IN MEMORIAM 
 FLOR1AN CAJORI 
 
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HIGHER ALGEBRA 
 
 BY 
 
 JOHN F. DOWNEY, M.A., C.E. 
 
 PROFESSOR OF MATHEMATICS IN THE UNIVERSITY OF MINNESOTA 
 
 ?XKc 
 
 NEW YORK-:. CINCINNATI. -CHICAGO 
 
 AMERICAN BOOK COMPANY 
 
Copyright, 1900, by 
 JOHN F. DOWNEY. 
 
 ^DOWNEY'S. HIGH. ^ALG. 
 
 CAJORI 
 
 
PREFACE 
 
 This work is designed as a text-book in universities, colleges, 
 and technical schools, the first fifteen chapters being also' adapted 
 to use in high schools and academies by students who have some 
 knowledge of elementary algebra. 
 
 The demonstrations constitute one of the characteristic features 
 of the book. While most of our text-books on Algebra state with 
 great clearness the theorems and rules, few of them, especially 
 in the earlier parts, give the demonstrations in a way that enables 
 a student to reproduce them. Usually illustration, explanation, 
 and general demonstration are so intermingled that the student 
 is not able to gather up and give in logical form just what con- 
 stitutes the proof. In this work the plan is that which gives so 
 much definiteness to our teaching in Geometry: each general 
 principle is followed by a concise, logical demonstration, contain- 
 ing only the reasoning necessary to establish it, while all illus- 
 trations and explanations by special cases are given in separate 
 articles. The student thus soon learns to know what is demanded 
 in a general proof, and to distinguish between rigorous demon- 
 stration and verification or illustration by a special case. With- 
 out any loss of conclusiveness in reasoning, the methods employed 
 have permitted, in many cases, much shorter and more easily 
 followed demonstrations than those usually given. 
 
 Another characteristic feature is the substitution of short 
 processes for many of the long and tedious ones in common use. 
 As mathematical operations, at best, involve much drudgery, all 
 practical means of shortening the work should be made available 
 to the student. The few short processes given in our text-books 
 are reserved until the student has formed a habit of using the 
 long processes, and, consequently, he never gains a practical use 
 of even these few. In this book short processes are introduced 
 
 3 
 
 911285 
 
4 PREFACE 
 
 at the beginning of the respective subjects, and are used wherever 
 applicable. For example, a very expeditious method of multiply- 
 ing by a factor of the form x ± a or x ± ay is fully explained, 
 and is applied to successive multiplication, to putting together 
 the factors that constitute the highest common divisor and the 
 lowest common multiple, to forming an equation having given 
 roots, and to various other operations. Again, the equally expe- 
 ditious method of dividing by a factor of the form x ± a or 
 x ± ay, instead of being confined to finding the commensurable 
 roots of higher equations, is fully explained in Division, and is 
 applied to successive division, to factoring polynomials, to find- 
 ing the highest common divisor and the lowest common multiple, 
 to reducing fractions to their lowest terms, to finding the com- 
 mensurable roots of higher equations, and to various other opera- 
 tions. The student thus becomes very familiar with the processes, 
 and expert in their use; and he not only escapes much tedious 
 labor that has no disciplinary value, but secures greater accuracy 
 in his results, the shorter processes diminishing the liability to 
 errors. 
 
 The methods given of obtaining at once the square root and 
 the cube root of polynomials, without writing any intermediate 
 steps, will be found not only a great relief from the tedious 
 processes given in our books, but an exhilarating exercise in 
 which students will take great interest. 
 
 The subject of Maxima and Minima of Functions is presented 
 in a fuller and more systematic way than heretofore, and the 
 application made to practical problems cannot fail to interest the 
 student. 
 
 While those properties of higher equations which serve no 
 useful purpose have been excluded, the subject has received a 
 fuller treatment than is common. The reader will find many 
 features presented in new and simpler ways, with everything 
 leading toward the easiest and most expeditious methods of 
 finding the roots of numerical higher equations. The process of 
 finding the roots of equations having only even or only odd 
 powers will be seen to be % remarkably brief. 
 
 Differentiation of algebraic and logarithmic functions is intro- 
 
PREFACE 5 
 
 duced, because it enables us to give in their true relations and 
 with their proper significance differential coefficients, which are 
 usually disguised under the name of "derived polynomials"; 
 because it enables us to include Taylor's Formula, and thus avoid 
 the usual cumbersome and unsatisfactory method of demonstrat- 
 ing the binomial formula and the logarithmic series ; because it 
 puts into its true relation f'(x) in the theorem for multiple roots 
 and in Sturm's Theorem; and because in his subsequent course 
 in mathematics the student will not use the methods which dif- 
 ferentiation avoids, and will constantly use the methods which 
 it introduces. Differentiation is given by the fluxionary system, 
 as furnishing the clearest and most satisfactory conceptions and 
 the simplest demonstrations. 
 
 Many exercises and problems have been given in all the dif- 
 ferent subjects. These will be found well graded, and neither 
 so difficult as to discourage the student nor so easy as to afford 
 little discipline. To guide the student in the application of prin- 
 ciples to numerical examples, such suggestions, observations, and 
 model solutions as long experience in teaching the subject has 
 shown to be needful have been given. While many of the exer- 
 cises and problems are new, free use has been made of the various 
 collections available. 
 
 The subject of Determinants is not included, because inexpen- 
 sive texts devoted exclusively to the subject can be readily pro- 
 cured by the comparatively small number of students who have 
 occasion to enter upon the subject at this stage of their progress. 
 The chapter on the Theory of Functions will be found of more 
 value to the student than would be an elementary treatment of the 
 Loci of Equations, as it gives him needed practice in the inter- 
 pretation of analytical results, and thus prepares him for the 
 subject of Loci when he reaches it in Analytical Geometry. 
 
 Assistance has been received from so many sources that no 
 attempt is made to name them. While in many respects the 
 book is a wide departure from the texts of the day, nothing has 
 been made different for the sake of novelty. W T hether new or 
 old, the methods which long experience with large classes has 
 proved to be the best have been given. 
 
6 PREFACE 
 
 The book is put forth with the hope that it will not only serve 
 well the usual purpose of a text on Algebra, but that it will, in 
 addition, cultivate in students much greater facility in the expla- 
 nation of principles, and that it will relieve them from a great 
 burden of worse than needless details in operations, which many 
 generations of students have been required to carry. 
 
CONTENTS 
 
 CHAPTER PAGE 
 
 I. Definitions and Notation 15 
 
 Quantity 15 
 
 Algebra 15 
 
 The Symbols of Quantity 16 
 
 The Symbols of Operation 17 
 
 The Symbols of Relation 18 
 
 The Symbols of Aggregation 19 
 
 The Symbols of Abbreviation 19 
 
 Algebraic Expressions 19 
 
 Positive and Negative Quantities 20 
 
 Axioms, Theorems, etc. 20 
 
 II. Addition 22 
 
 Addition of Monomials 22 
 
 Addition of Polynomials 23 
 
 Addition of Partially Similar Terms 25 
 
 Addition of Compound Terms 25 
 
 III. Subtraction 26 
 
 Rule for Subtraction 27 
 
 Signs of Aggregation as related to Addition and Subtraction 28 
 
 IV. Multiplication 30 
 
 The Sign of the Product 31 
 
 Multiplication of Monomials 31 
 
 Multiplication of Polynomials 32 
 
 Homogeneous Polynomials 33 
 
 Multiplication by Detached Coefficients .... 31 
 
 Short Methods of Multiplication 36 
 
 By a factor of the form x ± a 36 
 
 Successive Multiplication 39 
 
 When the factors are of the form x ± a and x± b . . 40 
 
 The Square of the Sum and of the Difference ... 42 
 
 The Product of the Sum and Difference .... 42 
 
 The Square of Any Polynomial 43 
 
 The Sum of Two Groups by the Difference ... 43 
 7 
 
8 CONTENTS 
 
 PAGE 
 
 V. Division 45 
 
 The Signs of the Quotient 45 
 
 Division of Monomials 45 
 
 Exponent 45 
 
 Transference of Factors from Dividend to Divisor, and vice 
 
 versa 46 
 
 Division of Polynomials 47 
 
 Short Methods of Division 49 
 
 Horner's Synthetic Division .50 
 
 When the divisor is of the form x ± a . . . .53 
 
 Successive Division 57 
 
 When the divisor is separable into factors of the form 
 
 x ± a 58 
 
 VI. Factoring 59 
 
 Monomial Factors 59 
 
 Trinomial Squares 59 
 
 Difference of Two Squares 59 
 
 Six-term Squares 59 
 
 Trinomials of the Form x 2m + px m + q 61 
 
 Trinomials of the Form ax 2m + bx m -f c . . . .62 
 
 Separating into Groups 63 
 
 The Difference of the Same Powers 65 
 
 The Sum of the Same Powers . . ■ . . .65 
 
 Binomial Factors by Trial 66 
 
 VII. Highest Common Divisor and Lowest Common Multiple . 71 
 
 Highest Common Divisor 71 
 
 Polynomials readily factored 71 
 
 One of Two Polynomials readily factored .... 74 
 
 Polynomials not readily factored 75 
 
 Lowest Common Multiple 77 
 
 VIII. Fractions 79 
 
 Keduction of Fractions 80 
 
 To Lowest Terms 81 
 
 To Integral or Mixed Forms . . . . . .82 
 
 Mixed Quantity to Fractional Form 83 
 
 To Lowest Common Denominator 84 
 
 Addition and Subtraction of Fractions 85 
 
 Multiplication of Fractions 86 
 
 A Fraction by an Integer 86 
 
 A Fraction by a Fraction 87 
 
CONTENTS 
 
 IX. 
 
 Division of Fractions . 
 
 A Fraction by an Integer 
 
 A Fraction by a Fraction 
 Simplification of Fractions 
 
 Complex Fractions 
 
 Terminated Continued Fractions 
 
 Fractions with Negative Exponents 
 
 Theory of Exponents, Involution, and Evolution 
 Theory of Exponents .... 
 Fractional and Negative Exponents . 
 
 (abc . . .) n = a n b n c n 
 
 To affect a Monomial with Any Exponent 
 
 Involution 
 
 Monomials 
 
 Binomials — Binomial Theorem 
 Cube of a Polynomial 
 
 Evolution 
 
 Monomials 
 
 Index a Composite Number 
 
 Square Root of Polynomials by Inspection 
 
 Trinomials 
 
 Only Two Powers of One Letter . 
 
 Any Other Perfect Square 
 Cube Root of Polynomials by Inspection 
 
 Quadrinomials .... 
 
 Only Three Powers of One Letter . 
 
 Any Other Perfect Cube 
 Any Root of Any Quantity 
 
 X. 
 
 Surds and Imaginaries . 
 Surds .... 
 
 Reduction of Surds . 
 Removal of a Factor 
 Diminishing the Degree 
 Rational Quantity to Radical Form 
 Introduction of Coefficient 
 To Equivalent Surds of Same Degree 
 Rationalizing a Denominator . 
 The Rationalizing Factor of a Binomial 
 
 Addition and Subtraction of Surds 
 
 Multiplication of Surds 
 
 Division of Surds .... 
 
 Surd 
 
10 
 
 CONTENTS 
 
 XI. 
 
 XII. 
 
 XIII. 
 
 Involution of Surds . 
 Evolution of Surds 
 Imaginary Quantities 
 General' Form . 
 The Imaginary Element 
 Index Composite 
 Conjugate Imaginaries 
 Modulus of a Quadratic Imaginary 
 Addition and Subtraction of Imaginaries 
 The Different Powers of ^/~=T and ^/^H 
 Multiplication and Division of Imaginaries 
 Use of Imaginaries 
 
 Simple Equations 
 
 One Unknown Quantity 
 Axioms .... 
 Rule for Solution 
 Some Practical Suggestions 
 Equations with Surd Terms 
 Problems .... 
 Two Unknown Quantities 
 Elimination 
 
 By Addition or Subtraction 
 By Comparison 
 By Substitution 
 Some Practical Suggestions 
 Problems 
 Several Unknown Quantities 
 Problems .... 
 
 Inequalities .... 
 Operations that do not change the 
 Operations that change the Sense 
 Caution 
 
 Sense 
 
 Ratio, Proportion, and Variation 
 
 Ratio 
 
 Operations that do not change a Ratio 
 Examples and Problems 
 
 Proportion 
 
 An Equation from a Proportion 
 A Proportion from an Equation 
 Operations that do not destroy a Proportion 
 
CONTENTS 11 
 
 PAGE 
 
 Examples and Problems 184 
 
 Variation 187 
 
 Ratio given by a Set of Values 189 
 
 From a Variation to an Equation, and vice versa . .190 
 
 From a Variation to a Proportion 190 
 
 Examples and Problems 190 
 
 XIV. Progressions 195 
 
 Arithmetical Progression 195 
 
 Formulae 196 
 
 Examples and Problems 197 
 
 Geometrical Progression . . . . . . . . 200 
 
 Formula 201 
 
 Examples and Problems 203 
 
 Harmonic Progression 205 
 
 XV. Quadratic Equations 208 
 
 Pure Quadratics 208 
 
 Rule for solving 208 
 
 Character of the Roots 209 
 
 Problems 209 
 
 Affected Quadratics 211 
 
 Rule for solving 211 
 
 Character of the Roots . . . . m . . .212 
 
 Problems 213 
 
 Solution by Factoring 216 
 
 Formation of an Equation having Given Roots . .217 
 
 Introduction and Loss of Roots 218 
 
 XVI. Some Higher Equations 222 
 
 Pure Equations 222 
 
 Equations in the Quadratic Form . ... . . 223 
 
 Equations with Integral Roots 226 
 
 XVII. Equations of the Second Degree with Two Unknown 
 
 Quantities 230 
 
 One of the Equations of the First Degree .... 231 
 All Terms but One each of First Degree . . . .232 
 
 One of the Equations Homogeneous 232 
 
 Both Homogeneous except in Absolute Term . . . 233 
 
 Symmetrical Equations 238 
 
 Problems 240 
 
12 CONTENTS 
 
 PACE 
 
 XVIII. Theory of Functions 242 
 
 Maxima and Minima of Functions . 242 
 
 Quadratic Functions 245 
 
 Reciprocals of Quadratic Functions 247 
 
 Algebraic Functions of Any Form 248 
 
 Problems 252 
 
 Zero, Infinity, and Indeterminate Forms .... 255 
 
 The Forms -£ and ^ 255 
 
 co 
 
 The Forms - and — 255 
 
 GO 
 
 Discussion of Functions and Problems .... 257 
 
 Interpretation of Negative Results 257 
 
 Interpretation of Imaginary Results .... 258 
 
 Examples and Problems 259 
 
 XIX. Differentiation of Algebraic Functions .... 263 
 
 Constant Terms 263 
 
 The Product of a Constant and a Variable . . * . . 263 
 
 A Polynomial ' . . 264 
 
 The Product of Two Variables 264 
 
 The Product of Several Variables 265 
 
 A Fraction 265 
 
 A Variable having a Constant Exponent .... 266 
 
 Second, Third, etc., Differential 268 
 
 First, Second, Third, etc., Derivative ..... 268 
 
 Partial Derivatives .270 
 
 XX. Development of Functions 272 
 
 Indeterminate Coefficients 274 
 
 The Law of the Series 275 
 
 Decomposition of Fractions 278 
 
 Taylor's Formula 282 
 
 The Binomial Formula 285 
 
 XXI. Logarithms 288 
 
 Logarithm of a Product 289 
 
 Logarithm of a Quotient ....... 290 
 
 Logarithm of a Power 290 
 
 Logarithm of a Root . 290 
 
 Law of Characteristics 291 
 
 The Mantissa and the Decimal Point 292 
 
 The Differential of a Logarithm 292 
 
 The Logarithmic Series , 293 
 
CONTENTS 
 
 13 
 
 PAGE 
 
 Relation of Logarithms in Different Systems . . . 294 
 
 The Logarithmic Series rendered Convergent . . . 295 
 
 Computation of Logarithms 295 
 
 Tables of Logarithms . . . ••■'-. . . . 296 
 
 Values of m and e ....... 296 
 
 Exponential Equations ....... 297 
 
 Compound Interest by Logarithms , . . . . 297 
 
 Present Worth of an Annuity 298 
 
 Explanation of Tables 299 
 
 XXII. Indeterminate Equations 
 Problems 
 
 305 
 309 
 
 XXIII. Theory of Equations and Solution of Numerical Higher 
 Equations . 
 Reduction to the Normal Form . 
 Test for Roots .... 
 
 Short Method of Substitution . 
 
 Test for Roots by Division 
 Number and Character of the Roots 
 
 Normal Equation has no Fractional Roots 
 
 An Equation has n Roots 
 
 Formation of an Equation from Roots 
 
 Imaginary Roots in Conjugate Pairs 
 
 Negative Roots 
 
 Descartes's Rule of Signs 
 Solution for Commensurable Roots 
 Equations with only Even or only Odd Powers 
 Relation of Roots to Coefficients . 
 Situation of Roots 
 Multiple Roots .... 
 
 Method by H. C. D. of /(x) and /'(a) 
 
 Better Methods 
 Solution for Incommensurable Roots 
 
 Horner's Method 
 
 Diminution of Roots by a Given Quantity 
 
 Approximate Value of a Fractional Root 
 
 Principle of Horner's Method . 
 Observations on Horner's Method 
 Examples and Problems 
 The nth Root of a Number . 
 
 Sturm's Theorem and Method . 
 The Theorem for Situation of Roots 
 
 310 
 311 
 313 
 314 
 314 
 315 
 315 
 316 
 316 
 316 
 317 
 318 
 320 
 322 
 323 
 325 
 328 
 328 
 329 
 331 
 331 
 331 
 334 
 335 
 336 
 337 
 339 
 340 
 341 
 
14 CONTENTS 
 
 PAGE 
 
 Recurring or Reciprocal Equations 347 
 
 Binomial Equations 350 
 
 Cardan's Solution of the General Cubic Equation . . 351 
 
 Descartes's Solution of the General Biquadratic Equation . 353 
 
 XXIV. Series 354 
 
 Convergency of Series 354 
 
 Scale of Relation of a Recurring Series .... 360 
 
 The nth Term of a Series 363 
 
 Summation of a Series ....... 365 
 
 By the Method of Decomposition 366 
 
 By the Scale of Relation 370 
 
 By the Method of Differences 372 
 
 Piles of Spherical Shot 373 
 
 Interpolation 375 
 
 XXV. Permutations and Combinations 379 
 
 Permutations 379 
 
 Problems 381 
 
 Combinations . 383 
 
 Problems 387 
 
 Answers . 389 
 
HIGHER ALGEBRA 
 
 CHAPTER I 
 DEFINITIONS AND NOTATION 
 
 1. Quantity is amount or extent, and is expressed in terms of a 
 unit of the same kind. 
 
 Thus, 10, 5 bushels, 6 tons, 50 miles, 7 years, 100 dollars, m square feet, 
 n cubic yards, are quantities, the units in order being the abstract number 1, 
 1 bushel, 1 ton, 1 mile, 1 year, 1 dollar, 1 square foot, 1 cubic yard. 
 
 2. Two or more quantities of the same kind are Commensurable 
 or Incommensurable with reference to one another according as 
 they can or can not be measured with the same unit. 
 
 Thus, if the sides of a rectangle are 3 feet and 4 feet respectively, the 
 diagonal is 5 feet, and is, therefore, commensurable with the sides ; but the 
 diagonal of a square is incommensurable with its sides, being V2 times one 
 of the sides. 
 
 3. An Incommensurable Quantity, without comparison with an- 
 other quantity, is one that cannot be exactly expressed in the 
 decimal notation. 
 
 Thus, V5 is an incommensurable quantity, being 2 plus a decimal fraction 
 which never terminates. 
 
 4. Algebra is that branch of pure mathematics which treats of 
 numbers as expressed by symbols having general values, and of 
 the nature, transformations, and use of equations. 
 
 5. Algebra differs from arithmetic in the following important 
 particulars : 
 
 15 
 
}'o HIGHER ALGEBRA 
 
 1st. In arithmetic values are counted in only one direction 
 from 0; while in algebra, by means of positive and negative 
 quantities, values are counted in two opposite directions from 0. 
 
 2d. Arithmetical quantities, denoted by figures, have each a 
 single, definite value ; while algebraical quantities, represented by 
 letters, may have any value we choose to assign to. them. These 
 quantities can be recognized anywhere in the operation, and the 
 results are, therefore, general formulae instead of special answers. 
 
 3d. In arithmetic a problem is solved by analyzing it, step by 
 step ; while in algebra the conditions are expressed in equations 
 involving one or more unknown quantities, usually representing 
 the answer or answers. 
 
 6. Algebra employs five kinds of symbols, viz. of quantity, of 
 operation, of relation, of aggregation, and of abbreviation. 
 
 7. The Symbols of Quantity commonly employed are the fol- 
 lowing : 
 
 1st. The Arabic figures. 
 
 2d. The letters of the Roman alphabet, known quantities being 
 represented by the leading letters, and unknown quantities by the 
 final letters. 
 
 Similar quantities employed in the demonstration of a theorem 
 or the solution of a problem are often represented by the same 
 letter with different accents, as a', a", a'", etc., read u a prime," 
 " a second," " a third," etc., or by the same letter with different 
 subscripts, as a lf a 2 , a 3 , etc., read " a sub-one," " a sub-two," " a 
 sub-three," etc. 
 
 Initial letters are sometimes used, as r for radius, s for sum, d 
 for difference, etc. 
 
 The Greek -* is used for the ratio of the circumference of a 
 circle to its diameter. 
 
 3d. Zero, 0, used to denote not only the absence of value, but 
 also a quantity that is less than any assignable value. 
 
 4th. Infinity, oo, used to represent a quantity that is greater 
 than any assignable value. 
 
DEFINITIONS AND NOTATION 17 
 
 8. The Symbols of Operation in algebra are those common to all 
 branches of mathematics, the principal ones being the following : 
 
 1st. The sign of addition, + , read "plus." 
 2d. The sign of subtraction, — , read "minus." 
 
 3d. The sign of multiplication, x, read "times," "into," or 
 " multiplied by." 
 
 Multiplication is indicated also by a dot between the quantities. 
 
 The quantities between which multiplication is indicated are 
 called Factors, and the result of the multiplication is called the 
 Product. 
 
 4th. The sign of division, ■+-, read "divided by." 
 Division is indicated also by writing the dividend above, and 
 the divisor below, a line, in the fractional form. 
 
 5th. The sign of evolution, -y/, called the radical sign, and read 
 " the square root of," the square root, or second root, being one of 
 the two equal factors into which a number is conceived to be 
 resolved. The 3d, 4th, or nth root, by which is meant one of the 
 3, 4, or n equal factors into which a number is conceived to be 
 resolved, is indicated by writing 3, 4, or n in the vertex of the 
 angle of the sign, this number being called the Index. 
 
 9. Besides the above signs, or symbols of operation, there are 
 certain positions of quantities with reference to other quantities 
 that indicate operations, the principal ones being the following : 
 
 1st. A Coefficient, which is a quantity written beside another 
 quantity to show how many times the latter is taken. 
 
 When algebraic quantities are written in succession with no 
 sign between them, their product is signified, and any factor, or 
 the product of any number of factors, is the coefficient of the 
 product of the remaining factors. 
 
 Thus, in 5 mnx, 5 is the coefficient of mnx, 5 m is the coefficient of nx, 
 and 5 mn is the coefficient of x. 
 
 The name is usually applied to the numerical factor, and when 
 this is 1, it is not written. 
 
 DOWNEY'S ALG. — 2 
 
18 HIGHER ALGEBRA 
 
 2d. An Exponent, which is a quantity written at the right of 
 and above another quantity, its meaning being as follows : 
 
 (a) When an exponent is a positive integer, it indicates that the 
 quantity affected by it is to be taken as a factor as many times 
 as there are units in the exponent. The result of the multiplica- 
 tion is called a Power ; hence a positive integral exponent is said 
 to indicate a power. 
 
 Thus, a • a = a 2 , read " a square," " a 2d power," or " a 2d " ; 
 
 a • a • a = a 3 , read "a cube," "a 3d power," or "a 3d" ; 
 a • a • a • a = a 4 , read " a 4th power," or " a 4th " ; 
 a • a • a ••• to n factors = a n , read " a nth power," or " a nth." 
 
 (b) When an exponent is a positive fraction,* the numerator 
 indicates a power and the denominator a root of the quantity 
 affected by it. 
 
 Thus, 32^ is the 4th power of the 5th root of 32, or the 5th root of the 4th 
 power of 32. 
 
 (c) When an exponent is negative, it indicates the reciprocal of 
 what it would indicate if it were positive, the reciprocal of a 
 quantity being 1 divided by that quantity. 
 
 Thus, a-2 = 1 and a 2 &" 3 = -• 
 
 a 2 W 
 
 10. The Symbols of Relation are the following : 
 
 1st. The sign of equality, =, read "equals," or "is equal to." 
 
 2d. The signs of inequality, > and <, read respectively, "is 
 greater than," and " is less than." 
 
 The signs, =£, >, and <, read respectively, "is not equal to," 
 "is not greater than," and "is not less than," are also employed 
 to some extent. 
 
 3d. The sign of geometrical ratio, : , read " to," or " is to." 
 
 * It will be shown in Art. 147 that if we assume the index law as proved 
 for positive integral exponents to be general, the meaning of positive frac- 
 tional exponents and negative exponents must be as defined above. Until 
 that article is reached, their meaning will be treated as a matter of definition. 
 
DEFINITIONS AND NOTATION 19 
 
 4th. The sign of equality of ratios, : :, read "as," or "equals." 
 
 Thus, a:b :tc:d is read either " a is to 6 as c is to d," or " the ratio of 
 a to b equals the ratio of c to tV 
 
 5th. The sign of variation, oc, read " varies as." 
 
 11. The Symbols of Aggregation are the parentheses ( ), the 
 
 brackets [ ], the braces \ j, and the vinculum , and all 
 
 indicate that the algebraic expression included is to be treated 
 as a whole. 
 
 Thus, a — (o + c), a — [6 + c], a — {b + c}, and a — b + c all indicate 
 that the sum of b and c is to be subtracted from a. 
 
 The vinculum is sometimes vertical. 
 
 Thus, a 
 b 
 c 
 
 x is the same as a + b + c x x. 
 
 The line between the numerator and denominator of a fraction 
 has the effect of a vinculum. 
 
 b -4- c 
 Thus, a -£— is the same as a — $(b + c). 
 
 12. The Symbols of Abbreviation are the following: 
 
 1st. The signs of deduction, .*. and v, read respectively, "there- 
 fore " or " hence," and " since " or " because." 
 
 2d. The sign of continuation, •••, read "and so on," or "and so 
 on to." 
 
 Thus, 1 + 3 + 5 + 7 + -. -is read " 1 plus 3 plus 5 plus 7 and so on," and 
 a + ar + ar 2 + ••• ar n ~* is read M a plus ar plus ar 2 and so on to ar"- 1 .'* 
 
 3d. The factorial sign, L, which indicates the product of all the 
 integers from 1 to the number written in the angle inclusive. 
 
 Thus, [8 means 2 x 3, [_5 means 2 x 3 x 4 x 5, [n means 2 x 3 x 4 ••• n. 
 
 13. An Algebraic Expression is a quantity, simple or made up 
 of parts, expressed in algebraic symbols. 
 
20 HIGHER ALGEBRA 
 
 14. The Terms of an algebraic expression are the parts con- 
 nected by the signs -f- and — . 
 
 15. An algebraic expression is called a Monomial, a Binomial, 
 a Trinomial, a Quadrinomial, or a Polynomial, according as it 
 consists of one, two, three, four, or many terms. 
 
 The term Polynomial is applied also, in a general way, to any 
 algebraic expression consisting of more than one term. 
 
 16. Similar Terms are terms having the same literal quantities 
 affected with the same exponents. 
 
 Similar terms can differ, therefore, only in their numerical 
 coefficients. 
 
 17. Positive and Negative Quantities. The signs -f and — , 
 besides indicating the operations of addition and subtraction, are 
 used to distinguish quantities of opposite character. 
 
 Thus, if distance east is considered + , or positive, distance west must be 
 considered — , or negative. If gains are +, losses are — . If temperature 
 above is + , temperature below is — . If the latitude of a place north of 
 the equator is -f , that of a place south of the equator is — , while that of a 
 place on the equator is 0. If a force tending to move a body in one direction 
 is +, a force tending to move it in the opposite direction is — . And, in 
 general, quantities which contribute to one result being considered positive, 
 those which contribute to the opposite result must be considered negative. 
 
 As quantities on one side of are -f, and those on the opposite 
 side of are — , positive quantities are sometimes said to be 
 greater than 0, and negative quantities less than 0. 
 
 For example, if a man possesses nothing and owes one hundred dollars, he 
 must earn one hundred dollars before his capital can be expressed by 0, and 
 we say he has a hundred dollars less than nothing. 
 
 An increase in the numerical value of a positive quantity 
 increases its algebraic value; but an increase in the numerical 
 value of a negative quantity decreases its algebraic value. 
 
 18. An Axiom is a truth which is assumed as self-evident. 
 
 19. A Theorem is a formal statement of a truth requiring 
 proof. 
 
DEFINITIONS AND NOTATION 21 
 
 20. The Hypothesis of a theorem consists of the conditions on 
 which it is affirmed. 
 
 21. A Problem is a question proposed for solution. 
 
 22. The Solution of a problem is the process of obtaining the 
 result sought. 
 
 23. A Rule directs how to proceed in solving problems that 
 belong to the same class. 
 
 24. A Demonstration is the course of reasoning by which the 
 truth of a theorem, or the correctness of a solution or a rule, is 
 established. 
 
 25. A Corollary is an inference from a preceding theorem, dem- 
 onstration, or solution. 
 
 26. A Scholium is a remark upon some feature of what has 
 preceded. 
 
CHAPTER II 
 ADDITION 
 
 27. The Algebraic Sum of several quantities is the excess of the 
 positive over the negative, or of the negative over the positive 
 quantities. 
 
 Thus, if a man's assets are $ 5000 and his liabilities $ 3000, the excess of 
 his assets over his liabilities is $ 2000. Now, if we consider assets positive 
 and liabilities negative, the algebraic sum of his assets, + $ 5000, and his 
 liabilities, -$3000, is +$2000. 
 
 Again, if a force of 50 pounds is exerted to move a body in one direction, 
 and a force of 30 pounds to move it in the opposite direction, the effective 
 force, i.e. the aggregate or the algebraic sum of the two forces, is 20 pounds. 
 
 In the same way the algebraic sum of + 8, +3, — 5, + 2, and — 4 is 4, 
 and the algebraic sum of +7, — 6, — 9, and +1 is — 7. 
 
 28. Algebraic Addition is the process of finding the algebraic 
 sum of several quantities. 
 
 29. Prob. To add monomials. 
 
 Rule. 1st. If the quantities are similar {Art. 16), find the 
 algebraic sum of the coefficients and annex the common literal part. 
 
 2d. If the quantities are dissimilar, tvrite them in succession with 
 their signs unchanged. 
 
 Dem. 1st. Let it be required to add 
 
 5 x*y, 7 x 2 y, — 4 x?y, and — 2 x*y. 
 
 By Art. 9, 1st, 5 x 2 y is 5 times x*y, and 7 x 2 y is 7 times the 
 same quantity. Now, 5 times any quantity and 7 times the same 
 quantity are 12 times that quantity, giving, in this case, 12 arfy. 
 By Art. 17, — ^x 2 y is so opposed in character to + 12 arfy as to 
 destroy or neutralize 4 of the 12 times v?y, giving -f 8 x 2 y. Simi- 
 larly, — 2 x 2 y destroys 2 of the 8 times x 2 y, giving for the sum 
 of the four terms + 6 x 2 y, in which the common literal part is 
 
 22 
 
ADDITION 
 
 23 
 
 simply annexed to the algebraic sum of the coefficients. The 
 same reasoning applies to any other set of quantities. 
 
 2d. Let it be required to add +3a, + 4 6, — 2 c, and — d. 
 
 Now, 3 a and 4 b are not respectively 3 and 4 times the same 
 quantity ; hence the sum will not be 7 times either quantity, and 
 we can only indicate the addition, giving for the sum of the two 
 3 a + 4 b. For the same reason the addition of — 2 c and — d to 
 this sum can only be indicated, the minus signs being preserved, 
 since these negative terms neutralize a part of the positive sum. 
 Hence the entire sum is 
 
 3a + 4:b-2c-d, 
 
 in which the given quantities are written in succession with their 
 signs unchanged. 
 
 30. Cor. Adding a negative quantity is the same as subtracting 
 a numerically equal positive quantity. 
 
 31. Prob. To add polynomials. 
 
 Rule. Write the quantities so that similar terms shall be in the 
 same vertical column. Add the columns separately and connect 
 their sums by the resulting signs. 
 
 Dem. The columns may be added separately by Art. 29, 1st. 
 The sums being dissimilar, the addition of them, according to 
 Art. 29, 2d, can only be indicated by connecting them by the 
 resulting signs. 
 
 32. In writing several similar terms in column for convenience 
 in adding, much time is saved by writing the literal factors only 
 once. The plus signs also may be omitted, inasmuch as when no 
 sign is written, the plus sign is understood. 
 
 7 a%W 
 -3 
 -5 
 2 
 6 
 9 
 
 
 ' + 7 a 3 b 2 c* 
 
 
 - 3 aWc* 
 
 
 - 5 a 3 & 2 c* 
 
 Instead of . 
 
 + 2 « 3 fc 2 C* 
 
 
 - aVS-d 
 
 
 + 9«3ft2 c 4 
 
 
 + 4 a*b 2 c* 
 
 ' 7 
 
 a 3 6V . ' 
 
 -3 
 
 
 -5 
 
 
 2 
 
 or { 
 
 -6 
 
 
 
 
 
 [ 4 
 
 a 8 6- 2 c* 
 
 4 aWc* 
 
24 HIGHER ALGEBRA 
 
 While the first operation, exclusive of the answer, contains 48 
 figures, letters, and signs, either of the others contains but 15. 
 
 33. Note. Although the principles of fractions will be treated 
 in their proper place, we shall from the beginning assume such 
 knowledge of them as comes from the study of Arithmetic. 
 
 EXAMPLES I 
 Add the following : 
 
 1. 2 a^ — 3 x 2 y — 4 xy 2 -f 6 y 2 , 4 x 2 y — xy 2 — 5 a? 3 , 4 x 2 — 5 x*y — y 3 
 + 2x>-3y 2 , x 2 + 8f-2y 2 + 7xy 2 , 2 x*y + 3 xy 2 - 3 y*, and 
 2tf-4;y s + 3y 2 + x 2 . 
 
 Operation 
 
 2 x 2 - 3 x?y - 4 xy 1 + 6 y 2 
 
 4 - 1 - 5 x 3 
 
 4-5 -3 2-?/3 
 
 1 7-2 8 
 
 2 3 -3 • 
 
 1 3 2-4 
 
 8 x 2 - 2 x 2 y + 5 xy 2 + 4 y°- - x* 
 
 2. m 2 — 3 mn -\- 2 n 2 , 3 n 2 — m 2 , and 5 mn — 3n 2 + 2 m 2 . 
 
 3. 2ab-3ax 2 -\-2a 2 x, 12 ab + 10 ax 2 - 6 a 2 x, and aar 3 -8a& 
 — 5 a 2 #. 
 
 4. a? 4 — 4 a^i/ -f 6 x?y 2 — 4 sci/ 3 + y 4 , 4 afy — 12 x?y 2 + 12 sc?/ 3 — 4 ?/ 4 , 
 6 #y — 12 xy 3 + 6 ?/ 4 , 4 ajz/ 3 — 4 ?/ 4 , and y\ 
 
 5. 5 a 2 ^ + 4 a 2 fo» 2 + m#y and 10 a 2 ex 2 — 2 arbx 2 4- 6 mx*y 2 . 
 
 6 . 3a6 2 -4a 2 6 + a 3 ,.5a& 2 -4ac 2 -c 3 , and 2a 2 6- 7a& 2 - 6ac. 
 
 7. 2 a 2 4- 5 ab — xy, — 7 a 2 + 3 ab — 3 xy, — 3 a 2 — 1 ab + 5 xy, 
 and 9 a 2 — ab — 2 xy. 
 
 8. 5 aW - 8 a 2 b 3 + x*y + ^ 2 , 4 a 2 6 3 - 7 a 3 6 2 - 3 ^ 2 + 6 afy, 
 3a 3 b 2 + 3a 2 b 3 -3x 2 y + 5xy 2 , and 2 a 2 6 3 - a 3 b 2 - 3 x*y - 3 xy 2 . 
 
 9. a 2 -6 2 +3a 2 7;-5a& 2 , 3a 2 -4a 2 &+3 6 3 -3a& 2 , a 3 +b 3 +3a 2 b, 
 2a 3 -4,b 3 -5ab 2 , 6a 2 b + 10ab 2 , and -6a 3 -7 ci'b + Aab 2 + 2b 3 . 
 
 10. 2 m 2 4- ^ ?i 2 — | m 2 7i 4- 3 mn 1 — mn, -| n 2 — 2 mn 2 4- 3 m 2 ?i, 
 
 iii 2 — £ 7i 2 4- 4 mn 4- J m 2 w, and i mn a 4- 3 m 2 — 3 mn 4- 2 w 2 . 
 
ADDITION 25 
 
 11. 2a^-4x* + r 5 , 5 x?y — ab + x* , 4ar-ar t , and 2x$— 3+2 a£ 
 
 12. ax + 2by + cz, Vc+Vy + Vz, _ 3Vy — 2VaJ + 3Vs, 
 4 cz — 3 ax — 2 by, and 2 ax — 4Vy — 2Vz. 
 
 34. Literal terms that are similar with reference to only part 
 of their factors, may be united into one term with a polynomial 
 coefficient. 
 
 Thus, 7 ax, — 2 bx, and 3 ex are similar with reference to x only, being, 
 respectively, 7 a, — 2 b, and 3 c times x, and the sum is (7 a — 2 b + 3 c) 
 times X, or (7 a — 2 6 + 3 c)x. 
 
 35. Compound terms that have a common compound, or poly- 
 nomial factor may be added with reference to that factor. 
 
 Thus, b(x 2 — y), Z(x 2 — y), and — 2(x 2 — y) are similar with reference to 
 the quantity x 2 — y, being, respectively, 5, 3, and — 2 times x 2 — y, and the 
 sum is (5 + 3 — 2) times (x 2 — y), or 6(x 2 — y). 
 
 EXAMPLES II 
 Add the following: 
 
 1. ax, bcx, — 3bx, —2 ex, and 4 x. 
 
 2. ax — 2 by, 2bx—3 by, and cy — ax. 
 
 3. ax + 2 6y — 4 z, 3 y — cz, and 4 x — 6 # + 2 cz. 
 
 4. a (x -f y) -f 6 (x — y) and m (x + y) — n (x — y). 
 
 5. a + 6 Vl — c 2 and a — b Vl — c 2 . 
 
 6. 5 V# — ?/, 5 a Vc — y,— 3vx — y, and — a Ve — y. 
 
 7. | Va 2 — «, — -§ Va 2 — a?, and Va 2 — #. 
 
 8. ^L + ?_26 ? i^-2 + 86, and --^-™_3&. 
 
 Vie 2/ Vc 2/ Ve 2/ 
 
 9. (a+b-c)^/m i -x 2 , (a-b + c)Vm 2 -x 2 , and 3aVm 2 -^. 
 
 10. (a + b - c)</x T ^-f, (a - b + c)W - ?, and 
 
 (b + c — c^Vic 2 — 2/ 2 . 
 
CHAPTER III 
 SUBTRACTION 
 
 36. The Algebraic Difference of two quantities is their difference 
 with regard to their character as positive and negative, the order 
 of subtraction being independent of their magnitudes. 
 
 While in finding the arithmetical difference the less quantity is 
 always taken from the greater, both quantities being regarded as 
 positive, in finding the algebraic difference either quantity may 
 be taken from the other, and one or both of the quantities may be 
 negative. 
 
 Thus, suppose that a merchant who had $ 500 bought goods to the amount 
 of $ 700, and we are asked to state how much money he had left. In an 
 arithmetical sense the question is absurd, inasmuch as he not only had 
 nothing left, but went in debt $ 200. In an algebraic sense, however, the 
 question is not absurd, the answer being — $200. 
 
 Again, if one man has $ 500 and owes nothing, and another man owes 
 8200 and has nothing, the difference of their possessions, a debt being 
 regarded as a negative possession, is $ 700. 
 
 37. Algebraic Subtraction is the process of finding the algebraic 
 difference between two quantities. 
 
 38. It is seen that subtraction is the inverse of addition, the 
 problem being to find one of two quantities when the other and 
 the sum of the two are given. 
 
 39. The terms Minuend and Subtrahend are used in Algebra as 
 in Arithmetic, but Addition, Subtraction, Sum, and Difference (or 
 Remainder) have a more general signification. In Arithmetic 
 addition always produces an increase, and subtraction a decrease ; 
 but in Algebra addition may produce a decrease and subtraction 
 an increase. 
 
 26 
 
SUBTRACTION 27 
 
 40. Prob. To subtract one quantity from another. 
 
 Rule. Conceive the signs of all the terms of the subtrahend to 
 be changed, then proceed as in addition. 
 
 Dem. Let it be required to subtract m — n from a. 
 
 If the whole of m be subtracted from a, the result will be 
 a — m. Now since the quantity subtracted is too large by », the 
 remainder is too small by n : hence to make it what it should be, 
 we add n to the result, giving a — m + n. In this operation we 
 see that we have changed the signs of the subtrahend and added 
 the result to the minuend. 
 
 41. Cor. Subtracting a negative quantity is the same as adding 
 a numerically equal jyositive quantity. 
 
 EXAMPLES III 
 From 3 a 3 - 4 a 2 b + 5 ab 2 -6 b 3 take a 3 + 2 a 2 b - 3 ab 2 - 8 b\ 
 
 Operation 
 1 2 -3 -8 
 
 2 a 3 - 6 a 2 6 + 8 ab 2 + 2 b* 
 
 2. From 3x 3 -2x 2 -x-l take 2x*-3a? + x + l. 
 
 3. From x 2 -f 2 xy + y 2 take x 2 — 2 xy -f y 2 . 
 
 4. From 1 + 3x + 3x 2 + x 3 take 1 - 3z + 3a 2 - ic 3 . 
 
 5. From 10ar 5 -8a 2 + 6a; + 4 take 7 a- 5 + 5ar> - 3 x + 7. 
 
 6. From 3 a - 3 a 2 + 1 + a 3 take 1 - a 3 - a - a 2 . 
 
 7. From a 2 + 2 a& -f b 2 take the sum of — a 2 -f- 2 at — 6 2 and 
 -2a 2 + 2fc 2 . 
 
 8. From 2 x 4 -|- 4 a 3 ?/ — 5 x*y 2 — 3 .r?/ 3 -f- ?/ 4 take x 4 — x?y — 3 x?y 2 
 + 6xy s + y\ 
 
 9. From the sum of 2a— 3 6 + 4 d and 2 6 + 4 c — 3d take 
 the sum of 3c — 4a — 4 fr — 2d and 3 a — 2 c. 
 
 10. From 1 + 3Vie + 3a: + Va? take 1 - 3 Vx + 3 a; — Vx?. 
 
28 HIGHER ALGEBRA 
 
 11. From x^ -\- 2 x^if 5 -f y z take x 5 — 2 x%y* + y* 
 
 12. From 5(x + y) + 3(x-y) take 2 + y) - 3 (a — y). 
 
 ' 13. From a + ?>Vl — c 2 take a — b Vl — c 2 . 
 
 14. Subtract l^a + ar' from J Va + a£ 
 
 15. Subtract (b — a — c) Vo 2 + ?/ 2 from (a — & + (^Var 2 4- 2/ 2 . 
 
 16. Subtract b yjx + y — a^/x — y from dV# + y + & V# — y. 
 
 17. What must be added to 4^—3^+2 to produce Ax 3 
 + 7x-6? 
 
 18. To what must X s + 3 arfy + 3<n/ 2 + y 3 be added to produce 
 3a?-x 2 y-2xy 2 + 4:f? 
 
 19. What must be subtracted from 5a — 46 + 3c to leave 
 2a + 2b-c? 
 
 20. From what must V# + y — bvx -\-y — d(x — y) be sub- 
 tracted to leave 5 V# + 2/ + a V* + ?/ + c ( x — if) ? 
 
 SIGNS OF AGGREGATION AS RELATED TO ADDITION 
 AND SUBTRACTION 
 
 42. Theorem. 1st. A parenthesis, or other sign of aggregation, 
 may be removed without changing the signs of the inclosed terms, if 
 it is preceded by +, and by changing the signs of the inclosed, terms, 
 if it is preceded by — . 
 
 2d. Conversely, terms may be inclosed within a parenthesis, or 
 other sign of aggregation, without changing their signs, if it is pre- 
 ceded by ■{-, and by changing their signs, if it is preceded by — . 
 
 Dem. 1st. A sign of aggregation indicates that the inclosed 
 terms are to be treated as a whole (Art. 11) ; hence, if preceded 
 by + , all the inclosed terms are to be added, which does not 
 change their signs (Art. 29), and if preceded by — , all the terms 
 are to be subtracted, which changes their signs (Art. 40). 
 
 2d. The truth of the converse is seen from the fact that the 
 removal of the sign of aggregation by the preceding principle 
 would restore the expression to its original form. 
 
SIGNS OF AGGREGATION 29 
 
 43. Sch. When signs of aggregation occur within other signs 
 of aggregation, they may be removed in succession, and it is 
 usually expedient to begin with the innermost. 
 
 Thus, a- [b + {c-(d-e -/)}] 
 
 = a-[6 + {c-(d-e+/)}] 
 = a- [6 + { C -rt + e-/}] 
 = a-[b + c-d + e-f] 
 = a — b — c+d — e+f. 
 
 EXAMPLES IV 
 Perform the following indicated operations : 
 
 1. 2x 2 -\5x-(x 2 + 2x-5)\. 
 
 2. 4a-3&-{2& + 3-(2a + & + l)}. 
 
 3. 7o-[3o- {4a-(5a-2)}]. 
 
 4. ^_z_(2a;-(3^ + 4-(z-l))). 
 
 5. 2 m — [3 m -[m-(2?»i-3m + 4)|-(5m- 2)]. 
 
 6. 4 z - (12x - 2 y) - \2 z - (10 x + 4y)- (2 x-6y)\. 
 
 7. x - [y + z - \x - (- x - y) + z\] + \z - (2 x - y)\. 
 
 In the following unite terms with reference to similar factors, 
 preceding each polynomial coefficient by — : 
 
 8. ax 3 + 2bx i -x-3x i + 4:x 2 + 5x-cx 2 -2dx. 
 
 9. ax 2 + arx 3 — bx 2 — 5 x 2 — co 3 . 
 
 10. oa 3 — 2 c\c — [frc 2 — j c» — c?a? — (dx 3 + 3 ex 2 ) \ — (ex 2 — bx)]. 
 
CHAPTER IV 
 MULTIPLICATION 
 
 44. Multiplication is the process of finding a quantity which 
 shall be as many times a specified quantity, or such a part of that 
 quantity, as is represented by a specified number. 
 
 The terms Multiplicand, Multiplier, Product, and Factors are 
 used as in Arithmetic. 
 
 It follows from the definition that multiplication by a positive 
 number is the repeated addition of the whole or a part of the 
 multiplicand to ; while multiplication by a negative number is 
 the repeated subtraction of the whole or a part of the multiplicand 
 from 0. 
 
 771 
 
 Thus, a x b is a repeated b times, and a x — is one nth part of a 
 
 repeated m times ; while a x(—b) is a subtracted b times, and a x ( — — ) 
 
 is one nth part of a subtracted m times. 
 
 The only difference between - x m and ax- is that in the former 
 
 n n 
 
 one nth part of a is represented as already taken, and this nth part is to be 
 
 repeated m times ; while in the latter one nth part of a is first to be taken, 
 
 and then this nth part is to be repeated m times. 
 
 45. Theorem. The product has the same sign as the multiplicand 
 when the multiplier is positive, and the opposite sign when the 
 multiplier is negative. 
 
 Dem. Repeated addition of the whole or a part of the multipli- 
 cand would not change the sign (Art. 29), while repeated subtrac- 
 tion of the whole or a part of the multiplicand would change the 
 sign (Art. 40). 
 
 46. Cor. i. When there are but two factors, like sig7is give +, 
 and unlike signs give — . 
 
 30 
 
MULTIPLICATION 31 
 
 47. Cor. 2. The product of an even number of negative factors 
 is -f , and of an odd number —. 
 
 48. Theorem. The exponent of a quantity in the product is the 
 sum of the exponents of the same quantity in the multiplicand and 
 multiplier. 
 
 Dem. ' 1st. When the exponents are positive integers. 
 Let it be required to multiply a m by a'\ By definition (Art. 
 9, 2d) 
 
 a m x o" = aaaa • • • to m factors x aaaa • • • to n factors 
 = aaaaaa • • • to m + n factors = a m+n . 
 
 2d. When the exponents are positive fractions. 
 
 TO p 
 
 Let it be required to multiply a" by a q . 
 
 m mq 
 
 Now a H — a nq \ for while the denominator of the exponent in 
 the latter indicates that a is resolved into q times as many 
 factors as in the former, the numerator indicates that q times as 
 
 p np 
 
 many of these factors are taken. In like manner a q = a nq . 
 Therefore, 
 
 m p mq np 1 1 
 
 a" x a q = 0^X0,^ = (aF q ) mq x (aF q ) np , 
 which, by the first part of the demonstration, is 
 
 1 mq+np m p 
 
 ( a n~qjmq+»P = a »? = a « + 9. 
 
 3d. When the exponents are negative. 
 Let it be required to multiply a~ m by a~ n . 
 By definition (Art. 9, 2d) 
 
 a~ m x or* = — X — 
 
 a m a n 
 
 Now, as fractions are multiplied by multiplying the numera- 
 tors together for a new numerator, and the denominators together 
 for a new denominator, we have by the first part of the demon- 
 stration, 
 
 cr w x a~ n = — x — = -=- = a~ n ~ n . 
 
32 HIGHER ALGEBRA 
 
 EXAMPLES V 
 
 Show that the following results are true, whether the multipli- 
 cation is before or after performing the operations indicated by 
 the exponents : 
 
 l. 81* x 81* = 19683. . 2. 16-* x 16"* = fc 
 
 3. 25~* x 25^ = 1. 4. a~ 2 x a 3 = a. 
 
 Find the products of the following : 
 5. 4 a 2 b s and 5 a 3 b 2 . 6. 12 cr*bc 2 and 6 a 4 b~ 2 c. 
 
 7. 14afy^ and —3 x'hfK 8. — 20a£y* and — 8aA/ _ i 
 
 9. 5 a& 2 , - 2 a 2 o 3 , - be, and 3 a 2 6c 2 . 
 
 10. 3 ax, — 3aV, 4 6?/, — ?/ 3 , and 2#y. 
 
 11. 7 a?* x*y, — 2 a?2/3 ? an d 3 *%». 
 
 12. 3a 2 6- 3 c'", -4a6c*, 2a~ 3 oc- 1 , and 9a 3 6 2 . 
 
 13. 6afy*, -5x 2 y, —x n y m , 3x~Y, and — 2x 2m y 5n . 
 
 49. Prob. To multiply one polynomial by another. 
 
 Kule. Multiply each term of the multiplicand by each term of 
 the multiplier and add the products. 
 
 Dem. Let it be required to multiply any polynomial by 
 a-\- b — c. 
 
 If we multiply it by a and then by b and add the results, we 
 shall have taken it a + b times. But the multiplier required that 
 it be taken only a + b minus c times. Hence from a + b times the 
 given polynomial we must deduct c times the same polynomial. 
 But subtracting this product is the same as adding it with its 
 signs changed, and these are changed by regarding the minus 
 sign of c in multiplying. 
 
 50. The Degree of a Term is the sum of the exponents of its 
 literal factors. 
 
 51. The Degree of a Polynomial is the same as that of its term 
 of highest degree. 
 
MUL TIPLICA TION 33 
 
 52. A polynomial is said to be Arranged with reference to a cer- 
 tain letter when the exponents of that letter increase or decrease 
 in the successive terms. 
 
 53. A Homogeneous Polynomial is one in which all the terms 
 are of the same degree. 
 
 54. Theorem. The product of two homogeneous polynomials is 
 itself homogeneous. 
 
 Dem. If the degree of every term of the first polynomial is m 
 and of the second ?i, then the product of any term of the first by 
 any term of the second will be of the (m + n)th degree. Hence 
 the product will be homogeneous. 
 
 55. If multiplicand and multiplier are arranged with reference 
 to the same letter, the product will have a similar arrangement, 
 and the work will be more systematic than it otherwise would be. 
 When terms are missing from an arranged multiplicand, their 
 places should be supplied by terms with coefficient zero. For 
 convenience in adding, similar terms of the partial products are 
 written in the same column. The literal factors should be ivritten 
 but once, as the writing of a coefficient under another sufficiently 
 indicates that the literal factors are the same. 
 
 EXAMPLES VI 
 Multiply the following : 
 
 1. 6 x 3 - 4 tfy - 2 xy 2 + 3 y 3 by 2 x 2 - 3 xy + 4 y 2 . 
 
 Operation 
 
 6x 3 - 4x 2 y- 2xy 2 + 3y 3 
 2x 2 - Sxy + 4 y 2 
 
 9xy* 
 
 8 
 
 
 12 x 5 - $x*y- 4x 3 y 2 + 6x 2 y s 
 - 18 12 6 - 
 24 - 16 
 
 12 y 5 
 
 12 x 5 - 20 x*y + 32 xty 2 - 4 x 2 y 3 - 17 x\f + 12 y* 
 
 2. 3 a 2 - 4 ab + 6 b 2 by 2 a 2 - 5 ab + 3 b 2 . 
 
 3. 3 x 2 + 2 x - 4 by x 2 - 3 x + 2. 
 
 DOWNEY'S ALG. 3 
 
34 HIGHER ALGEBRA 
 
 4. 2 ar 3 + 4 - 3 x - 3 x 2 by 3 x - 2 + 4 a* 
 
 5. # 4 — #y 4- y 4 by ar* 4- y 2 . 
 
 6. £ 2 -M?/ + 2/ 2 by tf — xy + y 2 . 
 
 7. m 4 + n 4 -f p 4 — m 2 n 2 — m 2 p 2 — n 2 p 2 by m 2 + ri 2 4- 1> 2 . 
 
 8. a m — a n + a 2 by a m — a. 
 
 9. 4 a 2 + 9 6 2 + c 2 + 3 6c + 2 ac - 6 a& by 2 a + 3 b - c. 
 
 10. a 2 + b 2 4- c 2 — ab — ac — 6c by a + b 4- c. 
 
 11. x 4 - 2 x 2 4- 3 x - 5 by x 3 - 2 a 2 + 3 x - 4. 
 
 Operation 
 
 a 4 40z 3 -2x 2 43x - 5 
 a3_2z 2 43;e -4 
 
 x 1 4 x 6 - 2 x 5 + 3 a 4 - 5 x 3 
 
 -2 4-6 10 x 2 
 
 3 _ 6 9-15* 
 
 - 4 8 - 12 20 
 
 x 7 - 2 x 6 4 x 5 + 3x 4 - 17a 3 4 27 x' 2 - 27 g + 20 
 
 12. 2 a 4 4- 3 x 2 - 4 by x 3 4-5 ar* - 6 x 4- 3. 
 
 13. 3 x 4 - 4 afy 2 - 2 ajy 8 + 5 ?/ 4 by 2 x 2 - xy 4- 3 y 2 . 
 
 14. m 5 — 7 m 3 4~ 4 m — 8 by m 3 — 3 m + 5. 
 
 15. * 4 + 3* 3 -2*4-5by* 3 -2^4-7^-3. 
 
 16. a^ 4- 4 ar 3 - 3 by 3 x 4 - 2 x 2 + 4. 
 
 17. 4 a* — 3 x 2f, y" 4- 6 a: p ?/ 9 4- 2 t/ 3 * by 2 a^ 4- 3 y q . 
 
 MULTIPLICATION BY DETACHED COEEFICIENTS 
 
 56. In examples like most of the foregoing, in which the 
 terms of both multiplicand and multiplier contain the same 
 letters, if the arrangement be made the same in both, the multi- 
 plication can be effected by using the coefficients alone, since 
 the literal factors in the product will follow the same law of 
 arrangement. 
 
MULTIPLICATION BY DETACHED COEFFICIENTS 35 
 
 EXAMPLES Vn 
 Multiply the following : 
 
 1. 6 ar 3 - 4 x*y - 2 a# 2 + 3 y 3 by 2 x 2 - 3 a# + 4 # 2 . 
 
 Operation 
 
 6_ 4- 2 + 
 
 2- 3+ 4 
 
 3 
 
 9 
 
 8 
 
 
 12- 8- 4 
 - 18 12 
 
 24- 
 
 6 
 
 6- 
 16- 
 
 12 
 
 12-26 32-4-17 12 
 Prod., 12 x b - 26 x*y + 32 x 3 y 2 - 4 xV - 17 xy* + 12 y 5 . 
 
 2. x* + 2 a; - 4 by ar* - 1. 
 
 Operation 
 
 1+0+2-4 
 1+0-1 
 
 10 2-4 
 
 -1 0-2 4 
 
 1 o 1-4-2 4 
 
 Prod., x 5 + x 8 - 4 x 2 - 2 x + 4. 
 
 3. 3 x 2_ a . 4 . 2 by 3ar J +-2a;-2. 
 
 4. ar'-S^ + Saj-l by af— 20 + 1 
 
 5. 3a 2 + 4:ax-5x 2 by 2 a 2 - 6 az + 4 x*. 
 
 6. 27ar J + 9afy + 3a*/ 2 + ^ by 3a;-?/. 
 
 7. 2ar } + 6ar J -4a;-3 by 3^-4^ -0 + 5. 
 
 8. m 3 +- 2 m 2 n + 2 ra?i 2 by ?n, 2 — 2 mn + 2 n* 
 
 9. a; 2 + 9a; + 20, 0*-70 + 12, and x 2 -2x-15. 
 10. 0* — 3*4.1, a^ + aj + l, and af— «+l. 
 
 When the student writes down the polynomials to be multiplied, 
 he should write them in the proper position for multiplying, and 
 
36 HIGHER ALGEBRA 
 
 then not make the unnecessary repetition of detaching the coeffi- 
 cients, but proceed at once as follows : 
 
 6 x 3 - 4x 2 y - 2 xyt 
 
 x 2 - Sxy + 4y 2 
 
 + 
 
 3 y* 
 
 
 12-8 - 4 
 
 - 18 12 
 
 24 
 
 
 6 
 
 6 
 
 16 
 
 - 9 
 
 - 8 
 
 12 
 
 12 x 5 - 26 x 4 y + 32 x s y 2 - 4 x 2 y* - 17 xy* + 12 y b 
 
 11. ar 5 - 2 ax 2 + 2 a 2 z - 3 a 3 by a 2 - 3ax + 2 a 2 . 
 
 12. #*+ 6afy + 12 xy 2 + 8y 3 by a?- 3afy-f 3a#* — if. 
 
 13. a j 4 -3ar , + a 2 + 4a;-6 by 2^-5^-3^ + 5. 
 
 14. or 5 -5a 4 + 13ar 3 -a; 2 -a + 2 by « 2 -2»-2. 
 
 SHORT METHODS OF MULTIPLICATION 
 
 57. In many special cases the process of multiplication may be 
 made very short, and in others the products may be written by 
 inspection. These methods are of so much importance and save 
 so much labor that they are here given by special rules and theo- 
 rems. The student should make himself thoroughly familiar with 
 them. 
 
 58. Prob. To multiply by a binomial of the first degree, the 
 coefficient of whose first term is unity, when multiplicand and multi- 
 plier contain but one letter, or when they contain two letters and are 
 homogeneous. 
 
 Rule. Arrange the multiplicand in descending powers of the first 
 letter of the multiplier, supplying the places of any missing terms by 
 terms with coefficient zero. Call the numerical part of the second 
 term a; multiply the coefficient of the first term of the multiplicand 
 by a, and add the product to that of the second; multiply the coeffi- 
 cient of the second term of the multiplicand by a, and add the prod- 
 uct to that of the third, and so on. The coefficient of the first term 
 of the multiplicand and these sums in order will be the coefficients 
 of the product, and the arrangement of the letters will be the same 
 as in the multiplicand, the degree being one greater. 
 
sin HIT METHODS OF MULTIPLICATION 37 
 
 Dem. Let it be required to multiply 
 
 3 x* — 2 x"y — 5 x 2 y~ + 4 xif — 6y 4 by x-\-3y. 
 
 By the process of Ex. 1, page 33, we have 
 
 3 a; 4 - 2 afy - 5 afy 2 + 4 ay 5 - 6 y A 
 x + 3y 
 
 3x? — 2 x 4 y — 5 x*y 2 -f- 4 a* 2 ?/ 3 — 6 #?/ 4 
 
 9 _ 6 -15 12 -18?/ 5 
 
 3 ic 5 + 7 afy - 11 afy 2 - 11 xFf + 6 xy* - 18 ^ 
 
 Since the coefficient of the first term of the multiplier is, by 
 hypothesis, unity, the coefficients of the terms of the first partial 
 product will always be the same as those of the multiplicand. 
 
 The coefficients of the terms of the second partial product are 
 a (in this case 3) times the coefficients of the terms of the multi- 
 plicand, and, being placed one term farther to the right to bring 
 similar terms together, we see that to obtain the coefficients of 
 the terms of the product, a (3) times the first coefficient of the 
 multiplicand is added to the second, a (3) times the second is 
 added to the third, and so on. Omitting the repetition of the 
 coefficients of the terms of the multiplicand in the first partial 
 product, the work stands as follows : 
 
 (x + 3y)(3x 4 -2x?y- 5a% 2 + Axf- G?/) 
 
 9 _ c -15 12 -18 
 
 3 7 -11 -11 6 -18 
 
 As shown before, the literal factors in the terms of the product 
 follow the same law of arrangement as in the multiplicand, and 
 since the multiplier is of the first degree, the degree of the prod- 
 uct is one greater than that of the multiplicand. Hence supply- 
 ing the literal parts we have for the product, 
 
 3 x 5 + 7 x 4 y - 1 1 xy -llxtf + G^-lSy 5 . 
 
 Only the a of the multiplier need be written ; and when the co- 
 efficients are small, as in this case, or so related that the sums are 
 
38 HIGHER ALGEBRA 
 
 small, the additions can be readily made without writing the in- 
 termediate products. 
 
 (3) 3x 4 -2x 3 y- 5afy 2 + ±xtf-§y 4 
 
 7 -11 -11 ^ 6 -18 
 
 Thus, 3 times 3 are 9, which added to — 2 gives 7 ; 3 times 
 — 2 are — 6, which added to — 5 gives —11; 3 times —5 are 
 — 15, which added to 4 gives — 11 ; 3 times 4 are 12, which 
 added to — 6 gives 6 ; 3 times — 6 are — 18, which added to 
 gives — 18. 
 
 Supplying the literal parts, we have for the product, 
 
 3 x 5 + 7 x 4 y - 11 x*y 2 - 11 afy 8 + 6 xy 4 - 18 y 5 . 
 
 EXAMPLES VIII 
 Multiply the following : 
 
 1. 4:X 5 -3x 4 y-Sx z y 2 + 2x 2 f-6xy 4 -15y 5 by x-6y. 
 
 Operation 
 
 (-6) 4a; 5 - 3x*y- 8x*y 2 + 2x 2 y s - 6xy* -15?/ 5 
 
 - 27 10 50 - 18 21 90 
 
 Prod., 4 x* - 27 x b y + 10 x 4 y 2 + 50 x*y* - 18 x 2 */* + 21 xy 5 + 90 y 6 
 
 If the student perform the multiplication in the usual way, 
 writing the partial products in full, he will find that, between 
 factors and product, he uses 70 figures, letters, and signs, while 
 by this process he uses but 14. 
 
 2. x 3 + 4 x-y + 7 xy 2 — 5 y* by x — 5y. 
 
 3. 2 x* - x*y + 7 tftf -20 xf + 75y 4 by x + 4#. 
 
 4. 4 m 5 — 8 m 4 w -f- 16 m 3 w 2 — 16 m 2 n z -f 8 mn 4 — 4 » 5 by m -j- 5 n. 
 
 5. 3 m 4 — 7 m 2 n 2 + 6n 4 by m-\-6n. 
 
 Operation 
 
 (6) 3 m 4 + m*n - 7 m%* + mn z +6n 4 
 
 18-7 -42 6 36 
 
 Prod., 3 wi 5 -(- 18 m*n - 7 m s n 2 - 42 w% 3 + 6 mn* + 36 w 1 
 
SHORT METHODS OF MULTIPLICATION 39 
 
 6. 5a 6 -6a 4 2/ 2 -4ay + 7?/ 6 by x-3y. 
 
 7. a 4 — 4 x?y + 3 xy 3 — 9 ?/ 4 by a + 7#. 
 
 8. a 4 -4a 3 & + 5a 2 6 2 + 7afc 3 -12& 4 by a + 12&. 
 
 9. 3ar 5 -40 4 + 6a 2 -5a + lO by a + 8. 
 
 Operation 
 
 (8) 3 x 5 - 4 x 4 + x 8 + 6 x 2 - 5 x + 10 
 
 20-32 6 43 -30 80 
 
 Prod., 3 x 6 + 20 x 5 - 32 x* + 6 x 3 + 43 x 2 - 30 x + 80 
 
 10. 2 a 4 + 3^ + 40^ + 5 x + 6 by a + 5. 
 
 11. 4a 4 -3a 3 -2a 2 -a-l by a-4. 
 
 12. ar 5 -3a 4 + 5a 8 + 2a 2 -6a;-8 by a; + 7. 
 
 13. ^-8x2 + 5 by « + 6. 
 
 14. 3 a; 4 + 7 by a; — 9. 
 
 15. 5 a 5 — 5 by a + 12. 
 
 16. ^ + 3^ + 6 by a-10. 
 
 59. Successive Multiplication. When several factors like those 
 of Art. 58 are to be combined into one product, the literal quanti- 
 ties should not be written until after all the multiplications have 
 been made. 
 
 EXAMPLES IX 
 Find the products of the following : 
 
 1. 3a 2 -2a + l, x + 2, x-1, and a + 3. 
 
 Operation 
 
 (2) 3x 2 - 2x +1 
 (-1) 4-3 2 
 
 (3) 1-7 5-2 
 
 10 - 4 - 16 13-6 
 
 Prod., 3x 5 + 10 x*- 4x*-16 x 2 + 13 x- 6 
 
 2. a 3 -2a 2 + 4a-8, a + 3, and a-4. 
 
 3. x 2 — 3a + 5, a + 1, and a — 7. 
 
 4. a + 4, a — 3, x-{-5, and a — 2. 
 
40 HIGHER ALGEBRA 
 
 5. 2x^-3x 2 + Ax-7, x + 2, x-3, and x -f 4. 
 
 6. x — 7, x + 1, # — 2, ic -f- 3, a? — 4. 
 
 7. 3x 2 -Ax + 2, x + 3, x-5, x-1, x + 2. 
 
 8. 3 x + 2> 3 ic — 2, x + 4, ic — 5, £ — 2, and x + 1. 
 
 9. or 2 + 2^ + 1, x + 3, x + 2, x — 2, x — 2, and x — 1. 
 
 60. Theorem. T%e product of two binomials having their first 
 terms the same is the square of the first (or common) term, plus the 
 algebraic sum of the second terms multiplied by the first (or common) 
 term, plus the product of the second terms. 
 
 Dem. The truth of the proposition is seen in the following 
 results, obtained by performing in the usual way the operations 
 indicated : 
 
 (a; + a) (x + b) = x 2 + (a + b) x + ab, 
 
 (x + a) (x — b) = x 2 + (a — b) x + a (— b), 
 
 (x — a) (x + b) = x 2 + (b — a) x + b (— a), 
 
 (x — a) (x — b) = x 2 + (— a — b) x + (— a) (— b). 
 
 EXAMPLES X 
 Multiply the following : 
 
 1. x + 8 by x — 5. 
 
 Sug. Omitting the intermediate steps, and writing at once the result by 
 the proposition, we have 
 
 (x + S)(x - 5) = x 2 + 3 x - 40. 
 
 2. x - 7 by x + 3. Ans. x 2 -±x- 21. 
 
 3. oj-15 by #-4. Am. a; 2 -19 # + 60. 
 
 4. a + 23 by a + 4. Ans. a 2 + 27a + 92. 
 
 5. x + 10 by x - 6. 6. x - 12 by x + 8. 
 7. m + 20 by m — 4. 8. m + 16 by m + 3. 
 9. x — 11 by x - 7. 10. y - 22 by ?/ + 4. 
 
 11. y + 90 by y - 10. 12. a + 50 by x - 25. 
 
SHORT METHODS OF MULTIPLICATION 41 
 
 13. z 2 + 9 by z 2 - 6. 
 
 Operation. Here z 2 takes the place of x in the demonstration, and 
 
 we have 
 
 (* 2 + 9) O 2 - 6) = 2* + 3 z 2 - 54. 
 
 14. a 2 - 8 by aj* - 6. 15. ^ - 12 by ar* + 7. 
 16. a? + 14 by a? - 5. 17. z 4 - 30 by z 4 + 5. 
 
 18. cub 3 — 11 by ax- 3 + 5. 
 
 Operation. Here ax 3 takes' the place of x in the demonstration, and 
 we have 
 
 (ax 3 - 11) (ax 3 + 5) = a 2 x 6 -6 ax 9 - 55. 
 
 19. «V + 6 by aV - 8. Ans. a 4 x* - 2 a?x 2 - 48. 
 
 20. ?>iar> - 13 by mx A + 3. 21. wV + 15 by m 2 z 4 - 6. 
 22. a¥ - 20 by a¥ +■ 4. 23. a¥ + 23 by a¥ - 3. 
 
 24. 3^ + 6 by 3.^-4. 
 
 Operation. Here 3x 2 takes the place of x in the demonstration, and 
 we have 
 
 (3x 2 + 6)(3 x 2 - 4) = 9x 4 + 2(3 x 2 ) - 24 = 9 x 4 + 6 x 2 - 24. 
 
 25. 2*-9 by 2a + 7. 26. 4^ + 11 by 4# 2 -8. 
 27. 3z 3 -12 by 3z 3 + 6. 28. 5z 2 -15 by 5z 2 -5. 
 
 29. x + 7 y by x — 3 ?/. 
 
 Operation. Here 7 y and — 3 y take the places of a and 6, respectively, 
 in the demonstration, and we have 
 
 (x + 7 y) (x - 3 y) = x 2 + 4 xy - 21 y 2 . 
 
 30. ic — 10 y by a — 6 #. 31. # -f 12 y by x + 4 #. 
 32. a; — 9 y by a; — 7 ?/. 33. a^ + 8 y by ar* — 4 #. 
 34. 2z 2 -5?/ by 2z 2 + 9y. 35. z 2 + 4/ by z 2 -8y 3 . 
 36. z*-lly 2 by z 3 + 7?/ 2 . 37. 3a? + 7tf by 3ar 3 -22/ 2 . 
 
 38. a; 2 + 5 m by a^-2w. 
 
 Operation. Here x 2 takes the place of x in the demonstration, and 5 to 
 and — 2 n of a and &, respectively, and we have 
 
 (x 2 + 5 to) (x 2 - 2 n) = x 4 + (5 to - 2 n)x 2 - 10 win. 
 
 39. x — Sy by a; + 3 z. 40. a; 2 -f 11 y by a^ 2 — 4 z. 
 
 41. Sx>+-ly by 3ar>-5z. 42. 5a 2 + 6m 3 by 5a 2 -3n 2 . 
 
42 HIGHER ALGEBRA 
 
 61. Theorem. The square of the sum of two quantities is equal 
 to the square of the first, plus twice the product of the two, plus the 
 square of the second. 
 
 Dem. By actual multiplication, we have 
 
 (x + y) 2 = (x + y) + y) = x 2 + 2 xy + y 2 . 
 
 62. Theorem. The square of the difference of two quantities is 
 equal to the square of the first, minus twice the product of the two, 
 plus the square of the second. 
 
 Dem. By actual multiplication, we have 
 
 (x - y) 2 = (x - y) (x - y) = x 2 - 2 xy + y 2 . 
 
 63. Theorem. TJie product of the sum and difference of two 
 quantities is equal to the difference of their squares. 
 
 Dem. By actual multiplication, we have 
 
 (as + y) (x - y) = x 2 - y 2 . 
 
 
 EXAMPLES XI 
 
 
 
 Square the following : 
 
 
 
 
 
 1. 1+x. 
 
 2. 
 
 x + 2. 
 
 3. 
 
 x 2 + y. 
 
 4. x — 5 y. 
 
 5. 
 
 3 a + 4 b. 
 
 6. 
 
 x n + 2. 
 
 m x , y 
 7. - ± £. 
 
 y x 
 
 8. 
 
 J£*-f af*. 
 
 9. 
 
 xi + iy-t. 
 
 10. x 2 + y 2 z*. 
 
 11. 
 
 7x?-3y 2 . 
 
 12. 
 
 a* — | a*b 3 . 
 
 13. x** — tf*. 
 
 14. 
 
 ax* — by 2 . 
 
 15. 
 
 5 a 3 x A -j- 4 b 2 y 3 . 
 
 16. ±x + 7x n+ \ 
 
 17. 
 
 7 as* -3y~\ 
 
 18. 
 
 2 a m 3 b n m 
 3b n 2 a m ' 
 
 Write the products of the following : 
 
 19. x 2 + 2 y by x 2 — 2 y. 20. 3m 2 + 5n 3 by 3 m 2 — 5 n 3 . 
 
 21. 1+f a by 1-f a. 22. 5x 3 + 3y 2 z by 5x*-3fz. 
 
 23. 2 #* + 3 ^ by 2^ — 3 yK 24. 9 ax+ Vox by 9 ax— aW. 
 
SHOUT METHODS OF MULTIPLICATION 43 
 
 64. Theorem. The square of any polynomial is the sum of the 
 
 squares of each of the terms, together with the algebraic sum of twice 
 each term into each of the terms that follow it. 
 
 Dem. When there are more than two terms, part of them may 
 be treated as constituting the first term and the others as the 
 second term of a binomial. Thus : 
 
 (a + b + c)» = [(a + b) + c]* « (a + h}* + 2 (a + b) c + c 2 
 
 = a* + 2ab + b 2 + 2ac + 2bc + c 2 , 
 
 or a 2 + b' 2 + c 2 + 2 ab + 2ac + 2 be. 
 
 Again, 
 
 ( a +b +c -d) 2 =[(a + b)i-(c-d)] 2 =(a + b) 2 +2(a+b)(c-d) + (c-(l) 2 
 
 = a 2 +2ab + b 2 +2ac-2ad-2bd + c 2 -2cd+d 2 , 
 
 or a 2 +b 2 +c 2 +d 2 +2 ab + 2 ac-2 ad + 2 be -2 bd-2 cd. 
 
 By inspecting the results of these operations we deduce the 
 important, time-saving theorem stated above. 
 
 EXAMPLES XII 
 Square the following : 
 
 1. a-b- c. 2. a 2 — b 2 + c. 
 
 3. m — n — p -f- q. 4. x 2 -\-2 y — u — • 3 v. 
 
 5. x* + a,* 2 -f a; -f 1. 6. a — b + c — cZ -|- e. 
 
 7. a; 2 — xy + y 2 . 8. ar* + a^y -f a:?/ 2 -f- y 3 . 
 
 9. 3a^ — 5y + 4**— 2*. 10. a 4 - 2 6 3 - 3 c 2 + 4 tf. 
 
 65. Theorem. The product of the sum of two groups of terms 
 by the difference of the same groups is equal to the difference of the 
 squares of the groups. 
 
 This is but an application of Art. 63, regarding the first group 
 as one term and the second group as the other term of a 
 binomial. 
 
44 HIGHER ALGEBRA 
 
 EXAMPLES XIII 
 Write the products of the following : 
 1. a + b — c by a — b + c. 
 Operation. This may be written, 
 {a + (b - c)}{a - (b - c)} = a 2 - (6 2 - 2 be + c 2 ) = a 2 - ft 2 + 2 6c - c 2 . 
 
 2. # — 3 ?/ 4- 5 z by x — 3y — 5 z. 
 
 3. a 2 4- y 2 + 33/ by x 2 -\-y 2 — xy. 
 
 4. 2x — ky — z by 2 x + 4 ?/ — z. 
 
 5. 4 a 4 - afy 2 + 3 y* by 4 a 4 + x*y 2 4- 3 ?/ 4 . 
 
 6. a* — x '4. 9 by x 2 - x - 9. 
 
 7. x 2 + x -7 by x 2 -x + 7. 
 
 8. 3 ar 2 - 4 a; 4- 5 by 3 x 2 + 4 a? 4- 5. 
 
 9. a + 2&4-3c4-dbya + 26-3c-riL 
 
CHAPTER V 
 DIVISION 
 
 66. Division is the inverse of multiplication, and is, therefore, 
 the process of finding either factor when the other factor and the 
 product are given. 
 
 The terms Dividend, Divisor, Quotient, Remainder, Numerator, 
 and Denominator are used as in Arithmetic. 
 
 67. Theorem. When dividend and divisor have like signs, the 
 quotient is +, and when unlike, — . 
 
 Dem. If d is the divisor and q the quotient, qd is the divi- 
 dend (Art. 66). Now by the law of signs in multiplication (Art. 
 46), if d and qd have like signs, q is + ; and if d and qd have 
 unlike signs, q is — . 
 
 68. Theorem. The exponent of a quantity in the quotient is the 
 exponent of this quantity in the dividend, diminished by its exponent 
 in the divisor. 
 
 Dem. Since the exponent of a quantity in the dividend, 
 which is the product of divisor and quotient, is the sum of the 
 exponents of that quantity in these factors (Art. 48), then, con- 
 versely, the exponent of a quantity in the quotient (one of the 
 factors) is the exponent of this quantity in the dividend (the 
 product), diminished by its exponent in the divisor (the other 
 factor). 
 
 69. Cor. i. Negative exponents arise from division ivhen the 
 exponent of a quantity in the divisor is greater than that of the 
 same quantity in the dividend. 
 
 70. Cor. 2. Any quantity icith the exponent is 1. 
 
 Dem. Let x represent any quantity and m any exponent. 
 Now x m -~ x m = 1. But by the law of exponents just established, 
 
 x m ~-x m = x m ~ m = x°. Hence x° = l. 
 45 
 
46 HIGHER ALGEBRA 
 
 71. Cor. 3. A factor may be transferred from dividend to 
 divisor (or from numerator to denominator of a fraction, ivhich is 
 the same thing), and vice versa, by changing the sign of its exponent. 
 
 -n, a ax° ax°~ m ax~ m 
 
 For 
 
 bx m bx m b b 
 
 EXAMPLES XIV 
 Divide the following : 
 
 1. 4 a 5 by 2 a 3 . 2. 20 ax 3 by 4 ax. 
 
 3. 32 afyV by 8 x?y 4 z 5 . 4. 27 x m by 9 x n . 
 
 5. 42 x m by 7 x~ n . 6. a 3 ^ by aaA 
 
 7. a| by a 2 z. 8. (ab) 2m by (06)—. 
 
 9. 18 or 3 by 3x*. 10. ra 4 af* by m 2 x~ 2 . 
 
 11. a 3 (a ~ x) 4 by a 3 (a — x) 2 . 
 
 12. 12m" 2 (a 2 -3a;) 3 by 3 m" 3 (a 2 - 3x)~ 2 . 
 
 13. 3 7 (or 2 - y 2 f by 3 4 (oj 2 - y 2 )\ 
 
 14. a 3 (2a + 4ar ? ) 2n - 3 by a(2x + 4ar 5 )"- 3 . 
 
 Free the following from negative exponents : 
 
 a 2 b~ 3 _ 3 afr"^ -1 ^ 4x m y~ n z~ p 
 
 x- 2 f ' 5a~ 2 d- 3 x- 2 ' lx~ n y m 
 
 72. Theorem. TJie quotient of the sum or difference of several 
 quantities is equal to the sum or difference of the quotients, the 
 divisor being the same. 
 
 Dem. Thus, x + y ~ z is equal to - + 1 - *, for the product 
 n n 71 n 
 
 of each by n is the same, viz. x -f y — z. 
 
 EXAMPLES XV 
 Divide the following : 
 
 1. 6 a 2 b 3 + 15 a A b 2 - 12 a 2 b 2 by 3 ab. 
 
 2. 28 x?y 4 - 84 x?y 5 + 63 xY by 7 ofy 3 . 
 
DIVISION 47 
 
 3. 15 ax* — 20 atx 2 -+- 5 <&z by — 5 ax. 
 
 4 . 40 aH,c - 24 aWc - 32 abc* by 8 abc. 
 
 5. 9 x 2 " 1 + 6 ar 5 '" — 12 a 4m by 3 x m . 
 
 6. 4 a"-*^-» - 6 a m+n 6 8+n by 2 a~ n b\ 
 
 7. 84 a 1( V - 12 a" 10 6 8 + 156 a 2 b 6 by 12 a" 10 . 
 
 8. a* -|- 3 a V — 2 x$ by ai 
 
 10. 6 (x - 2/)" -2 - 9 (x - y) n ~ l + 12 (a - 2/) M by 3 (x - y) n ~\ 
 
 73. Prob. To divide one polynomial by another. 
 
 Rule. Arrange dividend and divisor with reference to the same 
 letter. 
 
 For the first term of the quotient divide the first term of the divi- 
 dend by the first term of the divisor. 
 
 Multiply the divisor by this term of the quotient, and subtract the 
 product from the dividend. 
 
 Treat the remainder as a new dividend and proceed as before, 
 continuing the operation until there is no remainder, or until the 
 first term of the remainder is not divisible by the first term of the 
 divisor. 
 
 Dem. Since the dividend is the product of the divisor and 
 quotient (Art. 66), that term of the dividend which has the high- 
 est exponent of the letter of arrangement must be the product of 
 those terms of the divisor and quotient which contain the highest 
 exponents of the same letter. Hence, if we divide the first term 
 of the arranged dividend by the first term of the arranged divi- 
 sor, we shall obtain the first term of the quotient. 
 
 If the product of the whole divisor by the first term of the 
 quotient be subtracted from the dividend, the remainder must be 
 the product of the divisor by the sum of all the other terms of 
 the quotient ; hence the second term of the quotient may be found 
 from this new dividend as the first was found from the original 
 dividend. 
 
 If, finally, there is no remainder, the division is exact. If, in 
 the end, there is a remainder whose first term is not divisible by 
 
48 HIGHER ALGEBRA 
 
 the first term of the divisor, the quotient may be completed, if 
 desired, by adding this remainder over the divisor in the form 
 of a fraction. 
 
 74. Sch. The arrangement of the terms corresponds with the 
 succession of thousands, hundreds, etc., in decimal numbers, and 
 these operations of division are analogous to those of long division 
 in Arithmetic. 
 
 Example 
 
 Divide 6 x A - 13 ax* + 13 a 2 x? - 13 a 3 .T - 5 a 4 by 2 x 2 - 3 ax - a 2 . 
 
 Operation 
 6 x* - 13 az 3 + 13 a 2 x 2 _ ^ a s x _ 5 a 4 
 6 x* - 9 ax* - 3 a 2 x 2 
 
 \2x*- 
 3a; 2 - 
 
 - 3 ax - a? 
 
 - 2 ax + 5 a 2 
 
 - 4 ax 3 + 16 a 2 x 2 - 13 a s x 
 
 - 4ax s + 6a¥+ 2 a 3 x 
 
 
 10 a 2 x 2 - 15 aH - 5 a 4 
 10 a 2 x 2 - 15a%-5a 4 
 
 
 75. A simple inspection will show that the following may be 
 omitted from the operation as entirely unnecessary : 
 
 1. The first term of each product to be subtracted, because it 
 is always the same as the first term of that from which it is to 
 be subtracted. 
 
 2. The literal factors in the various terms of the products and 
 remainders, because they are always the same as those of the 
 terms of the dividend under which they stand. 
 
 3. The bringing down of terms from the dividend to the remain- 
 ders, because the subtractions can just as well be made while they 
 retain their original positions. 
 
 4. The signs except when — , because when no sign is written, 
 the + sign is always understood. 
 
 With these omissions the above operation contracts to the fol- 
 lowing : 
 
 6 x* - 13 ax* + 13 a 2 x 2 - 13 a*x -5 a* 
 
 | 2 x' 2 - 3 ax - a 2 
 
 - 9 - 3 
 
 Sx 2 -2ax + 5a? 
 
 - 4 16 
 
 
 6 2 
 
 
 10 -15 
 
 
 -15 -5 
 
 
SHORT METHODS OF DIVISION 49 
 
 When terms are missing from the dividend, their places must 
 be supplied by terms with coefficient zero. 
 
 EXAMPLES XVI 
 
 Divide the following : 
 
 1. i + 2x 2 -7x 4 -16x (i by 1 + 2x + 3ar J + 4ar } . 
 
 Operation 
 
 - I + Ox + 2x 2 + Ox* - 1 x* + Ox* - 1G& \ l + 2x + 3x 2 + 4s 8 
 
 2 
 
 3 
 
 4 
 
 
 -2 
 
 - 1 
 
 -4 
 
 
 
 -4 
 
 -6 
 
 -8 
 
 2 
 
 1 
 
 
 
 6 
 
 9 
 
 12 
 
 
 -4 
 
 -8 
 
 -12 
 
 
 
 -8 
 
 - 12 
 
 -16 
 
 2. 15x 4 -32x i + 50x i -32x + 15 by 3^-4^ + 5. 
 
 3. 6a*-31a? + 23a?-2x-48 by 3^-5^ + 6. 
 
 4. 2 a 7 6 - 5 a 6 6 2 - 11 a 5 b 3 + 5 a 4 6 4 - 26 a 3 6 5 + 7 a 2 // - 12 a& 7 by 
 
 SHORT METHODS OF DIVISION 
 
 76. The operation of division may be still further shortened 
 as follows : 
 
 If the signs of the terms of the divisor after the first are 
 changed, the signs of those terms of the various products which 
 were subtracted in the operations above will be changed, and the 
 subtraction changed to addition, since to subtract we change the 
 signs of the subtrahend and proceed as in addition. The opera- 
 tion at the bottom of page 48 would then become, 
 
 %x^-13ax i + 13a 2 x 2 -13a i x-ba A \ 2 x 2 + 3 ax + a 2 
 
 9 
 -4 
 
 3 
 16 
 
 - 6 
 
 - 2 
 
 5 
 
 3x i -2ax + 5a 2 
 
 
 10 
 
 -15 
 15 
 
 
 downey's alg. — 4 
 
50 HIGHER ALGEBRA 
 
 It will be seen that the result of the first addition to the term 
 13 a 2 x 2 is not used in obtaining a term of the quotient, and that 
 the result which is used, viz., 10 (with a¥ understood), can just 
 as well be obtained by adding 13, 3, and — 6 without first adding 
 13 and 3. The same remarks apply to the additions to the term 
 — 13a 3 x. With these omissions, and writing the terms of the 
 products as near as may be to the corresponding terms of the 
 dividend, the operation becomes, 
 
 6 x 4 - 13 ax 3 + 13 a 2 x 2 - 13 a s x - 5 a 4 \2x 2 + 3ax + a 2 
 9 3-2 5 3a 2 -2az + 5a 2 
 
 -4 -6 15 
 
 10 
 
 A convenient rule for this operation is the following : 
 
 Arrange dividend and divisor with reference to the same letter, 
 supplying the places of any missing terms of the dividend by terms 
 with zero coefficients. 
 
 Change the signs of all the terms of the divisor except the first. 
 
 Divide the first term of the dividend by the first term of the divisor 
 for the first term of the quotient. 
 
 Multiply the terms of the changed divisor except the first by this 
 term of the quotient and write the product under the corresponding 
 terms of the dividend, omitting the literal factors. 
 
 Add the first of these products to the second term of the dividend, 
 and divide this sum by the first term of the divisor for the second 
 term of the quotient. 
 
 Multiply the terms of the changed divisor except the first by this 
 second term of the quotient and write the products under the corre- 
 sponding terms of the dividend, omitting the literal factors. 
 
 Add the third term of the dividend to the two terms that stand 
 under it, and divide this sum by the first term of the divisor for the 
 third term of the quotient. 
 
 Continue this operation either until the division terminates or 
 until one of these sums contains a lower exponent of the letter of 
 arrangement than the first term of the divisor. In the latter case 
 this sum and the sums of the remaining terms of the dividend and 
 the terms that stand under them constitute the remainder. 
 
SHORT METHODS OF DIVISION 51 
 
 77. This is a modified form of Horner's Synthetic Division and 
 should be thoroughly mastered by the student. It requires but 
 little attention and practice to become familiar with it, while the 
 saving in time is very great. By comparing the last operation 
 with the usual process as given in the first solution it will be 
 seen that while that operation (not counting dividend, divisor, 
 and quotient) contains 80 figures, letters, and signs, this contains 
 but 15. 
 
 EXAMPLES XVII 
 Divide the following : 
 
 1. 10 a 4 - 27 «*a? + 34 aV - 18 ao 8 - 8 a 4 by 2a 2 - Sax + ix 2 . 
 
 2. x 4 -Sax i -Sa 2 x 2 -\-lSa ti x-Sa 4 by x 2 + 2ax-2a 2 . 
 
 3. 4y 5 -24?/ 5 + 60t/ 4 -80t/ 3 + 60?/ 2 -242/ + 4 by 2?/ 2 -4?/+2. 
 
 4. z 6 -5ar 5 + 15a 4 - 24ar } + 27z 2 - 13z + 5 by aj 4 - 2 a 8 + 4 aj 2 
 -2z + l. 
 
 5. 2 a'b - 5 a% 2 - 11 a 5 6 3 + 5 a 4 b 4 - 26 o 3 6 5 + 7 a 2 6 6 - 12 a& 7 by 
 a 4 -4a 3 & + a 2 6 2 -3a& 3 . 
 
 6. a 6 -3aV + 3«V-x 6 by a 3 -3a 2 x + Sax 2 -x\ 
 
 7. 9a? + 24x2/ + 122/ 2 + 30a* + 24yz + 9z 2 by x + 2y + 3z. 
 
 8. 8y 5 -22ay + 20afy 3 + xy-7aty + 6ar 5 by ±y 2 -3xy + 2x 2 . 
 
 9. 6ar 5 -7« 4 2/ + ary + 20^-22^ 4 + 8?/ 5 by 2a 2 -3a>y+ ±y 2 . 
 
 10. l + 2ar 5 + a; 6 + 2a,' 7 by l + a; + x 2 . 
 
 11. 16^ + 36^ + 81 by 4ar 2 -6x + 9. 
 
 12. 6a-20-2ar } + 15ar J -a 4 +2ar J by 5 + 2x*-±x-3x 2 . 
 
 13. 12 a 5 - 14 a 4 6 - 10 a 3 6 2 - a 2 6 3 - 8 a& 4 + 4 & 5 by 6a 3 -4a 2 6 
 -3a6 2 + 26 3 . 
 
 14. 14 x A + 45 xhj + 78 x 2 y 2 + 45xf + 2ly 4 by 2x> + 5xy + ly 2 . 
 
 15. 1 - a 6 by 1 + 2 a; + 2 a* + arl 
 
 Operation 
 
 
 l+Os + O^ + O^ + Oa^ + Oz 5 - 
 
 £6|l-2x-2x 2 -x 8 
 
 -2 -2 -1 2 -2 
 
 1 1 - 2 x + 2 x 2 - x 8 
 
 -2 4 4-4 2 
 
 
 2-4 2 
 
 
 -10 
 
 
 
52 HIGHER ALGEBRA 
 
 16. x s -|- if by x 4 — xhj -|- x 2 ?/ 2 — &y 3 -J- if. 
 
