blanks; B has 2 shares in a lottery containing 2 prizes and 6 blanks : compare their chances of success. Ans. 952 to 715. 9. Find the odds against throwing one of the two numbers 7 or 11 in a single throw with two dice. Ans. 7 to 2. 10. If from a lottery of 30 tickets marked 1, 2, 3, ... . four tickets be drawn, find the chance that 1 and 2 will be among them. Ans. T f-^. 11. Supposing that it is 8 to 7 against a person who is now 30 years of age living till he is 60, and 2 to 1 against a person who is now 40 living till he is 70 : find the chance that one at least of these persons will be alive 30 years hence. Ans. ff. 12. A party of 13 persons take their seats at a round table ; show that it is 5 to 1 against two particular persons sitting together. 13. The chance that A can solve a certain problem is one- fourth ; and the chance that B can solve it is two-thirds : find the chance that the problem will be solved if they both try. Ans. f . 14. The odds against A solving a certain problem are 4 to 3, and the odds in favor of B solving the same problem are 7 to 5 : what is the chance that the problem will be solved if they both try? Ans. Jf. 15. A lottery has 1 prize of $50, 2 prizes of $5 each, 400 EXAMPLES. 4 prizes of $1 each, and 13 blanks: find the expectation of the holder of 1 ticket. Ans. $3.20. 16. In three throws with a pair of dice, find the chance of having doublets one or more times. Ans. 1 — (|) 3 . 17. Find the chance of throwing double sixes once or oftener in 3 throws with a pair of dice. Ans. 1 — (ff) 3 . 18. If on an average 9 ships out of 10 return safe to port, find the chance that out of 5 ships expected at least 3 will arrive. Ans. $$£§§• 19. If four cards be drawn from a pack, find the chance that they will be marked one, two, three, four, of the same 4- 14 suit. Ans. . 52 -51- 50 -49 20. Two persons, A and B, engage in a game in which A's skill is to B's as 2 to 3. Find the chance of A's win- ning at least 2 games out of 5. Ans. . 21. As chance of winning a single game against B is § : find the chance of his winning at least 2 games out of 3. Ans. -j%V 22. What is the chance of throwing at least 2 sixes in 6 throws with a die? Ans. Hfti- 23. A person goes on throwing a single die until it turns up ace. What is the chance (1) that he will have to make at least ten throws ; (2) that he will have to make exactly ten throws? Ans. (1) (f) 9 ; (2) i 24. A die is to be thrown once by each of four persons, A, B, C, D, in order, and the first of them who throws an ace is to receive a prize. Find their respective chances, and (5) the chance that the prize will not be won at all. Ans. A's chance £, B's -fa, C's ■££$, D's ^sftfr, (5) T 6 ^- 25. What is the chance that a person with two dice will throw aces exactly four times in six trials? Ans. 7 ., - , : , , r l 'i u- GENERAL EQUATION OF THE n"' DEGREE. 497 CHAPTER XXIV. THEORY OF EQUATIONS. 233. General Equation of the n th Degree. — The general form of an equation of the n ih degree with one unknown quantity is x n + p.x"- 1 + p 2 x n - 2 + + Pn-jc + p H = 0, (1) where n is a positive integer, and the coefficients p v Pv p„ are either positive or negative, integral or fractional. Tlie coefficient of a? may always be made unity, as above, by dividing the equation by the coefficient of x n . The term p n may be considered as the coefficient of x°, and is called the absolute term. This equation is called the General Equation of the n"' Degree, because it is the form to which every Algebraic equation can be reduced. Unless the contrary is specified, the n coefficients p v jD 2 , p n will always be supposed rational. When one quantity is so connected with another that no ehange can take place in the latter without producing a corresponding change in the former, the former is called a function of the latter. Any Algebraic expression which contains x is an Algebraic function of x, and may be denoted by f(x).* Thus, the first member of (1) is a function of x, and may be denoted by f(x) ; then we may represent the general equation of the n ,h degree by f(x) = 0. Any value of x which, being substituted for a*, satisfies the equation, i.e., makes f(x) equal to 0, is called a root of the equation (Art. 53). We shall assume that every equation of the form f(x) = has a root, real or imaginary. The proof of this theorem is * This is read " a function of x," or " the / function of x," or more briefly, " th« / of x." 498 DIVISIBILITY OF EQUATIONS. given in special treatises on the Theory of Equations; but it is beyond the range of the present work. An equation of the third degree with one unknown quan- tity is called a cubic equation. An equation of the fourth degree with one unknown quantity is called a biquadratic equation. 234. Divisibility of Equations. — (1) If a is a root of the equation f(x) = 0, then f(x) is exactly divisible by x — a. For, divide f(x) by x — a until a remainder is obtained which does not contain x ; let Q be the quotient, and R the remainder if there be one. Then, since the dividend is equal to the product of the divisor and the quotient, added to the remainder, (x - a) Q + R = 0. But x = a; therefore (x — a)Q = 0, and .*. R = 0. That is, x — a divides f(x) without a remainder. (2) Conversely, if f(x) is exactly divisible by x — a, then a is a root of the equation f(x) = 0. For, let Q be the quotient when f{x) is divided by x — a ; then (x - a)Q = 0. This equation is satisfied when x = a, and therefore a is a root of the equation f(x) = 0. 1. Prove by this method that 2, 3, and 1 are roots of the equation x 3 - 9x 2 + 26a; - 24 = 0. 2. Prove that 4 is a root of the equation x s + a,a _ Ha - 24 = 0. 3. Prove that —1 is a root of the equation x i _ 38a 8 + 210a; 2 + 538a) + 289 = 0. (8) To find the remainder when any rational function, oj x is divided by x — a, where a is any constant. NUMBER OF ROOTS. 499 Let f{x) denote any rational function of x ; divide f(x) by x — a until the remainder is independent of x ; let Q denote the quotient, and R the remainder ; then f{x) = (x- a)Q + R. Since R does not involve x it will remain unaltered whatever value we give to x ; put x = a, then /(«) * = x Q + R ; now Q is finite for finite values of x ; hence R = /(«) . 235. Number of Roots. — Every equation of the n"' degree has n roots, and no more. We have seen (Art. 140) that a quadratic equation has two roots. In the same way, a cubic equation has three roots, and so on. Denote the given equation by f(x) = 0, where (Art. 233) f(x) = x n + pjof- 1 + p.^"" 2 + • • • + Pn-i® + P« = 0- ( l ) The equation f{x) = has a root, real or imaginary ; let this be denoted by a; then, by (1) of Art. 234, f(x) is divisible by x — a, so that (1) becomes fix) = im-a)f x ix) = .... (2) where f(x) is a rational function of (//- — 1) dimensions. Now (2) may be satisfied by making either factor equal to 0; hence f x (x) = 0; and the equation fix) = must have a root, real or imaginary ; let this be denoted by b ; then fix) is divisible by x — 6, so that (2) becomes fix) = ix- a)ix - b)f,ix) = 0, . . . (3) where fix) is a rational function of (n — 2) dimensions. Continuing in this way, the equation fix) — will ulti- mately be resolved into n binomial factors of the form x — a, x — 6, x — c, etc. Hence, calling the last root I, (1) may be written in the form fix) = ix - a)ix - b)ix -c) ix - I) = . (4) * /(«) is the expression obtained by substituting a for x in /(x). 