 17. a 8 + a?b- + a 4 6 4 + a 2 6 6 + 6 8 by a 4 + a?b + a 2 6 2 + a& 3 + b 4 . 
 
 18. a 6 -21ary + 24a;?/ 5 -8y 5 by rf-Sxy + if. 
 
 Sug. From the terms given, the arrangement is seen to be with reference 
 to the descending powers of one letter and the ascending powers of the other. 
 The dividend should, therefore, be written 
 
 se 6 + x b y + x*y 2 - 21 x s y* -f x 2 y* + 24 xy h - 8 y*. 
 
 19. Aa 4_j a s_ Ja * + | a + v by |a 2 -a-f. 
 
 20. 36^-6^-4^ + i2/-hi2/ 2 + J by 6*~}y-'$ 
 
 21. a&0* + 2 (a - 6) iB 3 - (a 2 + 4 - b 2 ) x 2 + 2 (a + b) x - ab by 
 bx* + 2 x — a. 
 
 22. 6 a 4 - 25 afy 4- 16 x*y 2 -17 xif + 3oy 4 by 2 x-7 y. 
 
 Sug. When the divisor is a binomial, as in this example, the operation 
 takes the following form : 
 
 6 x 4 - 25 x*y + 16 s 2 ?/ 2 - 17 xy* + 35 y 4 | 2 a; + 7 y 
 
 21 - 14 7 - 35 3 x s - 2 x 2 y + zy 2 - 5 y a 
 
 - 4 2-10 
 
 23. a 4- 9 x* - 1 4- 3 x 2 by 3 a? - 1. 
 
 24. 7 a? - 24 x 2 4- 58 a - 21 by 7 a - 3. 
 
 25. 2 a 4 4- 27 aft 8 - 81 6 4 by a + 3 6. 
 
 26. 6 or 5 - 11 or* 4- 3 x 2 4- 4 x - 4 by 3 a 2 - 4. 
 
 27. 6 a 4 - 96 by 3 a - 6. 
 
 28. a 1+n + a M 6 4- ab n 4- & 1+n by a* 4- b n . 
 
 29. m m+1 4- nm m -\- amn an -f cm an+1 by m -f w. 
 
 30. 6 a 4 - 13 a 3 6 4- 9 a 2 b 2 + 10 a& 3 - 8 b 4 by 3 a - 2 6. 
 
 31. 10ar , -7ic 4 2 / + 8« 3 ?/ 2 -9xy+10^ 4 + 28 2/ 5 by 5x + 4y. 
 
 78. "R7ien ^e coefficient of the first term of a binomial divisor of 
 the first degree is unity, the operation becomes exceedingly simple. 
 
SHORT METHODS OF DIVISION 53 
 
 The process is of so much importance and is of such frequent 
 occurrence in subsequent parts of the work that a special rule is 
 here given. 
 
 79. Prob. To divide by a binomial of the first degree when the 
 coefficient of the first term is unity. 
 
 Rule. Call the coefficient of the second term of the divisor, with 
 its sign changed, a; multiply the coefficient of the first term of the 
 dividend by a, and add the product to that of the second; multiply 
 this sum by a, and add the product to the coefficient of the third term 
 of the dividend, and so on. When the last sum is zero, the division 
 is exact; otherwise, the last sum, with the literal part of the last 
 term of the dividend, will be the remainder, and the coefficient of the 
 first term of the dividend, and the other sums in order, will be the 
 coefficients of the quotient, the arrangement of the letters of the quo- 
 tient being the same as in the dividend, and the degree being one less 
 than that of the dividend. 
 
 Note. This rule may be deduced as a special case of the last 
 process ; but it is deduced below directly from the ordinary proc- 
 ess of " long division," with which the student has become famil- 
 iar in Arithmetic and Elementary Algebra, in order that the 
 relation to that process, and at the same time the great saving in 
 labor and time, may be seen. 
 
 Dem. By the ordinary process of " long division " we have, 
 
 2 3^-3 ^y-13a,Y+14aY-2 ^-12^ 1 x -Sy 
 2x 5 -6x 4 y 2x A +3x i y-4:X 2 y 2 
 
 3x<y-13x>y* +2a*»+4jf 
 
 Sx 4 y- 9x*y 2 
 
 - Ixy+Ux 2 !? 
 
 - 4a?if +12 aty 
 
 2x>tf-2xy A 
 2x?y 3 -6xy 4 
 
 ±xy 4 -\2tf 
 ±xy 4 -\2tf 
 
54 
 
 HIGHER ALGEBRA 
 
 Omitting the wholly unnecessary parts, without changing the 
 position of the parts that remain, this becomes, 
 
 2 a 5 -3 
 
 -6 
 
 3 
 
 x 4 y - 13 
 
 -9 
 -4 
 
 xY + 14 
 
 12 
 
 2 
 
 ry 
 
 -6 
 4 
 
 a^ 4 - 12 
 
 12 
 
 
 
 y 5 | a?-3y 
 2x 4 + 3x 3 y-±x 2 y 2 
 + 2xtf + ±y 4 
 
 Now since the coefficient of the first term of the divisor is, by 
 hypothesis, unity, the coefficients of the terms of the quotient will 
 be the same as the coefficients of those terms which are divided by 
 x to obtain them (in this case the same as in the terms 2 x 5 , 3 x 4 y, 
 — 4 arty 2 , 2#y, and 4:xy 4 ). Hence the coefficient of the first term 
 of the dividend and the coefficients of the various remainders in 
 order will be the coefficients of the terms of the quotient. Any 
 one of these remainders is found by multiplying the one before 
 it (which is the same as the coefficient of the corresponding term 
 of the quotient) by the coefficient of the second term of the divi- 
 sor, and subtracting the product from the coefficient of the next 
 term of the dividend. By changing the sign of the second term 
 of the divisor this subtraction will be changed to addition. The 
 degree of the quotient is one less than that of the dividend, 
 because the divisor is of the first degree. Hence writing in place 
 of the divisor only the coefficient of the second term with its sign 
 changed, omitting from its usual place the quotient, since its 
 coefficients are but a repetition of numbers written before, and 
 writing the products and sums as near as may be to the terms of 
 the dividend with which they belong, the operation becomes, 
 
 2x?-3x 4 y- 13 xY + 14 tftf 
 6 9-12 
 
 3 - 4 
 
 2 xy 4 - 12 y 5 [£ 
 6 12 
 
 4 
 
SHORT METHODS OF DIVISION 55 
 
 Supplying now the literal parts, we have for the quotient, 
 
 2 x 4 + 3 afy - 4 x 2 y 2 + 2 xif + 4 y 4 . 
 
 When the coefficients are small, as in this case, or so related 
 that the sums are small, the additions can be readily made with- 
 out writing the products. 
 
 2 x 6 - 3 x 4 y - 13 x % f + 14 xhf - 2 xy A - 12 t/ 5 |_3 
 3-4 2 40 
 
 Thus, 3 times 2 are 6, which added to — 3 gives 3 ; 3 times 3 
 are 9, which added to — 13 gives — 4 ; 3 times — 4 are — 12, 
 which added to 14 gives 2; 3 times 2 are 6, which added to —2 
 gives 4 ; 3 times 4 are 12, which added to — 12 gives 0. Supply- 
 ing the literal parts, we have for the quotient, 
 
 2 x 4 + 3 x % y -4:X 2 y 2 + 2xif + 4: y\ 
 
 When the last sum is anything other than zero, this sum 
 (including the literal part) is the remainder, since the operation 
 by which it is obtained is the same as subtracting the product of 
 the last terms of the divisor and quotient from the last term of 
 the dividend. 
 
 80. By comparing the last operation with the usual process 
 by " long division," as given in the first operation, it will be seen 
 that while that operation (not counting dividend, divisor, and 
 quotient) contains 93 figures, letters, and signs, this contains 
 but 6. 
 
 EXAMPLES XVIII 
 
 Divide the following : 
 
 1. x* — 11 x?y + 41 xy 2 — 50 y 8 by x — 5y. 
 
 Operation 
 
 x* - 11 xhj + 41 xy 2 - 50 if [5 
 - 6 11 5 
 
 1, — 6, and 11 are the coefficients of the terms of the quotient 
 and 5 is the coefficient of the remainder. 
 
 Ans. x 2 — 6 xy -f- 11 y 2 , with remainder 5 y\ 
 
56 HIGHER ALGEBRA 
 
 2. 7ar 5 -3afy-2a2/ 2 -40.v 3 by x-2y. 
 
 3. x 4 + 4:X i y-13x i y 2 -28xy 3 + 60y 4 by x-2y. 
 
 4. 2x 4 + llx i y + 9x 2 y 2 -8xy 3 + 16y 4 by * + 4y. 
 
 5. 12 x 4 - 48 arty + 11 arfy 2 - 45 ay^ 7 ?/ 4 by #-4?/. 
 
 6. x i + 4x 4 y-2x i y 2 -18x 2 y !i -6xy* + 9y 5 by x + 3y. 
 
 7. 3ar , + 2xV-21a% 2 -14ary } + 36ay + 24?/ 5 by a-2y. 
 a 3^-25^i/-15ar ? 2/ 2 -31ic 2 2/ 5 + 38^ 4 -18^ by x-9y. 
 9. 4x 6 -24arV+23xy-14arY+afy 4 -25ay 5 -25?/ 6 by x-5y. 
 
 10. x 6 -2x'V-45xy+22arY+24a,V+2ay+12?/ 6 by x+6y. 
 
 11. 2ar 5 -9x 4 + llar 3 -20x + 6 by x - 3. 
 
 Operation 
 
 2 x 5 - 9 x 4 + 11 x 3 + x 2 - 20x + 6 |_3 
 
 -3 2 6-20 
 
 <4*a. 2 x 4 - 3 x 3 + 2 x 2 + 6 x - 2. 
 
 12. a; 4 + ar 3 - 50 a 2 - 40 x - 14 by * - 7. 
 
 13. 3^ + 85^-3^-45 by x + 3. 
 
 14. 8^-8^-60^ + 400 by a? + 4. 
 
 15. 5aj 4 -46ar } + 48ar 2 + 7a;-56 by a; -8. 
 
 16. a? -20 a?- 20a? 2 -20 a; -25 by a?-5. 
 
 17. a*-13a** + 15a?-30a*-75a; + 36 by a; -12. 
 
 18. 2x 4 + 35ar 5 + 80ar J + 75a; by a; + 15. - 
 
 19. a^ + 3ar 5 -7ar 4 -10ar 3 + 6x 2 +9a;-10 by x-2. 
 
 20. ^-25^ + 47 ar J -21ar 2 -43a;-69 by a; -23. 
 
 21. ^-64 by x-2. 
 
 Operation 
 
 x 6 + x 6 + x* + x 3 + x 2 + x - 64 |_2 
 2 4 8 16 32 
 
 Ans. x 5 + 2 x 4 + 4 x 3 + 8 x 2 + 16 x + 32. 
 
 22. a; 4 — 81 by x — 3. 
 
 23. x^ + if by x+?/. 
 
 24. a 6 — 729 1/ 6 by x—3y. 
 
SHORT METHODS OF DIVISION 57 
 
 81. Successive Division. In dividing successively by several 
 binomials of the first degree when the coefficient of the first term 
 of each is unity, the letters should not be written until after the 
 last division. It must be noted that the degree is reduced by 
 one for each division. 
 
 EXAMPLES XIX 
 Divide successively the following : 
 
 1. x 5 -3z 4 -9ar J + 21ar 2 -10z + 24 by a ._ 2j x + 3, and x — 4 
 
 Operation 
 
 x b _ 3 X i _ 9 X 3 + 21 & - 10 x + 24 | 2 
 
 _ 1 _ ii _ i _ 12 J -3 
 
 -4 1-4 | 4 
 
 10 j^i X 2 + o x + 1, or x- + 1. 
 
 2. a 4 -ar 5 -19ar> + 49#-30 by x-3 and z + 5. 
 
 3. x* + 5x i -30x i -80x + 224: by » — 2, #-4, and x + 4, 
 
 4. ar s + z 4 -31ar J + 68ar J -54a; + 63 by x-3 and x + 7. 
 
 5. ^-7^4-5 ^+55^-126^+72 by *-l, a?-2, z-3, z+3, 
 and a; — 4. 
 
 6. x 6 - 14 a; 4 + 49^-36 by »— 1, a+1, z-2, z+2, and x-3. 
 
 
 Operation 
 
 
 
 
 x 6 + 0z 5 -14x 4 + 
 
 0«3 
 
 + 49z 2 + Ox- 
 
 -36 [ 
 
 i 
 
 
 1 -13 - 
 
 13 
 
 36 36 
 
 I 
 
 -l 
 
 
 -13 
 
 
 
 36 
 
 ! 
 
 2 
 
 
 2 - 9 - 
 
 18 
 
 
 
 1 
 
 -2 
 
 
 0-9 
 
 
 
 
 
 3 
 
 
 3 
 
 
 
 
 
 Ans. x -f 3, 
 
 7. x 5 - 10 x 4 - 3 ar* +258 x 2 - 550 « - 200 successively by x - 4, 
 x — 5, and x + 5. 
 
 82. Theorem. fTAen the dividend is the square of the first term 
 of the divisor, plus or minus twice the product of the first and second 
 terms, plus the square of the second, the quotient is the same as the 
 divisor. 
 
 This is a consequence of Arts. 61 and 62. 
 
58 HIGHER ALGEBRA 
 
 83. Theorem. When the divisor is the sum or difference of two 
 quantities arid the dividend is the difference of the squares of these 
 quantities, the quotient is the difference or sum of the same quantities. 
 
 This is a consequence of Art. 63. 
 
 EXAMPLES XX 
 
 Divide the following : 
 
 1. S6x 4 ^-36xy + df by 6^ + 3^. 
 
 2. 49 a 8 - 70 a 4 b 5 + 25 b 10 by 7 a 4 - 5b 5 . 
 
 3. 16a 6 -64fc 4 by 4a 3 -8& 2 . 
 
 4. Six 5 — Ay 2 by 9x% + 2y. 
 
 5. 121a 6 -264ary+144y 2 by 11 X s -12/. 
 
 6. 64 a 4 b 2 + 112 a?b(? + 49 a 2 c* by 8 a 2 b + 7 ac 3 . 
 
 84. TF/ien the divisor is readily separated into binomial factors of 
 the first degree having unity for the coefficient of the first term of each, 
 the division is most expeditiously made by dividing successively 
 by the factors, as in the process of Art. 81. 
 
 EXAMPLES XXI 
 
 Divide the following : 
 
 1. ^ + 3a; 4 - 20ar 3 -60« 2 + 64^ + 192 by ^-6^ + 8. 
 
 Operations 
 
 The factors of x 2 — 6 x + 8 being x — 2 and x — 4, we divide successively 
 by these factors, thus : 
 
 5c 5 + 3 x 4 - 20 x> - 60 x 2 + 64 x + 192 [2 
 5-10-80-96 0[4 
 
 9 26 24 
 
 .4ms. x 3 + 9 z 2 + 26 x + 24. 
 
 2. ^ 4 -3x 3 -7^+15x + 18 by a 2 -6a? + 9. 
 
 3. ^ + 3x 3_ 7a . 2 _27^_i8 by x 2 -9. 
 
 4. 0? + 3 as 4 -17 a?- 27^ 2 +52x + 60 by a? + 2x-15. 
 
 5. o? 6 -6;c 4 -4ar ? +9.T 2 +12a,' + 4 by ^-4^ + 4. 
 
 6. ^-7^ + 17^-28^+37^-20 by x 2 -5x + 4. 
 
 7. 3x 5 + x 4 -5x i -6x 2 -17x-U by x 2 -x-2. 
 
CHAPTER VI 
 FACTORING 
 
 85. A Factor of a quantity is a quantity that will divide it 
 without a remainder. 
 
 86. The Factors of a quantity are those quantities which multi- 
 plied together produce it. 
 
 87. To Factor a quantity is to separate it into its factors. 
 
 88. A Prime Quantity is one which has no integral factors 
 except itself and unity. 
 
 89. A Composite Quantity is one which has integral factors 
 other than itself and unity. 
 
 90. Theorem. Any monomial factor which occurs in every term 
 of a polynomial can be removed by dividing each term of the polyno- 
 mial by it. 
 
 Thus, a 3 x + a 2 x 2 — a¥ = a 2 x (« + «- cfix 2 ). 
 
 91. Theorem. If two terms of a trinomial are p>ositive, and the 
 remaining term is ± twice the product of their square roots, the tri- 
 nomial is the square of the sum or difference of these square roots. 
 
 This is a consequence of Arts. 61 and 62. 
 
 92. Theorem. The difference between two quantities is equal to 
 the product of the sum and difference of their square roots. 
 
 This is a consequence of Art. 63. 
 
 93. Theorem. If a polynomial of six terms consists of three 
 perfect squares and three double products of their square roots, 
 taken in pairs, the polynomial is the square of the algebraic sum of 
 these sq)iare roots. 
 
 This is a consequence of Art. 64. 
 
 Arts. 169 and 170 furnish the best means of factoring poly- 
 nomials that are perfect squares. 
 
 69 
 
60 HIGHER ALGEBRA 
 
 EXAMPLES XXII 
 
 Factor the following : 
 1. a 4 & + 3a 6 6 4 + 4a 3 & 3 . 2. 3a 4 -6a 5 -12a 2 + 9a 3 . 
 
 3. 12 a 4 b 7 - 20 a 3 b* + 16 ab 7 - 8 a 5 b s . 
 
 4. 10 xHf + 5 fcty* — 15 a?%* — 20 as*y*. 
 
 5. a 2 + 6«&-f9 6 2 . 6. a 4 - $ a?b + 16 b 2 . 
 
 7. a^ — afy + Jy 2 . 8. x 4 — Ay 2 . 
 
 9. 3ar — 48 2/ 4 . 10. Sx'-lSxy 4 . 
 
 11. a 4 -?/ 8 . 12. 81 A 2 -1. 
 
 13. 1 — 10 xy 2 -f 25 x*y 4 . 14. 4 m 2 # 4 + 4 mnxPy -f n ? i/ 2 . 
 
 15. (^ + 2/) 2 -f2(^ + 2/) + l- 16. x>-(x-y)\ 
 
 17. a 2 -fc 2 -c 2 + 26c. 18. (a? + 2/) 4 + 4(x + 2/) 2 + 4. 
 
 19. (a 2 + 2/) 2 -6V(a 2 + 2/)+9z 2 . 
 
 20. (2x 2 + 3^-8) 2 -(^-3^-8) 2 . 
 
 21. a 2 + b 2 + c 2 -2ab + 2ac-2bc. 
 
 22. ^ + !/ 2 + 9 + 2^?/ — 6a; — 62/. 
 
 23. 4 x 2 — 4 #1/ + y 2 -f- 12 #z — 6 2/2 + 9 z 2 . 
 
 24. 9a 4 + 4?/ 2 -f 16z 2 - 12x 2 y-2Ax 2 z + 16yz. 
 
 25. 16 x 6 + 24 a«y + 9 y 4 - 40 ^ - 30 y 2 z + 25 z 2 . 
 
 26. One factor of x 4 — 7 a? + 6x 2 + 23x — 15 being # - 5, what 
 is the other ? 
 
 Sug. Divide as in Art. 79. 
 
 27. One factor of X s — 6 x 2 y + 15 #?/ 2 — 18 1/ 3 being x — Sy, what 
 is the other ? 
 
 28. Two factors of 2x 4 -\-5x? — 13x 2 + 16 being a? + 4 and 
 a? -h 1, what is the third ? 
 
 29. One factor of X s — Ax 2 — 35 # + 150 being a? + 6, what are 
 the other two ? 
 
 30. One factor of 4 X s — 32 a^ + 69 x — 45 being a? — 5, what are 
 the other two ? 
 
FACTORING 61 
 
 94. Prob. To factor a trinomial of the form a* 2 "* +px m + q. 
 
 Solution. We have seen (Art. 60) that 
 
 (x m + a) (af* + b)= x 2 " 1 + (a + b) of + ab. 
 
 Hence if in an expression of the form x 2 ™ + px m + q, q can be 
 resolved into two factors, as a and b, snch that their sum, a + &, 
 is equal to p, then the factors are x m + a and x m + 6. 
 
 EXAMPLES XXIII 
 Factor the following : 
 
 1. a 2 + 7 a; + 10. 
 
 Here 7 a; is the product of the square root of the first term, and the sum 
 of two factors, 2 and 5, of the last term. Hence the factors of x 2 + Ix -f 10 
 are x + 2 and x + 5. 
 
 2. ^ + 4^-21. 
 
 Here 4 a; 2 is the product of the square root of the first term, and the sum 
 of two factors, 7 and — 3, of the last term. Hence the factors of oj 4 +4x 2 — 21 
 are x + 7 and x — 3. 
 
 3. ^_4aj_32. 4. # 2 -8a + 15. 
 5. ^ + 10^ + 9. 6. x 2 + 6x-72. 
 7. a* + 12 a; -45. 8. ar 9 -16a,'-80. 
 9. a? 2 + 13 a; + 30. 10. x 2 -6x-55. 
 
 11. aj* + 5a? , -14 12. z 4 -17x 2 + 72. 
 
 13. * 4 + 10aj 2 + 24. 14. a 4 + 6z 2 -27. 
 
 15. z 6 -5ar J -84. 16. a 8 + 8 a! 4 -9. 
 
 17. ^ + 2/) 2 + 2(a + 2/)-35. 18. (ar J +3 2 /) 2 +13(ar 2 +3 7/)+42. 
 
 19. a 2 + 4a&-12 6 2 . 
 
 Here 4 a& is the product of the square root of the first term, and the sum 
 of two factors, 6 b and —2 b, of the last term. Hence the factors of 
 a 2 + 4 ab - 12 b 2 are a + 6 5 and « - 2 ft. 
 
 20. ar J -9a*?/-362/ 2 . 21. a^ + llafy 2 + 28# 4 . 
 22. x*-9x?y 2 + l$y*. 23. z 4 + 25 afy + 100 ?/ 2 . 
 24. ary + 3a*/z 2 -282 4 . 25. a; 4 - afyV - 56 y 4 z\ 
 
62 HIGHER ALGEBRA 
 
 95. Prob. To factor a trinomial of the form ax 2m + bx m + c. 
 
 Solution. ax 2m + bx m + c = - [(aaf) 2 + 6 (aaf) + ac], since 
 
 ft 
 
 dividing and multiplying by the same quantity does not change 
 
 the value of the expression. The part within the brackets may 
 
 now be factored by the process of the last article if b is the sum 
 
 of two factors of ac. 
 
 As the part within the brackets is to be divided by a, we may 
 
 divide one of its factors by the whole of a, or one of them by one 
 
 factor of a, and the other by the remaining factor of a, and thus 
 
 have two integral binomial factors of the original trinomial. 
 
 EXAMPLES XXIV 
 Factor the following : 
 
 1. 6x 2 + 7x-20. 
 
 Operation. 6 x 2 + 7 x - 20 = J [(6 x) 2 + 7 (6 x) - 120]. 
 
 Here 7 (6x) is the product of the square root of the first term and the sum 
 of two factors, 15 and — 8, of the last term. Hence the factors of the part 
 within the brackets are 6x+15 and 6 x — 8. Dividing the first of these by 
 3, one of the factors of 6, and the other by 2, the remaining factor of 6, we 
 have for the factors of the original trinomial 2 x + 5 and 3 x — 4. 
 
 2. 3x 2 -2«-5. 
 
 Operation. 3 x 2 - 2 x - 5 = \ [(3 x) 2 - 2 (3 x) - 15] 
 
 = i (3 x - 5)(3 x + 3) = (3 x - 5)(x + 1). 
 
 3. 12 x 2 -5 z-2. 
 
 Operation. 12 x 2 - 5 x - 2 = T \ [(12 x) 2 - 5 (12 x) - 24] 
 
 = & (12 x - 8) (12 x + 3) = (3 x - 2) (4 x + 1). 
 
 4. 6£ ? + 5a-4. 5. 10 a; 2 -11 ^ + 3. 
 6. 8 a? + 26 a? + 21. 7. 6 a? + 13 a> - 15. 
 
 8. 12 a; 2 -25 a + 12. 9. 15 a? + 14 <e .- & 
 
 10. 6 £ 2 — 7 a;# — 3 y 2 . *. 
 Operation. 6 x 2 - 7 xy - 3 */ 2 = \ [(6 x) 2 - 7 y (6 x) - 18 y 2 ] 
 
 = K6x-9?/)(6x+2?/) = (2x-3?/)(3x+!/) 
 
 11. 6 x 2 - 19 a# + 10 y 2 . 12. 10 « 2 - 3 xy - 18 f. 
 13. 7 a 4 + 25 a 2 + 12. 14. 14 a 4 + 25 a 2 + 6. 
 
FACTORING 63 
 
 15. (>a 4 + 7aV/-3b\ 16. 8 a 6 + 14 a*b 2 - 15 b\ 
 
 17. 15 a 6 - 23 arty 2 - 28 y 4 . la o.t- 8 + 11 a; 4 - 12. 
 
 19. 8^-6^-35. 20. 10 x 10 - 31 a?Y + 24 /'. 
 
 96. The methods of Arts. 94 and 95 will give the factors of 
 any trinomial of the form ax 2 ™ -f bx m y n -f- cy 2n when the factors are 
 rational. The finding of such factors when irrational or imagi- 
 nary requires the solution of a quadratic equation. 
 
 97. A polynomial having more than three terms may sometimes 
 be factored by first removing monomial factors from two or more 
 groups of terms. 
 
 Thus, let it be required to factor am + an + bm -f- bn. 
 
 am -\- an -f bm + bn = a (*» + n) -f b (m + n). 
 
 We now see that the binomial factor m + n is common to the 
 two terms, being contained a times in the first and b times in the 
 second. Hence the factors are m -f n and a + b. 
 
 98. Wlien a part of a polynomial can be factored by any of the 
 preceding methods, we may sometimes find the third terms of trino- 
 mial factors as follows : 
 
 Let it be required to find the factors of 
 
 6 x 2 - 5 xy - 2 x + 43 y - 2 1 y 2 - 20 . 
 
 This has the form of the product of two trinomial factors, each 
 having terms in x and y, and a term containing neither. 
 
 The factors of the part 6 x 2 — 5xy — 21 y 2 are, by Art. 95, 
 Sx — 7y and 2x + 3y. 
 
 The product of the terms not containing x and y is — 20. Hence 
 
 — 20 must be resolved into two factors such that the sum of the 
 products obtained by multiplying one of these factors by 3 x and 
 the other by 2 x shall be — 2 x; also the sum of the products 
 obtained by multiplying one of these factors by — 7 y and the 
 other by 3 y shall be 43 y. By trial these are found to be 5 and 
 
 — 4. Hence 
 
 6 x 2 - 5 xy - 2 x + 43 y - 21 f - 20 = (3 x - 7 y + 5) (2 x + 3 y - 4) . 
 
64 HIGHER ALGEBRA 
 
 EXAMPLES XXV 
 
 Factor the following : 
 
 1. ax* -2 ax 2 -3 x + 6. 
 
 Operation, ax 3 -2 ax 2 -3x+6 = ax 2 (x-2) -3(x-2) = (x-2)(ax 2 -3). 
 
 The result may just as well be obtained by placing the first and third 
 terms in one group, and the second and fourth in the other. 
 
 2. x 2 — mx — nx-\- mn. 3. a?x + ax -f a s y + ay. 
 
 4. 5a^-20a^-2a; + 8a. 5. 28 a 2 -21 a^ + 24a-18 2/. 
 
 6. x i — arfy + #z 2 — yz 2 . 7. a&a£ + bxy — axy — y 2 . 
 
 8. 9 a 2 + 6 ab - 15 ac- 10 be. 
 
 9. 3^-9i»y + 5a; 2 -15^ 2 -2^ + 62/ 2 - 
 
 10. x 2 y 2 -4;X 2 + 3xy 2 -12x- 10 y 2 + 40 (four factors). 
 
 11. 10 x>+ 11 a# - 14 xz - 6 / + 17#b - 12 z 2 . 
 
 Operation. This has the form of the product of two trinomials, each 
 having terms in x, y, and z. 
 
 By Art. 95, the factors of the part 10 x 2 + 11 xy — 6 y 2 are 5 x — 2 y and 
 2x + Sy. 
 
 The product of the terms not containing x and y is — 12 z 2 . Hence — 12 z 2 
 must be resolved into two factors such that the sum of the products obtained 
 by multiplying one of these factors by 5 a; and the other by 2 x shall be 
 — 14 xz ; also the sum of the products obtained by multiplying one of these 
 factors by — 2 y and the other by 3 y shall be 17 yz. By trial these are 
 found to be 3 z and — 4 z. Hence the factors sought are 5a; — 2y + 3z and 
 2x + 3y-4z. 
 
 12. x 2 + 3xy + 5x + 2y 2 + 8y + 6. 
 
 13. 2 x 2 - 5 xy - 12 y 2 - 9 xz - 8 yz + 4 z 2 . 
 
 14. 6 x 2 - 4 a?y - 12 xz + 9 a# - 6 # 2 - 5 yz + 6 z 2 . 
 
 15. 12 x 2 4- 14 ^ - 11 x - 10 y 2 + 25 y - 15. 
 
 16. 6 a 2 - 7 ^ - 3 ?/ 2 - 9 x + 30 y - 27. 
 
 17. 4 x 2 - 2 as - 9 y 2 - 27 yz - 20 z 2 . 
 
 18. 15 x 4 + sty* - 19 a,- 2 z - 6 2/ 4 + 19 y 2 z - 10 z 4 . 
 
FACTORING 65 
 
 99. Theorem. The difference of any two quantities is a divisor 
 of the difference of the same powers of the quantities. 
 
 The sum of two quantities is a divisor of the difference of the 
 same even powers, and the sum of the same odd powers of the 
 quantities. 
 
 Dem. 1st. Let x and y be any quantities, and n any positive 
 integer. Then x — y divides x n — y n . 
 
 Supplying the missing terms, changing the sign of the coeffi- 
 cient of the second term of the divisor, and proceeding with the 
 division as in Art. 79, we have 
 
 x n + x n ~ l y + x n ~y + x n ~y y n [1 
 
 111 
 
 It is seen that the quantity to be added to the last term is 1, 
 giving zero, and, consequently, that the division terminates, the 
 quotient being 
 
 x"- 1 + x n ~ 2 y + x n ~ 3 y 2 + af~y • • • -f- y n ~\ (1) 
 
 2d. Let n be even. Then x + y divides x n — y n . 
 Proceeding as before, we have 
 
 fl^+G«^y+-Oaf-y+0ir-y-" -y n \ -1 
 
 -i i -i 
 
 It is seen that if n is even, the quantity to be added to the last 
 term is 1, giving zero, and, consequently, that the division termi- 
 nates, the quotient being 
 
 x n-\ _ x n-2y _|_ x n-3 y 2 _ ^n-y . . . _ y n-l ( 2 ) 
 
 3d. Let n be odd. Then x + y divides x n + y n . 
 Proceeding as before, we have 
 
 x« + afty -f af-y + x n ~Y • •• + y n | — 1 
 -1 1 -1 
 
 It is seen that if n is odd, the quantity to be added to the last 
 term is — 1, giving zero, and, consequently, that the division 
 terminates, the quotient being 
 
 x n-l _ x n-2 y + x n-3 y 2 _ ^-y ... + yn-l^ 
 
 which is the same as (2). 
 downev's alg. — 6 
 
66 HIGHER ALGEBRA 
 
 100. Sen. When the sum or difference of the same powers of 
 two quantities is given, these theorems enable us to determine 
 whether it can be factored, and, if so, what one of the factors is. 
 Then the other can be written at once by the form (1) or (2). 
 
 EXAMPLES XXVI 
 Factor the following : 
 
 1. x? + y\ 
 
 Operation. Since this is the sum of the same odd powers of two quan- 
 tities, it is divisible by the sum of the quantities. Hence x + y is one factor, 
 and the other, as given in form (2) above, is x x — x*y + x 2 y 2 — xy % + y*. 
 
 2. x* — y 6 . 
 
 Operation'. x? — y 6 = (x 3 + y 3 ) (x* — y % ) 
 
 = (x + y) O 2 - xy + y 2 ) (x - y) (x 2 + xy + y 2 ) • 
 
 3. x G + f. 
 
 Operation. x« + y 3 = (z 2 ) 3 + y* = (x 2 + y) [(z 2 ) 2 - {x 2 )y + y 2 ] 
 = (x 2 + y)(x* -x 2 y + y 2 ). 
 
 4. 243 -a 5 . 
 
 Operation. 243 - a 5 = 3* - a s = (3 - a) (8« + 3 3 a + 3 2 a 2 + 3 a 3 + a 4 ) 
 = (3 -a) (81 + 27 a + 9a 2 + 3 a 3 -fa*). 
 
 5. x* — y 4 . 6. af — y 5 . 
 
 7. a 6 -64. 8. 27flf + y. 
 
 9. 32^-1. 10. 27 a 3 + 125 ?/ 6 . 
 
 11. 8^ 6 - 216. 12. a 9 + 6 6 . 
 
 101. Prob. To find by trial the binomial factors of a polynomial 
 of the form Ax n + Bx n ~ l + Cx n ~ 2 ••• +L, or of a homogeneous poly- 
 nomial of the form Ax n -f Bx 1l ~ l y + Cx n ~ 2 y 2 • •• + 7^/ M . 
 
 Solution. The short method of successive division, given in 
 Art. 81, affords a most useful method of factoring such expres- 
 sions. Since the product of the factors of a polynomial will 
 produce the polynomial, it is evident from the process of multi- 
 plication that the product of the last terms of the factors will 
 
FACTORING 67 
 
 produce the last term of the polynomial. Hence in rinding by 
 trial these factors we need to try for the last terms only those 
 numbers that are exact divisors of the last term of the polynomial. 
 For example, let it be required to rind the factors of 
 
 x 5 - 3 x A - 5 x* -\-15x 2 + ix - 12. 
 
 The binomial factors of the first degree are each x ± some num- 
 ber which is a factor of 12. A trial of the smaller factors of 12 
 results in the following : 
 
 3x 4 -5.^ 
 
 + WX 2 
 
 + 4 x - 
 
 -12| 
 
 1 
 
 ■2 -7 
 
 8 
 
 12 
 
 0| 
 
 2 
 
 -7 
 
 -6 
 
 
 
 I 
 
 3 
 
 3 2 
 
 
 
 
 1 
 
 -1 
 
 2 
 
 
 
 1 
 
 -2 
 
 
 
 
 
 
 
 It will be remembered that this is a short process of dividing 
 successively by x — 1, x — 2, x — 3, x -+- 1, and x -f- 2. Hence, 
 as each division is exact, the factors are x — 1, x — 2, x — 3, 
 x + 1, and x + 2. 
 
 102. Sch. It will usually be found expedient to find the posi- 
 tive numbers first, as above. When all the coefficients of any 
 row are plus, none but negative numbers need be tried, as the 
 successive additions of positive products to positive numbers 
 could not produce zero. Before writing down the various sums, 
 it is best, if the coefficients are not too large, to run through 
 mentally with a factor of the last term, and see whether the last 
 addition gives zero. 
 
 103. Note. If the student has forgotten the short process of 
 division as given in Art. 79, with its application in successive 
 division as given in Art. 81, he should go back and thoroughly 
 master it, as it is of frequent and important use in subsequent 
 parts of the work. 
 
68 HIGHER ALGEBRA 
 
 EXAMPLES XXVII 
 Factor the following : 
 
 1. rf - x 4 - 13 x 3 + 13 x 2 + 36 x - 36. 
 
 Operation 
 
 & - x 4 - 13 x 3 + 13 x 2 + 36 x - 36 1 1 
 
 0-13 36 0| 2 
 
 2-9-18 | 3 
 
 5 6 \-2 
 
 3 |_-3 
 
 
 The factors are x - 1 , x — 2, x — 3, x + 2, and x + 3. 
 
 2. 3 a* 4 + 13 a? - 117 a? - 243. 
 
 Operation 
 
 3x 4 + 13x 3 + 0x 2 -117x-243| 3 
 
 22 66 81 | — 3 
 
 13 27 
 
 The factors are x - 3, x + 3, and 3 x 2 + 13 x + 27. 
 
 3. ar 5 - 6 a; 2 + 11 a; -6. 4. ar* + 5a; 2 + 3a;-9. 
 
 5. ar*- 6 ar> + 13 a;- 10. 6. a 6 + 8 x 2 + 17 a? + 10. 
 
 7. a? } -13ar 2 + 49a;-45. 8. ^-15^+74^-120. 
 
 9. a; 4 + 2^-3^-4^ + 4. JO. x 4 - 10 ar 3 + 35 x 2 - 50 x + 24. 
 
 11. aj 4 -2a^-25aj 2 +26a;+120. 12. ^-4^ + 8^-8 a; -21. 
 
 13. a*-6rf + 5a* + 12x-60. 
 
 14. a?-a**-9a? + 5a? + 16aj-12. 
 
 15. aj 5 -9aJ 4 + 25a?-15aj«-26aj + 24. 
 
 16. ar 5 -4a! 4 -5a? + 20a* + 4aj-16. 
 
 17. 2a 4 -9ar } + 4a: 2 + 21a;-18. 
 
 18. 5ar 5 +7a; 4 -21ar } -lla; 2 + 32a;-12. 
 
 19. 3aJ 5 -2a* 4 -41a 8 + 56aj a -4a; + 48. 
 
 20. a; 6 - 14 a; 4 + 49 a; 2 -36. 
 
 21. a; 6 -4ar i -3a; 4 + 24ar J -10ar 2 -32a; + 24. 
 
 22. a? 7 + 5a^ + 6a?-6a; 4 -15ar J -3a; 2 + 8a; + 4. 
 
FACTORING 69 
 
 EXAMPLES XXVIII 
 Factor the following : 
 
 1. 7fg 2 y-28f 2 gy 2 + 42f s gy. 2. afy 8 - 7 tfy 4 + 12 xy 5 . 
 3. m 4 — n 4 . 4. 1 — 2 V# + a,*. 
 
 5. 256 a 4 + 544 a 2 + 289. 6. 1-c 3 . 
 
 7. X 2_ a ,_ 72 . 8. y e '-z 4 . 
 
 9. ^4-^-17^ + 15. 10. z 4 -9.t 2 -90. 
 
 u. ?5_^. + i§. 12 . ^-2. 
 m 1 mx 2 x 4 b 2 a 2 
 
 13. 125 + 64 a 3 . 14. ^-15^ + 47^ + 63. 
 
 15. a 2 + 23 a + 22. 16. a? + b\ 
 
 17 . c G-d c '. 18. c~ G -d- G . 
 
 19. 1-13^ + 22^. 20. 4X 2 + 8 x + 3. 
 
 21. a - b~ 6 . ■ 22. or 2 + 6 xy - 16 ?/ 2 . 
 
 23. --6- 10 . 24. 507ra 4 +1326m 2 n*+867n 3 . 
 x 4 
 
 25. ^a 4wi -2 2 T « 2w ^ 2n+2 + ¥ 1 T^ n+4 - 26 - 3a + 3 6-6Va6. 
 
 27. a 5 + & 5 . 28. 15 a + 5 a# — x — 3. 
 
 29. ^+3^-15 ^-19 a + 30. 30. 12aV*-12a 2 x* + 3a 2 . 
 
 31. a:y-29^ 2 + 54. 
 
 32. 21 abed — 28 cda;?/ + 15 abmn — 20 mnxy. 
 
 33. 2z 2 -13a + 6. 34. 3^-12^ 2 -4?/ 2 + l. 
 
 35. ^-14^+32^+95^+63. 36. ar 9 -a-9900. 
 
 37. x 2 + ax + # + a. 38. x 4 — 11 x?y + 18 y 2 . 
 
 39. 6aj 3 -7ax 2 -20a 2 a;. 40. ^-20^+30^+19^-30 
 
 41. x 2 " 1 + 31 x m - 32. 42. af— a£ — 2#— 2. 
 
 x 10 ?/ 100 
 
 «>«[!+g-2«* 
 
70 HIGHER ALGEBRA 
 
 45. 72 cdV - 84 cd*m 2 + 96 c 2 cl 2 m 2 . 46. 10 a 4 +- 79 x 2 - 8. 
 
 47. 2x 7 y + 54:xy 4 . 48. 6 <e° + 19 afy 2 — 7 ?/ 4 . 
 
 49. x 6 -x- 5 -20x' 4 + 50ar J -ll^ 2 -49a,-H-30. 
 
 50. 9 x 2 -\- 4 y 2 -\- z 2 — 12 xy -\- 6 xz — 4 yz. 
 
 51. 3» G -17^ + 23x 4 + 9x 3 -14a; 2 -4a-24. 
 
 52. 9 a 2 + 3 xy - 2 y 2 - 3 a + 13 y - 20. 
 
 53. x 2 +12 ay — 14 as + 36 ?/ 2 - 84 2/2 + 49 z 2 . 
 
 54. 2 x 7 - 3 a* - 14 x 5 + 20 ^ + 28 a* 1 - 37 x 2 - 16 a? + 20. 
 
 55. 10^ + 16a# + 13a;z-8?/ 2 -102/z-3z 2 . 
 
 56. Is a: 17 + y 17 divisible by x + y? by x — y ? 
 
 57. Is X s5 + 2/ 119 divisible by a* - y 7 ? by ar 5 + / ? 
 
 58. Write by form (2), Art. 99, the quotient of a 10 + b w divided 
 by a 2 + b\ 
 
CHAPTER VII 
 
 HIGHEST COMMON DIVISOR AND LOWEST COMMON 
 MULTIPLE 
 
 SECTION I — HIGHEST COMMON DIVISOR 
 
 104. The Highest Common Divisor of two or more algebraical 
 quantities is the quantity of highest degree that will exactly 
 divide each of them. 
 
 The abbreviation h. c. d. or H. C. D. is often used for highest 
 common divisor. The name highest common factor, with the 
 abbreviation h. c. f. or H. C. F., is also used. 
 
 105. Quantities are said to be prime to each other when they 
 have no common factor other than unity. 
 
 106. Note. As applied to literal quantities, highest common 
 divisor is preferred to greatest common divisor, since, when 
 numerical values are assigned, the quantity having the higher 
 degree may be the smaller. 
 
 Thus, for positive values, if a > 1, a z > a ; but if a < 1, a % < a. Again, 
 for values of x and y greater than 1, if x > y, x 2 — y' 2 > x — y ; but if x < y, 
 
 107. Theorem. The highest common divisor of two or more quan- 
 tities is the product of their common prime factors. 
 
 This follows directly from the definition. 
 
 EXAMPLES XXIX 
 Find the highest common divisor of each of the following: 
 1. 72 aW, 84 a*b 2 c, and 180 aW. 
 
 Solution. 72 a 4 W = 2 3 • 3 2 a*bW, 
 
 84 a 8 6 2 c =2».8 • 7 a?b% 
 
 180 aWc* = 2 2 • 3 2 • 5 a 6 6 4 c8. 
 
 71 
 
72 HIGHER ALGEBRA 
 
 The factors of highest degree common to all are 2 2 , 3, a 3 , 6 2 , and c. 
 Hence the h. c. d. is the product of these, or 12 a?b 2 c. 
 
 2. 48a 2 6 4 , 204 a?b 2 , and 228 a s b s . 
 
 3. 81 oY* 4 , 123 aYs 8 , and 315 aY- 
 
 4. 6 m\x — ?/) 3 and 9 m\x — yf. 
 
 5. aY + 2 ar 3 ?/ 2 and aY — 4 a; 4 ?/ 2 . 
 
 6. a; 2 4- 6x + & and a? 4- 3 a; 4- 2. 
 
 7. a; 2 + x — 6 and x 2 — 4. 
 
 8. a; 2 — 1 , x 2 — 3x-\-2, and a; 2 -f 6 a; — 7. 
 
 9. ar 3 4-1, a*-l, and &'—2x — 3. 
 
 10. 2ar*-3a;-2 and 4x 2 + 8^ + 3. 
 
 11. a 2 + 2aH & 2 , a 2 - & 2 , and a 3 4- b 3 . 
 
 12. a? 3 — a?, x 3 4- 9^ 2 — 10a?, and xP — x. 
 
 13. ar* 4- 3 afy 4- 2 ar# 2 and a; 4 4- 6 ar 3 ?/ 4- 8 a; 2 ?/ 2 . 
 
 14. ^_ x _42, x 2 -4b-60, and ar 2 4- 12^4-36. 
 
 15. 2ar 2 -7a;4-3 and 3b 2 -7b-6. 
 
 16. a; 4 -2ar 3 -13ar 2 +38a;-24 and ^-4^-7^ + 34^-24. 
 
 Operation. We proceed as follows, by the process of Art. 101 : 
 
 X 4 _ 2 x 3 - 13 x 2 + 38 x - 24 j_J_ x* - 4 a 3 - 7 a 2 + 34 x - 24 j 1 
 
 - 1 - 14 24 ) 2 -3-10 24 | 2 
 
 1-12 13 _1_12 14 
 
 4 [-4 3 [-3 
 
 
 
 The common factors are thus seen to be x — 1 and x — 2. Hence, multi 
 plying by the process of Art. 60, the h. c. d. is x 2 — 3 x + 2. 
 
 17. x 3 4- 2 sc 2 4- a; 4- 2 and ar 4 — 4 as 8 — b — 2. 
 
 18. a; 3 4- 4 x 2 - 8 a; 4- 24 and x 4 - x 3 4- 8 x - 8. 
 
HI0HB8T COMMON DIVISOR 73 
 
 19. 2ar 3 + x 2 - x - 2 and 6ar 3 - 4ar> + 2a; - 4. 
 
 20. x 4 -5x i -\-5x 2 -x-12 and a; 4 - 2x>- 12a? + 11 a; + 20. 
 
 21. ar 3 - 13a; + 12 and a 4 + 3 or 3 + 12 x - 16. 
 
 22. ^-4^-7 ^+34 z-24 and a^-G^+ar 3 +36^-20^-48. 
 
 23. a^-1 and x 4 + ar 3 - 9a,- 2 + 10a: - 8. 
 
 24. 12 x 4 - 24 afy + 12 ar 2 ?/ 2 and 8 x*y 2 - 24 afy 8 + 24 xy 4 -8y>. 
 
 25. a 4 - a s b - a 2 b 2 -2 b 4 and 3 a 3 - 7 a 2 b + 3 a£ 2 - 2 6 3 . 
 
 26. a 4 -5a 3 6 + 5a 2 fr 2 -a& 3 -12& 4 and a 4 - 2 a 3 6 - 12 a 2 b 2 + 
 lla& 3 + 20 6 4 . 
 
 27. 3ar , + 2a; 4 -47ar 3 + 10ar 2 + 128a;-96 and 3 ar 5 - 10 a; 4 - 31 ar 3 
 + 94 ar + 16ar-96. 
 
 28. ar 5 -5a; 4 -15ar 3 + 65ar 2 + 74a;-120 and V - 4 x 4 - 16 a^ 
 + 46^4- 63 a; -90. 
 
 Operation. By the process of Art. 101, the common factors are found to 
 be x — 5, x + 3, at + 2, and a — 1. After writing by Art. 60, the product 
 of the first two, we proceed as follows, by the process of Art. 59 : 
 
 (2) 1 -2 -15 
 (-1) -19 -30 
 
 _1 _io _n 30 
 
 Hence the h. c. d. is a* - x s - 19 x 2 - 11 x + 30. 
 
 29. x*-5x 4 + 7ar 3 -7ar 2 + 16a;-12 and x 5 - 8 x 4 + 26 x* - 46 ar 2 
 + 45 a;- 18. 
 
 30. ar 5 + 2ar 4 -15ar 3 -8ar 2 + 68a;-48 and ar 5 + 8 x 4 + 15 ar 3 - 
 20a; 2 -76 a; -48. 
 
 . 31. x* + 6x* + 6x* -16a,- 2 -15a; + 18 and a,- 5 + 2 a; 4 - 10 a,- 3 - 
 Sx 2 + 33a; -18. 
 
 32. 3ar 5 +a; 4 -llx 3 +3ar J +8a,'-4 and 3a,- 5 +7 x 4 -3 ar'-llarM- 4. 
 
 33. 2 y ; + 5.r 5 - 6a; 4 -4ar 3 + 2a; 2 -29a; + 30 and 2ar 5 -7a; 4 + 
 4ar 3 -4ar 2 -2a; + 15. 
 
74 HIGHER ALGEBRA 
 
 108. When one of the polynomials is readily factored and the 
 other not, ice may find by tried what factors of the first are contained 
 in the second, and thus obtain the h. c. d, of the two. 
 
 EXAMPLES XXX 
 Find the h. c. d. of each of the following: 
 
 1. 3*+ if and 5x 4 - 18afy + I2xhf -7 xif - 0>y\ 
 
 The factors of x* + ?/ 3 (form (2) of Art. 99) are x + y and x 2 - xy + y*. 
 By trial the second of these is found to be a factor of the second polynomial. 
 
 2. x* — 1 and 2 ar 3 — x 2 — x — 3. 
 
 3. 9«*-16 and 9^-15^ + 10^-8. 
 
 4. x 3 + 2 afy + 4 #?/ 2 + 3 if and a* 1 + afy + 4 a,- 2 ?/ 2 + .ry 3 + 3 y\ 
 
 5. a^ + T/ 5 and 7 x 6 - 10x*y + lOtfy 2 -10x 2 f + 10xy A - 3f. 
 
 109. Note. When the polynomials, or one of them, can be 
 readily factored, the processes of Arts. 107 and 108 of finding the 
 h. c. d. are the most expeditious. Otherwise, the process of Art. 
 Ill, similar to that used for numbers in Arithmetic, may be 
 employed. 
 
 110. Theorem. 1st. A divisor of a quantity is a divisor of any 
 multiple of that quantity. 
 
 2d. A common divisor of two quantities is a divisor of their sum 
 and also of their difference. 
 
 The first is self-evident. 
 
 For the second, let d be a common divisor of a and b, and 
 q and q' the respective quotients. Then 
 
 \a = qd, 
 
 and b = q'd, ■ 
 
 whence a±b = qd ± q'd = (q ±q')d. 
 
 Hence d is a divisor of a ± b. 
 
HIGHEST COMMON DIVISOR 75 
 
 111. Prob. To find the h. c. d. of two polynomial* which are not 
 readily resolved into their prime factors. 
 
 Rule. Having arranged the polynomials with reference to the 
 same letter, remove, and reserve as factors of the h. c. d., any mo- 
 nomial factors common to both polynomials, and reject from each 
 polynomial all other monomial factors. 
 
 Divide the reduced polyyiomial of higher degree in the letter of 
 arrangement by the other (if both are of the same degree, either may 
 be used as the dividend), first multiplying the dividend, if necessary, 
 by any number that will avoid a fraction in the quotient. 
 
 Reject from the remainder any monomial factors and divide 
 the former divisor by this reduced remainder, first multiplying } 
 if necessary, by any number that will avoid a fraction in the 
 quotient. 
 
 Continue the process either until there is no remainder, or until 
 the letter of arrangement disappears from the remainder. In the 
 latter case the two reduced polynomials are prime to each other, and 
 in the former case the product of the last divisor and the reserved 
 common monomial factors ivill be the h. c. d. of the given polynomials. 
 
 Dem. Rejecting from or introducing into either polynomial 
 or any remainder a monomial factor will not affect the h. c. d., 
 since the h. c. d. is composed of those factors only that are common 
 to the two polynomials. 
 
 If A and B represent the reduced polynomials, q, q', etc., the 
 successive quotients, and R, R', etc., the successive remainders, 
 and if we suppose the third remainder to be 
 0, the work (except rejecting and introducing B)A(q 
 factors) will stand as in the margin. qll 
 
 A divisor of A and B is (Art. 110) a divisor ft)B(q' 
 
 of A — qB, or R ; and a divisor of B and R is q'R 
 
 a divisor of B - q'R, or R'. Therefore any . R^Riq" 
 
 divisor of A and B is a divisor of R'; and q"R' 
 
 since R' is its own highest divisor, it is the 
 
 highest common divisor of A and B. Hence 
 the product of R' (the last divisor), and the reserved common 
 monomial factors is the h. c. d. of the given polynomials. 
 
76 TI1GUER ALGEBRA 
 
 EXAMPLES XXXI 
 Find the h. c. d. of the following : 
 
 1. 8^-48afy + 84afy 2 -40ary and 
 
 6a% - 48afy 2 + 126 ofy 3 - 120 afy 4 + 24 xy 5 . 
 Operation 
 
 Removing 4 x 2 from the first and 6 xy from the second, reserving 2 ac, the 
 h. c. d. of these, as a part of the h. c. d. of the polynomials, we have left 
 
 2 %* - 12 x 2 y + 21 xy 2 - 10 # 3 and x 4 - 8 x 3 y 4- 21 x 2 ?/ 2 - 20 x?/ 3 + 4 y 2 . 
 
 Multiplying the second by 2 to avoid fractions, and using Synthetic 
 Division (in which the signs of all the terms of the divisor except the first 
 must be changed), we have 
 
 2 x 4 - 16 x*y + 42 x 2 y 2 - 40 xy* + 8 y* | 2 x 3 + 12 x 2 y - 21 xy 2 + 10 y 3 
 12 -21 10 -20 x-2y 
 
 -4 -_24 42 
 
 - 3 12 - 12 
 
 .'. the remainder is — 3 x 2 + 12 xy — 12 ?/' 2 . 
 
 Rejecting — 3, we have for the new divisor x 2 — 4 xy + 4 y 2 . 
 
 2x 3 - 12x 2 */ + 21 xy 2 - 10y 3 [ x 2 + 4xy - 4y 2 
 _8 - 8 16 2x-4y 
 
 - 4 -10 
 
 - 3 ~6 
 
 .-. the remainder is —3x + 6y. 
 
 Rejecting — 3, we have for the next divisor x — 2y. 
 
 x 2 -4xy + 4y 2 \_2 
 -2 
 
 Hence, in dividing by x — 2 y there is no remainder. Multiplying this 
 last divisor by the common factor 2x reserved at the beginning, we have for 
 the h. c. d. of the original polynomials 2 x (x — 2 y) = 2 x 2 — 4 xy. 
 
 2. 3^ + 9afy-6a W 2 -62 / 3 and 24 a? + 6 x 2 y - 12 xy 2 - 18 f. 
 
 3. 21a?-32« 8 -54a>-7 and 21^-4^-15^-2. 
 
 4. 10^ + ^-9^ + 24 an d 20a 4 - 17.r 2 + 48a-3. 
 
 5. 2a! 4 -7<B 3 + ll<B 2 -8a; + 2 and 2 a; 4 + a? - 9^ 2 + Sx - 2. 
 
 6. 4^+14x- 4 +20a; 3 +70a; 2 and 8 af+28^- 8 x 5 - 12 a,* 4 +56 a?. 
 
LOWEST COMMON MULTIPLE 77 
 
 112. Prob. To find the h. c. d. of three or more polynomials. 
 Rule. Find the h. c. d. of any two of the given polynomials, then 
 
 find the h. c. d. of this h. c. d. and a third polynomial, and so on 
 until all have been used. 
 
 Dem. If A, B, C, etc., represent the given polynomials, and 
 M the h. c. d. of A and B, M contains all the factors common to 
 A and B (Art. 107). If N represent the h. c. d. of M and C, N 
 contains all the factors common to M and C, and consequently 
 all the factors common to A, B, and (7; and so on. 
 
 EXAMPLES XXXn 
 Find the h. c. d. of the following : 
 
 1. x^+llx + ZO, 2a 2 + 21;c + 54, and 9ar 3 + 53x? - 9x - 18. 
 
 2. x^-la? + 5x 2 + 31 x -30, x 4 - 9 ar 3 + 21 x 2 + x - 30, and 
 x 4 -8x i + 6x 2 + 72x-135. 
 
 3. 10ar 5 + 10ary + 20afy, 20^ + 2^, and 4 y* + 12 x-y + 4 ar 3 ?/ 
 + 12ar/. 
 
 SECTION II — LOWEST COMMON MULTIPLE 
 
 113. The Lowest Common Multiple of two or more algebraical 
 quantities is the quantity of lowest degree that is exactly divis- 
 ible by each of them. 
 
 The abbreviation 1. c. m. or L. C. M. is often used for lowest 
 common multiple. 
 
 114. Prob. To find the 1. c. m. of two or more algebraical 
 quantities. 
 
 Rule. Multiply one of the quantities (the one of highest degree 
 when they differ in degree) by all the factors of the others that are 
 not found in it. 
 
 Dem. Let the quantities be represented by A, B, C, etc. As 
 any one of the quantities, A for example, is its own lowest 
 multiple, the 1. c. m. of all the quantities must contain it as a 
 factor. As the 1. c. m. contains also B, C, etc., it must contain all 
 the factors of these quantities ; hence any of these factors that 
 are not found in A must be introduced. 
 
78 HIGHER ALGEBRA 
 
 115. Soh. In applying this rule, if the factors of the poly- 
 nomials are not readily found, we may find the h. c. d. in the usual 
 way, and then find the other factors by dividing each polynomial 
 by the h. c. d. 
 
 EXAMPLES XXXIII 
 Find the 1. c. m. of the following : 
 
 1. x 3 -2x 2 -5x + 6, ^-3 ^-z + 3, and ar ? + 4a 2 + *— $ 
 Solution. Factoring by the method of Art. 101, we have 
 
 x*-2x*-bx + Q=(x- 1)0-3)0 + 2), 
 x s -Sx 2 - £ + 3 =0-1)0 ~ 3 )0 + 1). 
 x* + 4x 2 + x-6=(x-l)(x + 2)(x + 3). 
 
 The second and third polynomials contain one factor each not found in 
 the first. Hence multiplying the first polynomial by these factors, using the 
 method of Art. 59, we have for the 1. c. m. 
 
 3 -2x 2 - 5x + 6)0 + l)0 + 3)=:« 5 + 2ic4 - 10x 3 -20x 2 + 9x+ 18. 
 
 2. x 2 + 2, x 2 - 2, and a; 4 - 4. 
 
 3. x 2 -y 2 , (x + y) 2 , (x - y) 2 , and (x + yf. 
 
 4. x>-9 and a 2 + 10 a -f 21. 
 
 5. x 2 -2x-lo and a 2 -4a -21. 
 
 6. £ 2 -3#-70 and a) 3 -39# + 70. 
 
 7. x 2 - 9, x 2 + x - 12, and x 2 + 2 x - 15. 
 
 8. x 2 + x — 2, a 2 — x — 6, and a,* 2 — 4 # +- 3. 
 
 9. 2^-7^-4, 4a 2 + 10a; + 4, and 6x? -1 x -5. 
 
 10. a 3 - 2 a 2 - 5 a + 6, a 3 - 3 a 2 - a +- 3, and a 3 + 4 a 2 + a-6. 
 
 11. ar 3 -2a; 2 2/+4a!?/ 2 -8^, ar } - r -2afy+4a;?/ 2 +8 2/ 3 , and ar-4?/ 2 . 
 
 12. a^ + Sa^ + Sa^-Sz-e, ^ + 6^ + 11^4-6, and f + 4f 
 + a;-6. 
 
 13. a 4 - 1, a 3 + a 2 + a + 1, a 3 — a 2 + a — 1, and a 2 + 1. 
 
 14. a 4 -10a; 2 + 9, a; 4 + 10 ar> +- 20 a; 2 -10 a; -21, anda 4 + 4ar 3 
 -22a 2 -4a; + 21. 
 
 15. 6o 2 -5a*/ + 2a;-62/ 2 + 232/-20 and 8 x 2 - 14 a^ + 22 x 
 + 3/ -13 2/ + 12. 
 
CHAPTER VIII 
 FRACTIONS 
 
 116. An Algebraic Fraction is an indicated operation in division 
 
 711 
 
 when written in the form — or m/n. 
 
 n 
 
 In this form the dividend, written above or before the vinculum, 
 
 is called the Numerator, the divisor is called the Denominator, and 
 
 the quotient is called the Value of the Fraction. 
 
 117. When the denominator is a positive integer, it indicates, 
 as in case of arithmetical fractions, the number of equal parts 
 into which a unit is conceived to be divided, and the numerator 
 indicates the number of these parts taken ; but it would be absurd 
 to represent a unit as divided into, say 5| equal parts, or into — 4 
 
 equal parts. Now in such a fraction as — , m and n are un re- 
 ft 
 
 stricted in value, and are not necessarily positive integers. Hence 
 the denominator of an algebraic fraction does not necessarily indi- 
 cate into how many equal parts a unit is conceived to be divided. 
 
 118. A quantity is said to have the Integral Form when it has 
 no part in the fractional form. 
 
 Thus, 3 mn, 2x — 3y, ax 3 - bx 2 y -f cy 3 , are in the integral form. 
 
 119. A quantity is said to have the Mixed Form when it con- 
 tains terms in both the integral and the fractional form. 
 
 Thus, a + — , x 2 - y + x ~ y , have the mixed form. 
 n x 2 — y 
 
 120. A Proper Fraction is one which cannot, without the use of 
 negative exponents, be reduced to the integral or mixed form. 
 If its numerator and denominator contain a common letter, the 
 numerator is of lower degree in this letter than the denominator. 
 
 79 
 
80 HIGHER ALGEBRA 
 
 Thus, x + y , — x ~ + ° — , are proper fractions. 
 
 121. An Improper Fraction is one whose numerator is not of 
 lower decree in a common letter than its denominator. 
 
 Thus, 
 
 x+5/ , 1 \ z 3 + 3a; 2 -4z + 5/ .. , - , Sx -10 
 
 £(=1— JLA ^ + 3^-4^ + 5/ 
 6\ x + <$) z 2 -2x + 3 V 
 
 K + 6\ x + ti/ x 2 -2x + S \ x 2 -2x + 3 
 
 are improper fractions. 
 
 122. A Simple Fraction is a fraction whose numerator and 
 denominator are both in the integral form. 
 
 123. A Complex Fraction is a fraction having its numerator or 
 its denominator or both in the fractional or mixed form. 
 
 124. Reduction is the operation of changing the form of a quan- 
 tity without changing its value. 
 
 125. The Lowest Common Denominator of several fractions is the 
 lowest common multiple (Art. 113) of their denominators. 
 
 , 126. A fraction is in its Lowest Terms when its numerator and 
 denominator are prime to each other. 
 
 REDUCTION OF FRACTIONS 
 
 127. Theorem. Multiplying or dividing both numerator and 
 denominator of a fraction by the same quantity does not change 
 the value of the fraction. 
 
 Dem. Represent numerator, denominator, and quotient by », 
 d, and q, respectively. 
 
 Since divisor times quotient equals dividend (Art. 66), we have 
 
 d x q = n. 
 
 Since multiplying or dividing a factor of a product multiplies 
 or divides the product, multiplying or dividing d and n in the 
 above equation by the same quantity will not change q. 
 
 128. Cor. Changing the signs of all the terms in a fraction does 
 not change the value of the fraction. 
 
 For this is equivalent to multiplying or dividing both numera- 
 tor and denominator by — 1. 
 
FRACTIONS 81 
 
 129. Prob. To reduce a fraction to its lowest terms. 
 
 Rule. Reject from numerator and denominator all factors com- 
 mon to both; or, what is the same thing, reject from both their 
 h. c. d. 
 
 This follows from Arts. 127 and 107. 
 
 130. Note. The student should endeavor to find the factors 
 of numerator and denominator, and after finding them he should 
 cancel those that are common, avoiding whenever possible the 
 laborious method of Art. Ill of finding the h. c. d. In none of 
 the following is it necessary to resort to this method. In the 
 last seven he should factor by the method of Art. 101. 
 
 EXAMPLES XXXIV 
 
 Reduce the following to their lowest terms : 
 
 60 a&V 420 a^c 2 " 
 
 84 abW " 630 ab 2 c n ' 
 
 3 210 ahk* n 4 bx + x 2 
 
 ' 330 aWd* ' db + ax 
 
 x 2 -^ 6 3 a*b - 6 a 2 b 2 
 
 5 X 4 _ y 4 ' 4a 2 & 2 -8ttfc 3 ' 
 
 x 2 _l_ 7 x _|_ 10 8 x 2 - x -12 
 
 " aj» + 4s-5" ' tf + x-20 
 
 ±x*-4:X + l 10 v? + 5 x - 14 
 
 ' 4ar 3 -3a + l' ' x* + 10 a + 21 
 
 ii a+*Y - i2. a3 +^ ■ 
 
 ' (1 — ic 2 ) 2 a 2 -\-2ax + x 2 
 
 ar»-27 14 a 2 - 7 x + 10 , 
 
 ' x*-2x-£ tf-Sx + G 
 
 x 2 -^ 2 16 l-5a + 6a 2 , 
 
 ' (x + xyf ' l-7a + 12a 2 
 
 afr-afl* 18 (a + &)'-(? . 
 
 downey's alg. — 6 
 
82 HIGHER ALGEBRA 
 
 19 (a + bf - ( c + df 2Q 6^-5^-4 
 
 (a + c) 2 - (b + d) 2 ' 6 a; 2 + a; -12* 
 
 ^ (fr + c + d) 2 - ft 2 22 s»-7s + 6 
 
 (a - &) 2 - (c + d) 2 ' tf- 6 ar* + 11 a: -6* 
 
 Scg. Factor by the process of Art. 101. 
 
 ar 3 - 8 x 2 + 19 x - 12 
 
 23. 
 
 24. 
 
 25. 
 
 26. 
 
 27. 
 
 28. 
 
 ar* - 10 ar> + 29 2 - 20 
 
 2 a; 4 - jg - 9 x* + 13 g - 5 
 
 7 x 4 - 26 x* + 36 jb 2 - 22 a; + 5* 
 
 gg - 8 ar* + 21 x - 18 
 
 3x 4 -22ar J + 53a 2 -42a/ 
 
 a 4 + 10 ar» + 35 ar* + 50 a- + 24 
 ic 4 + a? -19 a? 2 - 49 a;- 30 
 
 2 or 5 - 9 a; 4 + 8 a^ + 15 x 2 - 28 g + 12 
 3 X s - 19 x 4 + 45 a? - 49 x 2 + 24 a; - 4* 
 
 6 ar* - 19 a; 4 + 6 ar* + 36 ar* - 44 a; + 15 
 
 8 ic 5 - 18 x 4 - 3 ar 3 + 37 a^ - 33 x + 9 ' 
 
 131. Prob. To reduce an improper fraction to an integral or 
 mixed form. 
 
 Rule. Divide the numerator by the denominator, continuing the 
 division either until it terminates, or until the remainder is of lower 
 degree than the denominator. 
 
 EXAMPLES XXXV 
 
 Reduce the following to integral or mixed forms: 
 
 1 6a?-4s g -18a?-10 
 2s 8 -4s-l 
 
 Operation. Dividing by the method of Art. 76, we have 
 
 G x 3 - 4 x 2 - 18 x - 10 I 2a 2 + 4s-f 1 
 12 3 4 . 8x + 4 
 
 8 16 
 
 1 -6 
 
 Hence the result is 3 x + 4 -j g 
 
 2 se' 2 — 4 x — 1 
 
4 ^ x J "*" J ~ 9 5. 
 
 2af- 
 
 ■ 4 a 2 ?/ 4- 4 #?/ 2 
 
 -25 2/ 
 
 
 a» — 3jf 
 
 
 12 x 3 
 
 -8^ + 4a;- 
 
 -5 
 
 
 4ar + 3 
 
 
 2 a 3 - 
 
 - 3 ar 9 - 5 
 
 
 FRACTIONS 83 
 
 2 6 ar 9 - 12 a; -2 3 3 a 3 - 8 or + 6 a + 3 
 
 3x ' x- 2 
 
 Sue In such examples as the 3d and 4th employ the method of Art. 101. 
 
 m 2 4- n 2 4- 2 mn — x — y 
 m + n 
 
 32 ar 8 + 243 
 2a?4-3 
 
 8^+16^-10^-28^4-11 
 x 2 — x—1 2xr + x — 3 
 
 10 a 4 +7 ar*-ll arM- 13 x- 3 „ % 4 +4 afy 4-6 afy 2 -f 4 ay 3 4- ?/ 4 
 ' ~ 2o 3 4-7ar i 4-5a;-2 X s + 3 x*y + 3xy 2 + f 
 
 132. Prob. To reduce a mixed quantity to the fractional form. 
 
 Rule. Multiply the integral part by the denominator of the 
 fractional part, to the product add the numerator of the fractional 
 part, and, place the sum over the denominator of the fractional part. 
 
 Dem. The value of the integral part is not changed by multi- 
 plying it by the denominator of the fractional part and then 
 indicating its division by the same quantity. The two numera- 
 tors may now be added and written over the common denomi- 
 nator, since the quotient of the sum is equal to the sum of the 
 quotients, the divisor being the same (Art. 72). 
 
 133. Cor. An integer may be reduced to the form of a fraction 
 with any denominator by multiplying it by that denominator and 
 then indicating its division by the same quantity. 
 
 EXAMPLES XXXVI 
 
 Reduce the following to the fractional form : 
 
 a 3x — 4: 
 
 3. a - x -( a + *y. 4. *-*„ + ?-£*-. 
 
 a—x %+y 
 
84 HIGHER ALGEBRA 
 
 5. a 3 + a 2 6 + a& 2 + & 3 + -^-. 6. 1 ■ 
 
 a — b 2 be 
 
 7 . S.2+ ll* + 22 . 8 _ 6 3a»ft + 2ay-y 
 
 or 2 + 7 a? + 10 a 2 - b 2 
 
 134. Prob. To r reduce fractions having different denominators to 
 equivalent fractions having the lowest common denominator. 
 
 Rule. Divide the 1. c. m. of all the denominators by the denomi- 
 nator of each fraction, and multiply its numerator by the quotient. 
 
 Dem. The object of the division is merely to find the factor 
 by which the denominator of the fraction must be multiplied to 
 produce the lowest common denominator. As the numerator is 
 multiplied by the same factor, the value of the fraction is not 
 changed. 
 
 135. Cor. When the denominators have no common factors, the 
 lowest common denominator is the product of all the denominators, 
 and each numerator is multiplied by the denomiyiators of all the 
 other fractions. 
 
 EXAMPLES XXXVII 
 
 Reduce the following to equivalent fractions having the lowest 
 common denominator : 
 
 - x y z n 5 1 2 
 
 a + b a-b a 2 -b 2 2-2ar* 1-x 2 3#-3 
 
 x x 2 
 
 x y 
 
 x + y x*-y 2 x*-y 4 a 2 -b 2 (a + b) 2 (a-b)* 
 
 K 2x Sy 4z . 1 x x 2 
 
 x 2 — xy + y 2 x 3 + y 3 x + y ' x — y (x — y) 2 (x — yf 
 
 1 2 3 
 
 tf + x — ti ^4- 5 a + 6* a.- 2 — 4' 
 
 11 1 
 
 x + 2 ar* + 4a,- + 4 x*+6x* +12x + 4 
 
 9 (a?-2)(a?Hrl) (s+2)(a?-l) (a?+2)(a?+l) (x-2)(x-l) 
 x?+x-2 xF-x-2 ^-3^+2 ' a?+3x+2 
 
ADDITION AND SUBTRACTION OF FRACTIONS 85 
 
 ADDITION AND SUBTRACTION OF FRACTIONS 
 
 136. Prob. To add or subtract fractions. 
 
 Rule. Reduce the fractions to equivalent fractioyis having 
 the lowest common denominator, write the sum or difference of the 
 numerators over the lowest common denominator, and reduce the 
 result to its simplest form. 
 
 Dem. After the first step, which does not change the values 
 of the fractions (Art. 134), we apply the principle that the sum 
 or difference of the quotients is equal to the quotient of the sum 
 or difference, the divisors being the same (Art. 72). 
 
 EXAMPLES XXXVIII 
 
 Perform the operations indicated in the following : 
 
 a 2 + x 2 a 2 x _ x 3 -f- a 4 x 
 x + a 2 x — a 2 x 2 — a 4 
 
 Operation. Reducing to equivalent fractions having a common denomi- 
 nator, we have 
 
 x 3 - a 2 x 2 + a 2 x - a* a 2 x 2 + a*x _ sc 8 4 a*x _ a 2 x - a 4 _ a 2 (x - a 2 ) _ a 8 
 x' 2 - a 4 x 2 - a* x 2 - a* ~ x 2 — a 4 x 2 — a 4 x + a 2 
 
 1 +-L- 3 * + * 
 
 x -f y x — y x -\- x 2 x — x 2 
 
 a 2 + ab + b 2 a 2 -ab+b 2 _ m 2x + x*-x* 
 
 4. » • 5. 
 
 a 4-6 a — b 4 x 2 — 4 aj 8 + a; 4 2 # — ar* 
 
 6- 2 .. + 3 
 
 ar* + a? 4- a? 4- 1 ar* — sc 2 4- a; — 1 
 
 % — ct a 2 4- 3 ax x + a 
 a? 4- a a 2 — x 2 x — a 
 
 . 1 , 1 2 
 
 a; — 3 a 2 — 5 a! -(- 6 a^ — 6 # + 8 
 
 i , 7 + »'+-s; 
 
 #4-3 a? 2 — x — 12 a; — 4 
 10 3a^-8 bx + 1 2 
 
 a^_l aj 8 4-aJ + l*-l 
 
86 HIGHER ALGEBRA 
 
 x — 3 x — 2 x — 1 
 
 11. 
 
 12. 
 
 a 2 — 3 x + 2 x 2 — 4 a* + 3 a; 2 — 5 a; + 6 
 
 10a; 2 "2 1 
 
 (1 + a 2 ) (1 - 4 x 2 ) 1 + a* 1 - 2 a' 
 
 13 ft + 2 1 3 2 
 
 14. 
 
 a 2 — 1 x 4- 1 a; — 1 
 
 tf 4 + ^ + 1 SB 2 — X -f 1 it* 2 + 3 + 1 
 
 15 
 
 11 a-(a- 2 + 3) 
 
 
 x* _ 1 (a- + 1)« (& _ l)^ ( x . ._ iy 
 
 16. 
 
 1 05 + 1 X 2 + X + l t 
 
 a + 3 3*-3flS + ? ^ + 27 
 17 x 2 -2x-15 a» 2 +7a-+12 z 2 -16 
 
 ar 3 -5a* 2 -9a+45 ar'+3a- 2 -9a;-27 ar { -4a- 2 -9a-+36 
 
 MULTIPLICATION OF FRACTIONS 
 
 137. Theorem. 1st. Multiplying the numerator of a fraction 
 multiplies the fraction. 
 
 2d. Dividing the numerator of a fraction divides the fraction. 
 3d. Multiplying the denominator of a fraction divides the fraction. 
 4th. Dividing the denominator of a fraction multiplies the fraction. 
 
 Dem. The numerator is dividend, or quantity to be measured, 
 and the denominator is divisor, or measure. Now multiplying 
 the quantity to be measured multiplies the number of times it 
 contains the measure, and dividing the quantity to be measured 
 divides the number of times it contains the measure. 
 
 Likewise, multiplying the measure divides the number of times 
 it is contained in the quantity to be measured, and dividing the 
 measure multiplies the number of times it is contained in the 
 quantity to be measured. 
 
 138. Prob. To multiply a fraction by an integer. 
 
 Rule. 1st. Cancel all factors common to the integer and the 
 denominator. 
 
MULTIPLICATION OF Fli ACTIONS 87 
 
 2d. Multiply the numerator by the remaining factors of the 
 integer. 
 
 This follows from Arts. 127 and 137. 
 
 139. Prob. To multiply a fraction by a fraction. 
 
 Rule. 1st. Cancel all factors common to either numerator and 
 either denominator. 
 
 2d. Take the product of the remaining factors of the numerators 
 for the numerator of the result, and the product of the remaining 
 factors of the denominators for the denominator of the residt. 
 
 Dem. Let it be required to multiply - by — . these fractions 
 
 b n 
 
 being supposed to be simple, since if the given fractions are not 
 simple, they may be made so without changing their values. 
 
 Multiplying by m gives (Art. 137) ^. But we were to mul- 
 tiply, not by m, but by one nth. of m; hence this product must 
 be divided by n, which gives (Art. 137) — . 
 
 The rejection of all factors common to the numerator and 
 denominator will evidently leave the result the same as it would 
 have been had they been rejected before the multiplication. 
 Their rejection before multiplying leaves simpler quantities to 
 multiply. 
 
 EXAMPLES XXXIX 
 
 Multiply the following : 
 
 ± x 2 -j-2x-S b 4s 2 -12a-40 
 
 x 2 + 2 x - 3 4 x 2 - 12 x - 40 
 
 Operation. — — x — 
 
 x* + 5x + G 3x 2 -18z+15 
 
 = (x + S)(x -l) y 4(x- 5)(x + 2) = 4 
 (s + 3)(x + 2) 3(x-5)(x-l) 3 
 
 2. §££- by 21a^. 3. - v ^— , by x-y. 
 
 28 x K y l x- — y l 
 
 ^ + 4 5 _3 by ^ + 2z-3. 
 
 ^ + 4a;-21 J ^._7# + 6 J 
 
 Sue Factor the denominator of the 5th by the method of Art. 101, first 
 supplying the missing term, Ox 2 . 
 
88 HIGHER ALGEBRA 
 
 3o^ G^c 3 10 ac 2 ? a , a 2 + 6 3 
 
 4 c 4 5 a 6 9 6 3 a 4 -6« ^ a 2 - W 
 
 _ 9 a,* 2 — 1 , sb 2 + 5 sb _ a- — \ * a fi — 1 
 
 8. by — ! 9. by 
 
 x i -2ox J 3sb-1 (a + 1) 2 J (a 2 - a) 2 
 
 ;v & + 3x-iB . 2sb 3 -4sb 2 
 10. — ! by 
 
 a? _ 8 sb + 12 J af - 6 
 
 __ 5x + W ^ 3x-9^ 8x 2 -2 
 
 8 a? - 4 10 sb + 5 3 sb 2 - 27 
 
 , _ 1 — it* 2 1 — w 2 
 
 12. X — n X [ 1 + 
 
 1 + ?/ x + x 2 
 
 13. x2 ~ a; - 6 by _£-2»-8 
 
 sb 2 + 4sb + 4 a; 2 -7 sb + 12 
 
 14 x 2 + ?/ 2 b xy - y 2 
 x 2 — xy x 4 — if 
 
 " +:-+ Uy 
 
 iB" — ?/- 
 
 2\x — y % -\- yj ~ J v?y + sb?/ 2 
 
 16 rf + X y-2tf b or - 7 sb// + 12 ?/ 2 
 sb 2 — 5 #2/ + 4 y 2 x 2 + 5 xy + 6y 2 
 
 17 sb 2 y 2 , (sb 2 — ?/ 2 ) 2 
 
 " SB 2 -/ SB 2 + ?/ 2 y (f^ff+XJ+tff 
 
 18 sb 3 + 4sb 2 + 4sb + 3 , ar 3 - 5 sb 2 + 5 sb + 2 . 
 ' sb 3 - 4 sb 2 + 2 sb + 1 y X i -X 2 -X-2 
 
 19 3 x 4 - 17 sr 3 + 27 sb 2 - 7 sb - 6 b 2 sr 3 + 7 sb 2 + 7 sb + 2 
 
 ' 2 sb 4 + 13 sr 3 + 28 sb 2 + 23 sb + 6 J x? - 6 sb 2 + 11 x - 6 " 
 
 20 sb 5 - 3 SB 4 - 5 X s + 15 SB 2 + 4 SB - 12 . 
 
 sb 4 - sb 3 - 19 sb 2 + 49 .b- 30 " y 
 
 aj« - 2 sb 3 - 13 sb 2 + 38 sb - 24 
 
 a* _ 2 sb 4 - 10 ar 3 + 20 sb 2 + 9 sb - 18 
 
 21 10s*r i -15sB 3 -14sB 2 + 21 b Bar 8 - 9sb 4 + 5sb - 15 
 15sb 7 -21sb 4 + 25st 3 -35 y 2x i - 6sb 2 - 3sb + 9 ' 
 
DIVISION OF FRACTIONS 89 
 
 DIVISION OF FRACTIONS 
 
 140. Prob. To divide a fraction by an integer. 
 
 Rule. 1st. Cancel all factors common to the integer and the 
 numerator. 
 
 2d. Multiply the denominator by the remaining factors of the 
 integer. 
 
 This follows from Arts. 127 and 137. 
 
 141. Prob. To divide a fraction by a fraction. 
 
 Rule. Invert the divisor and proceed as in multiplication. 
 
 Dem. Let it be required to divide - by — , these fractions 
 
 b n 
 
 being supposed to be simple, since if the given fractions are not 
 simple, they may be made so without changing their values. 
 Dividing by m gives (Art. 137) — . But we were to divide, 
 
 not by m, but by one nth. of m ; hence, since the divisor used is n 
 times what it should be, the quotient is one nth of what it should 
 
 be, and must be multiplied by n, which gives (Art. 137) — . We 
 
 observe that in the operation we have inverted the divisor and 
 proceeded as in multiplying a fraction by a fraction. 
 
 EXAMPLES XL 
 
 Divide the following : 
 
 1. ^-^ by Ua 2 b 2 . 2. <*- 9 f by ar> + 3y. 
 
 6xy- x + ±y 
 
 3 a -.V by # + xy + y 2 4> « °_ by a" - b m . 
 
 3 ax ab 
 
 5 t^ll by a ~ b • 6 * 2 + 10 * + 21 by £=£ 
 
 ' a + 2b J 3a + 6b ' ^ - 4 or* + 3 a J x* - x 2 
 
 7. *~y* by *=!'• 8. x --± by (a + x y\ 
 
 tf _|_ yS *> rg _|_ y Q X aX 
 
 s» + 4gy + 4yg b xy + 2y 2 4 (a? - ah) , C>ab 
 
 x-y y tf-xy ' b(a + bf J d 2 -b 2 
 
90 HIGHER ALGEBRA 
 
 11. a 2 - b 2 - c 2 - 2 be by ^L±Jl±l. 
 J a+b-c 
 
 2sf±lSx±W , 2 a; 2 + 11 a; + 5 
 4a; 2 -9 y 4a; 2 - 1 
 
 13 ar* - 14 x - 15 b ar 9 - 12 a; -45 
 x 2 - 4 a; - 45 1J x 2 - 6 x - 27 ' 
 
 14. 5 - a * - 19 ^ by 3 - a ~ 5x . 
 
 15. ^ + 2 .V | x by x + ty ® 
 
 x + y y y x + y 
 
 16. f ~ 4 by 
 
 16 ar 3 — a; 2 -f 4 a; — 4 
 17 a? -7 a; + 6 a^ + 3^ 
 
 a! 4 -3s 8 -7s 2 + 27a;-18 * a; 4 - 10 a; 2 + 9 
 18 2a; 4 + 3ar 3 -14ar*-9a; + 18 b 2a; 4 - 9ar» + 4ar* + 21 x - 18 
 aJ 4 + a?-7aj 2 -13*-6 J a; 4 + 6 ar 3 + 13a; 2 + 12 x + 4 ' 
 
 SIMPLIFICATION OF COMPLEX FRACTIONS 
 
 142. Prob. To reduce a complex to a simple fraction. 
 
 Rule. Multiply numerator and denominator of the complex 
 fraction by the 1. c. m. of the denominators of the partial fractions. 
 
 Dem. This does not change the value of the complex fraction 
 (Art. 127), and it removes the partial denominators, since they 
 are factors of the quantity by which all the partial fractions are 
 multiplied. 
 
 143. Note. A complex fraction may be simplified also by 
 uniting the terms of the numerator and denominator separately, 
 and then dividing the former result by the latter. 
 
 EXAMPLES XLI 
 
 Reduce the following to simple fractions : 
 
 a_ c a b 
 
 b d b a 
 
 1. 
 
 f*h ab 
 
SIMPLIFICATION OF FRACTIONS 91 
 
 a + h + 'l <i±A + >±=± 
 a a — 1 «. + 1 
 
 6 a — 1 a + 1 
 
 ^ x + y x a + 4 
 *• ; — • 6. — — • 
 
 a + 1 - 
 
 x—yx a + 4 
 
 a , b , c 
 
 m — n . m* 
 
 + _ + I^__L" + 
 
 6c ac ao m + « 
 
 7. -• 8. 
 
 aft ac , 6c m 2 nt*» -f n 3 
 
 c b a m — n (ra — n) 2 
 
 144. Prob. To simplify a terminated continued fraction. 
 Rule. Begin at the bottom and simplify backwards and upwards, 
 performing, step by step, the operations indicated. 
 
 EXAMPLES XLII 
 Simplify the following : 
 
 1 a 
 
 l. 
 
 * + -~T b-r-—^- 
 
 1 
 
 y+- d + 
 
 3. 
 
 e 
 
 i 
 
 x + 2 
 
 a' — 
 
 .+*-£±i a + 
 
 *+ >' 
 
 a + 1 
 
 1 + 7 
 
 5. T^^-b"). 
 
 b 
 a x + 
 
 1 _a 1+*±I 
 
 & 3 -a 
 
 145. Prob. To free a fraction from negative exponents. 
 
 Rule. If the quantity affected with a negative exponent is a 
 factor of the numerator, transfer it to the denominator, or if a 
 
92 HIGHER ALGEBRA 
 
 factor of the denominator, transfer it to the numerator, and change 
 the sign of the exponent. 
 
 If it is a term or part of a term of a polynomial, transfer it to 
 the denominator of the term in which it stands, changing the sign of 
 the exponent, and then reducing the resulting complex fraction to a 
 simple one; or, what is the same thing, without this transference 
 multiply both numerator and denominator by the same quantity 
 affected ivith a numerically equal positive exponent. 
 
 Dem. The rule is a direct consequence of the signification of 
 a negative exponent (Art. 9), viz. the reciprocal of what the ex- 
 pression would be if the exponent were positive. 
 
 EXAMPLES XLIII 
 Free the following from negative exponents : 
 
 ± a- 2 b- 3 c\ 
 x~ x y~ n ' 
 
 3 . ( x ~y)~\ 
 
 x~ 3 + y~ s 
 
 (ar 1 - y~ 2 )~ 2 (x~ 2 -iy 
 
 2 
 
 a~' 
 
 l -b- 
 
 -3 
 
 
 
 a~ { 
 
 5 +6- 
 
 -2 
 
 
 4 
 
 xy 
 
 - 3 (a~ 
 
 2 -b 
 
 - 2 ) 
 
 
 z~ 
 
 2 (a" 3 
 
 ■ + b~ 
 
 3 ) 
 
 6. 
 
 x-'' 
 
 m (x 2 - 
 
 -</)" 
 
 ■n 
 
CHAPTER IX 
 
 THEORY OF EXPONENTS, INVOLUTION, AND EVOLUTION 
 
 SECTION I — THEORY OF EXPONENTS 
 
 146. We may either define fractional and negative exponents, 
 as in Art. 9, and prove the index law of multiplication to be 
 general, as in Art. 48, or we may assume the index law as proved 
 for positive integral exponents to be general, and determine what 
 fractional and negative exponents should mean. 
 
 147. Theorem. If the index law of multiplication is made 
 general, 
 
 1st. The numerator of a fractional exponent denotes a power, 
 and the denominator a root. 
 
 2d. A negative exponent denotes the reciprocal of ivhat it would 
 denote if positive. 
 
 Dem. 1st. By definition (Art. 9, 2d (a)), 
 
 
 (ic n ) n = X" • x n • of ••• to n factors. (1) 
 
 By the index law of multiplication (Art. 48, 1st) this is 
 
 — H 1 — ton terms ,_, 
 
 x n n n = x m . (2) 
 
 m 
 
 Therefore, (x 7l ) n = x m . (3) 
 
 Extracting the nth root of these equals, we^have 
 
 m 
 
 x" = y/gr. (4) 
 
 2d. By the index law of multiplication (Art. 48), 
 
 x 11 • x~ n = t? s 1 (Art. 70). 
 93 
 
94 HIGHER ALGEBRA 
 
 Dividing these equals by x n , we have 
 
 x~ n = — • 
 x n 
 
 EXAMPLES XLIV 
 
 Express the following without fractional and negative ex- 
 ponents : 
 
 1. a¥. 2. aW. 3. 5(ax)~*. 
 
 i 
 
 2 _ 5x~$ - 7*x " 
 
 O. • D. 
 
 7. 
 
 3 x* 7 2/* 6* y? 
 
 m 
 
 3a 2 6" 2 Q 5~ 2 ^V f 
 
 ■ • O. ■ ' * w. 
 
 5* a"*6* fSfi 4 -i y n 
 
 Express the following without radical signs and negative ex- 
 ponents : 
 
 io. v^^?. ii. tv&y?. i2. vzw^r*. 
 
 JLo. — — _ • Xrk, — — _ ^^z — * 
 
 Vsc v^2/" 2 V5 Va V# -3 V?/ 2 
 
 15. m ^gy^L - 16. ^/(^T67V(a-6)- 3 . 
 
 tyx t Vy~* 
 
 1? _ ^(a 2 + 2 5)\ 18 _ V(^-y 2 )-^ 
 
 V(x + i/)- 3 * ^/(^ + 2/ 2 )^ 
 
 148. Theorem. The product of several quantities affected collec- 
 tively by an exponent is equal to the product of the several quantities 
 affected separately by the same exponents. 
 
 Dem. 1st. When the exponent is a positive integer. 
 (abc~-) n 
 
 = {abc '")(abc >-)(abc ...)... to n factors, 
 
 = (aaa ••• to n f actors) (bbb ••• to n f actors) (ccc ••• to n factors) • ••> 
 
 = a n b n c n •••. 
 
THEORY OF EXPONENTS 95 
 
 2d. When the exponent is a positive fraction. 
 
 m to m to to to 
 
 (a^&V ...)* =(a - )"(6*)"(c*)" ••• by the 1st case, 
 
 = a^c" 1 —by (3) of Art. 147, 
 
 = {abc •••)"* by the 1st case. 
 
 Extracting the nth root of these equals, we have 
 
 TO TO TO TO 
 
 a n b n c n ~> = (abc •••)•. 
 
 3d. When the exponent is negative. 
 
 By Art. 71, and the 1st case of this demonstration, 
 
 (abc ..•)"" = = = a n b- n c- n .... 
 
 v J (abc •••)" a n b n c n ••• 
 
 149. Cor. By the proposition, 
 
 fa\ n ( IV „ 1 a n 
 
 [-) ± ox 7 =a n x— = — ; 
 
 \6y V &y h n b n ' 
 
 also, \h = \ a h~* a 
 
 la n\ ~i „/- v^ 
 
 150. Sch. The above theorem and corollary are briefly stated 
 thus : The power of the product equals the product of the powers, 
 and the root of the product equals the product of the roots (the 
 indices being the same) ; also the power of the quotient equals the 
 quotient of the poivers, and the root of the quotient equals the quotient 
 of the roots (the indices being the same). 
 
 The student needs to be cautioned against supposing that the 
 power or root of the sum or difference equals the sum or differ- 
 ence of the powers or roots. 
 
 151. Prob. To affect a monomial with any exponent. 
 
 Rule. Multiply the exponent of each of the factors by the given 
 exponent. 
 
 Dem. Let it be required to affect af* with the exponent n, m 
 having any value, integral or fractional, positive or negative. 
 
96 HIGHER ALGEBRA 
 
 1st. When n is a positive integer. 
 
 (x m ) n = x^x m x m ••• to n factors, 
 
 = x n 
 
 riff nn 
 
 /y.w+m+mH — to n terms 
 
 2d. When n is a positive fraction, as — • 
 
 p 
 By Art. 147, (x m ) q is the gth root of the pth. power of x m , i.e., 
 
 (x n y =V(x m y, 
 
 _ -fyo/mp^ \yj the 1st case of this demonstration, 
 
 mp 
 
 = a?« , by Art. 147. 
 
 3d. When n is negative. 
 
 By Art. 147, and the 1st case of this demonstration, 
 
 (x m )~ n = -i— = — = x~ mn . 
 
 Finally, by Art. 148, what has been proved above for x m applies 
 to any number of factors of any monomial. 
 
 EXAMPLES XLV 
 Perform the following indicated operations : 
 l. (a 3 6 2 ) 2 . 2. (a-V 2 c*) 2 . 3. (ah~k3f. 
 