500 RELATIONS BETWEEN COEFFICIENTS AND ROOTS. Therefore the equation /(a;) = has n roots, since /(a;) is equal to when x has any of the values a, b, c, I. And the equation cannot have more than n roots, for if we give to x a value different from any of the n values a, b, c, I, all the factors of (4) are different from zero, and therefore f(x) cannot be zero. In the above investigation some of the quantities, a, b, c, .... may be equal ; in this case, however, we shall suppose that the equation still has n roots, although two or more of them are equal. Thus, factoring the equation x* - x 2 - 8x + 12 = 0, it becomes (x - 2) (re - 2) (a: + 3) = 0, and hence the three roots are 2, 2, — 3 ; and the equation has two equal roots. It follows from this theorem that if we know one root of an equation, we may by division reduce it to another equation of the next lower degree, which contains the remaining roots. Hence, when all the roots of an equation but two are known, it may be reduced to a quadratic by division, and the remain- ing roots found by the rules for quadratics. 1. One root of the equation x 3 - 9a;' 2 + 2Gx — 24 = is 3 : find the other roots. Ans. 2, 4. 2 . Two roots of the equ ation x 4 - 1 2a- 3 + 48a; 9 - 68a; +15 = are 3 and 5 : find the other roots. Ans. 2 ± y3. 3. Three roots of a; 5 + 6a; 4 - 10a; 3 - 112a; 2 - 207a; -110 = are —1, —2, —5 : find the other roots. Ans. 1 ± ^12. 236. Relations between Coefficients and Roots. — To investigate the relations betiveen the coefficients and the roots of any given equation. If we form an equation with the two roots, a and b, we have (Art. 235) (x - a) {x - b) = ; and performing the multiplication indicated, this becomes a? — (a + b)x + ab = 0. RELATIONS BETWEEN COEFFICIENTS AND ROOTS. 501 Similarly, if we form an equation with the three roots, a, b, and c, we have x 3 — (a + b + c).r + (afi + ac + 6c) a; — a&c = ; and so on. Comparing this with the general form (see (1) of Art. 233), we see that Pj = — (a + b + c) ; p. 2 = (ab + ac + 5c) ; p 3 = —abc. If we add another root, d, we find the result to be p x = — (o + 6 + c + d) ; p a = (aft + ac + ad -f 6c + bd + cd) ; p 3 = — (abc + a&d + acd + &cd) ; p i = a&cd". Hence, generally, we have the following Rule. 2%e coefficient of the second term of the general equation is the sum of all the roots with their signs changed. The coefficient of the third term is the sum of the products >f all the roots, taken tioo and two. The coefficient of the fourth term is the sum of the products of all the roots, taken three and three, with their signs changed, etc. The last term is the product of all the roots ivith their signs changed. Note. — It will be seen that these relations include those of quadratics given in Art. 140. If the second term of an equation is wanting, the sum of the roots will be 0. The absolute term of an equation is divisible by every rational root. If the absolute term is wanting, one root at least will be 0. Form the equation whose roots are 1. 3, 4, -5. Ans. x z - 2a; 2 - 23x- + GO = 0. 2. 3, -2, 7. x 3 - 8x 2 + x + 42 = 0. 3. 2, 3, 5, -G. aj* - 4x 3 - 29a 8 + 156a; - 180 = 0. 502 FRACTIONAL ROOTS — IMAGINARY ROOTS. 237. Fractional Roots. — An equation having unity for the coefficient of the first term, and all the other coefficients integers, cannot have a root which is a rational fraction. If possible, let -, a rational fraction in its lowest terms, b be a root of x" + p.af- 1 + p«x»-- + + p n _ lX + p n = 0, (1) where all the coefficients are integers. Substituting - for x in (1), multiplying by b"' 1 , and trans- posing, we have |- = ~(p 1 a n - i + p. 2 a"~ 2 b + p s a n ~ s lr + + pj*" 1 ). That is, we have a fraction in its lowest terms (Art. 225) equal to the sum of a number of integers, which is impossi- ble. Therefore (1) cannot have a # root which is a rational fraction. Note. — This proposition only proves that the rational roots of the equation must be integers. The equation may, however, have irrOr tional fractions as roots. Thus, in the equation ,r 3 + a; 2 — 11a; + 10 = 0, one of the roots is 2, and the other two are i^_. Such roots as 2 these, whose values cannot be exactly expressed, are called incom- mensurable. 238. Imaginary Roots. — In an equation with real coefficients imaginary roots occur in pairs. Denote one of the imaginary roots of such an equation by a + hi (Art. 219) ; then if a + hi be a root of the equation, a — hi will also be a root. For, if a + hi be substituted for x in the equation, the result will consist, (l)of real terms which involve the odd and even powers of a and the even powers of bi, and (2) of imaginary terms which involve the odd powers of bi. DESCARTES' RULE OF SFGNS. 503 Representing the sum of all the real terms by A, and the sum of all the imaginary terms by Bi, we have A + Bi = 0. But in order that this equation may be true, we must have A = and B = (Art. 220). If we now substitute a — bi for x in the given equation, the result will differ from the preceding only in the signs of the imaginary terms. Hence we obtain A — Bi. But since -4 = and B = 0, we have A - Bi = 0. Thus, a — bi satisfies the equation, and therefore it is also a root. In like manner it may be shown that in an equation with rational coefficients, surd roots enter in pairs ; i.e., if a +\b is a root, then a — ^b is also a root. Hence, every equation of an odd degree with real coefficients has at least one real root. Since the product of a pair of conjugate imaginary factors is a real quadratic factor (Art. 220) , it follows that, in an equation of an even degree, all the roots may be imaginary. 1 . One root of the equation x 3 — 15a: + 4 = is 2 + \3 : find the other roots. Ans. 2 — V ^, — 4. 2. One root of x i - 3a; 2 - 42»-40 = is -^3-^-31) : find the other roots. Ans. -£(3 + V— 31), 4,-1. 3. One root of 6a.- 4 - 13a; 3 - 35a; 2 - x + 3 = is 2 - V^ : find the other roots. Ans. 2 + \^3, — $, — }. 239. Descartes' Rule of Signs. — An equation f(x) =0 cannot have more positive roots than there are variations of sign in f(x), or more negative roots than there are variations of sign in f(-x). A variation of sigu is said to occur when the signs of two successive terms of a series are unlike; and a permanence, when the signs are alike. 504 DESCARTES' RULE OF SIGNS. (1) Suppose that the signs of the terms in any equation are -f- H 1 1 1 — j i' 1 which there are four per- manences and seven variations. We shall show that if we introduce a new positive root a into this equation, which we do by multiplying it by x — a (Art. 235) , whose signs are -\ — , there will be at least one more variation of sign in the product than in the original equation. Writing down only the signs of the terms in the multiplication, we have + + -- + + - + - + - + + -- + + - + + ± - T + - T T + - + - + A double sign is placed where the sign of any term in the product is ambiguous. It will be seen by inspection (1) that each permanence in the multiplicand is replaced by an ambi- guity in the product, (2) that the signs before and after an ambiguity or a set of ambiguities are unlike, and (3) that a variation is introduced at the end. Hence, taking the ambiguous sign either way, the intro- duction of the positive root + a has not increased the num- ber of permanences, while it has added at least one variation.* Therefore as each factor, corresponding to a positive root, must introduce at least one new variation, the whole number of positive roots cannot exceed the number of variations of sign of f(x) . (2) To prove the second part of the rule, put — x for x ; then the variations in f(x) become •permanences in /(—a;), and the negative roots of the equation f(x) = are the positive roots of the equation /( —a?) = (see Art. 242) . But from (1) the number of positive roots of /(— ») = cannot * If the original polynomial ends with a permanence, an ambiguity will result in the final polynomial. Thin ambiguity furnishes a variation either with the preceding or with the final sign just added. So that, taking the most unfavorable case, there is one variation added. DERIVED FUNCTIONS. 505 exceed the number of variations in f(—x). Therefore the whole number of negative roots of the given equation f(x) =0 cannot exceed the number of variations in f(—x) = 0. The following results follow from Descartes' Rule of Signs : (I) Since in a complete equation the number of variations and permanences is equal to the degree of the equation, therefore, when the roots are all real, the number of positive roots is equal to the number of variations, and the number of negative roots is equal to the number of permanences. (II) If the coefficients of a complete equation are all posi- tive, the equation has no positive root ; thus, x 3 + Gx~ + H# + 6 = cannot have a positive root. (III) If the coefficients of a complete equation are alter- nately positive and negative, the equation has no negative root : thus the equation x*- — 5a; 3 + 9a; 2 — 7x + 2 = cannot have a negative root. (IV) In an incomplete equation, imaginary roots may sometimes be discovered. Thus in the equation 3a; 4 + 12a; 2 + 5a; - 4 = 0. there is one variation, and therefore there is but one positive root. Again f{—x) = 3x* + 12a; 2 — 5a; — 4 = ; here there is but one variation, and therefore but one negative root. Therefore the other two roots must be imaginary. 1. Show that the equation 2a; 7 — a; 4 + 4a; 3 — 5 = has at least four imaginary roots. 2. Show that the equation x 9 -f- 5a; 8 — a; 3 + 7x + 2 = has at least four imaginary roots. 240. Derived Functions. — To find the value off(x+h), when f(x) is a rationed integral function of x. Let f(x) represent the general equation of the ?t th degree (Art. 233), so that f{x) = af + p x x n ~ l + iW- 2 4- + p„_,a; + p n (1) 506 DERIVED FUNCTIONS. If we substitute x + h for a (1) becomes f(x+h) = (x+h) n +p 1 (x+h)*~ l + . . +p n - l (xj-h)+p, (2) Expanding each term of the second member by the bino- mial theorem and arranging the result in ascending powers of h, we have f(x + It) = X n + pi??- 1 + ptf?~ 2 + ....+ p^jO! + P« + fe[%aJ»- 1 + (n - 1)^0;"-- -f- (to - 2)p.jX n ~ z + . . .] + £[»(« - 1)«"- 2 + (« - 1)(" - 2)/^"- 3 + . . .] LA (3) We see that the first line of this expansion is the original function f(x) , and if we denote the coefficient of /* in the second line by f'(x) ; the coefficient of — in the third line by f"(x), etc., we have f{x+h) =f(x) + hf(x) +£/*(») + £/"(*) + • • +--/"(*)• [2. [3_ [»L The functions /'(x), /'(»), fix) are called thej^AsZ, second, third, .... derive* dr functions * off(x) ; the given function /(.c) is sometimes called the primitive function. Examining the coefficients of /i, — ... in (3) , we see that each derived function is obtained from the preceding by the following rule : Multiply cadi term by the exponent of x in that term, and then diminish the exponent b;/ unity. Note. — When the student becomes acquainted with the elements of the Differentia] Calculus he will see that each derived function is the differential coefficient with respect to x of the immediately preced- ing derived function, and that the above development of J\.t + h) is only a particular case of Taylor's Theorem. * Called also derived polynomials, or simply derttattoot. EQUAL ROOTS. 507 Find the successive derived functions of the following : 1 . 2a 4 - x 3 - 2a 2 + 5x — 1. Ans. 1st 8a 3 - 3a 2 - 4a + 5, 2d 24a 2 - 6x - 4, 3d 48a - 6, 4th 48. 2. a 4 - 8a 3 + 14a 2 + 4a - 8. 1st 4a 3 - 24a 2 + 28a + 4, 2d 12a 2 - 48a -f 28, 3d 24a - 48, 4th 24. 241. Equal Roots. — If an equation has r roots equal to a, then the first derived function will have r — 1 roots equal to a. Let /(a) and /'(a) denote the equation and its derived function respectively, and let <£(a) be the quotient when /(a) is divided by (a — a) r ; then /(a) = (a — a) r lace it equal to zero, and solve the resulting equation. If f(x) be divided by the G. C. D. of f(x) and /'(as), the depressed equation will contain the remaining roots of f(x) = 0. Note. —Since /"(as) is the first derived function of f'(x), f'"(x) the first derived function of f"(x), and so on, we see that, if the equa- tion ./'(.r) = has r roots equal to a, the equation /'(as) = must have r — 1 roots = a, the equation /"(as) = must have r — 2 roots equal to a, and so on. By means of this principle we may sometimes dis- cover the equal roots of an equation with less trouble than by the ride above given. 1. Solve the equation a 4 — lias 8 + 44a; 2 — 76.? + 48 = 0, which has equal roots. Here /(as) = a; 4 - liar 5 + 44a; 2 - 76a; + 48, f(x) = Ax 3 - 33a; 2 + 88.r - 76 ; and by the rule (Art. 74) we find that the G. C. D. of /(.»•) and f'{x) is (as — 2) ; hence (as — 2) a is a factor of f(x) ; and .-. f(x) = (x-2y-(x*-7x+l2) = (x-2y(x-3)(x-i) ; thus the four roots are 2, 2, 3, 4. Solve the following equations which have equal roots: 2. X s - 2as 2 - 15as + 30 = 0. Ans. 3, 3, - I 3. x A - 9.r- + !./• +12 = 0. 2, 2, -1, -3. 4. as 4 - Gx 3 + 12a; 2 - 10a; + 3 = 0. 1,1, 1, 3. TRANSFORMATION OF EQUATIONS — DEFINITION. 509 TRANSFORMATION OF EQUATIONS. 242. Definition. — An equation is said to be transformed when it is changed into another equation whose roots bear some assigned relation to those of the given equation. Hem. — Various transformations of this kind can be effected with- out knowing the roots of the given equation; and examples occur in which a root of the transformed equation, and hence the correspond- ing root of the given equation is more easily found than by solving the given equation directly. Such transformations arc especially useful in the solution of cubic equations. Problem I. — To transform an equation into another ichose roots are those of the given equation with contrary signs. Let f(x) — be the given equation. Put — y for .r, so that when x has any particular value, y has numerically the same value but with the contrary sign ; then the required equation is f(—y) = 0. Thus, iff(x)=X n + 1 ) 1 X n - 1 + 2W n - 2 + ■ • +Pn-lV + Pn=0, then f(-y) = (-yf+ Pl (-yY-' t +p 2 (-y) n --+ . . -p n -i?J+P,-0 ; and since the even powers of —y are positive and the odd powers negative, the above equation becomes t ~ PlV n ~ l + Plf~- + ■ • • • ± Pn-\V T Pn- Hence the transformed equation may be obtained from the given equation by changing the sign of every alternate term beginning with the second. Note. — The above rule assumes that the given equation is com- plete (Art. 34) ; if any terms are wanting, the equation may be ren- dered complete by introducing the missing terms with zero for the coefficient of each of them. 1. Transform the equation 3a 4 — 4x s — Ax + 7 = into another whose roots are numerically the same with contrary signs. + 1'n = 0. 