 4. (3 2 afy-*) 3 . „ 5. (5x^sr l )\ 6. (6*«yW) 5 . 
 
 WJ \VyV \ ®>Vi J 
 
 10. (oafy 6 )*. 11. (25a 4 #V 2 )*. 12. ' 
 
 \27 by*, 
 13. (5^-y)" 3 . 14. (a~%-y)"i 15. (32 a 5 b- l0 c^)\ 
 
 16. (-64ic 6 2/V w )i 17. {V(afV) 8 } *• 18 (^ m9 2/^7 ? - 
 
BINOMIAL THEOREM 97 
 
 SECTION II — INVOLUTION 
 
 152. Involution is the process of raising quantities to required 
 powers. 
 
 153. Prob. To raise a monomial to any required power. 
 
 Rule. Use the numerical coefficient as many times as a factor 
 as there are units in the degree of the power, and multiply the expo- 
 nent of each literal factor by the exponent of the required power, 
 writing the minus sign before the result only when an odd power of 
 a negative quantity is required. 
 
 This is but an application of the definition of a power (Art. 9), 
 affecting a monomial with any exponent (Art. 151), and the law 
 of signs in multiplication (Art. 47). 
 
 154. Factorial n is the product of all the integral numbers from 
 1 to n inclusive, and is written \n.* 
 
 Thus, [3 = 2-3, [5=2.3-4.5, |w = 2 • 3 • 4 ... n. 
 
 155. Prob. To raise a binomial to any required power. 
 
 Formula, x, y, and m being any quantities whatever, positive 
 or negative, integral or fractional, 
 
 (x+y)*» = x>»+mx<»->y+ m ( m ~ 1 )^-y+ Mm-l)(m-2) xtn _ y 
 + m(m-l)(m-2)(m-3) ^_ y + ^ 
 
 This is Newton's Binomial Formula, the demonstration of which 
 is given in a subsequent part of the work. From an inspection 
 of the formula we deduce the following theorem, called 
 
 THE BINOMIAL THEOREM 
 
 In the expansion of a binomial affected with any exponent, the 
 exponent of the first letter begins in the first term with the exponent 
 
 * The notation n ! is also employed. 
 downey's alg. — 7 
 
98 TITO HER ALGEBRA 
 
 of the binomial, and in each succeeding term decreases by 1 ; while 
 the exponent of the second letter begins in the second term with 1, 
 and in each succeeding term increases by 1. 
 
 Tlie coefficient of the first term is 1 ; that of the second, term is the 
 exponent of the binomial; and if the coefficient of any term be mul- 
 tiplied by the exponent of the first letter in that term and divided by 
 the exponent of the second letter increased by 1, the result will be the 
 coefficient of the next term. 
 
 156. Cor. i. The expansion of a binomial terminates when m 
 is a positive integer, and the number of terms is m -j- 1. 
 
 For the coefficient m(m — l)(m — 2) ••• (m — m), which is in 
 the (m + 2)th term, and all subsequent coefficients, are 0. 
 
 157. Cor. 2. When m is a positive integer, the coefficients equally 
 distant from the extremes are numerically equal. 
 
 For the expansion of (b + a) m has the same value as the expan- 
 sion of (a + b) m , but the terms occur in the reverse order. Hence 
 beyond the middle, which is a term or a sign, according as m is 
 even or odd, the coefficients need not be computed. 
 
 158. Cor. 3. If the sign of the second letter is minus, and m is 
 a positive integer, the signs of the terms of the expansion will be 
 alternately plus and minus. 
 
 For the odd powers of a minus quantity are minus. 
 
 159. Note. At this stage of his progress the student should 
 learn the above theorem and familiarize himself with its use in 
 raising binomials to powers, i.e. for positive, integral exponents. 
 He should be careful to note that, while such a form as (3 a 2 — 2 b s ) 5 
 may be expanded by the theorem, the laws apply, not to the expo- 
 nents and coefficients of a and b in the full expansion, but to 
 those of 3 a 2 and — 2 b s , regarded as the terms of the binomial. 
 These should be kept in parentheses through the expansion, the 
 indicated operations being performed afterward. It must not be 
 forgotten that an odd power of a negative quantity is plus and 
 an even power minus. 
 
J « 
 
 J. 
 
 
 -/ /ft- 
 
 
 f 
 
 <-&-' 
 
 
 BINOMIAL THEOREM 
 
 EXAMPLES XL VI 
 Expand the following by the Binomial Theorem : 
 
 1. (3x 2 -2tfy\ v__ _ ___~ 
 
 Solution. Here 3 x 2 takes the place of the first letter, and ' — 2 y z of the 
 second letter of the formula. ~*~c* L & r 
 
 Hence we have by the theorem, 
 
 (3 x 2 -2 ifY = (8 x°') 5 + &(3 x 2 ) 4 (- 2 ?/ 3 ) + 10(3 x 2 ) 3 (- 2 y*) 2 . 
 
 + 10(3 X 2 ) 2 (- 2 *»)« + 5(3 x 2)(_ 2 y 3)4 + (_ 2 y3)5 
 
 = 243 x 10 - 810 xV + 1080 x 6 ^ - 270 x*y» + 240 x 2 y 12 - 32 y 15 
 
 Since wt = 5, the expansion will contain 6 terms (Art. 156), and the coeffi- 
 cients after the middle sign are the same as those before it, but in the reverse 
 order, and may be written without further computation (Art. 157). 
 
 2. (x-\-y)\ 3. (a -b) 5 . 4. (2x*-yy. 
 
 5. (l-x 2 ) 4 . 6. (a + by. 7. (x*-y 2 y. 
 
 8. (x* + 2y) 4 . 9. (l+2ar 9 ) 5 . 10. (a* + 2 fc^" 1 ) 4 . 
 
 11. (.*;-i) 6 . 12. (-J/a+V6) 8 . 13. (ix- 2 -2aV) 6 . 
 
 14. (.k - g + z) 4 . 15- (V^ + l-Va-l) 4 . 16. (x + V^-l) 6 . 
 
 160. A special theorem for squaring a polynomial has been 
 given in Art. 64. It is convenient to have also a special theorem 
 for cubing a polynomial. 
 
 161. Theorem. The cube of any polynomial is the algebraic sum 
 of the cubes of all the terms, three times the square of each term into 
 each of the other terms, and six times each group that can be formed 
 icith three terms each as factors. 
 
 Dem. When there are more than two terms, part of them may 
 be treated as constituting the first term, and the others as the 
 second term of a binomial, thus : 
 
 (a + b + cy=[(a + b)+cy=(a+by+3(a + b) 2 c+3(a + by+c* . 
 
 = a 3 -f 3 d 2 b+3 a6 2 +6 s + 3 a 2 c+(5 abc+3 b 2 c+3 ac 2 +3 bc 2 +(? 
 =a s +b s +c*+3cfb+3a 2 c+3b 2 a+3b 2 c+3c 2 a+3c 2 b + 6abc. 
 
100 HIGHER ALGEBRA 
 
 Again, (a+b + c—<Xf=[(a + b) + {c—d)J 
 
 = (a-|-6) 8 +3(a+6) 2 (c-d) + 3(a+6)(c-d) 2 +(c-d) 8 
 
 =a 3 +6 3 +c 3 -d 3 +3a 2 6+3a 2 c-3a 2 d + 3& 2 a+3 6 2 c 
 
 -3 tfd+3 c 2 a+3 c 2 6-3 c 2 d+3 d 2 a+3 d 2 b+3 d 2 c 
 
 + 6 abc—6 abd—6acd—6bcd. 
 
 By inspecting the results of these operations we deduce the 
 important theorem stated above. 
 
 EXAMPLES XLVII 
 
 Cube each of the following : 
 1. 4£ 2 + 2#-3. 
 
 Operation. It is best first to write in order the different powers of x, 
 and then supply the coefficients in the order of the rule, thus : 
 
 x 6 x 5 x 4 x 3 x 2 x 
 
 64 96 -144 8 - 36 54 - 27 
 
 48 - 144 108 
 
 64 x 6 + 96 x 5 - 96 x 4 - 136 x 3 + 72 x 2 + 54 x - 27 
 
 2. 2x 2 -3x-\-5. 3. 6a?-4xy + 2y 2 . 
 
 4. 4a?-2<B 2 + 3»-f-5. 5. x 4 -2x* + 3x 2 -x + 2. 
 
 SECTION III — EVOLUTION 
 
 162. Evolution is the process of extracting roots of quantities. 
 In extracting any even root of a quantity there are always two 
 
 results, differing only in sign, since even powers of both positive 
 and negative quantities are plus. For the present purposes it is 
 customary to write only the positive root, or in case of polyno- 
 mials, the root that begins with a positive term. 
 
 163. Prob. To extract any root of a monomial. 
 
 Rule. Extract the required root of the coefficient and divide the 
 exponent of each letter by the index of the root. 
 
SQUARE ROOT OF POLYNOMIALS 101 
 
 Dem. This is but an application of Art. 151, since to extract 
 the square root is to affect a quantity with the exponent \, the 
 
 cube root -J, the nth root -, etc. 
 
 164. Prob. To extract a root whose index is a composite number. 
 
 Rule. Extract successively the roots indicated by the prime fac- 
 tors of the index. 
 
 Dem. By Arts. 147 and 151, 
 
 2_ l i - 
 
 m m Vx = x mn = (af)» = V^a>; 
 
 j_ l i .. 
 
 also m Vx = x mn = (x n ) m = V Vaj. 
 
 Hence the 4th root is the square root of the square root ; the 
 6th root is the cube root of the square root, or the square root of 
 the cube root ; the 8th root is the square root of the square root 
 of the square root ; the 9th root is the cube root of the cube root ; 
 and so on. 
 
 SQUARE ROOT OF POLYNOMIALS 
 
 Case I 
 
 165. When the square is a trinomial. 
 
 Rule. Arrange with reference to one letter, extract the square 
 root of the extreme terms, and connect the results by the sign of the 
 middle term. 
 
 This follows from Arts. 61 and 62. The expression is not a 
 perfect square unless the middle term is twice the product of the 
 square roots of the extreme terms. 
 
 Case II 
 
 166. When the square contains only tivo powers of any one of its 
 letters. 
 
 Rule. Arrange ivith reference to any letter which has only two 
 poioer*. regard the terms not containing that letter as the third term 
 of a trinomial, and proceed as in the preceding case. 
 
102 HIGHER ALGEBRA 
 
 EXAMPLES XLVI1I 
 Find the square root of each of the following : 
 
 1. 9 a 4 + 4 b 2 + c 6 -f 12 a 2 b - 6 a 2 c 3 - 4 6c?. 
 
 Solution. Here we have but two powers of each letter, viz., a 4 and a 2 , 
 b' 2 and &, c 6 and c 3 . Arranging with reference to a (any other letter would 
 do as well), we have 
 
 9 a 4 + (12 b - 6 c 3 )a 2 + (4 b 2 - 4 6c 3 + c 6 ). 
 
 The square root of the first term is 3 a 2 , and of the last term 2 b — c 3 ; 
 while the middle term is plus twice the product of these square roots. Hence 
 the square root is 3 a 2 + 2 b — c 3 . 
 
 2. a 2 + & W - 2 abVd 3 + bYd* + 2 abed - 2 6W. 
 
 Solution. Here a is the only letter having but two powers. Arranging 
 with reference to this, we have 
 
 a 2 + (2 bed - 2 6W 8 )a + (& 2 c 2 d 2 - 2 6W* + & 4 c 4 d 3 ) 
 
 or a 2 + (2 bed - 2 & 2 c 2 #)a + (bed - MAP)*. 
 
 Hence the square root is a + 6ccZ — &&&. 
 
 3. 4 a 2 + 9 & 4 - 12 aft 2 - 24 & 2 d 3 + 16 ad s + 16 d\ 
 
 4. 16 sc 4 4- 16 oH 4- ?/ 2 — 8 ary 4- 4 z 2 — 4 ys. 
 
 5. 25z 4 + 20^-4#z 2 + 4/ + z 4 -10a; 2 z 2 . 
 
 6. 9x^z 2 + 12xY-4:tfz-\-4y 4 -Gx i z. 
 
 7. 16 a 4 4- 24 a 2 b + 9 b 2 - 16 a 2 c 3 4- 4 c 6 + 8 a 2 d 4- d J - 12 &c 3 + 
 6 6d-4c 3 d. 
 
 Solution. Here each letter has but two powers. Arranging with refer- 
 ence to a and, at the same time, arranging with reference to b the terms that 
 do not contain a, we have 
 
 16 a* + (24 b - 16 c 3 + 8 d)a 2 + [9 V 2 + (6 d - 12 c 3 )& + (4 c 6 - 4 c 3 d + <P)]. 
 
 The square root of the first term being 4 a 2 and of the last term (the part 
 in the brackets) 3 b — 2 c 3 + d, and the middle term plus twice the product 
 of these square roots, the square root of the whole expression is 
 
 4 a 2 + 3 b - 2 c 3 + d. 
 
 8. 9 a 6 - 6 a 3 & 2 - 12 a 3 c + 24 a*d 4- 6 4 4- 4 &*c - 8 & 2 cJ 4- 4 c 2 - 16 cd 
 4- 16 d 2 . 
 
SQUARE ROOT OF POLYNOMIALS 103 
 
 9. 36 a B + 30 « 3 c - 24 aV - 60 crtF + 4 6 4 + 20 6 2 d 4 - 12 b 2 c + 9 c 2 
 -30ctf 4 + 25d 8 . 
 
 10. 125 « 9 + 225 aftr - 150 a«c + 135 <W - 180 a 3 6 2 c + 60 a 3 c 2 + 
 27 6 6 - 54 6 4 c + 36 W - 8 c 3 . 
 
 Case III 
 167. Any perfect square not included in Cases I and II. 
 Solution. By Art. 64 we have, after arranging in descending 
 powers of x, 
 
 (ax n + bar- 1 + cx n ~ 2 + dx n ~ 3 -\ ) 2 
 
 = a*s* + 2 abx 2 ' 1 - 1 + (b 2 + 2 ae) x 2n ~ 2 + (2 6c + 2 ad) x 2 ^ 3 -f- .... 
 
 In squaring a polynomial of the nth degree a term in ar n can 
 result only from the square of the first term of the root; hence if 
 we take the square root of the first term of the arranged square, we 
 shall have the first term of the square root. 
 
 A term in x 2 "" 1 can result only from twice the first term of the 
 root by the second ; hence if we divide the second term of the ar- 
 ranged square by tvnce the first term of the root, we shall obtain the 
 second term of the root. 
 
 Terms in x 2 ' 1 - 2 can result only from the square of the second 
 term of the root and twice the first by the third ; hence if we sub- 
 tract from the third term of the arranged square the square of the 
 second term of the root and divide the remainder by twice the first 
 term of the root, we shall obtain the third term of the root. 
 
 Terms in x 2 "' 3 can result only from twice the second term of 
 the root by the third and twice the first by the fourth ; hence if 
 we subtract from the fourth term of the arranged square twice the 
 second term of the root by the third and divide the remainder by 
 twice the first term of the root, we shall obtain the fourth term of the 
 root. 
 
 In the same way four terms may be obtained from the other end 
 of the square, i.e., obtained as they would be if the square were writ- 
 ten the other end foremost, the signs being left temporarily ambiguous. 
 
 In practice it is better to obtain terms alternately from the two 
 ends, continuing the operation only until reaching d term that is 
 numerically the same, from whichever end obtained. The sign of 
 
104 HIGH Eli ALGEBRA 
 
 this term as obtained from the first end will determine the signs of 
 the terms obtained from the last end. 
 
 168. Sch. 1. If the square of the second term of the root is 
 similar to the fourth term of the given polynomial instead of the 
 third, the third term of the root is found by dividing the whole 
 of the third term of the given polynomial by twice the first term 
 of the root. 
 
 If the square of the second term of the root is similar to a 
 missing term of the given polynomial between the second and 
 third terms, the third term of the root is found by subtracting 
 the square of the second term of the root from 0, the coefficient 
 of the missing term, and dividing the remainder by twice the first 
 term of the root. 
 
 169. Sch. 2. Test. If the overlapping term (i.e., the term con- 
 taining the same power of x from whichever end obtained) is not 
 numerically the same when found from each end, the polynomial 
 is not a perfect square. When it is the same, to ascertain whether 
 the polynomial is a perfect square, it is necessary simply to note 
 how those terms not used in the operation are made up from the 
 square of the supposed square root. 
 
 170. Note. The advantage of this method over that employed 
 for extracting the square root of numbers is its brevity, the square 
 root being written at once, without writing any intermediate 
 steps. The written work in the model solutions below is merely 
 for explanation and can all be omitted in practice. 
 
 EXAMPLES XLIX 
 Find the square root of each of the following: 
 1. 25a 4 -30ar 3 + 49a 2 -24a + ±6. 
 Solution. If this is a perfect square, 
 
 first term of sq. rt. = V25& 4 = 5 x 2 , 
 second term of sq. rt. = — 30 x 3 -=-2(5 x 2 ) = —3 se, 
 last term of sq. rt. = VlO = ± 4, 
 next to last term of sq. rt. = - 24 x -=- 2 ( ± 4) = T 3 x. 
 
SQUARE ROOT OF POLYNOMIALS 105 
 
 This term (=F 3x) is numerically the same as the second, and to make it 
 the same in sign the upper signs must be used. Hence the" square root is 
 5 X 2 _ 3 x + 4. 
 
 Test. In the operation all the terms of the given polynomial have been 
 used except 49 x 2 . By reference to Art. 64 it will be seen that in squaring 
 5 x 2 — 3 x + 4, terms in x' 2 can result only from the square of the second term 
 and twice the first by the third. This gives (- 3x) 2 + 2(5 x) (4) = 49 x 2 , 
 showing the given polynomial to be a perfect square. 
 
 2. 49a; 4 + 28^ -38a,* 2 -12a; + 9. 
 
 3. 36a- 6 - 36a? -3a- 4 + 6a,- 3 + a* 2 . 
 
 4. 16 -40a; + 89ar*-80ar } + 64a* 
 
 5. 8a; + 4 + a; 4 + 4ar 3 + 8a* 2 . 
 
 6. o; 4 -ar 3 + -+4a;-2+4- 
 
 4 or 
 
 7. 36« 4 -48ar> + 52afy 2 -24#?/ 3 + 92A 
 Solution. If this is a perfect square, 
 
 first term of sq. rt. = V36x* = 6 x 2 , 
 
 second term of sq. rt. = — 48 xhj + 2(6 x 2 ) = — 4 xy, 
 
 last term of sq. rt. = V9 y* = ± 3 y 2 , 
 
 next to last term of sq. rt. = — 24 x?/ 3 -h 2( ± 3 y 2 ) = T 4 xy. 
 
 We see that there are but three terms and that the plus sign of \A)y* 
 must be used. Hence the square root is 6 x 2 — 4 xy + 3 y 2 . 
 
 Test. (4 xy) 2 + 2(6 x 2 ) (3 y 2 ) = 52 x 2 y 2 . 
 
 8. 25a? 4 -30arfy-31a% 2 + 24a^ + 16?/ 4 . 
 
 9. 9z 4 + 24afy-14afy 2 -40a; < 7- 3 + 25?/ 4 . 
 
 10. 16a 8 _32afy 2 -40 xy + 56z?f + 49/. 
 
 11. « 8 + 2ar 5 + 2a- 4 + ar J + 2a; + l. 
 
 12. x«-§x 5 + 9 as* — 10 a; 8 + 30 x* + 25. 
 
 13. 49 x™ -56x' + 16 x° + 126 av - 72 a; 3 4- 81. 
 
 14. 9x r > + 6a?*- 11 a; 1 + 20 ar* + 12 a 2 - 16a; 4- 16. 
 
106 HIGHER ALGEBRA 
 
 Solution. If -this is a perfect square, 
 
 first term of sq. rt. = V9 x 6 — 3 x 3 , 
 second term of sq. rt. = 6 x 5 -f- 2(3 x 3 ) = x 2 , 
 third term of sq. rt. = [ - 11 x 4 - (x 2 ) 2 ] - 2(3 x 3 ) = - 2 x, 
 last term of sq. rt. = Vl6 = ± 4, 
 next to last term of sq. rt. = — 16 x •*- 2( ± 4) = T 2 x. 
 
 This term (T2x) is numerically the same as the third, and to- make it 
 the same in sign the upper signs must be used. Hence the square root is 
 
 3x 3 + x 2 -2x + 4. 
 
 Test. All the terms of the given polynomial have been used except 20 x 3 
 and 12 x 2 . In squaring 3 x 3 + x 2 — 2 x + 4, terms in x 3 can result only from 
 twice the first by the fourth and twice the second by the third, giving 
 
 2(3x 2 ) (4) + 2(x 2 )( - 2 x) = 20 x 3 . 
 
 Terms in x 2 can result only from the square of the third and twice the 
 second by the fourth, giving 
 
 (-2x) 2 + 2(x 2 )(4)=12x 2 . 
 
 Hence the given polynomial is a perfect square. 
 
 15. 9X 6 - 24x 5 + 28 x 4 -46a? + 44a 2 - 20x + 25. 
 
 16. 16x 6 -40x 5 + 41x 4 -44x 3 + 34x 2 -12x + 9. 
 
 17. 49 x 6 + 42 x 5 - 19 x 4 - 68 x 3 - 20 x 2 + 16 x + 16. 
 
 18. 36 tf - 48 afy - 20 afy 8 + 84 xhf - 31 x 2 ?/ 4 - 30 xy 5 + 25 if. 
 
 19. 64 x s - 16 x 5 - 47 x 4 - 90 x 3 + 21 x 2 + 36 * + 36. 
 
 20. 121 x 6 - 66 afy + 119 xy + 168 xY - 29 x 2 ?/ 4 + 90 x,?/ 5 + 81 tf. 
 
 21. 9 x 6 + 6 x>y - 11 xSf + 20 x 3 ?/ 3 + 12 x 2 y* - 16 xy 5 + 16 /. 
 
 22. 4x° + 16x 3 -40x 3 + 16x + 4. 
 
 Solution. If this is a perfect square, 
 
 first term of sq. rt. = V4 x 6 = 2 x 3 , 
 second term of sq. rt. = 16 x 5 ■*- 2(2 x 3 ) = 4 x 2 , 
 third term of sq. rt. = [0 - (4 x 2 ) 2 ] -*■ 2(2 x 3 ) = - 4 x, 
 last term of sq. rt. s= VI = ± 2, 
 next to last term of sq. rt. = 16 x ■+■ 2( ± 2) = ± 4 x. 
 
8QUABS HOOT OF POLYNOMIALS 107 
 
 This term (±4x) is numerically the same as the third, and to make it 
 the same in sign the lower signs must be used. Hence the square root is. 
 2 x 3 + 4 x 2 - 4 x - 2. 
 
 Test. All the terms of the given polynomial have been used except 
 — 40 x 3 and a missing term in x 2 . In squaring 2 x 3 + 4 x 2 — 4 x - 2, terms 
 in x 3 can result only from twice the first by the fourth and twice the second 
 by the third, giving 
 
 2(2 x 3 ) ( - 2) + 2 (4 x 2 ) ( - 4 x) = - 40 x 3 . 
 
 Terms in x 2 can result only from the square of the third term and twice 
 the second by the fourth, giving 
 
 (_4x) 2 + 2(4x 2 )(-2)=0. 
 Hence the given polynomial is a perfect square. 
 
 23. ar s + 2a 6 + 2a~ 5 + 3a: 4 + 2ar J + 3ar J + 2a; + l. 
 
 24. 25^ + 20 a 7 - 6 x 6 - 4. ^ + 61 a 4 + 24. ^-12^ + 36. 
 
 25. 16a,- 10 - 8 a? + 16« 7 + 17 x 6 - 28 ar 5 + 14 ar 5 - 8 ar* -12 a? + 9. 
 
 Solution. If this is a perfect square, 
 
 first term of sq. rt. = Vl6 x 10 = 4 x 5 , 
 second term of sq. rt. = — 8 x 8 -*• 2(4 x 5 ) = — x 8 , 
 third term of sq. rt. = 16 x 7 -5- 2(4 x 5 ) = 2 x 2 , 
 last term of sq. rt. = V9 = ± 3, 
 next to last term of sq. rt. = — 12 x ■*■ 2( ± 3) = T 2 x, 
 third from last term of sq. rt. = [ - 8 x 2 - ( T 2 x) 2 ] + 2( ± 3) = T 2 x 2 . 
 
 This term (=F 2 x 2 ) is numerically the same as the third, and to make it 
 the same in sign the lower signs must be used. Hence the square root is 
 4 x 5 - x 3 + 2 x 2 + 2 x - 3. 
 
 As the square of the second term of the root is similar, not to the third 
 term of the given polynomial, but to the fourth, the whole of the third term 
 must be divided by twice the first term of the root. 
 
 26. a; 8 - 6 a; 7 + 5 a* + 22 u" - 18 x* - 44 x* + 9 x 2 + 40 x + 16. 
 
 27. 25x ,0 -30a^ + 20x 7 -31a^ 
 
 + 48 x> + 28 a- 4 - 52 x*+ 40 x* - 48 x + 36. 
 
 28. 16 x™ -48 a; 10 + 16 0* + 12 a,- 8 
 
 - 24 x 7 + 112 a: 6 - 12 x 5 - 99 x 4 + 36 a?- 54 3? + 81 
 
 29. 16 a- 10 -40 a* + 41 a,- 8 + 4. x 7 
 
 - 74 a* + 96 ar 5 - 45 x 4 - 24 a* + 54 a? - 36 x + 9. 
 
108 HIGHER ALGEBRA 
 
 Solution. If this is a perfect square, 
 
 first term of sq. rt. = VW x 10 = 4 x 5 , 
 second term of sq. rt. = — 40 x 9 -r- 2(4 x 5 ) = — 5 x 4 , 
 
 third term of sq. rt. = [41 x 8 - ( - 5 x 4 ) 2 ] + 2(4 x 5 ) = 2 x 3 , 
 fourth term of sq. rt. = [4 x 7 - 2 ( - 5 x 4 ) (2 x 3 )] - 2 (4 x 5 ) = 3 x 2 , 
 last term of sq. rt. = V9 = ± 3, 
 next to last term of sq. rt. = — 36 x -=- 2( ± 3) = q= 6 x, 
 third from last term of sq. rt. = [54 x 2 - ( T 6 x) 2 ] -*■ 2( ± 3 x) = ± 3 
 
 This term (± 3x 2 ) is numerically the same as the fourth, and to make it 
 the same in sign the upper signs must be used. Hence the square root is 
 4 x 5 - 5 x* + 2 x 3 + 3 x 2 - 6 x + 3. 
 
 30. 49 a; 10 4 14 a; 9 -27 a; 8 -60 a: 7 
 
 - 46 x« 4 80 x 5 + 38 x 4 + 4 a? - 31 a? - 30 a? + 25. 
 
 31. 16a 10 -16a 9 -52^ + 52a 7 
 
 4- 29 a,* 6 - 102 a; 5 + 55x 4 + 106 a 3 - 47 x 2 + 16 a + 64. 
 
 32. 9 a: 12 -30 a; 11 4 25 a; 10 -24 a? 8 4 52 a; 7 
 
 - 50 x G + 50 a,- 5 + 16 x 4 - 16 X s + 44 a* -20 x + 25. 
 
 33. 36 a; 12 - 48 x 11 -8x 10 + 28 x 9 + 56 x 8 - 80 a; 7 
 
 - 79 a; 6 + 78 ar> 4 47 x 4 - 44 ar> - 61 x 2 + 42 a; + 49. 
 
 34. 49 x 14 + 70 x 13 y - 17 x 12 y 2 - 44 ^y 4 111 x w y 4 4 142 afy" 
 
 - 35 a^/ 6 + 24 xhf + 128 afy 8 4 32 afy 9 - 28 afy 10 
 + 80 x*y n 4 52 arfy 12 - 24 a,y 3 4 36 y 14 . 
 
 Sug. The signs of the last four terms can be determined by noting what 
 sign of the fifth term of the root must be used in producing the fifth term of 
 the given polynomial. 
 
 CUBE ROOT OF POLYNOMIALS 
 
 Case I 
 171. When the cube is a quadrinomial. 
 
 Rule. Arrange with reference to one letter and take the algebraic 
 sum of the cube roots of the extreme terms. 
 
 Dem. By the binomial formula or by actual multiplication 
 
 we have 
 
 (a ± b) 3 = a 3 ±3 a 2 b 4 3 ab 2 ± b 3 , 
 
CUBE BOOT OF POLYNOMIALS 109 
 
 in which it is seen that the extreme terms of the quadrinomial 
 are the cubes of the respective terms of the cube root. 
 
 Test. If the quadrinomial is a perfect cube, its second term 
 will be three times the square of the first term of the root multi- 
 plied by the second, and its third term will be three times the 
 first term of the root multiplied by the square of the second. 
 
 EXAMPLES L 
 
 Find the cube root of each of the following : 
 1. 8aj» -36 3* + 5±x -27.. 
 Solution. If this is a perfect cube, 
 
 first term of cu. rt. = V8x 3 = 2 x, 
 second term of cu. rt. = V - 27 = — 3. 
 Hence the cube root is 2 x — 3. 
 Test. We observe that 
 
 3(2z)2(-3) = -36xV 
 and 3(2x)(-3) 2 = 54x. 
 
 2. 64x 6 + 96afy + 48ay + 82/ 3 . 
 
 3. 216 a* + 540 a? + 450 x + 125. 
 
 4. 125^-75^y + 15^-?/ 6 . 
 
 5. 64 x 3 - 144 x*y 2 + 108 xy 4 - 21 y\ 
 
 Case II 
 
 172. When the cube contains only three powers of any one of its 
 letters. 
 
 Rule. Arrange with reference to any one of the letters which has 
 only three powers, regard the terms not containing that letter as the 
 fourth term of a quadrinomial, and proceed as in the preceding 
 case. 
 
 EXAMPLES LI 
 
 Find the cube root of each of the following : 
 
 l. 8 a 6 - 36 a 4 b + 54 a 2 b 2 -27b 3 + 48 aV - 144 a 2 b<? + 108 6V 
 + 96 a 2 c 4 - 144 be 4 + 64 c 6 . 
 
110 HIGHER ALGEBRA 
 
 Solution. Here each letter has but three powers. Arranging with 
 reference to a, and, at the same time, arranging with reference to b the 
 terms that do not contain a, we have 
 
 8 a 6 + ( - 36 b + 48 c 2 ) a 4 + (54 b 2 - 144 be 2 + 96 c 4 ) a 2 + (- 27 6 3 
 
 + 108& 2 c 2 -144 6c 1 + 64c 6 ). 
 
 Since this is a quadrinomial, if it is a perfect cube, 
 
 first term of cu. rt. = V8 « 6 = 2 a 2 , 
 
 second term of cu. rt. = V - 27 & 3 + 108 6 2 c 2 - 144 6c* + 64 c 6 
 = _ 3 b + 4 c 2 . 
 
 As the intermediate terms conform to the test named in Case I, the given 
 polynomial is a perfect cube, and the cube root is 2 a 2 — 3 b + 4 c 2 . 
 
 2. a 3 4 6 3 + c 3 + 3 a 2 b + 3 a 2 c 4 3 ab 2 + 3 oc«+ 6 abc 4 3 6 2 c + 3 6c 2 . 
 
 3. 27 or 3 4 54 afy - 27 a 2 z 2 + 36 xy 2 - 36 a^-J- 9 xz A + 8 y 5 - 12 fz 2 
 
 4 6 2/z 4 — z 6 . 
 
 4. 61 x 6 - 96 x 4 f + 144 x 4 z 4 48 arty 6 - 144 tftfz -f 108 arz 2 - 8 */ 9 
 
 + 362/ 6 z-54^2 2 + 27z 3 . 
 
 5. 125 w 3 + 150 z<; 2 a; - 225 wrfy - 300 w 2 z + 60 wx 2 - 180 ivxy 
 
 - 240 tans + 135 wy 2 + 360 t<$2 4 240 wz 2 + 8 ar 3 - 36 a*y 
 
 - 48 a& 4- 54«?/ 2 + 144 ^z 4 96 a* 2 -27 f- 108 # 2 z 
 -144^ 2 -64z 3 . 
 
 Case III 
 173. Any perfect cube not included in Cases land II. 
 Solution. By Art. 161 we have, after arranging in descend- 
 ing powers of x, 
 
 (ax n 4 bx n ~ l 4 cx n ~ 2 4 dx n ~ 3 4 • • -) 3 
 = aV"4-3 a^- 2 +(3 a6 2 43 a 2 c) 3*- 2 4-(& 8 +6 a&c43a 2 cT)ar }w - 3 4 •••• 
 
 In cubing a polynomial of the nth. degree, a term in ar 3 * can 
 result only from the cube of the first term of the root ; hence, if 
 we take the cube root of the first term of the arranged cube, we shall 
 have the first term of the cube root. 
 
CUBE BOOT OF POLYNOMIALS 111 
 
 A term in a,* 3 " -1 can result only from three times the square 
 of the first term of the root by the second ; hence, if we divide 
 the second term of the arranged cube by three times the square of 
 the first term of the root, ice shall obtain the second term of the 
 root. 
 
 Terms in ic 3 " -2 can result only from three times the first term 
 of the root by the square of the second and three times the 
 square of the first by the third; hence, if we subtract from the 
 third term of the arranged cube three times the first term of the root 
 by the square of the second and divide the remainder by three times 
 the square of the first term of the root, ice shall obtain the third term 
 of the root. 
 
 Terms in a 3 " -3 can result only from the cube of the second 
 term of the root, six times the product of the first three terms, 
 and three times the square of the first term by the fourth ; hence, 
 if we subtract from the fourth term of the arranged cube the cube of 
 the second term of the root and six times the product of the first 
 three terms of the root, and then divide the remainder by three times 
 the square of the first term of the root, we shall obtain the fourth 
 term of the root. 
 
 In the same way four terms of the root may be obtained from the 
 other end of the cube, i.e. obtained as they would be if the cube were 
 written the other end foremost. 
 
 In practice, it is better to obtain terms alternately from the two 
 ends, continuing the operation only until the terms meet in the 
 middle of the root. 
 
 174. Sch. 1. There are the same kinds of exceptions here as 
 noted in Sch. 1 of Case III for square root. 
 
 175. Sch. 2. Test. To ascertain whether the given poly- 
 nomial is a perfect cube, it is necessary simply to note how 
 those terms not used in the operation are made up from the cube 
 of the supposed cube root. 
 
 176. Note. It is to be observed that, as in square root of 
 polynomials, the cube root can be written at once, without writing 
 any intermediate steps. 
 
112 HIGHER ALGEBBA 
 
 EXAMPLES LH 
 Find the cube root of each of the following : 
 
 1. 64 x & + 96 x 5 - 96 x 4 - 136 x? + 72 x 2 + 54 x - 27. 
 Solution. If this is a perfect cube, 
 
 first term of cu. rt. = ^(34 x*> = 4 x 2 , 
 second term of cu. rt. = 96 x 5 h- 3(4 x 2 ) 2 = 2x, 
 last term of cu. rt. = \/— 27 = — 3. 
 
 There can be no other terms in the cube root, as there can be no terms 
 between 2 x and — 3. Moreover, next to the last term of the cube root is 
 54 z -f- 3(— 3) 2 = 2x, the same as already obtained for the second term. 
 Hence the cube root is 4 x 2 + 2 x — 3. 
 
 Test. In the operation all the terms of the given polynomial have been 
 used except the third, the fourth, and the fifth. By reference to Art. 161 it 
 will be seen that in cubing 4 x 2 + 2 x — 3, terms in x 4 can result only from 
 three times the square of the first into the third and three times the square 
 of the second into the first, giving 
 
 3(4 x 2 ) 2 (- 3) + 3(2 x) 2 (4 x 2 ) = - 96 x 4 . 
 
 Terms in x 3 can result only from the cube of the second and six times the 
 product of the three, giving 
 
 (2 x) 3 + 6(4 x 2 ) (2 x)(- 3) = - 136 x 3 . 
 
 Terms in x 2 can result only from three times the square of the second into 
 the third and three times the square of the third into the first, giving 
 
 3(2x) 2 (- 8) + 80- 3)2(4 x 2 )= 72 x 2 . 
 
 Hence the polynomial is a perfect cube. 
 
 2. 8 x 6 - 36 x*y + 114 a*tf - 207 a?f + 285 x 2 y 4 - 225 xtf + 125 f. 
 
 3. 8 x 12 - 36 x w + 66 X s - 63 x« + 33 x A - 9 x 2 + 1. 
 
 4 . 8 - 12a + 42 x 2 - 61 tf+ 87 a** - 105^ + 87a) 6 - 66 x 7 + 36^ 
 
 Solution. If this is a perfect cube, 
 
 first term of cu. rt. = \/8 == 2, 
 second term of cu. rt. = — 12 x ■*■ 3(2) 2 = — x, 
 last term of cu. rt. = \/— 8 cc 9 = — 2 x s , 
 next to last term of cu. rt. = 36 x 8 -=- 3(- 2 x 3 ) 2 = 3 x 2 . 
 
CUBE ROOT OF POLYNOMIALS 113 
 
 As there can be no other powers of x between the extreme terms 2 and 
 
 - 2 x 3 , there are but four terms, and the cube root is 2 — x + 3 x' 2 — 2 x 3 . 
 
 5. 8 x 9 - 36 afy + 66 afy 2 - 87 ./'//' + 105 x y 4 - 87 afy 5 + 61 afy 6 - 
 42 afy 7 -+- 12 a;?/ 8 — 8 ?/ 9 . 
 
 6. 64x 9 -96« 8 + 192a; 7 4-88x 6 -96ar i + 366^ + 147^-15ar , + 
 225 a; + 125. 
 
 7. x 12 - 6x 11 + 21 a 10 - 47a 9 + 81 a 8 - 108 a 7 + 126 z G - 117 a 5 + 
 99 x 4 - 61 ar> + 42 x 2 - 12 x + 8. 
 
 Solution. If this is a perfect cube, 
 
 first term of cu. rt. = \ / x^ = x 4 , 
 
 second term of cu. rt. = — 6 a; 11 -h 3(x 4 ) 2 = — 2 x 3 , 
 
 third term of cu. rt. = [21 x 10 - 3(x*)(- 2 x 3 ) 2 ] + 3(x*) 2 = 3 x 2 , 
 
 last term of cu. rt. = V8 = 2, 
 
 next to last term of cu. rt. = — 12 x -^ 3(2) 2 = — x. 
 
 There can be no other powers of x between the extreme terms x* and 2. 
 Moreover, the third term from the last is [42 x 2 - 3(2) (- x) 2 ] + 3(2) 2 = 3 x 2 , 
 the same as already obtained for the third term from the other end. Hence 
 the cube root is x* — 2 x 3 + 3 x 2 — x + 2. 
 
 8. 64z 12 + 144a; 11 + 12 x w - 261 x? + 66 X s + 387^ + 82^- 171a 5 
 + 666 a 4 + 513a* 5 - 54.x 2 - 324a; + 216. 
 
 9. 27 x 15 - 54 x 14 + 63 x l * + 64 x' 2 - 204 x n + 87 x w + 187 a 9 -414 X s 
 
 - 36 x 7 + 433 x 6 - 192 x 5 - 192 x 4 + 408 x? + 165 x 2 - 225 x - 125. 
 
 EXAMPLES LIII 
 Solve the following by the principle of Art. 164 : 
 
 1. The 4th root of 
 
 16 a 4 - 96 a?x + 216 tfx 2 - 216 ax 3 + 81 x\ 
 
 2. The 6th root of 
 
 a 6 _ 6 a 5 b + 15 a 4 b 2 - 20 a 3 b 3 + 15 a 2 b 4 -6ab 5 + b 6 . 
 
 3. The 6th root of 
 
 729 - 2916 x 2 + 4860 x 4 - 4320 x 6 + 2160 a 8 - 576 a 10 + 64 x' 2 . 
 
 4. The 8th root of x 3 - 16 x 7 y + 112 afy 2 - 448 afy 3 + 1120 afy 4 
 
 - 1792 afy 5 + 1792 afy 6 - 1024 a?y 7 + 256 y 8 . 
 
 downey's alg. — 8 
 
114 HIGHER ALGEBRA 
 
 177. Prob. To extract any root of any quantity. 
 
 Solution. By the Binomial Theorem or by actual multiplica- 
 tion we have 
 
 (a + &) 2 =a 2 + (2a + 6)6, 
 
 (a + bf = a 3 + (3 a 2 + 3 ab + 6 2 )&, 
 
 (a + 6) 5 = a 5 + (5a 4 + 10a 8 6 + 10 a 2 6 2 + 5 a& 3 + b 4 ) b, 
 
 etc., etc., etc. 
 
 In reversing the process it is seen that the required root of the 
 first term of the given polynomial (or of the first period of the 
 given number) will be the first term (or figure) of the root. 
 
 Could we divide the rest of the polynomial (or number) by the 
 part in the parenthesis, we should obtain b, the next term (or 
 figure) of the root. Only the first term (or figure) of this divisor 
 is known until b is found. In case of polynomials, however, only 
 the first term of the divisor is needed (and in case of numbers 
 this first figure is much the larger part, since it is tens with ref- 
 erence to b as units). After the second term (or figure) of the 
 root is found, the divisor can be completed. 
 
 If the root contains more than two terms (or figures), we may 
 form a new trial divisor by considering the whole of the root now 
 found as the first term (or figure), and proceed as before. 
 
 It is assumed that the student is familiar with the reason, as 
 given in Arithmetic, for pointing off the number into periods of 
 as many figures each as indicated by the index of the root, begin- 
 ning at the decimal point. 
 
 178. Note. As already shown, the square root and cube root 
 of polynomials can be written at once by inspection. In the 
 extraction of roots of numbers the work is greatly facilitated by 
 the use of logarithms, as shown in a subsequent part of the work. 
 
CHAPTER X 
 
 SURDS AND IMAGINARIES 
 
 SECTION I — SURDS 
 
 179. A Radical is an indicated root of a quantity. 
 
 180. A Surd, or Irrational Quantity, is a radical quantity whose 
 indicated root cannot be exactly extracted. 
 
 181. A surd is Quadratic, or of the Second Degree, Cubic, or of 
 the Tliird Degree, Biquadratic, or of the Fourth Degree, etc., 
 according as its index is 2, 3, 4, etc. 
 
 182. Similar Surds are expressions containing the same surd 
 factor. 
 
 Thus, 2v / 3a, 5 m V3 a, (a 2 — m 2 ) VSa are similar surds. 
 
 183. To Rationalize an Expression is to free it from surds. 
 
 184. The Rationalizing Factor is the factor by which a surd 
 must be multiplied to rationalize it. 
 
 185. To Rationalize the Denominator of a Fraction is to free 
 the denominator from surds without changing the value of the 
 fraction. 
 
 186. The Simplest Form of a Surd is the form in which the 
 smallest possible integral number is left under the radical sign. 
 
 187. To Simplify a Surd is to reduce it to its simplest form. 
 
 REDUCTION OF SURDS 
 
 188. The substance of all demonstrations in Reductions is to 
 show: 
 
 1st. That the operation does not change the value of the ex- 
 pression. 
 
 2d. That the operation produces the required form. 
 
 115 
 
116 
 
 HIGHER ALGEBRA 
 
 189. Prob. To simplify a surd when the quantity under the rad- 
 ical sign contains a factor of which the indicated root can be taken. 
 
 Rule. Write the required root of this factor as the coefficient of 
 the indicated root of the other factor. 
 
 This is because the root of the product equals the product of 
 the roots (Art. 150). 
 
 EXAMPLES LIV 
 Simplify the following : 
 
 1. 5aVl47ajy. 
 Operation. 5 a Vl47 x 2 y 5 = 5 a V49xV x '3y =a 35 axy 2 VS y. 
 
 2. V32. 
 5. 2V98. 
 
 3. V75. 
 6. 3VI25. 
 
 4. V72. 
 7. 4V180. 
 
 a V48a^ 3 . 
 
 9.. V63a 5 b\ 
 
 10. 2cVl25aW. 
 
 11. 4V108ay. 
 
 14. V56a : W. 
 
 12. 3aV245 6V. 
 15. 3VS!^y 4 . 
 
 13. 6aV200 
 
 m°x*. 
 
 16. 5V135a&V. 
 
 17. aV432cW\ 18. 5 6 V128 a 4ra 6 3 "+ 6 . 19. 7 V -108 a 6 6 7 c 9 . 
 
 20. 2V343ajy. 
 23. Va 3 -f a 2 b. 
 
 21. 3 V720 a 2 6-V. 22. 5V448tt 3 6" 6 c 7 . 
 
 25. V20 a 5 6' 2 - 32 a 6 b 3 . 
 
 27. V3^--6^y + 3^*. 
 
 24. Vl8a?-27afy. 
 
 26. V2x 4 + 4^ + 2icV 
 
 28. V24a 4 -32a 3 6. 
 
 190. Prob. To simplify a surd when the quantity under the rad- 
 ical sign is a perfect power of the degree indicated by a factor of the 
 index. 
 
 Rule. Extract the root indicated by this factor of the index and 
 write the result as a surd whose index is the other factor. 
 
 This rule is simply an application of Art. 150. 
 
SEDUCTION OF SURDS 117 
 
 EXAMPLES LV 
 
 Simplify the following : 
 
 1. 4343. 
 
 Operation. -^343 = V\^H = V7. 
 
 2. </49. 3. </169. 4. 4225 a 4 & 2 . 
 5. 4'49. 6. ^216. 7. 4l000 a*b 6 . 
 8. 481. 9. a/625 a s b 4 . 10. 464. 
 
 11. 4- 729 aW.. 12. 41728 a 3wi + 6 . 13. 2^/32 a 5 b~ w . 
 
 14. 4l96a 6 6 4 . 15. SK/Sl^b 8 . 16. 4729 a 6 &«. 
 
 17. 4l6a^ + 64»2/ + 64 2/ 2 . 
 
 18. V64 x 3 - 576 x 2 y + 1728 as^ 2 — 1728/. 
 
 191. Prob. To reduce a rational quantity to the radical form. 
 
 Rule. Raise the quantity to the degree of the required radical 
 and place the result under the corresponding radical sign. 
 
 As the operation performed is the inverse of the operation 
 indicated, the value of the quantity is not changed. 
 
 192. Prob. To introduce under the radical sign the coefficient of 
 a surd. 
 
 Rule. Raise it to the power of the same degree as the surd and 
 multiply the result by the quantity under the radical sign. 
 
 Dem. The coefficient being reduced by the last rule to the 
 same form as the surd, we have the product of two like roots, 
 which is equal to the same root of the product (Art. 150). 
 
 EXAMPLES LVI 
 
 1. Reduce to radicals of the 2d degree, 
 
 3 ab 2 , 7 xhf, 11 xy\ and 2 x - 3 f. 
 
 2. Reduce to radicals of the 3d degree, 
 
 4 aPy, 5 a¥, — 3 a~ 2 bc 2 , and 6 xy 2 z~ 3 . 
 
118 HIGHER ALGEBRA 
 
 In the following introduce the coefficients under the radical 
 sign or within the parenthesis : 
 
 3. 3V2. 4. 4«V#a. 5. 2VJ. 
 
 6. \ VS. 7. 3^/3. a ^25. 
 
 
 f \ a?J " m — » >m + a 
 
 15. '*{*-?)*. 16- ^(l-J)- 17 - ( a - 1 )V5 +1 - 
 
 193. Prob. To reduce surds of different degrees to equivalent 
 ones of the same degree. 
 
 First Rule. Write for the common index the 1. c. m. of all the 
 indices, and raise each quantity under the radical sign to a power 
 ivhose degree is the factor by which its root index must be multiplied 
 to produce this 1. c. m. 
 
 Second Rule. Write the surds with fractional exponents, reduce 
 these exponents to a common denominator, and express the results in 
 the radical form. 
 
 Dem. By either rule the value of each surd remains the same, 
 since we extract a certain root and then raise the surd to the 
 corresponding power. 
 
 194. Sch. Surds are readily compared in magnitude by first 
 reducing them to equivalent surds of the same degree. 
 
 EXAMPLES LVII 
 
 Reduce the following to equivalent surds of the same degree : 
 1. V3#, -y/2x 2 , and Vox :{ . 
 
 OPERATION 
 
 1/bx* = ^(SajS)* = v"126jR 
 
REDUCTION OF SURDS 119 
 
 Or we may proceed thus : 
 
 y/2x* = 2^ x$ = 2& x& = y/Wx\ 
 ■ W& = &x* = 5T2 x & = Wl6?. 
 2. V2 and -^3. 3. 1/2 and ^/2. 
 
 4. V2 and V^. 5. y/t<? and \/3i, 
 
 6. V3 and W. 7. #5* and ^&» 
 
 8. Va, "v^6, and Vc. 9. V3a, V2 6, and V6c. 
 
 10. 3, VS, and </Il. 11. a 3 , W, and #?. 
 
 12. Va + 6 and V« — b. 13. V2 a, V2 a, and V2 a. 
 
 14. avfaaF* 2^^, and 4-y/12¥ 3 . 
 
 15. 5\/l0¥ 3 , 7^4^, and 6^7a-y. 
 Determine which is greater, 
 
 16. V7 or Vl8. 17. ^2 or </3. 
 
 18. V3 or </l6. 19. ^5 or </l4. 
 
 20. V& 3 or V# 4 . 21. -y/x* or V^ 5 - 
 
 195. While the value of a numerical surd cannot be determined 
 exactly, it may be determined to any required degree of accuracy 
 by the use of decimal places in the root. To find the approxi- 
 mate numerical value of V7 -*- V5 we may proceed thus : 
 
 ^I^ 2 - 645751 - = 1.183215-. 
 
 V5 2.236068 ••• 
 
 This, however, involves three tedious operations, viz., extracting 
 the square root of 7, extracting the square root of 5, and divid- 
 ing the first result by the second. By multiplying numerator 
 and denominator by V5, the second of these operations may be 
 avoided, and the third replaced by a much shorter one. Thus, 
 
 V7 = V7 x V5 = V35 = 5.916079 - _ 11832 i5 .... 
 V5 ' V5 V5 5 5 
 
120 HIGHER ALGEBRA 
 
 This suggests as suitable form for all surd fractions equivalent 
 fractions with rational denominators (Arts. 178 and 180). 
 
 196. Conjugate Surds are quadratic surds which differ only in 
 the sign of one of their terms. 
 
 Thus, Va + V& and Va — Vb, a + V& and a - Vb, Va + b and Va — b, 
 Va + Vb + Vc and Va + Vb — Vc, etc., are general forms of conjugate 
 surds. 
 
 197. Prob. To rationalize the surd denominator of a fraction. 
 Rule. 1st. WJien it contains a monomial quadratic surd, multi- 
 ply both terms of the fraction by this surd. 
 
 2d. When it contains a monomial surd of any other degree, as 
 -\/x m , in which m <n, multiply both terms of the fraction by -\/x n ~ m . 
 
 3d. When it contains a binomial quadratic surd, multiply both 
 terms of the fraction by the conjugate siird. 
 
 4th. When it contains a trinomial quadratic surd, multiply both 
 terms of the fraction by one of the surds conjugate to this, and then 
 by the surd conjugate to the reduced product of these two surds. 
 
 Dem. 1st. Va X Va = a, a rational quantity. 
 
 2d. -\/x m x -Vx n ~ m = -\/x n = x, a rational quantity. 
 
 3d. ( Va ± V&) ( Va =F Vb) = a — b, a rational quantity. 
 Also (a ± Vo) (a T V&) = a 2 —b, a rational quantity. 
 
 4th. ( Va ± V6 ± Vc) ( Va ± V& T Vc) = ( Va ± Vb) 2 - c 
 = a + b — c ± 2 Va6 ; and (a + b — c ± 2 Va6) (a + b — cT 2VaFj 
 = (a + 6 — c) 2 — 4 ab, a rational quantity. 
 
 EXAMPLES LVIH 
 
 Reduce the following to equivalent forms having rational 
 denominators : 
 
 1. 3Vg 
 
 4V7a 
 
 ±VTx *VYx" VTx 28x ' " 28 x 
 
 n 3V5 3V5 v V7x 3\/35x 3 ,^— 
 Operation. = x = , or V35 x. 
 
REDUCTION OF SURDS 
 
 121 
 
 2. 
 
 6 x 
 
 5</3a? 
 
 Operation. 
 4 
 
 <; 
 
 6* x ^ = 6^ = 2^ ^ 
 
 3. 
 
 V3 
 
 3V2 
 2V3* 
 
 5v/3x 2 5S!/3x 2 v/3 2 x 
 1 
 
 V2* 
 6V6ft 
 
 15 x 
 
 2Vft 2 + ft,y + y 2 
 3 Vft — # 
 
 12. uVf 
 
 15. 8VJ. 
 
 18. 
 
 21. 
 
 3 V3 x 
 
 2l/2x*' 
 
 Sab 
 
 v 27 a 4 
 
 22. 18 \7- 
 
 9a 
 16/ 
 
 25. *r* 
 
 28. 
 
 V(2-ft) 2 
 5 
 
 V5-V2 
 Operation. — 
 
 4. 
 
 7x 
 
 2V5ft 
 8 
 
 10. 
 
 7vTz 
 
 aVx 2 — xy + y' 2 
 
 b~vx-t-y 
 
 13. 20aJp-- 
 *5ft 
 
 16. 
 
 8 « V4~ft 
 
 7 6^/2^ 
 19. 7</f. 
 
 Sug. Write in the form 
 
 Vft^2 
 11. 3V|. 
 
 </3 
 17. ^| 
 
 20. 12\/|. 
 3 aft 
 
 V3%* 
 
 23. 
 
 16 ft 
 
 ^64^ 
 
 26 . ^+y° 
 
 V(ft + 2/) 3 
 
 24. 
 
 27. 
 
 1-ft 2 
 VWft 2 " 
 ft 4 — ?/ 4 
 
 29. 
 
 31. 
 
 33 
 
 3-V2 
 
 5+V3 
 5-V3* 
 
 Vs 2 + 1 - g; 
 Vft 2 + 1 + ft 
 
 5__ = 5 x v^±^ = 51^ 5 _±V?l = 5 C V5+V2) 
 
 V5 V2 V5-V2 V5+V2 6-3 
 
 30. ^£±^S 
 
 Vft— Vy 
 
 32. 
 
 34. 
 
 VI + ft 2 — ft 
 Vft + l+Vg — 1 
 Vft + 1 — Vft — 1 
 
122 HIGHER ALGEBRA 
 
 __ Vx 2 + x + Var* — x „ cc 2 H- a 4- xVa? + a 
 
 35. — ^^^2 » 3d. ^r:^^^ • 
 
 V& 2 4- 5 — V& 2 — a as + Va,* 2 -f a 
 
 37 .T + l+V^TT ^ a + ^-vV + frV 
 
 x + l-Va^-l a-ta + Vc^ + fc 2 ^ 
 
 39. — ■* 
 
 Operation. S x V5 + V5 + Vg = 8(V6 + V5 + V§) 
 
 V5 + V3 - V2 V5 + V3 + V2 3 + >/15 
 and 
 
 6(V5+V3+V2) x 3-V15 = 6(-2V3 + 3V2-v / 30) = ^30 + 2 V5-3 V2. 
 
 3+VT5 3-VI5 9 ~ 15 
 
 8 1 
 
 40. 41. 
 
 V3+V2 + 1 2V2+V3+V5 
 
 198. Prob. To find the factor which will rationalize any given 
 binomial surd. 
 
 Solution. The general form of such surd being y/cf ± V&*, 
 we have, 
 
 r « nr m» 1 1 
 
 Vo" r ± Vb 1 = a* ± 6" = a™ ± b™ = (a™)™ 1 ± b ms )™. 
 
 Now, with the lower sign in all cases and with the upper sign 
 when mn is even, a nr — b ms , which is rational, is divisible by 
 
 (a nr ) mn ± (b ms ) mn (Art. 99). When mn is odd, a nr + b™, which is 
 
 rational, is divisible by (a nr ) mn + (b ms ) mn (Art. 99). In either case 
 the quotient, which can be written by forms (1) and (2), Art. 99, 
 will be the rationalizing factor. 
 
 EXAMPLES LIX 
 
 Find the rationalizing factor of each of the following : 
 
 1. V3+^5. 
 
 Operation. V3 + y/E = 3* + 5* = 3^ + 5^ = (33) * + (5 2 ) K 
 
 Now 3 3 — 5 2 is the difference of the same even powers (the 6th powers) of 
 
 the terms of (8*)^ +<#)*, or 3^ + 5* and hence is divisible by 8*4-6* 
 (Art. 99). The quotient, which may be written by form (2), Art. 99, is the 
 
ADDITION AND SUBTRACTION OF SURDS 123 
 
 factor by which 3 2 -f 5*, or >/3 + \/b, must be multiplied to produce 3 3 — 5 2 , 
 or 2, a rational quantity. 
 
 (33 _ 52) - (3* + 5*) 
 
 = (3*)* - (3*)* (5^) + (3*)3 ( 5 *)2 _ ( 3 I)2 ( 5 1)8 + ( 3 i) ( 5 i)4 _ ( 5 1)6 
 
 = VJP - 9^5 + V3« #P - 15 + V3 v/5* - #P. 
 2. ^3-^2. 3. 2+^4. 
 
 4. V2-^5. 5. </?-$?. 
 
 ADDITION AND SUBTRACTION OF SURDS 
 
 199. Prob. To add or subtract surds. 
 
 Rule. Reduce each surd to its simplest form; then to the alge- 
 braic sum of the coefficients of the similar surds annex the common 
 surd part, and indicate the addition or subtraction of the dissimilar 
 surds. 
 
 This is but an application of the principles of Arts. 29 and 40. 
 
 EXAMPLES LX 
 Perform the following indicated combinations : 
 
 1. V27+3VI08 + 2V75. 
 
 Operation. V27 = V9 x 3 = 3V3 
 
 3VT08 = 3V36 x 3 = 18 y/S 
 
 2V75 =2V25o^ = 10V3 
 
 Ans., 31v/3 
 
 2. 3V50 + 4V72. 3. 2V75 + 3V245. 
 4. 5 V147 + 4 Vi08. 5. G V175 -f 7 V252. 
 6. 4V150-3V96. 7. 5V396-V176. 
 
 8. VT08 + 3 V27— 2 V48. 9. |Vl80 + ^V80- |Vi25. 
 
 10. V54 + 3v/l28. 11. 5^108-3^32. 
 
 12. 3 6^/128 a 4 b 2 - a^/Wab 1 . 13. 3 xJ/M2lc + 5</l62^. 
 
124 HIGHER ALGEBRA 
 
 14. bVWrf + xVMlrf -4^162^. 
 
 15. 7aV2SS~a s -7a 2 -\/512a 3 . 
 
 16. 3^/486^ + 2-^/64^. 
 
 17. 13^324¥ 4 -2V338aT 4 . 
 
 18. 2V| + 3V|+V^. 
 Operation. 2Vf = 4\/| = 4Vf = $v / 6 
 
 V^ = i Vf = f Vf =|V6 
 19. 6V| + 7V24. 20. 2V| + |V240. 
 
 - ^+#-# - ^i-i2vi. 
 
 23. Vf + Vi + i-v/9. 24 . Vfl + Vj + fv/sI. 
 
 25. Vf+VS-Vf. 26. Vf+V9-Vf|. 
 
 27. 3Vl47-jVi-2V^. 28. 18^ + -v/243-6^/I 
 
 29. 7 6 V18 a 4 - 27 a 2 6 2 - 2 a V8 a 2 6 2 - 12 b\ 
 
 30. 6V8a 3 6+l6a 4 + aV27fc 4 + 54a& 3 . 
 
 31. « = + 
 
 a -f- Va 2 ^^ 2 a — Va 2 — a 2 
 
 32 V a + # + Va — & Va + a;-Va-a; 
 Va + # — Va-« Va + a? + s/a — x 
 
 33. // a/3+V2 V | / V3-V2 V 2 
 
 W3-V2/ W3+V2/ 5-2V6* 
 
MULTIPLICATION OF SURDS 125 
 
 MULTIPLICATION OF SURDS 
 
 200. Prob. To multiply surds. 
 
 Rule. If the surds are of different degree, reduce them to equiva- 
 lent surds of the same degree; then to the product of the coefficients 
 annex the common root of the product of the quantities under the 
 radical sign, and reduce the result to its simplest form. 
 
 This is but an application of the principles of Arts. 193 and 148. 
 
 201. Sch. 1. If the quantities under the radical sign are 
 integral, the surds should first be simplified if not already in the 
 simplest form, as the factors that may be removed are much more 
 readily detected before the multiplication than after. Thus, we 
 may write, 
 
 V28 x V72 = V2016 = Vl44 x 14 = 12 VH ; 
 
 but after multiplying, considerable inspection is required to find 
 the largest square factor, and the difficulty is greater with larger 
 numbers. We should proceed thus : 
 
 V28 x V72 = 2 V7 x 6 V2 = 12 Vl4. 
 
 202. Sch. 2. When the quantities under the radical sign are 
 fractions, it is usually best not to simplify, but to employ cancel- 
 lation as far as possible. Thus, 
 
 Vf x V| x V^ = V| x | x V- = V3. 
 
 203. Sch. 3. Much work is often saved by resolving the 
 quantities under the radical sign into such factors as will show 
 those that are common, even when the surds are in the simplest 
 form, as this reveals any factors that may be removed after the 
 multiplication. Thus, we may write, 
 
 V42 xV77=V3234=V49 x66 = 7V66; 
 but much inspection is saved by writing 
 
 V42 xV77=V i nr6 xV7 x 11 = V7 2 x 66 = 7 V66. 
 Again, 
 
 T05xV70=V7x5x3xV7x5x2=V7 2 x5 2 x6 = 35V6. 
 
126 HIGHER ALGEBRA 
 
 204. Sch. 4. If multiplicand or multiplier or each of them 
 is a polynomial, the sum of the partial products must be taken, 
 as in the multiplication of rational polynomials. 
 
 EXAMPLES LXI 
 Perform the following multiplications : 
 
 1. V3 by Vl2. 2. V2l by VT. 
 
 3. V20 by VS. 4. V9 by y/S. 
 
 5. -V36x> by J/6x. 6. 3Vl5 by V50. 
 
 7. 4V6 by 3V12. 8. Jjxy- by tyify. 
 
 9. 4^/36 by V48. 10. 2VH by V2l. 
 
 11. V60 by V30. 12. V30 by V35. 
 
 13. 7V35 by V65. 14. V39 by V91. 
 
 15. V231 by VIM. 16. V9l by V182. 
 
 17. S/| by Vf. 18. Vf by Vf 
 
 19. 5V24a 2 6 by 7&V32«. 
 
 Solution. As the product of the same roots equals the root of the prod- 
 uct (Art. 148), if these surds are first reduced to the same degree, the multi- 
 plication can be performed. 
 
 Now, 5^24 a*b = 5^8 x 3a*b - 10y/S¥b = lOVWaW 
 
 1bVWa = lbVlQx2a = 28bV2a = 2$by/8<? 
 
 Product, 280 ab\Zl2ab* 
 
 20. 4V2 by 5V3. 21. 2V24 by 3^3. 
 
 22. </2tfb by Vl6^6. 23. Vl2 by y/3. 
 
 24. V3^ by V27^. 25. V7 x V2 x </E. 
 
 26. VjxVfxVV-- 
 
 r, 3 /T /2 6 /27 6 /T 6 /23 6 /33 . 
 
 Operation. V2 X \5 X A/Y = ^ X % x V| =1 - 
 
MULTIPLICATION OF SURDS 127 
 
 27. Vf by 3\/|. 28. 6V^ by \/2. 
 
 29. 72x^X^X^1. 30. Vf X a/I X </f . 
 
 31. VT+x by -v/l-ha?. 32. Va— 6 x Va+fc X Va* 11 ^. 
 
 33. 3V6-4V3 + 3 by 2V6 + 5V3. 
 
 Operation. 3\/6 - 4>/3 + 3 
 
 2V6 + 5V3 
 
 36-24V2+6V6 
 -60 + 45V2 + 15v/3 
 
 24+21\/2 + 6V6+ 15V3 
 
 34. 5-2V3 by 4 + 3V3. 
 
 35. 2V^ + 3V2 by 6V^-V2. 
 
 36. V5+V3-V2 by V5.+ V2. 
 
 37. 5 + 3V2 by 5-3V2. 
 
 Sue In multiplying the sum of two quantities by the difference, as in 
 the 37th, always apply the principle of Art. 63. 
 
 3a 3V5 + 2V3 by 3V5-2V3. 
 
 39. 2VII-5 by 2V14 + 5. 
 
 40. 7a^-4aV3a by 7a* + 4aV3a. 
 
 41. V5-V3-V2 by V5 + V3-V2. 
 
 42. V6 + V3+V5 by V5 + V3-V6. 
 
 43. 5V8 + 6VI2-2V20 by 7V2-3V3 + 4V5. 
 
 44. 3a-3Vab + 2b by 3a + 3Vab + 2 6. 
 
 Sec In solving examples like the last four, the terms should be so 
 grouped as to give the product of the sum and difference of two quantities. 
 
 45. 3Va + Va-9a by 3Va-V*-9a. 
 
 46. V&--Vxy + -y/tfby Vx + Vy. 
 
 47. ^5-2^6 by 3^4-^36. 
 
128 HIGHER ALGEBRA 
 
 DIVISION OF SURDS 
 
 205. Prob. To divide surds. 
 
 Rule. If the surds are of different degree, reduce them to equiva- 
 lent surds of the same degree; then to the quotient of the coefficients 
 annex the common root of the quotient of the quantities under the 
 radical sign, arid reduce the result to its simplest form. 
 
 This is but an application of the principles of Arts. 193 and 150. 
 
 206. Sch. When the division gives rise to a fraction with 
 a surd in the denominator, this denominator should be rational- 
 ized. 
 
 EXAMPLES LXII 
 
 Perform the following divisions : 
 
 1. 18V6 by 3V10. 
 
 Operation. ^VS = gjjf = Q 13 = 6 ^ 
 
 3V10 >10 >5 5 
 
 2. 8 a V6o" by ±J/Ta~ 2 . 
 
 Operation. ^^ = 2«:^ = 2a^ = 2^B4lF. 
 
 3. 6V12 by 3V3. 4. 10V15 by 2.VK. 
 
 5. V32 by V8. 6. V42 by V84. 
 
 7. ^80^ 2 by ^5^. 8. a 2 </l92 by aVs. 
 
 9. V^T? by V^+^. 10. a/(^-3?/) 3 by V(^-32/) 3 . 
 
 11. V30 by V42. 12. Vl5 by S^15. 
 
 13. 6 by ^6. 14. -^81 a 2 by 3 a. 
 
 15. VlO by v®. 16. V2 by ^8. 
 
 17. -\/^ by a^. 18. aWl2^6 by ab 2 J/2lab. 
 
 19. V| by VJf. 20. Vfi by Vff. 
 
 21. ^ ty ^t- 22 5-VI35 by V5. 
 
INVOLUTION OF 8UBDS 123 
 
 23. GV32-12V24 + 3V5 by 3V2. 
 
 24. Vl4 - V35 by V2 - V5. 
 
 25. 3-^3+^3 by VS. 
 
 26. 2 V3 -f 7 V2 by 5 V3 - 4 V2. 
 
 Bug. Write as a fraction and rationalize the denominator. 
 
 27. 5V5 + 3V2 by V5 + V5. 
 
 28. a + b — c + 2 Va£ by Va + y/h — Vc. 
 
 30. V5-J-2V6-V5-2V6 by V5 + 2V6 +V5 -2V6. 
 
 INVOLUTION OF SURDS 
 
 207. Prob. To raise a monomial surd to any power. 
 
 Rule. Either divide the index of the root by the exponent of the 
 power, or raise the quantity under the radical sign to the required 
 power; then annex the result to the required power of the coefficient, 
 and reduce to the simplest form. 
 
 This is but an application of the principles of Arts. 148 
 and 151. 
 
 208. Cor. A surd is raised to a power whose exponent is the 
 index of the root by simply removing the radical sign. 
 
 EXAMPLES LXIH 
 
 Perform the following indicated operations : 
 
 1. (6V3*)*. 2. (-6Vp) 2 . 3. (±y~5xhjY. 
 
 4. (SV-2ab 2 ) 5 . 5. (-iVEaf. 6. (Sy/Uay. 
 
 7. (-Z/SSaWf. 8. (2^8^) 4 . 9. (</2aWy. 
 
 10. (-s/laWf. 11. (^50^/) 4 . 12. {^V-21d i W)\ 
 
 13. (^64 afy 2 ) 3 . 14. (V7-V5) 2 . 15. (3+V6) 2 . 
 
 16. (V3-V6) 2 . 17. (^2--\/3) 3 . 18. (^3+s!/2) 3 . 
 
 downey's alg. — 9 
 
130 HIGHER ALGEBRA 
 
 EVOLUTION OF SURDS 
 209. Prob. To extract any root of a monomial surd. 
 
 . Rule. Either extract the required root of the quantity tinder the 
 radical sign, or multiply the index of the given root by the index of 
 the required root, then annex the result to the required root of the 
 coefficient, and reduce to the simplest form. 
 
 This is but an application of the principles of Arts. 148 
 and 151. 
 
 EXAMPLES LXIV 
 Perform the following indicated operations : 
 1. -yJ{VaVf). 2. V( 36 V257?^). 
 
 3. */(- 27^04oW). 4. -*/(a w h 5 J/3X a'0 M ). 
 
 5. -ty(8V7ax). 6. V(^ 1024 *¥> 
 
 7. ^a~Va. 8. vWsi/2. 
 
 9. .£/(43 xy*V43xf). 10. Vx\x-y) 5n . 
 
 11. V(v49-70^ + 25x 4 ). 12. ^/(64«%' ] V64a :i 6 3 ) ; 
 
 210. Theorem. The square root of a binomial, one of ivhose 
 terms is rational and the other a quadratic surd, can be found when- 
 ever the rational term is separable into two parts the product of whose 
 square roots is half of the surd term. 
 
 Dem. Let a±b-\/c be such that a=m-\-n, and ±-J-&Vc=Vmn. 
 Then a ± bVc = m + n ± 2^/mn = (Vm ± Vw) 2 . Therefore the 
 square root is Vm ± y/n. 
 
 EXAMPLES LXV 
 Extract the square root of each of the following : 
 
 1. 27-10V2. 
 
 Solution. 27 - 10 V2 = 25 - 10 v2 + 2 = (5 - \/2)2. Hence the square 
 root is 5 — V2. Of course, the negative of this rQot, viz., V2 — 5, is equally 
 admissible. 
 
IMAGI NAMES 131 
 
 2. 12-2V35. 
 
 Solution. 12 - 2 V35 = 7-2 V35 + 5 = ( VI - V5) 2 . Hence the square 
 root is V7 — V5. 
 
 3. 4 + 2V3. 4. 3 + 2V2. 5. 7-2V10. 
 
 6. 8-4V3. 7. 30 + 10V5. 8. 8 - 2V15. 
 
 9. 70 -30 VS. 
 
 Solution. 70 - 30>/5 = 25 - 30 V5 + 9 x 5 = (5 - 3\/5) 2 . Hence the 
 square root is 5 — 3 V5. 
 
 10. 18 + 8V5. 11. 11-4V6. 12. 49 + 12V5. 
 
 211. Note. The method of inspection here given is of limited 
 application. A general method, applicable to all perfect squares 
 of the kind mentioned in the theorem, requiring, however, the 
 solution of two equations, one of the first and the other of the 
 second degree, will be given as an application of the process of 
 solving such equations. 
 
 SECTION II — IMAGINARY QUANTITIES 
 
 212. An Imaginary Quantity is an expression which contains an 
 indicated even root of a negative quantity. 
 
 Thus, V- 1, V- x 2 , \/— 12, 3±2v / ^~4, a ± bV— 1 are imaginary 
 quantities. 
 
 213. All quantities not imaginary are called Real Quantities. 
 
 214. Expressions which contain both real and imaginary terms 
 are often called Complex Quantities. 
 
 215. When the radical sign affects a negative quantity, we can- 
 not regard it as indicating a possible arithmetical operation. For 
 example, V— 9 is neither + 3 nor — 3, for neither multiplied by 
 itself, i.e. squared, will produce — 9, but + 9 instead. Since im- 
 aginary quantities occur frequently in mathematical investigations 
 and lead to important results, we need to consider what meaning 
 
132 HIGHER ALGEBRA 
 
 should be attached to them in order that they may obey the 
 ordinary laws of algebra. 
 
 The expression Va is understood to be such that Vax Va=a. 
 Now if we agree that V— « is such that, in the same way, 
 V— axV- a = — a, we shall find that imaginary quantities 
 obey the algebraic laws already established. This agreement 
 would forbid our first multiplying together the quantities under 
 the radical sign, observing the law of signs, and then extracting 
 the square root ; for in that ease we would have V— a X V— a 
 = Va 2 = ± a. The agreement limits us to the minus sign, as it 
 should, since the square root of a quantity multiplied by itself 
 should produce the original quantity. 
 
 216. Theorem. Every monomial imaginary can be reduced to the 
 form b^s/— 1, in which b may be either rational or surd. 
 
 Dem. The general form of a monomial imaginary is mV- k, 
 in which n is even. Now, since the root of the product equals the 
 product of the roots, mV- k=m-\/k(— l) = m~\/k-\/— l = b-\/ — 1, 
 where b = m-\/Tc. 
 
 217. Theorem. Every 'polynomial containing some real terms and 
 some imaginary terms of the same degree can be reduced to the form 
 a ± b^J — 1, in which a and b may be either rational or surd. 
 
 Dem. This is evident from the fact that all the real terms can 
 be combined into one (it may be a polynomial) and represented by 
 a, and the imaginary terms into another represented by ± b~V— 1. 
 
 218. Cor. The general form of an imaginary of the second degree 
 is a ± b V— 1, in which b is rational or surd, and a is rational, 
 surd, or 0. 
 
 219. The Imaginary Element is some even root of — 1 which 
 renders an expression imaginary. 
 
 Thus, in V^x = VxV^l, bV^x 2 = 5 x\/^T, 
 
 2 + V^4 = 2+ 2V"^T, V^S2 = 2v^2 = 2\/2v/^T, 
 the imaginary elements are \/— 1, V— 1, y/— 1. 
 
 V— 1, sometimes called the Imaginary Unit, is often represented by i. 
 
IMAGINARIES 133 
 
 220. Theorem. When the index of the imaginary element is a 
 composite number and one of tfie factors odd, the value is not changed 
 by rejecting the odd factor. 
 
 Dem. In the form bV — 1 let the factors of n be p and q, 
 q being odd. Then b V^T = b *V-1 = b^V^l = bV^l, since 
 any odd root of — 1 is — 1 . 
 
 Thus, 3v^nr = 3Vv / ^l = 3v^T, 
 
 
 221. Conjugate Imaginaries are imaginaries of the second degree 
 which differ only in the sign of the imaginary part. 
 
 Thus, a + frV^HE and a - 6V— 1, 5 + 3v^l and 5 - SV-T, mV^l 
 
 and — mV— 1, 4V— 1 and — 4V— 1, etc., are conjugate imaginaries. 
 
 222. Theorem. Both the sum and the product of two conjugate 
 imaginaries are real. 
 
 Dem. b V^l + (- 6V^T) = 0, 
 
 and a+bV^T + a — bV^l = 2a. 
 
 Also bV^l x (- &V^T) = - b\- 1)= 6 2 , 
 
 and (a + 6 V^l) X (a - &V-T) = a 2 + b 2 , 
 
 the last being the product of the sum and difference of two 
 quantities, which is the difference of their squares. 
 
 223. The Modulus of a quadratic imaginary of the form a±&V— 1 
 is the positive square root of a 2 + b 2 . 
 
 Each of the imaginaries that constitute a conjugate pair has 
 the same modulus, and this common modulus is seen to be the 
 positive square root of the product of the two. 
 
 224. Prob. To add or subtract imaginaries. 
 
 Rule. Reduce each to the form in which the imaginary element 
 is expressed by the smallest, possible index (Art. 220). If the imagi- 
 
134 HIGHER ALGEBRA 
 
 nary element is then the same in each, combine its coefficients and 
 write the result as the coefficient of the common imaginary element; 
 if the imaginary element is not the same in each, indicate the addi- 
 tion or subtraction. 
 
 • If the imaginary element is the same in each, the problem is 
 simply that of uniting quantities of the same kind. 
 
 EXAMPLES LXVI 
 
 Combine the following : 
 
 1. 6+V :r 4 and S+V^. 
 Operation. 6 + v^Ti = 6 + 2\/^T 
 
 Sum, 14 + 5V^T 
 
 2. 2-^8 and 5a/^2. 
 
 Operation. 2v^8 = 2 Vv^8 = 2V- 2 = 2v / 2v/~^T 
 
 Sum, 7V2V-1 
 
 3. 4V- r 27 and 6V-32. 
 
 Operation. 4 V- 27 = 4 V9 x 3 x(- 1) = 12 Vs V^T, 
 6V-32 = 6Vl6 x 2 x(- l)=24V2v/^T. 
 
 Here, although the imaginary element is the same in both quantities, 
 they contain dissimilar surds. Hence we cannot unite into one term, but 
 have 
 
 12V3V"^I + 24\/2V^Tr= 12(V3 + 2V2)\/^T. 
 
 4. 5 a + 3 6 V 11 ^ and 3 a + 2 6 V 11 !. 
 
 5. 15V- 16 and GV"^. 6. V- 225 and V-169. 
 
 7. y-9a 2 an d V-4ar>. a 3 V- 9 and -V-49. 
 9. SV" 17 !^ and 2V^48. 10. V-289 and -V-U. 
 
 11. aV-8 and V-18a 2 . 12. 7+-4V-27 and 3-2V-12. 
 13. \^Tl6 and -J/~^l. 14. 4 + 2^/^ and 2 - </^a. 
 
 15. ty^32 and V^8. 
 
 16. V^il-v^T ^-V^12l + ^-243. 
 
IM AGIN ARIES 
 
 135 
 
 225. Prob. To determine the different powers of V— 1 and 
 
 </- 1. . 
 
 Solution. (V— l) l = V— 1, 
 
 (V-l) 2 =-1, by Art. 215, 
 
 ( V=T) 3 = ( V^T) 2 V^T = - V^T, 
 (V=3)*=[(V^T)T =1, 
 
 (V :=: T) 5 =(V-i)V : ^T = V- 1, 
 
 (V^T) fi = (V^T)V^T = - 1, 
 ( V^T) 7 =( V=T) 6 V==T = - V- i , 
 
 (V=1) 8 =[(V=I)<]* =h 
 etc. 
 
 It is thus seen that we have a succession of the results, V— 1, 
 _ i ? — V— 1, and 1, or that the only forms assumed by the 
 different powers of V— 1 are ±V— 1 and T 1. 
 
 (^/^T) 3 = (</3i)*,J/:ri = v=T</- 1, 
 
 (^)W1, 
 
 (</^)"=(</^T)^^l = -</- 1, 
 (<^T)«= (</=Iy(l/=lf= - V^T, 
 (</- 1)'= (^/- l) 6 ^=~1= - V=1</^T, 
 
 (^-l)"=[«/^l) 4 J=l- 
 etc. 
 
 It is thus seen that Ihe only forms assumed by the different 
 powers of \^^1 are ± \' r — T, ± V— T, ± V^T^^T, and T 1. 
 
 226. Prob. To multiply or divide imaginaries. 
 
 Rule. Reduce to the form in which the root index of the imagi- 
 nary element is the same in each imaginary term', and multiply or 
 divide as in case of other quantities, observing (he principles of 
 Art. 225 for the product or quotient of the imaginary elements. 
 
136 
 
 HIGHER ALGEBRA 
 
 EC AMPLE 3 LXVII 
 Multiply the following : 
 
 1. 4V^ by 2V^. 
 Operation*. 4 V^3 = 4 V3 V^T 
 
 2V^2 = 2V2V^1 
 
 2. 7\/-32 by 5</~- 
 
 Operation. 
 
 Product, 8 V6 ( - 1 ) = - 8 V6. 
 
 '^2. 
 
 "32 = 7 </l6\/^2 --= 14 ^2 v>^l 
 
 5V^2= 5^'2'v/^l 
 
 Product, 1QV2(V- l) 2 , 
 or 70V2V^T 
 
 3. V— a,* 2 by V— ?/ 2 
 
 5. _2V-25 by 3V-36. 
 
 4. 5 V— 5 by 4V — 3. 
 
 6. V^2 by V^8. 
 8. 3^/^Yl by 4v cr 3. 
 
 7. V-196 by V-27. 
 
 9. UV^ + V^bj^V^l. 10. 4-j-2V^4 by 5 - 3 V^I. 
 
 11. 1 + V^l by 1-V-X 12. 12+3 V^ by 12-3 V^f. 
 13. 3-J/^9 by 4^~8. 
 
 Operation. 
 
 3#- 9 = 3J/qV^1 = 3Vi 
 
 5 4 / 
 
 4^/_ 8 = 4\/8\/- 1 = 4V2 V- 1 = 4 v/2(\/-~l)* 
 
 Product, 12V0(v- l) 3 =12v / 6V-l V-l. 
 14. 3a/^2 by 5V^. 15. 3a/^2 by 5^3. 
 
 Find the value of each of the following : 
 16. (5V^2) 2 . 17. (2V^3) 3 . 18. (4^2) 4 , 
 
 19. (3V^2) 5 . 20. (2^2) 6 . 21. (V^7) 7 . 
 
 Find the modulus of each of the following : 
 22. V2+V^2. 23. V2-V^2. 24. 3 + 2\/^3. 
 
 25. 3_2V^3. 26. 4-3V := l. 27. 4 + 3V :=: l. 
 
MAG I NAMES 
 
 137 
 
 Kationalize the denominators of the following : 
 
 28. 
 
 30 
 
 3-V=2 
 2 
 
 V-9 
 
 Divide the following: 
 32. 6V-16 by 2V z: 4. 
 
 GV^nTi 24 
 
 29. 
 
 31. 
 
 Operation. 
 
 4V- 1 
 
 2V-4 
 
 The imaginary elements cancel. 
 
 33. 2V :r T by v^2. 
 
 n oc .» ™™ 2\/^T 2(\/^T)^ 2</^T 
 Operation. =s— ? — = 
 
 V32+2V-3 
 
 V"-^2-2V^3* 
 
 3V^2 + 2V^5 
 
 3V :r 2-2V^5 
 
 = 6. 
 
 = ^8v^l = ^T8. 
 
 y/~2 ^2C/-~i ^2 
 
 34. 16V^"l) by 2 v 7 ^4. 35. tV^i by 2V3& 
 
 36. 1 by V—I. 37. 36 by 12V :=r I. 
 
 38. 63V- 16 by V-81. 
 
 40. 4+V^2 by 2-V^2. 
 
 39. V-16 by V-4. 
 
 Operation. When there is more than one term In the divisor, it is usually 
 best to write the quotient as a fraction and then rationalize the denominator. 
 Thus, 
 
 4+>/ z ~ 2 =z 4 + V2>/3T _ (4 + V2Vin)(2 + V2V^-"1) 
 2 _ V^2 2 - V2 V^T 4 + 2 
 
 6 
 
 41. 1 + V^l by 1 -V— 1. 
 
 43. a+V-« by a — V— #• 
 
 44. s-i+^Zlby - 
 
 42. 1 by 3— 2V := ~3. 
 
 45. Simplify a+V-ft ^a 
 
 2z-l+V-3 
 ^6 
 
 a — V—b a + V— 6 
 
138 HIGHER ALGEBRA 
 
 Extract the square root of each of the following : 
 
 46. 5 + 12V^T. 
 
 Operation. 5 + 12Vj-l = 9 + 12v^l - 4 = (3 + 2>/^~l) 2 . Hence the 
 square root is 8 -f 2V— 1. 
 
 47. -5 + 12V-1. 
 
 Sua. Write in the form 4 + 1SV^-T - 9. 
 
 48 . _7_24V^T. 
 
 Sug. Write in the form 9 - 24 V^l - 16. 
 
 49. 7 + 12V^4. 50. 21-20V ::: T 
 51. -9 + 20V :=: l. 52. 35-12V ::= T. 
 53. 40 + 42V :=r I. 54. 33-56V^l. 
 55. 27 + 36V :=: T. 56. 24-70\/^l. 
 
 57. Simplify —-£ — 4- 
 
 58. Simplify 
 
 2-5V-1 2+5V-1 
 
 V3 — a; + V^2 V^^x — V^5 
 
 227. In Algebra imaginary quantities have their greatest prac- 
 tical use in showing that, arithmetically, the conditions of prob- 
 lems in the solution of which they occur cannot be fulfilled. For 
 example, let it be required to divide 10 into two parts whose prod- 
 uct shall be 40. Proceeding in the usual way we find the two 
 parts to be 5 -f- V— 15 and 5 —V— 15. These imaginary results 
 show that the requirements of the problem cannot, in the arith- 
 metical sense, be fulfilled. It is easy to show that the largest 
 product of any two parts of 10 is 25. Nevertheless, these imagi- 
 nary values do satisfy the algebraic requirements of the problem, 
 the sum of 5 + V— 15 and 5 —V— 15 being 10 and the product 
 40. Imaginary quantities occur as well, and always in conjugate 
 pairs, as roots of an equation which also has one or more real 
 roots. 
 
IMAGTNARIES 139 
 
 Imaginaries also have a practical use in Algebra in showing the 
 limits of a variable quantity ; for example, in finding, by algebraic 
 methods, the maximum or minimum value of a function. 
 
 In Analytical Geoinetry imaginaries have a frequent and im- 
 portant use in showing the limits of loci represented by certain 
 forms of equations. 
 
 In a branch of Mathematics known as Quaternions imaginary 
 quantities have a graphical and definite signification. If any 
 magnitude be represented by a, then — a represents an equal 
 magnitude in the opposite direction. Now ax(— 1) = — a; 
 hence — 1, regarded as an operator, causes a reversal. We have 
 seen that «xV-l X V— 1 = — a; hence V— 1, regarded as an 
 operator, causes a reversal when repeated. 
 
 Now any magnitude may be represented graphically by a length 
 laid off on a straight line. Suppose this length to be a. Then 
 — a would be the same length in the opposite direction. Now we 
 may, if we choose, regard V— 1 as being an operator which turns 
 a through a right angle, because, when repeated in the same direc- 
 tion, it causes a reversal. Thus, just as a x V— 1 x V— 1 = — a, 
 a turned successively through two right angles gives — a ; and 
 V— 1, represented by i in Quaternions, is the operator which, 
 when repeated, causes this reversal. 
 
CHAPTER XI 
 SIMPLE EQUATIONS 
 
 SECTION I — SIMPLE EQUATIONS WITH ONE UNKNOWN 
 
 QUANTITY 
 
 228. An Equation is an assertion by means of a mathematical 
 symbol that two expressions have the same value. This symbol 
 is the Sign of Equality (Art. 10). 
 
 229. The Members of an Equation are the expressions connected 
 by the sign of equality, the one on the left being called the First 
 Member and the one on the right the Second Member. 
 
 230. An Identical Equation, called also an Identity, is an equa- 
 tion that is true for all values of the letter or letters involved.* 
 
 Thus, (x + a) (x — a) = x 2 — a 2 is an identity. 
 
 231. An Equation of Condition, usually called simply an Equa- 
 tion, is an equation that is true only for a limited number of 
 values of the letter or letters involved. 
 
 Thus, x + 5 = 9 is true only f or * = 4 ; and x 2 — 6 x = 7 is true only for 
 x = 7 and x — — 1. 
 
 232. The Unknown Quantity is the letter whose value is re- 
 quired. 
 
 233. An Absolute Term of an equation is a term which does not 
 contain an unknown quantity. 
 
 234. A Numerical Equation is one in which the known quantities 
 are represented by numbers. 
 
 * Some writers use the sign = to indicate that two expressions are identi- 
 cally equal. The sign is not used in this work, as the distinction is not 
 thought to be of sufficient importance. 
 
 140 
 
SIMPLE EQUATIONS 141 
 
 235. A Literal Equation is one in which some or all of the known 
 quantities are represented by letters. 
 
 236. The Degree of an Equation is the same as that of its term 
 of highest degree (Art. 50), the unknown quantity or quantities 
 first being freed from fractional and negative exponents. 
 
 Thus, ox 2 + bx = c is of the second degree, 3 x 2 y — bxy + 2x + 6y 2 = 12 is 
 
 of the third degree, xy = m is of the second degree, - + - = 5 (which, by mul- 
 
 x y 
 
 tiplying both members by xy, becomes y + x — 5 xy) is of the second degree, 
 
 x 2 + 3 x — 5 x -2 — 3 = (which, by multiplying both members by x' 2 , becomes 
 
 x* 4- 3 x 8 — 5 —3 x 2 = 0) is of the fourth degree. 
 
 237. A Simple or Linear Equation is one of the first degree. 
 
 238. Equations of one unknown quantity are called Quadratic, 
 Cubic, Biquadratic, or Quintic, according as they are of the second, 
 third, fourth, or fifth degree. 
 
 239. Any equation above the second degree is called a Higher 
 Equation. 
 
 240. A Root of an Equation is a value of the unknown quantity 
 which renders the equation true. 
 
 241. To solve an Equation is to find its root or roots. 
 
 242. An equation is said to be Satisfied, and a supposed root 
 Verified, when, on substituting for the unknown quantity the sup- 
 posed root, the equation becomes an identity. 
 
 243. In solving equations the following axioms are employed : 
 
 1. If the same quantity or equal quantities be added to or sub- 
 tracted from both members of an equation, the equality of the mem- 
 bers will not be destroyed. 
 
 2. If both members of an equation be multiplied or divided by 
 any quantity that is not equal to 0, the equality of the members will 
 not be destroyed* 
 
 * For the introduction and loss of roots by multiplying or dividing by 
 a factor containing the unknown quantity, see Art. 351. 
 
142 HIGHER ALGEBRA 
 
 To show that the equality of the members may be destroyed 
 
 by dividing both members by a quantity that is equal to 0, let us 
 
 take the equation 
 
 5x-15 = 2x-6, (1) 
 
 which is seen to be satisfied for x = 3. 
 By factoring we have 
 
 5(x-3) = 2(x-3). (2) 
 
 Now if we should divide both members of (2) by x — 3, we 
 should have the absurd result 5 = 2. The two members of (2) 
 are equal, not because of the relation of 5 to 2, but because 
 x — 3 = 0, and 5 times is the same as 2 times 0. In dividing 
 by x— 3 we remove the element that makes the two members 
 equal. 
 
 244. Prob. To solve a simple equation ivith one unknown quan- 
 tity. 
 
 Rule. 1. Clear the equation of fractions, if it have any, by 
 multiplying each term by the 1. c. m. of all the denominators. 
 
 2. Transpose the unknown terms to the first member and the 
 known terms to the second member, by changing the sign of each 
 term transposed. 
 
 3. Unite similar terms, and divide both members by the coefficient 
 of the unknown quantity. 
 
 Dem. 1. Multiplying by the l.c.m. of all the denominators 
 does not destroy the equality of the members (Art. 243, 2), and 
 it clears of fractions because in multiplying any one of the frac- 
 tions by this 1. c. m. the denominator of that term is canceled by 
 one factor of the multiplier. 
 
 2. When a term is dropped from one member, it is subtracted 
 from that member. Now if it is written with its sign changed in 
 the other member, it is subtracted from that member also, and 
 the equality of the members is not destroyed (Art. 243, 1). 
 
 3. Dividing by the coefficient of the unknown quantity, after 
 uniting terms, does not destroy the equality of the members 
 (Art. 243, 2), and it leaves in the first member simply the un- 
 known quantity; hence the second member is its value, or the 
 root of the equation. 
 
SIMPLE EQUATIONS 143 
 
 245. Cor. The signs of all the terms of an equation may be 
 changed without destroying the equality of the members. 
 
 For this is the same as multiplying or dividing by — 1. 
 
 246. Sen. Equations having terms of higher degree than the 
 first often reduce to simple equations by the disappearance of 
 these terms in collecting. 
 
 SOME PRACTICAL SUGGESTIONS 
 
 247. 1. When there are several integral terms and but few 
 fractional terms, time is saved by collecting terms before clear- 
 ing of fractions. 
 
 2. In clearing of fractions the student must be careful to 
 change the signs of the terms in the numerator of a fraction that 
 is preceded by the minus sign (Art. 11). 
 
 3. In simple cases the terms may be^transposed and united at 
 the same time. 
 
 4. Factors which appear in both members of the equation 
 should be canceled, and equal terms with opposite signs in the 
 same member of the equation, or the same sign in opposite mem- 
 bers of the equation, should be stricken out. 
 
 5. When a fraction has a polynomial numerator and monomial 
 denominator, it is often better to separate the fraction into parts 
 by dividing each term of the numerator separately by the 
 denominator, 
 
 6. It is often expedient to clear of the smaller or simpler 
 denominators first, and after each step to see that by transposi- 
 tion, uniting terms, etc., the equation is kept in as simple a form 
 as possible. 
 
 EXAMPLES LXVIU 
 Solve the following : 
 
 ., o , 2#-10 Sx 5a;-14 
 
 1. o. x H = 
 
 3 2 4 
 
 Solution. Clearing of fractions by multiplying every term by 12, the 
 1. c. m. of the denominators, we have 
 
 36 x + 8 x - 40 = 18 x - 15 x + 42. 
 
144 HIGHER ALGEBRA 
 
 Observe that the minus sign before the last term of the given equation 
 denotes that the whole term is to be subtracted. Hence when the vinculum 
 is removed in clearing of fractions, we must either still indicate this subtrac- 
 tion by writing — (15 x — 42), or perforin it by changing the signs. 
 
 Transposing, 36 x 4- 8 x — 18 x -f 15 x — 42 + 40. 
 
 Collecting terms, 41 x — 82. 
 
 Dividing by the coefficient of x, x = 2. 
 
 x — a _ (x — b) 2 
 2 ~ 2x-a 
 
 Solution. Performing the indicated involution and clearing of fractions, 
 
 2 x°- - 2 ax - ax 4 a 2 = 2 x* - 4 bx + 2 6 2 . 
 
 dropping 2 x 2 from both members, transposing and uniting, 
 
 4 bx - 3 ax = 2 6 2 - a 2 , 
 
 or (4 6 - 3 a)x = 2 6 2 - a 2 . 
 
 . _ 2 6 2 - a 2 
 45 -3a' 
 
 The answer would be equally correct if written 
 
 s = *- 2 *. Why? 
 3a-46 
 
 3. 8a;-5(4ar+3) = 25-8aj. 4. 5(aj- 2) - 6 (a + 4) = 21. 
 
 5. ^^ + ^ = 20-^i5. 6. 2x- 3x + 7 = - + l. 
 
 11 ^ 
 
 6 x — 4 9 _ 18 — 4 # 
 3 3 
 
 x 4- 1 X — 1 4 
 
 
 2 ' 
 
 fr ~ 
 
 2 
 
 
 3 
 
 X + 4: 
 
 19x-3 4 
 
 -7 
 
 a? 
 
 
 2 
 
 8 
 
 12 
 
 
 
 2 
 
 5 2 
 
 
 
 SB 
 
 -2 
 
 SB + 2 x 2 - 
 
 -4 
 
 
 2 
 
 a + 3 
 
 _4x + 5 
 
 
 
 3 
 
 x — 4 
 
 6a>-l 
 
 
 a 
 
 
 
 
 g 
 
 -1 x-2_x-5 
 
 ar- 
 
 6 
 
 10. 
 
 35 — 1 £ -f- 1 X 2 — 1 
 
 I 6x+7 7a?-13 = 2o;+4 
 9 6a?+3 3 
 
 3x 2x 2ar } -5 
 
 13. v XJiz = ac4-^. 14. 
 
 2a;+3 2x-3 4a^-9 
 
 . 2(a?-7) a?-2^a;+< 
 X -2 x-3 jc-6 x-7 * x 2 +3x-28 a,--4 x+7 
 
 16 2(»-7) + x = 2 == x ± 3 m 
 
17. 
 
 SIMPLE EQUATIONS 145 
 
 2x + 1 2x-l _ 9# + 17 
 
 2^-16 2x + 12 a? -2 a;- 48 
 
 „ „ a — b,b — c a — c 
 
 18. 1 = 
 
 x — c x—a x 
 
 19. (a 2 + s) (ft 2 + a?) = (aft + z) 2 . 
 
 20. (a'4-a) 4 -(x-a) 4 -8a.^ + 8a 4 = 0. 
 
 21. (a?-l) 3 + .^ + (z + l) 3 = 3a;(a; 2 -l). 
 
 22. (x-l)(x + 2)(x-3) = x 2 (x-2) + 2(x + 4). 
 
 248. Theorem. 7/* a; + V# = a + Vft, tot itf/uc/i a; awrf a are 
 rational and V# and Vft «wrd, ^e rational and surd terms are 
 separately equal. 
 
 Dem. Transposing, we have 
 
 x — a = Vft — Vy . 
 
 Since a rational quantity cannot equal a surd, this equation 
 can be true only when 
 
 x — a = and Vft — V# = 0, 
 whence # = a and Vft = V#. 
 
 249. Theorem. 7/" x + V — # = a -f V— ft, wi w/wc/i a; and a are 
 rad and V— y and V— ft imaginary, the real and imaginary terms 
 are separately equal. 
 
 Dem. Transposing, we have 
 
 # — a = V— ft — V — y. 
 
 Since a real quantity cannot equal an imaginary one, this equa- 
 tion can be true only when 
 
 # — a = and V — ft — V— .y = 0, 
 whence # = a and V— 6 = V— 2/. 
 
 250. Many equations containing surds become simple equa- 
 tions after being freed from surds and reduced. No general rule 
 can be given for solving such equations, as different cases must 
 
 downey's alg. — 10 
 
146 HIGHER ALGEBRA 
 
 have different treatment. Much depends on the student's insight. 
 When inspection does not suggest what steps will free the equa- 
 tion from, surds, the student must resort to trial, being careful to 
 preserve the equality of the members. We give here a few sug- 
 gestions. Some of the operations, as will be shown in Art. 351, 
 cause roots to be lost or extraneous roots to be introduced. 
 
 1. By making a surd term constitute one member of an equa- 
 tion, its radical sign will disappear when both members are raised 
 to a power of the same degree as the surd. A repetition of the 
 process (having treated the most complicated surd first and 
 reduced as much as possible) will cause a second radical sign to 
 disappear, and so on. 
 
 2. When a surd denominator is similar to the surd numerator 
 of another fraction or to another term, it is usually best to multi- 
 ply both members by this denominator. 
 
 3. It is sometimes best to rationalize a surd denominator, espe- 
 cially when it differs only by a sign from the numerator of the 
 same fraction. 
 
 4. Sometimes a surd factor, or a factor containing a surd term, 
 can be removed either from both terms of a fraction or from both 
 members of the equation. 
 
 EXAMPLES LXIX* 
 
 Solve the following : 
 
 1. Va-32 = 16-Va. 2. - = c. 
 
 3. VS + V^7 21 6 *- 9 
 
 -Vx — 7 V 5 x + 3 
 
 5. Vfl -|- s/x -f- va — V# = Vx. 
 
 6. V(l + a) 2 + (1 - a)x + V(l - a) 2 + (1 + a)x = 2a. 
 
 7. V113+V[7+V(3+V^)]S = 4. 
 
 I Vl+ v/ 3+V6« = 2. 
 
 * Most of the examples of this set are from Olney's University Algebra. 
 
SIMPLE EQUATIONS 147 
 
 9 V6x-2 = ^6x-9 
 VWx + 2 ±V6x + 6 
 
 16 a; — 3 
 
 11 a ~ ^ a2 ~ ®* —h 12 ax ~ 1 _ a _j_ Voal 
 
 a -f- Va 2 — x 2 VaaJ-f-1 2 
 
 13 3 Va; - 4 = 15 + 3 Va; Vox -{- V& _ Va + V& 
 
 2+Vx 40+Va; Va«-V6 V& 
 
 15 Va- vg-Vft 2 -aa; _ , Vm + Vra — y _ t_ # 
 
 Va+^a--y/a 2 -ax Vm-Vm-y m 
 
 17 a + ? + ^ 2 aa; + a; 2 _ &2 18 Va; + 1 — Va; — 1 _ 1 t 
 a + ^-V^aaJ + aj" ' VaT+T + -y/x^l 3 
 
 19. Va + x + Va — a; = &. 
 
 2a 1 + a? + V2a? + a?' =a V2 + a; +Vjp . 
 1 + x — V2aT+a? V2 — x — Va; 
 
 ^ V3a; + l4-V3a =1 
 V3a;+1-V3a; 
 
 22. _I = + _=! -2& 
 
 Va — a; + Va Va~^le — Va x 
 
 23 a + 2a;+Va 2 -4x 2 := g^. 
 tt + 2a;-Va 2 -4ar J a 
 
 24 18(7a;-3) ^ 250V2a; + l 
 
 2x + l ' 3V7a;-3 
 
 PROBLEMS LEADING TO SIMPLE EQUATIONS WITH ONE 
 UNKNOWN QUANTITY 
 
 251. The solution of a problem by algebraic methods consists 
 of two distinct parts: 1st. Forming the equation or equations, 
 which consists in expressing the conditions of the problem in 
 algebraic language. 2d. Solving the equation or equations. 
 
148 HIGHER ALGEBRA 
 
 252. In solving an algebraic problem involving but one un- 
 known quantity, we usually proceed as follows : 
 
 1. We represent the unknown quantity by one of the final 
 letters of the alphabet. 
 
 2. We form an equation by indicating the operations that 
 would be necessary to verify the result if it were known. 
 
 3. We solve this equation. 
 
 253. While we usually represent the number sought by x, it is 
 sometimes advantageous to represent it by some multiple of x. 
 For example, if three numbers whose sum is 36 are in the ratio of 
 2, 3, and 4, we may avoid fractions by letting 2x, 3x, and 4 x be 
 the numbers, giving the equation 2 x + 3 x -f- 4 x = 36. In other 
 cases it may be advantageous to represent by x some number 
 (time, for example) from which the number sought (distance, for 
 example) is readily found. 
 
 EXAMPLES LXX 
 
 1. A bicyclist made a trip at the rate of 13 miles an hour. His 
 return by a road 10 miles longer at the rate of 16 miles an hour 
 required 15 minutes more time. What was the first distance ? 
 
 Solution. Let x = the first distance ; 
 
 then x + 10 = the second distance, 
 
 — = the first time, 
 13 
 
 and £-E — = the second time. 
 
 16 
 
 Now, since this second time is longer than the first by \ of an hour, 
 
 x_ _ x + 10 _ I 
 13 16 4* 
 
 Clearing of fractions, 16 x = 13 x + 130 - 52. 
 
 Transposing and collecting, 3 x = 78. 
 
 Therefore x = 26. 
 
 2. A can do in 20 days a piece of work which B can do in 
 12 days. A began the work, but after a time B took his place, 
 and the whole work was finished in 14 days from the beginning. 
 How long did each work ? 
 
SIMPLE EQUATIONS 149 
 
 Solution. Let x —■ the no. of days A worked ; 
 then 14 — x = the no. of days B worked ; 
 
 ^V = the part of the work A did in 1 day, 
 ^ = the part of the work B did in 1 day, 
 
 — = the part of the work A did in x days, 
 
 14 — x 
 
 ■ — = the part of the work B did in 14 — x days. 
 
 Now the sum of what A and B did in their respective times was the whole 
 work, or 1. Hence 
 
 x_ 14 — x _ , 
 20 12 ' 
 
 Clearing of fractions, 3 x + 70 — 5 x = 60. 
 
 Transposing and collecting, — 2 x = — 10. 
 
 Therefore x = 5, 
 
 and 14 - x = 9. 
 
 3. A post is I in the earth, ^ in the water, and 13 feet in the 
 air. Find the length of the post. 
 
 fit D 
 
 4. A post is — in the earth, — in the water, and a feet in the 
 
 r n ' q 
 
 air. Find the length of the post. 
 
 5. If an outward trip is made at the rate of 30 miles an 
 hour, the return trip at 18 miles an hour, and the whole time is 
 10 hours, what is the distance ? 
 
 6. When a is taken from the numerator of a fraction whose 
 numerator is b less than its denominator, the value of the fraction 
 
 becomes — . What is the original fraction ? 
 n 
 
 7. When 5 is taken from the numerator of a fraction whose 
 numerator is 3 les§ than its denominator, the value of the fraction 
 becomes f . What is the original fraction ? 
 
 Sue Let a — 5, b — 8, m = 3, and n = 7, and substitute in the formula 
 obtained from the solution of the preceding example. We thus obtain at 
 once from the general case the result for the special case. 
 
150 BIG HER ALGEBRA 
 
 8. A man having completed f of his journey, finds that after 
 traveling 30 miles farther only f of the journey remains. Find 
 the length of the journey. 
 
 9. A man rows down a river for 2 hours at a rate that would 
 take him 4 miles an hour in still water ; then, resting, he floats 
 with the current for half an hour. He then rows back to the 
 starting place in 3 hours. Find the rate of the current. 
 
 10. A grocer drew 14 gallons of syrup from a cask which had 
 lost J part by leakage, and found that f remained. Find the 
 capacity of the cask. 
 
 11. A and B can together do in 12 days as much work as A can 
 do alone in 20 days. In how many days could B alone do the 
 same amount of work ? 
 
 12. A cistern can be filled by the first of three pipes in 1J hours, 
 by the second in 2\ hours, and by the third in 5 hours. In what 
 time can they together fill the cistern ? 
 
 13. A can do in 15 hours a piece of work which B can do in 
 25 hours. After A has worked for a certain time, B completes 
 the job, working 9 hours longer than A. How many hours did 
 A work ? 
 
 14. A cistern which contains 2400 gallons can be filled in 
 15 minutes by three pipes, the first of which lets in 10 gallons 
 per minute, and the second 4 gallons less than the third. How 
 many gallons does the third let in per minute ? 
 
 15. A tank can be filled by one of two pipes in 24 minutes, 
 and by the other in 30 minutes, and emptied by a third in 20 min- 
 utes. In what time will the tank be filled if all are left open ? 
 
 16. A can row 4 and B 3 miles an hour in still water. A is 
 14 miles farther upstream than B, and they row toward each 
 other till they meet, 4 miles above B's starting place. Find the 
 rate of the current. • 
 
 17. On a sum of money borrowed, annual interest is paid at 
 5%. After a time $200 is paid on the principal, and the inter- 
 est on the remainder is reduced to 4%. By these changes the 
 annual interest is lessened one third. What is the sum borrowed ? 
 
SIMPLE EQUATIONS 151 
 
 18. A grocer has two kinds of coffee, one at a cents, and the 
 other at b cents a pound. How much of each must he take to 
 make a mixture of n pounds at c cents a pound ? 
 
 19. A grocer has two kinds of tea, one at GO cents and the 
 other at 90 cents a pound. How much of each must he take to 
 make a mixture of 120 pounds at 80 cents a pound ? 
 
 Sue. Solve by substituting in the formula obtained from the solution of 
 the preceding example. 
 
 20. If, for a given distance, the rate is 3 miles an hour over 
 the first half, what must it be over the second half to make the 
 average rate 4 miles an hour ? 5 miles an hour ? G miles an hour ? 
 
 21. The hind wheels and fore wheels of a carriage have circum- 
 ferences 1G and 14 feet respectively. How far has the carriage 
 advanced when the fore wheels have made 51 revolutions more 
 than the hind wheels ? 
 
 22. A train leaves A at 11 a.m. for B, at the rate of 25 miles 
 an hour. Another train leaves C at noon and runs through A to 
 B at the rate of 35 miles an hour, arriving at B 24 minutes later 
 than the first train. The distance from C to A being 21 miles, 
 find the distance from A to B. 
 
 23. An artesian well supplies a manufactory. The water is 
 drawn out each week day from 3 a.m. to G p.m. twice as rapidly 
 as it runs into the well. If the well contained 2250 gallons of 
 water on Monday morning and was just emptied at G p.m. a week 
 from the following Thursday, how many gallons flowed into the 
 well per hour ? 
 
 24. Two men, A and B, 57 miles apart, travel toward each 
 other, A at the rate of 6 miles an hour and B at the rate of 5 
 miles an hour, B starting 20 minutes later than A. How far will 
 each have traveled when they meet ? 
 
 25. Find the time between 3 and 4 o'clock when the hands of 
 a watch are opposite each other. 
 
 Solution. Let x = the time past 3 ; then x is also the number of minute 
 spaces passed over by the minute hand, and — is the number passed over 
 by the hour hand, since it moves fa as fast. Now, x in spaces is made up of 
 
152 . ' HIGHER ALGEBRA 
 
 three parts, viz., 15 spaces from the XII mark to the III mark, — from the 
 
 III mark to where the hour hand is, and 30 from where the hour hand is to 
 where the minute hand is ; i.e., 
 
 whence x = 49^. 
 
 26. Find the time between 3 and 4 o'clock when the hands of a 
 watch are together, and the tkne when they are at tight angles to 
 each other. 
 
 27. Find the time between 4 and 5 o'clock when the minute 
 hand of a watch is 18 minute spaces ahead of the hour hand. 
 
 28. A bicyclist, at 12 miles an hour, went to meet another who 
 started half an hour later, at 15 miles an hour, from a place 114 
 miles distant. In how many hours after the second started did 
 they meet, and how far did each go ? 
 
 29. The sum of the two digits of a number is 7. If be sub- 
 tracted from the number, the digits will be interchanged. What 
 is the number ? 
 
 30. The left-hand digit of a number composed of six digits 
 is 1. If the 1 be removed to units' place, the other digits remain- 
 ing in the same order as before, the new number will be 3 times 
 the original number. Find the number. 
 
 Sue Let x be the number exclusive of the left-hand digit. 
 
 31. Accommodation trains leave A for B at intervals of a hours,, 
 running m miles an hour. Express trains run from B to A at 
 the rate of n miles an hour. At what intervals of time does an 
 express train meet the accommodation trains ? 
 
 32. The duty on a certain article is reduced from $ 2.25 per 
 hundredweight, and in consequence of this reduction the con- 
 sumption is increased one half, but the revenue falls one third. 
 Find the duty per hundredweight after the reduction. 
 
 33. A man walking from a town A to another B at the rate of 
 4 miles an hour, starts one hour before a coach which goes 12 
 miles an hour, and is picked up by the coach. On arriving at B, 
 he observes that his coach journey lasted 2 hours. Find the 
 distance from A to B. 
 
SIMPLE EQUATIONS 153 
 
 34. A hare, 50 of her leaps ahead of a hound, takes 4 leaps to 
 the hound's 3 ; but 2 of the hound's leaps equal 3 of the hare's. 
 How many leaps must the hound take to catch the hare ? 
 
 35. A bicyclist starts on the road with a message at a rate 
 which will require 6 hours for its delivery. At the halfway 
 point it is taken by another bicyclist who rides 3 miles an hour 
 faster than the first, and thus the message is delivered half an 
 hour earlier than it would have been had the first continued. 
 Find the rate of the first bicyclist and the whole distance. 
 
 36. A can do in 30 days a piece of work which B can do in 
 20 days. A begins the work, but after a time B takes his place, 
 and the whole work is finished 25 days from the beginning. How 
 long did A work ? 
 
 37. A can do in 20 days a piece of work which B can do in 30 
 days. A begins the work, but later B takes his place and finishes 
 it, working 10 days longer than A. How long did A work ? 
 
 38. A man starts on a bicycle ride at the rate of 10 miles an 
 hour, intending to be back in 2 hours. Owing to a breakdown 
 he walks back at the rate of 4 miles an hour, and finds him- 
 self 11 hours late. Find how far he went. 
 
 39. The capacity of the second of four casks is £ of the first, 
 the third is f of the second, the fourth is -^ 6 - of the third, and 
 the first holds 15 quarts more than the third and fourth. How 
 many quarts does each hold ? 
 
 40. A man invested part of $ 2550 in 3% stocks and the rest 
 in railroad shares of $ 25 each, which pay annual dividends of 
 $ 1.00 per share. The stocks cost him $ 81 on a hundred, and 
 the railroad shares $ 24 per share. His income from each source 
 is the same. Find the number of railroad shares and the amount 
 of each investment. 
 
 41. An express train runs 15 miles an hour faster than an 
 accommodation train, and occupies T 9 $ as much time in running 
 120 miles. The express train loses at intermediate stations half 
 as much time as does the accommodation train, and the latter 
 loses as much time as it would require in running 20 miles. 
 Find the rate of each train. 
 
154 HIGHER ALGEBRA 
 
 SECTION II — SIMULTANEOUS SIMPLE EQUATIONS WITH 
 TWO UNKNOWN QUANTITIES 
 
 254. Simultaneous Equations are such as are to be satisfied for 
 the same values of the unknown quantities. They express differ- 
 ent relations of the same unknown quantities and arise from 
 different conditions of the same problem. 
 
 If we have a single equation containing two unknown quanti- 
 ties, we may assume any value we please for one and find such 
 value for the other as will satisfy the equation. Not so, however, 
 if the values are to satisfy at the same time two different equations, 
 such as come from different conditions of the same problem. If 
 the equations are of the first degree, only one set of values will 
 satisfy both. We are now concerned with the methods of finding 
 these values. 
 
 255. Elimination is the process of deducing from a set of two 
 or more simultaneous equations containing as many unknown 
 quantities a new set in which the number of equations and the 
 number of unknown quantities shall be diminished by at least one. 
 
 256. Three methods of elimination are in common use, viz., by 
 Addition or Subtraction, by Comparison, and by Substitution. 
 
 257. The rules for elimination by these different methods are 
 here given and illustrated. The student should be able to show, 
 first, why these operations give true equations, and second, why 
 they eliminate one of the unknown quantities. 
 
 After eliminating one of the two unknown quantities from two 
 equations, the resulting equation is to be solved for the remain- 
 ing unknown quantity. Usually the other is found by substitut- 
 ing this value in the simpler of the two given equations, and 
 solving the resulting equation. 
 
 258. Prob. To eliminate by Addition or Subtraction. 
 
 Rule. 1st. If the coefficients of the unknown quantity to be elim- 
 inated are not numerically the same, make them so by multiplying 
 each by that number which will give for the product their 1. c. m. 
 
SIMULTANEOUS SIMPLE EQUATIONS 155 
 
 2d. If the signs of these coefficients are unlike, add the equations; 
 if alike, subtract one equation from the other. 
 
 ILLUSTRATIVE EXAMPLE 
 
 Sol- {T + l y = T 
 [4x- Sy = 4. 
 
 Multiplying the first by 2 and the second by 3 and subtracting, 
 
 12 x + 14^ = 58 
 
 12s- 9y=12 
 
 23 y = 46 
 
 ••• y = 2. 
 
 Substituting this value of y in the second of the given equations and 
 
 solving for x, 
 
 4x -6 = 4, 
 
 4z=10, 
 
 x = 2%. 
 
 Had we multiplied the first equation by 3 and the second by 7 and added, 
 y would have been eliminated. 
 
 After finding the value of one of the unknown quantities, the other should 
 be found by substituting this in the equation containing the smallest numbers. 
 
 259. Prob. To eliminate by Comparison. 
 
 Rule. 1st. Find from each equation the value of the same un- 
 knoivn quantity in terms of the other and known quantities. 
 2d. Place these two values equal. 
 
 Solve 
 
 ILLUSTRATIVE EXAMPLE 
 
 7 x + 1 y = 68, 
 
 r 7 x + 11 y = m 
 
 \9x -4y = 33. 
 
 />Q 11.. 
 
 From the first, x = = 
 
 33 + 4y 
 
 From the second, x 
 
 Hence, 
 
 9 
 68-lly 33 + 4 y 
 
 7 9 
 
 612- 99?/ = 231 + 28 y, 
 - 127 y =-381, 
 y = 3. 
 
156 HIGHER ALGEBRA 
 
 Substituting this value of y in the second of the given equations and 
 solving for as, 
 
 9x- 12 = 33, 
 
 9x = 45, 
 
 x = 5. 
 
 260. Prob. To eliminate by Substitution. 
 
 Rule. 1st. Find from one of the equations the value of one of 
 the unknown quantities in terms of the other and known quantities. 
 
 2d. Substitute this value for the same unknown quantity in the 
 other equation. 
 
 ILLUSTRATIVE EXAMPLE 
 
 lx-2y = l, (1) 
 
 3x + 5y = 59. (2) 
 
 From(l), y^^f 1 ' & 
 
 Substituting this value of y in (2) and solving for x, 
 
 7ic — 1 
 
 Solve 
 
 3s + 5[ 
 
 59, 
 
 2 / 
 
 6z + 35z-5 = 118, 
 
 41 x = 123, 
 
 x = S. 
 Substituting this value of x in (3), 
 
 SOME PRACTICAL SUGGESTIONS 
 
 261. 1. When using the method by Addition or Subtraction, if 
 one of the coefficients of one of the unknown quantities is itself 
 the 1. c. m. of these coefficients, eliminate that unknown quan- 
 tity, inasmuch as this requires that only one of the equations be 
 multiplied. Otherwise, eliminate the unknown quantity whose 
 coefficients require the smallest multipliers to make them the 
 same. 
 
 2. After finding the value of one of the unknown quantities, 
 the other should be found by substituting this in the equation 
 containing the smallest numbers, or, in case of literal equations, 
 the simplest coefficients. 
 
SIMULTANEOUS SIMPLE EQUATIONS 
 
 157 
 
 3. When either of the equations, after reduction, contains in 
 more than one term the unknown quantity to be eliminated, it is 
 not expedient to use the method by Addition or Subtraction, inas- 
 much as no monomial multiplier will make the coefficients alike. 
 
 4. When the unknown quantity to be eliminated occurs in a 
 monomial denominator in both equations, do not clear of frac- 
 tions before eliminating. Such equations may thus be solved by 
 the methods for simple equations, although they are of the second 
 degree. 
 
 EXAMPLES LXXI 
 
 Solve the following, using all the different methods of elimina- 
 tion : 
 
 i'3x — 4y = 2, 
 [ 7 x - 9 y =* 7. 
 
 11 x - 7 y = 37, 
 8x + 9*/ = 41. 
 
 2 x + 7 y = 41, 
 
 3 x + 4 y = 42. 
 
 |5a> 
 \l x 
 
 5 x - 3 y = 4, 
 
 12 7/ = -10. 
 
 5 j3z-4*/ = 18, 
 {3x + 2y = 0. 
 
 f 15 a + 7 y = 29, 
 1 9 x + 15 y = 39. 
 
 \2x-5y = -21, 
 13 x - 4 y = 120. 
 
 f 22 x -f 15 y = 9, 
 18 a + 25 y = 71. 
 
 18 x - 20 y = 44, 
 17 x - 15 y = 26. 
 
 10. 
 
 17 x - 69 2/ = - 103, 
 14 « - 13 y = - 41. 
 
 11. 
 
 2x , 3y 7 
 T + 4 -2' 
 
 x 2y 
 4 5 
 
 11 
 
 2' 
 
 12 
 
 a - 1 y - 1 
 5 7 
 
 0, 
 
 ,2x-3 22/ + 13 
 
 = 0. 
 
 13. 
 
 10 x 
 
 Sy- 
 
 ^ = -17. 
 
 14. 
 
 6 + x-y = 7 } 
 1 — x — y 4 
 2a; + 3# = -l. 
 
158 
 
 HIGHER ALGEBRA 
 
 15. 
 
 17. 
 
 19 
 
 21. 
 
 23. 
 
 25. 
 
 2x 
 3 
 
 5y 
 12 
 
 Sx 
 
 f 
 
 4 
 
 23 
 
 2 
 
 = 2, 
 
 I 
 
 .i 
 
 aj + y o 
 ax + by = c, 
 a'x + b'y = c'. 
 
 (Hc)a;+(6-c)?/=2a&, 
 (a -f- c) a — (a — c)y=2 ac. 
 
 10 9 
 
 SB 2/ 
 8_15 
 
 x y 
 
 = 4, 
 
 9 
 
 2" 
 
 £-1 = 2. 
 
 ax by 
 
 a b 
 bx ay 
 
 a + b, 
 
 x y 
 
 20. 
 
 22. 
 
 24. 
 
 26. 
 
 16. 
 
 18. \ 
 
 x — 3 y y — 3 g 
 
 2 2 
 
 30 
 
 
 ( ax — by = 2 a6, 
 
 I 2 6a; + 2 ay = 3 6 2 - a 2 . 
 (a-j-6)a;— (a— b)y=3ab, 
 (a—b) x— (a-\-b)y =ab. 
 
 _5__7 = 29 
 3a; y 9' 
 
 3 , JL = _2. 
 
 u 4?/ 8 
 
 6a; 
 
 «'/ 
 
 = 0, 
 
 aa; by a 2 
 
 a + b 
 
 x 
 
 ay 
 
 ab 
 
 a; ?/ 
 
 b 
 
 — , 
 
 a 
 g 2 +3«6 
 a + b 
 
 PROBLEMS LEADING TO SIMPLE EQUATIONS WITH TWO 
 UNKNOWN QUANTITIES 
 
 EXAMPLES LXXII 
 
 1. The daily pay of 5 men and 4 boys is $ 19, and the daily 
 pay of 2 men exceeds that of 3. boys by $ 3. What is the daily 
 pay of each man and boy ? 
 
 Solution. Let x — the daily pay of each man, 
 
 and y = the daily pay of each boy ; 
 
 then 6 x + 4 y = 19, 
 
 and 2 x - - 3 y = 3. 
 
 From these equations we find x = 3 and y = 1. 
 
SIMULTANEOUS SIMPLE EQUATIONS 
 
 159 
 
 Solution. 
 and 
 
 Then 
 and 
 
 Let 
 
 2. The sum of the two digits of a number is 13, and the num- 
 ber diminished by that formed by reversing the digits is 27. 
 Find the number. 
 
 x = the digit in tens' place, 
 y = the digit in units' place. 
 10 x + y = the number, 
 
 10 y + x = the number formed by reversing the 
 digits. 
 Hence, x + y = 13, 
 
 and I0x + y -(\0y + x) = 2~. 
 
 From these equations we find x = 8 and y = 5. 
 Hence the number is 10 x 8 + 5 = 85. 
 
 3. An income of $ 120 a year is derived from a sum of money 
 invested, partly in 3J per cent stock and partly in 4 per cent 
 stock. If the stock be sold when the first is at 108 and the second 
 at 120, the sum realized will be $ 3672. Find each investment. 
 
 Solution. 
 and 
 Then 
 
 Let 
 
 income from the first, 
 
 x = the 3£ per cent stock, 
 y = the 4 per cent stock. 
 
 100 
 
 4y 
 
 100 
 
 108 a; 
 
 100 
 
 income from the second, 
 selling price of the first, 
 
 and 
 
 Hence, 
 
 and 
 
 y~ = selling price of the second. 
 
 100 
 
 ^ + il=120, 
 100 100 
 
 M£ + !20j/ = 3672. 
 
 100 100 
 From these equations we find x = 2400 and y = 900. 
 
 4. One kind of wine is worth 72 cents a quart, and another 
 40 cents. How much of each must be put into a mixture of 
 50 quarts that shall be worth 60 cents a quart? 
 
 5. If 6 pounds of sugar and 10 pounds of tea cost $ 6.30, and, 
 at the same price, 10 pounds of sugar and 6 pounds of tea cost 
 $ 4.10, what is the price of each per pound ? 
 
160 HIGHER ALGEBRA 
 
 6. A farmer bought 120 acres of land for $13,200, paying 
 $ 80 an acre for part of it, and $ 120 an acre for the remainder. 
 Find the number of acres in each part. 
 
 7. If 5 be added to the numerator of a certain fraction, its 
 value becomes j ; and if 5 be subtracted from its denominator, its 
 value becomes f . Find the fraction. 
 
 8. A crew that can row 12 miles an hour downstream finds 
 that it takes twice as long to row a given distance upstream. 
 Find the rate of the current and the rate of the crew in still 
 water. 
 
 9. A certain number is equal to 4 times the sum of its two 
 digits, and if 18 be added to it, the digits will be reversed. What 
 is the number ? 
 
 10. A and B can do a piece of work in 9 hours. After working 
 together 7 hours, B finishes the work in 5 hours more. In how 
 many hours could each do the work ? 
 
 11. In an informal ballot a resolution was adopted by a 
 majority of 10 votes ; but in the formal ballot one fourth of those 
 who had before voted for it voted against it, and the resolution 
 was lost by a majority of 6 votes. How many voted each way in 
 the formal ballot ? 
 
 12. When a is added to the greater of two numbers, the sum is 
 m times the less ; but when b is added to the less, the sum is n 
 times the greater. Find the numbers. 
 
 13. When 4 is added to the greater of two numbers, the sum is 
 3^ times the less ; but when 8 is added to the less, the sum is ^ 
 the greater. Find the numbers by substituting in the results of 
 the preceding example. 
 
 14. If two trains, 100 miles apart, approach each other, they 
 will meet in 2 hours ; but if they run in the same direction, the 
 slower train leading, they will be together in 10 hours. What is 
 the rate of each train ? 
 
 15. A father has two sons, one 4 years older than the other. 
 Six years ago the father's age was 6 times the joint ages of his 
 sons, while 2 years hence his age will be twice the joint ages of 
 his sons. Find the age of each. 
 
SIMULTANEOUS SIMPLE EQUATIONS 161 
 
 16. 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. 
 
 17. A man rows 30 miles down a river and then back to the 
 starting place, his whole time being 12 hours. He finds that he 
 can row 5 miles with the stream in the same time as 3 against it. 
 Find his time down and up respectively. 
 
 18. In an alloy of silver and copper — of the whole -\-p ounces 
 
 -. m 
 
 was silver, and - of the whole — q ounces was copper. How 
 n 
 
 many ounces were there of each ? 
 
 19. A man invests $ 5100, partly in 3| per cent stock at 90 and 
 partly in 4 per cent stock at 120, and receives from the two 
 investments $ 185 a year. How many shares of each stock does 
 he buy ? 
 
 20. A man sculls in 1 hr. 20 min. a certain distance down a 
 stream which runs at the rate of 4 miles an hour. In returning 
 it takes him 4 hr. 15 min. to reach a point 3 miles below his start- 
 ing place. How far did he scull down the stream, and at what 
 rate could he scull in still water ? 
 
 21. A tank is supplied by two pipes. If the first be opened 6 
 minutes and the second 7 minutes, the tank will be filled ; or if 
 the first be opened 3 minutes and the second 12 minutes, the tank 
 will be filled. In what time will each pipe fill the tank ? 
 
 22. A cistern is supplied by three pipes, two of which are of the, 
 same size. When they are all open, T 5 ^ of the cistern is filled in 
 4 hours ; but if one of the equal pipes be closed, J of the cistern 
 is filled in 10| hours. In how many hours would each pipe fill 
 the cistern ? 
 
 23. An alloy of tin and lead, weighing 40 pounds, loses 4 
 pounds when immersed in water. It is found that 10 pounds of 
 tin lose 1.375 pounds when immersed in water, and 5 pounds of 
 lead lose .375 pounds. How many pounds of each metal are in the 
 alloy ? 
 
 downey's alg. — 11 
 
162 BIGHER ALGEBRA 
 
 24. A man invests $ 17,200 in 3 per cent bonds at 90 and 5 per 
 cent bonds at 108, and his incomes from the two investments are 
 the same. Find the amount of each investment. 
 
 25. A dairyman mixed with 100 quarts of morning's milk the 
 night's milk, from which the cream had been skimmed, selling 
 the mixture for 5 cents per quart and the cream for 20 cents per 
 quart. By this fraudulent means he realized $ 1.80 more than 
 he would had he sold at 5 cents per quart both the morning's 
 milk and the night's milk without skimming. How much 
 skimmed milk and how much cream did he sell? 
 
 26. B has $ 1000 more capital than A, invests it at one per 
 cent more, and receives $ 80 more income. C has $ 500 more 
 capital than B, invests it at one per cent more, and receives $ 70 
 more income. Find A's capital and rate. 
 
 27. Two men, A and B, are employed on a piece of work. A 
 works 1| days by himself, when B joins him, and they complete 
 T 8 3 of the work 2\ days later ; they then find that 2 days more 
 will be required for them to finish the work. In how many days 
 could each do the work ? 
 
 28. If a pieces of one kind of money make a dollar, and b pieces 
 of another kind make a dollar, how many pieces of each kind 
 must be taken to have c pieces in a dollar ? 
 
 29. The distance from A to B along a railway is 70 miles, the 
 first 10 miles being level, the next 35 sloping upward, and the 
 rest level. A train starting from A runs half the distance in 62 
 minutes, and the whole distance in 1 hour 52 minutes. Find the 
 rates of the train on the level ground and on the up grade. 
 
 30. Two cyclists, A and B, ride a race, the course being from 
 P to Q, a distance of 12 miles, and back. A gives B a start of 15 
 minutes, and meets him on his return journey 1680 yards from Q, 
 afterward winning the race by 1 minute. Find the rate of each 
 cyclist, assuming it uniform. 
 
 31. Two men received $ 96 for a piece of work which they could 
 do together in 30 days. When half the work was done, one of 
 them stopped 8 days and the other 4 days. They completed the 
 work in 35^ days from the beginning. How long would each 
 require to do the work, and how many dollars should each receive ? 
 
SIMULTANEOUS SIMPLE EQUATIONS 163 
 
 32. A man in a rowboat is at a distance a from two barges at 
 the instant when they are passing each other, one coming toward 
 him and the other going away from him, the two having the same 
 rate. Show that if b and b' are the distances he rows before 
 meeting one and overtaking the other, 
 
 ? = !+!. 
 
 a b b' 
 
 33. A and B formed a partnership. A invested $ 20,000 of his 
 own money and $ 5000 which he borrowed ; B invested $ 22,000 of 
 his own money and $ 8000 which he borrowed at the same rate of 
 interest as was paid by A. At the end of a year A's share of the 
 profits was $ 1750 more than the interest on his $ 5000, and B's 
 was $2000 more than the interest on his $8000. What rate of 
 interest did they pay, and what rate per cent did they realize on 
 their investment ? 
 
 34. An ingot of metal which weighs n pounds loses p pounds 
 when weighed in water. This ingot is itself composed of two 
 other metals, which we may call A and B. Now n pounds of A 
 lose q pounds when weighed in water, and n pounds of B lose r 
 pounds when weighed in water. How much of each metal does 
 the original ingot contain ? 
 
 35. A body moves with uniform velocity from A to B, 323 
 feet, and, without stopping, returns. A second body leaves B 13 
 seconds after the first leaves A and moves toward A with uniform 
 velocity. The first body meets the second 10 seconds after the 
 latter starts, and, in returning to A, overtakes the second body 45 
 seconds after the latter starts. Find the velocity of each body. 
 
 SECTION III — SIMULTANEOUS SIMPLE EQUATIONS WITH 
 SEVERAL UNKNOWN QUANTITIES 
 
 262. Prob. To solve several simultaneous simple equations with 
 as many unknown quantities. 
 
 Rule. 1st. Eliminate the same unknown quantity from different 
 pairs of the given equations, thus forming a set of equations inde- 
 pendent of this unknown quantity, and one less in number than the 
 given equations. 
 
164 
 
 HIGHER ALGEBRA 
 
 2d. From these, in like manner, eliminate another unknown quan- 
 tity, and so continue till an equation with but one unknown quantity 
 is found. 
 
 3d. Find the value of this unknown quantity and substitute it in 
 one of the two equations of the next preceding set. Solve this equa- 
 tion and substitute the two values now found in one of the three equa- 
 tions of the next preceding set. Continue this jwocess till all the 
 unknown quayitities are determined. 
 
 263. Sch. 1. It is usually best to combine the equation hav- 
 ing the smallest coefficients with each of the other equations of 
 the same set. 
 
 264. Sch. 2. If any equation of any set does not contain the 
 unknown quantity we are eliminating, this equation is written 
 unchanged in the next set. 
 
 l. 
 
 EXAMPLES LXXIII 
 Solve the following : 
 
 (3x + 4y — z = 8, 
 x + 2y-4;Z = -7, 
 2x-5y-7z = -29. 
 
 2x + 3y-4z = 8, 
 3x-±y + 2z = 3, 
 4:X — 2y—'3z = 5. 
 
 7. 
 
 {2x-3y = 4, 
 4x-3z=2, 
 4?/ + 2z=-3. 
 
 w -\-3x — y — z = 7, 
 2w — 2x + y + 3z = 8, 
 3w — x + y — 4« = 8, 
 4w + a; — y — 2z = 7. 
 
 4. 
 
 2a?-f3y + 3 = 17, 
 2 x 4- 2 y + z === 14, 
 
 x + 3y + 2z = Vd. 
 
 12x — ky + z = 3, 
 x—y — 2z — — l, 
 [5x-2y = 0. 
 
 y + z 
 
 x + 
 
 85, 
 
 6. ^ + ^ = 85, 
 
 z + 
 
 x + y _ 
 
 = 85. 
 
 2x + 3y + z = 23, 
 2u + 3x + y = 25, 
 u + x + 3z = 24, 
 [3u + 2y + 2z = 36. 
 
SIMULTANEOUS SIMPLE EQUATIONS 
 
 165 
 
 u-2x = -13, 
 x-3y=13, 
 y-±z = 5, 
 
 10. 
 
 ou 
 
 23. 
 
 u + 5x — 7 y + z = 2, 
 2u + 7x-3y-3z = 2 y 
 4x-2y = 2, 
 ix + 5y — 2z = 2. 
 
 Although the following are not of the first degree with reference to x, y, 
 and z, they are of the first degree with reference to the reciprocals of these 
 
 We may, therefore, regard , -, and - as the unknown quanti- 
 se y z 
 
 quantities 
 
 ties and solve accordingly, i.e., eliminate without first clearing of fractions, 
 
 il- 
 
 ia. 
 
 15. 
 
 fl 1 
 
 1 _ 
 
 z 
 
 3, 
 
 1.-14 
 
 x y 
 
 1 
 
 z ~ 
 
 •2, 
 
 M- 
 
 lx y 
 
 1_ 
 
 z ~ 
 
 :1. 
 
 [1+1- 
 
 x y 
 
 2 
 = 3' 
 
 
 1 1_ 
 
 x z~ 
 
 3 
 4' 
 
 
 [y z 
 
 5 
 = 6' 
 
 
 a b 
 x y~ 
 
 = 1, 
 
 
 6 + °- 
 
 y z 
 
 = 1, 
 
 
 c a 
 
 . z x~ 
 
 :1. 
 
 
 12. 
 
 14. 
 
 16. 
 
 f2 
 
 X 
 
 1 
 
 + y- 
 
 2; 
 
 2 
 
 y 
 
 3_ 
 
 z ~ 
 
 -2, 
 
 l 
 
 . x 
 
 1 
 
 + ~z = 
 
 4 
 3* 
 
 1 
 
 X 
 
 1 
 
 + - y + 
 
 z 
 
 2 
 
 X 
 
 3 
 
 z 
 
 5 
 . x 
 
 * 
 
 5 = 4. 
 
 z 
 
 1 
 
 X 
 
 + A 
 
 9 
 
 "5' 
 
 1 
 
 y 
 
 +& 
 
 5 
 
 "3' 
 
 l 
 
 .z 
 
 4sc 
 
 7 
 4* 
 
 In solving the next two examples, divide each equation by the unknown 
 quantities in the second member, and then eliminate fractional terms. 
 
 17. 
 
 y + z = 2yz, 
 x + z = 3 xz, 
 x + y = 4xy. 
 
 yz+k xz—3 xy=2xyz, 
 
 \ 3yz—2xz-\-6xy=4:xyz, 
 
 5 yz—6 xz—3 xy=xyz. 
 
166 
 
 HIGHER ALGEBRA 
 
 In solving the next two examples, let xyz constitute one member of each 
 equation, divide each equation by xyz, and then eliminate fractional terms. 
 
 ' 19. xyz = 2 (xz -f- xy — yz) = 3 (xz -f- yz — xy) = 4 (xy -\-yz — xz). 
 
 20. xyz = a(yz — xz — xy) = b (xz — xy — yz) = c (xy — yz — #z). 
 
 When one of the unknown quantities is wanting in each of the given equa- 
 tions, and all the coefficients (after simplifying the equations) are unity, the 
 work may be made very short by adding all the equations, dividing by the 
 common coefficient, and then subtracting from the resulting equation each of 
 the given equations in turn. Likewise, when all of the unknown quantities 
 are present in each equation, and all of the coefficients (after simplifying the 
 equations) except one are unity, and this one coefficient is repeated with 
 successive unknown quantities in successive equations, the work may be 
 made very short by adding all the equations, dividing by the common coeffi- 
 cient, and then subtracting the resulting equation from each of the given 
 equations in turn. Solve in this manner the following : 
 
 (x + y = 9, 
 21. I x 4- * = 10, 
 I y + z = 11. 
 
 22. 
 
 23. 
 
 25. 
 
 27. 
 
 fw + x + y = 3, 
 w -\- x + z = — 4, 
 iv + y + z = 2, 
 x + y + z=-l. 
 
 v + w + x -j- y = 10, 
 v + iv -f x -f- z = 11, 
 v + w + y -f z = 12, 
 v + x + y + z = 13, 
 w + x + y -f z = 14. 
 
 3w + x + y + z = l4:, 
 
 w + 3x + y + z = l2, 
 
 w 4- x ' + 3y + z = 16, 
 
 [w + x + y + 3z = 18. 
 
 24. 
 
 26. 
 
 28. 
 
 w + x + y = l, 
 w -f it- -f z = 6, 
 iu + ?/ + z = 9, 
 x + y + z = 8. 
 
 w + x + y = Q, 
 iv -f x -(- z = 6, 
 w + ?/ + z = — 1, 
 # -f- ?/ + z = — 2. 
 
 f u -f- v -f- i<; + x -f ?/ = 3, 
 u + v + w -f- x + z = 8, 
 w + v-j-w-f2/-fz = 4, 
 ** + v -f- x + y + k « 9, 
 m -f- to + x -f- ?/ + z = 6, 
 
 lv + w4-aj-r-y4-2! = 5. 
 
 (w-f-# + ?/-|-4z = — 9, 
 w-f-#-f-4?/-|-z = 3, 
 w + 4#-f-y-f-z = — 6, 
 4w + x-f-?/-f-z = 12. 
 
SIMULTANEOUS SIMPLE EQUATIONS 
 
 167 
 
 29. 
 
 2w+2x+2y+3z=39, 
 2tv-{-2x-\-3y-\-2z=37, 
 2w+3x+2y+2z=35, 
 3w+2x+2y+2z=33. 
 
 ' v + w -f- x + y + 2 z = 52, 
 
 v + w + g + 2 ?/ + z = 50, 
 
 31. {t> + to+2a?+y+s=48, 
 
 v 4- 2 w + a; + y + z = 46, 
 1 2v + M7 + aj + y + z = 44. 
 
 30. 
 
 fw4-5a;4-2/4-z = 22, 
 iu -f x + y 4- 5 z = 26, 
 5w + x-f?/ + z = 18, 
 
 . ic 4- a; 4- 5 y 4- 2 = — 18. 
 
 ' v 4- w 4- 6 #4- 1/4-2= 40, 
 6v-M«4-a;4-2/4-z=36, 
 32. j v+w+ajH-6y+2=lj 
 
 v+w-f-o;+?/-|-62;=31. 
 
 PROBLEMS LEADING TO SIMPLE EQUATIONS WITH SEVERAL 
 UNKNOWN QUANTITIES 
 
 EXAMPLES LXXIV 
 
 1. The sum of the three digits of a number is 12 ; the digit in 
 the tens' place is \ the sum of the other two, and the number ex- 
 pressed by the two left-hand digits is 7 times the digit in units' 
 place. Find the number. 
 
 2. For $ 8 I can buy 2 pounds of tea, 10 pounds of coffee, and 
 20 pounds of sugar ; or 2 pounds of tea, 5 pounds of coffee, and 
 30 pounds of sugar; or 3 pounds of tea, 5 pounds of coffee, 
 and 10 pounds of sugar. . What are the prices ? 
 
 3. Three cities, A, B, and C, not in the same straight line, are 
 connected by straight roads. The distance from A to C by way 
 of B is 82 miles, from B to A by way of C is 97 miles, and 
 from G to B by way of A is 89 miles. Find the distances be- 
 tween the cities. 
 
 Sug. In solving the equations, proceed as directed for examples 21 to 2G 
 of the preceding set. 
 
 4. The total capacity of 3 casks is 1440 quarts. Two of 
 them are full, and one is empty. To fill the empty one requires 
 the contents of the first and ^ the contents of the second, or the 
 contents of the second and ^ the contents of the first. Find the 
 capacity of each cask. 
 
168 HIGHER ALGEBRA 
 
 5. A farm was rented for 2 years for a fixed money payment 
 and 400 bushels of wheat and barley. The first year wheat was 
 70 cents per bushel and barley 50 cents, and the entire rent was 
 $ 1000. The second year wheat was 50 cents per bushel and 
 barley 45 cents, and the entire rent was $ 950. Find the amounts 
 of money, wheat, and barley paid as rent each year. 
 
 6. When 3 partners began business, A had $ 2000 more than 
 twice as much capital as B, and C had $ 500 less than A and B 
 together. The first year A gained as much as B's capital, B 
 gained as much as A's capital, and C gained as much as A's and 
 B's capital together, whereupon each had the same sum. Find 
 how much each had at first, and interpret the results. 
 
 7. The capacity of 3 casks is 344 gallons, and all are full. 
 Fifty gallons are used from the first; then i of what is in the 
 second is poured into the first, and £ of what is in the third is 
 poured into the second. After these changes, the first contains 
 10 gallons more than the second, and the second 10 gallons more 
 than the third. Find the capacity of each cask. 
 
 8. In walking along a street on which electric cars are running 
 at equal intervals from both ends, I observe that I am overtaken 
 by a car every 12 minutes, and that I meet one every 4 minutes. 
 What are the relative rates of myself and the cars, and at what 
 intervals of time do the cars start ? 
 
 9. Four towns, A, B, C, and D, connected by rail, are at the 
 vertices of a quadrilateral. A commercial traveler, in making 
 the rounds of these towns, observes that in going from A by way 
 of B and C to D, the fare, at 2 cents per mile, is $ 1.22, and in 
 going from A by way of D and C to B, it is $ 1.10 ; while from A 
 by way of B to C, it is the same as from A by way of D to C, and 
 from B by way of A to D, it is 40 cents less than from B by way 
 of C to D. Find the distances. 
 
 10. A, B, C, and D engage to do a certain work. A and B can 
 do it in 12 days, A and D in 15 days, and D and C in 18 days. 
 B and C begin the work together, after 3 days are joined by A, 
 and after 4 days more by D. Then all working together they 
 finish it in 2 days more. How long would each have required to 
 do the entire work ? 
 
CHAPTER XII 
 INEQUALITIES 
 
 265. An Inequality is an assertion by means of a mathematical 
 symbol that two expressions have different values. This symbol 
 is the Sign of Inequality > or <, read "greater than" or "less 
 than," according as the opening is to the left or right. 
 
 266. The Members of an Inequality are the expressions con- 
 nected by the sign of inequality, the one on the left being called 
 the First Member, and the one on the right the Second Member. 
 
 267. Of two positive quantities, the greater is the one which is 
 numerically the greater; while of two negative quantities the 
 greater is the one which is numerically the less. A negative 
 quantity is less than a positive quantity, regardless of their 
 numerical values. In general, when a — b is positive, a>b, and 
 when a — b is negative, a<b. 
 
 268. Two inequalities exist in the same sense or exist in an 
 opposite sense, according as their signs of inequality are alike or 
 opposite. 
 
 269. Theorem. An inequality will continue to exist in the same 
 sense, 
 
 1st. By adding equals to both members, or subtracting equals 
 from both members. 
 
 c 2d. By multiplying or dividing both members by equal positive 
 quantities. 
 
 3d. By raising both members to the same odd power. 
 
 4th. By raising both members to the same even power, if both 
 members are positive. 
 
 5th. By extracting the same root of both members, if when the 
 index is even, only the positive roots be compared. 
 
 169 
 
170 HIGHER ALGEBRA 
 
 Dem. Each of the operations of the 1st and 2d increases or 
 decreases both members alike, either by the same amount or 
 in the same ratio, and does not change the signs of the members. 
 Therefore, that member which was greater before any of the 
 operations named is greater after. 
 
 Since the same power or root of the numerically greater of two 
 quantities is numerically the greater, and the operations of the 3d, 
 4th, and 5th do not change the signs of the members, that member 
 which was greater before any of the operations named is greater 
 after. 
 
 270. Cor. i. An inequality may be cleared of fractions without 
 changing the sense in which it exists. 
 
 This is simply multiplying both members by the same positive 
 quantity. 
 
 271. . Cor. 2. A term, by changing its sign, may be transposed from 
 one member of an inequality to the other without changing the sense 
 in ivhich it exists. 
 
 This is simply subtracting the same quantity from, or adding 
 the same quantity to, both members. 
 
 272. Theorem. An inequality will be made to exist in an opposite 
 sense, 
 
 1st. By changing the signs of both members. 
 
 2d. By multiplying or dividing both members by the same negative 
 quantity. 
 
 3d. By raising both members to the same even power, if both 
 members are negative. 
 
 4th. By comparing the negative even roots. 
 
 Dem. The members are affected numerically as in the pre- 
 ceding demonstration, but the signs are changed. Therefore, that 
 member which was greater before any of the operations named is 
 less after. 
 
 273. Theorem. If two or more inequalities exist in the same 
 sense, 
 
 1st. The sums will exist in the same sense as the given inequalities. 
 2d. The products will exist in the same sense as the given inequali- 
 ties, if their members are all positive. 
 
INEQUALITIES 171 
 
 Dem. In the 1st case the greater member of any one of the 
 inequalities is increased by a greater amount or decreased by a 
 less amount than the less member ; and in the 2d case the greater 
 member is increased in a greater ratio or diminished in a less ratio 
 than the less member. 
 
 274. Caution. The student is cautioned against taking the 
 differences or the quotients of two inequalities that exist in the 
 same sense, inasmuch as the differences and quotients will exist in 
 the same sense as, or the opposite sense from, the given inequali- 
 ties, according to the order of subtracting or dividing, or will give, 
 with certain relations, an equation. 
 
 275. Note. In determining which of two expressions is the 
 greater it is sometimes best to place both signs, *S, read "greater 
 or less," between them and then operate on the inequality in such 
 ways as not to change the sense, until it becomes apparent which 
 is the greater member. This will show which of the two signs to 
 use in the original inequality. 
 
 EXAMPLES LXXV 
 
 It is understood that the letters of each example are positive 
 and unequal. 
 
 1. Which is greater, the sum of the squares of any two quanti- 
 ties or twice their product ? 
 
 Solution. We have 
 
 
 
 a 2 + b 2 >2ab. 
 
 Transposing, 
 
 a 2_ 2a &-f &2>o, 
 
 
 (a-b) 2 >0. 
 
 But (a — b) 2 is necessarily plus and, therefore, greater than 0. Hence, 
 since the operations leave the inequality in the same sense as the first, 
 
 • a 2 + 6 2 > 2 ab. 
 
 2. Which is greater, half the sum of two quantities (their 
 arithmetical mean) or the square root of their product (their 
 geometrical mean) ? 
 
172 HIGHER ALGEBRA 
 
 Solution. We have ^-±-^ > Vab. 
 
 2 < 
 
 Squaring, etc., a 2 + 2 ab + 6 2 J 4 «&, 
 
 a 2 -2a& + ft2>o, 
 
 («-6) 2 <0. 
 
 But (a — 6) 2 > 0, because the square of any quantity is plus. Hence, since 
 the operations leave the inequality in the same sense as the first, 
 
 3. Show that a 2 -f b 2 + c 2 > afr + ac 4- 6c. 
 Solution. By Ex. 1, a 2 + b 2 >2ab, 
 
 a 2 + c 2 > 2 ac, 
 6 2 + c 2 > 2 6c. 
 Adding and dividing by 2, a 2 -f 6 2 4 c 2 > «6 + ac 4 6c. 
 
 4. Which is greater, — or , if x < a ? 
 
 a- + r a-\-x 
 
 5. Show that any fraction plus its reciprocal is greater than 2. 
 
 6. If a, b, c are such that the sum of any two is greater than 
 the third, show that a 2 -f b 2 -f- c 2 < 2 (a& -4- ac 4- be). 
 
 7. If a 2 + b 2 + c 2 = 1, and m 2 + n 2 + r 2 = 1, show whether 
 am + £m -f cr is greater or less than 1. 
 
 8. Show that 
 
 (a + b — c) 2 +(a + c — b) 2 + (6 + c — a) 2 > a/> 4- be + ac. 
 
 9. Which is greater, a 3 4- & 3 or a 2 £> + a& 2 ? 
 
 10. Prove that (a& 4- xy) (ax + %) > 4 abxy. 
 
 11. Which is greater, 2 X s or sc 4- 1, if x > 1 ? 
 
 12. Find the limits of x determined by the conditions 
 
 x —Z . x — 4 . A -, a; — 10 - x -\- 5 1 
 
 > 9, and < — — 
 
 95' 69 6 
 
INEQUALITIES 173 
 
 13. The double of a number diminished by 5 is greater than 
 25, and the triple of the number diminished by 7 is less than 
 the double increased by 13. Find the limits of the number. 
 
 14. A man wishes to make a purchase for $ 14, but has not 
 enough money. If he borrows, one third as much money as he 
 now has, he will be able to make the purchase and have more 
 money left than he now lacks. How much money has he ? 
 
 15. The daily pay roll of a contractor, who pays masons $ 4.40 
 a day and carpenters $ 3.60 a day, is between $ 104 and $ 112, 
 and there are 3 more masons than carpenters. Find the number 
 of each. 
 
 16. A is 24 years old and B is 15. What is the shortest time 
 after which A's age will be less than If times B's age ? 
 
 17. If only 5 pupils be seated on each bench in a recitation 
 room containing fewer than 6 benches, 4 pupils will be without 
 seats ; but if 6 pupils be seated on each bench, some seats will be 
 unoccupied. Find the number of benches. 
 
 18. The sum of two whole numbers is 25. If the greater be 
 divided by the less, the quotient will be less than 3J ; and if the 
 less be divided by the greater, the quotient will be greater than £. 
 What are the numbers ? 
 
CHAPTER XIII 
 
 RATIO, PROPORTION, AND VARIATION 
 
 SECTION I — RATIO 
 
 276. Ratio is the relative magnitude of two quantities of the 
 same kind, and is measured by the quotient of the first by the 
 second. 
 
 The ratio of two quantities is expressed either by writing a 
 colon between them or by writing them in the fractional form. 
 
 Thus the ratio of a to 6 is written either a : b or -• 
 
 b 
 
 277. The Antecedent or First Term of a ratio is the first of the 
 two quantities compared, and the Consequent or Second Term is 
 the second. 
 
 278. A ratio is a Ratio of Greater Inequality, Less Inequality, 
 or Equality, according as it is greater than, less than, or equal to, 
 unity. 
 
 279. The Duplicate, Sub-duplicate, Triplicate, and Sub-triplicate 
 Ratios of two quantities are the ratios of the squares, square 
 roots, cubes, and cube roots of those quantities respectively. 
 
 280. A Compound Ratio is the ratio of the products of the cor- 
 responding terms of two or more simple ratios. 
 
 281. Theorem. 1st. A ratio is not changed by multiplying or 
 dividing both its terms by the same quantity. 
 
 2d. A ratio is multiplied by multiplying its antecedent or dividing 
 its consequent. 
 
 3d. A ratio is divided by dividing its antecedent or multiplying 
 its consequent. 
 
 174 
 
RATIO 175 
 
 Since a ratio is simply a fraction, or an indicated division, 
 these follow from Arts. 127 and 137. 
 
 282. Theorem. A ratio of greater inequality is diminished, and 
 a ratio of less inequality is increased, by adding the same positive 
 quantity to both its terms; i.e., any ratio is made more nearly equal 
 to unity by adding the same positive quantity to both its terms. 
 
 Dem. Let m : n, or — , be the given ratio, and a the quantity 
 n 
 
 to be added to both its terms. Then we have 
 
 m > m + a ^ . 
 
 n < n+a' ' 
 
 Whichever member is greater will still be greater after both 
 are multiplied by n (n + a), giving 
 
 mn + am ^ mn + an. (2) 
 
 Since mn is common to both members, the first member of (2) 
 will be greater or less than the second, according as am is greater 
 or less than an, or m greater or less than n. Hence the first 
 member of (1) is greater or less than the second, according as m 
 is greater or less than n. 
 
 EXAMPLES LXXVI 
 Find the values of the following ratios ; 
 1. x 2 — 7 x + 10: x — 2. 2. #+5:o2 + 3a;-10. 
 
 3. x *-y*:x i -x 2 y4-xy 2 -y\ 4. 12 (a - b) 2 : 8 (a 2 - b 2 ). 
 
 5. 3x 3 -2x 2 -19x-6:3x i + 4:X 2 -5x-2. 
 
 6. The duplicate ratio of 13 : 39. 
 
 7. The sub-duplicate ratio of 121^-726 a; +1089 : a^-Gz+a 
 
 8. The triplicate ratio of } : | • 
 
 9. The sub-triplicate ratio of 729 : 1728. 
 
 10. Is a 2 — x 2 : a 2 + x 2 greater or less than a — x.a + x ? 
 
 11. Is x 3 + i/ 3 : x 2 + y 2 greater or less than x 2 + y 2 : x + y ? 
 
176 HIGHER ALGEBRA 
 
 12. Is X s — y* : (x 4- yf greater or less than x 2 -f- xy -f- y 2 : x — #, 
 a; being greater than y ? 
 
 13. If 6 be added to each of two numbers that are in the ratio 
 of 5 : 7, the sums will be in the ratio of 7:9. Find the numbers. 
 
 14. If 8 be subtracted from each of two numbers that are in 
 the ratio of 2:5, the remainders will be in the ratio of 2:9. 
 Find the numbers. 
 
 15. What must be subtracted from each term of 13 : 25 to make 
 it 1:3? 
 
 16. A certain ratio becomes J when 4 is added to both its terms, 
 and I- when 2 is subtracted from both its terms. Find the ratio. 
 
 17. What quantity subtracted from each term of the duplicate 
 ratio of m : n will give the triplicate ratio of m : n ? 
 
 18. If 5 gold coins and 30 silver ones are worth as much as 10 
 gold coins and 10 silver ones, what is the ratio of their values ? 
 
 19. If the wages of 5 boys and 6 girls is J of the wages of 6 boys 
 and 9 girls for the same time, what is the ratio of their wages ? 
 
 20. The sides of a triangle are in the ratio of 3, 4, and 5, and 
 the perimeter is 480 yards. Find the sides. 
 
 21. The ratio of a father's age to his son's is 7 : 2, and the father 
 is 30 years older than the son. Find the age of each. 
 
 22. A fox makes 4 leaps while a hound makes 3 ; but 2 of the 
 hound's leaps equal 3 of the fox's. Find relative rates of running. 
 
 23. If in working out road tax 6 men and 4 teams are counted 
 as much as 10 men and 1 team, what is the ratio of wages for 
 men and teams ? 
 
 24. A bequest of $ 900 wa,s divided among three sons, and after 
 their shares had increased by $ 10, $ 15, and $ 20 respectively, 
 the sums were in the ratio 4:5:6. Find the shares. 
 
 25. The weights of two loads are in the ratio of 4 to 5. Parts 
 of the loads in the ratio of 6 to 7 being removed, the remaining 
 weights are in the ratio of 2 to 3, and the sum of the weights is 
 then 10 tons. What were the weights at first ? 
 
Piwpoinio.x 177 
 
 26. In a college boat race, crew A pull 15 strokes to 14 strokes 
 of crew B ; but 28 strokes of crew B are as effective as 33 of crew 
 A. Which is the faster crew, and in what ratio ? 
 
 27. Find the gear of a bicycle whose wheels are d inches in 
 diameter, whose front sprocket has m teeth and whose rear sprocket 
 li;is n teeth, it being understood that the "gear'' of a bicycle is 
 the diameter of a wheel one revolution of which would advance it 
 as far as one revolution of the pedal advances the bicycle. 
 
 Solution. Let x = the gear. 
 
 One revolution of a wheel whose diameter is x would advance the wheel by 
 the amount of its circumference, ttx. 
 
 The circumferences of the sprocket wheels are in the ratio of their number of 
 
 teeth, m : n. Hence — = the number of revolutions of the bicycle wheels to 
 
 M 
 one revolution of the pedal, and - x wd = the distance the bicycle advances 
 
 for one revolution of the pedal. Therefore, 
 
 m 7 
 irx = — ?rf7, 
 n 
 
 md 
 whence x = 
 
 N 
 
 28. Find, by substituting in the formula of the last example, 
 the gear in each of the following cases : 
 
 (a) d = 2S in., m = 18, n = 7. 
 
 (b) d = 2S in., m = 21, n = 7. 
 
 (c) d = 30 in., m = 20, n = 8. 
 
 (d) d = 28 in., m = 22, n = 8. 
 
 SECTION II — PROPORTION 
 
 283. A Proportion is an equality of ratios. 
 
 The equality is indicated by the sign of equality or by the 
 double colon. If the ratio a : b equals the ratio c : d, the propor- 
 tion may be written in any one of the three ways, 
 
 a:b: : c : d, a : b — c : d, or - = -• 
 ' b d 
 
 In any form the proportion is read, " a is to b as c is to pL" 
 downey's alo. — 12 
 
178 ITTGTIER ALGEBRA 
 
 284. Four quantities are Directly Proportional when the ratio of 
 two of them is equal to the ratio of the other two taken in the 
 same order. 
 
 Thus, the times being the same, any two distances are directly propor- 
 tional to the corresponding rates. 
 
 285. Four quantities are Inversely or Reciprocally Proportional 
 when the ratio of two of them is equal to the ratio of the other 
 two taken in the inverse order. 
 
 Thus, the distances being the same, any two times are inversely or recip- 
 rocally proportional to the rates. If T is the time at the rate li, and t the 
 time at the rate r, then 
 
 T:t::r:B, or T : t : : — : -• 
 li r 
 
 The same relation is expressed by saying, "The times are in the inverse, 
 or reciprocal, ratio of the rates," or "The times vary inversely, or recipro- 
 cally, as the rates." 
 
 286. The Extremes of a proportion are its first and last terms. 
 The Means of a proportion are its second and third terms. 
 
 287. When the means of a proportion are the same quantity, 
 this quantity is called a Mean Proportional between the other two 
 quantities, and the last term is called a Third. Proportional to the 
 other two quantities. 
 
 Thus, in a : b : : b : c, b is a mean proportional between a and c, and c is a 
 third proportional to a and b. 
 
 288. A Fourth Proportional to three quantities is the fourth term 
 of a proportion whose other three terms are the three quantities 
 taken in their order. 
 
 289. A proportion is taken by Inversion when the terms of 
 each ratio are written in inverse order. 
 
 290. A proportion is taken by Alternation when the means are 
 interchanged, or when the extremes are interchanged. 
 
 291. A proportion is taken by Composition when the sum of the 
 terms of each ratio is compared with either term of that ratio, the 
 
PROPORTION 179 
 
 same order being observed ; or when the sum of the antecedents 
 and the sum of the consequents are compared with either ante- 
 cedent and its consequent. 
 
 292. A proportion is taken by Division if difference be substi- 
 tuted for sum in the last definition. 
 
 293. A Continued Proportion is a succession of equal ratios in 
 which each consequent is the antecedent of the next ratio. 
 
 Thus, a :b : :b : c: : c :d: : d : e is a continued proportion. 
 
 294. Theorem. In any proportion the product of the extremes 
 equals the product of the means. 
 
 Dem. Let the proportion be 
 
 a : b : : c : d. 
 
 This is the same as ^ = - (Art. 283). 
 
 o a 
 
 Clearing of fractions, ad = be. 
 
 295. Cor. i. A mean proportional between two quantities is 
 equal to the square root of their product. 
 
 For if a:b: :b:c, 
 
 b 2 = ac, 
 whence b = sfac. 
 
 296. Cor. 2. Either extreme of a proportion equals the product 
 of the means divided by the other extreme; and either mean equals 
 the product of the extremes divided by the other mean. 
 
 297. Theorem. If the product of two quantities equals the prod- 
 uct of two others, the quantities of one product may be made the 
 extremes and the quantities of the other product the means of a 
 proportion. 
 
 Dem. Let ad = be. 
 
 Dividing by M, £=|; 
 
 o a 
 
 that is, a : b : : c : d. 
 
180 HIGHER ALGEBRA 
 
 Writing the first equation in the form 
 
 bc = ad, 
 
 and dividing by ac, - = -; 
 
 that is, b:a:: d:c. 
 
 By dividing by other combinations of the letters other forms 
 can be obtained ; but in each the quantities of one product will be 
 the extremes, and the quantities of the other product the means. 
 
 298. Theorem. Proportionals result from taking a proportion 
 (a) by inversion, {b) by alternation, (c) by composition, (d) by 
 division, (e) by composition and division. 
 
 Dem. Let the given proportion be * 
 
 a:b :: c: d. (1) 
 
 Then proportionals result from taking this 
 
 (2) 
 (3) 
 
 (a) By inversion, 
 
 b: a: : d: c. 
 
 From (1), 
 
 a _c 
 b~d 
 
 Dividing unity by each member, 
 
 
 b __d. 
 
 — 5 
 a c 
 
 that is, 
 
 b: a: : d: c. 
 
 (b) By alternation, 
 
 ( a: c: :b: d, 
 [d: b: : c : a. 
 
 From (1), 
 
 ad = be. 
 
 Dividing by cd, 
 
 a b 
 c~d' 
 
 that is, 
 
 a:c: :b:d. 
 
 Dividing (4) by ab, 
 
 d _c m 
 b a' 
 
 that is, 
 
 d: b: : c: a. 
 
 (4) 
 
PROPORTION 
 
 181 
 
 or 
 
 
 ' a-\- b : b : : c -\-d 
 
 4 
 
 
 a + b'.awc + d: 
 
 c, 
 
 (c) By composition, 
 
 a + c: a: :b + d 
 
 &, 
 
 
 a + c : c : : b + d: 
 
 d, 
 
 
 and other forms. 
 
 
 Adding unity to both members of (2), 
 
 
 - + 1=-+1, 
 
 
 
 a + b _c + d , 
 b d 
 
 
 (5) 
 
 that is, a + 6 : b : : c + d : d. 
 
 Let the student demonstrate the other three forms given. 
 Other forms are obtained by taking these four forms by inversion 
 and by alternation. 
 
 or 
 
 
 r a — b : b : : c — d : d, 
 
 
 a — b :a:: c — d: c, 
 
 (d) By division, 
 
 a — c : a : : b — d : b, 
 
 
 a — c : c : : b — d : d, 
 
 
 and other forms. 
 
 Subtracting unity from both members of (2), 
 
 «-l= c --l, 
 b d 
 
 a—b c—d 
 
 b d ' 
 
 at is, c 
 
 i — b:b : :c — did. 
 
 (6) 
 
 Let the student demonstrate the other three forms given. 
 Other forms are obtained by taking these four forms by inversion 
 and by alternation. 
 
 ( a + b : a — b : : c + d: c — d, 
 
 ] a -f- c: a — c::b + d:l 
 and other forms. 
 a + b _c + d. m 
 a — b c — d 1 
 that is, a -\- b : a — b : \ c -\- d : c — d. 
 
 (e) By composition 
 and division, 
 
 Dividing (5) by (0), 
 
 d, 
 
182 HIGHER ALGEBRA 
 
 Let the student demonstrate the othor form given. Other 
 forms are obtained by taking these two forms by inversion and 
 by alternation. 
 
 299. Theorem. If four quantities are in proportion, proportionals 
 result from taking equimultiples, (a) of the terms of a couplet, (b) of 
 the antecedents, (c) of the consequents, (d) of all the terms. 
 
 Dem. Let the given proportion be 
 
 a : b : : c : d, (1) 
 
 Then proportionals result from taking equimultiples, 
 (a) Of the terms of a couplet. 
 
 From(l), l = c - 
 
 o d 
 
 Multiplying numerator and denominator of the first member 
 
 by m, 
 
 am _c m 
 bm d' 
 
 that is, am:bm::c: d. 
 
 Let the student demonstrate the other cases. 
 
 300. Theorem. The same powers or roots of proportionals are 
 proportional. 
 
 Let the student demonstrate. 
 
 301. Theorem. The products or the quotients of the corresponding 
 terms of two (or more) proportions are proportional. 
 
 Dem. Let the given proportions be 
 
 a : b : : c : d, 
 
 a) 
 
 (2) 
 
 and 
 
 m:n::p: q. 
 
 These are the same as 
 
 a _c 
 b~d 
 
 and 
 
 m p 
 n q 
 
PROPORTION 183 
 
 Multiplying (1) by (2), j^ = %\ 
 
 that is, am :bn : : ep : dg« 
 
 From the given proportions, 
 
 Dividing (3) by (4), 
 
 or 
 
 ad -- 
 
 = be, 
 
 mq -- 
 
 = np. 
 
 ad 
 mq 
 
 _bc 
 
 ~ np 
 
 a d 
 
 — X - = 
 m q 
 
 b c 
 
 = - x - 
 
 n p 
 
 a b 
 
 c d 
 
 m' n ' 
 
 " p'q 
 
 (3) 
 
 (4) 
 
 Hence, by Art. 297, 
 
 302. Theorem. In a series of equal ratios the sum of all the ante- 
 cedents is to the sum of all the consequents as any antecedent is to its 
 consequent. 
 
 Dem. If a:b :: c: d:: e : f:: g : h, etc., (1) 
 
 then a + c + e+g + etc. : b + d +f-\-h + etc. : : a : b or c : d, etc. 
 We have the identity, ab = ba. 
 
 From (1) we have ad = be, 
 
 af=be, 
 
 ah = bg, 
 
 etc., etc. 
 
 Adding, a{b + d +/-f h + etc.)= b{a + c + e + g + etc.). 
 Hence, by Art. 297, 
 
 a + c + e + g + etc. :b + d +f+ h + etc. ::a:b. 
 
 303. When a proportion is given and we wish to determine 
 whether some other proposed relation involving the same quan- 
 tities is true, we may proceed in any one of several ways. For 
 example, let the given proportion be 
 
 a : b : : c : d, 
 
 and let the problem be to determine whether 
 
 a -f- b : b : : c -f d : d. 
 
184 HIGHER ALGEBRA 
 
 1st. We may proceed as in the demonstration of this case, 
 Art. 198, c. 
 
 2d. The proposed proportion is true if it can be shown that the 
 
 product of the extremes equals the product of the means. This 
 
 would give 
 
 ad -f bd = be -J- bd, 
 
 or ad = be, 
 
 which is seen from the given proportion to be true. 
 
 3d. In the given proportion we may represent each of the 
 equal ratios, a : b and c : d, by r, giving 
 
 a -, c 
 
 - = r and - = r, 
 
 b d 
 
 or a = br and c = dr. 
 
 Substituting these values in the proposed form, we have 
 
 br + b : b : : dr + d:d, 
 
 in which the ratios are seen to be equal, each being r + 1 : 1. 
 
 Any of the forms of the preceding theorems may be tested in 
 this way. 
 
 EXAMPLES LXXVII 
 
 1. Find the ratio of x to y in 7 x — 5 y : 4: x — 3 y: : 5 : 2. 
 
 2. "From a:b:: c : d deduce 
 
 5 a + S b : 5 a — 3 & : : &c + 3 d : 5 c — 3 d 
 
 3. From x : y : : 4 : 7, find the ratio of # — 4 to y — 7. 
 
 4. Find the mean proportional between x 2 and if — -- 
 
 y *^ 
 
 5. Find x from x 2 +5 x+6 : a^ + lO tf+21 : : x 2 -± : 4^+8 a;- 32. 
 
 6. The third proportional to two numbers is 48, and the mean 
 proportional between them is 6. Find the numbers. 
 
 7. If a, b, c, d are in continued proportion, prove that b + c is 
 a mean proportional between a -f 6 and c + c/. 
 
 8. From a:b :: c:d deduce a— c: b — d:: Va 2 -f- c* : V& 2 + d 2 . 
 
 9. From |a-x;ia + a;::6-?/:6 + ?/ deduce 2 x\ y:\a\b. 
 Sue. Take by division and composition, divide second couplet by 2, and 
 
 take by alternation. 
 
PROPORTION 185 
 
 10. If a : b : : e:d : : e:f, prove that 
 
 a + 3 c + 2 e : a - e : : 6 + 3 d + 2/: 6 -/. 
 
 11. What operations on x : y : : a : b will produce 
 
 sc 2 : a 2 : : a 2 -f y 2 : a 2 + 6 2 ? 
 
 12. If a : 6 : : » : q, prove that a 2 + b 2 : : : p 2 + q 2 : — 
 
 a + b p + q 
 
 13. What quantity added to each of the quantities a, b, c, d y 
 will make them proportionals ? 
 
 14. If four quantities are proportionals, show that there is no 
 quantity which, being added to each, will leave the sums propor- 
 tionals. 
 
 15. If the fourth proportional to a, b, c is the same as that to 
 a, b'j c', show that b : b' : : c' : c. 
 
 16. Iia:b::c:d, prove that a 2 + b 2 : c 2 + d 2 : : (a + &) 2 : (c + tf) 2 - 
 
 17. If a : b : : c : d, prove that ab -f cd is a mean proportional 
 between a 2 + c 2 and b 2 + d 2 - 
 
 18. What quantity subtracted from each of the quantities 
 a, 6, c will leave the remainders in continued proportion ? 
 
 19. If x be to y in the duplicate ratio of a to 6, and a be to b 
 in the sub-duplicate ratio of a + x to a — y, prove that 
 
 2 x : a : : x — y : y. 
 
 20. A farmer's crop of wheat was to his crop of oats as 2:3. 
 His neighbor raised 50 bushels more of each, and his crop of 
 wheat was to his crop of oats as 5:7. How many bushels of 
 each did the first farmer raise? 
 
 Sug. If in solving such problems proportions are used, place die product 
 of the extremes in each equal to the product of the means, and solve the 
 equations in the usual way. In this example but one unknown quantity is 
 necessary, as we may represent by 2 x and 3 x the number of bushels of wheat 
 and oats respectively. 
 
 21. Divide 14 into two such parts that the greater divided by 
 the less shall be to the less divided by the greater as 16 to 9. 
 
186 HIGH Eli ALGEBRA 
 
 22. Two vessels contain respectively 15 and 27$ gallons. How 
 many gallons must be transferred from one to the other that the 
 amounts may be in the ratio 2:3? 
 
 23. It is required to find a number such that the sum of its 
 digits is to the number itself as 4 to 13, and the difference of its 
 digits is to the number expressed when the digits are inter- 
 changed as 2 to 31. 
 
 24. Two numbers having the same two digits are to each other 
 as 5 : 6. What are the numbers ? 
 
 25. Find two numbers such that their sum, difference, and 
 product may be as the numbers «s, d, and p, respectively. 
 
 26. Find two numbers whose difference is to the difference of 
 their squares as m : n, and whose sum is to the difference of their 
 squares as a : b. 
 
 27. A bin contains chop-feed composed of corn and oats. A 
 second bin contains 6 bushels more of each, and has 7 bushels of 
 corn to every 6 bushels of oats. A third bin contains 6 bushels 
 less of each than the first, and has 6 bushels of corn to every 5 
 bushels of oats. How many bushels of each does the first bin 
 contain ? 
 
 28. The force of the earth's attraction is inversely as the square 
 of the distance from the center. At the surface this force is ex- 
 pressed by the number 32.16. By what is it expressed at the 
 moon, whose distance from the center of the earth is 60 radii of 
 the earth ? 
 
 29. The velocities of bodies revolving around another body are 
 inversely proportional to the squares of the distances. If the 
 velocity is v when the distance is r, what is it when the dis- 
 tance is r' ? 
 
 30. Two men have equal capital in business, and one is losing 
 as fast as the other is gaining. A third man's capital is less byp 
 dollars, but is rapidly increasing. Show that if m and n are the 
 respective gains of the third man at the times when he has the 
 same amounts as the others, then 2 mn =_p(m -f- ri). 
 
VARIATION 187 
 
 31. Before noon, a clock which is too fast and points to after- 
 noon time, is turned back 5 hours and 40 minutes to the true 
 time; and it is observed that the time before shown is to the 
 true time as 29 to 105. Find the true time. 
 
 32. Areas of circles are as the squares of their diameters. 
 Two circular metallic plates, each an inch thick, whose diameters 
 are 6 and 8 inches respectively, are melted and cast into a single 
 circular plate, 1 inch thick. Find its diameter. 
 
 33. The volumes of spheres are as the cubes of their radii. 
 Find the radius of a sphere whose volume is equal to the sum of 
 the volumes of three spheres whose radii are 3, 4, and 5, re- 
 spectively. 
 
 34. The rates of an express train and an accommodation train 
 differ by 15 miles an hour, and their times of making a distance 
 of 180 miles are as 9 : 14. The express train loses by stoppages 
 only half as much time as the accommodation train, and the 
 latter thus loses as much time as it would require in running 30 
 miles. Find the rates. 
 
 35. Two passengers have together 560 pounds of baggage, and 
 are charged for the excess above the weight allowed 62 cents and 
 $ 1.18 respectively. A third passenger has as much baggage as 
 the other two, and is charged $ 2.30 for excess. How much 
 baggage is each passenger allowed without charge ? 
 
 SECTION III — VARIATION 
 
 304. When the ratio of two quantities is constant, either is 
 said to vary directly as the other, or simply to vary as the other. 
 
 Thus, if 
 
 x = my, 
 
 in which m is constant, any change in y causes x to change in the same ratio : 
 if y be doubled, x will be doubled ; if y be tripled, x will be tripled ; if y be 
 halved, x will be halved ; and so on. 
 
 This relation between x and y is written 
 
 XQC?/, 
 
 and is read " x varies as y." 
 
188 HIGHER ALGEBRA 
 
 Distance traveled at a given rate varies as the time, the constant ratio 
 
 being the given rate. If d, r, and t represent distance, rate, and time 
 
 respectively, 
 
 d = rt, 
 
 and dcct. 
 
 Amount earned in a given time varies as the daily wages, the constant ratio 
 
 being the given time. If a, t, and ic represent amount, time, and daily wages 
 
 respectively, 
 
 a = tw, 
 
 and a x w. 
 
 The circumference of a circle varies as its radius, the constant ratio being 
 2 7r. If c and r represent the circumference and radius respectively, 
 
 c = 2 7rr, 
 and cccr. 
 
 305. When the ratio of a quantity to the reciprocal of another 
 is constant, either is said to vary inversely as the other. 
 
 ' ThUS ' if *=* 
 
 in which m is constant, any change in y causes x to change in the inverse ratio 
 (Art. 285) : if y be doubled, x will be halved ; if y be tripled, x will be made 
 one third as great ; if y be halved, x will be doubled ; and so on. 
 This relation between x and y is written 
 
 20C 1 , 
 
 y 
 
 and is read, " x varies inversely as ?/." 
 
 The time required to travel a given distance varies inversely as the rate, 
 the constant ratio being the given distance. If t, d, and r represent time, 
 distance, and rate respectively, 
 
 and t x - . 
 
 r 
 
 The time req lired to earn a given amount varies inversely as the daily 
 wages, the constant ratio being the given amount. If t, d, and w represent 
 time, amount, and daily wages respectively, 
 
 w 
 
 and U-, 
 
 w 
 
VARIATION 189 
 
 306. When the ratio of one quantity to the product of two or 
 
 more other quantities is constant, the first is said to vary jointly 
 
 as the others. 
 
 Thus, if 
 
 x = myz, 
 
 in which m is constant, any change in y and z causes x to change in a ratio 
 represented by the product of these changes ; if y be doubled and z tripled, 
 x will be made six times as great ; and so on. 
 This relation between x and yz is written 
 
 xctyz, 
 
 and is read, "x varies jointly as y and z." 
 
 Distance traveled varies jointly as the time and rate, the constant ratio 
 being 1. 
 
 The area of a triangle varies jointly as its base and altitude, the constant 
 ratio being \. If s, b, and h represent area, base, and altitude respectively, 
 
 s= \bh, 
 and s x bh. 
 
 307. When the ratio of one quantity to the quotient of a 
 second divided by a third is constant, the first is said to vary 
 directly as the second and inversely as the third. 
 
 Thus, if 
 
 V 
 x — m—, 
 
 z 
 
 in which m is constant, then 
 
 1 
 which is read, "x varies directly as y and inversely ass."' 
 
 If time, distance, and rate are all variable, the time varies directly as the 
 distance and inversely as the rate. 
 
 In doing work, the time varies directly as the amount of work and 
 inversely as the number of workmen employed. 
 
 308. Theorem. If one set of corresponding values of the vari- 
 ables of a variation be given, the constant ratio becomes known. 
 Dem. Let the variation be 
 
 xcKy. (1) 
 
 By definition (Art. 304) the ratio of x to y is constant. Let 
 this constant ratio be m. Then 
 
 x ^ my. (2) 
 
190 HIGHER ALGEBRA 
 
 Or we may say, since x varies as y, x is always a certain num- 
 ber of times y, as m times y. 
 
 Let a and b be corresponding values of x and y. Substituting 
 
 these in (2), we have 
 
 a = mb, 
 
 whence m = — 
 
 b 
 
 The same reasoning applies to the other forms of variation. 
 
 309. Cor. To pass from a variation to an equation, a constant 
 factor, the ratio, is introduced; and, conversely, to pass from an 
 equation to a variation, constant factors are omitted. 
 
 310. Caution. When two or more variations occur in the 
 same problem, the same constant must not be used twice in pass- 
 ing from variations to equations ; for, while in each variation the 
 ratio is constant, it will not do to assume that it is the same. 
 
 311. Theorem. A variation may always be expressed as a pro- 
 portion, and is, in fact, simply a contracted proportion. 
 
 Dem. By definition the expression 
 
 xocy 
 signifies that whatever value x may have, y has a corresponding 
 value, such that the ratio of x to y is constant. Letting x' and x" 
 be two values of x, and y' and y" the corresponding values of y, 
 the ratio x' : y' is the same as the ratio x" : y"; that is, 
 
 x' : y' : : x" : y", 
 or, if we choose (Art. 298, b), 
 
 x' :x"::y': y". 
 The same reasoning applies to the other forms of variation. 
 
 EXAMPLES LXXVHI 
 
 1. If x oc y and y cc -, show how x varies with reference to z. 
 
 z 
 
 Solution. Since 
 
 xccy, 
 
 we have (Art. 309) 
 
 x = my; 
 
 and since 
 
 ycc-> 
 z 
 
 we have 
 
 n 
 
 y = r 
 
 (1) 
 
 (2) 
 
VA It I ATI ON 191 
 
 Eliminating y from (1) and (2), 
 
 mn 
 
 whence (Art. 309) % «-, 
 
 2. If x oc and ?/ oc z, show how # varies with reference to ?/. 
 
 3. If x oc 2/ 2 and y oc z 2 , show how z varies with reference to z. 
 
 4. If x 2 x z and ?/ 2 oc z, prove that a# ac z. 
 
 5. If a* oc y and g = 14 when y = §, what is the value of g in 
 terms of y ? 
 
 Solution. Since xxj/, 
 
 we have (Art. 300) x = my. 
 
 Substituting the given corresponding values, 
 
 14 = % m, 
 whence m = 35. 
 
 Since m is constant, when we have found its value for one pair of corre- 
 sponding values of x and ?/, we have it for all corresponding values of x and 
 
 y. Hence 
 
 x = 35?/. 
 
 6. If a; oc y, and x = 20 when y = 2\, what is the value of x in 
 terms of y ? 
 
 7. If x oc -, and x = 13 when y = 4, what is the value of x in 
 
 y 
 terms of y ? 
 
 8. If a; oc — , and x = 8 when y = h what is the value of x in 
 
 y 2 
 terms of ?/ ? 
 
 9. If # oc y, and x = V2 when 2/ = 3, what is the value of y 
 when a; = 32 ? 
 
 1st Solution. Since xccy, (1) 
 
 we have (Art. 309) x = my. (2) 
 
 Substituting the first pair of values, 
 
 12 = 3 m, 
 whence . m = 4. 
 
 Substituting this and the second given value of x in (2), 
 
 32 = 4*/, 
 whence y = 8. 
 
192 HIGHER ALGEBRA 
 
 • 
 
 2i> Solution. Since £xy, 
 
 we have (Art. 311) . x' : y' : : a" : ?/". 
 
 Substituting the given values, 
 
 12:3:: 32 : y", 
 whence y" — 8. 
 
 Caution. In solving such examples, students sometimes sub- 
 stitute given values directly in the variation, thus : " 12 oc 3, etc." 
 As 12 and 3 cannot vary, it is absurd to write x between them. 
 
 10. If xccy, and x = 15 when y = 2±, what is the value of x 
 when y = 4 ? 
 
 11. If x x -, and x = 10 when ?/ = 3, what is the value of y 
 
 y 
 
 when x = 6 ? 
 
 12. If x- x — , and x = 4 when ?/ = 3, what is the value of x 
 
 y- 
 
 when y = 2 ? 
 
 13. If a? x ?/2, and # = 24 when y = 2 and z = 3, what is the 
 value of y when a: = 8 and z = 4 ? 
 
 14. If x x £, and a; = 12 when y = 8 and 2 = 2, what is the 
 
 2 
 
 value of 2 when x = 9 and y — 12 ? 
 
 15. If a? varies directly as y and inversely as the square of 2, 
 and a; = 6 when y = 8 and 2 = 2, find the value of x when 2/ = 18 
 and 2 = 3. 
 
 16. The volume of a sphere varies as the cube of its radius. If 
 the volume of a soap bubble is 5.888 when its radius is 1 inch, 
 what is its volume when its radius is 1\ inches ? 
 
 Solution. Since v x r 3 , 
 
 we have (Art. 309) v = mr z . 
 
 Substituting the given values of v and r, 
 
 m = 5.888. 
 Substituting this value of to and the second value of r, 
 v = 5.888 (|) 3 = 19.872. 
 
VARIATION 193 
 
 Or, if we choose, we may use a proposition, thus : 
 
 V : v' : : r 3 : r' 3 , 
 5.888 : v' : : 1 : ^, 
 whence V = 19.872. 
 
 17. If a metal ball whose radius is 2 inches weighs 6 pounds, 
 what is the weight of a ball of the same metal whose radius is 
 4 inches ? 
 
 la The distance fallen by a body from rest varies as the 
 square of the time of falling. If a body falls 257J feet in 4 sec- 
 onds, how far will it fall in 6 seconds ? 
 
 19. Amount of illumination varies directly as the intensity of 
 the light and inversely as the square of the distance from the 
 light. What must be the intensity of a light to give at the dis- 
 tance of 75 feet 3 times the illumination of one whose intensity 
 is 10 and distance 50 feet ? 
 
 20. The volume of a pyramid varies jointly as its base and 
 altitude. A pyramid whose base is 9 feet square and whose 
 height is 10 feet contains 10 cubic yards. What must be the 
 height of a pyramid with a base 3 feet square in order that it 
 may contain 2 cubic yards ? 
 
 21. The volume of a right circular cone varies jointly as its 
 height and the square of the radius of its base. If the volume of 
 a certain cone is 94 cubic inches, what is the volume of another 
 cone twice as high and the radius of whose base is half as great ? 
 
 22. The volume of a gas varies as the absolute temperature 
 and inversely as the pressure. When the temperature in a given 
 case is 260 and the pressure 15, the volume is 200 cubic inches. 
 What will be the volume when the temperature becomes 390 and 
 the pressure 18 ? 
 
 23. The pressure of the wind on a plane area varies jointly as 
 the area and the square of the velocity of the wind. If the 
 pressure on 1 square foot is 1 pound when the velocity of the 
 wind is 16 miles an hour, what is the velocity of the wind when 
 the pressure on 2 square yards is 50 pounds ? 
 