510 PROBLEM II. We may write the equation thus, 3^4 _ 4a j + 0-t ,2 _ 4a? + 7 = 0, then the transformed equation by the rule is 3ar + 4x 3 + Oar + Ax + 7 = 0, or 3a; 4 + 4a: 3 + 4a; +7=0. 243. Problem -II.— To transform an equation into an- other whose roots are equal to those of the proposed equation multiplied by a given factor. Let m denote the given factor ; put y = mx, so that when x has any particular value, the value of y is m times as large ; then x = y~. m Substituting this in the general equation (Art. 233) we have Multiplying by m", we have y n + 2hmy n ~ l + p{nvy n ~ 2 + . . + p n -i>n n -hj + p n m n = 0. Hence we have the following rule : Mxdtiply the second term by the given factor, the third term by this factor squared, and so on to the last term. The chief use of this transformation is to clear an equa- tion of fractional coefficients. 1. Remove fractional coefficients from the equation 2a; 3 - far -^4-^ = 0. Multiplying the second term by m, the third by //<--, etc., by the rule, we get 2a; 3 - |maj a — Jm 3 * + ^.m s = 0. By properly assuming m we may make the coefficients of the transformed equation all integers. Thus, if m = 4, all the terms become integral, and on dividing by 2 we obtain a» _ 3aJ - x + 6 = 0, in which the roots arc four limes as great as those of the given equation. PROBLEMS III. AND IV. 511 If the roots are to be divided by any number, we obtain the transformed equation by dividing the second term by the given number, the third term by its square, and so on. 2. Find the equation whose roots are three times as great as those of as 4 + 7a; 2 — 4x + 3 = 0. Ans. if + G3// 2 - 108// + 243 = 0. 3. Find the equation whose roots are one-half those of x- - 2x + 3 = 0. Ans. y* — y + f = 0. 244. Problem III. — To transform an equation into another whose roots are the reciprocals of the roots of the given equation. Let f(x) = be the given equation ; put y = -, so that when x has any particular value, the value of y is the recip- rocal of that value ; then x — -, and the required equation Thus, if /(») = ; x n +p l x n - 1 +p& n -' 2 + . . + p n ^ x x+ p n =0, men f( I \ = I + J>L. + J^ + . . + &=l + Pn = ; W y y y v that is, p„y n + p„_,//" -1 + P»-2?/ re_2 + • • + P$ + 1 = ° Hence, to effect the transformation, reverse the order of the coefficients. Transform the following equations into others whose roots are the reciprocals of the roots of the given equations. 1. x* + x 3 + 3x + 2 = 0. Ans. 2// 4 + 3// 3 + y + 1 = 0. 2. a: 3 - 6a; 2 + 11a? — 6 = 0. 6// 3 - ll// 2 + 6// -1 = 0. 3. x« - 5a,- 5 - 7a; 4 + Qx s + 13a; 2 - 9a; + 25 = 0. Ans. 25?/ 6 - 9// 5 + 13// 4 + 6// 3 - If - 5// + 1 = 0. 245. Problem IV, — To find the equation whose roots are the squares of those of a proposed equation. Let f(x) = be the given equation ; put // = a* 2 , so that when x has any particular value, the value of y is the square 512 PROBLEM V. of that value ; then x = \y, and the required equation is ftfy) = o. P^xample 1. Find the equation whose roots are the squares of those of the equation x 8 + 2h$~ + jV + Pi = 0. Put x = \y, and transpose, Of +Pi)^y = -{Piv +p 3 )\ squaring, (y 2 + 2p$ + p?)y = p?y 2 + 2piP& + pi- .-. y s + {2p, - p{ 2 )y 2 + {p.? - 2p lPs )y - p 2 = 0. 2. Find the equation whose roots are the squares of the roots of the equation x~+hx— 7 = 0. Ans. y 2 — 39?/ +49 = 0. 246. Problem V. — To transform an equation into another whose roots are less than those of the proposed equa- tion by a given quantity. Let the given equation be f(x) = 0, or Po x n + p^f 1 + p,p?- 2 + ....+ p n _ x x + p n = 0, (1) and let h be the given quantity. Put y = x — h, so that when x has any particular value, the value of y is less by h; then x = y + A, and the required equation is Po(y + ^" + Pi(V + >0" _1 + Pa(2( + k)"~ 2 + • • + ft = 0, which, when arranged in descending powers of y, becomes by (3) of Art. 240 whose roots are less by h than those of ( I ) . If n is small, the transformation by this method is easily effected; but for equations of higher degrees it is quite tedious. The following method is simpler. Denoting the coefficients of the terms of (2) by q u <].,, . . . 5„, it may be written Qoy n + 7.2T' 1 + ^"" 2 + • • • + r h-xV + % = °- (8) in which the coefficients g„ q,, . . . q„ are unknown, whoso values are to be determined. HORNER'S METHOD OE SYNTHETIC DIVISION. 513 For y substitute its value x — ft, and (3; becomes q (x - h)»+ r h (.v - hy- 1 + . ..+ g B _ 1 (a; - h) + q n = 0, (4) which is identical with (1), since the value's of x are the same in (1) and (4). Dividing (4) by (x — h) we obtain the quotient q (x - h)"- 1 + q,(x - 7i) n ~ a + • • • • Qn-i, • ( ; }) with a remainder q„. Similarly, q n _ x is the remainder found by dividing (5) by x — h, and the quotient arising from the division is q (x - ?i) n ~ 2 + q,{x - h) n - 3 + . . . + Qn-*; and so ou, until all the coefficients of (3) are obtained. The last quotient is q , and is obviously equal to j)„. Hence we have the following rule : Divide f(x) by x — h, and the remainder will be the absolute term of the transformed equation. Divide the quotient thus found by x — h, and the remainder will be the coefficient of the last term but one of the transformed equation; and so on. By dividing f(x) by x + h we shall transform f(x) = iuto an equation whose roots are greater by h. 247. Horner's Method of Synthetic Division. — This is an abridged method of dividing by detached coeffi- cients. The work of transforming an equation may be abridged by writing down only the coefficients of the several terms, zero coefficients being used to represent terms corre- sponding to powers of x which are absent. In writing down the divisor, the sign of the last term may be changed ; this enables us to replace the process of subtraction by that of addition at each successive stage of the work. The first term of the divisor is usually omitted. The following example is an illustration. 1. Find the quotient and remainder when 3x 7 — a; 6 + 31a 4 -J- 2 lie + 5 is divided by a; + 2. 514 HORNER'S METHOD OF SYNTHETIC DIVISION. The work stands as follows : Coefficients. 3-1 + 31 + 21 + 5 |-2 ' _ 6 + 14 - 28 - 6 + 12 - 24 + 6 3 _ 7 + 14 + 3 - 6 + 12 - 3 + 11 Explanation. — In this work, 3 is the first term of the quotient ; _C> is the product of —1 and 3; the sum of —1 and —6 is —7. the second term of the quotient; 14 is the product of —2 and —7; the sum of 14 and is 14, the third term of the quotient, and so on. The last term 11 is the remainder. Thus, supplying the powers of x, the quotient is 3x 6 — 7x 5 + 14.f 4 + 3x 8 — 6x a + I2i — 3, and the remainder is 11. We shall now illustrate the use of Horner's method in the following transformation. 