 24. If x -f y cc x — y, prove that x 2 + y 2 oc xy. 
 
 downey's alg. — 13 
 
194 HIGHER ALGEBRA 
 
 25. Given y =p + q, in which pccx, and q cc — When x = 1, 
 
 x 
 
 y = 6 ; and when x = 2, ?/ = 5. Find the value of y in terms of as. 
 
 26. Given # = a +^:> +</, in which a is constant, p&y, and 
 5 cc 2/ 2 . When g = 6, 17, 34, ?/ = 1, 2, 3, respectively. Find the 
 value of a? in terms of y. 
 
 27. Given that s oc t 2 when / is constant, and 8 oc/ when t is 
 constant; also 2s = / when t = 1. Find the value of s in terms 
 of / and t. 
 
 28. By Kepler's third law the square of a planet's period of 
 revolution around the sun varies as the cube of its mean distance 
 from the sun. If r is the distance of a planet whose period is t, 
 what is the distance of a planet whose period is t' ? 
 
 29. Attraction varies directly as the mass of the attracting 
 body and inversely as the square of the distance from its center. 
 The mass of the sun being 332,000 times that of the earth, and its 
 radius 109.4 times that of the earth, what is the weight of a 
 body at the surface of the sun as compared with its weight at the 
 surface of the earth ? 
 
 30. The mass of the sun being 1047.35 times that of Jupiter, 
 and the distance of Jupiter from the sun being 414 times the 
 distance from Jupiter to his outer satellite, what is the attraction 
 of the sun on Jupiter as compared with Jupiter's attraction on 
 his outer satellite ? 
 
CHAPTER XIV 
 
 PROGRESSIONS 
 
 SECTION I — ARITHMETICAL PROGRESSION 
 
 312. A Series is a succession of terms proceeding according to 
 a definite law. 
 
 313. A series is Increasing or Decreasing, or Ascending or De- 
 scending, according as the terms increase or decrease. 
 
 314. An Arithmetical Progression is a series in which each term 
 is greater or less than the preceding term by a constant quantity, 
 called the Common Difference. 
 
 If we regard each term as being obtained by the addition of 
 the common difference to the preceding term, the common differ- 
 ence is plus in an increasing, and minus in a decreasing, arith- 
 metical progression. 
 
 Thus, 1, 3, 5, 7, etc., is an increasing arithmetical progression, in which 
 the common difference is 2. 
 
 15, 10, 5, 0, — 5, — 10, etc., is a decreasing arithmetical progression, in 
 which the common difference is — 5. 
 
 a, a + d, a + 2d, a + 3d, etc., is the general form of an arith- 
 metical progression, the common difference being d, which may 
 be either plus or minus. 
 
 315. In the treatment of arithmetical progression, five ele- 
 ments are involved : 
 
 1. The first term, a. 
 
 2. The last term, I. 
 
 3. The common difference, d. 
 
 4. The number of terms, n. 
 
 5. The sum of the terms, S. 
 
 195 
 
196 HIGHER ALGEBRA 
 
 316. Arithmetical Means between any two quantities are the 
 terms that lie between them in an arithmetical progression. 
 
 When there is but one intermediate term, it is called the Arith- 
 metical Mean of (or between) the other two. 
 
 317. Theorem. TJie arithmetical mean of two quantities is equal 
 to half their sum. 
 
 Dem. Let a, b, c be in arithmetical progression. 
 
 By definition b — a = c — b, 
 
 whence b = ^ (a -f c). 
 
 318. Theorem. TJie formula for the last, or nth, term of an 
 arithmetical progression in terms of the first term, the common dif- 
 ference, and the number of terms, is 
 
 I = a-\- (n — 1) d. 
 
 Dem. In the general form of an arithmetical progression, 
 
 a, a -f- d, a -{-2d, a -{-3d, a -f 4 d, etc., 
 
 it is seen that each term is formed by adding to a the product of 
 d and a number which is 1 less than the number of the term; 
 that is, 
 
 I = a -\- (n — l)d. 
 
 319. Theorem. The formula for the sum of an arithmetical pro- 
 gression in terms of the first term, the last term, and the number of 
 terms, is 
 
 S = l 
 
 a + Z\ 
 
 Dem. Taking the sum of the terms of the general form of an 
 arithmetical progression and the sum of the same terms in the 
 reverse order, we have 
 
 # = a +0 + d) + (« + 2ff)+ --(l-2d) + (l-d) + l, (1) 
 
 and S = I + (I - d) + (I - 2 d) + • • . (a + 2 d) + (a + d) + a. (2) 
 
 Adding (1) and (2), 
 
 2 S = (a + 1) + (a + 1) + (a + 1) + ■ • • (a + + (a + 1) + (a + 1). 
 
ARITHMETICAL PROGRESSION 197 
 
 It is seen that (a 4-/) is taken as many times as there are 
 terms. Hence 
 
 2 S = (a + n, 
 
 or S 
 
 m 
 
 320. Any one of the four quantities involved in either of the 
 
 equations, 
 
 I = a -f- (w — 1) d, 
 
 £ 
 
 (t> 
 
 may be found in terms of the other three ; and by combining the 
 two equations, any one of the five quantities involved may be 
 eliminated, and any one of the remaining four found in terms of 
 the other three. The twenty formulae on page 198 result. They 
 are convenient, but not necessary, as all cases may be solved by the 
 two fundamental formulae, either directly or by first finding an 
 intermediate element. The student should reserve, until after he 
 has taken the subject of Quadratic Equations, the development of 
 numbers 2, 11, 18, and 20. 
 
 EXAMPLES LXXIX 
 
 1. Find the 20th term and the sum of 20 terms of 1, 4, 7, 
 10, etc. 
 
 2. Find the 21st term and the sum of 21 terms of 3, 7, 11, 
 15, etc. 
 
 3. Find the 36th term and the sum of 36 terms of 12, 10, 8, 
 6, etc. 
 
 4. Find the 10th term and the sum of 10 terms of 3, 2. \, 
 If, etc. 
 
 In each of the following, find the two elements not given : 
 5. a = 1, d = 5, n = 24. 6. 1 = 71, d = 5, n = 15. 
 
 7. 0=13, 1 = 73, rc = ll. 8. o = 10, 1 = 87, d = 7. 
 
 9. d = 4, n = U, s = 812. 10. a= -5, ?i = 19, s=-950. 
 
 ll.a = 7, 1 = 143, s = 1350. 12. l=-§, n = 19, 8 = 0. 
 
198 HIGHER ALGEBRA 
 
 Formula in Arithmetical Progression 
 
 Number 
 
 Given 
 
 Required 
 
 Formulae 
 
 4. 
 
 a, d, n 
 
 a, d, S 
 
 a, 7i, S 
 
 d, n, S 
 
 a, d, I 
 
 a, n, I 
 d, 7i, I 
 
 l=za + (n — l)cl, 
 
 I = - ^d ±V \2 dS +(a - \df\ 
 
 2S 
 
 a, 
 
 1 = 
 
 S (n-l)d 
 n 2 
 
 S 
 
 \n\2a+(n-l)d\, 
 
 2 2d ' 
 
 (* + «)§ 
 \n\2l-(n-l)d\. 
 
 9. 
 10. 
 
 11. 
 
 12. 
 
 13. 
 14. 
 15. 
 16. 
 
 d, 7i, I 
 
 d, n, S 
 
 d, I, S 
 
 n, I, .S 
 
 a, 71, I 
 
 a, n, S 
 
 a, /, S 
 
 n, I, S 
 
 l-(7i-l)d, 
 
 S (n - l)d 
 n 2 ' 
 
 = i d ± V(7 + i df - 2 dS, 
 
 I — a 
 
 w (w — 1) 
 I 2 -a 2 
 2S-l-a 
 2(nl-S) 
 
 7% (n — 1) 
 
 17. 
 18. 
 19. 
 20. 
 
 a, d, I 
 
 a, d, S 
 a, I, S 
 d, I, 8 
 
 d 
 
 + 1, 
 
 ± V(2 a-ri)'+8 dS-2 q+rf 
 2d 
 
 = 2S ; 
 
 = 2 Z + ^ ± V(2 I + d) 2 -^WdS 
 2d 
 
ARITHMETICAL PROGRESSION 199 
 
 13. Insert 3 arithmetic means between 73 and 193. 
 Sue First find <?, n being 5. 
 
 14. Insert 11 arithmetic means between — 49 and 59. 
 
 15. Find the first term of an arithmetical progression whose 
 59th term is — 2\, and 60th term — If. 
 
 16. Find the first term of an arithmetical progression whose 
 2d term is \, and 55th term 5.8. 
 
 17. Find the sum of 24 terms of an arithmetical progression 
 whose 13th term is 25, and 19th term 37. 
 
 18. Find the sum of 12 terms of the arithmetical progression 
 whose 1st term is 38, and 4th term 86. 
 
 19. Between what two terms of 3, 8, 13, etc., does 391 lie ? 
 
 20. If a body falling from rest descends a feet the first second, 
 3 a the second, 5 a the third, and so on, how far will it fall during 
 the tth second, and how far in t seconds ? 
 
 21. If a body falling from rest descends 16^ feet the first 
 second, 48 J; feet the second, 80 T % feet the third, and so on, how 
 far will it fall in 30 seconds ? 
 
 22. A man training for a mile race ran over the course every 
 day for 21 days. His time the first day was 8 minutes^ but this 
 was gradually diminished, his whole time for the 21 runs being 
 133 minutes. What was the average daily diminution of time, 
 and what was his time the last day ? 
 
 23. Some fence posts are to be carried from a pile to the holes 
 where they are to be set. How far must a laborer walk, carrying 
 one post at a time, to place them in the 50 post holes, which are 
 in a straight line and 8 feet apart, the first one being at the pile ? 
 
 24. A man bought an estate which yielded $ 1500 profit the 
 first year. His personal expenses for the first year were $ 1250. 
 His income from the estate increased $ 100 yearly, and his 
 personal expenses increased $ 125 yearly. After how many 
 years were his personal expenses equal to the income from his 
 estate ? 
 
200 HIGHER ALGEBRA 
 
 SECTION II. — GEOMETRICAL PROGRESSION 
 
 321. A Geometrical Progression is a series in which each term 
 is eqnal to the preceding term multiplied by a constant quantity, 
 called the Ratio. 
 
 The ratio is greater than unity in an increasing and less than 
 unity in a decreasing geometrical progression. 
 
 Thus, 2, 4, 8, 10, etc., is an increasing geometrical progression in which 
 the ratio is 2. 
 
 12, 4, |, |, 5 4 7 , etc., is a decreasing geometrical progression in which the 
 ratio is ^. 
 
 a, ar, ar 2 , ar 3 , ar A , etc., is the general form of a geometrical 
 progression, the ratio being r. 
 
 322. In the treatment of geometrical progression, five elements 
 are involved: 
 
 1. The first term, a. 
 
 2. The last term, I. 
 
 3. The ratio, r. 
 
 4. The number of terms, n. 
 
 5. The sum of the terms, S. 
 
 323. Geometric Means between any two quantities are the terms 
 that lie between them in a geometrical progression. 
 
 When there is but one intermediate term, it is called the 
 Geometric Mean of (or between) the other two. 
 
 324. Theorem. The geometric mean of two quantities is equal to 
 the square root of their product. 
 
 Dem. Let a, b, c, be in geometric progression. 
 
 By definition - = -, 
 
 a b 
 
 whence b = V«c. 
 
 325. Sch. The geometric mean of two quantities is the same 
 as the mean proportional between them (Arts. 287 and 295). 
 
GEOMETRICAL PROGRESSION 201 
 
 326. Theorem. The formula for the last, or nth, term of a 
 geometrical progression in terms of the first term, the ratio, and the 
 number of terms, is 
 
 l = ar n ~\ 
 
 Dem. In the general form of a geometrical progression, 
 
 a, ar, ar 2 , ar 3 , ar*, etc., 
 
 it is seen that each term is formed by multiplying a by r affected 
 with an exponent 1 less than the number of the term ; that is, 
 
 I = ar n ~ l . 
 
 327. Theorem. The formula for the sum of a geometrical pro- 
 gression in terms of the first term, the ratio, and the number of 
 terms, is 
 
 ar n — a 
 
 S = 
 
 r-1 
 
 Dem. Taking the sum of the terms of the general form of a 
 geometrical progression and multiplying this series by r, we 
 have 
 
 S = a + ar + ar 2 + ••• ar n ~ 3 + ar n ~ 2 + ar n ~ l , (1) 
 
 and rS = ar -f ar + *..«**-* 4. ar n - 2 + ar n ~ 1 + ar n . (2) 
 
 Subtracting (1) from (2), 
 
 rS — S = ar n — a, 
 
 whence ^ = ar "~ a . 
 
 r — 1 
 
 328. Theorem. The formula for the sum of a geometrical pro- 
 gression in terms of the first term, the last term, and the ratio, is 
 
 a Ir — a 
 
 Dem. From l = ar n ~ 1 
 
 we have lr = ai M . 
 
 Substituting this value of ar n in the formula for S, Art. 327, 
 c Ir — a 
 
202 HIGHER ALGEBRA 
 
 Formula in Geometrical Progression 
 
 Number | Given 
 
 a, r, n 
 
 a, r, jS 
 
 a, n, S 
 
 r, n, S 
 
 Required 
 
 FOHMULuE 
 
 I = at*" 1 , 
 
 r 
 l(S - l) n ~ l -a(S~ ay- 1 = 0, 
 l== (r- l)^"- 1 
 r n — 1 
 
 a. r, n 
 a, r, I 
 a, n, I 
 
 S 
 
 s = 
 s = 
 s = 
 s = 
 
 r — 1 
 rl — a 
 
 r-1 
 
 n —l /IT, n ~ 1 /~^i 
 
 v < — v a 
 
 n— 1/7 B— 1/~~ 
 
 v t — v a 
 
 10. 
 
 11. 
 12. 
 
 13. 
 14. 
 15. 
 16. 
 
 17. 
 18. 
 19. 
 20. 
 
 r, n, I 
 
 r, n, S 
 
 r, I, S 
 
 n. I, S 
 
 a, n, I 
 
 a, n, S 
 
 a, I, S 
 
 n, I, S 
 
 a, r, S 
 
 a, I, S 
 r, h S 
 
 a = — ? 
 
 r n — l 
 
 a = rl-(r-l)S, 
 
 a(S-a) n 1 -l(S-l) n - 1 = 0. 
 
 
 r n r H = 0, 
 
 a a 
 
 r = 
 
 r «_ 
 
 
 £-/ 
 
 -f 
 
 ^ — z 
 
 0. 
 
 log? -log a -j 
 logr 
 _ log [a + (r - 1)3] - log a 
 log r 
 log Z — log a 
 
 \og(S-a)-\og(S-l) 
 
 + 1, 
 
 _ logZ-log[/r-(r-l)£] | L 
 
 logr 
 
GEOMETRICAL PROGRESSION 203 
 
 329. Theorem. The formula for the sum of an infinite, de- 
 creasing geometrical x>rogression is 
 
 s= a 
 
 1-r 
 
 Of" 
 
 — a 
 
 r - 
 
 -1 ' 
 
 Ir- 
 
 -a 
 
 Dem. Since in a decreasing geometrical progression r is less 
 than unity, the term ar n , in the formula 
 
 c ar n — a 
 
 *> = 7~> 
 
 r— 1 
 
 has no appreciable value when n is infinite. Hence this formula 
 becomes 
 
 o_ — a _ a 
 ~r-l~r~r' 
 
 330. Any one of the four quantities involved in any one of the 
 equations, 
 
 I = ar n " l , 
 
 JS = 
 
 s = 
 
 r — 1 
 
 may be found in terms of the other three ; and by combining the 
 first with one of the others, any one of the five quantities in- 
 volved may be eliminated, and any one of the remaining four 
 found in terms of the other three, except in a few cases involving 
 higher literal equations. These last can be solved when numeri- 
 cal quantities are substituted for the literal. The twenty formulae 
 on page 202 result. The student should reserve until after taking 
 the subject of Logarithms the development of the last four. 
 
 EXAMPLES LXXX 
 
 1. Prove the following properties of a geometrical progression : 
 
 (a) The alternate terms, or any terms separated by the same 
 intervals, are in geometrical progression. 
 
 (b) The products of the terms by the same quantity are in 
 geometrical progression. 
 
204 HIGHER ALGEBRA 
 
 (c) The same powers of the terms are in geometrical pro- 
 gression. 
 
 (d) The reciprocals of the terms are in geometrical progression. 
 
 (e) Any term is a mean proportional between any two terms 
 separated from it by the same intervals. 
 
 (/) The product of any odd number, p, of consecutive terms 
 is equal to the pth power of the middle one. 
 
 (g) The product of any two terms is equal to the product of 
 any other two terms separated from these by the same intervals 
 in opposite directions. 
 
 2. Find the 12th term and the sum of 12 terms of 1, 2, 4, 8, etc. 
 
 3. Find the 10th term and the sum of 10 terms of 1, 3, 9, 27, etc. 
 
 4. Find the 11th term and the sum of 11 terms of ^, Jg, J, 1, 
 etc. 
 
 5. Find the 11th term and the sum of 11 terms of 3, — 6, 12, 
 - 24, etc. 
 
 In each of the following find the two elements not given : 
 
 6. r = 2, w = 8, £ = 1275. 7. r = 3, n = 9, I = 26,244. 
 8. a = - |, n = 7, r = - J. 9. I = 256, n = 9, r = 2. 
 
 10. a = -2,?i = 6, Z = 2048. 11. a = 2, n = 7, I = 145. 
 12. a = 1, I = 81, r = 3. 13. Z = 160, r = 2, S = 315. 
 
 14. Insert 3 geometric means between 17 and 4352. 
 Sug. First find ?% n being 5. 
 
 15. Insert 4 geometric means between 26 and 6318. 
 
 16. Find the 1st term of a geometrical progression whose 5th 
 term is 336, and 9th term 5376. 
 
 17. Find the sum of 8 terms of the geometrical progression 
 whose 4th term is 108, and 7th term 2916. 
 
 18. Find the 11th term of a geometrical progression whose 7th 
 term is 192, and 10th term - 1536. 
 
 19. The sum of the first 8 terms of a geometrical progression 
 is 17 times the sum of the first 4 terms. Find the ratio. 
 
HARMONIC PROGRESSION 205 
 
 20. The 1st term of a geometrical progression is 3, and the 
 sum of the first 3 terms is one eighth of the sum of the next 
 3 terms. Find the ratio. 
 
 21. The sum of the 1st and 2d terms of a geometrical progres- 
 sion is 30, and the sum of the 4th and 5th is 1920. Find the first 
 5 terms of the progression. 
 
 22. Find the sum of 1, £, \, |, etc., to infinity. 
 
 23. Find the value of 1 -f- \ + i -f ^ 7 + etc., to infinity. 
 
 24. Find the value ofl — ^ + i — 2V + etc., to infinity. 
 
 25. Find the value of .423. 
 
 Sug. .423 = .4232323 . . . =.4 + .023 + .00023 + etc. 
 
 26. Find the value of .27. 
 
 27. Find the value of .3i2. 
 
 28. Each term in a certain infinite decreasing geometrical pro- 
 gression is equal to the sum of all that follow it. Find the ratio. 
 
 29. What is the distance passed through before coming to rest 
 by a ball which falls from a height of 50 feet and at every fall 
 rebounds half the distance ? 
 
 30. A "letter chain" is started for the benefit of a public 
 charity, three letters, each numbered 1, being sent out by the 
 starter with the request that each of the recipients remit 10 cents 
 and send out three other letters, each numbered 2, with a similar 
 request, and so on, until the numbers reach 25. Should all com- 
 ply, («) How much would be realized for the charity ? (b) What 
 would be the entire outlay for postage at 2 cents for each letter ? 
 (c) With a uniform distribution, how many times would each of 
 the 75 million inhabitants of the United States respond ? 
 
 SECTION III. — HARMONIC PROGRESSION 
 
 331. A Harmonic Progression is a series of terms whose recipro- 
 cals are in arithmetical progression. 
 
 Thus, 1, \, f, ^, ^5, etc., are in harmonic progression, because their 
 reciprocals, 1, 4, 7, 10, 13, etc., are in arithmetical progression. 
 
206 HIGHER ALGEBRA 
 
 The general form of a harmonic progression is -, , — , 
 
 -^ a a + d a + 2d 
 
 etc., in which a is the first term of an arithmetical pro- 
 
 a + 3d 
 
 gression, and d the common difference. 
 
 Most problems concerning quantities in harmonic progression 
 are solved by treating the arithmetical progression obtained by 
 taking the reciprocals of the quantities that are in harmonic 
 progression. 
 
 332. If the lengths of strings of the same substance, size, and 
 tension be proportional to the terms of a harmonic progression, 
 any two of these strings vibrating together produce harmony of 
 sound ; hence the term harmonic. 
 
 333. Harmonic Means between any two quantities are the terms 
 that lie between those quantities in a harmonic progression. 
 
 When there is but one intermediate term, it is called the Har- 
 monic Mean of (or between) the other two. 
 
 334. Theorem. The harmonic mean of two quantities is twice 
 their product divided by their sum. 
 
 Dem. If a, b, c are in harmonic progression, -, -, - are, by 
 
 a b c 
 definition, in arithmetical progression ; hence 
 
 11^11 
 
 b a c b 
 
 whence b = — '■ — 
 
 a + c 
 
 335. Theorem. If three quantities are in harmonic progression, 
 the difference between the first and second is to the difference between 
 the second and third as the first is to the third. 
 
 Dem. If a, b, c are in harmonic progression, -, -> - are, by 
 
 a b c 
 definition, in arithmetical progression ; hence 
 
 1_1 = 1_1 
 
 b a c b 
 Clearing of fractions, ac — bc = ab — ac, 
 
 or c(a — b) = a(b — c) ; 
 
 therefore (Art. 297) a — b : b — c : : a : c. 
 
HARMONIC PROGRESSION 207 
 
 336. Theorem. The formula for the last, or nth, term of a har- 
 monic progression in terms of the first term, the second term, and the 
 number of terms, is 
 
 ab 
 
 I 
 
 b + (n—l)(a — b) 
 
 Dem. If a, b, etc., are in harmonic progression, -, -, etc., are, 
 
 d b 
 by definition, in arithmetical progression ; hence 
 
 b a ab 
 Substituting in V = a' + (» - l)d (Art. 318), 
 
 1-1 , T (n !)<* - o - o +(* - i)(o - b) 
 
 la ab ab 
 
 whence I = 
 
 b + {n - 1) (a - b) 
 
 337. Note. No formula for the sum of a harmonic progression 
 is known. 
 
 EXAMPLES LXXXI 
 
 1. Find the 23d term of J, \, y 1 ^, y 1 ^, etc. 
 
 2. Find the 4th and -7th terms of a harmonic progression 
 whose 2d term is \, and 5th term y 1 ^. 
 
 3. Find the harmonic mean of 21 and 42. 
 
 4. Insert 3 harmonic means between 14 and 42. 
 
 Sue Find 3 ' arithmetic means between T a ¥ and fa an d take their 
 reciprocals. 
 
 5. Insert 3 harmonic means between -J^ and fa. 
 
 6. If a, b, c, d are in harmonic progression, show that 
 
 ab : cd : : a — b : c — d. 
 
 7. Show that the geometric mean of two numbers is also the 
 geometric mean of their arithmetic and harmonic means. 
 
CHAPTER XV 
 QUADRATIC EQUATIONS 
 
 338. Quadratic Equations (Art. 238) are distinguished as Pure 
 (called also Incomplete) and Affected (called also Complete). 
 
 339. A Pure Quadratic Equation is an equation which contains 
 no power of the unknown quantity but the second. 
 
 Thus, ax 2 + b = cd and 3 x 2 = 108 are pure quadratic equations, or pure 
 quadratics. 
 
 340. An Affected Quadratic Equation is an equation which con- 
 tains both the first and the second powers of the unknown 
 quantity. 
 
 Thus, x 2 — 4 x = 12, 2 x 2 + 7 x - 18 = 0,* ax 2 + bx = c, are affected quad- 
 ratic equations, or affected quadratics. 
 
 341. A Root of an Equation is a quantity which, substituted for 
 the unknown quantity, satisfies the equation. 
 
 PURE QUADRATICS 
 
 342. Prob. To solve a pure quadratic equation. 
 
 Rule. By clearing of fractions, transposing, uniting terms, and 
 dividing by the coefficient of the square of the unknown quantity, 
 reduce the equation to the form x 1 = m ; then extract the square root 
 of both members, giving the double sign to the second member of the 
 residt. 
 
 Dem. All of the operations of reducing to the form x 2 = m, 
 and the extraction of the square root as well, affect both mem- 
 bers alike, and, consequently (Art. 243), do not destroy the 
 equality of the members. As these operations leave in the first 
 member simply the unknown quantity, the equation is solved. 
 
 208 
 
QU ABU AT [C EQUATIONS 209 
 
 That the result should have the double sign is evident from 
 the fact that a quantity has two square roots numerically equal, 
 but with opposite signs. 
 
 343. Cor. The roots of a pure quadratic equation are both rational 
 or both surd, both real or both imaginary. 
 
 EXAMPLES LXXXn 
 Solve the following : 
 
 1. 11 a,- 2 -44 = 5^ + 10. 2. 5ar 2 -9 = 2a,- 2 -f-G6. 
 
 5 8 5 
 
 3. + 2) 2 = 4z + 5. 
 
 4 — x 3 4 + x 
 
 5. _i-- + —5—-X25. 6. 2(x+3)(x-3) = (x+iy-2x. 
 
 J. — L X L -j- — X 
 
 7. x 2 — ax -j- b = ax (x — 1). 8. xy/a? -+- (J = a; 2 4- 1. 
 
 rt « . V« 2 — a,* 2 ar _• . / 2 , n 2 a 2 
 
 9. - + — = — 10. x H- Va J 4- a; 2 
 
 x x b Va 2 + a^ 
 
 11 2 + 2 =a? . 12. ^ + l+VaV ZL l =; to 
 
 PROBLEMS LEADING TO PURE QUADRATIC EQUATIONS 
 EXAMPLES LXXXHI 
 
 1. Find two numbers in the ratio of 2 to 5, the sum of whose 
 squares is 261. 
 
 2. Find three numbers which shall be to one another as m, n, 
 and p, and the sum of whose squares shall be $. 
 
 3. Divide 21 into two such parts that the square of the less 
 shall be to that of the greater as 4 to 25. 
 
 4. The sides of two square rooms are in the ratio of 2 to 3,. 
 and the larger room requires 20 square yards more of carpet than 
 the smaller. Find a side of each. 
 
 5. Two square plats of ground contain 272 square rods, and 
 a side of the larger is as much greater than 10 rods as a side of 
 the other is less than 10 rods. Find a side of each. 
 
 Downey's alg. — 14 
 
210 HIGHER ALGEBRA 
 
 6. A rectangular field whose length is li times its breadth 
 contains 9 acres. Find, the length, in rods, of each side. 
 
 7. An army was formed, with 5 more men in file than in rank ; 
 but when the form was changed so that there were 845 more men 
 in rank than before, there were but 5 ranks. Find the number 
 of men in the army. 
 
 8. A boat's crew can row in still water at the rate of 9 miles 
 an hour. If it take the crew 2\ hours to row 9 miles up a river 
 and back to the starting point, what is the rate of the current ? 
 
 9. The distances through which a body falls being as the 
 squares of the times, and the distance fallen during the first 
 second being lGy 1 ^ feet, in what time will a body fall 500 feet ? 
 In what time will it fall a mile ? 
 
 10. The mass of the earth is 332,000 times that of the earth, 
 and the distance between the two bodies is 93,000,000 miles. 
 How far from the earth, between the earth and the sun, is the 
 point of equal attraction, the law of attraction being that it 
 varies directly as the mass and inversely as the square of the 
 distance ? 
 
 Sue In this and some of the following problems, to avoid an affected 
 quadratic equation, take the square root before performing indicated opera- 
 tions. 
 
 11. The intensities of two lights are as 7 : 17, and their dis- 
 tance apart 132 feet. Where in the line of the lights are the 
 points of equal illumination, assuming that the amounts of 
 illumination are to each other directly as the intensities and 
 inversely as the squares of the distances ? 
 
 12. The loudness of one church bell is three times that of 
 another. If the amount of sound varies directly as the loudness 
 and inversely as the square of the distance, where on the line of 
 the two will the bells be equally well heard,. the distance between 
 them being a ? 
 
 13. A girl worked two square pieces of worsted work of the 
 same kind, the edges of one being an inch longer than those of 
 the other ; one took 12^ skeins and the other 18 skeins. How 
 long were the edges of the smaller piece ? 
 
QUADRATIC EQUATIONS 211 
 
 14. From two towns m miles apart, two persons, A and B, 
 started at the same time and traveled toward each other. When 
 they met, A, the faster traveler, had gone n miles, the time on 
 the road being equal to the difference of their rates. Find their 
 rates. 
 
 15. A and B are two stations 300 miles apart. Two trains 
 start simultaneously from A and B, each to the opposite station. 
 The train from A reaches B 9 hours, and the train from B reaches 
 A 1 hours, after they meet. Find the rate of each train. 
 
 16. Two travelers, A and B, started at the same time from 
 two different places, C and D respectively, and traveled toward 
 each other. When they met, it appeared that A had gone 30 miles 
 more than B ; also that A could reach D in 4 days, and B could 
 reach C in 9 days. Find the distance from C to D. 
 
 17. Two bicyclists start at the intersection of two roads at 
 right angles to each other and ride, one on each road, at rates of 
 12 and 16 miles an hour respectively. In how many minutes 
 will they be 8 miles apart ? 
 
 AFFECTED QUADRATICS 
 344. Prob. To solve an affected quadratic equation. 
 Rule. Reduce to the form x 2 +px = q (in ivhich p and q may be 
 positive or negative, integral or fractional) ; then the roots are half 
 of the coefficient of the second term taken with the opposite sign, ± 
 the square root of the sum of the square of this half coefficient and 
 the absolute term. 
 
 Dem. All of the operations of reducing to the form 
 
 x* +px = q, 
 
 viz., clearing of fractions, transposing, uniting terms, and divid- 
 ing by the coefficient of the second power of the unknown quantity, 
 affect both members alike, and, consequently (Art. 243), do not 
 destroy the equality of the members. 
 
 P 2 
 Now if -j be added to both members of the equation, giving 
 
212 HIGHER ALGEBRA 
 
 the equality will not be destroyed and the first member will be a 
 perfect square, since the middle term is twice the product of the 
 square roots of the other two (Arts. 61 and 62). This operation is 
 called completing the square. Extracting the square root of both 
 members, which does not destroy the equality, we have 
 
 I=±v? 
 
 tflF+ffi 
 
 or 
 
 which corresponds with the statement in the rule. 
 
 345. Cor. i. If the first term of the roots is numerically greater 
 than the radical term, both roots have the sign of the first term; if 
 numerically less, one root is plus and one minus. If q is negative 
 
 and numerically greater than £-, both roots are imaginary. If q is 
 
 p 2 
 negative and numerically equal to -j, the two roots are equal. 
 
 346. Cor. 2. The sum of the two roots is — p and the product is 
 
 -q- 
 
 EXAMPLES LXXXIV 
 Solve the following : 
 
 1. Sx 2 -29 = 18 a? -8. 
 
 Solution. Transposing, uniting terms, and dividing by the coefficient of 
 x 2 , this becomes 
 
 x*-Qx = 7. 
 
 The student should not complete the square — that has been done once for 
 all in the demonstration of the rule — but should write the roots at once from 
 the formula obtained by solving the general case. The roots being half of the 
 coefficient of the second term taken with the opposite sign, ± the square root 
 of the sum of the square of this half coefficient and the absolute term, we 
 have from 
 
 x 2 - 6 x = 7, 
 
 ac = 8±4 = 7or-l. 
 
 The full form is 
 
 x = 3 ± VPT^ =3±4 = 7or-l; 
 
QUADRATIC EQUATIONS 213 
 
 but when the numbers are small, the operations are readily performed 
 mentally. 
 
 2. 15-5^=12^. 
 
 Solution. Reduced to the required form, this becomes 
 
 
 
 x 2 + 5 x = 0, 
 
 henc 
 
 e 
 
 » = -f ±1 = 1 or -6. 
 
 3. 
 
 x 2 — 4 x = 5. 
 
 4. a; 2 + 6 a? = - 5. 5. x 2 - 10 or = 11. 
 
 6. 
 
 x 2 + 8 x = 9. 
 
 7. >— 12 * =—11. 8. x 2 - 2 x = 48. 
 
 9. 
 
 x 2 — 6 x = 16. 
 
 10. 3 aV- + 36 = 24 x. 11. a; 2 — 4 x = 60. 
 
 12. 
 
 p* _ 7 a = 8. 
 
 13. ar 2 - 30 x = 64. 14. ar - 12 a: = 28. 
 
 15. 
 
 x 2 - 12 a = 45. 
 
 16. x 2 - 8 x = 33. 17. 6 a,- 2 - 21 a; = 12. 
 
 
 18 * -§ 
 
 a? — 1 g a¥ 2 aa; 6 2 ^ 
 
 
 x-1 2 
 
 
 o * + 2 - 
 
 8 -1 x 2 \ a2 ~ b2 x-l 
 
 
 5 3 
 
 a; -f- 4 «6 
 
 
 /7 2 
 
 22 ra* 2 -4- — 
 
 * a _o ax o 3 **»-«&_» 
 
 
 c 
 
 _za*. 23. 3ft _ 2a; _ 4 
 
 PROBLEMS LEADING TO AFFECTED QUADRATIC EQUATIONS 
 
 347. When a problem gives rise to a quadratic equation, the 
 student should note whether both values obtained by the solution 
 of the equation are admissible and capable of interpretation. A 
 problem often contains restrictions, expressed or implied, which 
 cannot be incorporated in the equation, and hence the equation 
 may contain a root that will not conform to this restriction. For 
 example, in a problem involving as the unknown quantity the 
 digits of a number, sheep, cattle, men, etc., any fractional root 
 must be rejected. In some problems negative roots must be 
 rejected. Imaginary roots indicate incompatibility of conditions 
 in the problem (Art. 227). 
 
 EXAMPLES LXXXV 
 1. Jn a number consisting of two digits the tens' digit is 1 less 
 than the square of the units' digit, and when 45 is subtracted from 
 the number, the digits change places. Find the number. 
 
214 HIGHER ALGEBRA 
 
 Solution. Let x = the units' digit ; 
 
 then (x 2 — 1) = the tens' digit, 
 
 and we have from the conditions, 
 
 10(x 2 - 1) + x - 45 = 10 x + x 2 - 1, 
 
 or x 2 — x = 6, 
 
 whence x = J ± f = 8 or — 2. 
 
 The root — 2 must be rejected, as the digits of a number must be positive. 
 Hence the units' digit is 3, the tens' digit is 3 2 — 1 = 8, and the number is 83. 
 
 2. Two boys are coaching for an examination in Greek. One of 
 them reads a page the first day, and each succeeding day reads 
 one more page than the day before. The other, beginning 5 days 
 later, reads 12 pages a day. In how many days will they have 
 read the same number of pages ? 
 
 Solution. Let x = the first boy's time ; 
 then " x — 5 = the second boy's time, 
 
 and (x — 5)12 = the number of .pages read by the second boy. 
 
 The number of pages read by the first boy is the sum of an arithmetical 
 progression in which the first term is 1, the number of terms x, and the last 
 term x; i.e. (Art. 319), 
 
 (^)»=( L f> 
 
 Hence 
 
 or 
 
 and 
 
 Both roots of the equation are admissible. The second boy overtakes the 
 first in 8 days and is overtaken by him in 15 days (from the beginning), since 
 the first is reading at an increasing rate. 
 
 3. Divide 48 into two such parts that their product may be 432. 
 
 4. Divide 24 into two such parts that their product may be 35 
 times their difference. 
 
 5. For a journey of 108 miles, G hours less would have sufficed, 
 had the traveler gone 3 miles an hour faster. At what rate did 
 he travel ? 
 
 c 
 
 **)■- 
 
 (x - 5) 12, 
 
 x- 
 
 ! - 23 x = 
 
 : - 120, 
 
 X = 
 
 . 23 7 _ 
 " 2 2 
 
 : 15 or 8. 
 
QUADRATIC EQUATIONS 215 
 
 6. The length of a rectangle is 10 feet more than the breadth, 
 and the area is 600 square feet. Find the length and breadth of 
 the rectangle. 
 
 7. If a bar of iron weighing 60 pounds be drawn out 3 feet 
 longer, it will weigh 1 pound less per linear foot. Find its length. 
 
 8. A man bought two farms for $ 2800 each. The larger con- 
 tained 10 acres more than the smaller, but he paid $ 5 more per 
 acre for the smaller than for the larger. How many acres did 
 each contain ? 
 
 9. A man paid $ 300 for a drove of sheep. By selling all but 
 10 of them at a profit of $ 2.50 each, he received the amount he 
 paid for all the sheep. How many sheep did he buy ? 
 
 10. It takes a boat's crew 4 hours and 12 minutes to row 12 miles 
 down a river and back. If the rate of the current is 3 miles an 
 hour, at what rate can the crew row in still water ? 
 
 11. Two steamers ply between the same two ports a distance 
 of 420 miles. One goes £ mile per hour faster than the other, 
 and is 2 hours less on the voyage. At what rates do they go ? 
 
 12. The plate of a mirror, 18 inches by 12, is to be surrounded 
 by a plain frame whose surface shall be equal to that of the 
 glass. Find the width of the frame. 
 
 13. A man bought some sheep for $ 360, and his neighbor 
 bought 6 more for the same amount, paying $ 5 less for each. 
 How many did the first man buy, and what was the price of each? 
 
 14. A man bought shares in a company for $ 375. A later 
 investor, after the shares had declined $ 6.25 each, bought for 
 the same amount five more than did the first man. How many 
 shares did the first man buy ? 
 
 15. A battalion of soldiers, when formed into a solid square, 
 presents 16 men fewer in the front than when formed into a 
 hollow square four deep. Required the number of men. 
 
 16. Two vessels, one of which sails faster than the other by 2 
 miles an hour, start together for different ports. The faster ves- 
 sel completes its voyage of 1152 miles 1 day later than the other 
 completes its voyage of 720 miles. What is the rate of the faster 
 
 vessel ".' 
 
216 HIGHER ALGEBRA 
 
 17. A regiment received orders to send 216 men on garrison 
 duty, each company sending the same number of men ; but before 
 the detachments marched, three entire companies were sent on 
 other service, and it was then found that each remaining company 
 would have to send 12 men additional to furnish the required 216. 
 How many companies were in the regiment, and how many men 
 did each remaining company send on garrison duty ? 
 
 18. Two trains simultaneously leave A and B, which are 81 
 miles apart, and pass each other in 1 hour. The train from A 
 reaches B 27 minutes earlier than the one from B reaches A. 
 Find the time of each train. 
 
 19. A boat goes along a straight reach of a canal at 6 miles an 
 hour. A person living 4 miles from the canal sets out, three 
 quarters of an hour before it is due at it nearest point to his resi- 
 dence, to catch the boat. If he goes 4 miles an hour, find how far 
 below the nearest point of the canal is the point toward which he 
 must direct his course, in order that he may reach it just with the 
 boat. 
 
 20. A starts at 10 a.m. to walk from P to Q, and B starts at 
 10 : 24 a.m. to walk from Q to P. They meet 6 miles from Q. B 
 stops 1 hour at P, and A stops 2 hours and 54 minutes at Q, and 
 returning they meet midway between P and Q at 6 : 54 p.m. 
 Find the distance from P to Q. 
 
 348. Theorem. If all the terms of an equation that is integral 
 with reference to the unknown quantity be transposed to the first 
 member, and the resulting polynomial be resolved into factors con- 
 taining the unknown quantity, the values obtained by placing these 
 factors in turn equal to are the roots of the. original equation. 
 
 Dem. By hypothesis the second member of the equation is 0. 
 Now since any finite quantity multiplied by is 0, the placing of 
 any one of the factors equal to will render the first member 
 0, and the equation will be satisfied. Hence the values which 
 render the factor will be roots of the original equation. 
 
 If the factors are of the first degree in the form x — a, x— b, 
 x — c, ••• x — I, the equation is 
 
 (x — a) (x — 6) (x — c) • • • (x — I) = 0, 
 
QUADRATIC EQUATIONS 217 
 
 and from x — a = 0, x — b = 0, x — c = 0, etc., we have by trans- 
 position x = a, x = b, x = c, etc. 
 
 349. Cor. Conversely, to form an equation having given roots, 
 it is but necessary to subtract the roots from x and place the product 
 of the remainders equal to 0. 
 
 Thus, to form an equation whose roots are 1, 8, and — 5, i.e., an equation 
 which is satisfied when x = 1, x = 3, and x = — 5, we have by transposition 
 £ — 1=0, x — 3 = 0, and x + 5 = 0, and by multiplying these equations 
 together (using the method of Art. 50), x 3 + x 2 — 17 x + 15 = 0. 
 
 If but two roots be given, the required quadratic may be written at once by 
 Art. 346. Thus, if the roots are 2 and 3, 
 
 p = - (2 + 3) = - 5, q = - (2 x 3) = -6, 
 
 and the equation is x 2 — 5 x = — G. 
 
 EXAMPLES LXXXVI 
 Solve the following by Art. 348. 
 
 1. x 2 — 5 x = — 6. 
 
 Solution. Transposing, x 2 — 5 x + 6 = 0. 
 Factoring, (x - 2) (x - 3) = 0. 
 
 This is satisfied when x — 2 = 0, and when x — 3=0, giving x =2 and x=3. 
 
 2. x 2 -5x = U. 
 
 Solution, x 2 - 5 x - 14 = (x + 2) (x - 7) = ; .-. x = - 2 or 7. 
 
 3. a 2 - 6 <b + 5 = 0. 4. ar 2 - 5 £ = 24. 5. ar*-6x + 9 = 0. 
 6. ar* = 49. 7. a 2 - 7 a; = 0. 8. a? — 12 a? = -36. 
 
 Form the equations whose roots are the following, either writ- 
 ing the products by Art. 00, or the equations at once by Art. 346 : 
 
 9. 4 and - 2. 10. 7 and 3. 11. — 8 and 5. 
 
 12. -6 and -9. 13. y/S and— VS. 14. 2+V5and2-V5. 
 
 15. By a misprint in an examination paper the absolute term 
 of a quadratic equation in the form x 2 +px = <y is made -f- instead 
 of — , thus making the roots 12 and — 2. What are the roots of 
 the equation in the copy ? 
 
218 HIGHER ALGEBRA 
 
 16. Two boys attempt to solve a quadratic equation. After 
 reducing it to the form x 2 + px = q, one of them has a mistake 
 only in the absolute term, and finds the roots to be 1 and 7 ; the 
 other has a mistake only in the coefficient of x, and finds the roots 
 to be — 1 and — 12. Find the roots of the correct equation. 
 
 17. Find the relation that must exist between p and q in order 
 that one root of x 2 -f px -f q — may be double of the other. 
 
 18. What, condition will make one root of x 2 -{- px = q the 
 reciprocal of the other? 
 
 350. Art. 348 has not been introduced in connection with 
 affected quadratic equations for the purpose of replacing the 
 method of solution given in Art. 344, but for the purpose of 
 showing how roots are sometimes gained and sometimes lost in 
 transforming equations, and for providing against such gain or 
 loss. 
 
 351. Theorem. Multiplying an equation by a factor containing 
 the unknown quantity introduces into the resulting equation the root 
 or roots obtained by placing the factor equal to ; and, conversely, 
 removing from an equation a factor containing the unknown quantity 
 removes from the equation the root or roots obtained by placing the 
 factor equal to 0. 
 
 Dem. 1st. The terms all being transposed to the first mem- 
 ber, the second member is 0; and if the introduced factor be 
 made equal to 0, the product will be 0, and the new equation 
 will be satisfied. Hence the values which render the introduced 
 factor are roots of the new equation. 
 
 2d. If an equation be divided by a factor containing the 
 unknown quantity, it loses those values of the unknown quantity 
 which, by rendering that factor 0, satisfy the original equation. 
 
 352. Sch. In clearing of fractions an equation having the 
 unknown quantity in the denominator of some of the terms, to 
 avoid the introduction of roots that do not belong to the original 
 equation, sometimes called Extraneous Roots, care should be taken 
 to multiply by the lowest common denominator; and when the 
 
QUADRATIC EQUATIONS 219 
 
 degree of an equation is reduced by dividing out a factor, the 
 roots obtained by placing this factor equal to must be retained 
 as roots of the original equation. 
 
 ILLUSTRATIVE EXAMPLES 
 
 1. Let it be required to find the roots of 
 
 5 2 = 1 
 
 x 2 — 1 x + 1 it' — 1 
 
 When we clear of fractions by multiplying by the 1. c. m. of the denomi- 
 nators and reduce, we have 
 
 3(x-2)=0, 
 
 and x — 2 is the only factor that can be placed equal to 0, giving x = 2 as the 
 only root. 
 
 When we clear of fractions by multiplying by the product of all the 
 denominators and reduce, we have 
 
 3(x 2 -l)(x-2)=0, 
 
 which is satisfied not only for x — 2 = 0, but also for x 2 — 1 = 0, giving x — 2, 
 1, or — 1. The roots 1 and — 1 do not belong to the original equation, but 
 entered by the introduction of the extra factor x 2 — 1 in clearing of fractions. 
 
 2. Let it be required to find the roots of 
 
 a 2 _ G k - 7 = 3 V ar* - 6 a - 7 . 
 
 Dividing by vx 2 — 6 x — 7, we have 
 
 Vx 2 - 6 x - 7 = 3. 
 
 Squaring both members and solving, we have x = 8 or — 2. These num- 
 bers satisfy the original equation, but so also do 7 and — 1. How, then, did 
 these two roots escape ? 
 
 The original equation may be put in the form* 
 
 Vx 2 - 6 x - 7( Vx 2 - Ox - 7 - 3)= 0, 
 
 and this equation is satisfied by placing either factor equal to 0, and each of 
 these equations gives two roots. Two of them, 7 and — 1, were lost in divid- 
 ing out the factor Vx 2 — 6x — 7. 
 
 If we choose, we may free the equation of radicals, giving an equation of 
 the 4th degree, viz., 
 
 ^4 _ 12 x 3 + 13 x 2 + 138 x + 112 = 0, 
 and factor by Art. 101, giving 
 
 (a;_8)(x + 2)(x-7)(x + l)=0, 
 the four roots of which are seen to be 8, — 2, 7, and — 1. 
 
220 HIGHER ALGEBBA 
 
 353. In verifying a radical equation the double sign of a radi- 
 cal term is not always admissible. In the last illustrative ex- 
 ample, if 8 be substituted for x in the equation 
 
 a* ^ 6 x _ 7 s» 3 Vaj 2 -6a>-7, 
 the first member becomes 9 and the second member becomes 
 
 3V9 = 3(±3) = ±9; 
 
 and, as -j- 9 cannot equal — 9, only the upper sign is admissible. 
 The same is true when — 2 is substituted ; but, when 7 or — 1 is 
 substituted, both signs are admissible, as in each case each mem- 
 ber vanishes. In such an expression as x — 2 = V# + H there 
 are really two equations, viz., 
 
 a-2 = + Va + ll and x — 2 = — Vs-j- H. 
 
 Each gives, when squared and reduced, x 2 -f- x = 12 ; whence, 
 x = 3 or — 4. One of these values satisfies the first of the given 
 equations, the other satisfies the second, and both satisfy the 
 equation that results from squaring. 
 
 354. When, to free it of radicals or for other purpose, an equa- 
 tion is squared, there is no way of indicating in the resulting 
 equation itself that there is any restriction in the matter of 
 signs. For example, if we square both members of x = 3, giving 
 x 2 = 9, and then solve this equation, we obtain x = 3 or — 3. 
 If, however, we transpose before squaring, we have x — 3 = 0, 
 x 2 — 6x + 9 = 0; whence, x 2 — 6 x = — 9 and x = 3 ± = 3. 
 Hence, additional roots may or may not be introduced into an 
 equation by squaring it. It follows that tvhen an equation is 
 rendered rational by squaring it, some of the roots may have to be 
 rejected. A trial of the roots in the original equation will de- 
 termine what ones are to he retained. 
 
 EXAMPLES LXXXVII 
 
 Solve the following, rejecting extraneous roots : 
 1. V5 x + 1 = V# + 1 + 2. 
 
 2. V7 x + 14 = V2 x + 6 + V# -f 4. 
 
QUADRATIC EQUATIONS 221 
 
 3. 2Vx 2 -Vx + 18 - Var - 4 a; - 12 
 
 4. V2 a: 2 + 7x- d-Vx 2 - 5 a; + 4 = Vx 2 - 1. 
 
 5. V2 a,- 2 + 10 x + 8 - Var> + 6 x + 5 = Va? + 1. 
 
 6. 2Va; + V4aj+V7aj + 2 = l. 
 
 7. Vaa; + &* + Vbx + a 2 = a — 6. 
 
 8. V(a?-l)(aj-2)+V(a;-3)(a;-4)=V2. 
 
 9. 2Va^ + -4 : = 5. 10. *-Vs + l = _5, 
 
 Va? a; + VaT+T H 
 
 1 + 6 Va? Va? 
 
 12. — 1 + I -& 
 
 V2 4-a?-V2 V2^^" + V2 x 
 
 13. ^±^^ = (x-2) 2 . 14. Vl 
 
 Va^-9 1+vT+a" 1-VT^ 
 
 15. V 4 - hV '2^?^ = ^±i. 16. a?+Va^I 4 _a ; -V^l =98t 
 2 x—^/x 2 —! x4-Var*— 1 
 
CHAPTER XVI 
 
 SOME HIGHER EQUATIONS 
 
 PURE EQUATIONS 
 
 355. A Pure Equation is an equation in which the unknown 
 quantity is affected with but one exponent. 
 
 356. Prob. To solve any pure equation. 
 
 Rule. Reduce to the form in which the affected unknown quan- 
 tity with coefficient unity shall constitute one member of the equation ; 
 then perform upon both members the operations necessary to make 
 the exponent of the unknown quantity unity. 
 
 Dem. 1st. When the form after reduction is x m = a. 
 
 By extracting the mth root of both members we have x — Va. 
 
 m 
 
 2d. When the reduced form is x n = a. 
 
 By extracting the mth root of both members and raising to the 
 ?ith power both members of the resulting equation, we have 
 
 x = ( Va) n . 
 
 3d. When the reduced form is x m = a. 
 
 By Art. 147, 2d, this is the same as — = a: whence x m = -, and 
 
 ' ' x m ' a 
 
 = \!a 
 
 357. Note. Wlien both evolution and involution are neces- 
 sary, as in freeing of a fractional exponent, it is expedient to use 
 evolution first, thus avoiding large numbers. 
 
 It will be shown in a subsequent part of the work that an 
 equation has as many roots as indicated by its degree. In the 
 following examples only those roots that are obtained by the 
 above process are required. 
 
 222 
 
EQUATIONS IN THE QUADRATIC FORM 223 
 
 3. a* = 216. 
 6. a* = 243. 
 9. </**• = 256. 
 
 
 EXAMPLES LXXXVIII 
 
 Solve the following 
 
 
 1. a 4 = 256. 
 
 2. 0>=s243; 
 
 4. fflaSl. 
 
 5. a*=625. 
 
 7 ™-3 — 2 7 
 
 o T ~i — 8 2,. 
 
 358. Tlie ride of Art. 356 applies when, instead of a simple 
 quantity, as x, there is a group of terms affected as a whole with an 
 exponent. 
 
 EXAMPLES LXXXIX 
 Solve the following : 
 
 1. a 6 -6a 4 + 12a 2 -8 = 0. 
 
 Solution. This is not a pure equation with reference to x, but is a pure 
 equation with reference to X 1 — 2. Kegarding x 2 — 2 as the unknown quan- 
 tity, we have 
 
 x 6 - 6 x* + 12 x 2 - 8 = (x 2 - 2) 3 = 0, 
 
 whence x 2 — 2 = 0, and x = ± V2. 
 
 2. (o 3 -6) 5 = 32. 3. (x 2 - Gx - 2) 3 = 125. 
 4. (a? + 2)* = 81. 5. (x- 2 + i)" 3 = 8. 
 
 6. X s + 9x 2 + 27z + 27 = 0. 7. a 6 -12 a? 4 + 48 a 2 = 64. 
 
 8. a,* 4 -10ar } + 35x 2 -50a + 25 = 0. 
 
 9. 64 a 6 + 96^ - 96 x 4 - 136 a 3 + 72 or* + 54 a - 27 = 0. 
 10. (a + 3) 6 =(2a 2 + 10a + 4) 3 . 
 
 EQUATIONS IN THE QUADRATIC FORM 
 
 359. An equation is in the Quadratic Form when the unknown 
 quantity has but two exponents, one of which is twice the other. 
 
 360. Prob. To solve an equation in or reducible to the quadratic 
 form. 
 
 Solution. The reduced general form being x 2m +px m = q, we 
 may write it (x m ) 2 + px m = q, which is an affected quadratic equa- 
 
224 HIGHER ALGEBRA 
 
 tion with reference to x m . Regarding x m as the unknown quantity, 
 we have, by Art. 344, 
 
 * m = -T>±\ ] j + q, 
 
 mM 
 
 whence x = \ — ^ ± \l~ + q. 
 
 EXAMPLES XC 
 
 Solve the following : 
 
 1. x 6 + 7ar> = 8. 
 
 Solution. Writing the equation in the form 
 
 (x 3 ) a + 7 x 3 = 8, 
 
 it is seen to be an affected quadratic, not with reference to x, but with refer- 
 ence to x 3 . Regarding a 3 as the unknown quantity and solving by Art. 344, 
 we have 
 
 x 3 =-f ±f = 1 or -8, 
 
 whence x — 1 or — 2. 
 
 2. x 4 -5x 2 = -4:. 3. x*-6x 2 = -5. 4. x 4 -8x* = 9. 
 
 5. a 4 -2oV = -144. 6. x G -{-6x i = l(j. 7. x?-x% = 56. 
 
 8. x* + 5 J = 22. 9. a^ + a;* = 756. 10. x* + — = — • 
 
 2 a* 4 
 
 11. ox" 4 -11 x 2 = -6. 12. a# + sr* = 1056. 
 
 13. A- + A = 18 - 14. 3 *"#? + ^=16. 
 
 15. x* — x~* = f . 16. x- s + x? = *f.. 
 
 361. The solution of Art. 360 applies when, instead of a simple 
 quantity, as x, there is a group) of terms having as a whole only two 
 exponents, one of which is twice the other. 
 
 EXAMPLES XCI 
 Solve the following: 
 
 1. a r J _4a-6Vx 2 -4x-5 = -3. 
 Solution. Subtracting 5 from both members, we have 
 x 2 - 4 x - 5 - 6vV 2 - 4x- 5 = - 8. 
 
EQUATIONS IN THE QUADRATIC FORM 225 
 
 Here we have a group of terms, minus 6 times the square root of the 
 group, or, what is the same thing, a group of terms with the exponent 1, 
 minus 6 times the same group with the exponent \. Hence, regarding 
 Vx' 2 — 4 x — 5 as the unknown quantity and solving as an affected quadratic 
 (Art. 344), we have 
 
 Vx 2 - 4 x - 5 = 3 ± 1 = 4 or 2. 
 
 Squaring, etc., x' 2 — 4 x — 5 = 1C or 4, 
 
 x 2 - 4 x = 21 or 9, 
 
 x = 2±5or2± Vl3 ; 
 
 .-. x as 7 or - 3 or 2 ± \/l3. 
 
 The last two roots do not satisfy the original equation. 
 
 2. 2x* - 5x -2V2x>- 5a = 15. 
 
 3. ^-2a-|-6Va 2 -2a; + 5 = ll. 
 4. 3 (a + 7)* + 3 (a + ?)"* = 10. 5. #-3VaT+T6 = -6. 
 
 6. aj l -V&*-9 = 21. 
 
 7. 2ar 2 -5z + 5V2ar J -5a; + 6 = 33. 
 
 8. a2 + 3z-V2ar J + 6a; + l = l. 
 
 9. Vx + 16 -f -</« + 16 = 6. 
 
 10. a 2 +Va 2 -7a; + 8 = 7a + 4. 
 
 11. 2« 2 -7a; + 2V2^-7a;+.6=-6. 
 
 12. 2ar J + 6V2ar J -3a + 2 = 3z + 14. 
 
 13. 2a> 2 -4;c + 3Var i -2z + 6 = 15. 
 
 ar* 'a? 
 
 14. f x -^j+7fx--\ = 12^. 15. ^ + 1 + 3: + - 
 
 16 ^-Sx-S 1 = 6 
 
 7 ^-Saj-a i 
 
 17. a j*/ f l+JLY_(3{B l + ») = 70. 
 
 J^+Jl-i=:Z. 19- l+Jl-^=Jl+*. 
 
 * x * x \ a; \ a 
 
 ia 
 
 downey's alo. — 15 
 
226 HIGHER ALGEBRA 
 
 EQUATIONS WITH INTEGRAL ROOTS 
 
 362. An equation containing one unknown quantity is said to 
 be in the Normal or Typical Form when the exponents are all 
 positive integers, the coefficient of the highest power is 1, and 
 the other coefficients are integers. 
 
 This form is called normal or typical because, as shown in a 
 subsequent part of the work, every equation having rational coef- 
 ficients can be reduced to it. 
 
 363. Prob. To find the integral roots of an equation in the 
 typical form. 
 
 Solution. If the polynomial resulting from transposing all 
 the terms to the first member has integral factors of the first 
 degree, they may be found by the process of Art. 101. Now, by 
 Art. 348, the roots obtained by placing these factors equal to 
 are roots of the original equation. 
 
 If all the roots but two are found by this process, the remain- 
 ing two, even if surd or imaginary, may be found by placing the 
 remaining factor, which is quadratic, equal to 0. 
 
 364. Cor. Some or all of the roots may be equal. 
 
 Note. In solving the following examples carefully observe 
 the instructions of Art. 102. 
 
 EXAMPLES XCII 
 
 Find the roots of the following : 
 
 1. x>-x A - 13a? + 13a? 2 + 36a;- 36 = 0. 
 
 Operation. Proceeding as in Art. 101, using the smaller divisors of 30 
 (only factors of the absolute term need be tried in any case), we have the 
 following : 
 
 x h _ x \ _ 13 x s + 13 x -2 + 36 x - 36 I 1 
 
 0- 
 
 -13 
 
 
 
 36 
 
 | 2 
 
 2- 
 
 - 9 
 
 -18 
 
 
 
 Li 
 
 5 
 
 6 
 
 
 
 
 LzJ 
 
 S 
 
 
 
 
 
 |-3 
 
 
 
 
 
 
 
EQUATIONS WITH INTEGRAL ROOTS 
 
 227 
 
 5. ar* + 2x 2 -23# -60 = 0. 
 
 Therefore the roots are 1, 2, 3, — 2, — 3. This is because the factors are 
 x — 1, x — 2, x — 3, x + 2, and x + 3, and when these are placed equal to 
 (Art. 348), we have by transposition 
 
 x = 1, x = 2, x = 3, x = - 2, x = - 3. 
 2. ic 3 - 9ar> + 26z -24 = 0. 
 4. ^-8x 2 -fl3x-6 = 0. 
 
 6 . ar 3 -a; 2 -8.T + 12 = 0. 7. ar 5 - 5x 2 - 8a; + 48 = 0. 
 
 8. ar 3 + 8ar J + 20a; + l6 = 0. 9. ar 3 -13ar + 47a; -35 = 0. 
 
 10. a; 4 -3ar 3 - 14a; 2 + 48a; -32 = 0. 
 
 11. rf-lla? + l&x-8 = 0. 
 
 Sug. Supply the missing term, with coefficient 0. 
 
 12. a; 4 -45ar J -40o; + 84 = 0. 
 
 13. a? 4 + 13 a? + 33 or 8 + 31 a; + 10 = 0. 
 Query. Why is it unnecessary to try any positive numbers ? 
 
 14. a*-3ar 5 + 6ar } -3ar J -3a; + 2 = 0. 
 
 
 
 Operation 
 
 
 
 x«-3x £ 
 
 + x 4 + 6 x 3 - 
 
 -3x 2 
 
 -3x 
 
 + 21 1 
 
 -2 
 
 -2 
 
 4 
 
 1 
 
 -2 
 
 01 i 
 
 -1 
 
 -3 
 
 1 
 
 2 
 
 
 
 l_l 
 
 
 
 -3 
 
 -2 
 
 
 
 
 [=1 
 
 -1 
 
 -2 
 
 
 
 
 
 bi 
 
 -2 
 
 
 
 
 
 
 L^ 
 
 
 
 
 
 
 
 
 Hence the roots are 1 , 
 
 1,1, 
 
 -1,-1 
 
 ,2. 
 
 
 
 15. a?-2a; 4 -15ar 3 + 8a; 2 + 68a; + 48 = 0. 
 
 16. x" - 13 a; 4 + 67 a? - 171 x 2 + 216 a; - 108 = 0. 
 
 17. a^ 5 + 3ar 5 -6a; 4 -6ar 3 + 9ar J + 3a;-4 = 0. 
 
 18. a; 7 + 5a^ + 6ar 5 -6a; 4 -15ar 3 -3ar J + 8a; + 4 = 0. 
 
 19. ar i -8a; 4 + 21ar 3 -16a,- 2 -10a; + 12 = 0. 
 
228 HIGHER ALGEBRA 
 
 Operation 
 
 X 5 _ 8 x ± + 21 x* - 16 x 2 - 10 x + 12 (_J_ 
 -7 14 -2-12 0(_2 
 
 -5460 L§ 
 
 -2 -2 
 
 Further trial by this process fails to give the other two roots. (There 
 must be five roots (Art. 357), since the equation is of the fifth degree.) 
 However, the factors of the polynomial are x — 1, x — 2, x — 3, and 
 x 2 — 2 x — 2, the last row of numbers, 1, — 2, — 2, furnishing the coefficients 
 for the last one (Art. 81). Placing the quadratic factor equal to (Art. 348), 
 
 we have 
 
 x 2 - 2 x - 2 = 0, 
 
 or x 2 -2x = 2, 
 
 whence x = I ± VS. 
 
 Hence the roots are 1, 2, 3, 1 ± V3. 
 
 20. ^-6^ + 10^-8=0. 21. a 3 -3 a 2 + £ + 2 = 0. 
 
 22. aj* -6 ar* + 24 x -16 = 0. 23. a 4 - 4 a? - 8x + 32 = 0. 
 
 24. ^-9^ + 17^ + 27^-60 = 0. 
 
 25. ar 5 -3a 4 -9ar 5 + 21a? 2 -10:r + 24 = 0. 
 
 26. .^ 5 -4x 4 -16ic 3 + 112^ 2 -208« + 128 = 0. . 
 
 27. x 6 -7a,- 5 + lla; 4 -7^ + 14a 2 -28a; + 40=:0. 
 
 365. The method of Art. 363 applies also when the coefficient 
 of the first term is not 1 ; but in that case, if there is no mono- 
 mial factor common to all the terms, the rational roots are not all 
 integral divisors of the absolute term. We should find first any 
 integral roots the equation may contain, and find the remaining 
 root or roots by placing the remaining factor equal to 0. 
 
 EXAMPLES XCIII 
 Find the roots of the following : 
 1. 3x 4 -8^-llx 2 + 28a?-12 = 0. 
 
 Operation 
 
 3 X 4 _ 8 x 3 - 11 x 2 + 28 x - 12 J 1 
 
 - 5 - 16 12 | 3 
 
 4-4 1 -2 
 
 -2 
 
EQUATIONS WITH INTEGRAL ROOTS 229 
 
 The last factor of the polynomial is seen to be 3 x — 2 ; and by Art. 348, 
 3 x - 2 = 0, 
 whence x = §. 
 
 Hence the roots are 1, 3, — 2, f . 
 2. 6^-17 z 4 -25ar 3 + 55x 2 + 39a-18 = 0. 
 
 Operation 
 
 G x 5 - 17 x 4 - 25 x 3 + 55 x 2 + 39 x - 18 | 2 
 
 _ 5 _ 35 _ 15 9 | 3 
 
 13 4-3 | -1 
 
 7-3 
 
 By Art. 348, 6 x 2 + 7 x - 3 = 0, 
 
 whence x 2 + £ x = \, 
 
 and x = - T 7 3 ± H = i <* - !• 
 
 Hence the roots are 2, 3, — 1, |, — f . 
 
 3. 5^-22^ + 15^ + 18 = 0. 
 
 4. 6ar J -29.T 2 -6a>+5 = 0. 
 
 5. 6z 4 + 5ar l -25ar J -10a; + 24=:0. 
 
 6. 4af-28z 4 + 57ar J -8x 2 -67.T + 30 = 0. 
 
 366. By Art. 349, produce the equations whose roots are the 
 following, performing the multiplications, as far as practicable, 
 as in Art. 59 : 
 
 1. 1, - 3, 4. 2. 1, 3, - 2, - 4. 
 
 3. V2, -V2, -1, 3. 4. 1, 2, 2, -3, 4. 
 
 5. 2, 3, 4, - 1, - 5. 6. 1, 3, - 2, - 2, - 2. 
 
 7. -3, 2 + V^l, 2-V == ~l. a ±V :r 2, ±V5. 
 9. 1 ±V^2, 2±V ::: 3. 10. f, 2, V3, -VS. 
 
 Solution, (x - § ) (x - 2) (x - V3) (x + V3) = 0, 
 
 (2x-3)(x-2)(x 2 -3) = 0, and 2 x* - 7 x 3 4- 21 x - 18 = 0. 
 
 2 x — 3 
 Query. Why may the denominator of — — — be dropped ? 
 
 U9 4.1 3 12 1 _I _1 _3 
 
CHAPTER XVII 
 
 SIMULTANEOUS EQUATIONS OF THE SECOND DEGREE 
 WITH TWO UNKNOWN QUANTITIES 
 
 367. A Homogeneous Equation is an equation in which all the 
 terms are of the same degree.* 
 
 Thus; 8 x 2 - 2 xy = 4 y 2 , x 2 - 4 xy + 5 y 2 = 0, ax* + bx 2 y + cxy 2 + dy s = 0, 
 are homogeneous equations. 
 
 368. A Symmetrical Equation is an equation in which the 
 unknown quantities may change places without affecting the 
 equality. 
 
 Thus, 
 Sx 2 + 3y' 2 - x - y = 5, 2 x 2 + 2 y 2 - 3 xy = 7, ax 2 + ay 2 + &x?/ + cz + ty =d, 
 are symmetrical equations. 
 
 369. Theorem. The solution of two equations of the second degree 
 with two unknown quantities requires, in general, the solution of a 
 biquadratic. 
 
 Dem. Two general equations of the second degree with two 
 unknown quantities (such equations must provide for all terms 
 that can possibly occur) have the forms 
 
 ax 2 + bxy + af + dx + ey + f =0, (1) 
 
 a'x 2 + b'xy + c'y 2 + d'x + e'y +/' = 0. (2) 
 
 From(l) ,, -»»±*± J Q*^ ^&±a±L 
 
 2 a ^ 4a 2 a 
 
 The substitution of this in (2) will give terms in y 2 and a radical of 
 the second degree. Then the rationalization of this equation will 
 require the squaring of a polynomial containing y 2 , and the result- 
 ing equation will be of the fourth degree. 
 
 * Some writers apply the name to equations in which all the terms except 
 an absolute term are of the same degree. 
 
 230 
 
SIMULTANEOUS EQUATIONS 231 
 
 370. Although the elimination of one of the two unknown 
 quantities from two equations of the second degree results, in 
 general, as shown above, in a biquadratic, there are many cases in 
 which, owing to the absence of some of the terms, the solution 
 may be effected by the use of a quadratic, or by the use of two 
 simple equations with two unknown quantities. The most useful 
 of the various methods are here given. 
 
 Case I 
 
 371. When one of the equations is of the first degree. 
 
 Rule. Find from the simple equation the value of one. of the 
 unknown quantities in terms of the other and known quantities. 
 Substitute this in the other equation and solve in the usual way the 
 resulting quadratic. 
 
 Dem. The only feature of this rule needing proof is that the 
 equation resulting from eliminating one of the unknown quanti- 
 ties is a quadratic. 
 
 The general form of an equation of the second degree with two 
 unknown quantities is 
 
 ax 1 + bxy + cy- + dx + ey +/= 0. 
 
 The general form of an equation of the first degree, with two 
 unknown quantities, is 
 
 mx + ny + p = 0. 
 
 x From the latter, x = ~ m J~V , 
 
 m 
 
 which, substituted in the former, gives no term containing a 
 higher power of y than the second. 
 
 EXAMPLES XCIV 
 Solve the following : 
 
 7 x 2 - 8 xy = 159, f x-- 2 xy-y 2 = l, 
 
 ox + 2y = l. ' \x + y = 2. 
 
 f * + ? ' = 4 ' 1^+4^ + 23 + 37=17, 
 
 - + - = 1- ' [2x-y = 0. 
 
 x y 
 
232 HIGHER ALGEBRA 
 
 a . b 
 
 x y 10 
 
 x y 
 
 a 2 b 2 n 
 
 Although the last two examples contain no equation of the first degree 
 with reference to x and y, the first equation of each is of the first degree 
 with reference to the reciprocals of x and y, and the second equation of each 
 contains the squares of these reciprocals. We may, therefore, regard - [or 
 
 in the 0th - ) as the unknown quantity, and solve accordingly. 
 x) 
 
 Case II 
 
 372. When each equation has only one term of the second degree 
 and these terms are similar. 
 
 Rule. Eliminate the terms of the second degree, and combine 
 the resulting equation of the first degree with either of the original 
 equations, as in Case I. 
 
 EXAMPLES XCV 
 
 Solve the following : 
 
 t f _j_ 3 x _ 4 y = 6, | 6 x 2 + 9 x - 2 y = 15, 
 
 (2xy-2x + ±y = ±, ( 3xy + 6x - 3y = 10, 
 
 { xy -+- x + 6 y = 6. 1 6 xy + 2 # — y = 10. 
 
 Case III 
 
 373. WVie/i o?ie o/ tfie equations is homogeneous. 
 
 Rule. Find from the homogeneous equation the value of one 
 unknown quantity in terms of the other and knoivn quantities. 
 Substitute this value in the other equation and solve the resulting 
 quadratic. 
 
 Dem. The general form of a homogeneous equation of the sec- 
 ond degree with two unknown quantities being 
 ax 2 + bxy + cy 2 = 0, 
 by , y 
 
 we have x = — 7 ^-± ~Vb 2 — 4 ac, 
 
 2a 2a 
 
 or x = ~(— b±-Vb 2 — 4ac). 
 
 2a x J 
 
SIMULTANEOUS EQUATIONS 233 
 
 As this is of the first degree with reference to y, no term higher 
 than the second degree can result from substituting this value of 
 x in an equation of the second degree. 
 
 EXAMPLES XCVI 
 Solve the following : 
 
 x 2 + y 2 + 2x = 12, | x 2 + 5xy + 6y 2 = 180, 
 
 3ar> + 2xy-y 2 = 0. * [ x 2 + xy - 6y 2 = 0. 
 
 f2s"-2a; + y = 13, 4 
 
 2x 2 -2x + y = 13, • ($x 2 -6xy + y 2 = 0, 
 
 4. 
 
 Case IV 
 
 x 2 + f + x = 6. 
 
 374. When both equations have an absolute term, but are other- 
 wise homogeneous. 
 
 FIRST METHOD 
 
 Rule. Eliminate the absolute terms and then proceed as in 
 Case III. 
 
 (x 2 -xy + y 2 = 21, (1) 
 
 Example. \ 
 
 \2xy-y 2 = 15. (2) 
 
 Solution. Multiplying (1) by 5 and (2) by 7, and subtracting, we have 
 
 5x 2 - bxy + 5 j/ 2 = 105 
 14 xy - 7 y 2 = 105 
 
 6 x 2 - 19 xy + 12 y 2 = 
 This solved for x gives 
 
 x = tty±Uy = zy°rU' (3) 
 
 Substituting the first of these values in (2) and solving, we have 
 
 y = ± V3. 
 
 Substituting these values of y in that part of (3) used in finding them, we 
 have 
 
 x = 3 y = ± 8 \/3. 
 
 Substituting the other value of x, viz., $ y, in (2) and solving, we have 
 
 y =±5. 
 
 Substituting these values of y in *A«i par* of (3) wsed m finding them, we 
 have 
 
 x = £ y = ± 4. 
 
Example. 
 
 234 HIGHER ALGEBRA 
 
 It should be noted that the values of x and y occur in pairs, thus : 
 
 Ja; = 3V3, Jx = -3V3, jx = 4, fx = -4, 
 
 \y = y/3. \ y = -VS. U = 5. If = - 5. 
 
 SECOND METHOD 
 
 375. Rule. Find from one of the given equations the value of 
 one of the unknown quantities in terms of the other and known 
 quantities. Substitute this value in the other equation and solve in 
 the usual way the resulting equation, which will always have the 
 quadratic form (Art. 359). 
 
 x 2 -xy + y 2 = 21, (1) 
 
 2xy-f=15. (2) 
 
 Solution. From (2) x = 15 + fc (3) 
 
 2y 
 
 Substituting this value of x in (1) and reducing, we have 
 f« - 28 y* = - 75, 
 whence y 2 = 14 ± 11 = 25 or 3, 
 
 and y = ± 5 or ± V3. 
 
 These values of y substituted in (3) give 
 
 x=±4 or ±3>/3. 
 
 THIRD METHOD 
 
 376. Rule. Assume x = vy, and substitute in both equations. 
 By elimination form an equation involving only v, and, solve for v. 
 Simple substitution will then determine x and y. 
 
 Dem. Taking the most general form of the equations in ques- 
 tion, we have 
 
 ax 2 -f bxy + cy 2 = m, 
 
 dx 2 -f exy -\-fy 2 = n. 
 
 Now if x = vy, (1) 
 
 where v is simply the ratio of x to y, 
 
 av 2 y 2 + bvy 2 + cy 2 = m, 
 
 dv 2 y 2 + evy 2 +fy 2 = n ; 
 
 whence y 2 = ™ = 2 (2) 
 
 av 2 + bv + c dvr + ev+f 
 
SIMULTANEOUS EQUATIONS 235 
 
 As this equation is of the second degree with reference to v, it 
 can always be solved. 
 
 When the value of v is substituted in either of the equations 
 (2), y becomes known ; and when v and y are substituted in (1), 
 x becomes known. 
 
 Example. 
 
 [2xy — y 2 = 15. 
 
 Solution. Assuming x = vy, (1) 
 
 and substituting in both equations, we have 
 
 21 15 
 
 y v -2 _ v + i 2 v - l' 
 whence v = 3 or f . 
 
 Substituting the first value of v in (2), 
 
 2 v — 1 
 whence y = ± V3. 
 
 Substituting in (1), x = vy =± Z\/S. 
 
 Substituting the second value of v in (2), 
 
 
 2/ 2 = ~ 15 T = 25, 
 
 2 v — 1 
 
 whence 
 
 y = ±5. 
 
 Substituting in (1), 
 
 x = i>y = ± 4. 
 
 Solve the following : 
 
 EXAMPLES XCVII 
 
 far 9 + ^ = 15, jar 9 + ^+4^ = 6, 
 
 f^ + ^4-22/ 2 = 74, (^+7/ + l=3a*/, 
 
 (2) 
 
 3 ' J2a 8 + 2xy + tf = 73. *' 1 2(ojy + 4) = 3y*. 
 
 jar> + ^ + 2/ 2 = 52, . r^-2^-7/^31, 
 
 U-i/-ar = 8. * 1 1^ + 2*2/ -2/ 2 = 101. 
 
 377. Many equations falling under Case IV., viz., equations 
 having an absolute term, but otherwise homogeneous, admit of 
 shorter solutions than by any of the three methods given above. 
 
236 HIGHER ALGEBRA 
 
 1st. When any multiple of one equation added to or subtracted 
 from the other gives a perfect square. 
 
 Rule. If two simple equations are thus obtained, finish the 
 solution by any of the methods of elimination. If but one simple 
 equation is obtained, finish the solution as in Case I. 
 
 1 xy = 28. (2) 
 
 Solution. Adding twice (2) to (1), we have 
 x 2 + 2 xy + y 2 = 121, 
 whence x + y = ±11. (3) 
 
 Subtracting twice (2) from (1), we have • 
 
 x 2 - 2 xy + y 2 = 9, 
 whence x — y = ± 3. (4) 
 
 The combination of (3) and (4) gives 
 
 x = ± 7 or ±4, 
 y = ± 4 or ±7. 
 
 The same process is often applicable to equations above the 
 
 second degree. 
 
 (x* + 3x*y* = 28, (1) 
 
 Example. < 
 
 lary + 47/ 4 = 8. (2) 
 
 Solution. The addition of (1) and (2) gives 
 x i + 4 x 2 y 2 + 4y 2 = 36, 
 whence x 2 + y 2 = ± 6, 
 
 and y 2 = ± 6 - x 2 . (3) 
 
 Substituting this in (I) and reducing, we have 
 x^9x 2 = - 14, 
 whence z 2 = ±f±f = ±7 or ±4, 
 
 and x = ± \/l or ± V-~T, or ± 2 or ± 2V- 1. 
 
 By substituting these values of x in (3), y may be found. 
 
 2d. When the first members of the two equations have a common 
 factor containing the unknown quantity. 
 
 Rule. Divide one equation by the other, canceling the commoyi 
 factor. Then proceed as in Case I. 
 
SIMULTANEOUS EQUATIONS 237 
 
 (x 2 + xy = 35, (1) 
 
 Example. , ,, /fV 
 
 (<*.+ /*> 14 (2) 
 
 Solution. The equations may be put in the forms 
 
 *(x + 20=35, (3) 
 
 y(x + y)= 14. (4) 
 
 Dividing (3) by (4), we have 
 
 x_35 
 
 2/~14' 
 
 whence x = -— f~ (5) 
 
 14 
 
 Substituting in (2) and solving, 
 
 V = ±2. 
 
 Substituting these values of y in (5), 
 
 * = ± 5. 
 
 The same process is applicable to equations above the second 
 degree whose first members have a common factor containing the 
 unknown quantity. 
 
 a? + 3f = 28, (1) 
 
 Example. 
 
 * + 2/ = 4. (2) 
 
 Solution. Dividing (1) by (2), we have 
 
 x 2 - xy + y 1 = 7. (3) 
 
 Proceeding with (2) and (3) as in Case I., we find 
 
 x = 1 or 3, 
 y = 3 or 1. 
 
 EXAMPLES XCVIII 
 Solve the following by Art. 377 : 
 
 , x 2 -?f = 12, m {x 2 + y 2 = n, 
 
 (x 2 + y 2 = 
 
 [ xy = 35. 
 
 xy + y 2 = 12. I xy 
 
 3 ^x 2 y + xy 2 = 30, 4 j^ + ^91, 
 
 ' [x + y = 5. ' \x + y = l. 
 
 r x 2 -±y 2 = 9, (x i + 3xy = 10, 
 
 5 ' \xy + 2y 2 = 3. * \xy + 4y 2 = 6. 
 
 (x s -tf = 875, (x A -\-x 2 y 2 -\-y 4 = 133 J 
 
 7 " [x 2 + xy + y 2 = 175. * 1 x 2 - xy + y 2 = 7. 
 
238 HIGHER ALGEBRA 
 
 Case V 
 
 378. When the equations are symmetrical, and not included in 
 any of the preceding cases. 
 
 Rule. Assume x = u + v and y = u — v, and substitute in both 
 equations. Then reduce and eliminate v. 
 
 Dem. Since x and y are by hypothesis involved alike (Art. 
 368), u -\- v and u — v will be involved alike. Hence for every 
 plus term containing an odd power of v there will be an equal 
 negative term. Therefore only even powers of v will remain, and 
 v can be eliminated. 
 
 The method is not limited to equations of the second degree. 
 
 EXAMPLES XCIX 
 Solve the following : 
 
 , t xy(x + y) = 30, (1) 
 
 s 
 
 X s + f = 35. (2) 
 
 Solution. Assume z = u + v, (3) 
 
 and y = u — v. (4) 
 
 Substituting these values in (1) and (2) and reducing, we have 
 
 u s - uv 2 = 15, (5) 
 
 and 2 m 3 + 6 uv* = 35. (6) 
 
 Adding 6 times (5) to (6), we have 
 
 8 u* = 125, 
 whence u = f . 
 
 Substituting this value of u in (5) and solving, we obtain 
 
 Substituting these values of u and v in (3) and (4) , we obtain 
 
 x - | ± \ = 3 or 2, 
 
 and y — f T i = 2 or 3. 
 
 In this example a shorter process is to add 3 times the first equation to the 
 second, extract the cube root of the result, giving x + y — 5, and substitute 
 this value in the first equation, giving xy = 6. Then the equations x HJ Jf as 6 
 and xy = 6 are easily solved. 
 
SIMULTANEOUS EQUATIONS 
 
 239 
 
 a* -f y* + * + y » 20. 
 
 2ar 9 + 22/ 2 = 5^, 
 4(a* + y)=»y. 
 
 | a* 4 — 9 ar 2 ^ 2 
 1 a; + w = 4. 
 
 9*Y + y« = l, 
 
 5. 
 
 2/ 
 
 '^ 2 + 2/ 2 — « — y = 14, 
 
 . a# 4- x 4- y = 14. 
 
 MISCELLANEOUS EXAMPLES 
 EXAMPLES C 
 
 1. 
 
 Solve the following: 
 
 'a* + xy = 12, 
 I xy 4- 2/ 2 = 2. 
 
 [ a 2 - ?/ 2 = 45, 
 I a? — y = 3. 
 
 |3ajy-a;-5y = 8, 
 I a?y + a; — 3 # = 4. 
 
 5. 
 
 6 
 
 ' \xy = 12. 
 
 32, 
 
 xy 
 
 x 2 + y- 4- x 4- y = 18, 
 
 11. 
 
 13. 
 
 15. 
 
 17. 
 
 r a 4- y 4- V # -f i/ = 6, 
 U + ^lft 
 
 a- 2 + 3a# = 28, 
 a*/ 4- 4 1/ 2 = 8. 
 
 x 2 -Q>xy + y 2 -2x+l0y = -12, 
 x 2 + 6xy-7y 2 = Q. 
 
 x 2 -y 2 = 60, 
 a^ = 16. 
 
 x 2 + xy = 12, 
 it*?/ — 2 2/ 2 = 1 - 
 
 (x* + y* = 97, 
 I a* 4- ?/ = 5. 
 
 a? 2 y 2 
 ab 
 
 10. 
 
 12. 
 
 + - 2 =io, 
 y 2 
 
 3. 
 
 x 2 4- xy = 15, 
 xy-f = 2. 
 
 x*-tf = 26, 
 
 a* 2 ?/ — ar?/ 2 = 6. 
 
 3a- 2 + a-?/-2?/ 2 =162/, 
 
 x 2 — 2 2Z 2 = 4 1/. 
 
 14 
 
 16. 
 
 18 
 
 ■{ 
 
 an/ 
 
 a; 4- Vxy + y = 19, 
 aj» + xy + / = 133. 
 8 ar 3 - 27 ^ = 271, 
 2a--3y = l. 
 x 2 + 3 a*/ + 2 ?/ 2 = 63, 
 
 8a# + 4y 2 = 171. 
 
 1^4-3 
 13a* 2 4- 
 
240 HIGHER ALGEBRA 
 
 { tf + tf = 152, 
 I x 2 — xy + y 2 = 19. 
 
 20. 
 
 | & + y 2 + 3xy- ±(x+y)=- 3, 
 I xy 4- 2 (a? 4* y) = 5. 
 
 PROBLEMS LEADING TO SIMULTANEOUS EQUATIONS OF THE 
 SECOND DEGREE WITH TWO UNKNOWN QUANTITIES 
 
 EXAMPLES CI 
 
 1. 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 dimensions of the first rectangle. 
 
 2. The area of a rectangular field is 300 square rods, and the 
 length of its diagonal is 25 rods. Find the sides. 
 
 3. A man bought some horses for $ 1250. At another time he 
 bought 3 more than before, paying $ 25 less apiece, and they cost 
 him $ 1300. How many horses did he buy the first time, and at 
 what price ? 
 
 4. The fore wheel of a buggy makes 6 revolutions more than 
 the hind wheel in going 120 yards ; but the fore wheel of a coach, 
 each of whose wheels is larger in circumference by 1 yard respec- 
 tively, makes only 4 revolutions more than the hind wheel in going 
 the same distance. What is the circumference of each buggy 
 wheel ? 
 
 5. In walking to the summit of a mountain a man's rate during 
 the second half of the distance is \ mile per hour less than during 
 the first half, and he reaches the summit in 5J hours. He de- 
 scends in 3| hours at a uniform rate, which is 1 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. 
 
 6. Two trains start at the same time from two places, A and B, 
 168 miles apart, and travel toward each other. They pass in 1 hr. 
 52 min., and the first reaches B \ hour before the second reaches 
 A. Find the speed of each train. 
 
SIMULTANEOUS EQUATIONS 241 
 
 7. A crew, rowing at half their usual speed, row 3 miles down 
 stream and back in 2 hr. 40 min. At full speed they can go over 
 the same course in 1 hr. 4 min. Find the rate of the crew and of 
 the current. 
 
 a A courier, riding at a uniform rate, left a station. Five 
 hours afterward a second followed, riding 3 miles an hour faster. 
 Two hours after the second a third started at the rate of 10 miles 
 an hour. They all reached their destination at the same time. 
 Find the distance ; also the rate of the first. 
 
 9. A and B are two towns situated 18 miles apart on the same 
 bank of a river. A man goes from A up to B in 4 hours, 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 faster than 
 in going, and covers the whole distance in 3£ hours. Find the 
 rates of rowing and walking. 
 
 10. A man arrives at the railroad station nearest his home 1\ 
 hours before the time at which he has ordered his carriage to 
 meet him. He sets out at once to walk at the rate of 4 miles an 
 hour, and meeting his carriage when it has traveled 8 miles, 
 reaches home 1 hour earlier than he had originally expected. 
 How far is his home from the station, and at what rate was his 
 carriage driven ? 
 
 i>o\vm:y's alg. — 16 
 
CHAPTER XVIII 
 
 THEORY OF FUNCTIONS 
 
 SECTION I — MAXIMA AND MINIMA OF FUNCTIONS 
 
 379. An Arbitrary Constant is a constant to which any value 
 may be assigned, but which maintains the same value throughout 
 the same operation or discussion. 
 
 An Absolute Constant is a constant which admits of no change. 
 
 Thus, in the formula for the area of a circle, wr 2 , r is an arbitrary constant, 
 since it may have any value we choose to give it ; but t , always having the 
 same value, viz., 3.14159 approximately, is an absolute constant. When we 
 assign to r any particular value, as 5, the radius becomes an absolute constant. 
 
 Arbitrary constants are represented by the leading letters of 
 the alphabet, and absolute constants by figures or by letters which 
 always stand for the same numbers. 
 
 380. A Variable is a quantity which may have in the same 
 operation or discussion any value within the limits determined by 
 the conditions. 
 
 Thus, if y = V25 — z' 2 , and this is the only required relation between x and 
 y, we may give to x any values whatever between — 5 and + 5 and find 
 corresponding values of y. Hence x and y are variables. If x is made less 
 than — 5 or greater than + 5, y is imaginary. Hence the limits of x are — 5 
 and + 5. 
 
 Variables are represented by the final letters of the alphabet. 
 
 381. Constants and variables are not the same as known and 
 unknown quantities, although the notation is the same. In the 
 simultaneous equations 5 x -f 2 y = 7 and 7 x- — Sxy = 159, x and 
 y have two, and only two, values each, and hence they are constants ; 
 but if x and y are required to fulfill only the one condition ex- 
 pressed in the first equation, 5 x + 2 y = 7, we can give to one of 
 them any value we please, and can find for the other such value 
 
 242 
 
MAXIMA AND MINIMA OF FUNCTIONS 243 
 
 as will make the equation true. In this case x and y are 
 variables. 
 
 382. A Function of a Variable is any expression which depends 
 upon that variable for its value. 
 
 Thus, y/^b — x 1 is a function of x, inasmuch as it changes when x changes. 
 Any expression containing x is a function of x. If we have y = V25 — x' 2 , 
 we say y is a function of x. 
 
 383. A Function of Two or More Variables is any expression 
 which depends upon those variables for its value. 
 
 Thus, interest on a loan of money is a function of the principal, the rate of 
 interest, and the time ; the volume of a cone is a function of the radius of the 
 base and the altitude ; the distance passed over by a body moving from rest 
 with a uniformly accelerated velocity (D = \ft 2 ) is a function of the acceler- 
 ation and the time. 
 
 384. Notation. A function of x is represented by/(#). This is 
 employed not only to represent any function of a single variable, 
 but also to represent a specified function to avoid repetition of the 
 function itself. In this expression /is not a factor, but simply an 
 abbreviation of the word function. 
 
 When different functions of the same variable are brought into 
 the same discussion, the notation 
 
 /(»)> /(*), /"(*)> /iO)> M*), *(*), *'(*)* etc., 
 is employed. These are read, "/function of x," "/' function of 
 x," "f" function of x" "f function of aj," "/ 2 function of a?," 
 "</> function of x," " <f>' function of x," etc. 
 
 To represent the same function of different variables, the nota- 
 tion f{x), f{y), f(z), /( - x), etc., is employed. 
 
 To indicate that a constant has been substituted for the variable 
 in a function, the notation f(a), /(2), /(0), etc., is employed. 
 
 Thus, if /O) = 2 x 3 - 3 x 2 - 4 x -f 5, 
 
 then /(*/)= 2 y» - 3 y 2 - 4 y + 5, 
 
 /CO = 2 z* - 3 z 2 - 4 z + 5, 
 /( - x) = - 2 x 3 - 3 x 2 + 4 x + 5, 
 f{a) = 2 a 3 - 3 a 2 - 4 a + 5, 
 /(2) = 2 • 2 3 - 3 • 2 2 - 4 • 2 + 5 = 1, 
 /(0) = 5. 
 
244 HIGHER ALGEBRA 
 
 To represent a function of several variables, the notation 
 f(x, y), f(x, y, z), etc., is employed. These are read, "function of 
 x and y," " function of x, y, and z," etc. 
 
 In the expression F(x, y), x and y are understood to be inde- 
 pendent of each other ; but if the expression occurs as part of an 
 equation, as f(x, y) = 0, x and y are dependent. If the equation 
 can be solved for one of the variables, say y, we shall have 
 
 y = <j>(x). 
 
 385. An Increasing Function is a function that increases when 
 its variable increases, and decreases when its variable decreases. 
 
 A Decreasing Function is a function that decreases when its 
 variable increases, and increases when its variable decreases. 
 Thus, y = r 3 , y = x h + x, y = mx + b, are increasing functions ; 
 
 y ss -x 3 , y = -i y = a- x, 
 x 
 
 are decreasing functions ; while y = x 2 is an increasing function for positive 
 
 values of a*, and a decreasing function for negative values of x. 
 
 386. A Rational Integral Function of x is a function in the form 
 
 ax n + bx*' 1 + cx n ~ 2 H /, 
 
 in which all the exponents are positive integers. 
 
 387. Functions are linear, quadratic, cubic, etc., according as 
 they are of the first, the second, the third, etc., degree. 
 
 388. The notation ] M signifies that m is to be substituted for 
 the variable in the function after which it is written. 
 
 Thus, * 3. ' means the value of the function when 3 is substi- 
 
 x l — x — 6 J 3 
 tuted for x. 
 
 389. A Maximum Value of a Function is that which is greater 
 than the immediately preceding and succeeding values. 
 
 A Minimum Value of a Function is that which is less than the 
 immediately preceding and succeeding values. 
 
 Thus, for larger and larger values of x in the function 6 x 2 — x s , we have 
 the following results : 
 
 Values of a;, - 3, - 2, - 1, 0, 1, 2, 3, 4, 5, 6, 7, etc. 
 
 Values of /(a), 81, 32, 7, 0, 5, 16, 27, 32, 25, 0, - 49, etc. 
 
MAXIMA AND MINIMA OF FUNCTIONS 245 
 
 It is seen that for increasing values of «, f(x) at first decreases until it 
 reaches when x — 0, then increases until it reaches 32 when x = 4, and 
 then decreases for all larger values of x. Hence is a minimum value of 
 /(«), and 32 is a maximum value of f(x). We say, also, that x = renders 
 f{x) a minimum, and x = 4 renders /(x) a maximum. 
 
 390. Note. The greatest value of a function is not necessarily 
 a maximum, and the least value is not necessarily a minimum, in 
 the mathematical sense of these words. For example, the least 
 possible value of the function 5 + -y/x — 3 (using only the positive 
 sign of the radical) is 5, and this is when x = 3. This, however, 
 is not a minimum, since it is not less than both the preceding 
 and the succeeding values — indeed, there is no preceding value, 
 for if x is made less than 3, the function is imaginary. 
 
 Again, a minimum of a function may be larger than a maxi- 
 mum of the same function. Thus, let the function -y/x 2 -f 9 be 
 represented by y, giving vV + 9 = y, or x = ± -y/y* — 9. In 
 this form it is seen that y can have no values between — 3 
 and + 3, as these would render x imaginary, but can have all 
 values beyond these limits. Hence — 3, being the largest of all 
 the negative values of y, is a maximum, and +3, being the 
 smallest of all the positive values of y, is a minimum. In this 
 case, then, the minimum value of the function is larger than the 
 maximum. 
 
 391. A full discussion of the subject of maxima and minima 
 of functions requires the aid of the Differential Calculus ; but the 
 maxima and minima of many functions may be determined by 
 methods that are purely algebraic, and the same methods apply 
 to numerous interesting and important practical problems. 
 
 Case I — Quadratic Functions 
 
 392. Theorem. 1st. A quadratic function, ax- -f bx + c, has a, 
 maximum if a is — , and a minimum if a is + . 
 
 2d. TJie value of x which renders the function a maximum or a 
 
 b 
 minimum is 
 
 h 2 
 
 3d. The maximum or minimum value of the function is c 
 
 4 a 
 
246 HIGHER ALGEBRA 
 
 Dem. 1st. Let ax 2 + bx -f- c = y. 
 Solving for x, we have 
 
 b 1 
 
 x = — — ± — V& 2 + 4 ay — 4 ac. (1) 
 
 Z a Z a 
 
 If a is — , the term 4 ay will have the opposite sign from?/; 
 hence y can decrease without limit (4 ay becomes + when y 
 decreases below 0), but can increase only to the value beyond 
 which the quantity under the radical sign would become — , mak- 
 ing x imaginary. This limit is a maximum value of y. 
 
 If a is +j the term 4 ay will have the same sign as y\ hence y 
 can increase without limit, but can decrease only to the value 
 beyond which the quantity under the radical sign would become 
 — , making x imaginary. This limit is a minimum value of y. 
 
 2d. In either case the limit, which makes the function a maxi- 
 mum or a minimum, is where 
 
 5* + 4 ay - 4 ac = 0, (2) 
 
 which gives -, 
 
 x = — - — 
 2a 
 
 3d. The maximum or minimum of the function may be found 
 either by substituting this value of the variable in the given func- 
 tion, or by solving (2) for y, which gives 
 
 b 2 
 
 y = C- — • 
 
 4a 
 
 393. Cor. The value of the variable which renders a pure 
 quadratic function, ax 2 + c, a maximum or a minimum is 0, and 
 the maximum or minimum value of the function is c. 
 
 This is because x = = — = 0, and y=c— — - = c — ---- = c. 
 
 2a 2a ' * 4a 4a 
 
 EXAMPLES CII 
 
 Examine the following functions for maxima and minima values : 
 
 1. ^-4^ + 10. 
 
 Solution. By 1st and 2d parts of the theorem, 
 
 2a 2 
 
 renders/ (x) a minimum, since a is +• 
 
MAXIMA AND MINIMA OF FUNCTIONS 247 
 
 By the 3d part of the theorem, 
 
 / (x) at a min. = c - — = 10 - - 6 = 6. 
 4 a 4 
 
 Or / (x) at a min. = x- - 4 x + 10] 2 = 6. 
 
 2. _2a 2 + 6a?-5. 
 
 Solution. x = = = li 
 
 2a -4 2 
 
 renders/ (x) a maximum, since a is — . 
 
 /(x) at a max. = — 2 x 2 + 6 x — 5] ij = — |. 
 
 3. 6 a? — a? 1 . 4. »*— 8aj-hl9. 
 5. 3af+12»+12. 6. 7 + 8x-2^. 
 
 7. 5^-20^ + 20. 8. 16^-2^+18. 
 
 9. 6«* + ll. 10. - 3x2-30; + 7. 
 
 Case II — Reciprocals of Quadratic Functions 
 
 394. Theorem. The value of the variable which renders any 
 function a maximum or a minimum, renders the reciprocal of that 
 function a minimum or a maximum. 
 
 Dem. This is seen from the fact that when f(x) increases or 
 decreases, —— decreases or increases. 
 
 m 
 
 395. Sch. The reciprocal of a rational integral function ap- 
 proaches as a limit as the variable approaches ± oo. This limit 
 is not a minimum or a maximum in a mathematical sense (Art. 
 389), but is merely the least or the greatest value possible. 
 
 EXAMPLES CIII 
 
 Examine the following functions for maxima and minima 
 values : 
 
 1. * 
 
 x*-2x-\ 
 
 Solution. By Art. 392, 
 
 2 a 2 
 
248 HIGHER ALGEBRA 
 
 renders the denominator a minimum. Hence (Art, 394) x = 1 renders the 
 reciprocal function a maximum. 
 
 f(x) at a max. = = — -. 
 
 1 -57 
 
 4-6^-3^ 2^ + 6^-5 
 
 18 m n -? 4 
 
 x + 7 3 + 20x-5a? 
 
 Case III — Algebraic Functions of Any Form 
 
 396. Prob. To examine for maxima or minima values algebraic 
 functions of any form. 
 
 Rule. 1st. Place the function equal to y, and solve for the 
 variable in terms of y. 
 
 2d. If the result does not involve a radical of even degree, the 
 function has 'neither a maximum nor a minimum. 
 
 3d. If the result involves a radical of even degree, place this radi- 
 cal part equal to and solve for y. Each resulting real value of y 
 will be a maximum or a minimum according as an increase or a 
 decrease would give an imaginary value of the variable. 
 
 Dem. If after expressing the variable in terms of y there is no 
 radical of even degree, it is evident that y, which is the function, 
 can increase or decrease without limit, every value of y giving a 
 real value of the variable. 
 
 But if there is a radical of even degree, and the quantity under 
 the radical sign is capable of becoming negative for any value of y, 
 y can increase or decrease only to the limit beyond which this 
 quantity would become negative, making the variable imaginary. 
 This limit, which is reached where the radical part equals 0, is a 
 maximum or a minimum according as an increase* or a decrease 
 would give an imaginary result. 
 
 397. Sch. 1. The method has its limitations in our inability 
 to solve equations of all forms. 
 
 398. Sch. 2. Case III includes Cases I and II, but the special 
 theorems of those cases give the result with less labor. 
 
MAXIMA AND MINIMA OF FUNCTIONS 249 
 
 EXAMPLES CIV 
 
 Examine for maxima and minima values the following func- 
 tions : 
 
 1. a (x — b) 4 + c. 
 
 Solution. Let a(x — b) A + c = y. 
 
 Then ( X -by = £^, 
 
 y a 
 
 and x = b ± 
 
 ' a 
 
 In this form it is seen that y can increase without limit, hut can decrease 
 only to y = e, which is, therefore, a minimum value. This makes the radical 
 part 0, and leaves x = b. 
 
 2. b + (x — a) 3 . 
 
 Solution. Let b + (x — a) 3 = y. 
 
 Then x = a -f Vy — b. 
 
 In this form it is seen that y can increase and decrease without limit. 
 Therefore the function has neither a maximum nor a minimum. 
 
 In the first example, if y is made less than c, we have the 4th root of a 
 negative quantity, which is imaginary ; but in this example no value of y 
 will give an imaginary result, as an odd instead of an even root is involved. 
 
 x 2 - 2 x + 13 
 
 4 x - 12 
 
 Solution. Let s 2 -2:e+13 
 
 4x-12 y 
 
 Then ' x* - (2 + 4 y) x = - 12 y - 13, 
 
 and x = I + 2 y ±V4 y 2 — X y - 12. 
 
 By the 3d part of the rule, for a maximum or minimum 
 
 4 y 2 - 8 y - 12 = 0, 
 
 whence y = 3 or — 1. 
 
 Any values of y between 3 and — 1 give imaginary values for x. From a 
 large value y can diminish to 3, and from a small value (a negative value) y 
 can increase to — 1. Hence 3 is a minimum value of the function and — 1 
 is a maximum. [See Ait. 390.] 
 
250 HIGHER ALGEBRA 
 
 The corresponding values of x are 
 
 £ = l + 2?/ = 7 and — 1. 
 
 2x-x 2 y 
 2w 6 
 
 4 3« 2 - 
 2x- 
 
 -6 
 
 X 2 
 
 SOLUTION 
 
 . Let 
 
 Then 
 
 
 and 
 
 
 3 + y 3 + y 
 
 1 
 
 V^2 + 6 y + 18 . 
 
 3 + 2/ 3 + y 
 
 For a maximum or a minimum (3d part of rule), 
 
 f + 6 y + 18 = 0, 
 
 whence y = — 3 ± 3V— 1. 
 
 This imaginary result shows that our supposition that there is a limit 
 beyond which y cannot go (a limit beyond which the quantity under the 
 radical sign would become - , giving an imaginary value of x) is absurd 
 (Art. 227). Hence the function has neither a maximum nor a minimum. 
 
 . l-3x 
 
 5. 
 
 x 2 -2x 
 Solution. Let 
 
 Then 
 
 x 2 - 2 x ' 
 a . = 2 1 _-_3 ±iv/(2y _ 3)2 + 4j 
 
 2 
 
 As y can increase and decrease without limit, the function has neither a 
 maximum nor a minimum. This may also be shown by placing the radical 
 part equal to 0, which gives an imaginary result. 
 
 6. x*-6x 2 -7. 
 
 Solution. Let £ 4 — 6 x 2 — 7 = y. 
 
 Then x 2 = 3 ± Vy + 16, 
 
 and x = ± V3 ± y/y + 16. 
 
 In this form it is seen that y cannot be less than — 16. This is, therefore, 
 a minimum, and gives x = ± V3. 
 
 Again, when the minus sign of Vy + 16 is used, the largest value of y is 
 that which makes Vy + 16 = 3, which gives y=—7, a maximum, and 
 x = 0. 
 
MAXIMA AND MINIMA OF FUNCTIONS 251 
 
 7. b + (x — a)\ 
 Solution. Let b + (x — a)^ = y. 
 
 Then x = a + V(y - b)' 2 . 
 
 By 2d part of the rule, the function has neither a maximum nor a 
 minimum. 
 
 8. |5-|.- 9. b + (x-a)i 
 
 10. 2 -^- 11. 2^». 
 
 x 2 or 
 
 £±2*±1. is. (2 a* -art*. 
 
 a? + 2x + 7 K } 
 
 12. 
 
 14. 2 a; 4 - [6x 2 44. 15 
 
 x + 1 
 
 16. 
 
 x 2 + x + 1 
 (x-a)*. 17. 3^-12^-15. 
 
 399. If the maxima and minima values of each of two variables 
 involved in an equation are required, it is evident that we may- 
 solve for each of the variables in turn, and proceed as in Case III. 
 
 EXAMPLES CV 
 
 Examine for maxima and minima values each of the variables 
 in the following functions : 
 
 1. x* + 8y + U = 2y 2 + 4:X. 
 Solution. Arranging with reference to x, we have 
 x? -4x = 2y*-8y- 14, 
 
 whence x = 2 ± V2 y* - 8 y - 10. (1) 
 
 For the limits of y we have 
 
 21/2-82/ -10 = 0, 
 whence y = 5 or — 1. 
 
 Since y can have no values between 5 and — 1, but can have all values 
 beyond these limits, 5 is a minimum and — 1 a maximum value of y. 
 
 For both of these values of y equation (1) gives x = 2. 
 
 By solving for y in terms of x it will be found that x has neither a maxi- 
 mum nor a minimum. 
 
 2. x 2 + 3y 2 + 36 = 2x + 12y. 
 
 3. x 2 + 2y 2 = 4:X + 4y + W. 
 
 4. a* + y 2 = 2y+lo. 
 
252 HIGHER ALGEBRA 
 
 400. When the theory of maxima and minima is to be applied 
 to a practical problem, the first step is to obtain an algebraic 
 expression for the function that is to be a maximum or a mini- 
 mum. If this contains but one variable, we proceed as in one of 
 the three cases explained in this section. If it contains two 
 variables, one of them must be eliminated by given relations 
 before applying the process. 
 
 PROBLEMS IN MAXIMA AND MINIMA OF FUNCTIONS 
 EXAMPLES CVI 
 
 1. Divide m into two such parts that their product shall be a 
 maximum. 
 
 Solution. Letting x and m — x be the two parts, the function to be ex- 
 amined is 
 
 x (w — x) = - x 2 + mx. 
 
 By Art. 392, X = _A = _J?L = ™ 
 
 ' - 2a -2 2 
 
 renders /(a?) a maximum, since a is -. Therefore the two parts are equal. 
 
 2. A farmer having 40 rods of portable fence wishes to inclose 
 with it the largest possible rectangular sheep yard. What must 
 be its dimensions ? 
 
 Solution. Letting x and y represent the sides, the function to be ex- 
 amined is xy. (1) 
 Since this contains two variables, one of them must be eliminated. This 
 is done by means of the condition 
 
 2 x + 2 y = 40, 
 whence y = 20 - x. (2) 
 
 Substituting this in (1), the function to be examined becomes 
 x(20-x)=- x 2 + 20 x. 
 
 By Art. 392, x = -^- = - — = 10 
 
 J 2a -2 
 
 renders f(x) a maximum, since a is — . 
 
 Substituting this in (2), we have 
 
 y = 20 - x = 20 - 10 = 10. 
 
 Hence the yard must be a square. 
 
 3. Divide 12 into two such parts that the sum of their squares 
 shall be a maximum. 
 
MAXIMA AND MINIMA OF FUNCTIONS 253 
 
 4. A farmer wishes to fence off a 10-acre field in the form of 
 a rectangle of such dimensions as to require the least amount of 
 fence. Find its dimensions. 
 
 5. If it is specified that a purchaser is to have a rectangular 
 plot of ground of such dimensions that 3 times its breadth added 
 to 2 times its length shall equal 96 yards, what is the greatest 
 amount of land he can take ? 
 
 6. An aqueduct consists of two vertical walls surmounted by a 
 semicylindrical arch, the stone bottom being of the same thick- 
 ness as the walls. Find the dimensions such that with a given 
 amount of material the capacity shall be a maximum. 
 
 Solution. The capacity will be greatest when the area of a cross section 
 is a maximum. 
 
 Let x be the height of the side walls, 2 y the breadth, and p the perimeter, 
 which is constant, since the amount of material is given. Then the function 
 to be examined is 
 
 o 
 
 (1) 
 
 (2) 
 
 
 
 •*.+ =£ 
 
 
 To eliminate 
 lence 
 
 x, we have 
 
 p = 2x 
 x=P~ 
 
 + 2 y + iry, 
 
 2 y — try 
 2 
 
 
 Substituting 
 
 this in (1), 
 
 we have 
 
 
 
 
 py-2if 
 
 - Try 2 + 
 
 iry 2 _ 4 + 
 2 2 
 
 -y 2 + . 
 
 By Art. 392, 
 
 
 
 
 
 
 y = - 
 
 b _ 
 2a 
 
 P 
 
 p 
 
 
 -(4 + 7T) 
 
 4 + *- 
 
 aders/(y) a maximum, since a is 
 
 -. 
 
 
 Substituting 
 
 this in (2), 
 
 we have 
 
 
 
 
 2 
 
 P . 
 
 4 + 7T 
 
 irp _ 
 
 . P 
 
 
 2(4 + 7T) 
 
 4 + 7T 
 
 Hence the height of the side walls equals one half the breadth of the aqueduct. 
 
 7. A ship steaming north 12 knots an hour sights another ship 
 10 knots directly ahead steaming east 9 knots an hour. If each 
 keeps on her course, what will be the least distance between them, 
 and at what time will it occur ? 
 
 Sug. Let x be the time, and find in terms of x an expression for the square 
 of the distance between the ships. The distance will be a minimum when 
 the square of the distance is a minimum. 
 
254 HIGHER ALGEBRA 
 
 8. 1 have material enough for a stone wall 48 rods in length. 
 I can use for one side any desired portion of a wall already built. 
 What is the largest rectangular area I can inclose, and what are 
 its dimensions ? 
 
 Show that the relative dimensions would be the same, whatever 
 the amount of wall. 
 
 9. Find the side of the least square that can be inscribed in the 
 square whose side is m. Find also the distance of its corners 
 from the corners of the given square. 
 
 Sug. Let x and m — x be the distances from a corner of the inscribed 
 square to the adjacent corners of the given square. It will then be found that 
 the function to be examined is 
 
 2 x 2 - 2 mx + m 2 . 
 
 10. A circular piece of tin is to be utilized for the bottom of the 
 largest possible rectangular box of given depth. Find the dimen- 
 sions of the bottom. 
 
 Solution. As the depth is fixed, the volume of the box will be a maximum 
 when the area of the bottom is a maximum. 
 
 Representing by x and y the half sides of the bottom, the function to be 
 examined is 4 xy. Eliminating y by the relation x 2 + y 2 = r 2 , we have for the 
 function 
 
 
 
 
 4z 
 
 Vr 2 
 
 -X 2 . 
 
 Proceeding a 
 
 A in 
 
 Case III, 
 u 
 
 x 
 
 Art. i 
 == 4x ■ 
 
 ~4 
 
 396, 
 
 we have 
 
 
 Vr 1 - 
 
 -X 2 , 
 
 
 V ±\ 
 
 
 whence 
 
 V4r 4 - w 2 . 
 
 For a maximum or a minimum 
 
 
 
 
 
 
 4 r 4_ 
 
 ■ u 2 = 
 
 :0, 
 
 which gives 
 
 
 
 
 u - 
 
 r 2 
 = 2' 
 
 and 
 
 
 
 
 X - 
 
 r 
 
 ' V2* 
 
 Substituting 
 
 this value of 
 
 x in x 2 
 
 + y- 
 
 ■ = r 2 , we have 
 
 
 
 
 
 y-- 
 
 r 
 
 "vS* 
 
 Therefore the rectangle is a square. 
 
 11. In the last example, if the bottom is to be cut from a 
 semicircle of radius r, what must be its dimensions ? 
 
ZERO, INFIX ITY, AND INDETERMINATE FORMS 255 
 
 12. Find the greatest right triangle that can be constructed 
 upon a given line as a hypotenuse. 
 
 13. A carpenter wishes to make the largest possible rectangular 
 table top from a board 10 feet long, 3 \ feet wide at one end, and 
 1 foot wide at the other. Find the dimensions. 
 
 SECTION II — ZERO, INFINITY, AND INDETERMINATE 
 
 FORMS 
 
 401. The symbol is used not only to represent the absence of 
 value, but also to represent a quantity that is less than any 
 assignable value ; i.e., it may represent either absolute zero or an 
 infinitesimal. * 
 
 Although all infinitesimals are not equal, they are all repre- 
 sented by the same symbol. 
 
 402. The symbol oo, called Infinity, represents a quantity that 
 is greater than any assignable value. 
 
 Although all infinities are not equal, they are all represented 
 by the same symbol. 
 
 403. Prob. To interpret the forms - and — 
 
 Solution. If in the fraction -, x diminishes while a remains 
 
 x 
 
 constant, the quotient increases ; and finally when x becomes less 
 than any assignable value, the quotient becomes greater than any 
 
 assignable value. Hence ^ = oo • 
 
 Again, if in the fraction -, x increases while a remains con- 
 
 x 
 stant, the quotient decreases ; and finally, when x becomes greater 
 than any assignable value, the quotient becomes less than any 
 
 assignable value. Hence — = 0. 
 
 00 
 
 oo 
 
 404. Prob. To interpret the forms - and — , called indeterminate 
 
 forms. 
 
 x 
 Solution. In the fraction -, in which the variables are inde- 
 
 y 
 
256 HIGHER ALGEBRA 
 
 pendent of each other, if x and y diminish until they become less 
 
 than any assignable values, giving — , or increase until they become 
 
 greater than any assignable values, giving — , the fraction is inde- 
 terminate, the quotient is any number whatever, and the condi- 
 tions of the problem from which it results are fulfilled for every 
 value. 
 
 But if x and y are dependent and their relation is known, for 
 
 x 
 example, if y = mx, giving for the fraction — -, then however small 
 
 or however large x may become, the numerator will be — of the 
 
 1 m 
 
 denominator, and the quotient will be — Hence if the form - 
 
 or — results from the presence in both terms of a fraction of a 
 
 factor which reduces to or oo for a particular value of the varia- 
 ble, the value of the fraction may be determined by dividing out 
 this factor before evaluating. 
 
 EXAMPLES CVII 
 Find the values of the following : 
 _ a* + 5 a; — 2" 
 
 il- 
 
 x*-2x + 
 Solution. Substituting 1 for x, we have £ = go (Art. 403). 
 
 ar>-5; 
 
 2. 
 
 x + 6 "| 
 x - 8 J 2 
 
 x> + 2, 
 
 Solution. Substituting 2 for x, the fraction assumes the form £. We 
 therefore seek for a factor common to both terms of the fraction. 
 X 2 _ 5 x + 6 _ ( X - 3) (x - 2) _ x - 3 1 _ _ 1 
 
 + 4J2 6' 
 
 a?_2a; + 4 ~] x + 3 ~| 
 
 s + 1 • **±*±*±11. 
 
 x~^l i X J° 
 
 aj-r-lj-i 5x J 
 
INTERPRETATION OF NEGATIVE RESULTS 257 
 
 1 + 3 x l t 1Q ax~ 2 + bx- 1 + c l # 
 
 3 + 5x] x dx- 2 +ex~ l +fX 
 
 > _ a? - x + 1 1 12 0J 4 -5aj 8 +6jB 8 +4aj-8 
 
 a*-3a> + 2 Ji" ' Sa>*-13x 8 +30ai 2 -28a;+8 - 
 
 SECTION III — DISCUSSION OF FUNCTIONS AND 
 PROBLEMS 
 
 405. The Discussion of a Function or a Problem is interpreting 
 it for different values of the literal quantities which enter it. If 
 only arbitrary and absolute constants enter it, the discussion is 
 for different values of the arbitrary constants. If variables and 
 constants enter it, the discussion is for different values of the 
 variables alone. 
 
 INTERPRETATION OF NEGATIVE RESULTS 
 
 406. When the answer or one of the answers to a problem is 
 negative, it must be reckoned in the opposite direction from that 
 assumed as positive in the statement of the problem. For 
 example, let it be required to find the time when A will be (or 
 was) one and one half times as old as B, A's age now being 28 and 
 B's 20. Suppose we assume that this ratio of their ages will be 
 at some time in the future, say x years from now. By the condi- 
 tions of the problem we have 
 
 28 + * = f(20 + z), 
 
 whence x = — 4. 
 
 As we assumed time in the future +, and stated our equation 
 in accordance with this assumption, the negative result shows that 
 the time is in the past, viz., 4 years ago. 
 
 Had we assumed the event in the past, calling past time +, our 
 equation would have been 
 
 28 - x = f(20 - x), 
 
 whence x = 4 ; 
 
 i.e., 4 years in the direction we assumed as positive, or 4 years. ago. 
 The first answer obtained, — 4, is considered correct in the alge- 
 downey's alg. — 17 
 
258 HIGHER ALGEBRA 
 
 braic sense, the negative of future time being past time. To 
 obtain the correct answer in an arithmetical sense, the words will 
 be in the enunciation of the problem must be changed to was. 
 
 If, as in Ex. 1, page 213, the nature of the problem is such, 
 either as enunciated or with a suitable change of words, as was for 
 will be, loss for gain, decrease for increase, etc., as not to permit an 
 interpretation of a negative result, then the negative answer must 
 be rejected. 
 
 INTERPRETATION OF IMAGINARY RESULTS 
 
 407. As stated in Art. 227, an imaginary answer shows that, 
 arithmetically, the conditions of the problem cannot, even with a 
 change of the wording, be f ulfilled. For example, let it be required 
 to divide 12 into two such parts that the sum of their squares shall 
 be 54. Eepresenting the two parts by x and 12 — x, we have by 
 the conditions 
 
 x 2 + (12 - xf = 54, 
 
 whence x = 6 ± 3 V— 1, 
 
 and 12-ic = 6 T3V^I. 
 
 These imaginary results show that, arithmetically, 12 cannot be 
 divided into two parts the sum of whose squares is 54. If we 
 apply the test of Art. 392, we find that the sum of the squares of 
 the two parts is a minimum when each of the two parts is half of 
 the number. Hence the smallest value of the sum of the squares 
 is 6 2 -f 6 2 = 72. Nevertheless, algebraically, the conditions of the 
 problem are fulfilled by these results, since 
 
 the sum = 6 ± 3 V :=r l + 6 T 3 V :r I = 12, 
 
 the sum of the squares = (6± 3V :r l) 2 + (6 T S^/^lf = 54. 
 
 408. In discussing a function which has no connection with 
 a problem, the following features should usually be determined : 
 
 1. What values of the variable (or arbitrary constants) give 
 but one value to the function. 
 
 2. Between what limits of the variable the function has more 
 than one real value, and whether these values are numerically 
 equal or unequal. 
 
DISCUSSION OF FUNCTIONS 259 
 
 3. Between what limits of the variable the function is imaginary. 
 
 4. What value of the variable makes the function a maximum 
 or a minimum, and the value of the function at a maximum or 
 a minimum. 
 
 In discussing a problem, the above features and others, including 
 negative and indeterminate results, should be interpreted with 
 reference to the particular problem. Of course if the solution 
 involves an equation of only the first degree, there could be no 
 double, imaginary, maxima or minima values. 
 
 EXAMPLES CVIII 
 Discuss the following : 
 1. if = x 4 — x 2 . 
 Discussion. Solving for y, we have 
 
 y =± xy/x' 1 — 1. 
 
 1st. For x = 1 or x = 0, y has but one value, viz., 0. 
 
 2d. For all values of x < — 1 and > + 1, y has two real values, numerically 
 equal, but with opposite signs. 
 
 3d. For all values of x between — 1 and 0, and between and + 1, y is 
 imaginary. 
 
 4th. Of all the negative values of x, — 1 is the greatest, a maximum ; and 
 of all positive values of x, + 1 is the least, a minimum. 
 
 2. 2 ay 2 = x 3 -f- xf. 
 
 Discussion. Solving for y, we have 
 
 v = ± xV * 
 
 V2a — x 
 
 1st. For x = 0, y has but one value, viz. , 0. 
 
 2d. For x > and < 2 a, y has two real values, numerically equal, but 
 with opposite signs. 
 
 3d. For x > 2 a or < 0, y is imaginary. 
 
 4th. x = is a minimum, and x = 2 a is a maximum. 
 
 5th. For x = 2 a, y = <x>. 
 
 3. 3 x 2 + if + 4 xy - 12 x - 8 y + 21 = 0. 
 Discussion. Solving for y, we have 
 
 if- + (4 x - 8) y = - 3 x 2 + 12 x - 21, 
 or V = — 2x-f4± Vx' 2 — 4 x — 5. 
 
260 HIGHER ALGEBRA 
 
 1st. When x 2 — 4 x — 5 = 0, which gives x = 5 or — 1, y has but one 
 value, viz., — 6 for the former and 6 for the latter. 
 
 2d. When x 2 — 4 x — 5 >0, i.e., when it is positive, y has two real, un- 
 equal values. This gives, by solving the inequality, sc>5 or <— 1. 
 
 3d. When x 2 — 4x — 5<0, i.e., when it is negative, which gives x<5 
 and >— 1, y is imaginary. 
 
 4th. Since any values of x between — 1 and 5 give imaginary values for 
 y, while any values of x < — 1 or > 5 give real values for y, — 1 is a maxi- 
 mum value of x, and 5 is a minimum value of x. 
 
 a' — a 
 
 1 + aa' 
 Discussion. 1st. When a'>a, y is + , and when a' <.a, y is — . 
 
 2d. When a' = a, y = — — , = 0. 
 1 + a 2 
 
 a 
 
 3d. When 1 + aa' — 0, or a' — , y = — rr — = co. 
 
 5. xy- = ±a 2 (2a-x). 6. (x-l)y 2 = x 2 . 
 
 7. x 2 y 2 = a 2 (x 2 -f- 2/ 2 ). 8. x 2 y = 4 a 2 (2 a — ?/). 
 
 9. if = a 3 — x 2 . 10. ?/ 2 — 2 #2/ -f- a? 2 — 4 ?/ 4- x 4- 4 = 0. 
 
 11. Divide a into two parts whose product shall be p. 
 Solution and Discussion. Representing the two parts by x and a — x, 
 
 we have x (a — x) = p, 
 
 whence x = | ± \\/a 2 — 4p, 
 
 and a — x = - T I vV 2 — 4p. 
 
 m 
 
 1st. For 4p = a 2 , oj has but one value, viz., -. 
 
 2i 
 
 2d. For 4 j9 < a 2 , x has two real, unequal values. 
 
 3d. For 4p>a 2 , x is imaginary, showing an arithmetical inconsistency, 
 
 viz., naming a product larger than any parts of the given number will give. 
 
 4th. From a 2 — 4 p = 0, we have (Art. 393) p at a maximum = — , which 
 
 4 
 gives x = -; i.e., to give a maximum product each of the two parts must be 
 
 A 
 
 half of the given number. 
 
 12. Two couriers, A at rate a and B at rate b, are traveling on 
 an east and west road. At noon A is at M and B at N, c miles 
 apart. Find the time and the place of their being together. 
 
DISCUSSION OF FUNCTIONS 261 
 
 Solution and Discussion. Let x be the number of hours between noon 
 and the instant of their being together. Let time after noon and distance to 
 the east of M be regarded as positive. Then 
 
 ax = bx + c, 
 
 whence x = — - — = the time from noon, 
 
 a — b 
 
 and ax = ac = the distance from 31. 
 
 a — b 
 
 1st. If a, 6, and c are positive and a > 6, — - — . and ac are both -t- , 
 
 a — b a — b 
 
 showing that the couriers will be together after noon and to the east of 31. 
 
 The place is also east of N, since ac > c. 
 
 a — b 
 2d. If a, by and c are positive and a<b, — — and — — - are both -, 
 
 a — b a- b 
 
 showing that the couriers were together before noon and to the west of M. 
 
 3d. If a, b, and c are positive and a — 6, — - — and aG both become 
 
 a — b a — b 
 
 oo, showing that the couriers never have been and never will be together. 
 Their rates being the same, the distance between them is always c. 
 
 4th. If c = and a ^ 6, — - — and ac are both 0, showing that the 
 ^ a—b a—b 
 
 couriers are together at noon, Jfand iV now coinciding. 
 
 5th. If c = and a = fc, — ^— and ao both become — As numerators 
 a— b a — b 
 
 and denominators reduce to by two independent conditions, viz., c = and 
 a = 6, the values of the fractions are indeterminate (Art. 404), and represent 
 any values, showing that the couriers are together all the time. 
 
 Cth. If a and c are positive and b negative, the time becomes — - — and 
 
 a -\-b 
 the distance — - — Since these are both -}-, the couriers will be together 
 
 a + b 
 after noon and to the east of M. The place is between M and N, since 
 
 ac <c. B'a negative rate means that he is traveling to the west. 
 a + b 
 
 7th. If b and c are positive and a negative, the time becomes — - 
 
 _ —a — b 
 
 and the distance ^— , or — — — Since the time is — and the distance 
 
 —a—b a+b 
 
 -f, the couriers were together before noon and to the east of 31. 
 
 8th. If a and c are positive and b = 0, the time becomes - and the dis- 
 
 a 
 tance c. Hence the couriers will be together after noon and at c miles east of 
 M. The condition 6=0 means that B remains at N. 
 
 9th. If b and c are positive and a = 0, the time becomes - and the distance 
 
 b 
 0. Hence the couriers were together before noon and at M. 
 If c w, re made negative, it would locate N west of M. 
 
262 HIGHER ALGEBRA 
 
 Thus all the features of the problem are correctly revealed by the formulas 
 for the time and the distance, the analytic results agreeing with the conditions. 
 
 13. Divide a into two parts the difference of whose squares shall 
 be cl. 
 
 What is the character of the parts when d > a 1 ? Why has d neither a 
 maximum nor a minimum value ? 
 
 14. Two couriers, A and B, are traveling the same road, the 
 former at rate a, and the latter n times as fast. They are at two 
 places c miles apart at the same time. Find the time and the 
 place of their being together. 
 
 15. A cistern can be filled by one pipe in m minutes and by 
 another in n minutes, and can be emptied by a third in p minutes. 
 In what time will it be filled if all are left open at once ? 
 
 16. A and B, traveling the same road, were at two towns m 
 miles apart at the same time. On coming together it was found 
 that A, the faster traveler, had gone n miles, and that their time 
 on the road was equal to the difference of their rates. Find their 
 rates. 
 
 17. The loudness of one church bell is m times that of another. 
 If the amount of sound varies directly as the loudness and 
 inversely as the square of the distance, where on the line of 
 the two will the bells be equally well heard, the distance between 
 them being a ? 
 
CHAPTER XIX 
 DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 
 
 409. The Differential of a Variable or Function (Arts. 380 and 
 382) is its rate of change. 
 
 If the change is uniform, the rate of change is the same as the 
 actual change during the unit of time; but if the change is not 
 uniform, the rate of change at any instant is the change that 
 would take place during the next unit of time if it continued 
 uniform at what it is at that instant. 
 
 Thus, we say that a body falling from rest has at the end of the first 
 second a velocity of 32£ feet. By this we do not mean that it falls only 32£ 
 feet during the next second, but that if it continued to fall uniformly with the 
 velocity it has at that instant, it would fall 32£ feet during the next second. 
 The distance from the falling point is variable and the differential of that 
 variable at the end of the first second is 32£ feet. 
 
 410. The differential of a variable is indicated by writing d 
 before it. Thus, the differential of x is written dx, and is read 
 " differential a,*." 
 
 411. If a variable or function is increasing, its differential is 
 positive ; if decreasing, negative. 
 
 412. Theorem. Constant terms disappear in differentiating. 
 
 For since a constant admits of no change, the differential of 
 a constant is 0. 
 
 413. Theorem. The differential of the product of a constant and 
 a variable is the constant multiplied by the differential of the variable. 
 
 Dem. Let the function be ax. (1) 
 
 Whether x changes uniformly or not, let dx be the amount by 
 
 which it ivoidd increase in a unit of time if, at any instant, the 
 
 change should become uniform. Then the state of the function 
 
 at the end of the unit of time would be 
 
 a(x + dx) = ax -f adx. (2) 
 
 263 
 
264 HIGHER ALGEBRA 
 
 The difference between the state of the function at the begin- 
 ning of a unit of time and what, with a uniform change, it would 
 be at the end of the unit of time is the differential of the function 
 (Art. 409). Hence, subtracting (1) from (2), we have 
 
 d(ax) = adx. 
 
 414. Theorem. The differential of a polynomial is the algebraic 
 sum of the differentials of its terms. 
 
 Dem. Let the function be x -f y — z. (1) 
 
 Whether x, y, and z change uniformly or not, let dx, dy, and dz 
 
 be the respective amounts by which they would increase in a unit 
 
 of time if, at any instant, the change should become uniform. 
 
 Then the state of the function at the end of the unit of time 
 
 would be 
 
 x +dx + y + dy-(z + dz). (2) 
 
 The difference between the state of a function at the beginning 
 of a unit of time and what, with a uniform change, it would be 
 at the end of the unit of time is the differential of the function 
 (Art. 409). Hence, subtracting (1) from (2), we have 
 
 d (x + y — z) = dx -f dy — dz. 
 
 415. Theorem. The differential of the product of two variables is 
 the sum of the products of each into the differential of the other. 
 
 Dem. Let the function be xy. 
 
 Whether x and y change uniformly or not, let dx and dy be the 
 respective amounts by which they would increase in a unit of 
 time if, at any instant, the changes should become uniform. 
 
 Now if x alone were to change, the change in the product would 
 be ydx (Art. 413), and if y alone were to change, the change in the 
 product would be xdy (Art. 413), while if both x and y change, 
 the change in the product at a uniform rate would be the sum of 
 the changes due to these two causes, or ydx + xdy. Hence, 
 
 d(xy) = ydx 4- xdy* 
 
 * While this is not a rigorous demonstration, it is sufficient to show the 
 truth of the theorem. 
 
DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 265 
 
 416. Theorem. TJie differential of the product of several variables 
 is the sum of the products of the differential of each into all the others. 
 
 Dem. Let the function be xyz. 
 Let v =xy. Then xyz = vz. 
 
 By Art. 415, d(vz) = zdv -f vdz, (1) 
 
 and dv = ydx -f xdy. (2) 
 
 Substituting in (1) the values of v and dv, we have 
 
 d(xyz) = z(i/c?;c -|- ax%) -f #ydz, 
 
 = yzda? -f xzdy -f xydz. 
 
 The same method of reasoning will apply to any number of 
 variables. 
 
 417. Theorem. The differential of a fraction ivith variable nu- 
 merator and denominator is the denominator into the differential of 
 the numerator minus the numerator into the differential of the 
 denominator, divided by the square of the denominator. 
 
 Dem. Let the function be -, and represent this function by u, 
 
 giving y 
 
 x 
 u = — 
 
 y 
 
 Clearing of fractions, 
 
 yu = x. 
 
 Pifferentiating by Art. 415, 
 
 ydu + udy = dx, 
 
 . , dx dy 
 
 , dx — udy y ydx — xdy 
 whence du = - = — = - ^ — -• 
 
 y y y 
 
 418. Cor. i. Tfie differential of a fraction whose numerator is 
 constant is minus the numerator into the differential of the denomi- 
 nator, divided by the square of the denominator. 
 
 For if x = a, a constant, then da = 0, and dl -)= ~- 
 
 \yj y 
 
266 HIGHER ALGEBRA 
 
 419. Cor. 2. The differential of a fraction ivith a constant 
 denominator is the differential of the numerator divided by the 
 denominator. 
 
 For - = -x, and - is a constant factor: therefore, by Art. 413, 
 b b ' b J 
 
 d( -)=:d( -x\=-dx. 
 
 b J b 
 
 b 
 
 420. Theorem. The differential of a variable having a cojxstant 
 exponent is the product of the exponent, the variable with its expo- 
 nent diminished by one, and the differential of the variable. 
 
 Dem. Let the function be x n . 
 
 1st. When n is a positive integer. 
 
 Now x n = x-x* X'"to n factors. 
 
 Hence, by Art. 416, 
 
 d(x n ) = (x • x • • • to n — 1 factors) dx 
 + (x • x • • • to n — 1 factors) dx -\ to n terms, 
 
 = x n ~ l dx + a;" -1 dx-\ to n terms, 
 
 = nx n ~ x dx. 
 2d. When n is a positive fraction, as — 
 
 Let 
 
 p 
 y = x«. 
 
 By involution, 
 
 y q = x p . 
 
 By 1st case, 
 
 qy*- 1 dy = px p ~ x dx, 
 
 , , px*' 1 dx px*- 1 dx p p -\ 
 whence dy = i — =- = — x 9 dx. 
 
 3d. When n is negative. 
 
 Now x~ n == — 
 
 Differentiating by Art 418 and 1st case, 
 
 (,-)=^(i): 
 
DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 267 
 
 421. Cor. The differential of the square root of a variable is the 
 differential of the variable divided by twice the square root of the 
 variable. 
 
 For d hjx) = d(xh=l af* dx = -i^L. 
 
 EXAMPLES CIX 
 
 Differentiate the following : 
 
 1. y = 2x 3 -4:x 2 + 5x. 
 Operation. By Art. 414, 
 
 dy = d(2 x 3 ) - d(4 x 2 ) + d(5 x) 
 
 = 3-2 xHx - 2 • 4 xdx + 5 dx, by Arts. 413 and 420. 
 = (6x 2 -8x + 5)dx. 
 
 This means that y, or 2 x 8 — 4 x 2 -f 5 x, -changes 6 x 2 — 8 a; + 5 times as 
 fast as x. Hence, 
 
 when x = 0, y changes 5 times as fast as x ; 
 when x = 1, y changes 3 times as fast as x ; 
 when x = 2, y changes 13 times as fast as x ; 
 when x = 3, y changes 35 times as fast as x. 
 
 2. y = x? + 5x*-Q>x + T. 3. y = x- s -4x 2 + 25. 
 
 4. y = 7x 4 + 3x 2 -6xl. 5. ?/ = 2^ - oar 4 - 5r* + 9. 
 
 6. ?/ = 3(5-f-4ar } ) 5 . 
 
 Operation. Treating the part in the parenthesis as the variable, we have 
 dy = 5 . 3(5 + 4 x 8 )*3 • 4 x 2 dx = 180(5 + 4 x 3 )Wx. 
 
 7. y = 5 (2 + 3 a,- 2 ) 4 . 8. y = 4 (1 - 5 a? 2 ) 3 . 
 9. ?y = (a + ^)i 10. 2/ = 2(3-ar J )" 2 . 
 
 11. 7* as X 2 !/*. 
 
 Operation. By Art. 415, 
 
 du = y 3 <7(x 2 ) +x*d(y*) = 2 xy 3 dx + 3 x 2 J/ 2 (7y. 
 
 12. w = afy*. 13. w = ar(.y 2 -3). 
 14. u = (x — 1) (f + 2). 15. M = aj(a?-|-a)i 
 
 16. 2/=— t— 
 
268 HIGHER ALGEBRA 
 
 Operation. By Art. 417, 
 
 . _ x*d(x* - 3) - O 2 - 3)d(x 3 ) 
 
 
 y — 
 
 2 a;4^ _ 
 
 - 3 z 4 <fcc + 9 xHx _ 9 - z 2 ^ 
 
 17 V — 
 
 421, 
 
 X? X* 
 
 18. tt = -. 
 
 2/ 2 
 
 ' f (1 + a^) 3 
 
 19. y=V3H-5#. 
 
 Operation. By Art. 
 
 • 
 
 2a/3 + 5x 3 2V3 + 5x : 
 
 20. ?/ = VI + a?. 21. y = 4 V2 x — 3 3. 
 
 422. An Equicrescent Variable is one which changes uniformly, 
 i.e., one which has a constant rate. 
 
 423. A Second Differential is a differential of a differential, i.e., 
 the rate at which the rate is changing. 
 
 The differential of dy is written d 2 y, and is read, " second dif- 
 ferential y " ; while dx x dx, or the square of dx, is written dx 2 . 
 
 From the function y = ax 3 , 
 
 we have dy = 3 ax^dx = (3 adit) x 2 . 
 
 If we regard x as an equicrescent variable, dx is constant ; but 
 dy is variable, since it is equal to an expression containing a vari- 
 able x. 
 
 Hence we have 
 
 d (dy) = d 2 y = d [(3 adx) x 2 ^ = 3 adx x 2 #cZ# = 6 axdx 2 . 
 
 424. A Third Differential is the differential of the second dif- 
 ferential, and so on. 
 
 425. The First Derivative, or the First Differential Coefficient, 
 
 is the ratio of the differential of the function to the differential 
 
 of the variable, and is represented by — , or f'(x). 
 
 (XX 
 
DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 269 
 
 Thus, from y = x :i + 3 x 2 — 5 x, 
 
 we have dy = (3 x 2 + 6 x — 5)dx, 
 
 and ^ = 3 x 2 + 6 x - 5, 
 
 dx 
 
 in which 3 x 2 + 6 x — 5 is the first derivative. 
 The same thing is expressed by writing 
 
 /(x) = x 3 + 3 x 2 - 5 x, 
 
 and f'(x) = 3 x 2 + 6 x - 5. 
 
 426. The Second Derivative, or the Second Differential Coefficient, is 
 the ratio of the second differential of the function to the square of 
 
 the differential of the variable, and is represented by — % or f"(x). 
 
 427. The Third Derivative, or the Third Differential Coefficient, is 
 the ratio of the third differential of the function to the cube of 
 
 d 3 y 
 the differential of the variable, and is represented by — ^, or 
 
 f"\x), and so on. d ^ 
 
 EXAMPLES CX 
 
 Find the first derivative of each of the following : 
 
 1. ?/ = 2 + 3#-4ar J . 
 
 Operation. dy = 3 dx - 12 x 2 dx, 
 
 whence ^ = 3 - 12 x 2 . 
 
 dx 
 
 2. ?/ = ar 5 -3ar 2 + 4. 3. y = (x 3 - 5) 3 . 
 
 4. y = Var* — 2. 5. y = xr -f x -f ar 3 . 
 
 6. y = m • 7. y= m ■ 
 
 (1 + *) 2 F (* + a) 4 
 
 Find the second derivative of each of the following : 
 
 a 2/ = o(3 + 4z) 3 . 
 Operation. /'(x) = 60(3 + 4 x) 2 , 
 
 and /"(x) = 480(3 + 4x). 
 
 9. y = x 2 -6x + 7. 10. 2/ = 3» 4 -5^ + 10. 
 
 11. y = 3(2 + ox)\ 12. ?/ = (a + z)- 3 . 
 
 13. y =*_**-. 14. y = 5(2x 2 -3) 4 . 
 
270 HIGHER ALGEBRA 
 
 Find the first four successive derivatives of each of the 
 following : 
 
 15. y = (a + x)- 2 . 16. y = ^-- 
 
 a-\- x 
 
 17. y = 3(l + xy. 18. y = 3x i -lx i +3x 2 + 6x-7. 
 
 19. y = 2(5 — x)K 
 
 20. y = A + Bx + Car 9 -f- Dar 3 + j£af\ 
 
 428. A Partial Differential of a function of more than one variable 
 is its differential on the hypothesis that all the variables but one 
 become constant. 
 
 Thus, from u = 3 xy + 5 x — 4 y + 7, 
 
 we have d x u = 3 y&c -f 5 dx, 
 
 and c^w = 3 xdy — 4 dy, 
 
 the subscripts indicating with reference to which variable we have differ- 
 entiated. 
 
 429. A Partial Derivative, or a Partial Differential Coefficient, is 
 the ratio of the partial differential of the function to the dif- 
 ferential of that variable with reference to which we have 
 differentiated. 
 
 Thus, from the above function we have 
 
 — = 3 y + 5, 
 dx 
 
 and ^ = 3 x - 4, 
 
 dy 
 
 the denominators indicating with reference to which variable we have dif- 
 ferentiated. 
 
 430. Theorem. In an algebraic function of the sum of two vari- 
 ables, x and y, the first, second, third, etc., partial derivatives with 
 reference to x are equal respectively to the first, second, third, etc., 
 partial derivatives ivith reference to y.* 
 
 * This is true for any function of the sum of two variables, but, it is 
 proved here only for algebraic functions. 
 
DIFFERENTIATION OF ALGEBRAIC FUNCTIONS 271 
 
 Dem. All algebraic functions of the sum of two variables are 
 included in the form 
 
 u = m(x + y) n , 
 
 in which m and n are positive or negative, integral or fractional. 
 
 Now d x u — mn (x + y) n ~ l dx, 
 
 and d y ii = mn (x + y) n ~ l dy, 
 
 which forms differ only in the factors dx and dy ; but in forming 
 the derivatives these factors are divided out, giving 
 
 da du / . n„_, 
 
 — = — = mn (x -f «) . 
 dx dy K ^ y) 
 
 This is a new function of the sum of two variables, and, as 
 just shown, its first partial derivatives are equal; but these are 
 the second partial derivatives of the original function, giving 
 
 Similarly, 
 
 431. Let the student form the first three successive partial 
 derivatives of u = 2 (x + y) z , u = 6 x 4 + 3 y 3 , u = 5 (x -f- y)~' 2 , 
 u = -yjx + y, u = 5 (x — y) 7 , u = (x 2 + 3 y) 4 , u = (x -f y) *, and 
 observe that the law stated in the theorem holds for all straight 
 functions of the sum of two variables, while it does not for other 
 functions ; though in a straight function of the difference of two 
 variables the corresponding partial derivatives are equal except 
 in sign. 
 
 dru _ dru 
 dx 2 dy 2 
 
 
 d?u _ cPu d*u _ d 4 u 
 dx 3 dy* dx* dy 4 
 
 d n u _ d n u 
 dx n dy n 
 
CHAPTER XX 
 DEVELOPMENT OF FUNCTIONS 
 
 432. A Finite Series is a series (Art. 312) which terminates. 
 
 433. An Infinite Series is a series which does not terminate, but 
 has an endless succession of terms. 
 
 434. An infinite series is Convergent when, by taking more and 
 more terms, the sum approaches a finite limit; it is Divergent 
 when the sum does not approach a finite limit ; and it is Oscillating, 
 or Neutral, when the sum, though finite, does not approach a 
 determinate limit. 
 
 Thus, 1 + x + x 2 + x 3 + x* -f x 5 + ••• 
 
 is convergent when x is numerically less than 1, approaching the limit of 
 
 (Art. 329) ; divergent when x is 1 or greater than 1 ; oscillating when x is — 1, 
 the sum being or 1 according as the number of terms is even or odd. 
 
 435. Note. Convergent, when applied to a series, is not synony- 
 mous with decreasing or descending (Art. 313). While every 
 convergent series is decreasing, not every decreasing series is con- 
 vergent. Thus, 1, i, J, J, i, ••• is decreasing, but is not convergent, 
 as shown in Art. 583 (3). 
 
 436. To Develop or Expand a Function is to find a series whose 
 sum is equal to the function. The value of a developed function, 
 therefore, is either the sum of a finite series, or the limit of the 
 sum of an infinite convergent series. 
 
 437. There are many ways of developing functions. The stu- 
 dent is already familiar with development by division, involution, 
 and evolution. 
 
 Thus, by actual division, 
 
 — — =1 -x + x^^ + ^-x 6 , 
 1 4- x 
 
 272 
 
DEVELOPMENT OF FUNCTIONS 273 
 
 a finite series and equal to the function for all values of x ; by the binomial 
 theorem (Art. 155), 
 
 (1 + x) 5 = 1 + 5 x + 10a; 2 + 10 x* + 5x* + x 5 , 
 
 a finite series and equal to the function for all values of x ; and by extracting 
 the square root of 1 — x 2 we have 
 
 vrr^i-^-* 4 -* -..., 
 
 2 8 lo 
 
 an infinite series and true only for values of x numerically equal to or less 
 than 1, being divergent for all other values. 
 
 438. It often occurs, as in the last example, that a development 
 is true for certain values of the variable, but- not for all values. 
 
 Thus, by division, 
 
 = l+z + £ 2 +a; 3 + £ 4 + .... 
 
 1-x 
 
 Now when x is a proper fraction, either positive or negative, the series is 
 convergent, and, if we treat it as an infinite, decreasing geometrical progres- 
 sion, the formula 
 
 S = -2- (Art. 329) gives 
 
 1 - r ' " 1 - x 
 
 and the series is seen to be equal to the function. 
 
 When x is greater than 1, the function (the first member) is a negative 
 fraction, while the series (the second member) is positive and departs farther 
 and farther from the value of the function as more and more terms are taken. 
 
 When x is numerically greater than 1, but negative, the function is a posi- 
 tive fraction, while the series is positive or negative according as the number 
 of terms is odd or even, and departs farther and farther from the value of 
 the function as more and more terms are taken. 
 
 When 35=— 1, the function is £, while the series is 1 — 1 + 1 — 1 + 1-1+ •••, 
 which is or 1 according as the number of terms is even or odd. 
 
 When as.ss 1, both the function and the series become oo. 
 
 If the order of the terms in the divisor be reversed, we have 
 
 1 1111 
 
 - X + 1 X X 2 X s X* 
 
 and the series equals the function for values of x numerically greater than 1, 
 but not for values numerically less than 1. 
 
 439. The subject of Convergence/ is reserved for a subsequent 
 part of the work, and we shall in this chapter concern ourselves 
 only with the Development. 
 Downey's alg. — 18 
 
274 HIGHER ALGEBRA 
 
 SECTION I — INDETERMINATE COEFFICIENTS 
 
 440. Indeterminate (or Undetermined) Coefficients are assumed 
 constant coefficients whose values, not known at the outset, are 
 to be determined in the course of the demonstration of a theorem 
 or the solution of a problem. 
 
 441. Theorem of Indeterminate Coefficients. If tico series con- 
 taining the same variable are equal for all values of the variable that 
 render both series convergent, the coefficients of the same poivers of 
 the variable in the two series are equal. 
 
 Dem. Let 
 
 A + Bx + Cx 2 -f Dx> + etc. = A' -f B'x + C'x 2 + D'x* + etc. 
 
 for all values of x which render each series either finite or infinite 
 and convergent ; then A = A ', B = B', C= C, D = I)', etc. 
 
 Since the equation is true for all values of x which render each 
 series either finite or infinite and convergent, it is true when 
 x = 0. But this gives A = A'; and since A and A' are constant, 
 they have the same values, whatever the value assigned to x, so 
 long as each series is either finite or infinite and convergent. 
 
 Dropping these equals from the two members and dividing by x, 
 
 B+Cx + Dx 2 + etc. = B' + C'x + D'x 2 + etc. 
 
 As before, making x = 0, we have B = B'. Proceeding in the 
 same way, we have C = C, D = D\ etc. 
 
 Cor. If A -f Bx + Cx 2 -f Dx 3 -f etc. = for all values of x, 
 each of the coefficients A, B, C, etc., is 0. 
 
 For we may write 
 
 A + Bx + Cx 2 + Dx 3 + etc = + 0^ + 0^ + 0.^ + etc., 
 whence, by the theorem, A — 0, B = 0, (7=0, etc. 
 
 442. The theorem of Art. 441 gives a method, called the Method 
 of Indeterminate (or Undetermined) Coefficients, of developing 
 a function into a series. It consists in assuming a series in 
 ascending powers of the variable with unknown coefficients, and 
 then finding the values of these coefficients by equating those of 
 the same powers of the variable. 
 
INDETERMINATE COEFFICIENTS 
 
 275 
 
 Thus, let it be required to develop 
 
 2x 
 
 Solution. Assume 
 l-3x-a 
 
 1 - 2 x - x* 
 Clearing of fractions, 
 
 1-% X - X 2 = A + j5 
 -2A 
 
 = A + Bx + Cx' 2 + Dx* + .ETx 4 + etc. 
 
 -2B 
 
 - A 
 
 x 2 + D 
 -20 
 - B 
 
 £ 3 + # 
 
 -2D 
 
 - C 
 
 x* + etc. 
 
 Equating the coefficients of the same powers of x (Art. 441), 
 A = l; 
 
 B-2A = -3, whence B = -3 + 2A = -l; 
 C-2B- A = -l, whence C = -l + 2B + A=-2 
 D - 2 C — B = 0, whence D = 2C+B = -b; 
 E-2D-C = 0, whence E = 2D+ C = -12. 
 
 Substituting these values in the assumed series, we have 
 
 l-3x 
 
 = 1 -X-2X*- 5x 3 -12x l 
 
 etc. 
 
 1 _ 2 x - x 2 
 This can readily be verified by actual division. 
 
 443. In developing by the Method of Indeterminate Coefficients, 
 the Law of the Series becomes evident early in the progress of the 
 work, and then as many more coefficients as desired may be 
 obtained by simply combining, according to the discovered law, 
 the coefficients already found. 
 
 Thus, in the example of the last article, we have 
 
 D = 2C+B, 
 
 E = 2 D + C. 
 It is therefore evident that 
 
 F = 2 E + D, 
 
 G = 2F+ E, 
 
 and so on, each coefficient after the third being twice the preceding plus the 
 second preceding. 
 
 In all cases in developing rational fractions the law of the series 
 appears in all the equations beyond the one ichose second member is 
 the coefficient of the highest power of the variable in the numerator 
 
276 HIGHER ALGEBRA 
 
 of the given fraction; for beyond this the first members all have 
 the same form, and the second members are all 0. 
 
 The student should in each example note the law of the series. 
 
 EXAMPLES CXI 
 
 Expand by the Method of Indeterminate Coefficients the 
 following : 
 
 t l+2a; 2 2-3s 3 1+3 
 
 1 — X — X 2 1 -f cs + x 2 
 
 4. l±* 5. 5 + 2x . 6. 
 
 l + 2x-\-3x 2 1-5X + X 2 
 
 2-Sx 2 -hx i 1 + x + x 2 
 
 ' % _ x 1 4. as? " 1 + ar* " 2 - a - a 2 
 
 444. In developing a rational fraction we must assume a series 
 such that, after clearing of fractions, the second member shall 
 contain all the powers of the variable that are found in the first 
 member, which is the numerator of the given fraction Other- 
 wise it would be necessary to equate given finite coefficients with 
 0, which would be absurd. The exponent with which to begin 
 the assumed series may be determined by noting what actual 
 division would give for the first, or lowest, exponent. 
 
 1 — x 
 In developing — — we assume 
 
 — X -f- O #/ 
 
 Ax~ 2 + 3x^+0 + Dx + Ex 2 + Fx 3 + etc. ; 
 
 1-x+x 2 
 
 l + 2x 
 
 1-3^ 
 
 \ + 2x 
 
 2^ + 3^ 
 
 1/ 1 — x 
 or, if we prefer, we may write the expression in the form — f - — — 
 
 X" \li -\~ o X J 
 
 expand the part within the parenthesis by the use of the series 
 beginning with an absolute term, and then multiply each term 
 
 If in any case we inadvertently assume a wrong series, the 
 fact will appear by the absurd results obtained when the coeffi- 
 cients of the same powers of the variable are equated. If in this 
 example we should assume the series 
 
 A + Bx 4- Cx 2 + Dx* + Ex A + etc., 
 
INDETERMINATE COEFFICIENTS 277 
 
 clear of fractions, and equate the coefficients of the same powers 
 
 of x, we would have 1 = 0, —1 = 0, etc., which are absurd results. 
 
 2 x 2 — 3 X s 
 
 In expanding such a fraction as — ■ — , no absurdity would 
 
 J. — x 
 
 result in assuming the series A + Bx + Cx 2 + Dot? + Ex 4 -f etc. ; 
 but, inasmuch as A and 2? would in that case be equated with 0, 
 it is shorter to assume the series Ax 2 + Bx 3 -f Cx 4 + Dx 5 -f etc. ; 
 or, if we prefer, we may write the expression in the form 
 
 — — - ), expand the part within the parenthesis by the use 
 x — x j 
 
 of the usual series, and then multiply each term by x 2 . 
 
 EXAMPLES CXII 
 
 Expand by the Method of Indeterminate Coefficients the fol- 
 lowing : 
 
 2 2 l + x-x 2 „ 2a? 
 
 • 3^-4^ x-2x 2 + 3x 3 3-20T 2 
 
 1-2^-^ _ x-Srf-x* c 3-2x + x* 
 
 4. 5. 
 
 a^ + ^-x 4 \-2x-x 2 2x 3 -x 4 -2x 6 
 
 445. The Method of Indeterminate Coefficients may also be 
 employed for expanding a radical. If the quantity under the 
 radical sign is a perfect power of the degree of the index, the 
 expansion will give the root. If the quantity is a binomial, 
 the binomial theorem gives a more expeditious method of expan- 
 sion. 
 
 EXAMPLES CXIII 
 
 Expand by the Method of Indeterminate Coefficients the 
 following : 
 
 L Vl-2a; + 3^. 
 Solution. Assume 
 
 Vl - 2 x + 3 x* = A + Bx + Cx 2 + Dx* + Ex 4 + etc. 
 Squaring both members, we have 
 l-2x + S x 2 = A 2 + 2 ABx + B 2 x 2 + 2 AD x? + C 2 x* + etc. 
 
 x + B 2 
 
 x 2 + 2AD 
 
 2?+ C 2 
 
 + 2 AC 
 
 + 2BC 
 
 + 2AE 
 + 2BD 
 
278 HIGHER ALGEBRA 
 
 Equating the coefficients of the same powers of x (Art. 441), 
 A 2 = 1, whence ^4 = 1 (using only the plus sign) ; 
 
 2 AB = - 2, whence B = — = - 1 ; 
 • A 
 
 2 AC + B 2 = 3, whence (7 = 8 ~ B * = 1 ; 
 
 2^4 
 
 ~RC 1 
 
 2AD+2BC = 0, whence 2) = - — = 1 ; 
 
 ^4 
 
 2 JE 1 + C 2 + 2 BD = 0, whence # = ~ C ' 2 ~ 2 ^ = -» 
 
 2.4 2 
 
 Substituting these values in the assumed series, we have 
 
 VT — 2 x + 3 x 2 s= 1 — * -r -*r -r ■*>- - 
 This can readily be verified by extracting in the usual way the square root 
 of 1 - 2 x + 3 x 2 . 
 
 2. Vl + x + jc 2 . 3. Vl + x — x 1 . 4. V9 -f- x — 3 x 2 . 
 
 5. V4-4a + 13a 2 -6ar 5 + 9a; 4 . 6. VI + * +«*. 
 
 DECOMPOSITION OF FRACTIONS 
 
 446. We have seen (Art. 136) that several fractions may be 
 united into one fraction whose denominator is the lowest common 
 multiple of all the denominators. The converse operation of 
 separating a fraction into partial fractions is sometimes neces- 
 sary, especially in the operations of the Integral Calculus. The 
 Method of Indeterminate Coefficients furnishes a means of mak- 
 ing this separation. 
 
 If the numerator of the fraction to be separated is of higher 
 degree than the denominator, it should, by division, be made 
 lower. Thus, 
 
 2s s + 3s 2 -4a;-f 2 = . 8x-5 § 
 
 0?--2*;+l V-2*+l" 
 
 Case I 
 
 447. TT/ten ^e denominator is resolvable into equal or unequal 
 factors of the first degree. 
 
 Rule. Assume the given fraction equal to several fractioyis with 
 undetermined numerators, and whose denominators are all of the 
 divisors of the denominator of the given fraction. 
 
DECOMPOSITION OF FRACTIONS 279 
 
 Clear the equation of fractions and collect terms. 
 Equate the coefficients of the same powers of the variable, and 
 determine the values of the assumed numerators. 
 Substitute these values in the assumed fractions. 
 
 Dem. Since the given fraction is the sum of the partial frac- 
 tions, each denominator must be a divisor of the given denomi- 
 nator; and since any divisor is a possible denominator, all the 
 divisors must be used. 
 
 The numerator of an assumed fraction must contain only the 
 undetermined constant, since otherwise the assumed fraction 
 
 Aw J\a 
 
 would be capable of farther separation ; thus, = A + 
 
 x — a x — a 
 
 After clearing the equation of fractions, the coefficients of the 
 same powers of the variable are, by Art. 441, equal. The equa- 
 tions thus formed will furnish the values of the assumed 
 numerators. 
 
 EXAMPLES CXIV 
 
 Decompose the following into partial fractions : 
 
 ^ 3s* + 7a;-4 
 
 Solution. As the divisors of the denominator are x, x — 1, and x + 2, 
 
 we assume 
 
 3x 2 + 7x- 4 = A B , G 
 
 x* -\- x 2 -2x x x-l x + 2 
 Clearing of fractions and collecting terms, 
 
 3x' i + 1x-4=(A+ B + C)x' 2 +(A + 2B- C)x-2A, 
 
 whence (Art. 441) 
 
 A+ J5+ = 3, 
 
 A + 2B-C=1, 
 
 -2A = -4. 
 From these equations we have 
 
 ^4 = 2, B = 2, and C = - 1. 
 
 Substituting these values in the assumed partial fractions, we have 
 
 3 x 2 + 7 x - 4 _ 2 2 1 
 
 x* + x 2 -2x x x-l x + 2 
 
280 HIGHER ALGEBEA 
 
 3^-7^ + 6 
 (x - l) 3 
 
 Solution. As the divisors of the denominator are (x — l) 3 , (x — l) 2 , 
 and x — 1, we assume 
 
 3x 2 -7x+6_ A B C 
 
 (x-l) 3 (x-l) 3 (x-l) 2 x-l 
 
 Clearing of fractions and collecting terms, 
 
 3 x 2 - 7 x + 6 = Cx 2 + (B - 2 C)x + .4 - B + C, 
 
 whence (Art. 441) 
 
 = 3, 
 
 5-2C = -7, 
 
 A-B+ 0=6. 
 
 From these equations we have 
 
 C = S, B--1, xbA A = 2. 
 
 Substituting these values in the assumed partial fractions, we have 
 
 3 ** - 7 fc + 6 _ 2 1 + 3 
 
 3. 
 
 5. 
 
 (X - l) 3 (X - l) 3 (x - l) 2 ' x-l 
 
 x 2 -2 x + 1 
 
 x — x* x 2 — 2x 
 
 x -h "3 - » 4- 1 
 
 a* - a? - 2 a 2 - 7 a? + 12 
 
 7 ^ - 11 g 4- 26 2.x -13 
 
 (a?-3) 8 ' ' aj» + 10aj + 25° 
 
 3 a: 2 -4 5a? 2 -4o; 
 
 ' (^ + 1) 3 ' " (5x-2f 
 
 13 a; + 10 12 £ 
 
 ' 6x 3 -13x 2 -5x' ' fl^ + 6^ + 110+6 
 
 .^4-3^-8 3a?-lla? + 13a?-4 
 
 ' x(^ + 2) 2 ' ' (x>-x)(x-2f 
 
 Case II 
 
 448. When the denominator is resolvable into equal or unequal 
 quadratic factors. 
 
 Rule. Assume the given fraction equal to the sum of several 
 fractions whose numerators have the form Mx + N, and whose de- 
 
DECOMPOSITION OF FRACTIONS 281 
 
 nominators are all the quadratic divisors of the given denominator, 
 and proceed as in Case I. 
 
 Dem. The numerator of a partial fraction may contain a term 
 with the first power of the variable as well as an absolute term, 
 since such a fraction, being in the form 
 
 Mx + N Mx + N 
 
 or 
 
 x 2 + ax + b (x* + ax + b) n ' 
 
 in which, by hypothesis, x 2 + ax + b can not be factored, is not 
 capable of farther reduction. 
 
 Moreover, if x were not introduced into the numerator, on clear- 
 ing of fractions the degree of the second member would be lower 
 by two than the denominator of the given fraction, while the first 
 member (the numerator of the given fraction) may be lower by 
 only one. Then in equating coefficients, the coefficient of the 
 highest power of x in the first member, a given constant, would be 
 placed equal to 0, which is absurd. It is for a similar reason that 
 a constant term must be introduced into the numerator of each 
 assumed fraction. 
 
 EXAMPLES CXV . 
 
 Decompose the following into partial fractions : 
 6^ + 53 + 4 
 
 Solution. As the divisors of the denominator are x and x 2 + 2 x + 2, we 
 assume 
 
 6x 2 + 5x + 4 _A Bx + C 
 
 x 3 + 2 x 2 + 2x x x 2 + 2x + 2* 
 
 Clearing of fractions and collecting terms, 
 
 6x 2 + 5x + 4 = (>H-£)x 2 + (2A + C)x+2A, 
 
 whence (Art. 441) 
 
 A + B = 6, 
 
 2A+C=o, 
 
 2A = 4. 
 
 From these equations we have .4 = 2, B — 4, and C = 1. 
 Substituting these values in the assumed partial fractions, we have 
 
 6x 2 + 5x + 4 __2 4x-+ 1 
 
 x 3 + 2 x 2 + 2 x x x 2 + 2 x + 2 
 
282 HIGHER ALGEBRA 
 
 ar 3 4- x — 1 
 
 2. 
 
 (»* + 2) 
 
 Solution. As the divisors of the denominator are (x 2 + 2) 2 and x 2 + 2, 
 we assume 
 
 x 3 + a; - 1 _ ^4x+ ff Cx -f D 
 
 (x 2 + 2) 2 (x 2 + 2) 2 x 2 + 2 ' 
 
 Clearing of fractions and collecting terms, 
 x 3 + x - 1 = Cx 3 + Dx 2 + (A+ 2 C)x + B + 2 D, whence (Art. 441), 
 
 C=l, 
 I> = 0, 
 .4 + 2 C=l, 
 B + 2D=-l. 
 
 From these equations we have A = -l, B = — 1, 0=1, Z> = 0. 
 Substituting these values in the assumed partial fractions, we have 
 
 x 3 + x — 1 _ x + 1 x 
 
 (x 2 + 2) 2 ~ (x 2 + 2) 2 x 2 + 2* 
 
 4 x*-2x + Z 
 
 
 42 - 19 a 
 
 
 X 3 - 
 
 -4ar* + a; — 
 1 
 
 4 
 
 ar 5 - 
 
 _a^ + 2a;- 
 
 2 
 
 12 
 
 S» _ x + 10 
 
 
 6. 
 
 (a,- 2 + 1) 2 
 
 ar 5 -! 
 
 ( £C 2_2#_ | _l)(V +a ;_|_4) 
 
 a 3 + 2 q? + 2 
 a^-1 (a; 2 + 2) (a? + x + 2) 
 
 SECTION II — TAYLOR'S FORMULA 
 
 449. Taylor's Formula is a formula for developing a function 
 of the sum of two variables in terms of the ascending powers of 
 one of the variables. 
 
 450. Factorial n is the product of all the integral numbers 
 from 1 to n inclusive, and is written \n.* Thus, 
 
 [3 = 2-3, [5 = 2.3.4.5, [w = 2. 3.4... n. 
 
 451. Prob. To produce Taylor's Formula. 
 Solution. Assume 
 
 u =f(x + y) = A + By+Cy 2 + Dtf + Ey 4 -f Ftf -f etc., 
 
 * The notation n ! is also employed. 
 
TAYLOR'S FOJRMULA 283 
 
 in which A, B, C, etc., though independent of y, are dependent on 
 x, since x appears in the development only as involved in these 
 coefficients. These coefficients are, therefore, variable. 
 
 Obtaining the successive partial derivatives with reference to y 
 and equating them, according to Art. 430, with the corresponding 
 successive partial derivatives with reference to x, we have 
 
 ^ = B + 2 Cy + 3Dtf + 4^ + 5Ftf + etc. =^, 
 dy ax 
 
 ^ = 2 C+ 2 • 3 Dy + 3 . 4 Ey* + 4 . 5 jy + etc. = || 
 
 ^ = 2 • 3Z) + 2 • 3 • AEy + 3 . 4 . 5*y + etc. = ^, 
 
 — = 2 • 3 • 4 E + 2 • 3 • 4 • 5 Fv + etc. = — [, 
 dy* dx A 
 
 etc., etc., etc. 
 
 Since A, B, C, etc., are independent of y, if we obtain their 
 values for one value of y, we shall have them (in form) for all val- 
 ues of y. Representing by u' the value of u when y = 0, we have 
 
 a i t> du 1 n d 2 a'l n d hi* 1 j-, dVl , n . 
 
 A=w ' B =T*' c =^2' d= mw E =^\£ etc - 
 
 Substituting these values in the assumed series, we have 
 
 */ \ i , flu' , d?u'y 2 . dPu'if . d*u'y A . , n 
 
 u =/(* + j,) = »< + _, + _| + _| + _£ + etc., 
 
 which is Taylor's Formula. 
 
 452. Sen. As successive derivatives are represented by /', /", 
 /'", etc. (Arts. 425, 426, and 427), and as f(x + y) becomes f(x) 
 when y = 0, Taylor's Formula may be written 
 
 u =f(x + y) =f(x) + f'(x)y + /"(x)|- +f"(x)£+r(x)^ + etc. 
 
 This formula may be stated as a theorem as follows : 
 Taylor's Formula develops u — f(x -f y) into a series in which the 
 1st term is ivhat the function becomes when y = 0; the '2d term is 
 
284 HIGHER ALGEBRA 
 
 y times ivhat the 1st derivative becomes when y = ; the 3d term is 
 
 V 2 
 
 t^ times what the second derivative becomes when y = ; the 4:th term 
 
 is ~ times whoX the 3d derivative becomes when y = 0, and so on. 
 [3 9 ' 
 
 EXAMPLES CXVI 
 
 Expand by Taylor's Formula the following : 
 
 i. (x + y y. 
 
 Solution, u = (x + */) 5 , u' = x 5 , — = 6 x 4 , — = 4 . 5 x 3 , 
 K --* w " dx dx* 
 
 $2- = 3 • 4 • 5 x 2 , ~ = 2 • 3 • 4 • 5 3, — =2.3.4.5. 
 <Zx 3 dx 4 <7x° 
 
 Substituting these values in Taylor's Formula, we have 
 u = (x 4- 2/) 5 = a,* 5 -4- 5 x 4 y + 10 a? 3 ?/ 2 4- 10 afy 3 + 5 #?/ 4 -4- ^z 5 . 
 
 2. (a-?/) 7 . 3. (x + y)k 4. (a? - y)~\ 
 
 5. (a; -4- y) *. 6. Vx + y. 7. V(a>* -f ?/) 3 . 
 
 453. Taylor's Formula is much used for developing a function 
 of a single variable after the variable has taken an increment. 
 
 EXAMPLES CXVII 
 
 Expand by Taylor's Formula the following, after the variable 
 has taken an increment h. 
 
 1. 2x 3 -x 2 + 5x-ll. 
 
 Solution. Using the notation of Art. 452, 
 
 /(* + h) = 2(* + hf - (x + ny + 5 (x + 70 - 11, 
 
 /(x) = 2 x s - x 2 + 5 x - 11, f (x) = 6 x? - 2 x + 5, 
 /" (x) = 12 x - 2, /'" (x) = 12. 
 Substituting these values in Taylor's Formula, we have 
 
 2 x* - x 2 + 5 x - 11 + (6 x 2 - 2 x + 5) /* + (6 x - 1) h* + 2 7i 3 . 
 
 2. 3 a 5 - 2 x- 2 . 3. 2 x 4 - 4 x 3 + a 2 - 5. 
 
 454. In producing Taylor's Formula we made use of the prin- 
 ciple that in a function of the sum of two variables the correspond- 
 ing partial derivatives are equal. The formula will not, therefore, 
 
THE BINOMIAL FORMULA 285 
 
 directly expand such a function as (3 x 2 -f 4 y*) 5 , in which the cor- 
 responding partial derivatives, on account of the coefficients and 
 exponents of x and y, are not equal. Indirectly, however, by 
 substituting z for 3 x 2 and v for 4 y 3 , the formula will apply ; for 
 in that case the coefficients and exponents of x and y will not 
 enter the differentiation. Then, after expansion, the values of z 
 and v may be restored. 
 
 EXAMPLES CXVin 
 Develop by Taylor's Formula the following : 
 
 l. (rf + dy)*. 
 
 Solution. Substituting z for x 2 and v for 3y, and developing by Taylor's 
 Formula, we have 
 
 (* + «)* = 
 
 Restoring values 
 
 (s + v)$ = z^ + £ 2 h-±z t* + T V ^ V* - T f* ^ V + etc. 
 
 yy 2x 8 a* 16 a;* 128x7 
 
 2. (2^-f-^) 4 - 3- (x~* + 2y)-\ 4. (a;-2 2/ 2 )'. 
 
 455. All of the above functions, which are algebraic, can be 
 readily expanded without the use of Taylor's Formula. The 
 formula has its most important application in expanding func- 
 tions that are not algebraic. 
 
 SECTIOiV III — THE BINOMIAL FORMULA 
 
 456. The Binomial Formula and applications with positive 
 integral exponents have already been given in Arts. 155 and 159. 
 The proof and applications with any exponent have been reserved 
 for this part of the work. 
 
 457. Prob. To produce the Binomial Formula. 
 
 Solution. Applying Taylor's Formula (Art. 451) to the ex- 
 pansion of the form (x -\- y) m , we have 
 
 f(x + y) = (x + y)'», f(x) = x m , ' 
 
 f(x) = mx™ 1 , f'(x) = m(m — t) x m ~\ 
 f'"(x) = m (m - 1 ) (m - 2) x m -% 
 f\x) = m(m- 1) (m - 2) (m - 3) x m ~\ etc. 
 
286 HIGHER ALGEBRA 
 
 Substituting these values, we have 
 
 / x™ » , . i , m(m — 1) m 9 „ m(m — l)(m — 2) a _ 
 (a? + 2/) m = sc TO -f- mx m ~ x y -\ ^ J -x m ~hf -\ ± ^ ; -x 1l ~hf 
 
 , * m (m - 1) (m - 2) • • • (m - n + 2) +1 
 + -+— [n _ 1 * 2/ + -, 
 
 which is the Binomial Formula. 
 
 458. The Binomial Theorem. In the expansion of a binomial 
 affected with any exponent, the exponent of the leading letter begins in 
 the first term with the exponent of the binomial, and in each succeeding 
 term decreases by 1 ; while the exponent of the second letter begins in 
 the second term with 1 and in each succeeding term increases by 1. 
 
 The coefficient of the first term is 1 ; that of the second term is 
 the exponent of the binomial ; and if the coefficient of any term be 
 multiplied by the exponent of the leading letter in that term arid 
 divided by the exponent of the second letter increased by 1, the result 
 will be the coefficient of the next term. 
 
 This is deduced by inspection from the Binomial Formula. 
 
 EXAMPLES CXIX 
 Expand by the binomial theorem the following : 
 
 1. (x 2 -2y)%. 
 
 Solution. The theorem (458) applies, not to the exponents and coeffi- 
 cients of x and y in such a function as this, but to x 2 and — 2 y treated 
 as wholes. To avoid carrying over into following terms factors and signs 
 that do not belong there, it is best to inclose the separate factors in paren- 
 theses, using the plus sign between all the terms, and reducing afterward. 
 Thus, 
 
 (x 2_2y)f = (^ ) t + ( 2 )(a .2 ) 4 ( _2y)- f (-i)(a:2)-^(_2 2 /)2 
 
 + (A)(*T^(-2 2/) 3 + etc., 
 = x% - | x~hj - f aT^/ 2 - ff x~ T y s - etc. 
 
 2. (a -bf. 3. (x-y) 7 . 4. (l+x)\ 
 5. (1-2/) 5 . 6. (z-r-2/)- 2 . 7. (x-y)~ s . 
 
THE BINOMIAL FORMULA 287 
 
 8. {a-x)~\ 9. , or (* + */)- 4 . 
 
 10. — L_. ii. (l_a 2 )l 12. (2 + a 3 )*. 
 
 13. fx-r-X- 14 0«~ 2 -y*)~* 15. (V^ 3 + 4s!/a) 4 . 
 
 1 
 
 (1 - a 2 ) 3 
 
 19. (3a-^)- 3 . 20. (a 2 -6 2 )l 21. a 
 
 V& 2 — c 2 ^ 2 
 
 22. (1+Va) 4 + (1-V^) 4 - 23. (1+V5) 5 + (1-V5) 5 . 
 
 24. (1 + X — a 2 ) 4 . 
 
 Sug. Write in the form [(1 -f x) — a: 2 ] 4 , and regard 1 + x as one term of 
 a binomial, and — x' 2 as the other. 
 
 25. (a + b + c) 5 . 26. (1 - x + x 2 - ar 5 ) 4 . 
 
 16. Va + 2. 17. - - L _ 98 » 18. (a 2 -a; 2 
 
CHAPTER XXI 
 LOGARITHMS 
 
 459. The Logarithm of a Number is the exponent by which a 
 fixed number is affected to produce any required number. The 
 fixed number is called the Base. 
 
 Thus, let 4 be the base ; then 
 
 3 is the logarithm of 64 to base 4, since 4 3 = G4 
 2 is the logarithm of 16 to base 4, since 4 2 = 16 
 1 is the logarithm of 4 to base 4, since 4 1 = 4 
 is the logarithm of 1 to base 4, since 4° = 1 
 
 — 1 is the logarithm of \ to base 4, since 4 _1 = J 
 
 — 2 is the logarithm of j 1 ^ to base 4, since 4 -2 = T V 
 
 460. Logarithms are in one system or another according to the 
 base assumed. Only two systems are in common use, viz., the 
 system whose base is 10, called the Briggean, Briggs, or Common 
 System, and the system whose base is 2.71828 +, called the 
 Napierian, Natural, or Hyperbolic System. The Common System 
 is used to facilitate numerical calculations, and the Napierian 
 System is much used in abstract mathematical discussions. 
 
 When necessary to distinguish different systems, the bases are written as 
 subscripts to the abbreviation log. Thus, k — log a n signifies that k is the 
 logarithm of n in the system whose base is a. The student should clearly 
 understand that, by definition, k = log a n expresses the same relation between 
 k and n as does a k = n. 
 
 461. Theorem. The logarithm of 1 is in all systems. 
 For, by Art. 70, a = 1 for all values of a. 
 
 462. Theorem. In any system whose base is greater than 1 the 
 logarithm ofOis— oo. 
 
 For, a being the base, a -00 = — = 0. Therefore log = — oo. 
 
 a? 
 
 288 
 
LOGARITHMS 289 
 
 463. Theorem. Neither 1 nor any negative number can be used 
 us tlw fxisc of a system of logarithms. 
 
 For with 1 as a base we can represent no other number than 1 
 by its exponents, 1 with any exponent being 1. 
 
 Again, with a negative base, odd exponents would give only- 
 negative numbers, and the corresponding positive numbers would 
 have no logarithms. For example, with — 2 as a base, 3 would 
 be the logarithm of — 8, since (— 2) 3 = — 8 ; but + 8 would have 
 no logarithm, since it cannot be produced by affecting — 2 with 
 any exponent. 
 
 464. Theorem. Negative numbers, as such, have no logarithms. 
 
 For a negative number cannot be produced by affecting a 
 positive base with any exponent. 
 
 465. Sen. When negative numbers occur in computation by 
 logarithms, they are treated as if they were positive, and the sign 
 of the result is determined by the number of negative factors. 
 It is usual to write a subscript n after the logarithm of a negative 
 number. Thus, 1.425673 n indicates that the number of which 
 1.4U5G73 is the logarithm is, in itself, negative. 
 
 466. The most important use of logarithms is to facilitate the 
 multiplication, division, involution, and evolution of numbers 
 containing several figures. The processes depend on the follow- 
 ing principles. 
 
 467. Theorem. The logarithm of the product of two numbers is 
 the sum of their logarithms. 
 
 Dem. Let a be the base of the system, and p and q any two 
 numbers whose logarithms are x and y respectively. Then, by 
 definition, 
 
 p = a x and q = a y . 
 
 Multiplying together the corresponding members of these equa- 
 tions, we have 
 
 pq — a x x a y = a 
 
 *+v 
 
 that is, log jx/ = * + y = logjr? + log q. 
 
 downey's alg. — 19 
 
290 HIGHER ALGEBRA 
 
 468. Cor. In the same way it may be shown that 
 
 logpgr ••• = logp -f logq + logr -f .... 
 
 469. Theorem. The logarithm of the quotient of two numbers 
 is the logarithm of the dividend minus the logarithm of the divisor. 
 
 Dem. Let a be the base of the system, and p and q any two 
 numbers whose logarithms are x and y respectively. Then, by 
 
 definition, 
 
 
 
 
 
 
 
 Dividing, 
 
 we 
 
 have 
 
 P = 
 P. 
 
 --a x 
 
 a* 
 ~~a? 
 
 and q = 
 = a x ~ v ; 
 
 -- a y . 
 
 that is, 
 
 
 lOg^ 
 
 Q 
 
 = X- 
 
 -y- 
 
 = logp - 
 
 -lo 
 
 470. Theorem. The logarithm of a power of a number is the 
 logarithm of the number multiplied by the index of the power. 
 
 Dem. Let a be the base, and x the logarithm of p. 
 Then, by definition, p = a*. 
 
 liaising both members to any power, as the nth, we have 
 p n = (a x ) n = a nx ; 
 that is, log p n = nx = n logp. 
 
 471. Theorem. The logarithm of any root of a number is the 
 logarithm of the number divided by the index of the root. 
 
 Dem. Let a be the base, and x the logarithm of p. 
 
 Then, by definition, p = a x . 
 
 Extracting the nth root of both members, we have 
 
 X 
 
 ■\/p as -\/a x = a n ; 
 
 that i S , ] 0g -w;=-=^- 
 
 n n 
 
 472. It is thus seen that by the use of logarithms the opera- 
 tions of multiplication are replaced by those of addition, division 
 by subtraction, involution by a single multiplication, and evolu- 
 tion by a single division. 
 
LOGARITHMS 291 
 
 473. Only those numbers that are exact powers of the base 
 have integral logarithms; hence the logarithms of most numbers 
 consist of two parts, an integer and a decimal fraction. It is 
 found convenient to write a logarithm so that its fractional part 
 shall be positive, and its integral part positive or negative, as the 
 case may be. 
 
 474. The Characteristic of a Logarithm is its integral part, and 
 the Mantissa of a Logarithm its fractional part, when the logarithm 
 is so written that the fractional part is positive. 
 
 475. Prob. To ascertain the law of the characteristics of loga- 
 rithms in the common system. 
 
 Solution. Since 10° = 1, log 1=0; 
 
 since 10 1 = 10, log 10=1; 
 
 since 10 2 = 100, log 100 = 2 ; 
 
 since 10 3 = 1000, log 1000 = 3 ; 
 
 Also, 
 
 since 10" 3 = -i- = .001, log .001 = - 3 ; 
 10' 5 
 
 etc., etc., etc. 
 
 Thus it is seen, and this is true in any system, that when the 
 logarithms are in arithmetical progression, the numbers are in 
 geometrical progression. 
 
 Now the logarithm of any number between 1 and 10 is between 
 and 1, i.e., + a fraction ; the logarithm of any number be- 
 tween 10 and 100 is between 1 and 2, i.e., 1 + a fraction ; the 
 logarithm of any number between 100 and 1000 is between 2 and 
 3, i.e., 2-f a fraction ; and so on. Hence the characteristic of the 
 logarithm of a number > 1 is one less than the number of figures in 
 the integral part of the number. 
 
 Again, the logarithm of any number between 1 and .1 is be- 
 tween and — 1, i.e., — 1 + a fraction ; the logarithm of any 
 
 etc., etc., 
 
 etc. 
 
 since 10" 1 = — = .1, 
 10 ' 
 
 log .1=-1; 
 
 since 10" 2 = — = .01, 
 10 2 
 
 log .01 = -2; 
 
292 HIGHER ALGEBRA 
 
 number between .1 and .01 is between — 1 and — 2, i.e., — 2 + a 
 fraction ; the logarithm of any number between .01 and .001 is 
 between — 2 and — 3, i.e., — 3 -f a fraction ; and so on. Hence 
 the characteristic of a logarithm of a decimal fraction is negative, 
 and numerically 1 greater them the number of 0's before the first 
 significant figure. 
 
 476. Characteristics are not written in tables of logarithms, 
 but are to be supplied according to the above principles. 
 
 477. When the characteristic of a logarithm is negative, the 
 minus sign is written, not before it, but over it, since the charac- 
 teristic alone is negative. Thus log .00167 = 3.222716. 
 
 478. Theorem. Whatever change be made in the position of the 
 decimal point in a number, the mantissa of its logarithm in the 
 common system remains the same. 
 
 Dem. Changing the position of the decimal point is equiva- 
 lent to multiplying or dividing by some power of 10. Now 
 
 log (k x 10 B ) rs log* -f log 10 n = log k + n log 10 = log k + n, 
 log (k -i- 10 n ) = log k — log 10* = log k — n log 10 = log k — n. 
 
 Hence the logarithms of k, k x 10 n , and k -=- 10 n differ only by 
 an integer n\ i.e., the mantissa? of the logarithms of these num- 
 bers, which differ only in the position of the decimal point, are 
 the same. 
 
 479. Sch. The characteristic, in the common system, charac- 
 terizes the number in showing between what two consecutive 
 powers of 10 it lies, and depends, therefore, simply on the posi- 
 tion of the decimal point; while the mantissa, which means an 
 addition, is the part added to the characteristic to give the 
 approximate logarithm, and depends simply on the sequence of 
 figures. 
 
 480. Theorem. The differential of the logarithm of a variable is 
 the differential of the variable multiplied by a constant, called the 
 modulus of the system, divided by the variable. 
 
LOGARITHMS 293 
 
 In the Napierian system, the modulus being 1, the differential of 
 the logarithm of a variable is the differential of the variable divided 
 by the variable. 
 
 Dem. Let y = x n , (1) 
 
 n being any arbitrary constant. Then (Art. 420), 
 
 dy = nx n 1 dx = n - dx, 
 u x ' 
 
 dy dx 
 
 From (1) we have, by Art. 470, 
 
 log y = n log x. 
 Hence from Art. 413, whatever the differentials of logy and 
 
 d(logy)=nd(\ogx). (3) 
 
 Dividing (3) by (2) to eliminate n, 
 
 dQogy) = d(\ogx) 
 dy dx 
 
 V % 
 
 Hence the ratio of d(\ogy) to_^is the same as that of d(\ogx) 
 
 dx 
 
 to — , whatever the values of x and y. 
 
 Represent this constant ratio by m. Then 
 
 d (l ogy)= 1^1, andd(logz) = ^. 
 y x 
 
 481. The Logarithmic Series is the development of log(l + a?) 
 in ascending powers of x. It reveals several important properties 
 of logarithms, and, when rendered convergent, furnishes a means 
 of computing the logarithms of numbers. 
 
 482. Prob. To produce the logarithmic series. 
 Solution. Let u = log (z -f x) ; then (Art. 451) 
 
 u' 
 
 = logz 
 
 du' 
 ' dz~ 
 
 m 
 
 '■ > 
 z 
 
 dV_ 
 
 dz 1 
 
 m 
 
 dz* 
 
 2 m 
 
 d*u'_ 
 dz> 
 
 2 
 
 •3m 
 
 z* ' 
 
 etc. 
 
294 HIGHER ALGEBRA 
 
 Substituting these values in Taylor's Formula (Art. 451), we have 
 
 « = log(* + ») = log* + ™x-™,f + ££-»! £ + eto. 
 
 z z 2 2 z 3 3 z 4 4 
 
 Making z = 1, this becomes 
 
 log (1 + x) = m(x - 1 + 1 - 1 4 + etc.), 
 which is the logarithmic series. 
 
 483. Sch. 1. In the above series, if we let x remain the same 
 and express the logarithms of 1 + x in two different systems 
 whose bases are a and b, log a (1 -f x) is not equal to log 6 (1 + x). 
 But the series in the parenthesis is the same in the two cases ; 
 hence m is different in the two cases, i.e., m depends on the base 
 and characterizes the system. This factor, which is constant in 
 the same system but different in different systems, has been 
 named the modulus. 
 
 484. Sch. 2. It is evident that, in establishing a system of 
 logarithms, we may either select the modulus and let the base be 
 determined by the mutual relation, or select the base and let the 
 modulus be determined by the mutual relation. In the Napierian 
 system (so called) 1 was selected as the modulus, and this gave 
 for the base 2.718281828 -f- ; while in the Briggs system 10 was 
 selected as the base, and this gave for the modulus .43429448 + . 
 
 485. Theorem. TJie logarithms of the same number in different 
 systems are to each other as the moduli of those systems. 
 
 Dem. Letting a and a! be the bases, and m and m' the moduli 
 of two systems, we have, from Art. 482, 
 
 log (l + z) = m(z-^ + J-J + etc.), (1) 
 
 log„.(l + *) = m\x - 1 + 1 - J + etc.). (2) 
 
 Dividing (1) by (2), 
 
 log.(l + x) _ m 
 log, (1 + x) to' - 
 
 (3) 
 
LOGARITHMS 295 
 
 486. Cor. The logarithm of a number in the Napierian system 
 multiplied by the modulus of any other system will give the logarithm 
 of the same number in the latter system. 
 
 487. Prob. To render the Napierian logarithmic series con- 
 vergent. 
 
 Solution. As the logarithmic series (Art. 482) is divergent 
 for x > 1, it cannot, nntil transformed into a convergent series, be 
 used for computing the logarithms of numbers. 
 