2. Find the equation whose roots are less by 3 than those of the equation 2.x 4 — x 3 — 2x 2 + 5x — 1 =0. To effect this transformation we divide successively by « - 3, (Art. 246). Changing the sign of —3, the operation is as follows : 2 _ 1 _ 2 + 5 - 1 |+3 6 + 15 + 39 + 13 2 • ' _ 5 + 13 + 44 + 131. .'. q 4 = 181. 6 + 33 + 138 | 11 + 46 + 182. .'. q z = 182. 6 + 51 | 17 + 97. ••• q a = 97 - J] 23 = q x . Hence the required equation is 2?/ 4 + 23// 3 + 97?/ + 182y + 131 = 0. N OTE . — Horner's method is chiefly useful in numerical work. 3. Find the equation whose roots are less by 5 than the roots of a; 4 - lx 3 - 7sc a + 22a + 24 = 0. Ana. p. Thus f(x) is positive if x == or > p+\ . Therefore p + 1, or any number greater, when substituted for a; in (1), will make the first term numerically greater than the sum of all the others, and thus make (1) positive; that is, p + 1 is a superior limit of the positive roots of the equation f{x) = 0. (2) In an equation of the n th degree in its simplest form if p be the numerical value of the greatest negative coefficient, and x"~ r the highest power of x which has a negative coeffi- cient, 1 + y/p IS a superior limit of the positive routs. Let f(x) = be the given equation ; since all the terms which precede x"~ r have positive coefficients, it is evident that any value of x which makes x" - p(x"- r + .(•"-'- J + . . . -f- x- + .c + 1) INFERIOR LIMIT. 517 positive will make f(x) positive. That is, f(x) is positive for a positive value of x if x" — p- — is positive, (Art. 1G3). But if x > 1 jga-r + l _ j af-r+l x » _ p > x n — p -. Hence f{x) will be positive when this second member is positive ; or when x"(x — 1) — px n ~ r+1 is positive ; or when x r ~ 1 (x — 1) — p is positive; or when (x — l) r = or > p. Thus, f(x) is positive if x = or > 1 + \Jp ; therefore 1 + Vi : is a superior limit of the positive roots of the equation /(a) = 0. 251- Inferior Limit. — To find the inferior limit of the negative roots of an equation, change the sign of every alternate term beginning with the second ; this will change the signs of all the roots of the given equation (Art. 242). Theu the superior limit of the positive roots of this trans- formed equation, with its sign changed, will be the inferior limit of the negative roots of the given equation. 1. Find the superior limit of the positive roots of aj6 + 8x* - Ux 3 - 53a; 2 + 56a - 18 = 0. (1) By (1) of Art. 250 we have 53 + 1 = 54, as a superior limit. (2) By (2) of Art. 250 we have n = 5, n - r = 3, and ._ r __ 2. .-. 1 + V53, or 9 in whole numbers ; so that 9 is a superior limit. We see that (2) of Art. 250 gives us the smallest superior limit, and it is evident that (2) always gives a smaller limit than (1), except when r = 1. Find the superior limits of the positive roots of 2. x 4 - hx z + 37a: 2 - 3a; + 39 = 0. Ans. 6. 3. a; 4 + 11a; 2 - 25a; - 67 = 0. 6. 518 LIMITS OF SEPARATE ROOTS. 252. Limits between which the Roots Separately Lie. — If two numbers, when substituted for the unknown quantity in an equation, (jive results with contrary signs, one root at least must lie betiveen these numbers. Let all the real roots of the equation f(z) = be denoted by a, b, c, . . . I, arranged in the order of their magnitude, a being Algebraically the smallest ; and let A" be the product of the quadratic factors, containing the imaginary roots (Art. 238), which can never change their sign. Then (Art. 235) f(x) = (a; - a)(x - b)(x - c) ....(« - l)X . (1) Now suppose x to increase gradually from a quantity less than a up to some quantity greater than /, assuming in suc- cession every intermediate value. As long as x is less than a, each simple factor x — a, x — b, . . . of fix) is negative, and hence the product is positive or negative according an the number of simple factors is even or odd. When x becomes a, fix) becomes 0. When x becomes greater than a and less than b, the first factor x — o becomes positive while all the other factors remain negative. Hence as x changes from a value less than a to a value greater than a and less than b, the function fix) changes its sign. Also when x becomes b, f(x) becomes ; when x becomes greater than b and less than c, the second factor x — b becomes positive while the signs of all the other factors remain unchanged. Hence as x changes from a value less than b to a value greater than b and less than c, the f unction /(.e) again chauges its sign : and so on for each real root. When the numbers which give results with contrary signs differ by unity, it is evident that the integral part of a root has been found. Some limits betiveen which the roots must lie man be found by the folloioing rules : I. If two numbers, substituted for X in /(»), , f(0) = p n , /(-co) =-00. Therefore if p n is positive f(x) = has a root between and —oo, and if p n is negative f(x) = has a root between and +co. V. Everu equation of an even degree having its last term neg- ative has at least two real roots, one positive and one negative. For in this case, /( + oo) = +oo, /(0) = p n , /(-co) = +co; but p n is negative ; heuce /(a) = has a root between and + oo, and a root between and — oo. EXAMPLES. 1. Prove that 4 is a root of the equation x s + a-a _ ux + 56 = (Art. 234). Use Horner's method. 2. Find the remainder when x 5 — 4 a 4 + 7& 2 — 2x + 47 is divided by x — 5. Use Horner's method. Ans. 837. Solve the equations : 3. x* + 2x* — 41a 2 — 42a + 360 = 0, two roots being 5 and — G. Ans. 3, 5, —4, — G. 4. x A — IGa 3 + 86a 2 — 176a + 105 = 0, two roots being 1 and 7. Ans. 1, 3, 5, 7. 520 EXAMPLES. 5. 4.x 3 + 16a 2 - 9a - 36 = 0, the sum of two of the roots being zero. Arts, f , — §, —4. G. 4a 3 + 20a 2 — 23x + G = 0, the sum of two of the roots being 1. Ans. £, £, —6. 7. x 3 — 7a -f- 6 = 0, the sum of two of the roots being 3. Am. 1, 2, —3. 8. x- 4 — 2a; 3 — 21a; 2 + 22a; + 40 = 0, two roots being -4 and -1. Ans. 2, 5, — 1, -4. 9. Gx* — 29a; 3 + 40a; 2 — 7x — 12 = 0, two roots being | and |. Jn«. f , f, 1 ± ^2. Form the equations whose roots are 10. 2, 3, -5. Ans. a 3 — 19a; + 30 = 0. 11. -2, 4, 4. a; 3 — 6x 2 + 32 = 0. 12. 3, -4, 2 ± ^3. x* — 3a; 8 - 15a; 2 + 49a; - 12 = 0. 13. |, |, ±^3. Qx* - 13a; 3 - 12a 2 + 39a - 18 = 0. 14. 0, 0, 2, 2, - 3, - 3. a 6 + 2x 5 - 11a 4 - 1 2aj*+ 36a 2 = 0. 15. 2, 2, -2, -2, 0, 5. Ans. a 6 — 5a 5 — 8a 4 + 40a 3 + IGa 2 — 80a = 0. 16. ±4^3, 5 ± 2\^1. Ans. a 4 - 10a 3 - 19a 2 + 480a - 1392 = 0. 17. 1 ± ^=2, 2 ± V^3. Ans. a 4 - 6x 3 + 18a- - 26a + 21 = 0. Solve the equations : 18. ;i,- 4 -10a 8 +4a 2 - x — 6 = 0, one root being i±Lzd*. 2 Ans . 3, _|, L±_vE*. ' 3 2 19. 6a 4 - 13a 8 — 85a* - a- + 8 = 0, one root being 2-vfo Ans. -|, —5, 2±V3. 20. a 4 + -la- s -f-r».c-+2r— 2 = 0, one root being —1 +yT-\. Ans. — 1±V^2, -1±V^T. EXAMPLES. 521 21. x* + 4a; 3 + 6a; 2 + 4a: + 5 = 0, one root being V^— 1. Ans. ±V /:r l, -2 ± V-^l. 22. a; 5 — a; 4 + 8a; 2 — 9a; —15 = 0, two roots being s/3, and 1 - 2V^T. Ans. -1, ±V5, 1 ± 2V^1. 23 . Find the least possible number of imaginary roots of the equations(l)a; 4 +3a; 2 +5.«-7 = 0,(2)x 10 -4a; c +a; 4 -2x--3 = 0, and (3) a; 9 -a; 5 +a; 4 +a; 2 +l = 0. Ans.(l) Two, (2) Four, (3) Six. 24. Find the fourth derived function of a; 4 — 8a; 3 + 14a; 2 . Ans. 24. 25. If f(x) = a; 4 - 12a: 3 + 17a; 2 - 9a; + 7, find the value of f(x + 3). Ans. x* - 37a; 2 - 123a; - 110. 26. If f(x) = a; 4 + 10a: 3 + 39a; 2 + 88x- + 113, find the value of f(x - 4). Ans. a; 4 - 6a; 3 + 15a; 2 + 1. Solve the following equations which have equal roots. 27. a; 4 - 9a; 2 + 4a; + 12 = 0. Ans. 2, 2,-1, -3. 28. a; 4 + 2a; 3 - 3a; 2 - 4x + 4 = 0. -2, -2, 1, 1. 29. a; 5 - 13a; 4 + 67a; 3 - 171a; 2 + 216a; - 108 = 0. Ans. 3, 3, 3, 2, 2. 30. a; 5 -^-}-4a; 2 -3a;+2 = 0. Ans. -2, 1 ± \~ 3 , 1 ± ^~ 3 . 31. '8a; 4 + 4a; 3 - 18a: 2 + 11a; - 2 = 0. J, \, \, -2. 32. x 6 - 2a; 5 - 4a; 4 + 12a; 3 - 3ar - 18a; + 18 = 0. Ans. ±V3, ±^3, 1 ± V^l. 33. Transform (1) a; 3 - 4a; 2 + \x - \ = and (2) 3a; 4 — 5a; 3 + a; 2 — a; + 1 = into equations with integral coefficients, and unity for the coefficient of the first term. Ans. (1) y 3 - 24 f + 9y - 24 = 0, (2) f - bif + 3f - 9y + 27 = 0. 34. Transform (1) a; 3 + 3a: 2 -f x — $ = and (2) a? _ 9 X 2 _(_ ^ x . _ _i^ _ o into others whose roots are the reciprocals of the roots of the given equations. Ans. (1) a: 3 -3a; 2 -9a;-3 = 0, (2) a; 3 -42a; 2 +441a;-49 = 0. 522 EXAMPLES. 35. Find the equation whose roots are the squares of the roots of a 4 + a 3 + 2x 2 + a + 1 = 0. Am. f + 3f + if + 3y + 1 = 0. 36. Find the equation whose roots are the cubes of the roots of x s + 3a 2 +2 = 0. Am. y 3 + 33i/ 2 -f V2y + 8 = 0. Find the equations whose roots are each less by 3 than the roots of 37. x a - 27a - 3G = 0. Arts. y a + 9y 2 - 90 = 0. 38. a 4 -27a 2 -14a+120 = 0. ?/ 4 +12?/ 8 +27?/ 2 -68?/ = T;2 _ Wx + 1 = . x 5 - 11a; 4 + 17a 8 + 17a; 2 - 11a; + 1 = 0. ax e — bx 5 + ex 4 — cx~ + bx — a — 0. Note. — It is evident from the second definition of a reciprocal equation that, when the degree is even and the equal coefficients have unlike signs, the middle term is wanting. Theorem I. — A reciprocal equation of an odd degree has a root —1 token the equal coefficients have the same sign, and + 1 when they have opposite signs. Let aa? + bx"- 1 + cx n ~ 2 + . . . ± car ± 6a; ± a = (1) be a reciprocal equation of an odd degree. The number of terms is even, and the equation may be written in the form, a(x n ± 1) + 6(af- 2 ± l)a + c(x n ~ 4 ± l)x 2 +. . . = 0, (2) in which each exponent of x in the parentheses is odd. When we take the upper signs the equation is divisible by x -f 1 ; hence — 1 is a root : when we take the lower signs the equation is divisible by x — 1 ; hence +1 is a root. That is, —1 or +1 is a root according as the equal coelli- cients have the same or opposite signs. 526 RECIPROCAL EQUATIONS. Theorem II. — A reciprocal equation of an even degree has a root +1 and a root — 1 when the equal coefficients have opposite signs. Taking the lower signs of (2), and regarding n as even, we have a(x n - 1) + b{x"-' 2 - \)x + c(.-e"- 4 - l)x 2 + . . . = 0, (3) which is an equation of the specified form. As each expo- nent of a- in the parentheses is even, (3) is divisible by x 2 — 1 ; hence + 1 and — 1 are roots. Therefore by division an equation of this form may be depressed two degrees, and become a reciprocal equation of an even degree, with the equal coefficients having like signs (Art. 51). Theorem III. — A reciprocal equation of an even degree ivith its last term positive can be reduced to an equation of half that degree. Let the equation be ax 2 "+ bx* n - 1 + cx 2n ~ 2 + . . . kx"+ . . .+ cx 2 + bx -fa = ; (4) dividing by x n and collecting the terms, we have Put x + - = z ; then x 2 + -, = z 2 — 2 ; a; x 2 * + b = { x + 1 if-i*+l) = *- s *-> xi + h = (*" + .^) 2 ~ 2 = zi ~ 4z "' + 2 ; and generally, x" -\ = z n — nz"~ 2 + . . . . x n Hence, each of the binomials may lie expressed in terms of z, and by substituting in (5) we have an equation in z of the a th degree, i.e., of half the degree of (4), the given equation. BINOMIAL EQUATIONS. 527 1. Given 2x G -fa; 5 - 13a; 4 + 13a; 2 - x — 2 = 0. Here +1 and — 1 are roots by inspection, also see Theorem II. Dividing the first member by x' 1 — 1, we obtain 2a: 4 + a; 3 - 11a; 2 + x + 2 = 0. Divide by 2ar, put » + - = «, and we have x , + §-», .-. ,=§or-3. 1 5 Hence a; -| — = - or —3 ; x 2 .-. a? = 2 or | or i(- 3 ± V^5). Solve the following equations : 2. a; 4 - 5a; 3 + 6a; 2 - 5a; + 1 = 0. Am. 2 ± V&, ±\/^S 3. a j6_ a 8 + aJ*-aj a + oj-l = 0. ±1, ±^1, 1± * 3 - 255. Binomial Equations. — The general form of a binomial equation is x n ± k = 0, where & is a known quan- tity. The roots of this equation are all different, because x n ± k and the first derived function nx"- 1 evidently have no common . divisor. See Art. 241. If x n — k — we have x = n yjk ; i.e., x is equal to an n ih root of k. But x n — k = has n roots (Art. 235) . There- fore any Algebraic quantity * has n different n th roots. Let a denote one of the w th roots of k, so that k = a". Assume x = az ; then a; n — fc =_ becomes a n z n — a" = ; or z n — 1 = 0. Hence z = y/l, i- e -> 2 is ec l ual to an w ' h root of unity. And since x = az = a y/l = Y& from above, we have n y'& = o yl. * By an Algebraic quantity here is meant either a real or an imaginary quantity. 528 BINOMIAL EQUATIONS. Thus all the n a roots of any Algebraic quantity may be found by multiplying any one of them in succession by the values of the n" 1 roots of unity. (1) If n is odd, the equation af ± 1 = has a root —1 or +1 according as we take the upper or lower sigu, Art. 254, Theo. I. If x n ± 1 be divided by x ± 1 it will be depressed one degree, and the resulting reciprocal equatiou will contain the remaining n — \ roots. (2) If n is even, the equation af — 1 = has two real roots +1 and -1, Art. 254 Theo. II. If x n — 1 be divided by x 2 — 1, it will be depressed two degrees, and the resulting reciprocal equation will contain the remaining n — 2 roots. (3) If n is even, the equation x n + 1 = has no real root, since y/ — 1 is then impossible. Hence all the roots of this equation are imaginary. The following are some of the cases of binomial equations which can be solved by methods already given. For the general case, De Moivre's theorem in Trigonometry must be employed. t. Solve x b - 1 = 0. Since x 5 - 1 = (x - 1) (x* + a 8 + ar» + x + 1), we have x — 1 = ; or else x* + x s + x 2 + a; + 1 = 0. Therefore x = 1 ; and dividing the latter equation by xr and putting x -\ — = z we have x *• + * = !; ... « = =if£ Replacing the value of z and solving for x, we have x = -* + V^ ± jtf-10 _ 2V5, 4 o r x = -1 ~ V^ ± ^-10 + 2^5, or x = 1, for the 5 roots. CARDAN'S SOLUTION OF A CUBIC EQUATION. 529 2. Solve a; 4 - 1 = 0. Am. 1,-1, /— h -y 7 -^- 3. Solve a; 4 + 1 = 0. ±1 *V-1 . y/2 256. Cardan's Solution of a Cubic Equation.