 For the Napierian system, whose base is usually represented 
 by e, the series becomes, when x is positive, 
 
 />»2 /y& /v»4 a») 
 
 log.(l+*) = *-f + |-f + |-etc, (1) 
 
 and when x is negative, 
 
 />*2 /smO /-}%A /yj> 
 
 log.(l-x)=-«-|-J-|-|-ete. (2) 
 
 Subtracting (2) from (1), we have 
 log.(l + x) - log.(l - x) = log e t±| = 2(W| + !+ etc.Y 
 
 Letting ^ = — , 
 
 1 — x q 
 
 which gives by solving the equation 
 
 - p-q 
 
 and substituting, 
 
 P + q' 
 
 l0ge f = l og e p-\og e q 
 
 or 
 
 which is seen to be a convergent series. 
 
 488. Prob. To compute the logarithms of 1, 2, 3, 4, 5, etc. 
 
 Solution. It is necessary to compute the logarithms of prime 
 numbers only, since the logarithm of a composite number is equal 
 to the sum of the logarithms of its factors (Art. 467). 
 
296 HIGHER ALGEBRA 
 
 The logarithm of 1 is (Art. 461). 
 
 To find the Napierian log 2, in (A) make p = 2 and q = 1. 
 To find the Napierian log 3, make p = 3 and q = 2, and so on. 
 We thus have 
 
 log^log.^g+^ + ^ + ^ + ^+otc.) 
 
 0.69314718, 
 
 log.3 = log e 2 + 2(1 + ^ + ^ + -^+ oto.) = 1.09861228, 
 
 log. 4 = 2 log. 2 = 1.38629436, 
 
 108,6 = 108,4 + 2^ + ^ + ^ + ^ + 610.)= 1.60943790, 
 
 and so on. 
 
 The common logarithms of the same numbers may now be 
 found by multiplying the Napierian logarithms by the modulus 
 of the common system (Art. 486), 
 
 489. Tables of such logarithms have been prepared with great 
 care and are easily obtainable by the student and the practical 
 computer. Bremiker's for six places and Vega's for seven are 
 standard tables. For some refined astronomical calculations ten- 
 place logarithms are necessary, while for many purposes four- 
 place and five-place logarithms are sufficient. 
 
 490. Theorem. The modulus of any system is the reciprocal of 
 the Napierian logarithm of the base of that system, and in the 
 common system is .43429448 ; and the Napierian base is the number 
 of which the modulus of any system is the logarithm in that system, 
 and is 2.71821828. 
 
 Dem. In equation (3) of Art. 485, letting 1 + x = a = base, 
 we have 
 
 log_ a m 1 
 
 — ^- = -, or m = - 
 
 log e a 1 log e a 
 
 In the common system this becomes 
 
 m = — — = = = .43429448. 
 
 log, 10 log e 2 + log e 5 2.30258508 
 
LOGARITHMS 297 
 
 Again, by the same equation, 
 
 log 10 e ^7ft 
 log e e 1 
 or log 10 e = m log e e = m = .43429448. 
 
 Finding from a table of common logarithms the number of 
 which .43429448 is the logarithm, we have 
 e = 2.71821828. 
 
 491. An Exponential Equation is an equation in which the 
 unknown quantity occurs as an exponent. 
 
 492. Prob. To solve an exponential equation. 
 Solution. By definition the equation has the form 
 
 a x = b. 
 Taking the logarithms of both members, we have 
 x log a = log 6, 
 
 whence x = r-^— 
 
 log a 
 
 493. Prob. The principal p, rate r, and time t being given, to 
 find the amount a, at compound interest. 
 
 Solution. We have 
 
 for 1 year, a as p -f pr = p (1 + r), 
 for 2 years, a = p (1 + r) + p (1 + r) r = p (1 + r) 2 , 
 for 3 years, a = p(l + r) 2 +p(l + r) 2 r=p(l + r)*, 
 and so on. Hence, for £ years, 
 
 a = p(l+ry. (1) 
 
 Taking the logarithms of both members, we have 
 
 log a = log p + t log (1 + r). (2) 
 
 494. Cor. i. If the interest is compounded m times annually, 
 formulas (1) and (2) become 
 
 log a = log ?> -f mt log f 1 + — ]< 
 
 mt 
 
 i 
 
298 HIGHER ALGEBRA 
 
 Cor. 2. From (2) we deduce 
 
 log p = \oga-t log (1 + r), 
 
 log(l + r) = lo g a - lo g* , 
 
 log (1 + r) 
 
 495. Prob. To find the present worth, S, of an annuity, a, run^ 
 nina t years. 
 
 Solution. Money being worth r per cent, 
 
 the present worth of 1st payment = 
 
 1 + r 
 the present worth of 2d payment 
 
 (l + r) s 
 
 the present worth of 3d payment = 
 
 (1+r) 3 
 and so on. Hence the present worth of the whole is the sum of 
 
 a geometrical progression in which the first term is — - — , the 
 
 1 1 + r 
 
 ratio is , and the number of terms is t. Substituting in the 
 
 formula of Art. 327 and reducing, we have 
 
 q[(l + ry-l] 
 
 r (i + r y ' 
 
 whence log S = log a + log [(1 + r) 1 — 1] — log r — t log (1 + r). 
 
 EXAMPLES CXX 
 Express the following operations by logarithms : 
 
 a 2 — ar 2 
 
 Solution. y = J^^ = Ka + x)(a-x) \\ 
 
 * l + x \ l + x I 
 
 Hence, log y = $ [log {a + x) + log (a - x) - log (1 + x)]. 
 
 2. y = **(l-aO*. 3 y=\P 
 
 — a) 
 be 
 
 4. v/= '/a&'c 4 . 5 . y= l a* -a 4 l 
 
6. y = 
 
 LOGARITHMS 299 
 
 -y/x — x 2 
 
 '-V? 
 
 8 y -,/^'/3-i3' V ^-x-2 ' 
 
 Va Va 2 — 6 2 
 Differentiate the following : 
 
 10. y = log VI + «. 
 
 Solution, y = log Vl + x = log (1 + x) 2 = | log (1 + sc). 
 Hence, by Arts. 413 and 480, 
 
 dy mdx 
 
 2(1 + x) 
 
 Or, differentiating without first changing the form, we have (Arts. 421 
 
 and 480), 
 
 mdx 
 
 , 2 VI + x mdx 
 dy- 
 
 Vl + x 2(1 + x) 
 
 11. y = log ax. 12. y = log (1 — x). 
 
 13. y = logar\ 14. y = log- 
 ic 
 
 15. ?/ = log (a 2 - ar>). 16. y = log (1 + x 2 ) 2 . 
 
 Solve the following equations : 
 
 17. a hx = c. 18. a x = -- 
 
 19. ba cx = d. 20. 10 2 * - 6 • 10* = 7. 
 
 10*+' = 1000, f a*+" 
 
 21. J 22 
 
 10*"" = 100. 
 
 1 &*-» = n. 
 
 496. Following are two sample pages from a table of logarithms. 
 The characteristics are not written in the table, but are to be sup- 
 plied by the principles of Art. 475. 
 
 497. To find from the table the mantissa of the logarithm of a 
 number. 
 
 In all cases the decimal point is disregarded, as it has to do 
 only with the characteristic (Art. 478). The logarithms of the 
 
300 
 
 HIGHER ALGEBRA 
 
 Logarithms of Numbers 
 
 N 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 Diff. 
 
 770 
 
 880491 
 
 6547 
 
 6004 
 
 6660 
 
 6716 
 
 6773 
 
 6829 
 
 6885 
 
 6942 
 
 6998 
 
 57 
 
 771 
 
 7054 
 
 7111 
 
 7167 
 
 7223 
 
 7280 
 
 7336 
 
 7392 
 
 7449 
 
 7505 
 
 7561 
 
 1 6 
 
 772 
 
 7617 
 
 7674 
 
 7730 
 
 7786 
 
 7842 
 
 7898 
 
 7955 
 
 8011 
 
 80(57 
 
 8123 
 
 2 11 
 
 773 
 
 8179 
 
 8236 
 
 8292 
 
 8348 
 
 8404 
 
 84(50 
 
 851(5 
 
 8573 
 
 8629 
 
 8(585 
 
 3 17 
 
 774 
 
 8741 
 
 8797 
 
 8853 
 
 8909 
 
 8965 
 
 9021 
 
 9077 
 
 9134 
 
 9190 
 
 924(5 
 
 4 23 
 
 775 
 
 9302 
 
 9358 
 
 9414 
 
 9470 
 
 9526 
 
 9582 
 
 9638 
 
 9694 
 
 9750 
 
 980(3 
 
 5 29 
 
 776 
 
 9862 
 
 9918 
 
 9974 
 
 **30 
 
 **86 
 
 *141 
 
 •197 
 
 *253 
 
 ♦309 
 
 *365 
 
 6 34 
 
 777 
 
 890421 
 
 0477 
 
 0533 
 
 0589 
 
 0(545 
 
 0700 
 
 075(5 
 
 0812 
 
 0868 
 
 0924 
 
 7 40 
 
 778 
 
 0980 
 
 1035 
 
 1091 
 
 1147 
 
 1203 
 
 1259 
 
 1314 
 
 1370 
 
 1426 
 
 1482 
 
 8 4(5 
 
 779 
 
 1537 
 
 1593 
 
 1649 
 
 1705 
 
 1760 
 
 1816 
 
 1872 
 
 1928 
 
 1983 
 
 2039 
 
 9 51 
 
 780 
 
 892095 
 
 2150 
 
 2206 
 
 2262 
 
 2317 
 
 2373 
 
 2429 
 
 2484 
 
 2540 
 
 2595 
 
 56 
 
 781 
 
 2651 
 
 2707 
 
 2762 
 
 2818 
 
 2873 
 
 2929 
 
 2985 
 
 3040 
 
 3096 
 
 3151 
 
 1 6 
 
 782 
 
 3207 
 
 3262 
 
 3318 
 
 3373 
 
 3429 
 
 3484 
 
 3540 
 
 3595 
 
 3651 
 
 370(5 
 
 2 11 
 
 783 
 
 3762 
 
 3817 
 
 3873 
 
 3928 
 
 3984 
 
 4039 
 
 4094 
 
 4150 
 
 4205 
 
 4261 
 
 3 17 
 
 784 
 
 4316 
 
 4371 
 
 4427 
 
 4482 
 
 4538 
 
 4593 
 
 4648 
 
 4704 
 
 4759 
 
 4814 
 
 4 22 
 
 785 
 
 4870 
 
 4925 
 
 4980 
 
 5036 
 
 5091 
 
 5146 
 
 5201 
 
 5257 
 
 5312 
 
 5367 
 
 5 28 
 
 786 
 
 5423 
 
 5478 
 
 5533 
 
 5588 
 
 5644 
 
 5(599 
 
 5754 
 
 5809 
 
 5864 
 
 5920 
 
 6 34 
 
 787 
 
 5975 
 
 6030 
 
 6085 
 
 6140 
 
 6195 
 
 (5251 
 
 630(5 
 
 63(51 
 
 6416 
 
 6471 
 
 7 39 
 
 788 
 
 6526 
 
 6581 
 
 663a 
 
 6692 
 
 6747 
 
 6802 
 
 6857 
 
 6912 
 
 69(57 
 
 7022 
 
 8 45 
 
 789 
 
 7077 
 
 7132 
 
 7187 
 
 7242 
 
 7297 
 
 7352 
 
 7407 
 
 7462 
 
 7517 
 
 7572 
 
 9 50 
 
 790 
 
 897627 
 
 7682 
 
 7737 
 
 7792 
 
 7847 
 
 7902 
 
 7957 
 
 8012 
 
 8067 
 
 8122 
 
 55 
 
 791 
 
 8176 
 
 8231 
 
 8286 
 
 8341 
 
 8396 
 
 8451 
 
 850(5 
 
 8561 
 
 8(515 
 
 8670 
 
 1 6 
 
 792 
 
 8725 
 
 8780 
 
 8835 
 
 8890 
 
 8944 
 
 8999 
 
 9054 
 
 9109 
 
 9164 
 
 9218 
 
 2 11 
 
 793 
 
 9273 
 
 9328 
 
 9383 
 
 9137 
 
 9492 
 
 9547 
 
 9602 
 
 9656 
 
 9711 
 
 976(5 
 
 3 17 
 
 794 
 
 9821 
 
 9875 
 
 9930 
 
 9985 
 
 **39 
 
 **94 
 
 *149 
 
 *203 
 
 *258 
 
 *312 
 
 4 22 
 
 795 
 
 900367 
 
 0422 
 
 0476 
 
 0531 
 
 0586 
 
 0(540 
 
 0695 
 
 0749 
 
 0804 
 
 0859 
 
 5 28 
 
 796 
 
 0913 
 
 01)68 
 
 1022 
 
 1077 
 
 1131 
 
 1186 
 
 1240 
 
 1295 
 
 1349 
 
 1404 
 
 6 33 
 
 797 
 
 1458 
 
 1513 
 
 1567 
 
 1622 
 
 1676 
 
 1731 
 
 1785 
 
 1840 
 
 1894 
 
 1948 
 
 7 39 
 
 798 
 
 2003 
 
 2057 
 
 2112 
 
 2166 
 
 2221 
 
 2275 
 
 2329 
 
 2384 
 
 2438 
 
 2492 
 
 8 44 
 
 799 
 
 2547 
 
 2601 
 
 2655 
 
 2710 
 
 2764 
 
 2818 
 
 2873 
 
 2927 
 
 2981 
 
 3036 
 
 9 50 
 
 800 
 
 9013090 
 
 3144 
 
 3199 
 
 3253 
 
 3307 
 
 3361 
 
 3416 
 
 3470 
 
 3524 
 
 3578 
 
 54 
 
 801 
 
 3633 
 
 3687 
 
 3741 
 
 3795 
 
 3849 
 
 3904 
 
 3958 
 
 4012 
 
 4066 
 
 4120 
 
 1 5 
 
 802 
 
 4174 
 
 4229 
 
 4283 
 
 4337 
 
 4391 
 
 4445 
 
 4499 
 
 4553 
 
 4607 
 
 46(31 
 
 2 11 
 
 803 
 
 4716 
 
 4770 
 
 4824 
 
 4878 
 
 4932 
 
 4986 
 
 5040 
 
 5094 
 
 5148 
 
 5202 
 
 3 16 
 
 804 
 
 5256 
 
 5310 
 
 5364 
 
 5418 
 
 5472 
 
 5526 
 
 5580 
 
 5634 
 
 5688 
 
 5742 
 
 4 22 
 
 805 
 
 5796 
 
 5850 
 
 5904 
 
 5958 
 
 6012 
 
 60(56 
 
 6119 
 
 6173 
 
 6227 
 
 6281 
 
 5 27 
 
 806 
 
 6335 
 
 6389 
 
 6143 
 
 6497 
 
 6551 
 
 6604 
 
 6658 
 
 6712 
 
 6766 
 
 6820 
 
 6 32 
 
 807 
 
 6874 
 
 6927 
 
 6981 
 
 7035 
 
 7089 
 
 7143 
 
 7195 
 
 7250 
 
 7304 
 
 7358 
 
 7 38 
 
 808 
 
 7411 
 
 7465 
 
 7519 
 
 7573 
 
 7626 
 
 7680 
 
 7734 
 
 7787 
 
 7841 
 
 7895 
 
 8 43 
 
 809 
 
 7949 
 
 8002 
 
 8056 
 
 8110 
 
 8163 
 
 8217 
 
 8270 
 
 8324 
 
 8378 
 
 8431 
 
 9 49 
 
 810 
 
 908485 
 
 8539 
 
 8592 
 
 8646 
 
 8699 
 
 8753 
 
 8807 
 
 8860 
 
 8914 
 
 8967 
 
 53 
 
 811 
 
 9021 
 
 9074 
 
 9128 
 
 9181 
 
 9235 
 
 9289 
 
 9342 
 
 9396 
 
 9449 
 
 9503 
 
 1 5 . 
 
 812 
 
 9556 
 
 9610 
 
 9663 
 
 9716 
 
 9770 
 
 9823 
 
 9877 
 
 9930 
 
 9984 
 
 **37 
 
 2 11 
 
 813 
 
 910091 
 
 0144 
 
 0197 
 
 0251 
 
 0304 
 
 0358 
 
 0411 
 
 0464 
 
 0518 
 
 0571 
 
 3 16 
 
 814 
 
 0624 
 
 0678 
 
 0731 
 
 0784 
 
 0838 
 
 0891 
 
 0944 
 
 0998 
 
 1051 
 
 1104 
 
 4 21 
 
 815 
 
 1158 
 
 1211 
 
 1264 
 
 1317 
 
 1371 
 
 1424 
 
 1477 
 
 1530 
 
 1584 
 
 1637 
 
 5 27 
 
 816 
 
 1690 
 
 1743 
 
 1797 
 
 1850 
 
 1903 
 
 1956 
 
 2009 
 
 2063 
 
 2116 
 
 2169 
 
 6 32 
 
 817 
 
 2222 
 
 2275 
 
 2328 
 
 2381 
 
 2435 
 
 2488 
 
 2541 
 
 2594 
 
 2(547 
 
 2700 
 
 7 37 
 
 818 
 
 2753 
 
 2806 
 
 2859 
 
 2913 
 
 296(5 
 
 3019 
 
 3072 
 
 3125 
 
 3178 
 
 3231 
 
 8 42 
 
 819 
 
 3284 
 
 3337 
 
 3390 
 
 3443 
 
 3490 
 
 3549 
 
 3(502 
 
 3(555 
 
 3708 
 
 3761 
 
 9 48 
 
 N 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 Diff. 
 
LOG A HIT II MS 
 
 301 
 
 Logarithms of Numbers 
 
 N 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 Diff. 
 
 820 
 
 913814 
 
 3867 
 
 3920 
 
 3973 
 
 4026 
 
 4079 
 
 4132 
 
 4184 
 
 4237 
 
 4290 
 
 53 
 
 821 
 
 4343 
 
 4396 
 
 4449 
 
 4502 
 
 4555 
 
 4(508 
 
 4(5(50 
 
 4713 
 
 47(5(5 
 
 4819 
 
 1 5 
 
 822 
 
 4S72 
 
 4925 
 
 41177 
 
 5030 
 
 5083 
 
 5136 
 
 5189 
 
 5241 
 
 6294 
 
 5347 
 
 2 11 
 
 828 
 
 5400 
 
 6463 
 
 5505 
 
 5558 
 
 6611 
 
 5664 
 
 5710 
 
 5769 
 
 5822 
 
 5875 
 
 3 10 
 
 824- 
 
 5927 
 
 6980 
 
 6033 
 
 6085 
 
 6138 
 
 6191 
 
 6243 
 
 6296 
 
 6349 
 
 6401 
 
 4 21 
 
 825 
 
 6454 
 
 6507 
 
 6669 
 
 6612 
 
 6664 
 
 6717 
 
 6770 
 
 (5822 
 
 6875 
 
 (5927 
 
 5 27 
 
 826 
 
 6960 
 
 7033 
 
 7085 
 
 7138 
 
 7190 
 
 7243 
 
 7295 
 
 7348 
 
 7400 
 
 7453 
 
 6 32 
 
 827 
 
 7606 
 
 7668 
 
 7611 
 
 7663 
 
 771(5 
 
 7768 
 
 7820 
 
 7873 
 
 7925 
 
 7978 
 
 7 37 
 
 828 
 
 8030 
 
 8083 
 
 8135 
 
 8188 
 
 8240 
 
 8293 
 
 8345 
 
 8397 
 
 8450 
 
 8502 
 
 8 42 
 
 829 
 
 8555 
 
 8607 
 
 8669 
 
 8712 
 
 8764 
 
 8816 
 
 8869 
 
 8921 
 
 8973 
 
 902(5 
 
 9 48 
 
 830 
 
 919078 
 
 9130 
 
 9183 
 
 9235 
 
 92S7 
 
 9340 
 
 9392 
 
 9444 
 
 9496 
 
 9549 
 
 52 
 
 831 
 
 9601 
 
 9663 
 
 9706 
 
 975S 
 
 9810 
 
 9862 
 
 9914 
 
 9967 
 
 **19 
 
 ** 71 
 
 1 5 
 
 832 
 
 920123 
 
 0176 
 
 0228 
 
 0280 
 
 0332 
 
 0384 
 
 0436 
 
 0489 
 
 0541 
 
 0593 
 
 2 10 
 
 833 
 
 0645 
 
 0697 
 
 0749 
 
 0801 
 
 0853 
 
 0900 
 
 0958 
 
 1010 
 
 1002 
 
 1114 
 
 3 10 
 
 834 
 
 1166 
 
 1218 
 
 1270 
 
 1322 
 
 1374 
 
 1420 
 
 1478 
 
 1530 
 
 1582 
 
 1634 
 
 4 21 
 
 835 
 
 1686 
 
 1738 
 
 1790 
 
 1842 
 
 1894 
 
 1946 
 
 1998 
 
 2050 
 
 2102 
 
 2154 
 
 5 26 
 
 836 
 
 2206 
 
 2258 
 
 2310 
 
 2362 
 
 2414 
 
 2466 
 
 2518 
 
 2570 
 
 2622 
 
 2674 
 
 6 31 
 
 837 
 
 2725 
 
 2777 
 
 2821) 
 
 2881 
 
 2933 
 
 29S5 
 
 3037 
 
 3089 
 
 3140 
 
 3192 
 
 7 36 
 
 838 
 
 3244 
 
 3296 
 
 3348 
 
 3399 
 
 3451 
 
 3503 
 
 3555 
 
 3007 
 
 3658 
 
 3710 
 
 8 42 
 
 839 
 
 3762 
 
 3814 
 
 3865 
 
 3917 
 
 39(59 
 
 4021 
 
 4072 
 
 4124 
 
 4176 
 
 4228 
 
 9 47 
 
 840 
 
 924279 
 
 4331 
 
 4383 
 
 4434 
 
 4486 
 
 4538 
 
 4589 
 
 4041 
 
 4693 
 
 4744 
 
 51 
 
 841 
 
 4796 
 
 4848 
 
 4899 
 
 4951 
 
 5003 
 
 5054 
 
 510(5 
 
 5157 
 
 5209 
 
 5201 
 
 1 5 
 
 842 
 
 5312 
 
 5304 
 
 5415 
 
 6467 
 
 5518 
 
 5570 
 
 5(521 
 
 5073 
 
 5725 
 
 5776 
 
 2 10 
 
 843 
 
 5828 
 
 5879 
 
 5931 
 
 6982 
 
 6034 
 
 6085 
 
 (5137 
 
 0188 
 
 0240 
 
 6291 
 
 3 15 
 
 844 
 
 6342 
 
 6394 
 
 6445 
 
 (J497 
 
 6548 
 
 (5000 
 
 0651 
 
 (5702 
 
 6754 
 
 6805 
 
 4 20 
 
 845 
 
 6867 
 
 6908 
 
 6969 
 
 701.1 
 
 7002 
 
 7114 
 
 7165 
 
 7216 
 
 7268 
 
 7319 
 
 5 20 
 
 846 
 
 7370 
 
 7422 
 
 7473 
 
 7521 
 
 757(5 
 
 7027 
 
 7678 
 
 7730 
 
 7781 
 
 7832 
 
 6 31 
 
 847 
 
 7883 
 
 7935 
 
 7986 
 
 8037 
 
 8088 
 
 8140 
 
 8191 
 
 8242 
 
 8293 
 
 8346 
 
 7 30 
 
 848 
 
 8366 
 
 8447 
 
 8498 
 
 8649 
 
 8601 
 
 8652 
 
 8703 
 
 8754 
 
 8805 
 
 8857 
 
 8 41 
 
 849 
 
 8908 
 
 8969 
 
 9010 
 
 9061 
 
 9112 
 
 9163 
 
 9215 
 
 9266 
 
 9317 
 
 9368 
 
 9 46 
 
 850 
 
 929419 
 
 9470 
 
 9521 
 
 9572 
 
 9623 
 
 9674 
 
 9726 
 
 9776 
 
 9827 
 
 9879 
 
 50 
 
 851 
 
 9930 
 
 9981 
 
 **32 
 
 **83 
 
 *134 
 
 *185 
 
 *236 
 
 *287 
 
 ♦338 
 
 *389 
 
 1 5 
 
 852 
 
 9:50440 
 
 0491 
 
 0542 
 
 0592 
 
 (M543 
 
 0694 
 
 0745 
 
 0790 
 
 0847 
 
 0898 
 
 2 10 
 
 853 
 
 0949 
 
 1000 
 
 1051 
 
 1102 
 
 1153 
 
 1204 
 
 1254 
 
 1305 
 
 135(5 
 
 1407 
 
 3 15 
 
 854 
 
 1458 
 
 1509 
 
 1560 
 
 1610 
 
 1661 
 
 1712 
 
 1763 
 
 1814 
 
 18(55 
 
 1915 
 
 4 20 
 
 855 
 
 1908 
 
 2017 
 
 2068 
 
 2118 
 
 2169 
 
 2220 
 
 2271 
 
 2322 
 
 2372 
 
 2423 
 
 5 25 
 
 856 
 
 2474 
 
 2524 
 
 2575 
 
 2626 
 
 2677 
 
 2727 
 
 2778 
 
 2829 
 
 2879 
 
 2930 
 
 6 30 
 
 857 
 
 2981 
 
 3031 
 
 3082 
 
 3133 
 
 3183 
 
 3234 
 
 3285 
 
 3335 
 
 3380 
 
 3437 
 
 7 35 
 
 868 
 
 3487 
 
 3638 
 
 3589 
 
 3639 
 
 3690 
 
 3740 
 
 3791 
 
 3841 
 
 3892 
 
 3943 
 
 8 40 
 
 859 
 
 3993 
 
 4044 
 
 4094 
 
 4145 
 
 4195 
 
 4246 
 
 4296 
 
 4347 
 
 4397 
 
 4448 
 
 9 45 
 
 860 
 
 934496 
 
 4549 
 
 4599 
 
 4650 
 
 4700 
 
 4751 
 
 4801 
 
 4852 
 
 4902 
 
 4953 
 
 49 
 
 861 
 
 6008 
 
 5054 
 
 5104 
 
 5154 
 
 5205 
 
 5255 
 
 5300 
 
 5350 
 
 5400 
 
 5457 
 
 1 5 
 
 862 
 
 5507 
 
 6668 
 
 5008 
 
 5658 
 
 570! » 
 
 5759 
 
 6809 
 
 58(50 
 
 5910 
 
 59(50 
 
 2 10 
 
 803 
 
 6011 
 
 6061 
 
 6111 
 
 6162 
 
 6212 
 
 (52(52 
 
 6313 
 
 6363 
 
 0413 
 
 (5463 
 
 3 15 
 
 864 
 
 6614 
 
 6664 
 
 6614 
 
 6666 
 
 (5715 
 
 (57(55 
 
 6815 
 
 6865 
 
 (591(5 
 
 6966 
 
 4 20 
 
 865 
 
 7016 
 
 7066 
 
 7117 
 
 7167 
 
 7217 
 
 7267 
 
 7317 
 
 73(57 
 
 7418 
 
 7468 
 
 5 25 
 
 866 
 
 7.-» IS 
 
 7668 
 
 7618 
 
 7008 
 
 7718 
 
 7769 
 
 7S19 
 
 78(59 
 
 7919 
 
 7969 
 
 6 29 
 
 867 
 
 8019 
 
 8069 
 
 8119 
 
 8169 
 
 S219 
 
 8269 
 
 S320 
 
 S370 
 
 8420 
 
 8470 
 
 7 34 
 
 868 
 
 8620 
 
 8670 
 
 8620 
 
 8670 
 
 8720 
 
 8770 
 
 SS20 
 
 8870 
 
 .Si 120 
 
 S970 
 
 8 39 
 
 869 
 
 9020 
 
 IK 170 
 
 9120 
 
 9170 
 
 9220 
 
 9270 
 
 9320 
 
 9369 
 
 9419 
 
 9469 
 
 9 44 
 
 N 
 
 
 
 1 
 
 2 
 
 3 
 
 4 
 
 5 
 
 6 
 
 7 
 
 8 
 
 9 
 
 Diff. 
 
302 HIGHER ALGEBRA 
 
 numbers 78, 780, 7800, 7.8, .78, .078, .0078, all have the same 
 mantissa. 
 
 In all cases the first three figures of the number are found in 
 the column headed N, and the mantissa of the logarithm is found 
 in the same horizontal line with these. 
 
 If the number has but three figures, the mantissa is in the 
 column headed O. Thus log 793 = 2.899273. 
 
 If the number has four figures, the last four figures of the 
 mantissa are in the column headed with the fourth figure of 
 the number, and the two initial figures are in the column headed 
 O. The initial figures are never taken from a lower line unless 
 asterisks occupy one or more places of the last four figures of the 
 mantissa, in which case the places of the asterisks are to be sup- 
 plied with 0's. Thus, 
 
 log 8426 = 3.925621, and log 7945 = 3.900094. 
 
 If the number has more than four figures, the mantissa of the 
 logarithm for the first four figures is found as just explained, and 
 additions are made to it for the other figures by means of one of 
 the auxiliary tables in the margin, the one to be used being 
 headed with the difference between the mantissa just found and 
 the one next larger in the table. The correction for the 5th 
 figure of the number is as given in the auxiliary table ; the cor- 
 rection for the 6th figure is one tenth of that given in the same 
 table, and so on. Thus, to find the logarithm of 8576479, we pro- 
 ceed as follows : 
 
 Mantissa of log 8576 = .933285 
 
 Correction for 4 in 5th place = 20 
 
 Correction for 7 in 6th place = 35 
 
 Correction for 9 in 7th place = 45 
 
 Log 8576479 = 6.933309 
 
 The corrections are found in the auxiliary table headed 50, 
 this being the difference (in the last two decimal places) between 
 the mantissa taken and the next larger. 
 
LOGARITHMS 303 
 
 EXAMPLES CXXI 
 
 Find from the table the logarithms of the following numbers: 
 1. 798. 2. 8.62. 3. 7784. 
 
 4. 82.36. 5. .8127. 6. 8516. 
 
 7 78754. 8. 84.936. 9. .077284. 
 
 10. 8682.56. 11. 78.5716. 12. .00859465. 
 
 498. To find from the table a number whose logarithm is given. 
 
 We find in the table the next lower mantissa. The first three 
 figures of the number will be in the same horizontal line in the 
 column headed N, and the fourth figure at the top of the column 
 in which the last four figures of the next lower mantissa are 
 found. The remaining figures are found by one of the auxiliary- 
 tables. Thus, to find x when log x = 2.904567, we proceed as 
 follows : 
 
 The next lower mantissa is .904553, which corresponds to the 
 number 8027. The difference between this mantissa and the next 
 higher is 54 (in the last two decimal places), and this shows 
 which of the auxiliary tables to use. The difference between 
 log x and log 8027 is 14. This exact difference is not found in the 
 auxiliary table headed 54, but the next lower difference, 11, gives 
 2 for the 5th figure of the number. This leaves a difference of 
 14 — 11 = 3 still to be provided for. As the differences for the 
 6th figure are only one tenth of the differences in the auxiliary 
 table, we multiply our difference by 10 (this is the same as divid- 
 ing the differences in the auxiliary table by 10), giving 30. The 
 next smaller difference in the table is 27, which gives 5 for the 
 6th figure of the number. The characteristic 2 shows that 
 the number has three figures to the left of the decimal point. 
 Hence a = 802.725. 
 
 EXAMPLES CXXH 
 
 Find from the table the numbers of which the following are 
 the logarithms : 
 
 1. 3.919758. 2. 2.899137. 
 
 3. 0.912451. 4. 4.938343. 
 
304 HIGHER ALGEBRA 
 
 5. 1.902140. 6. 1.922995. 
 
 7. 2.888596., 8. 3.930272. 
 
 9. 5.934953. 10. 5.913814. 
 
 11. 3.903090. 12. 5.913578. 
 
 Solve the following by logarithms : 
 
 13. 7746 : 8334 :: 8027 : x. 
 
 14. 77.53 : 821.6 : : 8.097 : x. 
 
 80.75 x 82.79 x 791.7 
 8.692 x 770.4 
 
 16. 1 -v / 791673000. 17. 8* = 79846. 
 
 18. 7.94 s6 = 86978. 19. 8.6* = 7896.24. 
 
 Solution of 17th. Passing to logarithms, we have 
 x log 8 = log 79846, 
 
 . log 79840 4.902253 r , oe „ 
 
 whence x = — p = = 5.4283« 
 
 log 8 .90309 
 
CHAPTER XXII 
 INDETERMINATE EQUATIONS OF THE FIRST DEGREE 
 
 499. An Indeterminate Equation is an equation having two or 
 more unknown quantities, this equation expressing the only con- 
 dition imposed upon the unknown quantities. Hence, 
 
 A Set of Equations is Indeterminate when it contains more un- 
 known quantities than equations ; for by elimination the set may 
 be reduced to a single equation containing two or more unknown 
 quantities. 
 
 As shown in Art. 380, the unknown quantities of such an equa- 
 tion are variables and admit of an infinite number of values. 
 
 Thus, in the equation bx — 3 y = 11, 
 
 we may give to one of the variables any value we please and find for the other 
 such a value as will satisfy the equation. 
 Again, from the equations 
 
 3x-y + 4z = 22, 
 
 4 x + 3 y - 2 z = 19, 
 
 we have by eliminating z, 11 x + by = 60, 
 
 in which x and y admit of an infinite number of values. 
 
 500. A single equation having more than one unknown quantity 
 is indeterminate in a less general sense, if an additional condition 
 not capable of being expressed in an equation is imposed. 
 
 Thus, let it be required to find the positive integral values of x and y in 
 
 7x-f 6y=118. 
 
 The introduction of the condition that the values are to be positive integers 
 greatly restricts the number. In this equation 
 
 when x = 4, y = 18, 
 
 when x = 9, y = 1 1 , 
 
 when x = 14, y = 4, 
 downey's alg. — 20 305 
 
306 HIGHER ALGEBRA 
 
 and it may be shown that these are the only positive integers that will satisfy 
 the equation. 
 
 The discussion here will be confined to positive integral values of the 
 unknown quantities. 
 
 501. Any equation of the first degree containing two unknown 
 quantities can be reduced to one of the forms ax ± by = ± c, in 
 which a, b, c are positive integers. The form ax + by = — c is 
 not satisfied for any positive integral values of x and y, and the 
 form ax — by = — c is equivalent to by — ax = c ; hence we need 
 to consider only the forms ax ± by = c. 
 
 502. Theorem. The forms ax ±by — c have no positive integral 
 values of x and y, if a and b have a common factor not contained 
 in c. 
 
 Dem. Dividing both members by this common factor, the 
 second member is fractional for all values of x and y, while the 
 first member is integral for all positive integral values of x and y ; 
 hence the equations are not satisfied for such values. 
 
 503. Theorem. The number of positive integral values of x and y 
 in the form ax—by = c(a and b being prime to each other) is limited. 
 
 Dem. Solving for x, we have 
 
 c— by 
 
 x — £• 
 
 a 
 
 Now we can use only those positive integral values of y that 
 will render c — by positive (i.e., by > c) and divisible by a ; hence 
 the number of positive integral values of both x and y is limited. 
 
 504. Theorem. The number of positive integral values of x and 
 y in the form ax — by = c (a and b being prime to each other) is 
 infinite. 
 
 Dem. Solving for x, we have 
 
 c + by 
 
 x = — 
 
 a 
 
 No positive integral value of y will render x negative, and an 
 infinite number of such values will render c + by divisible by a ; 
 hence the number of positive integral values of both x and y is 
 infinite. 
 
INDETERMINATE EQUATIONS 307 
 
 EXAMPLES CXXIH 
 
 Find the positive integral values of x and y in each of the 
 following : 
 
 1. 5» + 9y = 37. 
 
 Solution. Solving for x in terms of y, we have 
 
 • Aii'fi^Jt (!) 
 
 5 
 Since x must be positive, 
 
 9*/>37, 
 
 whence 2/ > 4$, 
 
 and as ?/ must be integral, it cannot be greater than 4. 
 Now (1) may be written in the form 
 
 x = 7 - v + l^A£ = 7 -, J + 2±zl2. (2 ) 
 
 o 5 
 
 To make x integral *• must be integral. The only value of y not 
 
 5 
 
 i o ft 
 
 greater than 4 that will make ^ integral is 3. This substituted in either 
 
 5 
 (1) or (2) gives x — 2. 
 
 2. 8a; + 13 7/ =138. 
 
 Solution. Solving for x in terms of y, we have 
 
 x = ™- 1 *'. (1) 
 
 8 v J 
 
 Since x must be positive, 
 
 13 y > 138, 
 
 whence y >> 10. 
 
 Now (1) may be written in the form 
 
 x = n-y+'l^JL. (2) 
 
 When it is not easy to see by inspection all the values of y 
 that will render the fractional part integral, we may proceed as 
 
 follows : 
 
 2 5 « 
 
 Since — ^ must be an integer, any integral number of times this 
 
 8 
 quantity will be an integer. Let us multiply by the smallest number that 
 will give a remainder 1 (or - 1) on dividing the coefficient of y by the 
 denominator. This multiplier in this case is 5, and we have 
 
308 HIGHER ALGEBRA 
 
 It is now easy to see that 2 and 10 are the only two numbers not greater 
 
 an 10 tl 
 14 and 1. 
 Or 
 
 givini 
 
 than 10 that will make - integral, and the corresponding values of x are 
 
 8 
 
 Or, if we choose, we may represent ~ y by the undetermined integer ra, 
 
 = m, 
 
 8 
 whence y = 2 - 8 m, • (4) 
 
 and from (1) x = 14 + 13 m. (5) 
 
 It is seen that x and y are integral for all integral values of m, and (4) and 
 (5) constitute what is called the general solution in integers of the given 
 equation ; but (4) shows that only and negative values of m will give posi- 
 tive values for y, and (5) shows that m cannot be smaller than — 1. Hence 
 and — 1 are the only admissible values of m. 
 
 Now when m = 0, x = 14, and y = 2 ; 
 
 and when m = — 1, x = 1, and y = 10. 
 
 3. 13a -30?/ = 61. 
 
 Solution. From Art. 504 we know that x and y have an infinite number 
 of values. Solving for x in terms of y, we have 
 
 g = 6i+aoy = 4 + 2 + o±±g. 
 
 13 * 13 
 
 Multiplying the numerator of the last fraction by 10 and reducing to a 
 mixed quantity, we have 
 
 <*> + 40y = 64 . 3 y + i2+x 
 
 13 J 13 
 
 The values of y that make 12 + y integral are I, 14, 25, etc., and the cor^ 
 13 
 
 responding values of x are 7, 37, 67, etc. 
 
 4. 14#-21?y = 32. 
 
 As proved in Art. 502, the solution is impossible. 
 
 5. 3z + 7# = 10. 6. 3x + 5y = 26. 
 7. 5x-Uy=ll. 8. 2»-M# = 10. 
 9. 27 x + 18?/ = 47. 10. 4a;- 19?/ = 23. 
 
 11. 3# + 7?/ = 58. 12. 13a + 2?/ = 119. 
 
 13. 14#-5?/ = 7. 14. 2Sx-Soy = 23. 
 
 3?/-5z = -8, f20a-21?y = 38, 
 
 15. 
 
 |2a; + 3?/-52 = -8 J {20*-- 
 
 \5a>-y + 43 = 21. j 3?/ + 4 
 
 2=34. 
 
INDETERMINATE EQUATIONS 309 
 
 PROBLEMS LEADING TO INDETERMINATE EQUATIONS 
 EXAMPLES CXXIV 
 
 1. A man employed two squads of men, paying $3a day to 
 each of the first squad and $4 a day to each of the second. He 
 paid them all $ 06 a day. How many belonged to each squad ? 
 
 2. A dairyman paid $ 752 for cows, $ 37 each for Jerseys and 
 $ 23 each for Durhams. How many of each did he buy ? 
 
 3. In how many ways can a debt of $ 43 be paid with 2-dollar 
 and 5-dollar bills ? 
 
 4. A man invested $ 10,000 in town lots, paying $ 190 each 
 for those in one locality, and $ 130 each for those in another. 
 Find the number in each locality. 
 
 5. In how many ways can a debt of £ 50 be discharged with 
 guineas and 3-shilling pieces ? £ 51 ? 
 
 6. What fraction becomes J when its numerator is doubled 
 and its denominator is increased by 7 ? 
 
 7. How many 4-pound and 6-pound weights must be used to 
 weigh 45 pounds? 
 
 8. Divide 136 into two parts, one of which when divided by 
 5 leaves a remainder 2, and the other divided by 8 leaves a 
 remainder 3. 
 
 9. How many times each must a 7-inch stick and a 13-inch 
 stick be applied to measure 4 feet ? 
 
 10. A owes B a shilling. A has only sovereigns, and B has only 
 dollars, worth 4s. 3d. each. How can A most easily pay B ? 
 
 11. A wholesale grocer received $ 180 for 15 barrels of sirup, 
 the prices being $ 10, $ 11, and $ 13 per barrel. How many of 
 each kind did he sell ? 
 
 12. A farmer paid $ 160 for pigs, sheep, and calves. The pigs 
 cost | 3 each, the sheep $ 4 each, and the calves $ 7 each ; and the 
 number of calves was equal to the number of pigs and sheep 
 together. How many of each did he buy? 
 
CHAPTER XXIII 
 
 THEORY OF EQUATIONS AND SOLUTION OF NUMERICAL 
 HIGHER EQUATIONS 
 
 505. When a quadratic equation has the general form 
 x 2 +px = q, 
 we have seen (Art. 344) that 
 
 -f±Vf+* 
 
 in which the unknown quantity is expressed in terms of the 
 general coefficients, giving a formula by which it may be found 
 in all special cases. Solutions of general (i.e., literal) equations 
 of the third and fourth degree are known (see the methods of 
 Cardan and Descartes, Arts. 573 and 576) ; but these solutions 
 have in many cases difficulties which render them of little practi- 
 cal value, and they are seldom employed. For general equations 
 above the fourth degree no solution is known, i.e., no method of 
 expressing in terms of the literal coefficients, the values of the 
 unknown quantity, — indeed, it has been shown by Abel, a Nor- 
 wegian mathematician, that it is impossible thus to express, 
 algebraically, the values of the unknown quantity. So soon, 
 however, as numerical values are assigned to the coefficients, the 
 real roots may in all cases readily be found, either exactly or to 
 any required degree of approximation. 
 
 506. A Root of an Equation is a quantity which, substituted for 
 the unknown quantity, satisfies the equation. It may be com- 
 mensurable, incommensurable, or imaginary. 
 
 507. Commensurable Numbers are such as can be exactly ex- 
 pressed in the decimal notation. They are integers, common 
 
 310 
 
-REDUCTION TO THE NORMAL FORM 311 
 
 fractions, terminating decimals, and repeating decimals (the 
 latter, as shown in ex. 25, page 205, being reducible to equivalent 
 common fractions). 
 
 508. Incommensurable Numbers are such as cannot be exactly- 
 expressed in the decimal notation. 
 
 Thus, V2, 2 ± V3, y/E are incommensurable numbers. 
 
 509. A Multiple Root is a root that occurs more than once in an 
 equation. If it occurs twice, it is called a double root; if three 
 times, a triple root; if four times, a quadruple root, etc. 
 
 Thus, from x 5 - 4 x 4 + x 3 -f 10 x 2 - 4 x - 8 = 0, we have (Art. 101) 
 ( X _ 2 )(x - 2)(x - 2)(x + l)(x + 1)= 0, 
 giving (Art. 348) 2 as a triple root and — 1 as a double root. 
 
 We may say that an equation has the same root repeated several 
 times and call it a Multiple Root, or that it has several roots of 
 the same value and call them Equal Roots. 
 
 510. An equation is said to be in the Normal or Typical Form 
 when the exponents are all positive integers, the coefficient of the 
 first term (highest power) is 1, and the other coefficients are all 
 integers. 
 
 REDUCTION TO THE NORMAL FORM 
 
 511. Theorem. Every algebraical equation having rational co- 
 efficients can be transformed into an equation of the normal form in 
 terms of a new unknown quantity ivhich is a known function of the 
 original unknown quantity. 
 
 Dem. 1st. To make the exponents positive, the equation is mul~ 
 tiplied by the unknown quantity ivith a positive exponent equal 
 numerically to the largest negative exponent. This neither destroys 
 the equality of the members nor changes the value of the un- 
 known quantity. 
 
 2d. To make the exjjonents integral, each exponent is multiplied 
 by the 1. c. m. of the denominators of the fractional exponents, and 
 the unknown quantify is rrj>hici>d by another. If the equation is in 
 x, and if m is the 1. c. m. of the denominators of the fractional 
 
312 HIGHER ALGEBRA 
 
 exponents, the transformation consists simply in substituting y m 
 for x. While this changes the roots, the roots of the original 
 equation are known functions of the roots of the transformed 
 equation, viz., the mth power of them. 
 
 3d. To make the coefficients integral, and the first one, that of y n , 
 unity, we divide the equation by the coefficient of y n , multiply the 
 coefficient of y n ~ l by k, that of y n ~ 2 by k 2 , that of y n ~ 3 by fc 8 , and so 
 on, replacing y by z, and then take k of such value as will make each 
 coefficient integral. After the operations of 1st and 2d, and divid- 
 ing by the coefficient of the highest power, the equation, if 
 arranged, has the form 
 
 r+^2/ n - 1 + ^r- 2 + ^2/ n - 3 + " 
 
 ..)-* 
 
 Substituting y = -, this becomes 
 k 
 
 
 z n a z n l b z n ~ 2 c 2 n - 3 
 k n a' k n ~ l b' k n ~ 2 c' k n3 
 
 4-» 
 
 Multiplying by k n , 
 
 
 (1) 
 
 „ , ak ■ , , bk 2 n 9 , cZc 3 „ , , lk n _ /ox 
 
 Z + 'a 1 Z + "F Z + 7 r Z + '"T = °- (2) 
 
 Such value of k may now be taken (it should be the least 
 possible) as will remove the denominators. 
 
 z 
 
 We now have x = y m = f - j , 
 
 and the roots of the original equation become known when the 
 roots of the equation in z become known. 
 
 512. Cor. From (1) and (2) it is seen that to obtain an equation 
 whose roots are k times those of a given equation we have but to 
 multiply the coefficient of the 2d term by k, that of the 3d by k 2 , 
 that of the kth by Zc 3 , and so on to the absolute term, which is 
 multiplied by k n . 
 
 EXAMPLES CXXV 
 
 Transform the following into equations of the normal form and 
 express x in terms of the new unknown quantity : 
 
TEST FOR ROOTS 313 
 
 1. 12 a* - 3 af * - 8 a* + 10 a;* - 3 af* - 10 = 0. 
 Solution. Multiplying by x 7 , 
 
 12 x^ - 3 x$ - 8 x* + 10 x* - 3 - 10 x$ = 0. 
 Multiplying the exponents by 6 and replacing x by y, 
 
 12 y* - 3 y - 8 y 5 + 10 y 3 - 3 - 10 y 2 = 0. 
 Arranging and dividing by — 8, 
 
 y 5 - | y* - | y 3 + | y 2 + f */ + f = o. 
 
 Introducing A: as indicated in the 3d part of the demonstration and replac- 
 
 ing y by z, 
 
 Making k = 2 
 
 ^_3^^_^ 3 5F 2 3^ 3^_ 
 
 ^5 _ 3 Z i _ 5 ^ + 10 z * + 6 + 12 = 0, 
 
 in which 
 
 *=H!) 6 
 
 By Art. 363, it is found that — 2 is a root of this equation ; hence 
 
 y = =£■ = - \, and x=(- 1)« = 1. 
 Therefore 1 is the corresponding root of the original equation. 
 
 4. }~* t =* ** + * - 5. 2Vr^ = 4-3af* 
 
 1 + # _1 jc + 2 
 
 6. 3ar 5 -2x 4 -?ar 3 + 5^--^ + 1 = 0. 
 
 5 ar o a; o 10 
 
 TEST FOR ROOTS 
 
 513. Theorem. If f(x) be divided by x — a, the last remainder 
 irill be f(a); i.e., will be what f(x) becomes when a is substituted 
 for x. 
 
 Dem. Let <f>(x) be the quotient and r the last remainder. 
 Then, by the laws of division, 
 
 /(*)«(«-a)f(*) + r. 
 
314 HIGHER ALGEBRA 
 
 As this equation is true for all values of x, it is true when 
 x = a. Substituting a for x, we have 
 
 /(a)=r. 
 
 514. Sch. It follows from the last theorem that the short 
 method of dividing by x — a, given in Art. 79, becomes a short 
 method of substituting any number for x. 
 
 The operation of substituting a value for x is often called 
 evaluating. 
 
 EXAMPLES CXXVI 
 
 Find what the following become when the indicated values of x 
 are substituted : 
 
 1- x 5 -2x 4 -4:X i + 3x 2 -5x + 6 for x = 3. 
 Solution. By the method of division given in Art. 79, we have 
 x 5 -2x 4 -4x 3 + 3x 2 -5x + 6 [_3 
 1 _ i o -5-9 
 
 Therefore, by Art. 513, this polynomial becomes — 9 when 3 is substi- 
 tuted for x. 
 
 2. x* + a? ~ 17 x 2 - 10 x + 11 for x = 4. 
 
 3. ^-8^ + 18^-12^-14^-4 for x = 5. 
 
 4. 3a 6 -4af-6ar 5 -4ar J -10;e + 23 for x = 2. 
 
 Operation 
 3 x 6 - 4 x 5 + x 4 - 6 x 3 - 4 x 2 - 10 x + 23 |_2 
 2 4 2 0-10 3 
 
 Therefore this polynomial becomes 3 when 2 is substituted for x. The 
 same operation shows (Art. 79) that when the polynomial is divided by 
 x — 2, the quotient is 3 x 5 + 2 x 4 + 4 x 3 + 2 x 2 — 10, with a remainder of 3. 
 
 5. 5 x 5 + 12 x 4 + 6 x 3 - 3 a; 2 - 13 x + 9 for a; = - 2. 
 
 6. 4 a 5 - 17 x 4 + 18 x 2 - 20 for z = 4. 
 
 7. x 6 - 8 x* + 10 z 4 + 40 x 3 - 71 x 2 - 32 a + 60 for x = 5. 
 
 515. Theorem. 7/" /(.r) & divisible by x — a, a is a rootf o/ 
 . f(x) = ; and, conversely, if a is a root of f(x) = 0, f(x) is 
 
 divisible by x — a. 
 
NUMBER AND CHARACTER OF ROOTS 315 
 
 Dem. By Art. 513 if f(x) be divided by x — a, the remainder 
 is /(a). Now, if the remainder is 0, /(a) = 0, and the equation 
 is satisfied. 
 
 Again, using the notation of Art. 513, 
 
 '(x — a)+(x) 4- r =/(*). 
 
 Now if a is a root, f(x) is and x — a is 0, and the equation 
 becomes + r = 0. 
 
 Hence, the remainder being 0, f(x) is divisible by x — a. 
 
 516. Cor. If a is a root of f(x) = 0, a is a factor of the abso- 
 lute term. 
 
 If a is a root, there is no remainder when f(x) is divided by 
 x — a ; therefore, as readily appears from the process of division, 
 a is a factor of the absolute term. 
 
 NUMBER AND CHARACTER OF THE ROOTS 
 
 517. Theorem. An equation in the normal form cannot have a 
 fractional root, and hence the real roots of such an equation are 
 either integral or incommensurable. 
 
 Dem. Suppose -, a simple fraction in its lowest terms, to be a 
 root of the normal equation 
 
 x n + Ax n ~ l + Bx n 2 4- Cx n - S + ••• L = 0. 
 Substituting this value of x, we have 
 
 
 + - + £ = 0. 
 
 In this equation all the terms except the first are integral, and 
 the first is a simple fraction in its lowest terms, s and t being 
 prime to each other. But the sum of a simple fraction in its 
 lowest terms and a series of integers cannot be 0. Hence x 
 
 cannot equal -, a fraction. 
 
 518. Sch. It is readily seen that the above reasoning does not 
 apply if the coeificient of the first term is other than 1 ; for in 
 
316 HIGHER ALGEBRA 
 
 that case tlie coefficient might contain t, making the first term 
 integral instead of fractional when - is substituted for x. 
 
 519. Theorem. An equation of the nth degree has n roots and no 
 more* 
 
 Dem. Let a be a root of /(a?) = 0. Then f(x) is divisible by 
 x — a (Art. 515). Representing the quotient by <f>(x), we have 
 
 O-a)<£(»=0. (1) 
 
 This is satisfied not only for x — a = 0, but also for <f>(x) = 
 (Art. 348), and the degree of <f>(x) is one lower than that of fix). 
 If b is a root of <f>(x) = 0, it is also a root of f(x) = 0, as seen 
 from (1). Dividing <f>(x) by x — a, the degree will again be 
 diminished by 1. This process may be continued for n divisions 
 and no more. Hence f(x) = has n roots and no more. 
 
 520. Cor. i. To form an equation ivhose roots shall be a, b, c, 
 • •• I, we have 
 
 f(x) = (x — a) (x — b) (x — c) • • • (x — l) = ; 
 
 i.e., to form an equation having any roots whatever, ive have but to 
 subtract these roots from x and place the product of the remainders 
 equal to 0. 
 
 521. Cor. 2. The equation f(x) = may have 2, 3, or even n 
 equal roots, as in the above demonstration we may have 
 
 a = b, a = b = c, or a — b = c = • • • I. 
 
 522. Theorem. If an equation having only rational coefficients 
 has imaginary or quadratic surd roots, these roots enter in conjugate 
 pairs (Art. 221). 
 
 Dem. If /3 V — 1 is a root of f(x) = 0, x — /?V— 1 is a factor 
 of f(x) (Art. 515). Now the only factor by which x — /?V— 1 
 can be multiplied to produce a real and rational quantity is 
 x + fi-y/— 1, giving x 2 + /3 2 for the product. Hence (Art. 515) 
 — /?V— 1 is also a root. 
 
 * It is here assumed that every equation has at least one root. This has 
 been proved by Argand, Gauss, Cauchy, Clifford, and others. 
 
NUMBER AND CHARACTER OF ROOTS 317 
 
 Again, if a -+- /?V— 1 is a root of f(x) = 0, x — (a-{- /?V— 1) 
 is a factor of f(x) ; hence to give a real and rational product, 
 x — (a — /JV^T) must also be a factor, giving « — /?V— 1 as a 
 root, and (# — «) 2 + /3 2 as the product of the two factors. 
 
 It may be shown in the same way that if /?Vy is a root of 
 f(x) = 0, — ftVy must also be a root; and that if a + /?Vy is a 
 root of /(#) = 0, a — /? Vy must also be a root. 
 
 523. Cor. i. An equation of odd degree has at least one real 
 root. 
 
 524. Cor. 2. All the roots of an equation of even degree can be 
 imaginary only when the absolute term is positive. 
 
 For, as seen above, the product of the factors obtained by sub- 
 tracting conjugate imaginaries from x always gives a positive 
 absolute term. 
 
 525. Cor. 3. An equation of even degree whose absolute term is 
 negative has at least two real roots, and these have opposite signs. 
 
 526. Theorem. Changing the signs of the terms containing the 
 odd powers, or the even powers, or, if the equation is complete, the 
 alternate terms, changes the signs of the roots of an equation; i.e., 
 the positive roots of f{— x) = are the negative roots of f(x) = 0. 
 
 Dem. Let a, b, c, •••/ be the roots of an equation f(x) = 0; 
 then (Art. 520) 
 
 f(x) = (x- a)(x -b)(x- c) ... (x - I) = 0. 
 
 Replacing x by — x, we have 
 
 /(- x) = (- x - a) (- x - b) (- x - c) ... (- x - I) = 0, 
 
 or, removing — 1 from each factor, 
 
 /(- x) = (- l) n (x + a) (x + b) (p + e) - (a? + 1)?* 0. 
 
 By equating each of these binomial factors with (Art. 348), 
 the roots are seen to be — a, — b, — c, ••• — I; i.e., changing the 
 sign of x changes the signs of the roots. But changing the sign 
 of x changes the signs of the terms containing the odd powers, 
 and only these. 
 
318 HIGHER ALGEBRA 
 
 Changing the signs of the terms containing the even powers 
 also changes the signs of the roots, since this is the same as 
 multiplying or dividing both members by — 1 after changing the 
 signs of the terms containing the odd powers. 
 
 527. Sch. The absolute term may be regarded as the coeffi- 
 cient of x°, and it must be noted that is an even number. 
 
 528. A Permanence is a succession of two like signs in the con- 
 secutive terms of a polynomial ; a Variation, of two unlike signs. 
 
 Thus, in x 6 — 2 x s — 4 x* + 5 x 3 + 3 x 2 + x — 4, the signs being -j 
 
 + + H — , there are three permanences and three variations. 
 
 529. Descartes' Rule of Signs. An equation in the form f(x) = 0, 
 ivJiether complete or incomplete, cannot have more positive roots than 
 the number of variations in f(x), nor more negative roots than the 
 number of variations in f(—x). 
 
 Verification. Let the signs and missing terms of f(x) be 
 
 -f +0-+00 +, 
 
 giving 4 variations. 
 
 Let us now introduce a positive root by multiplying by x minus 
 
 this root (Art. 515), writing only the signs in the operation, as 
 
 follows : 
 
 + +0-+00 + 
 
 + +0-+00 + 
 
 0+-00+ + + 
 
 + ± +-0-TT + 
 
 The ambiguous signs ± and q= are placed where the sign to be 
 used must be determined by the relative numerical values of the 
 coefficients. 
 
 Counting, now, the fewest number of variations that can occur, 
 by taking that one of the ambiguous signs that will give, with the 
 preceding sign, a permanence, we have 5 variations, a variation 
 being necessarily introduced at the end. Thus each time a pos- 
 itive root is introduced by multiplying by x minus that root, at 
 least one variation is added. Hence an equation cannot have 
 more positive roots than the number of variations. 
 
NUMBER AND CHARACTER OF ROOTS 319 
 
 Again the positive roots of /(— x)= are the negative roots of 
 f(x)=Q (Art. 526). As just shown, /(— x) = cannot have more 
 positive roots than the number of variations in /(— x). Hence 
 f(x) sa cannot have more negative roots than the number of 
 variations in /(—#). 
 
 530. Cor. i. An equation whose terms are all positive has no 
 positive roots. 
 
 531. Cor. 2. An equation whose terms of ev&n power are all of 
 one sign and ichose terms of odd power are all of the contrary sign 
 has no negative roots. Why? 
 
 Hence a complete equation whose terms are alternately + and — 
 has no negative roots. 
 
 532. Cor. 3. An equation having only even powers, and these all of 
 the same sign, has no real roots. An equation having only odd 
 powers, and these all of the same sign, has no real root except 0. 
 
 533. Note. The truth of the above corollaries is readily seen 
 independently of Descartes' Rule of Signs. 
 
 534. Cor. 4. An incomplete equation has at least as many imagi- 
 nary roots as the difference between the degree of the equation and the 
 total number of variations in f(x) and f(—x). 
 
 For the total number of roots is the same as the degree of the 
 equation, and there cannot be more real roots than the total num- 
 ber of variations in f(x) and /(— x). 
 
 535. Cor. 5. A complete equation cannot have more negative roots 
 than its number of permanences. 
 
 The total number of variations and permanences is one less than 
 the number of terms, i.e., the same as the degree of the equation 
 or the total number of roots. Now there cannot be more positive 
 roots than the number of variations, and hence not more negative 
 roots than the number of permanences. 
 
 536. Note. Descartes' Rule and the corollaries deduced from 
 it are useful only in preventing, in some cases, a fruitless search 
 for roots of a particular sign after all of that sign have been 
 found, or for the situation of roots after the situation of all the 
 real ones has been found. 
 
320 HIGHER ALGEBRA 
 
 EXAMPLES CXXVII 
 
 Determine the greatest number of positive and negative, and 
 the least number of imaginary, roots in the following : 
 
 1. » 6 + 8ar 3 + 3oj-6 = 0. 
 
 Solution. By Art. 525 this equation has 2 real roots, 1 positive and 1 
 negative. Since there is but 1 variation, the equation cannot have more than 
 1 positive root. Changing the signs of the terms containing the odd powers 
 of «, we have 
 
 x e _ 8 x 3 - 3 x - 6 = 0. 
 
 Since this has but 1 variation, the original equation cannot have more than 
 1 negative root. But the equation, being of the 6th degree, has 6 roots ; 
 hence it has 4 imaginary roots. 
 
 2. a; 4 + 12a 2 + 5a -10 = 0. 
 
 3. 3^ + 4^ + 5=: 0. 
 
 4. ar 5 - 7 x 2 - G = 0. 
 
 5. x 6 - 7^-^ + 3 = 0. 
 
 6. 2a 6 -43 5 -5<B 8 + 3a 2 -5 = 0. 
 
 7. x« + 7x 5 -4:X 4 -5x i -3x-6 = 0. 
 a x c, -2x* + x 4 -x 2 -l = 0. 
 
 9. x n — 1, when n is odd. 10. x n — 1, when n is even. 
 
 11. x n + 1, when n is odd. 12. x n + 1, when n is even. 
 
 SOLUTION FOR COMMENSURABLE ROOTS 
 
 537. The process of finding the roots of an equation when all 
 the roots, or all but two of them, are integral, has been fully 
 explained in Chapter XVI. We now see from Art. 513 that the 
 short process of factoring there used is also a short process of 
 evaluating for, or substituting, any particular value of the unknown 
 quantity. A few additional examples are here given for further 
 practice in that important process. In applying it the student 
 should keep in mind the following : 
 
 1. When the equation is in the normal form, only factors of 
 the absolute term need be tried (Art. 516). 
 
SOLUTION FOR COMMENSURABLE HOOTS 321 
 
 2. It will be found expedient to try the small- positive factors 
 first. Before writing down the various sums, it is best, if the 
 coefficients are not too large, to run through mentally with a 
 factor of the last term and see whether the last addition gives 0. 
 
 3. If all the terms are positive, no positive numbers need be 
 tried (Art. 102) ; and if at any stage in the operation of evaluat- 
 ing, all the sums become positive, no more positive numbers need 
 be tried. 
 
 4. When all but two of the roots have been found, the remain- 
 ing two, whatever their values, may be found by solving the 
 resulting quadratic. The last four roots, even if surd or imagi- 
 nary, may be found if the alternate terms of the depressed equa- 
 tion are 0, since this depressed equation would then have the 
 quadratic form (Art. 359). 
 
 EXAMPLES CXXVm 
 
 Find all the roots of the following : 
 
 1. a°-4ar i -6a; 4 + 40ar ? -31ar J -36a: + 36 = 0. 
 
 Operation 
 s6 _ 4 ^5 _ e a* + 40 a* - 31 x* - 36 x + 30 | 1 
 
 -3 
 
 - 9 
 
 31 
 
 
 
 -36 
 
 0|_2 
 
 -1 
 
 -11 
 
 9 
 
 18 
 
 
 
 [1 
 
 1 
 
 - 9 
 
 - 9 
 
 
 
 
 Li 
 
 4 
 
 3 
 
 
 
 
 
 l-i 
 
 3 
 
 
 
 
 
 
 |-3 
 
 
 Or the last two roots may be found by solving the quadratic 
 x 2 + 4 x + 3 = 0, giving x = -2±l=-l and - 3. 
 
 2. ^-3^-14^ + 24 = 0. 3. x 5 -7x + 6 = 0. 
 
 4. ar 3 -15ar> + 74a;-120 = 0. 5. ^-7^-52 + 35 = 0. 
 
 6. z 4 -10z 3 + 24z 2 + 10x-25 = 0. 
 
 7. a* 4 - 10a? + 20a? + 10a; -21=0. 
 
 a x 5 - 5x* - 13ar> + Gox 2 + 36x- 180 = 0. 
 downey's alg. — 21 
 
322 HIGHER ALGEBRA 
 
 9. af + 2x 4 -3a?-3x 2 + 2x + l = 0. 
 
 10. x 5 - 6 x" + 15 a? - 26 x 2 + 36 a? - 24 = 0. 
 
 11. ar 5 + 2 a; 4 - 9 X s + 14 a; - 8 = 0. 
 
 12. ar 5 - 5 ar 3 + 6 x 2 - 14 a? + 12 = 0. 
 
 13. x 5 - l&x 4 + 85 a 3 - 225 x 2 + 274a>- 120 = 0. 
 
 14. af - 2 x 4 - 6 a; 8 + 12 x 2 + 9 a; - 18 = 0. 
 
 15. ar 5 — x 4 + 4 a? 3 — 4 x 2 + 4 a; — 4 = 0. 
 
 16. 4 ar 5 - 4 x 4 - 31 ar 5 + 10 a; 2 + 39 x - 18 = 0. 
 
 Suggestion. When the coefficient of the first term is not 1, first find all 
 the integral roots, and then, if necessary, transform the depressed equation 
 to the normal form. 
 
 17. 6 x 5 - 19 x 4 + 13 X s + 13 x 2 - 19 x + 6 = 0. 
 
 18. a; 6 - 3 x> - 3 x 4 + 15 a? - 6 a* 2 - 12 x + 8 = 0. 
 
 19. x c > - 4 a^ 5 - 4 x 4 + 20 X s - x 2 - 16 x + 4 = 0. 
 
 20. a; 7 - 2 a; 6 - 10 a- 5 + 28 x 4 + 5 ar 5 - 74 a; 2 + 76 a? - 24 = 0. 
 
 21. 12 x 6 - 20 x 5 - 85 a 4 + 92 x> + 97 a; 2 - 132 x + 36 = 0. 
 
 EQUATIONS WITH ONLY EVEN OR ONLY ODD POWERS 
 
 538. When 'an equation contains only even powers of x, we may- 
 regard x 2 as the unknown quantity and, without supplying the 
 missing odd powers, proceed as above and then extract the square 
 root of the numbers which reduce the function to 0. This is of 
 great importance in finding the roots of such an equation when 
 they are monomial surds or monomial imaginaries. 
 
 An equation containing only odd powers can, after dividing by 
 x (for in that case there could be no absolute term), be treated in 
 the same way. 
 
 EXAMPLES CXXIX 
 
 Find all of the roots of the following : 
 
 1. a 6 -14x 4 -f 49o 2 -36 = 0. 
 
 Solution. This may be written 
 
 (a 2 ) 8 - 14 (z 2 ) 2 + 49 (a 2 ) - 36 = 0. 
 
RELATION OF ROOTS TO COEFFICIENTS 323 
 
 Regarding x 2 as the unknown quantity, there are no missing terms. We 
 may, therefore, proceed as follows : 
 
 (x 2 ) 3 - 14 (x 2 ) 2 + 49(x 2 ) - 36 \_l 
 -13 36 |_4 
 
 - 9 |_9 
 
 
 Hence the factors are x 2 — 1, x 2 — 4, x 2 — 9, and the roots are ± 1, ±2, ±3. 
 
 2. x*-3x«-15x A + 19^ + 30 = 0. 
 
 Operation 
 
 x * - 3X 6 - 15 x 4 + 19x 2 + 30 | 2 
 
 1 
 
 -17 
 
 -15 
 
 ( 5 
 
 4 
 
 3 
 
 
 
 L=i 
 
 3 
 
 
 
 |-3 
 
 
 
 
 
 
 Hence the factors are x 2 — 2, x 2 — 5, x 2 + 1, x 2 + 3, and the roots are 
 ±V2, ±V5, ±V^T, ±V^S. 
 
 3. a *-7x 4 +Ux i -8 = 0. 4. x^-lx'+lQx 2 - 12 = 0. 
 
 5. x 6 -10a; 4 + 31^-30 = 0. 6. x 6 - 13 a 4 + 35 a; 2 + 49=0. 
 
 7. a?-16rf+85a?-150=0. 8. ^-15a; 4 +74ar-l20 = 0. 
 
 9. # 6 -7a; 4 + 7ar J + l5 = 0. 10. x 7 - 6 ^ + 3 ar* + 10 x = 0. 
 
 11. x- 8 - 14^ + 65 a 4 -124^ + 84 = 0. 
 
 12. z 8 -5;c 6 -10z 4 +20ar J + 24 = 0. 
 
 13. ar , -9« 7 +14ar i + 36ar 5 -72a: = 0. 
 
 14. x? + 11 x- 6 + 41 x* + 61 x 2 + 30 = 0. 
 
 15. z 10 -4a 8 -3# 6 + 22a; 4 -4a; 2 -24 = 0. 
 
 16. ^-3^-7^-4- 21^+10^-30 = 0. 
 
 17. x 7 - 3 x 6 - 10 x 5 + 30 x 4 -4- 31 ar 3 - 93 a 2 - 30 a 4- 90 = 0. 
 
 RELATION OF ROOTS TO COEFFICIENTS 
 
 539. Prob. To find the relation between the roots and the coeffi- 
 cients of an equation. 
 
324 HIGHER ALGEBRA 
 
 Solution. If a, b, c, are the three roots of an equation, we 
 have (Art. 520) 
 
 f(x) = (x — a) (x — b) (x — c) = 0, 
 or X s — (a -f & 4- c) # 2 4- («6 4- ac 4 6c) cc — a6c = 0. 
 
 If a, b, c, d, are the four roots of an equation, we have 
 f(afy = (x — a) (x — b) (x — c)(x — d) = 0, 
 or sc 4 — (a 4- 6 4- c + d) x 3 + (a6 + ac + ad ■+■ 6c + 6d -f- cd) x 2 
 — (abc 4- a6d 4- acd 4- 6cd) ic + abed = 0. 
 
 Generalizing, we have the following theorem : 
 
 The coefficient of x n ~ l is the sum of the roots with their signs 
 changed. 
 
 The coefficient of x n ~ 2 is the sum of the products of the roots taken 
 two and two. 
 
 The coefficient of x n ~ 3 is the sum of the products of the roots, with 
 their signs changed, taken three and three, and so on. 
 
 The absolute term is the product of all the roots with their signs 
 changed. 
 
 540. In forming an equation from its roots the student will find 
 it safer and more expeditious to proceed as directed in Art. 349. 
 
 Thus, to produce the equation whose roots are 4, 3, 2, — 2, — 1, we have 
 
 /(*) = (* - 4)(x - 3)(x - 2)(x + 2)(x + 1) = 0. 
 
 The product of the first two factors is found by the process of Art. 60 to be 
 x 2 — 7 x + 12. We then proceed as follows : 
 
 (_2) 1-7 12 
 
 (2) _9 26 -24 
 
 (1) -7 8 28-48 
 
 -6 1 36-20-48 
 
 Hence the equation is 
 
 X 5 _ 6 x 4 + x 3 + 36 x 2 - 20x - 48 = 0. 
 
 If some of the roots are imaginaries or quadratic surds, and 
 hence occur in conjugate pairs, the remainders should be multi- 
 plied together in pairs, so as to make use of the theorem for the 
 product of the sum and difference of two quantities. 
 
SITUATION OF ROOTS 325 
 
 Thus, to produce the equation whose roots are 2 ± V3, 1 ± V— 1, we have 
 
 /(x)=(x-2-V3)(x-2+\/3)(x-l -V^T)(x- 1 +V^T) = 0, 
 
 or /(x) = (x 2 -4x + l)(x 2 -2x + 2)=0. 
 
 The indicated multiplication should now be performed by the process of 
 Art. 56. 
 
 EXAMPLES CXXX 
 Produce the equations whose roots are the following: 
 1. 1,2,3, -4. 2. 2,5,-1,-3. 
 
 3. 1, 1, 2, ±V2. 4. 1, -1, ±V2, ± VS. 
 
 5. 1, 2, 3, 4, -1,-2. 6. ± V2, ± V5, ± V^I 
 
 7. if, -1, -3,-4. a if, -l, ±V3. 
 
 9. 2, -2, 1±V2, 2±V^T. 10. 3±V5, 2±V2, 3±V^3. 
 
 SITUATION OF ROOTS 
 
 541. Theorem. If f(x) has opposite signs tvhen two different 
 numbers are substituted for x, an odd number of real roots lie be- 
 tween the substituted numbers; while if fix) has the same signs for 
 both numbers, either no real roots, or an even number of real roots, 
 lie between them. 
 
 Dem. Let a, b, c, etc., in the order of their values, a being 
 least, be the real roots of fix) = 0, and let <£ (x) be the product 
 of the remainders obtained by subtracting the imaginary roots, 
 if any, from x-, then (Art. 520), 
 
 f(x) = (* -a)ix- b) ix - c) .- (<£ ix)) = 0. 
 
 If any number be substituted for x, all the above factors con- 
 taining roots less than the substituted number will be +, and all 
 containing roots greater than the substituted number will be — . 
 If now a second number be substituted for x, only those factors 
 containing roots between the substituted numbers will change 
 sign. Hence, since <f> ix) is always + (Art. 522), a change of sign 
 of the product would indicate that an odd number of real roots lie 
 
326 HIGUEB ALGEBRA 
 
 between the substituted numbers ; while no change of sign of the 
 product would indicate that no real roots or an even number of 
 real roots lie between the substituted numbers. 
 
 Or the truth of the theorem may be seen thus : 
 
 All values < are — and all values > are + . Now if one 
 value of x makes /(aj) — , i.e., too small, and another value of x 
 makes f(x)+, i.e., too large, there must be a value of x between 
 these two that will make f(x) 0, i.e., satisfy the equation. 
 
 In passing from a value too small to a value too large, i.e., 
 from — to +, or vice versa, f(x) might pass through three, five, 
 or some other odd number of times; but if for two different 
 values of x,f(x) has the same sign, it is because f(x) has either 
 not passed through at all, or passed through an even number of 
 times, leaving it on the same side of 0. 
 
 542. Cor. Whenever the substitution of two consecutive numbers 
 gives opposite signs to f(x), the numerically smaller of the two num- 
 bers is the integral part of the root that lies between them. 
 
 543. Sch. 1. While we may find the situation of negative roots 
 by evaluating f(x) for negative numbers, it is a little more con- 
 venient to evaluate f(—x) for positive numbers (Art. 526). Be- 
 sides, as will be seen later, in approximating beyond the integral 
 part the value of a negative root, it is necessary to employ /(— x). 
 
 544. Sch. 2. By taking the value of x large enough the first 
 term off(x) may be made larger than any or all of the other terms. 
 Hence, if a series of minus signs result from the substitution 
 of larger and larger numbers, the situation of a root will be found 
 by taking still larger numbers, as the function must eventually 
 become plus, the sign of the first term. 
 
 EXAMPLES CXXXI 
 
 Find the situation of all the real roots of the following : 
 
 1. a 4 -2ar J -13a 2 + 22a + 22 = 0. 
 
 Solution. Since f(x) has but two variations, the equation cannot have 
 more than two positive roots (Art. 529). Substituting for x, by Art. 513, 
 
8ITUATION OF ROOTS 
 
 327 
 
 the values 0, 1, 2, 3, etc., we have for /(as) the signs given in the margin, 
 showing that the positive roots lie between 2 and 8, 3 
 and 4. 
 
 Changing the signs of the alternate terms of f(x), 
 we have 
 
 /(- x) = x* + 2 x 3 - 13 & - 22 x + 22 = 0. 
 
 /(*) 
 
 X 
 
 /(-*) 
 
 + 
 
 
 
 + 
 
 -f 
 
 1 
 
 — 
 
 + 
 
 2 
 
 — 
 
 - 
 
 3 
 
 - 
 
 + 
 
 4 
 
 + 
 
 The positive roots of this equation are seen to lie 
 between and 1, 3 and 4; therefore (Art. 526) the negative roots of the 
 original equation lie between and —1,-3 and — 4. 
 
 2. « 3 -3x 2 -4ic + ll = 0. 3. x 3 -3x 2 -10x-\-5 = 0. 
 4. x* + Q>tf + 2x- 1 = 0. 5. x*- Gx 2 + 2x + 2 = 0. 
 
 6. « 4 -4^-3xH-23 = 0. 
 
 Sue Since f(x) has but two variations, the equation cannot have more 
 than two positive roots (Art. 529). 
 
 Since all the signs of f{—x) are +, /(— x) = can have no positive 
 roots (Art. 529); hence f(x)= can have no negative roots. It follows that 
 two of the four roots are imaginary. 
 
 7. ar 3 -2« 2 -10 = 0. 8. ar 3 -3a;-l=0. 
 
 9. x*-2x 2 -6x-2 = 0. 10. x 3 -5x 2 -4: = 0. 
 
 11. ar 3 + ar>- 10 = 0. 
 
 Solution. The position of the one positive root is easily found. 
 Changing the signs of the terms containing the even powers (Art. 526), 
 we have 
 
 /(-Z)=£ 3 -X 2 + 10=0. 
 
 This function can become only by having the single negative term numeri- 
 cally equal to the sum of the two positive terms. Now, if x < 1, the 10 alone, 
 without the x 3 , is more than enough to counterbalance the negative term ; 
 while if x> 1, the x 3 alone without the 10, is more than enough to counter- 
 balance the negative term. 
 
 Hence this equation can have no positive roots, and the original equation 
 no negative roots. 
 
 12. x s +x i +x-50=0. 13. ic 5 4-^+^-20=0. 
 
 14. 8^-36^+46^-15=0. 15. z 4 -10ar 3 +31 ^-30^+6=0. 
 16. » 4 -6ar J +4x 2 +18a;-21=0. 
 
328 HIGH Ell ALGEBRA 
 
 Solution. The one variation of /(— x) shows that the equation cannot 
 have more than one negative root. A negative root is readily located between 
 
 - 1 and - 2. 
 
 x I f Cx~) 
 Evaluating f(x) for positive numbers, the results are as in the _ J v ' 
 
 margin,- locating one root between 4 and 5. As all numbers >5 0—21 
 
 give positive results, the other two roots must either both beimagi- 1—4 
 
 nary, or both be situated between two consecutive numbers. It 2 — 1 
 
 is observed that both 1 and 2 very nearly satisfy the equation (caus- 3—12 
 
 ing f(x) to differ little from 0), suggesting that two roots may lie 4—13 
 
 between these numbers. Evaluating for 1.1, 1.2, 1.3, etc., we find 5 + 
 
 changes of signs between 1.5 and 1.6, and between 1.7 and 1.8, thus locating 
 
 the other two roots. 
 
 17. 2ar s -3a 2 -6# + 9 = 0. 
 
 Sug. Proceeding as in the preceding example, it is rendered probable 
 that there are two roots between 1 and 2. Evaluating for the tenths, 1.5 is 
 found to be a root. The best way to find the remaining two roots is to 
 depress the equation and solve the resulting quadratic. 
 
 Or this equation may be transformed into one of the normal form, in which 
 form it has one integral root. 
 
 18. ^-2^-16^ + 24^ + 48 = 0. 
 
 19. x*-4:X s -x 2 + 12x-6 = 0. 
 
 20. x 5 -6x 4 + 36x 2 -25x-30 = 0. 
 
 Sug. Remove the factor containing the integral root and use the depressed 
 equation. 
 
 21. x?-5x 4 + 3x 3 + 16x 2 -26x + ll = 0. 
 
 22. x 5 -2x 3 -x 2 -3x-l = 0. 
 
 23. .? 4 -l0ar 3 + 33a 2 -40cc + i4 = 0. 
 
 MULTIPLE ROOTS 
 
 545. Theorem. If an equation f(x) = has equal roots, they are 
 the roots, and the only roots, of the equation formed by placing equal 
 to the highest common divisor of f(x) and its first differential 
 coefficient. 
 
 Dem. Let a be one of the m equal roots of f(x) = 0. Then 
 (x — a) m is a factor of f(x). If we represent the product of the 
 other factors by <f> (x), we have 
 
 /(*) = (*- «r <H*) = o. a) 
 
MULTIPLE ROOTS 329 
 
 Representing the first differential coefficients of f(x) and <ft (x) 
 by f'(x) and <f>'(x) respectively, we have (Arts. 415 and 420) 
 
 f'(x) = m (x - a)— 1 <f> (x) + (x- a) m <f>'( x ) = °- ( 2 ) 
 
 Comparing (2) with (1), we see that (x — a)** - - 1 is the h. c. d. of 
 f(x) and f'(x), and that the equation, 
 
 (x - a) m ~ l = 0, 
 
 has m — 1 roots equal to a, and has no other roots. 
 
 Similarly, if a is one of the m equal roots and b one of the n 
 equal roots of f(x) = 0, we shall have 
 
 (x - a)" 1 - 1 (x - b)^ 1 = 0, 
 
 which has m — 1 roots equal to a and n — 1 roots equal to 6, and 
 has no other roots. 
 
 546. The work of finding the h. c. d. of f(x) and f'(x) is so great 
 that the process should be resorted to only when shorter processes 
 fail. It may be avoided in the following cases : 
 
 1. Multiple integral roots are found by the same operation 
 that gives other integral roots (Art. 537). 
 
 2. Multiple monomial imaginary and quadratic surd roots in 
 equations containing only even powers or only odd powers are 
 found as in Art. 538. 
 
 3. Multiple roots in an equation in which f(x) is a perfect 
 square are found from the equation obtained by extracting the 
 square root. 
 
 4. Finally, in obtaining multiple roots, it is never necessary to 
 find the h. c. d. of f(x) and f'(x) for equations below the 6th 
 degree, as seen from the following considerations : 
 
 (a) If a cubic has a multiple root, we must have either 
 
 (x - af = or (x - a) 2 (x-b) = 0. 
 
 In both forms, to give rational coefficients, the roots must be 
 commensurable. 
 
 (b) If a biquadratic has multiple roots we must have one of 
 the forms 
 
 (x - a) 4 = 0, (x - af (x -b)=0, (x - a) 2 (x -b)(x-c)= 0, 
 
 or (x — a) 2 (x — b) 2 = 0. 
 
330 HIGHER ALGEBRA 
 
 In the first three forms, to give rational coefficients, the multiple 
 roots must be commensurable. In the fourth form a and b can be 
 incommensurable if they are numerically equal with opposite 
 
 signs, giving 
 
 (x - a) 2 (x + a) 2 = (x 2 - a 2 ) 2 = 0. 
 
 In this case f(x) is a perfect square. 
 
 (c) If a quintic has multiple roots, we must have one of the 
 forms 
 
 (x - a) 5 = 0, (x - a) 4 (x - b) = 0, (x - a) 3 (x -b)(x-c) = 0, 
 
 (x - a) 8 (x - bf = 0, (x - a) 2 (x -b)(;x- c)(x - 0) = 0, 
 
 or (x-a) 2 (x-b) 2 (x-c) = 0. 
 
 In all but the last of these forms, to give rational coefficients, 
 the multiple roots must be commensurable. In the last form 
 a and b can be incommensurable if they are numerically equal 
 with opposite signs, giving 
 
 (x -a) 2 (x + a) 9 (x-c) = (x 2 - a 2 ) 2 (x-c)= 0. 
 
 In this case the commensurable root c may be removed (Art. 
 515), leaving a perfect square. 
 
 EXAMPLES CXXXII 
 
 Find the multiple roots of the following : 
 
 1. x«-±x 6 -x* + 16x 3 -5x 2 -12x-3 = 0. 
 
 Solution. Since the equation is of even degree and the absolute term 
 — , it has at least two real roots, and these have opposite signs (Art. 525). 
 As indicated in the margin, there are no changes of sign except between 
 1 and 2, and between — 1 and — 2. 
 
 /GO 
 
 o 
 - l 
 
 + 2 
 
 + 3 
 
 /(-*) 
 
 + 
 
 Now, if any of the remaining roots are real, they 
 must be situated either between these same numbers q 
 (a change of sign indicates a passage through an odd ^ 
 number of roots), or in pairs between other con- % 
 secutive numbers. We therefore apply the test for g 
 equal roots. 
 
 Obtaining the first differential coefficient, we have 
 
 f'(x) - 6 x b - 20x 4 - 4 x s + 48 x 2 - 10 X - 12. 
 
 Rejecting the factor 2, finding the h. c. d. of f(x) and/(x), and placing it 
 equal toO, we have x2 _ 2x _ 1 = , 
 
 or x = 1 ± V2. 
 
SOLUTION FOR INCOMMENSURABLE ROOTS 331 
 
 Therefore 1 + v/2 and 1 — V2 are each double roots. 
 
 Again, since x 2 — 2 x — 1 is the h. c. d. of /(x) and f(x), (x 2 - 2 x — l) 2 , 
 orx*-4x 3 + 2x 2 + 4x + l, is a factor of /(x) (Art. 545). The other factor 
 is found by division to be x 2 — 3, giving for the other roots x = ± V3. Hence 
 the six roots are 1 ± V2, 1 ± V2, ± VS. 
 
 2. a; 4 -8ar 3 + 18a; 2 -8a;-f- 1 = 0. 
 
 3. z 4 -12ar J + 50# 2 -84a; + 49 = 0. 
 
 4. ar } -7a; 4 + 8ar } + 28ar J -22a;-48=0. 
 
 5. x 6 -7x 4 -}-Wx 2 -12 = 0. 
 
 6. a; 6 -2ar 5 -6a; 4 + 8ar } + 12ar o -8a;-8 = 0. 
 1. ^-12^ + 45^-50^ = 0. 
 
 8. x«-8x 5 + 12x 4 + 24:X 3 -27x 2 -lGx-l=0. 
 
 9. a?-9a^ 5 + 30a; 4 -44ar 2 -f-24 = 0. 
 
 10. x 8 - 8x 7 + 16a; 6 + 16a,- 5 - 56a; 4 - 32ar* + 64a; 2 + 64a; + 16= 0. 
 
 SOLUTION FOR INCOMMENSURABLE ROOTS 
 Horner's Method 
 
 547. Having explained how to find the situation and, conse- 
 quently, the initial figures of the real roots of an equation, we 
 now proceed to exhibit a process for finding the remaining parts 
 of such roots, exactly if commensurable, and to any required 
 degree of approximation if incommensurable. This process is 
 called Horner's Method, having been published by W. G. Horner, 
 of Bath, England, in 1819. The method is based on the next two 
 theorems. 
 
 548. Theorem. If the first member of an equation in the form 
 f(x) = be divided by x — a, then the integral part of the quotient be 
 again divided by x— a, and so on, the successive remainders ivill be, 
 in inverse order, the coefficients of an equation whose roots are less 
 by a than those of the given equation. 
 
 Dem. Let the given equation be 
 
 Asf + Bx n ~ l + Cx n ~ 2 H Jx- + Kx + L = 0. (1) 
 
 Let x = x x + a. Then by substitution 
 
 A(x, + a) n + B(x x + a) n ~ l + C(x x + a) n ~ 2 + • • • J(x, + af 
 -f- K(x x + a) + L = 0. 
 
332 HIGHER ALGEBRA 
 
 If we should perform the indicated operations and arrange 
 with reference to x l} there would result an equation of the form 
 
 Ax? + B&f- 1 + C x x?~ 2 H J^ 2 + A>, + A = 0. (2) 
 
 The unknown quantity in (1), the given equation, is x, while 
 in (2), the transformed equation, it is x x , which was taken less by 
 a than x. Hence, the roots (which are the values of the unknown 
 quantity) of (2) are less by a than those of (1). 
 
 In this transformation all the coefficients except the first have 
 been changed. Restoring in (2) the value of x x , which is x — a, 
 we have 
 
 A(x - a) n + B x (x - a)^ 1 + C t (x - a) n ~ 2 + .... J % (x - af 
 
 + K 1 (x-a) + L l = 0* (3) 
 
 which must be identical with (1), though in different form. It is 
 seen that if we divide the first member of (3) by x — a, then 
 divide the quotient by x — a, and so on, the successive remainders 
 will be L lf K\, J\, •" Cp B l} A, the coefficients, in inverse order, 
 of (2), which is an equation whose roots are less by a than those 
 of the given equation. 
 
 EXAMPLES CXXXIII 
 
 1. Find the equation whose roots are each less by 2 than 
 
 those of 
 
 x* - 1 x> + llx 2 + 1 x -12 =0. 
 
 * The coefficients L\, K\, J\, ••• C\, B\, A are, respectively, the function, 
 
 .. „ . 7 . .. M derivative 3d derivative nth derivative .., m 
 its 1st derivative, , — , ••• : , with x 
 
 replaced by a. •— •— 
 
 Substituting Xi + a for x in f(x) = and developing by Taylor's For- 
 mula, using the form in Art. 452, we have 
 
 /(*) =f(x l + d) =/(«) +/'(«)*i+/"(«) ^ +/'"(«) f£ +/ iv («)^ + etc. = 0. 
 
 2 t? I 4 
 
 Hence Ll =f(a), K x =f>(a), J t =C&K '" &=£=% A =£&>. 
 
 2 [n — 1 \n 
 
 If X\ is a small fraction, for an approximate value the higher powers may 
 be neglected, and we have 
 
 f'(a)x\ -\-f(a)= 0, whence x\ — — .^r\' a PP rox ' mate ly- [See Art. 549.] 
 
SOLUTION FOR INCOMMENSURABLE ROOTS 333 
 
 Solution. By the theorem the successive remainders obtained by dividing 
 by x — 2, then dividing the quotient by x — 2, and so on, will be, in inverse 
 order, the coefficients sought. These remainders are obtained with great 
 facility by synthetic division, as follows : 
 
 x 4 - 7 x 3 + 11 x 2 + 7 x - 12 |_2 
 
 - 5 1 9 6 1st rem. 
 
 - 3 — 5 — 1 2d rem. 
 
 - 1 - 7 3d rem. 
 1 4th rem. 
 
 1 5th rem. 
 
 Hence the equation whose roots are each less by 2 than those of the given 
 
 equation is 
 
 Xi* + Xi 3 - 7 xi 2 - xi + 6 = 0, 
 
 or, omitting subscripts, x 4 + x 3 — 7x 2 — x + 6 = 0. 
 
 Let the student find by Art. 537 the roots of each equation, and see 
 whether those of the second are less by 2 than the corresponding ones of the 
 first. 
 
 2. Find the equation whose roots are less by 3.14 than those of 
 2x A -5x i -3x i + 4:X-23 = 0. 
 
 Solution. We may diminish the roots first by 3, then by .1 more, and 
 then by .04 more, without writing the intermediate equations, as follows : 
 
 2 x* - 5 x 3 - 3 x 2 + 4 x - 23 [_3 
 
 4 -11 (1) [A 
 
 67(1) 7.3192 
 
 6.192 - 3.0808 ( 2) [M 
 
 73.192 3.28970432 
 
 -5x 3 
 
 - 3 x 2 
 
 1 
 
 
 
 7 
 
 21 
 
 13 
 
 60(i) 
 
 19(1) 
 
 1.92 
 
 .2 
 
 61.92 
 
 19.2 
 
 1.94 
 
 .2 
 
 63.86 
 
 19.4 
 
 1.96 
 
 .2 
 
 65.82(2) 
 
 19.6 
 
 .7952 
 
 .2 
 
 66.6152 
 
 19.8( 2 ) 
 
 .7984 
 
 .08 
 
 67.4136 
 
 19.88 
 
 .8016 
 
 .08 
 
 68.2152(3) 
 
 6.386 - .39109568(3) 
 
 79.578(2) 
 2.664608 
 
 82.242608 
 2.696544 
 84.939152(3) 
 
 19.96 
 
 .08 
 20.04 
 
 .08 
 20.12(8) 
 
334 HIGHER ALGEBRA 
 
 The numbers marked (1), together with the first coefficient, which re- 
 mains unchanged, are the coefficients of the equation whose roots are less 
 by 3 than those of the given equation. 
 
 The numbers marked (2) , together with the first coefficient, are the coeffi- 
 cients of the equation whose roots are less by . 1 than those of the equation 
 whose coefficients are marked (1), and, consequently, less by 3.1 than those 
 of the given equation. 
 
 The numbers marked (3), together with the first coefficient, are the 
 coefficients of the equation whose roots are less by .04 than those of the 
 equation whose coefficients are marked (2), less by .14 than those of the equa- 
 tion whose coefficients are marked (1), and less by 3.14 than those of the 
 given equation. Hence the equation whose roots are less by 3.14 than those 
 of the given equation is 
 
 2 x 4 + 20.12 x* + 68.2152 x 2 + 84.939152 x - .39109568 «c 0. 
 