— The general form of a cubic equation is x 3 -\- ax 2 -f bx + c = ; but as we can always remove the second term from an equa- tion (Art. 248), we shall suppose the cubic reduced to the simple form, sb 8 + px + q = (1) Put x = y -\- z; then substituting in (1) we have y 3 + z 3 + {oyz + p) (y + z) + q = 0. Now since y and z are any two quantities subject only to the condition that their sum is equal to one of the roots of the given equation, we are at liberty to suppose further that they satisfy the equation 3yz + p = 0. Thus we obtain y s + z s + q = . # m (2 ) 3yz= -p . . (3) Substituting in (2) the value of z in terms of y from (3), we obtain the quadratic y 6 + qif = 27 Solving, we have y 3 = — - + \ A 2 + il 8 - .-. V 4 27- • -(4) — -5-i Since a; = y + 2, we have /l + 27' ' * • -(5) -H + \/M4i-^ (6) Thus the expression for x is the sum of two cube roots, and as every quantity has three cube roots, it would appear that by combining each of the three cube roots in the value of y, with each of the three cube roots in the value of z, 530 CARDAN'S SOLUTION OF A CUBIC EQUATION. we should obtain on the whole nine values of x. But a cubic equation can have only three roots (Art. 235), so that we are led to conclude that only three values will be admissible for x, which may also be shown as follows : From (3) we see that the cube roots are to be taken in pairs so that the product of each pair is rational. Hence if y, z denote the values of any pair of cube roots which fulfil this condition, and «, a 2 the imaginary cube roots of unity, then the only admissible pairs will be y and z, ay and a' 2 z, a-y and az, since the product of each of these three pairs and of no other pair is rational. Hence the roots of (1) are y + z, ay + a 2 z, a 2 y + az. When the real root has been found from (6), instead of expressing the other two roots by the same method, it is preferable to divide ( 1 ) by x minus the root and thus depress the equation to a quadratic (Art. 235). Note. — The above solution is commonly known by the name of Cardan's Solution, because it was first published by him in 1545. Cardan, however, was not the inventor; he obtained it from Nicholas Tartalea. The solution ought to be attributed to Nicholas Tartalea and Scipio Ferreus, who seem to have discovered it about the same time, about 1505, and independently of each other. (See historical note in Burnside and Panton's Theory of Equations). 1. Solve the equation x 3 — 15cc = 126. Put x = y + z, then y 3 + z 3 + (3?/z - lo)(y + z) = 126 ; put 3yz - 15 = 0, then y 3 + z 3 = 126. Substituting for z its value, we have the quadratic y* - 126y 3 + 125 = .-. y 3 = 125, z 3 = 1; .-. y = 5, 2 = 1. Thus x = y + z = 5 + 1 = 6. Dividing the given equation by x — 6 we get x 2 + 6x + 21 = 0, the roots, of which are —3 ± 2v— 3. Ik' nee the three roots are 6, —3 + 2>J — o, —3 — 2^—3. CARDAN'S SOLUTION OF A CUBIC EQUATION. 531 If L _|- 3L is negative the expressions for y 3 and z 3 are 4 27 imaginary, but when this is the case it may be shown that the roots of the cubic are all real and unequal. As there is no general Arithmetic or Algebraic method of finding the exact value of the cube root of imaginary quantities, Car- dan's solution of a cubic equation is of little practical use when the roots of the cubic are all real and unequal. This case is sometimes called the Irreducible Case of Car- dan's solution. The process by which we may obtain the roots of a cubic equation, in the irreducible case, is given in works on Trigonometry. This is a matter however of very little practical value. 2. Solve x 3 + 3a; 2 + 9x - 13 = 0. To remove the second term, we have h = — 1, (Art. 248). .-. x = y — 1 (Art. 246). Substituting this value of x and reducing, the transformed equation becomes y 3 + 6y = 20. Here p — 6, q = — 20 ; hence, substituting in (6) we get y = (10 + V^108)* + (10 - VT08)* = 2.732 - .732 = 2. By division, we find for the depressed equation y* + 2y + 10 = ; therefore the other two values of y are — 1 ± V — 9 ; and since x = y — 1, the corresponding values of a; are 1, — 2+3V — 1, -2 - SV^. 3. Solve x 3 - 9x + 28 = 0. Ans. 4, 2 ± \P^d. 4. Solve x 3 - 18.x = 35. 5, ~° * 3 - 5. Solve x s + 72x - 1720 = 0. 10, -5 ± 7V^3. Rem. — It is not thought worth while to introduce into an element- ary treatise like this any of the solutions of biquadratic equations which have been obtained by Ferrari, Descartes, Waring, .Simpson, 532 INCOMMENSURABLE ROOTS. Euler, and others. In each of these methods we have first to solve an auxiliary cubic equation ; and as in the, case of the cubic, the general solution is not adapted for writing down the solution of a given numerical equation. Practically the methods are worth but little, as the value of any real root in a numerical equation may be found to any required degree of accuracy by Horner's method of approximation. 257. Incommensurable Roots. — If a numerical equa- tion is found to contain no commensurable roots, or, if after the commensurable roots are removed the depressed equation is still of a higher degree than the second, the incommen- surable roots must then be found. The first operation is to find the integral parts of these roots. There are several ways of doing this, the most celebrated of which is by Sturm's Theorem; but all of them are so laborious in practice, that, in ordinary cases, it is generally easiest to proceed by trial, substituting entire numbers for x in the equation, until two consecutive numbers are found between which one or more roots must lie (Art. 252, 1.). The decimal parts of the roots must then be found by methods of approximation. Of these the most convenient is Horner's, by which we can always determine the numerical values of the real roots to any required degree of accuracy. 258. Horner's Method of Approximation. — This method depends on the successive transformation of the given equation, so as to diminish its roots at each transfor- mation (Art. 24G). Leta?+p 1 rf t - 1 +ptff l -*+. . . . + p«-i& + i>« = . (1) be the given equation, one of whose roots is required : and suppose we have found a to be the integral part of this root so that the required root lies between a and a + 1; and lei a', 3 6 d) .61 .167128 7.61 -.071872 (3 > .1 6.1 .62 .068273152 8.23 (2) -.003598848^ .1 6.2 .1264 8.3564 .1 .1268 6.3< 2 > 8.4832< 3 > .02 6.32 .050944 8.534144 .02 .051008 6.34 8.585152 (4) .02 6T36 (3) .008 6.368 .008 6.376 .008 6.384 (4 > Here the figure (1) shows where the first transformation ends, and the figure (2) shows where the second transform- ation ends, and so on. By thus marking the coefficients in each transformed equation, it is not necessary to rewrite it.* * The successive transformed equations, (1), (2), (3) written out would be (1) y3+6y2+7(/-l=0, (2) 8S+6.3s 8 +8.23e-.239=0, (3) m;3+6.36hP+8.4S32!C-.071S72=0 . .. 536 NEWTON'S METHOD OF APPROXIMATION. To find the second figure of the root we have — ( — 1)-=-7, so that .1 is the nearest number to be tried. To find the third figure of the root we have -(-.239) -~ 8.23, so that .02 is the nearest number to be tried. To find the fourth figure of the root we have -(-.071872) -- 8.4832, so that .008 is the nearest number to be tried. In each of these cases the number tried is found to be correct. After obtaining three decimal figures of the root the next three may be obtained by dividing the absolute term by the coefficient of x. If the coefficient of x in any of the transformed equations is zero, the next figure of the root in this case may be found by dividing the absolute term by the coefficient of or, and extracting the square root of the quotient. Eem. — Various suggestions have been offered with the view of shortening the work in the use of Horner's method. With respect to such suggestions, it may be well to quote the following which occurs in connection with one of them. 'But considering that the process is one which no person will very often perform, we doubt whether to recommend even this abridgment. All such simplifica- tions tend to make the computer lose sight of the uniformity of method which runs through the whole; and we have always found them, in rules which occur only now and then, to afford greater assistance mforgetting the method than in abbreviating it." Penny Cyclopaedia, article Involution. Find a root in each of the following equations : 2. x- 3 + 4a; 2 - 5a; - 20 = 0. Ans. 2.23608. 3. a; 3 + 4a; 2 - 9a; - 57.623625 = 0. 3.45. 4. x * _ 4 X * _ 8a; + 27 = 0. 3.6796. 259. Newton's Method of Approximation. — Find by trial two numbers a and b, one less and the other greater than a root of the equation f(x) = and very nearly equal to it; then at least one real root lies between these numbers (Art. 252), and suppose a to be nearer the root than b ; let a + h denote the exact value of the root, so that h is a APPROXIMATION BY DOUBLE POSITION. 537 small fraction which is to be determined. Substituting a+h for cc in the given equation, we have, by Art. 240, /(« + h) = /(«) + kf(a) + ~f\a) + = 0. LL Now since h is supposed to be a small fraction, 7t 2 , 7i s , . . . will be small compared with h ; if we neglect the squares and higher powers of h in the above equation, we shall have approximately /*=-/(a)-/». Applying this approximate value of h to the assumed root a gives us «-/(«)-/», as a new approximation to the roots of the given equation. Denote this new approximation by a v and then proceeding as before we obtain «i - /(«i) + / («i)> as a new approximation ; and so on. (See Todhunter's Theory of Equations) . The following example of Newton's method was selected by Newton himself : Find the value of x in the equation, a;3 _ 2x — 5 = 0. Ans. 2.09455148 nearly. 260. Approximation by Double Position. — Find by trial two numbers a and 6, one less and the other greater than a root of the equation f(x) = 0, and very nearly equal to it ; then at least one root lies between these numbers ( Art. 252), and suppose a to be nearer the root than b. Let A and B denote the results when a and b are separately substituted for x in f(x) ; and assume that the results, which may also be considered as the errors of the results, are pro- portional to the errors of the assumed numbers. Although this assumption is not strictly correct, yet if the numbers a and b are nearly equal to the root, the error of the assump- 538 APPROXIMATION BY DOUBLE POSITION. tion is small, and becomes less and less the further the approximation is continued. Thus we have A : B : : x — a : x — b. .'. A — B:b — a :: A:x — a. Hence we have approximately the following Rule. — As the difference of the errors is to the difference of the two assumed numbers, so is either error to the correc- tion of the corresponding number. This correction is to be added to the corresponding num- ber when it is too small, and subtracted when it is too great, and the result will be a nearer approximation to the true root. This result and another assumed number such that the root may lie between them, may now be used, for obtain- ing a second approximate value ; and so on. It is generally best to begin with two numbers which differ from each other only by unit}' ; and it is also best to use the smaller error. This method of approximation is applicable whether the equation is fractional, radical, or exponential ; it may there- fore be applied to many equations which cannot be solved by either of the preceding methods. 1. Find a root of the equation x 3 + a; 2 + x — 100 = 0. "When 4 and 5 are substituted for x in the equation, the results are — 1G and +55. Hence we have 1G + 55 : 5 - 4 : : 1G : x - 4 ; .-. x = 4.22. As the root is greater than 4.2, we now assume 4.2 and 4.3 ; substituting these for x in the given equation, aud proceeding as before, we obtain for a second approximation x = 4.264. Assuming 4.2G4 and 4.2G5 for as, and repeating the oper- ation, we obtain for a third approximation u; = 4.2644299 nearly. 2. Find a root of a; 3 + 30a - 420 = 0. Ana. G. 170103. EXAMPLES. EXAMPLES. 539 Solve the following equations : 1. x 3 - 3a; 2 - 46a; - 72 = 0. Ans. 9, -2, -4. 2. x a - 2x 2 — 4x + 8 = 0. 2, 2, -2. 3. X 4 + 4x 3 _ X 2 _ 16aJ _ 12 = 0. 2, -1, -2, -3. 4. a; 4 - 27a; 2 + 14a; + 120 = 0. 3, 4, -2, -5. 5. x 4 - 12a; 3 + 47a; 2 - 72a; + 36 = 0. 1, 2, 3, 6. 6 . ^ _ 9x 3 + l7x 2 + 27a; - 60 = 0. 4, 5, ±^3. 7. a; 5 + 5a; 4 + x 3 - 16a; 2 - 20a; - 16 = 0. Ans. 2, -2, -4, ~ 1 ^ Solve the following reciprocal equations : 8. a; 4 +5a; 3 +2a; 2 +5a;+l = 0. Ans. $(-5±V^T), ±V-1. 9. a; 4 - 3a; 3 + 3a; - 1 = 0. ±1, £(3 ± V^5). 10. a; 4 - 10a; 3 + 26a; 2 - 10a; + 1 = 0. 3 ±2^2, 2 ±^3. 11. 4a; 4 - 4a; 3 - 7a; 2 - 4a; + 4 = 0. 2, |, i(-3±v / -7). 12. 9a; 4 -24a; 3 -2a; 2 -24a; + 9 = 0. 3, A, £(-l±\^-8. 13. a; 6 - 11a; 4 + 17a; 3 + 17a; 2 - 11a? + 1 = 0. Ans. -1, |(9 ± ^77), |(3 ± \£). 14. x 5 - 5a; 4 + 9a; 3 - 9a; 2 + 5a; - 1 = 0. Ans. 1,£(1 ± /^3),$(3± ^5). 15. 4a; 6 - 24a; 6 + 57a; 4 - 73a; 3 + 57a; 2 - 24a; + 4 = 0. Ans. 2, 2, i,}, -1(1 ± V/-3). 16. a; 5 + 1 = 0. ^s. -1, i[l+V / 5±V / (+2V / 5-10)], i[l-V / 5±V(_2V / 5-10)]. 17. a; 8 - 1 = 0. ±!» ±^1, ±4 ± N^« Solve the following cubic equations : 18. a; 3 + 63a; - 316 = 0. Ans. 4, -2 ± 5V^3. 19. a; 3 + 21a; + 342 = 0. -6, 3 ± 4V^3. 540 EXAMPLES. 20. x s - 6a; 2 + 13a; - 10 = 0. Ans. 2, 2 ± V^l. 21. a; 3 + 6a; 2 - 32 = 0. 2, -4, -4. 22. a; 3 - Gar + 3a; - 18 = 0. 6, +V^3, -V^3. 23. x s - 15a; 2 - 33a; + 847 = 0. 11, 11, 7. 24. 28x 3 - 9x 2 +1 = 0. -£, i(2 ± V^). 25. 2a; 3 + 25a; 2 + 56.x - 147 = 0. -7,-7, 1$. Find a root by Horner's method in the following equations : 26. x 3 + 3z 2 + 5a; - 178 = 0. Ans. 4.5388. 27. x 3 + 11a; 2 - 102a; + 181 = 0. 3.21312. 28. a; 3 + x 2 - 500 = 0. 7.61728. 29. a; 3 + a; 2 + x - 100 = 0. 4.26443. 30. a; 3 + 10a; 2 + 6a; - 120 = 0. 2.8330665. 31. a; 4 - 8a; 3 + 20a; 2 - 15a; +.5 = 0. 1.284724. 32. a; 4 + a; 2 - 8a; - 15 = 0. 2.302775. 33. a; 5 + 4a; 4 - 3a; 8 + 10a; 2 - 2a; - 962 = 0. 3.385777. 34. a; 4 - 2a; 3 + 21a; - 23 = 0. 1.1574515. 35. a; 4 - 5a; 3 + 3a; 2 + 35a; - 70 = 0. 2.6457513. Find a root by Newton's method in the following equa- tions : 36. a; 3 + 12a; 2 - 18a; - 216 = 0. 4.2426407. 37. x 3 + a; - 3 = 0. 1.21341166. Find a root by the method of Double Position in the fol- lowing equations : 38. a; 3 + 2x - 20 = 0. 2.46954565. 39. a; 5 + 10a; 2 + 8a; - 120 = 0. 2.76834546. Arithmetic. Aids to Slumber. — First Series. Teachers' Edition. Oral Work — One to ten. 25 cards with concise directions. By Anna B. Badlam, Principal of Training School, Lewiston, Me., formerly of Rice Training School, Boston. 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