 3. Find the equation whose roots are less by 3.213 than those 
 of x 3 -f- 11 x 2 - 102 x + 181 = 0. 
 
 4. Find the equation whose roots are less by 2.85 than those 
 of a 4 -12x 2 + 12 a? -.3 = 0. 
 
 549. Theorem. When a root of an equation is a small fraction, 
 it is approximately equal to the absolute term divided by the coefficient 
 of the first power of x. 
 
 Dem. Let x 1} a small fraction, be a root of the equation 
 
 Ax n + Bx n ~ l + Cx n ~ 2 +.--Kx+L = 0. 
 Substituting, we have 
 
 As? + Bxf- 1 + Cx?- 2 + •• • Kx x +L = 0. 
 
 Now since x 1 is a small fraction, the terms with powers above 
 the first are so small that the equation will be little affected by 
 neglecting them and retaining only Kx t + L = 0, which gives 
 
 x x = , approximately. 
 
 K 
 
 Illustration. Consider the equation 
 
 9 X 3_ X 2 + 9 X _ i _o. 
 
 When x = 0, f(x) = -1, and when x = 1, f{x) = + 16. 
 Hence, there is a real root between and 1, and it is much nearer than 
 1, i.e., it is a small fraction. Then x 3 and x 2 are much smaller than x, and 
 
SOLUTION FOR INCOMMENSURABLE ROOTS 335 
 
 the equation will be little affected by neglecting the terras containing them, 
 giving 9 x — 1 = 0, whence X a £ = .11, approximately. Indeed, in this case, 
 the quotient, $, of the absolute term divided by the coefficient of the 1st 
 power of x is the exact root. 
 
 550. The principle of Horner's Method of finding a root of an 
 equation, exactly if commensurable, approximately if incommen- 
 surable, is this : Suppose the integral part of the root to have 
 been found by Art. 542. If by Art. 548 we find the equation 
 whose roots are less by this integral part than those of the 
 given equation, the corresponding root of this transformed equa- 
 tion will be the decimal part of the root of the given equation. 
 Now, by Art. 549, this part is approximately equal to the abso- 
 lute term of this transformed equation divided by the coefficient 
 of the 1st power of x in the same equation (i.e.; the first remain- 
 der divided by the second remainder obtained in successively 
 dividing f(x) by x minus the integral part of the root). Using 
 only the first decimal figure of the quotient, we may find the- 
 equation whose roots are less by this decimal figure than those of 
 the first transformed equation, and, as before, obtain another 
 figure of the root by dividing the absolute term of this second 
 transformed equation by the coefficient of the 1st power of x in 
 the same equation, and so on. 
 
 551. Reverting to example 2, page 333, we may see an application 
 of the method. By the principle of Art. 541, it is found that a real 
 root lies between 3 and 4, i.e., one root is 3 + a decimal fraction. 
 
 The equation whose roots are less by 3 has for its correspond- 
 ing root this decimal fraction, and this fraction, by Art. 549, is 
 approximately the first remainder divided by the second, i.e., 
 11 -5- 67 = .1, approximately, using only the first decimal figure. 
 
 The equation whose roots are less by .1 than those of the last 
 has for its corresponding root the remaining figures of the deci- 
 mal fraction, and this is approximately the new first remainder 
 divided by the new second remainder, i.e., 3.6808 -j- 79.578 (it is 
 sufficient to use 3.68 -r- 79) = .04, approximately. 
 
 For the third and probably the fourth figures of the decimal 
 part of the root we have .39109568 -- 84.939152 = .0046. Hence 
 the root to four decimal places is 3.1446. 
 
336 HIGHER ALGEBRA 
 
 For a farther application of the method, let it be required to 
 find to four decimal places the positive root of 
 
 a? + 2^ -9a; -22 = 0. x f(z) 
 
 As indicated in the margin, the one positive root lies between 
 3 and 4, i.e., it is 3 + a decimal fraction. We must, therefore, 1 
 diminish the roots by 3, and for a second figure of the root, divide 2 
 the first remainder by the second, and so on, the work being as 3 
 follows : 4 
 
 x 3 + 2x 2 - 9 x - 22 J3.1273 
 
 5 6- 4(i) 
 
 8 30(i) 3.111 
 
 11(D l.U - .889(2) 
 
 .1 31.11 .649128 
 
 11.1 1.12 - .239872(3) 
 .1 32.23(2) 
 
 11.2 .2264 
 .1 32.4564 
 
 11.3(2) .2268 1st figure = 3, 
 
 .02 32.6832(3) 2d figure = 4 + 30 = .1, 
 
 11.32 3d figure = .889 + 32.23 = .02, 
 
 .02 4th and 5th figures = .239872 4- 32.6832 = .0073 
 
 11.34 
 
 552. Carefully note the following observations : 
 
 1. As we find a root by starting with a value too small and 
 increasing it by annexing figure after figure, a change of sign of 
 the absolute term in any of the transformed equations, i.e., the first 
 remainder of any of the successive sets of divisions, would indicate 
 that we had passed beyond the value of the root (Art. 541), and that 
 the last figure must be diminished. The figure to be adopted in 
 any case is the largest number which, in the process of diminishing 
 the roots, will not make the sign of the absolute term different from 
 that of the first transformed equation. If the absolute term of the 
 first transformed equation is different from that of the given equa- 
 tion, it simply indicates that there is another root between and 
 the one we have started to find. 
 
 2. As the successive figures of the root are found by neglecting 
 all but the last two terms of an equation whose roots are less by 
 the part of the root already found than those of the given equa- 
 
SOLUTION FOR INCOMMENSURABLE ROOTS 337 
 
 tion (Art. 549), the quotient of the absolute term of a transformed 
 equation by the coefficient of the 1st power of x in the same equa- 
 tion is liable to be too large, though seldom so beyond the second 
 figure of the root. To avoid this for the second figure of the root, 
 we may solve, approximately, the quadratic obtained by neglect- 
 ing, not all but two, but all but three, of the terms of the first 
 transformed equation. This is also necessary if, in any case, the 
 next to the last term is 0. 
 
 3. Since x 1 = — -—, approximately (Art. 549), in which L is the 
 
 first and K the second remainder, K and L must have opposite 
 signs, for otherwise the quotient would be the amount to be sub- 
 tracted from, instead of added to, the part of the root already 
 found. If after the first transformation the first and second 
 remainders have like signs, the next figure of the root cannot be 
 found by division, but must be found by the same kind of trial 
 which gives the first figure. 
 
 4. If two or more roots have the same initial part, the next 
 figure of each must be found by trial, after which the process is 
 the same as for other cases. 
 
 5. Ordinarily the fourth decimal figure will be correctly given 
 by writing two figures of the third quotient. 
 
 6. The negative roots of an equation are found by finding the 
 positive roots of the equation obtained by changing the signs of 
 the odd or the even powers (Art. 526). 
 
 7. When all but one of the roots have been found, the remain- 
 ing one becomes known by adding the sum of the known roots to 
 the coefficient of the second term (the coefficient of # n_1 ) of the 
 given equation and changing the sign of the result (Art. 539). 
 
 EXAMPLES CXXXIV 
 
 Find to three or four decimal places the 
 real roots of the following : 
 
 1. x 3 -4:X 2 -Gx + 8 = 0. 
 
 Solution. As indicated in the margin, the posi- 
 tive roots are between and 1, 4 and 5, while the 
 negative root is between — 1 and - 2. 
 downey's alg. — 22 
 
 X 
 
 m 
 
 X 
 
 
 
 + 
 
 
 
 1 
 
 — 
 
 1 
 
 2 
 
 — 
 
 2 
 
 3 
 
 — 
 
 
 4 
 
 — 
 
 
 5 
 
 + 
 
 
 /(-*) 
 
338 Mgher Algebra 
 
 To find the negative root the operation is as follows : 
 
 -8 [ 1.8004 
 
 -9(1) 
 8.992 
 
 x s + 4 x 2 
 5 
 
 -6x 
 1 
 
 6 
 
 7(1) 
 
 5(1) 
 6.24 
 
 •008(2) 
 
 _ 11.24 
 
 7.8 6.88 Since 9-f 5 cannot give the 
 
 18.12 (3 ) 
 
 8.6 
 
 first decimal figure, we use 
 
 7x 2 + 5:s-9 = 0, (Obs. 3) 
 
 .g whence, x 2 + f x = f, 
 
 9.4 ( 3) and g~ * ±*L* SS .8. 
 
 The 2d quotient = .008 + 18.12 = .0004. 
 
 Let the student show that the root between 4 and 5 is 4.8922. y/^ 
 The sum of these two roots and the second coefficient is — .9082. Hence 
 the third root is .9082 (Obs. 7). 
 
 2. 8 a 8 -17 a 2 -16 a; + 34 = 0. 
 
 Solution. By trial the roots are seen to lie between 1 and 2, 2 and 3, 
 
 — 1 and — 2. The operation of finding the largest root is as follows : 
 8x 3 -17x 2 -16 a; + 34 | 2.125 
 
 - 1 -18 - 2 ( i) 
 
 15 12(i) 1.518 
 
 31(i) JU8 - .482(2) 
 
 .8 15.18 .382224 
 
 31.8 3.26 - .099776(3) 
 
 .8 18.44(2) .099776 
 
 32.6 .6712 000000 
 
 .8 19.1112 
 
 33.4(2) .6744 
 
 1st figure = 2, 
 
 .16 19.7856(3) 
 
 3^56 .1696 2d figure = 2^ 12 = .1, 
 
 .16 19.9552 3d figure = .482 + 18.44 = .02, 
 
 33. 72 4th figure = .099776 - 19.7856 = .005. 
 
 .16 As the absolute term reduces to 0, the exact root 
 
 33.88(3) has been found. 
 
 .04 The other two roots might also be found by Horner's 
 
 33.92 Method ; but since this root is commensurable, and its 
 
 exact value has been found, a much better way is to 
 depress the equation and solve the resulting quadratic, as follows : 
 
 8x 3 -17x 2 -16x + 34 LgJ 
 
 0-16 
 
SOLUTION FOR INCOMMENSURABLE ROOTS 339 
 
 Hence 8 x 2 - 16 = 0, whence x 2 =s 2, and x = ± V2. 
 
 3. a*-6a? + 5x + ll = 0. 4. ar 3 + 10ar> + 5a; -260 = 0. 
 
 5. 8a?-65x*+ 140 z-33=0. 6. ar 5 + 3x 2 + 4a; + 5 = 0. 
 
 7. ar» + a; = 1000. 8. 4ar*- 9ar*-52a; + 117 = 0. 
 
 9. ic 3 -3x 2 -3x = -18. 10. « 3 -3» 2 -4^ + 13 = 0. 
 
 11. x 4 + 4ar 3 -5ar J -18a; = 22. 12. x 4 -6x* + 3x> + 30x = 50. 
 
 13. ^+4^-4^-11x4-4=0. 14. 2a,- 4 +5ar 3 +4ar 2 +3a;=8002. 
 
 15. ar 5 -4x-2000 = 0. 16. ar 5 - 4 a; 4 +- 7 ar* - 863 = 0. 
 
 17. The radius of a sphere is 9 inches, and the volume of a 
 segment of one base cut from it is one fourth of that of the 
 sphere. Find its altitude, the volume of a spherical segment of 
 one base in terms of its altitude and the radius of the sphere 
 being ir(rh 2 --^\ 
 
 125 7T 
 
 18. The weight of a sphere 1 foot in diameter is — - — pounds. 
 When floated in a full vessel of water weighing 62.5 pounds per 
 
 cubic foot, it causes an overflow of -^- cubic feet of water. Find 
 
 16 
 
 the depth to which it sinks, the law being that the weight of 
 the water displaced equals the weight of the floating body. 
 
 19. A rectangular inner court, 100 feet long and 50 feet wide, 
 has its principal openings at diagonally opposite corners. Between 
 these openings matting 7 feet wide and with square ends, the 
 corners just touching the walls, is laid (i.e., the matting is a 
 rectangle inscribed within a rectangle). Find the length of the 
 matting. 
 
 553. The extraction of the nth root of any number, as a, is the 
 same as finding the positive, real root of the equation, 
 
 x n + Oaj"" 1 + 0x n - 2 -\ a = 0. 
 
 Hence Horner's Method will give to any required degree of 
 accuracy any root of a given number. By the principle of Art. 
 471, the use of a table of logarithms will give the result with far 
 less labor than will either the elementary method or Horner's 
 Method. 
 
340 HIGHER ALGEBRA 
 
 EXAMPLES CXXXV 
 
 Perform by Horner's Method the following indicated opera- 
 tions : 
 
 1- -\/29791. 2. V70444997. 
 
 3. ^5\6847. 4. 4/B. 
 
 Sturm's Theorem and Method 
 
 554. The change of sign of f(x) as we pass from one value of 
 x to another usually reveals with little difficulty the situation of 
 the real roots of an equation (Art. 541). In infrequent cases, 
 however, there are difficulties, the nature of them being this: 
 After locating, by Art. 541, all the real roots which the substitu- 
 tion of integral numbers will reveal, we are sometimes uncertain 
 whether, 1st, the remaining roots are imaginary, 2d, three or 
 higher odd number of real roots lie between two consecutive 
 numbers that cause /(a?) to change sign, or, 3d, two or higher even 
 number of real roots lie between two consecutive numbers that 
 do not cause f(x) to change sign. In the latter case the method 
 of ex. 16, p. 327, usually removes the uncertainty and locates the 
 roots if real ; but in exceptional cases the uncertainty remains or 
 is removed only after much labor. In 1829 Jacques Charles 
 Franqois Sturm (1803-1855), a Swiss mathematician, who after- 
 ward became a member of the French Academy, Professor of 
 Mathematics in the Polytechnic School and Professor of Mechan- 
 ics in the Faculte des Sciences in Paris, discovered a theorem by 
 means of which, in all cases, the number and situation of the real 
 roots of a numerical equation may be found. As its application 
 is laborious, it is used only as a last resort when the method of 
 Art. 541 fails or is not readily applicable. 
 
 555. If f(x) and its first derivative be treated as in the process 
 of finding the h. c. d , except that each remainder, before being used 
 as the next divisor, have its signs changed and that no negative 
 factors be introduced or rejected, these successive remainders 
 with their signs changed constitute what are called Sturm's Func- 
 tions, or the Sturmian Functions. 
 
SOLUTION FOR INCOMMENSURABLE ROOTS 341 
 
 Thus, let f(x) = x 3 - 4 x' 2 - x + 4 = 0. 
 
 The first derivative is 3 x* - 8 x - 1. Dividing x 3 -4z 2 -j; + 4 by 
 3 x 2 — 8 a; — 1, first multiplying the former by 3 to avoid fractions, as in the 
 process of finding the h. c. d. (Art. Ill), the first remainder of lower degree 
 than the divisor is found to be — 19 se 4- 16. Hence 19 x — 16 is the first 
 Sturmian Function. Similarly, the next remainder is found to be — 2025. 
 Hence 2025 is the second Sturmian Function. 
 
 556. Notation. As the first member of an equation is repre- 
 sented by f(x) and its first derivative by f(x), we shall represent 
 Sturm's Functions by fix), f/x), fix), etc. 
 
 Applying this notation to the above example, we have 
 
 f{x) =x*-4x 2 -x + 4, 
 f(x)=Sx*-$x-l, 
 Mx)= 19* -16, 
 / 2 (x)=2025. 
 
 557. Iff(x) and fix) have no common factor, the last Sturmian 
 Function does not contain x ; if they have a common factor, the 
 last Sturmian Function is their h. c. d. 
 
 558. Sturm's Theorem. If in fix), /'(#)? and Sturm's Functions 
 two different numbers be substituted, the difference in the number 
 of variations in the two cases ivill be the number of real roots of 
 fix) = that lie between the substituted numbers, multiple roots being 
 counted but once. 
 
 Dem. I. When f(x) = has no equal roots. 
 
 1st. Two consecutive functions cannot vanish, i.e., become 0, for 
 the same value of x, and, consequently, cannot change signs simul- 
 taneously. 
 
 Let the several quotients in the process of obtaining Sturm's 
 Functions be represented by q, q', q", q'", etc. ; then by the 
 principles of division we have 
 
 /(*)=/*(*)?-/,(*), a) 
 
 f(x)=Mx),,'-Mx), (2) 
 
 /i(*Wt(*)9" -/.(*), (3) 
 
 Mx)=f 3 (x)q'"-Mx), (4) 
 
 etc., etc., etc. 
 
342 HIGHER ALGEBRA 
 
 Now suppose that some value of x causes two consecutive 
 functions, as /,. (x) and f (x), to vanish. Then from (3), f (x) = ; 
 from (A),f 4 (x) = 0; and so on to the last function. But the last 
 function does not contain x (Art. 557), and cannot vanish for 
 any value of x. Hence two consecutive functions cannot vanish 
 for the same value of x, and, consequently, cannot change signs 
 simultaneously.* 
 
 2d. The changing of sign {for different values of x) of any function 
 after first, f(x), has no effect on the number of variations. 
 
 The last function cannot change sign for any value of x, as it 
 does not contain x, and the other functions can change signs only 
 by passing through 0. Now, when any function, as f 2 (x), becomes 
 0, (3) becomes f(x) = — f 3 (x), i.e., the adjacent functions have 
 opposite signs, and, by 1st, neither of these can change sign when 
 f,{x) changes. Hence, if for a value of a; a little less than that 
 which makes f,{x) = the signs Of f{x), f 2 (x), and f 3 {x), in order, 
 
 are + -\ , then for a value a little greater the signs in order 
 
 are H — , giving only one variation in either case. 
 
 3d. One variation, and only one, is lost when f(x) vanishes for 
 increasing values of x, i.e., when x increases through a root of 
 
 f(x) = 0. 
 
 Taking values a little less and a little greater than x and devel- 
 oping by Taylor's Formula, using the form of Art. 452, we have 
 
 f(x T h) =f(x) T f\x) h +f"(x) A' T /"'(*) | + etc. 
 
 Let a be a root of f{x) = 0. Then f{x) will vanish, and we 
 shall have 
 
 f(a Ti)=T /'(«) '* +/"(«) f T /'"(«) j| + etc. 
 
 By taking h small enough, the second member will have the 
 same sign as its first term. Hence when x is a little smaller than 
 
 * A function can change sign only by passing through or <x>. As the 
 functions under consideration are integral functions of as, they cannot become 
 oo for any finite value of a. 
 
SOLUTION FOR INCOMMENSURABLE ROOTS 343 
 
 a root a, /(a — h) and f'(a) have opposite signs, and when a; is a 
 little greater than a root a, f(a + h) and /'(a) have the same sign ; 
 i.e., there is a loss of one variation when x increases through a 
 root of f(x) = 0. 
 
 Now, as there is a loss of one variation whenever x increases 
 through a root of f(x) = 0, causing f(x) to change sign, and no 
 change in the number of variations for the change of sign of any 
 of the other functions, the difference in the number of variations 
 when any two numbers are substituted for x will be the number 
 of real roots between these numbers. 
 
 II. When f(x) = has equal roots. 
 
 When f(x) = has equal roots, the h. c. d. off(x) andf\x), ivhich 
 contains them (AH. 545), is a factor of all the functions, and, conse- 
 quently, does not affect the number of variations obtained from those 
 functions. 
 
 Since a divisor of two quantities is a divisor of their difference 
 (Art. 110), equation (1) shows that the h. c. d. of f(x) and/'(#) is 
 also a divisor of f(x) ; then equation (2) shows that it is a divisor 
 of f 2 (x) ; then equation (3) shows that it is a divisor of f(x) ; and 
 so on. If the exact value of a multiple root should be substituted, 
 the h. c. d. would be 0, and all the functions would vanish ; but 
 if the h. c. d. is + for any particular value of x, its presence in 
 all the functions will not affect the signs, and if it is — , it will 
 change all the signs, thus making no change in the number of 
 variations. But since the Sturmian Functions terminate with 
 the h. c. d. (Art. 557), the number of them will be less by the 
 degree of the h. c. d. than it would be if there were no multiple 
 roots' and the division were continued until a numerical re- 
 mainder should be reached. Now the number of times a multiple 
 root is repeated in the h. c. d. is one less than in the original 
 equation (Art. 545). Hence the multiple roots occur but once 
 each in the other factor of /(#), and the difference in the number 
 of variations when any two numbers are substituted in the func- 
 tions will be the number of real roots between these limits, each 
 multiple root being counted but once. The h. c. d. will give the 
 number and also the value of the equal roots (Art. 545). 
 
344 
 
 HIGHER ALGEBRA 
 
 559. Cor. Tlie number of variations lost as x increases from 
 — oo to is the number of negative roots, and the number lost as x 
 increases from- to + oo is the number of positive roots; while the 
 number lost as x increases from — oo to + oo is the total number of 
 real roots. 
 
 560. Sch. 1. When — oo or + oo is substituted, the sign of any 
 function is the same as the resulting sign of its first term. 
 
 561. Sch. 2. As only the sign of the last function is used 
 when there are no equal roots, the numerical value need not be 
 found. 
 
 EXAMPLES CXXXVI 
 
 Find by Sturm's Method the number and situation of the real 
 roots of the following : 
 
 1. a 4 -8ar 5 + 19ar 2 -12a; + 2 = 0. 
 
 The reason for applying to this example Sturm's Method instead of the 
 method in Art. 541 is this : All the terms of /(— x) are -f , showing that the 
 equation has no negative roots ; and evaluating /(x) for positive 
 numbers, the results are as in the margin, leaving it uncertain 
 whether the roots are all imaginary, two real and two imaginary, 
 or all real. We only know that if the roots are all real, two of 
 them lie between and 1, and two of them between 3 and 4, as 
 these numbers come the nearest to satisfying the equation. Evalu- 
 ating for the tenths between these numbers would, in this case, 
 remove the uncertainty, but we could not know that beforehand. 
 
 The application of Sturm's Method is as follows : 
 
 f(x) = x 4 - 8 x 3 + 19 x 2 - 12 x +. 2, 
 i/'(x) = 2 x 3 - 12 x 2 + 19 x - 6, 
 /i (x) = 5 x 2 - 20 x + 8, 
 / 2 (x)=x-2, 
 /s 0*0 =12. 
 
 X 
 
 /CO 
 
 /'CO 
 
 /iCO 
 
 /«(*) 
 
 Mx) 
 
 Variations 
 
 — CO 
 
 + 
 
 — 
 
 + 
 
 - 
 
 + 
 
 4 
 
 
 
 + 
 
 - 
 
 + 
 
 - 
 
 + 
 
 4 
 
 + CC 
 
 + 
 
 + 
 
 + 
 
 + 
 
 + 
 
 
 
 X 
 
 /(*) 
 
 
 
 + 2 
 
 1 
 
 + 2 
 
 2 
 
 + 6 
 
 3 
 
 + 2 
 
 4 
 
 + 2 
 
 5 
 
 + 42 
 
SOLUTION FOB INCOMMENSURABLE ROOTS 
 
 345 
 
 The loss of 4 variations between and x> shows that all the roots are real 
 and that they are all positive. Again, 
 
 X 
 
 /GO 
 
 /'(*) 
 
 ZiOO 
 
 /a GO 
 
 /«(*) 
 
 Variations 
 
 <r 
 
 + 
 
 — 
 
 + 
 
 - 
 
 + 
 
 4 
 
 i 
 
 + 
 
 + 
 
 - 
 
 - 
 
 + 
 
 2 
 
 2 
 
 + 
 
 - 
 
 - 
 
 
 
 + 
 
 2 
 
 3 
 
 + 
 
 
 
 - 
 
 + 
 
 + 
 
 2 
 
 4 
 
 + 
 
 + 
 
 + 
 
 + 
 
 + 
 
 
 
 Hence there are two real roots between and 1, and two real roots be- 
 tween 3 and 4. 
 
 Evaluating for the tenths between these limits, the roots are found to lie 
 between .2 and .3, .5 and .6, 3.4 and 3.5, 3.7 and 3.8. As the roots are now 
 separated, the remaining figures are found by Horner's Method. 
 
 2. x* + 6x 2 +10x-l=0. 
 
 3. x 4 -2ar 3 -6ar 9 + 10x + 5=0. 
 
 4. z 4 -2ar 3 -8ar>+12z + 12 = 0. 
 
 5. z 4 -2ar J + 4x-4=0. 
 
 6. 2x J -15a; 4 + 28ar' + 9ar J -64x + 42 = 0. 
 
 7. ^-8ar 5 + 18x 4 -2x 3 -49x 2 + 78x-42=:0. 
 
 8 . a*-4ar i -llx 4 + 46ar } -10ar 2 '-76a; + 56 = 0. 
 
 9. x 6 - 2ar> -5a? + &a? + 8a? -8^-4 = 0. 
 
 Solution 
 
 /(x) = x 6 -2x 5 -5x 4 + 8x 3 + 8x 2 -8x-4, 
 
 ] /'(x) = 3 x 5 - 5 x 4 - 10 x 3 + 12 x 2 + 8 x - 4, 
 
 /i(x) = 10 x 4 - 13 x 3 - 30 x 2 + 26 x + 20, 
 
 / 2 (x) = 27 x 8 - 10 x 2 - 54 x + 20, 
 
 / 3 (x)=x 2 -2, 
 
 / 4 (x) = 0. 
 
 As the last remainder is 0, /(x) and/'(x) have a common divisor, and the 
 given equation has multiple roots (Art. 545). As the h. c. d., x 2 — 2, occurs 
 
346 
 
 HIGHER ALGEBRA 
 
 in every function, its presence will make no change in the number of varia- 
 tions produced by these functions (II.) . Hence we evaluate the functions 
 as they stand. 
 
 X 
 
 /(*) 
 
 /'(*) 
 
 /i(«) 
 
 /*(*) 
 
 /«(*) 
 
 Variations 
 
 — CO 
 
 + 
 
 - 
 
 + 
 
 — 
 
 + 
 
 4 
 
 
 
 — 
 
 - 
 
 + 
 
 + 
 
 - 
 
 2 
 
 + co 
 
 + 
 
 + 
 
 + 
 
 + 
 
 + 
 
 
 
 - 1 
 
 + 
 
 + 
 
 - 
 
 + 
 
 - 
 
 3 
 
 -2 
 
 + 
 
 - 
 
 - 
 
 - 
 
 + 
 
 2 
 
 1 
 
 - 
 
 + 
 
 + 
 
 - 
 
 - 
 
 2 
 
 2 
 
 - 
 
 - 
 
 + 
 
 + 
 
 + 
 
 1 
 
 3 
 
 + 
 
 + 
 
 + 
 
 + 
 
 + 
 
 
 
 Hence the separate roots (each multiple root being counted but one) are 
 between and — 1, — 1 and — 2, 1 and 2, 2 and 3. 
 
 Placing the h. c. d. of f{x) and/(x) equal to (Art. 545), we have 
 
 x 2 - 2 = 0, 
 
 whence x = ± a/2. 
 
 Hence V2 and — V2 are each double roots. These are the roots that lie 
 between 1 and 2, and — 1 and — 2. 
 
 10. a 6 -4 s 5 + 12ar J -3ar 2 -8a;-l = 0. 
 
 11. x 6 — 2s 5 -4s 4 + 12ar 5 -3a; 2 -l8a; + l8 = 0. 
 
 12. a; 6 - 2 a 5 - 2 a! 4 + 12 s 3 -15 a; 2 -18a? + 36 = 0. 
 
 13. ic 6 -2ar 5 -8a; 4 + 8ar J + 20s 2 -8a;-16 = 0. 
 
 RECURRING OR RECIPROCAL EQUATIONS 
 
 562. A Recurring or Reciprocal Equation is an equation in the 
 normal form (Art. 510) in which the coefficients equidistant from 
 the two ends are numerically equal, the corresponding coefficients 
 having either all like or all unlike signs. 
 
 Thus, x* - 3 x s + 4 x 2 - 3 x + 1 = 0, 
 
 3x 5 -2x* + 5x 3 -5x 2 + 2x-3 = 0, 
 Ax n + B&- 1 + Cx n ~ 2 + .-• + Cx 2 + Bx + A = 0, 
 are recurring or reciprocal equations. 
 
RECURRING EQUATIONS 347 
 
 563. It is evident that when the corresponding coefficients of a 
 recurring or reciprocal equation of even degree, having, therefore, 
 an odd number of terms, have unlike signs, the coefficient of the 
 middle term is ; i.e., the middle term is wanting. 
 
 564. Theorem. The reciprocal of any root of a recurring equa- 
 tion is also a root. 
 
 Dem. If a satisfies the equation 
 
 Ax n + Bx*- 1 + Oaf 1 " 2 -\ \- Cx 2 + Bx + A = 0, 
 
 - will also satisfy it : for the substitution of the former gives 
 a 
 
 Aa n + Ba n - 1 + Ca n ~ 2 + ». + Ca 2 + Ba + A = 0, 
 
 and the substitution of the latter gives 
 
 which, when cleared of fractions, becomes 
 
 A + Ba + Ca 2 + ••• + Co— 8 + J^a' 4 " 1 + Aa n = 0, 
 
 and this is the same as the equation obtained by substituting a. 
 
 Sch. It is on account of this reciprocal relation of the roots of 
 recurring equations that they are called also Reciprocal Equations. 
 
 565. Theorem. A recurring equation of odd degree has -f 1 or 
 
 — 1 as a root, according as the cori'esponding coefficients have unlike 
 or like signs. 
 
 Dem. When the corresponding terms have unlike signs, the 
 substitution of -f 1 for x will cause them to cancel each other ; 
 and when they have like signs, the substitution of — 1 for x will 
 also cause them to cancel each other, since one of these terms is 
 an even and the other an odd power of x. 
 
 566. Theorem. A recurring equation of even degree, ivhose cor- 
 responding coefficients have unlike signs, has both + 1 and — 1 as 
 roots. 
 
 Dem. The equation has the form 
 
 X 2 " + Ax 2 "- 1 4- Bxr n 2 + Bx 2 - Ax - 1 = 0. 
 
348 HIGHER ALGEBRA 
 
 It is evident that both + 1 and — 1 will cause corresponding 
 terms to cancel. 
 
 567. A recurring equation is said to be in the Standard Form 
 when it is of even degree and its corresponding coefficients have 
 like signs. 
 
 568. Theorem. Every recurring equation not in the standard 
 form may be reduced to that form. 
 
 Dem. If the equation is of odd degree, it may be divided by 
 X — 1 or x '■+» 1, according as its corresponding coefficients have 
 unlike or like signs (Arts. 565 and 515), reducing it to the 
 standard form. 
 
 If it is of even degree, and its corresponding coefficients have 
 unlike- signs, it may be divided by x + 1 and x — 1, or x 2 — 1 
 (Arts. 566 and 515), reducing it to the standard form. 
 
 569. Theorem. Any recurring equation in the standard form 
 may be reduced to an ordinary equation of half the degree. 
 
 Dem. The equation has the form 
 
 x 2 " + Ax 2 "- 1 + Bx 2 " 2 ~.Mx n ~. Bx 2 + Ax + 1 = 0. (1) 
 
 Dividing by x n and grouping the terms, we have 
 
 ( xn + ~V) + A ( xn ~ l + ~hi) + B(x n ~ 2 +-^\ + — M& 0. (2) 
 
 Let x 4- - = i/ ; 
 
 x 9i 
 
 then (x +1V = x> + 2 + 1= y\ 
 
 whence , x 2 + ~ = y 2 — 2. 
 
 x 2 
 
 Similarly, x 3 -\- — = f — Sy, 
 
 x 4 + ± = y 4 -±y 2 + 2, 
 
 X* 
 
 x n + _=y n — ny n2 -\ 
 
 x n 
 
 and so on. 
 Hence 
 
RECURRING EQUATIONS 349 
 
 Now, if these values be substituted in (2), there will result an 
 equation of the nth degree, which is half that of the given 
 equation. 
 
 EXAMPLES CXXXVn 
 
 Solve the following : 
 
 1. ^-11^ + 17^-f 17x 2 -ll^ + l = 0. 
 
 Solution. By Art. 565, — 1 is a root of this equation ; hence the equa- 
 tion is divisible by x + 1 (Art. 515). 
 
 x 5 - 11 x* + 17 x :J + 17 x 2 - 11 x +1 |_-J. 
 - 12 29-12 1 
 
 The depressed equation is 
 
 X 4 _ 12 x3 + 29 x* - 12 x + 1 = 
 
 Reducing to an equation of half the degree by dividing by x 2 and regard- 
 ing x + - as the unknown quantity (Art. 569), we have 
 
 X 2 
 
 -12(« + i)=-»: 
 
 or, adding 2 to both members, 
 
 
 hir- 
 
 • 1*(* + I) = -27, 
 
 whence (Art. 361), 
 
 x+-=6±3=9 or 3. 
 
 X 
 
 From 
 
 x + - = 9 
 
 X 
 
 we have 
 
 x = i(9±Vri). 
 
 From 
 
 x + - = 3 
 
 X 
 
 we have x = $(3 ± V5). 
 
 Hence the roots are — 1, $(9 ± a/77), and J (3 ± VE). 
 
 2. x 4 -S aj s +4aj , -3a?+l=0. 3. ^-5^+ 6 ^-5 x+l=0. 
 
 4. x 4_ iC 3_|_ a ;_i = o. 5. z 4 + 7ar 3 -7a;-l=0. 
 
 6. x 4 -Ma^-19a 2 z 2 + 4« 3 # + a 4 = 0. 
 
 7. ax 4 -2x* + 2x-a = 0. 
 
 a 6^-lla; 4 -33ar 3 + 33^-f-ll«-6 = 0. 
 9. 3ar i -2a; 4 -r-5^-5x 2 -h2a;-3 = 0. 
 10. 4x 6 -24z 5 + 57x 4 -73ar J + 57 x 2 -24a? + 4 = 0. 
 
350 HIGHER ALGEBRA 
 
 570. A Binomial Equation is an equation in the form x n ± a = 0. 
 The n roots of the equation are called the n nth roots of T a. 
 
 For example, the solution of the equation X s — 1 = 0, or x? = 1, 
 gives the three cube roots of unity. 
 
 571. Theorem. Every binomial equation can be reduced to the 
 form y n ± 1 = 0, which may be regarded as a recurring equation 
 and treated accordingly. 
 
 Dem. In the form x n ± a = 0, let x n = ay n ; then the equation 
 becomes ay n ± a = 0, whence y n ± 1 = 0. 
 
 EXAMPLES CXXXVIII 
 Solve the following : 
 
 1. ^-1=0. 
 
 Solution. By Art. 565, 1 is a root of this equation ; hence the equation 
 is divisible by x — 1 (Art. 515). 
 
 11110 
 The depressed equation is 
 
 x* + x B + x 2 + X + 1 as 0. 
 Reducing to an equation of half the degree by dividing by x 2 and regarding 
 x + - as the unknown quantity (Art. 569), we have 
 
 or, adding 2 to both members, 
 
 K) 2 +KH- 
 
 whence (Art. 361), 
 
 x + l = -±.±±V5 = l(-l±Vl). 
 
 x 2 2 2\ I 
 
 Solving this equation for as, we have 
 
 x - £(- 1 ± V5 ± V- 10 =F 2V5). 
 
 These four roots and the one first obtained are the five 5th roots of unity. 
 The 5th roots of any other number are these roots multiplied by the real 5th 
 root of that number. For example, the five 5th roots of 32 (or the five roots 
 of the equation a* 5 — 32 = 0, or x 5 = 32) are the above roots multiplied by 2. 
 
GENERAL SOLUTION OP CUBlCS 351 
 
 2. ar*-l = 0. 3. ar , + l=0. 4. z 4 -l=0. 
 
 5. z 4 + l=0. 6. ar' + 1 = 0. 7. ^-243 = 0. 
 
 8. ^-1 = 0. 9. a* + l=0. 
 
 572. From ar 3 — 1 = (x — 1) (x 2 + a + 1) = 0, we obtain for the 
 three cube roots of unity 
 
 if -i + lV-3, -i-4V^3. 
 
 The two imaginary roots possess the peculiarity that each is 
 the square of the other. It is, therefore, customary to write as 
 the three cube roots of unity 1, <o, and <o 2 . 
 
 GENERAL SOLUTION OF CUBICS AND BIQUADRATICS 
 
 573. Cardan's solution of the general cubic equation 
 
 x 3 + px 2 + qx + r = 0. 
 
 To transform this into an equation lacking the second power, 
 we take 
 
 * = </-§ (1) 
 
 obtaining fy - f}+p(y - f Y+ qC V ~ f ) +r = 0, 
 
 or f + of +fq - £\y + — p 3 --pq + r=0, 
 
 which has the form y 3 + my + n = 0. (2) 
 
 To transform this into an equation of the quadratic form (Art. 
 359), we take 
 
 '—J? (3) 
 
 obtaining z 6 + nz 3 — — = 0, 
 
 which gives (Art. 360) 
 
 , n . hi* m 3 
 
 '27 
 
 II n , In 2 . m 3 
 = V- »±V4 +27- 
 
352 HIGHER ALGEBRA 
 
 Substituting this value of z in (3), we have 
 
 *l 11 In* 
 
 ,.l 2 ; m s t» 
 
 27 
 
 <V-1±>S 
 
 r 
 \4 + 27 
 
 Multiplying numerator and denominator of the last fraction by 
 
 n _. In 1 
 
 + 27' 
 
 and omitting double signs, since they give no more values than do 
 
 single ones, we have 
 
 = \-2 + \4 + 72 + ^"2~\4 + 27' 
 
 574. Cor. Cardan's formula fails when all the roots are real 
 and unequal. 
 
 Dem. If a is one of the roots, the other two may be found by 
 dividing the equation by x — a and solving the resulting quad- 
 ratic. These roots will, therefore, be embraced in the forms 
 I) _|_ Vc and b — Vc, in which, for real roots, Vc may be either 
 rational or surd. 
 
 Since the coefficient of y 2 is 0, we have (Art. 539) 
 
 - a - (b + Vc) - (b - Vc) = 0, 
 
 whence a — — 2 b. 
 
 Now the equation whose roots are —2 b, b -f Vc, and b — Vc 
 
 is (kit. 520) 
 
 2/ 3 -(3& 2 + c) 2 / + 2(& 3 -&c) = 0, 
 
 in which m = — (3 b 2 + c), 
 
 and n = 2(6 3 -&c). 
 
 When these values are substituted in Cardan's formula, instead 
 of obtaining real roots according to the hypothesis, we have the 
 imaginary roots 
 
 ^-(v-bcy^ 
 
 This is what is called the Irreducible Case. 
 
GENERAL SOLUTION OF BIQUADRATICS 
 
 353 
 
 575. Sch. Since the methods heretofore given are more expe- 
 ditious for solving numerical cubics, no examples are appended. 
 
 576. Descartes 1 solution of the general biquadratic equation 
 x* -+- ex 3 +f& + gx -f h as 0. 
 
 Transforming as in Art. 573 to remove the second term, we 
 have an equation of the form 
 
 y A +.nf + ty + i = o. 
 
 Now let us assume 
 
 tf +J?f + ty + I = (y 2 + my + n) (y 2 + py + q) = 0. 
 Developing and collecting terms, 
 
 a) 
 
 (2) 
 
 V* +jy 2 + Tcy + l = y* + m 
 P 
 
 f 4- n 
 
 y 2 + np 
 
 mp 
 
 mq 
 
 Q 
 
 
 + nq. 
 
 By Art. 441 we have 
 
 m + p = 0, n + mp + q=j, np + mq = k, nq = Z, 
 from which we obtain 
 
 ?i =o( m2 --+A <l=7>( m2 +-+j\ 
 
 2\ m J 2\ m J 
 
 Substituting these values of n and q in w</ = I, we have 
 
 m 6 + 2jm* + (/ _ 4r>m 2 - A; 2 = 0. 
 
 If in this equation we take 
 
 m = Vmj — }J, 
 
 we shall have a cubic equation, from which m x may be found by 
 Cardan's formula, and then m, n, p f and q from the relations 
 above. Substituting these values in the second member of (2), 
 and equating each of the factors with 0, we find the value of y. 
 
 downey's alg. — 23 
 
CHAPTER XXIV 
 SERIES 
 
 577. Having treated in Chapter XIV the simpler kinds of 
 series, and in Chapter XX of the development of functions into 
 series by the Method of Indeterminate Coefficients, Taylor's 
 Formula, and the Binomial Theorem, we now proceed to the con- 
 sideration of series in general. 
 
 SECTION I— CONVERGENCY OF SERIES 
 
 578. We have seen (Art. 438) that when a function is devel- 
 oped into an infinite series, the sum of the series does not equal 
 the function unless the series is convergent (Art. 434). It is 
 clear, therefore, that a series cannot be used for purposes of dem- 
 onstration unless it is known to be convergent. It hence often 
 becomes necessary to determine whether or not a series is con- 
 vergent. There is no universal test, but the convergency or 
 divergency of series can usually be determined by means of the 
 following theorems. 
 
 579. Theorem. The convergency or divergency of a series is not 
 affected by the addition or subtraction of a finite number of terms. 
 
 For the sum of these terms is finite and determinate. 
 
 580. Theorem. If a series all of whose terms are positive is con- 
 vergent, it is convergent when some or all of its terms are made 
 negative. 
 
 For the sum is greatest when all the terms are positive. 
 
 581. Note. In what follows, therefore, it will be understood 
 that all the terms are positive unless otherwise stated. 
 
 582. Theorem. A series is convergent if all its terms, or all after 
 a finite number, are less than the corresponding terms of a series that 
 
 354 
 
CONVERGENCE OF SERIES 355 
 
 is known to be convergent; and divergent if all its terms, or all after 
 a finite number, are greater than the corresponding terms of a series 
 that is known to be divergent. 
 
 The truth of this theorem is apparent. 
 
 583. To apply this theorem it is necessary to have several 
 standard series with which other series may be compared. The 
 following serve for many cases. 
 
 I. TJie geometrical series 
 
 1+ l + l + l + h + -' (1) 
 
 which may be written in the form 
 
 i I * I * I * I * I (2) 
 
 2 2 • 2 2 • 2 • 2 2 • 2 • 2 • 2 
 
 or a??y oMer decreasing geometrical progression, is convergent. 
 For proof see Art. 329. 
 
 II. The series l-\ ^ 1 h ••• is convergent when m > 1, 
 
 2 m 3 m 4" 1 
 
 and divergent when m ^ 1. 
 
 Dem. 1st. When m > 1. 
 
 We have 1 = 1, 
 
 1 + A/A 
 
 2 m 3 W 2* 
 
 etc., etc. 
 
 By adding, we have 
 
 1 + 4 + 4 + A + 4 + ^ + ^; + etc -< 1 +l; + ^ + etc - 
 
 »>"» £$"» ^.m 5»« (j»» j»» 2 4 
 
 But this last series is a geometrical progression whose ratio is 
 
 2 2 
 
 — -. Hence, since m > 1, making — < 1, the series is convergent. 
 
 2 m 2 m 
 
 2d. When m = 1, giving the series 
 
 ^2^3^4^5^ W 
 
356 HIGHER ALGEBRA 
 
 Grouping the terms thus, • 
 
 it is seen that each group is greater than \. Hence the series is 
 greater than 
 
 i +1+1+1+1+ -, 
 
 and is, therefore, divergent. 
 3d. When m < 1. 
 
 In this case each term after the first is greater than the corre- 
 sponding term of the series 
 
 which, as shown above, is divergent. 
 
 EXAMPLES CXXXIX 
 
 Determine whether the following series are convergent or 
 divergent : 
 
 Scg. Compare with (2). 
 
 2. 1 + 3 + - + - + - + - + -+--. 
 
 Solution. After the 5th term each term of this series is less than the 
 corresponding term of the geometrical progression 
 
 4-4 4-4.4 4.4.4.4 
 whose ratio is |. Therefore, by I, the given series is convergent. 
 
 3. 1+1 + 1 + 1 + 1 + .... 
 
 T 2 3 3 3 4 3 5 3 
 
 4^9^16 n 2 
 
 Sug. Compare with (3) . 
 
CONVERGENCT OF SERIES 357 
 
 584. Theorem. A series is convergent if, from the beginning or 
 after a finite number of terms, the ratio of each term to the preceding 
 term is less than some quantity which is itself less than 1. 
 
 Dem. 
 
 the series be 
 
 
 
 S= \-k + l + m + n-\ , 
 
 
 a) 
 
 =- + *H + ^ + -> 
 
 
 (2) 
 
 = ... + fcfi+i + ^+^ + .. 
 
 \ k kl Mm 
 
 ..). 
 
 (3) 
 
 Now if the ratio of each term of (1) to the preceding term is 
 less than p, we have from (3) 
 
 S <... + & (1+j, +p 2 + p 3 +...)• 
 
 Hence if p < 1, we have from Art. 329 
 
 1-p 
 
 and the series is convergent. 
 
 585. Cor. A series is convergent if from the beginning or after 
 a finite number of terms, the ratio of each term to the preceding term 
 is less than 1 and approaches as a limit. 
 
 For the ratio is then always less than a quantity which is itself 
 less than 1. 
 
 586. Sch. The series (3) of Art. 583 shows that it is not 
 sufficient in the theorem of Art. 584 to say that the ratio of 
 each term to the preceding term is less than 1. It is less than 
 1 in this series, but approaches 1 as a limit as the terms are 
 indefinitely increased. 
 
 587. Theorem. A series is divergent if, from the beginning or 
 after a finite number of terms, the ratio of each term to the preceding 
 term is equal to or greater than 1. 
 
 For the sum of the series is the sum of an infinite number of 
 finite terms. 
 
358 HIGHER ALGEBRA 
 
 588. Theorem. A series of numerically decreasing terms which 
 are alternately -\- and — is convergent. 
 
 Dem. Let the series be 
 
 S = a - b + c - d -f e -/H . (1) 
 
 This may be written in the forms 
 
 S = (a-b) + (c-d) + (e -f) + .-., (2) 
 
 and 8 = a - (b - c) - (d - e) - (/- g) . (3) 
 
 From (2) 8 > a - b, 
 
 and from (3) S < a. 
 
 Hence the series is convergent. 
 
 EXAMPLES CXL 
 
 Determine whether the following series are convergent or 
 divergent. 
 
 n& n& /y" 
 
 1. x + Q L+°L + a L + .... 
 • 2 T 3 ± 
 
 , 1 1,1 1,1 1 , 
 
 2 - 1 -2 + 3-4 + 5-6 + -- 
 
 
 3. The logarithmic series 
 
 
 x 2 . X s x 4 . x 5 
 "-2 + 3"4 + 5- 
 
 x« 
 6 
 
 1.1.1.1. 
 
 1.2 3- 45- 6 7- 8 
 
 2 2 2 2 3 2 4 2 n 
 
 5 ' rT2 4 "2T3 + 3T4 + 475 + '"n(n + l)" f '"' 
 
 Solution. The limit of the ratio of the (w + l)th term to the nth term is 
 
 2«+i 2 M 2 w 
 
 («+l)(« + 2)' r «(» + l) 
 
 Hence (Art. 587) the series is divergent. 
 
 6 . 1+1 + * + i +... + » + .... 
 2 2* 2* 2* 2" 
 
 -*"-1 =2 
 « + 2_L 
 
CONVERGENCY OF SERIES 359 
 
 7. — I 1 1 r • • • 
 
 x 1 + x 2 + x 3 + x 
 
 Sug. Compare with (3), Art. 583. 
 
 1 11 1 
 
 ■ l + x 1 + 2X 2 1+3^ "'"""l + naf 
 
 1 _1 11 
 
 ' l + a> 1 + 2* 1 + Sa? l + 4a? 
 
 10. l + l + I+I+...-lj+---. 
 
 2 3 2 4 3 n n ■ 
 
 11. 1 *__ + — ^ ^__l...... 
 
 1 + a 1 + 2 ct l+3a 
 
 12. The Exponential Series (Napierian) 
 
 Solution. The limit of the ratio of the (n + l)th term to the nth term is 
 x n+\ x n _x~\ _q 
 jft + 1 [n wJ„=x 
 Hence (Art. 585) the series is convergent for all values of x. 
 
 13. The Binomial Series 
 
 (1 + X ) m = 1 + mx + m ( m ~ V > x 2 + -. 
 If 
 
 m(m-l)..-(m-yi + 2) a;W _ 1 | ^ 
 
 |m — 1 
 
 Solution. The limit of the ratio of the (n + l)th term to the nth term is 
 m(m — 1) ••• (w — n -f l)x w w(w — !)••• (m — n + 2)x n ~ 1 
 
 n \ 11 J Jn=oo 
 
 If, then, x be numerically less than 1, the series will be convergent 
 (Art. 581). 
 
 It may be shown that the series is convergent when x = 1 , provided 
 m > — 1 ; also when x = — 1, provided m is positive. See C. Smith's Trea- 
 tise on Algebra, Art. 338. 
 
360 HIGHER ALGEBRA 
 
 i4 - 1+ \+i + i + - 
 
 ■l e ^ [ ^ j g I j ^ i 
 
 ' l.a T 2\3 T 8v4- *(» + !)* 
 
 16. The exponential series 
 
 a" = 1 + (log, a)x + (log, a) 2 | + (log, a) 3 j| + • • -. 
 
 17. The series whose 7*th or general term is — • 
 
 \n 
 
 18. The series whose nth term is 
 
 (n + l) n+1 
 
 SECTION II — SCALE OF RELATION OF A RECURRING 
 
 SERIES 
 
 589. A Recurring Series is a series in which, either, from the 
 beginning or after a finite number of terms, each term is equal to 
 the algebraic sum of the products of a fixed number of the pre- 
 ceding terms, multiplied, respectively, by certain quantities which 
 remain the same throughout the series. 
 
 590. A recurring series is of the First, Second, Third, etc., Order, 
 according as each term is derived from one, two, three, etc., of the 
 preceding terms. 
 
 Thus, by the method of indeterminate coefficients we found (page 275) 
 
 1 _ 3 x - x 2 = x _ _ 2 x2 _ g 3 _ 12 xi _ etc 
 
 1 -2x-x 2 
 
 In this series each term after the third is equal to 2 x times the preceding 
 term plus x 2 times the second preceding term. Hence it is a recurring series 
 of the second order. 
 
 591. If a recurring series of any order be written in the form 
 u x + u 2 + u 3 H h u^ 2 + u n _ x + v n + u n+1 + u n+2 + —, 
 
 we shall have by definition 
 
 u n =pxu n _ 1 + qtfu n _ 2 + rx*u n _ s + ••-, 
 whence u n — pxUn^ — gA n _ 2 — 7*A n _ 3 — • •• = 0, 
 
 which expresses the law of the series. 
 
SCALE OF RELATION 
 
 361 
 
 In this form 
 
 1 — px — qtf — rx*—-- 
 
 constitutes the Scale of Relation, or the Scale, and p, q, r, etc., are 
 the Constants of the Scale. 
 
 Thus, in the series of Art. 590 the scale of relation is 1 — 2 x — x 2 , and the 
 constants of the scale are 2, 1. 
 
 592. To extend a series some of whose terms are given, it is 
 sufficient to make use of the constants of the scale, since the 
 powers of the variable may be supplied by inspection. 
 
 593. Prob. To find the constants of the scale. 
 
 Solution. 1st. In a series of the first order. 
 Let the series be 
 
 a + bx + ex 2 + dx* + ex 4 +fx 6 + ••-, 
 and p the constant of the scale. 
 
 Then 
 
 f=l™, 
 
 whence 
 
 p = 
 
 f 
 
 The series is a geometrical progression. 
 2d. In a series of the second order. 
 Let p, q be the constants of the scale. Then 
 f=*pe + qd, 
 e=pd + qc, 
 from which p and q may be found. 
 
 3d. In a series of the third order. 
 
 Letp, q, r be the constants of the scale. Then 
 
 f=pe + qd + rc, 
 
 e=pd -\-qc+ rb, 
 
 d = pc-\-qb + ra, 
 
 from which p, q, and r may be found. 
 
 We may proceed in the same way for series of higher order. 
 
 When the order is not known, we may assume it of the second, 
 third, etc., order until the right order be found. If the order be 
 assumed too high, one or more of the constants will be ; and if 
 
362 HIGHER ALGEBRA 
 
 assumed too low, the error will become apparent in applying the 
 scale found. 
 
 594. Cor. To find the scale ive must have twice as many terms of 
 the series as there are constants in the scale. 
 
 EXAMPLES CXLI 
 
 Find the constants of the scale in each of the following, and 
 extend each series one term : 
 
 1. i + 4 x + 6 x 2 + 11 x* + 28 x* + 63 as 8 + 131 x« + • • •. 
 
 Solution. Assuming the series of the second order, we have 
 
 6p + 4g = ll, 
 
 whence P — H an d Q = f • 
 
 If the proper constants have been found, we shall have 
 lljp+ 6g = 28; 
 but ll(tf)+6(*)=19.7. 
 
 Hence the series is not of the second order. 
 Next assuming it of the third order, we have 
 
 6p + 4g + r = 11, 
 
 llj) + 6g + 4r = 28, 
 
 28p + Uq + 6r = 63, 
 
 whence p = 2, q = — 1, r = 3. 
 
 If the proper constants have been found, we shall have 
 
 63j> + 28g + llr = 131. 
 
 Now 63-2 + 28(-l)+ 11-3 = 131. 
 
 Hence the proper constants have been found. 
 To find the coefficient of the next term we have 
 
 131- 2 + 63(-l)+28- 3 = 283. 
 
 Hence the next term is 283 x" 1 . 
 
 2. 1 + 6^ + 12^ + 48a 3 + 120« 4 +"-. 
 
 3. l + 3a + 7a 2 + 17ar 5 + 41x 4 +—. 
 
 4. l-f-9x-15ar 9 4-57o 3 -159a 4 +-.-. 
 
 5. l + # + 2a 2 + 2x 3 + 3a; 4 + 3ar 5 + 4a; 6 + .... 
 
THE NTH TERM OF A SERIES 363 
 
 6. 2 + z-3z 2 + 2ar 5 + :K 4 -3ar 5 +.... 
 
 7. 3 + 5 a- + 7 x> + 13 ar 3 + 23a; 4 + 45*" + 87.T 6 + •••• 
 
 SECTION III — THE NTH TERM OF A SERIES 
 
 595. We have learned how to find the nth term of an arith- 
 metical series (Art. 318), a geometrical series (Art. 326), the 
 binomial series (Art. 457), and of a recurring series by extension 
 to the nth term by means of the constants of the scale of relation 
 (Art. 593). A useful method of finding the nth term of a less 
 simple series is by the Successive Orders of Differences. 
 
 596. The First Order of Differences of a series is the series 
 obtained by subtracting the 1st term of the given series from 
 the 2d, the 2d from the 3d, the 3d from the 4th, and so on. The 
 Second Order of Differences is the series obtained from the first 
 order as the first order is obtained from the given series. The 
 Third, Fourth, etc., Orders are obtained similarly. 
 
 Thus, given series, 1, 8, 27, 64, 125, etc., 
 
 1st order of differences, 7, 19, 37, 61, etc., 
 
 2d order of differences, 12, 18, 24, etc., 
 
 3d order of differences, 6, 6, etc., 
 
 4th order of differences, 0, etc. 
 
 597. Prob. To find the first term of any order of differences. 
 Solution. Let the series be a, b, c, d, e, ••• 
 
 1st order of diff., b — a, c — b, d — c, e — d, ••• 
 
 2d order of diff., c — 2 6 + a, d — 2c + b, e — 2d + c, •• • 
 3d order of diff., d - 3 c + 3 b - a, e - 3 d + 3 c - b, ••• 
 
 4th order of diff., e — 4 d + 6c — 46 + a, ••• 
 
 Denoting the first terms of the respective orders of differences 
 by D lf D 2 , Z>3, Z> 4 , etc., we have 
 
 D x = — a + b, 
 
 D 2 =a — 2b + c, 
 
 D 3 =-a + 3&-3c + d, 
 
 B 4 = a - 4 b + 6 c — 4 d + e, 
 etc., etc. 
 
364 HIGHER ALGEBRA 
 
 The coefficients in these terms are seen to be, numerically, those 
 
 of a developed binomial by the binomial theorem. Hence, when 
 
 n is even, 
 
 ^ , . n (n — 1) ft (ft — 1) (ft — 2) 7 , 
 D n = a-nb+-±— — } -c ^ -£* ld + • •, 
 
 and when n is odd, 
 
 r» , 7 w (n — 1) , ft (ft — 1) (ft — 2) 7 
 
 D n = -a + 7ib--^— — ^c+-^ -^ *d . 
 
 598. Cor. To find the 1st term of the nth order of differences, 
 n + 1 terms of the series must be given. 
 
 This is seen by inspecting the values of D lf D 2 , D 3 , D 4 , etc. 
 
 EXAMPLES CXLII 
 
 Find the first term of the specified order of differences in each 
 of the following: 
 
 1. 3d and 4th of 7, 12, 21, 36, 62, etc. 
 Solution. For the third order we have 
 
 Z> 3 =-« + 3&-3c + d = -7+3.12-3.21+36 = 2. 
 For the 4th order, 
 
 Z) 4 = a-46 + 6c-4d+e = 7-4.12 + 6.21-4.36 + 02 = 3. 
 In practice the shortest way is to find the successive orders by subtraction. 
 
 2. 3d of 1, 3, 6, 10, 15, etc. 
 
 3. 3d and 4th of 1, 8, 27, 64, 125, etc. 
 
 4. 3d and 5th of 1, 3, 3 2 , 3 3 , 3 4 , 3 5 , etc. 
 
 599. Prob. To find, by the successive orders of differences, the nth 
 term of a series. 
 
 Solution. From Art. 597 we obtain 
 b = a + D l} 
 c = a + 2D l + D 2 , 
 d = a + 3D l + 3D 2 + D 3 , 
 
 e = a + 4A + 6A + 4A + A, 
 etc., etc. 
 
SUMMATION OF A SERIES 365 
 
 It is seen that the coefficients of the nth terra of the series are 
 the coefficients of the (n — l)th power of a binomial. Hence, 
 writing n — 1 for n in the coefficients of the binomial formula 
 (Art. 457), we have 
 
 nth term = a + (* U 1) A + ( w ~ 1 H n ~ 2) Z> > 
 
 j (rc-l)(n-2)( m -3) A| 
 
 It is evident that the nth term of a series can be found exactly 
 only when the terms of some order of differences are 0. 
 
 EXAMPLES CXLIH 
 
 Find the terms specified in the following : 
 
 1. 12th term of 1, 5, 15, 35, 70, 126, etc. 
 
 Solution. Obtaining the successive orders of differences, we have 
 
 4, 10, 20, 35, 56, etc., whence D L = 4 
 
 6, 10, 15, 21, etc., whence D 2 -Q 
 
 4, 5, 6, etc. , whence Dz = 4 
 
 1, 1, etc., whence Z> 4 = 1 
 
 0, etc., whence Z> 5 = 0. 
 
 Substituting in the formula, we have 
 
 „ tUorm ! . „ , . 1L10 A . 11-10-9. , 11 -10.9-8 1Q( ., 
 
 nth term = 1 -f 11 . 4 -\ 6 -\ 4 H = 1365. 
 
 2 2-3 2-3-4 
 
 2. 12th term of 1, 3, 6, 10, 15, 21, etc. 
 
 3. 15th term of 1, 2 2 , 3 2 , 4 2 , etc. 
 
 4. 12th terra of 1, 4z, 6a,- 2 , liar 5 , 28a; 4 , 633^, etc. 
 
 SECTION IV — SUMMATION OF SERIES 
 
 600. We have learned how to find the sum of n terms of an 
 arithmetical series (Art. 319), and of a geometrical series (Art. 
 327), and the limit of the sura of an infinite decreasing geometrical 
 series (Art. 329). We proceed to develop methods for finding 
 the sum of series that are less simple, though there is no general 
 formula for summation. 
 
366 HIGHER ALGEBRA 
 
 601. The sign of summation is the Greek letter sigma, X 
 Written before the general term of a series it signifies the sum of 
 the series obtained by making n successively equal to 1, 2, 3, 4, etc. 
 
 Thus, V 1 = _!_ + _!_ + _J_ + _!_+.... 
 
 4n(n + l) 1-2 2-3 3- 44- 5 
 
 602. The Generating Function of a series is the function which, 
 when expanded (Art. 436), produces the series. 
 
 Thus, since * ~ 3 x ~ x * = 1 - X - 2 x 2 - 5 x 3 - 12 x 1 
 
 1 - 2 x - x 2 1 - 2 x - x 2 
 
 is the generating function of the series 1 — x — 2 x 2 — 5 x 3 — 12 x 4 — •••. 
 
 The generating function of an infinite series is the same as the 
 sum of the series when the series is convergent, but not when the 
 series is divergent (Arts. 436 and 438). 
 
 603. Prob. To find, by the method of decomposition, the sum of 
 a series whose general term has the form 
 
 q q 
 
 -—± or y 
 
 n(n +p) n(n + p) (n + 2 p) 
 
 or, in general, 
 
 n {n + p) (« + 2 p) • • • (n + rp) 
 Solution. By Art. 447, we have 
 
 q = A + B 
 
 n (n -f p) n n + p 
 
 1 1 
 
 whence A = - and B = 
 
 p p 
 
 Hence 
 
 p\n n+ p 
 
 n(n+p) 
 
 , y q_ = yv?__i_vvy?_y_L_ > i m 
 
 L*n{n+p) L*p\n n+p) p\L*n L^n+pJ 
 
 Now, when n is taken successively equal to 1, 2, 3, 4, etc., all 
 
 the terms of / - and / -^—, except the first p terms of the 
 L*n L*n+p 
 
SUMMATION OP A SERIES 367 
 
 former and the last p terms of the latter, will cancel, and the sum 
 
 will be -th the sum of these remaining terms. Hence 
 P 
 
 y g 
 
 L^nin + p) 
 
 = -1 1st p terms of / - — last p terms of / — - — ). (2) 
 p\ —> Amjn + pJ 
 
 The sum of an infinite number of terms is -th the sum of the 
 
 Zq Q 1 P 
 
 -, since — y-2 r = 0. 
 n *{n+P)X 
 
 Similarly, we have 
 
 Z g _ i rsp g y q v (3) 
 n(n+i>)(u + 2p) 2p\L*n(n+p) A*(n+p)(n+2p))' 
 
 and, in general, 
 
 ? =-(Y < ? 
 
 n (n +p) (?i + 2 p) • • • (n + rp) rp \L** n (n +p) • • • [ n + (r — l)p] 
 
 _y 2 v (4) 
 
 EXAMPLES CXLIV 
 
 Find, by the method of decomposition, the sum of the follow- 
 ing to n terms and to oo. 
 
 1.22.3 3.44.5 n(n + l) 
 
 Solution. Since q — 1 and p = 1, we have from (2) 
 
 5] 2 1st term of ^ - - nth term of ^J — — 
 
 ** n (» + p) ^ n T» n + 1 
 
 Hence A, - 
 
 1 n + 1 w + 1 
 
 and &o as 1st term of 5/- — 1« 
 
 The value of S x is also obtained from the value of S n by making n = 00, 
 giving 
 
 n + Uoo 
 
368 HIGHER ALGEBRA 
 
 Without the use of the formula we may proceed thus i 
 
 % _1 1 1 _\ 1 1 _1 1 
 
 1-2 1 2 2 • 3 2 3' 3 • 4 3 4 G °' 
 
 Hence the series becomes 
 
 (i_i\ + /l_l\ + /i_i\+...+/l__l_\ = ,__J_ = _S_. 
 
 i .32.43.5 
 
 Solution. Since q = 1 and p = 2, we have from (2) 
 
 5) 2 1/ i s t 2 terms of V - - last 2 terms of T -J— V 
 
 ^w(»+p) 2V ^n ^n + 2j 
 
 Hence «, = 1(1 + 1^I__!_U1(8_1 __1_\ 
 
 2\1 2 n n + «y 2\2 n n + 2/ 
 
 and ^ = i(?- X — Ul = f. 
 
 " 2\,2 n «,+ 2/J„ 4 
 
 3. 1 + 1 + _1_ + 1 +...+. 1 
 
 1-42.53-64.7 « (n + 3) 
 
 3-8 6,12 9-16 
 
 Sug. Write in the form 
 
 i2Vi-2 2.3 3.4 y 
 
 and compare with ex. 1. 
 
 5 1 1 1 ■ 1 I 1 . 
 
 ' 1-4 2-6 3-8 4-10 ' 
 
 6 1 I 1 I 1 I - I 1 I • 
 
 ' 1-3 3.5^5.7 (2n-l)(2n + l) ' 
 
 Solution. As (2) is not applicable, the denominator not having the 
 form n(n + J)), we have from (1), 
 
 V 2 =1(2—1 T — L_ ) 
 
 ^(2«-l)(2«+l) 2V-^2«-l -W2n+1/ 
 
 =1/1+1+1+. ? 1_I ! L_\. 
 
 2V1 a 6 2)1-1 3 5 2n-l 2n + 1/ 
 
SUMMATION OF A SERIES 369 
 
 Hence S n = \{ 1 - ^— A = — -^— -» 
 
 2\ 2n + l/ 2n + 1 
 
 and ^ = 2^nL = i- 
 
 . 2 2 2 2 
 
 ' 3.5" h 5.7" t "7.9 *~ 1 ~(2tt + l)(2n + 3) i ~'"' 
 
 8 . i + 2 3 _^_ ^ 
 
 1 1 
 
 \n + 1 [n l /i + l 
 
 9 _2 3, _4 SL.+ ... : ± "" + ! t 
 
 3-5 5-7 7-9 9-11 (2n + l)(2n + 3) 
 
 Solution. Since 5 = n + 1, a variable, andp = 2, we have from (1), 
 
 A(2n+l)(2n + 3) 2\^f2n + l -W2n + 3/ 
 
 = y2_3 , 4_ # n + 1 2 3 71 , ^ + 1 \ 
 
 2\3 5 7 2n+l 5 7 2w + l 2n + 3/ 
 
 2V3 2n + 3J 
 
 Hence, when w is even, 
 
 q -V 2 11 "+ 1 "UV 1 1 w + 1 V 
 n "2\3 2W + 3J 2\ 3"^2« + 3/' 
 
 and when n is odd, 
 
 * = l/2 n + l \ 
 n 2\3 2w+3y/ 
 
 1 1 1 1 . 
 
 ' 1.2.3" t "2.3.4" t "3.4.5 _h ' ' n(w + l)(w + 2) ^ ' 
 Sue. Use (3). 
 
 1.3.5^3.5.7^5.7.9^ 
 
 12. T"^^ + ^-4-T + 
 
 1-2.3 2.3.4 3.4.5 
 
 downey's alg. — 24 
 
370 
 
 HIGHER ALGEBRA 
 
 Solution. Since q = n + 3 and p = 1, we have from (3) 
 
 y g = i iy w+8 y n+3 ^ 
 
 ^f*(n+p)(n + 2p) 2\^n{n + \) <W( W + i)( n + 2)/ 
 1/ 4_ , _5_ _6_ . 7i + 3 
 
 2 2-3 3-4 "" »(n+l) 
 
 -V- 
 
 4 5_ >i+2 n + 3 \ 
 
 2-3 3-4 »(»+I) (n + l)(n + 2)j 
 
 = I/ r 2+-J- + J-+...+— i w + 3 ^ 
 
 2\ 2-3 3-4 w(n + l) (n + l)(n + 2)/' 
 
 and 
 13. 
 
 14. 
 
 .c 
 
 ■** 
 
 1 
 2" 
 
 1 
 
 n + 1 
 
 w + 3 
 
 
 
 (n + l)(n 
 
 + 2) 
 
 
 = !(«- 
 
 2V2 
 
 c« 
 
 2n + 5 \ 
 
 
 #00 
 
 _5 
 4' 
 
 
 
 
 
 + 
 
 + 
 
 + 
 
 1.2.32.3.43.4.54.5-6 
 
 3.4.54.5.65.6.7 
 
 + 
 
 15. 
 
 + 
 
 + 
 
 + .... 
 
 1-3 1.3.5 1.3.5.7 1.3.5...(2n + l) 
 
 + •••• 
 
 604. Prob. To find, by the scale of relation, the sum of n terms 
 of a recurring series. 
 
 Solution. Let the series be a +- bx -\- ex 2 +■ dar 3 +- •••, 
 
 whose scale of relation (Art. 591) is 1 — px — qx 2 . 
 
 Though this assumes that the series is of the second order, the 
 method is general. 
 
 Representing by S n the sum of n terms, we have 
 
 S n = a +- bx +- ex 2 +- dx* -\ h l®"' 1 - 
 
 Multiplying both members by the scale of relation and arrang- 
 ing according to the powers of x, we have 
 
 (1— px— qx 2 )S n =a+ b 
 —pa 
 
 X+- c 
 
 x' 2 +- d 
 
 —pb 
 
 —pc 
 
 — qa 
 
 -qb 
 
 x?-\ h l\x n ~ l 
 
 —pk\ —pi 
 
 - qj\ —g* 
 
 -qlx n+ \ 
 
SUMMATION OF A SERIES 371 
 
 Since, by Art. 591, the coefficients of a 2 , ar 3 , •••# n ~ 1 are 0, we 
 have 
 
 (1 — px — qx 2 ) S n = a+ b 
 — pa 
 
 x— pl\ x n 
 — qk\ — qlx n + l . 
 
 Hence S n = a + (b ~ pa) x ~ ^ + ^ )af ~ qlx " +1 
 
 1 — px — qx 2 
 
 605. Cor. The sum of an infinite convergent recurring series of 
 the second order is 
 
 & _ a+(b — pa)x 
 1 — px — qxr 
 
 This is because the last two terms of the expression for S n 
 approach as a limit. 
 
 606. Sch. The expression 
 
 a + (b — pd)x 
 1— px— qx 2 
 
 is the generating function (Art. 602), and the development of it 
 will reproduce the original series. 
 
 EXAMPLES CXLV 
 
 Find, by the scale of relation, the generating function of each of 
 the following : 
 
 1. l + 2» + 8a? l + 28aj 8 + 100aJ 4 + .... 
 
 Solution. We must first find the constants of the scale. From Art. 593 
 we have 
 
 S = 2p + q, 
 
 2S = 8p + 2q, 
 
 whence p = 3, q = 2. 
 
 Since 28 p + Sq = 100, 
 
 the proper constants have been found. 
 
 Substituting in S = « +(&-!»)* 
 
 I — px — qx 2 
 
 we have S= 
 
 1 - 3 x - 2 x 2 
 
372 HIGHER ALGEBRA 
 
 2. l + 9a-15ar + 57ar 3 -lo9a; 4 + .... 
 
 3. l + 2« + 3ar* + 5ar 3 + 100£ 4 +---. 
 
 4. 1 +5a + 9ar 9 + 13ar J + .... 
 
 5. 1 + 3x + 8^ + 22 ar^ 60a 4 +•••• 
 
 6 . 2 - 5a + 17a 2 -65^ + 257a 4 . 
 
 7. 1 + 3&*-{-o£ 2 + 7£ 3 H . 
 
 8. 3 + 5a + 7x 2 + 13a 3 + 23a 4 + 45a 5 + •••. 
 
 9. 1 + 3 a — x 2 — 5a 3 — 7 a 4 — a^ + lla^H . 
 
 10. 1 - 3 x + 5 x 2 + 5 a 8 + 13 x 4 + 61 a 5 + 181 a 6 + • • •. 
 
 607. Prob. To Jfndj &# £7ie method of differences, the sum of n 
 terms of a series. 
 
 Solution. Let the given series be 
 
 a, b, c, d, e, f etc., (1) 
 
 and let S represent the sum of n terms. 
 
 Then 8 is the (n + l)th term of the series 
 
 0, a, a + b, a + b -+- c, a + b + c + d, etc. (2) 
 
 Series (1) is the same as the first order of differences (Art. 596) 
 of series (2), the first order of differences of (1) is the same as the 
 second order of differences of (2), and so on. 
 
 Substituting in the formula for the nth. term (Art. 599) n + 1 
 for n, for a, a for D 1} D x for D 2 , etc., we have 
 
 c . n (n — 1) 7-v . n (n — 1) (n — 2) ^ . . 
 S = na-\ — ^— — L D x H — * -^ — i — L D 2 + etc. 
 
 This formula is applicable only when the series is such that all 
 the terms of some order of differences become 0. 
 
 EXAMPLES CXLVI 
 
 Find, by the method of differences, the sum of the following : 
 1. 1, 8, 21, 40, 65, etc., to 12 terms. 
 
 Solution. Obtaining the successive orders of differences, we have 
 
 1, 8, 21, 40, 65, ... 
 
 7, 13, 19, 25, - 
 
 6, 6, 6, -. 
 
 0, 0, ... 
 
PILES OF SPHERICAL SHOT 373 
 
 Substituting in the formula, 
 
 5=12xl + 1 ^ 12 - 1 ^ + 12 ( 12 -, 1 K 12 - 2 )6 = 1794. 
 
 2. 1, 3, 5, 7, etc., to 20 terms. 
 
 3. 4, 14, 30, 52, 80, etc., to 13 terms. 
 
 4. 1, 2, 3, 4, 5, etc., to 50 terms. 
 
 5. 1, 5, 15, 35, 70, 126, etc., to 30 terms. 
 
 6. 7, 14, 19, 22, 23, etc., to 9 terms. 
 
 7. The following, obtained as above, are useful in Physics : 
 
 2n =l+2 + 3 + 4H \-n 
 
 (n + 1) 
 
 2 ' 
 
 2n 2 = l 2 + 2 2 + 3 2 + 4 2 + ••• + n 2 = n ( n + 1 ^ 2n + 1 \ 
 
 6 
 
 2n 3 = l 3 + 2 3 + 3 3 + 4 3 + ... +n 3 = n2(n + *)* = (2n) 2 , 
 
 Sn 4 = 1 4 + 2 4 + 34 + 44 + ... + n 4 = ^ + l)(67i 3 + 9n 2 + n-l) 
 
 If n is a large number and only approximate results are sought, 
 all the lower powers may be omitted, and we have 
 
 *-!* *•-$ Sn«~f, **•-£ 
 
 2 3 4 5 
 
 and, in general, 
 
 ^ » w m+1 
 
 ra + 1 
 
 ^ n(n + l) 1-2 2.3 3-44.5 n(n+l) 
 
 2 2" r 2~ h 2~ t ~2~ r 2 
 
 = i[2 (n 2 + n)] = \ (Sn 2 + 2 M ) = » ( " + ff" + 2) - 
 
 PILES OF SPHERICAL SHOT 
 
 608. Prob. To find the number of balls in a complete pyramid 
 or wedge. 
 
 Solution. 1st. A triangular pyramid. 
 
 Let n be the number of balls in one side of the bottom course. 
 This will also be the number of courses. 
 
374 HIGHER ALGEBRA 
 
 The number of balls in the successive courses, beginning at the 
 top, is 
 
 1, 3, 6, 10, 15, etc. 
 
 Obtaining the successive orders of differences and substituting 
 in the formula of Art. 607, we have 
 
 o _ n(n+l)(n + 2) 
 
 2d. A square pyramid, 
 
 The number of balls in the successive courses is 
 
 l 2 , 2 2 , 3 2 , 4 2 , 5 2 , ... n\ 
 As before, we obtain 
 
 o w(n + l)(2n + l) 
 
 B 
 
 #d ^4 wedge ivith rectangular base ayid single row at top. 
 
 Let m' be the number in the top row. Then the next course, 
 being longer by 1, will contain 2(m' + l); the next, 3(m'+2); 
 the next, 4 (m' -f 3), and so on, giving the series 
 
 m', 2 m' + 2, 3m' + 6, 4m' + 12, .... 
 As before, we obtain 
 
 s _ n(n + 1) (3 m' + 2 n - 2) 
 6 
 If we let m be the number of balls in the length of the base, 
 we have 
 
 m' = m — n -f 1, 
 
 and the last formula becomes 
 
 p _ n(n + 1) (3 m — n -\- 1) 
 6 
 609. Sch. The number of balls in an incomplete pile is the 
 number in a complete pile having the same base, diminished by 
 the number in a complete pile whose base would be the next 
 course above the top course of the incomplete pile. 
 
 EXAMPLES CXLVII 
 Find the number of balls in each of the following : 
 
 1. A triangular pile of 15 courses. 
 
 2. A triangular pile of 20 courses. 
 
INTERPOLATION 375 
 
 3. An incomplete triangular pile of 15 courses, having 21 balls 
 in the top course. 
 
 4. An incomplete triangular pile whose bottom course has 
 15 balls on a side, and whose top course contains 28 balls. 
 
 5. A square pile of 15 courses. 
 
 6. An incomplete square pile whose bottom course has 20 balls 
 on a side, and top course 8 on a side. 
 
 7. A rectangular pile whose bottom course is 42 balls by 20. 
 
 8. A rectangular pile whose top row contains 23 balls. 
 
 9. An incomplete rectangular pile whose top course is 12 balls 
 by 20, and whose bottom course is 52 balls in length. 
 
 INTERPOLATION 
 
 610. Interpolation is the process of introducing between the 
 terms of a series other terms which conform to the law of the 
 series. 
 
 The most extensive use of interpolation is in finding inter- 
 mediate terms between those given in mathematical tables, and 
 in finding right ascensions, declinations, etc., for other times than 
 those given in the Nautical Almanac. 
 
 611. The Argument is the variable quantity on the value of 
 which the magnitude of the function depends. 
 
 Thus, in finding from the table on pages 300 and 301 the 
 logarithms of given numbers, the given numbers constitute the 
 argument. In finding from the Nautical Almanac the moon's 
 declinations for given times, the given times constitute the 
 argument. 
 
 When the changes of the argument are proportional to the 
 changes of the function, no formula is needed for interpolating. 
 
 612. Prob. To interpolate between two consecutive terms of a 
 series a term that shall conform to the law of the series. 
 
 Solution. Let p be the distance, in intervals, of the required 
 term t from the first term a. Then p is an improper fraction and 
 
376 
 
 HIGHER A LG Eli HA 
 
 is one less than the number of terms. Substituting p + 1 for n 
 in the formula for the ?<th term of a series (Art. 599), we have 
 
 p(p _ i) p(p _ i) (* _ 2) 
 
 = a + p A + | — z A + — nr^— 
 
 1? 
 
 |3 
 
 A+- 
 
 613. The value of t is more closely approximated as more 
 orders of differences are used, and is absolutely correct when the 
 terms of the next order of differences are 0. 
 
 For facilitating computation the values of the coefficients of 
 A? A A? an( l A> have been computed for every hundredth part 
 of an interval and arranged in a table. See Loomis's Practical 
 Astronomy, page 393. * 
 
 614. The formula of Art. 612 is for increasing values of the 
 argument. A similar formula could be deduced for decreasing 
 values of the argument. When one of these formulas gives a 
 positive error, the other usually gives a negative error. Bessel's 
 Formula is obtained by taking the half sum of these two for- 
 mulas. Representing the argument by A, and the function by F, 
 we have 
 
 Argument Function 
 
 A-2 
 A-l 
 A 
 
 -4 + 1 
 
 A + 2 
 A + 3 
 
 F 
 F' 
 
 F"' 
 F iv 
 
 1st diff. 
 
 A 
 
 A' 
 
 A"- 
 
 A" 
 
 A iv 
 
 2d diff. 
 
 A 
 
 J^2 
 
 A'" 
 
 3d diff. 
 
 I), 
 
 4th diff. 
 
 A 
 
 A' 
 
 ith diff. 
 
 IK 
 
 For interpolating between A and A + 1, F' is taken as the first 
 term, making p a proper fraction ; and the formula becomes 
 
 m 
 
 + 
 
 p(p + l)(p-l)(p -2) 
 I* 
 
 15 
 
 A-h 
 
INTERPOLATION 
 
 377 
 
 Bessel's table (see Loomis's Practical Astronomy, page 392) 
 gives the coefficients as far as the fifth differences for every 
 hundredth part of an interval. 
 
 For most purposes the first three terms of the above formula, 
 employing only first and second differences, are sufficient.* 
 
 EXAMPLES CXLVm 
 
 1. An eclipse of the moon occurs only at the time of opposition, 
 i.e., when the sun and the moon are on opposite sides of the earth. 
 For the eclipse of June 12, 1900, the Greenwich mean time of 
 opposition, as given by the Nautical Almanac, is 15 h. 31 m. 30 s., 
 and the right ascensions for the 14th, 15th, 10th, and 17th hours 
 are as given below. Find the right ascension of the moon at the 
 time of opposition. 
 
 Solution 
 
 Argument 
 
 Function 
 
 1st cliff. 
 
 2d diff. 
 
 3d diff. 
 
 hr. 
 
 hr. min. sec. 
 
 
 
 
 June 12, 14 
 
 17 19 56.00 
 
 min. sec. 
 
 2 24.94 
 
 sec. 
 
 
 15 
 
 17 22 20.94 
 
 .10 
 
 A 
 
 16 
 
 17 24 45.98 
 
 2 25.14 
 
 .10 
 
 u 
 
 17 
 
 17 27 11.12 
 
 
 
 In this case F" = 17 h. 22 m. 20.94 s., p = 31 m. 30 s. = .525 h., 
 A" = 2 m. 25.04 s. = 145.04 s., DJ = .1 s., and D 2 " = .1 s. Sub- 
 stituting in the formula of Art. 614, we have* 
 
 t = 7 h. 22 m. 20.94 s. + .525 x 145.04 s. + .525 (- .475) x .1 s. 
 = 17 h. 22 m. 37.074 s. 
 
 2. The Greenwich mean time of conjunction at the time of the 
 total eclipse of the sun on May 28, 1900, as given by the Nautical 
 Almanac, is 2 hours 57 minutes 2.7 seconds, and the right ascen- 
 sions of the sun for two noons before and two after are 4 hours 
 
 * For a fuller discussion of the subject of interpolation, see Loomis's 
 Practical Astronomy, Chauvenet's Practical Astronomy, Doolittle's Practical 
 Astronomy. 
 
378 HIGHER ALGEBRA 
 
 15 minutes 13.84 seconds, 4 hours 19 minutes 17.40 seconds, 4 
 hours 23 minutes 21.43 seconds, and 4 hours 27 minutes 25.90 
 seconds, respectively. Find the mean time of that phase of the 
 eclipse which occurs at the instant of conjunction, for New 
 Orleans, which is 6 hours west of Greenwich. 
 
 3. The right ascensions of Jupiter on four successive days at 
 noon being 10 hours 5 minutes 38.6 seconds, 10 hours 6 minutes 
 18.86 seconds, 10 hours 6 minutes 59.41 seconds, and 10 hours 7 
 minutes 40.24 seconds, respectively, find the right ascension for 
 midnight of the second day. 
 
 4. The cube roots of 60, 62, 64, and 66 being 3.91487, 3.95789, 
 4, and 4.04124, respectively, find the cube root of 63. 
 
 5. On Oct. 29, 1900, the altitude of the sun when on the 
 meridian at Minneapolis was 31° 32' 17. "17. The south declina- 
 tions of the sun at Greenwich apparent noon on Oct. 28th, 29th, 
 30th, and 31st were 13° 3' 49. "8, 13°23'53."2, 13° 43' 43. "9, and 
 14° 3' 21. "4, respectively. Find the latitude of Minneapolis, 
 which is 6 hours 12 minutes 56.8 seconds west of Greenwich, the 
 latitude being the complement of the sum of the meridian altitude 
 and the south declination. 
 
CHAPTER XXV 
 
 PERMUTATIONS AND COMBINATIONS 
 
 SECTION I — PERMUTATIONS 
 
 615. Permutations are the different orders in which tilings, taken 
 the same number at a time, can be arranged. 
 
 Thus, the permutations of the letters a, b, c, taken two at a time, 
 are ab, ba, ac, ca, be, cb, and taken three at a time, abc, acb, bca, 
 bac, cab, cba. 
 
 616. Theorem. The number of permutations of n things taken r 
 at a time is 
 
 „P r = n(n-l)(n-2)(w-3) ••• (n-r + 1). 
 
 Dem. Let n P 2 , n P 3 , n P 4 , • • • n P r be the number of permutations 
 of the n things, according as 2, 3, 4, ••• r things are taken at a 
 time. 
 
 Taken two at a time, each of the n things may be placed in 
 turn before each of the remaining n — 1 things, giving 
 
 n P 2 =7l(7l-l). 
 
 Taken three at a time, each of the n things may be placed in 
 turn before each of the (n — 1) (n — 2) permutations that may be 
 formed of the n — 1 remaining things taken two at a time, giving 
 
 n P,= n(n-l)(n-2). 
 
 Taken four at a time, ea.ch of the n things may be placed in 
 turn before each of the (n — 1) (n — 2) (n — 3) permutations that 
 may be formed of the n — 1 remaining things taken three at a 
 time, giving . nA = , l(n _ 1} ()( _ 2 ) („ _ 3). 
 
 The number subtracted from n in the last factor is seen to be in 
 each case 1 less than the number of things in each permutation ; 
 hence ^ = n ()i _ 1} (w _ 2) (n _ 3) ... (b _ r + X) 
 
 379 
 
380 HIGHER ALGEBRA 
 
 617. Cor. The number of permutations of n things taken all at a 
 time is n P n — [w. 
 
 In this case r = n. 
 
 618. Theorem. The number of permutations of n things taken 
 all at a time, ivhen p of one kind are alike, q of another kind alike, 
 
 and so on, is 
 
 \n 
 
 [p_ X \q X etc. 
 
 Dem. Let N be the number of permutations that can be 
 formed. 
 
 If in any one of these N permutations the p like things were 
 replaced by p unlike things (unlike one another and unlike the 
 other n — p things), by changing the order of these p unlike things, 
 leaving the other n — p things unchanged, this single permutation 
 would furnish [p permutations (Art. 617). The same change in 
 each of the N permutations would furnish N x \p_ permutations. 
 
 As the same reasoning applies to the q like things and to the 
 other sets of like things, the replacing of the different sets of 
 like things by unlike things would furnish N x \p_ X \q X etc. 
 permutations. But the n unlike things would furnish [n permu- 
 tations (Art. 617); hence 
 
 N X [p_ X \q_ X etc. = \n, 
 
 " [» 
 
 whence N = , r== 
 
 \jpx\qx etc. 
 
 619. Theorem. If p specified things are required to occupy speci- 
 fied places, the number of permutations is the same as of n—p> 
 things taken r — p at a time. 
 
 Dem. The only permutations possible are those arising from 
 changes of the n—p things, of which r—p are in each per- 
 mutation. 
 
 620. Cor. If the p things can be rearranged among themselves, 
 the number of permutations is \p times the number of permutations 
 of n—p things taken r—p at a time. 
 
 621. Theorem. The number of permutations ofn things taken all 
 at a time in a circle is \n — 1 . 
 
PERMUTATIONS 381 
 
 Dem. Since for every arrangement the tilings may all be shifted 
 the same number of places in either direction around the circle, 
 only relative positions, and not actual positions, are to be con- 
 sidered. Hence, if any one of the things remain in any one of the 
 positions, all possible permutations will be formed by permuting 
 the remaining n — 1 things among the remaining n — 1 positions, 
 giving | m — 1 permutations (Art. 617). 
 
 622. Cor. If each order of arrangement is limited to one direc- 
 tion around the circle, the number of permutations is \ \n — l . 
 
 It is to be noted that a right-hand arrangement as viewed from 
 one side of the circle is identical with the corresponding left-hand 
 arrangement as viewed from the other side of the circle. 
 
 EXAMPLES CXLIX ' 
 
 1. In how many different orders can 3 hats be hung on 8 
 hooks ? 
 
 2. How many different numbers of 3 figures each can be 
 formed from the digits 1, 2, 3, 4, 5 ? 
 
 3. How many different numbers of 2 figures each can be formed 
 from the 9 digits ? How many of 3 figures each ? Of 4 figures 
 each? Of 5 figures each? Of 6 figures each? Of 7 figures 
 each ? Of 8 figures each ? Of 9 figures each ? 
 
 4. In how many different orders can 6 persons be seated at a 
 dinner table ? 
 
 5. In how many different orders can 3 persons occupy 7 fixed 
 seats ? 
 
 6. In how many different orders can 4 players use 6 billiard 
 cues? 
 
 7. In how many different ways can 4 gentlemen select from 7 
 ladies partners for the waltz ? 
 
 8. In how many different orders can a single platoon of 8 
 soldiers be arranged in line ? 
 
 9. How many numbers between 1000 and 10000 can be formed 
 by use of the digits 1, 2, 3, 4, 5, 6 ? 
 
382 HIGHER ALGEBRA 
 
 10. In how many different orders can the letters of the word 
 Algebra be arranged? In how many the letters of the word 
 Mathematics f 
 
 11. How many different signals may be made with 4 different 
 colors, taken any number at a time ? 
 
 Sug. The number = 4 Pi 4- 4P2 + 4 Pz + 4P4. 
 
 12. In how many different orders can 8 boys, any number at a 
 time, enter a room? 
 
 13. In how many different orders can 12 members of a minstrel 
 troupe arrange themselves in line on the stage, the same two 
 always acting as end men ? 
 
 See Arts. 619 and 620. 
 
 14. A shelf contains 20 books, of which 4 are single volumes 
 and the others are in sets of 8, 5, and 3 volumes respectively. 
 Find the number of ways in which the books can be arranged on 
 the shelf, (a) when the volumes of each set remain in the order of 
 their number, (b) when the volumes of each set are together, but 
 in any order. 
 
 15. Find the number of permutations of the factors of a 2 b 3 c 4 . 
 
 16. In how many different orders can a football eleven play, if 
 the full-back, the half-back, the quarter-back, and the center-rush 
 always play in the same positions ? 
 
 17. Four of the crew of an eight-oared boat have trained to row 
 only on the stroke side and four to row only on the bow side. In 
 how many ways can the captain arrange his crew, (a) when the 
 stroke (the rower who sets the stroke) is any -one of the four on 
 the stroke side, (6) when the stroke is always the same man ? 
 
 18. Either A or B of a baseball nine must pitch, either C or D 
 must catch, while E, F, and G must play on the bases. In how 
 many ways can the captain play his team ? 
 
 19. In how many different orders can 6 persons be seated at a 
 round table ? 
 
 20. In how many different orders can 7 children stand in a 
 ring? 
 
com r> ix Ai 'ioxs 383 
 
 21. In how many different orders ean a host and 7 guests sit at 
 a round table, the host always having the guest highest in rank 
 on his right, and the guest next in rank on his left ? 
 
 22. In how many different orders can a party of 5 ladies and 
 5 gentlemen sit at a round table, the ladies and gentlemen sitting 
 at alternate places ? 
 
 Explain why this is [4 x |JS. 
 
 23. In how many different orders can 10 beads be strung for a 
 rosary ring ? 
 
 See Art. 622. 
 
 24. In how many different orders can 5 like pearls, 6 like 
 rubies, and 7 like diamonds be strung for a bracelet ? 
 
 25. If the number of permutations of G things is 360, how 
 many are taken at a time ? 
 
 26. If nJ P 2 = 30, what is n? 
 
 27. If n P n = 40,320, find n. 
 
 28. If n P 6 = 10 x n P,, find n. 
 
 29. If n P 5 = 12 x n P 3 , find n. 
 
 30. If n P r = 990, what is r ? 
 
 31. If 2n P :i = 100 x n P 2 , what is n ? 
 
 SECTION II — COMBINATIONS 
 
 623. Combinations are the different groups into which things, 
 taken the same number at a time, without reference to the order 
 of arrangement, can be formed. 
 
 Thus, the combinations of the letters a, b, c, d, taken two at a 
 time, are ab, ac, ad, be, bd, cd; and taken three at a time, abc, abd, 
 acd, bed. 
 
 While ab and ba are different permutations, they are the same 
 combination. 
 
 624. Theorem. TTie number of combinations of n things taken r 
 at a time is 
 
 _n(n--l )(n-2)(n-3)-(n-r + l) 
 
 [r 
 
384 HIGHER ALGEBRA 
 
 Dem. Let n G r be the number of combinations of the n things 
 taken r at a time. 
 
 By Art. 617 each combination of r things can have \r permuta- 
 tions. Hence the number of combinations is the number of 
 permutations divided by [r; that is (Art. 616), 
 
 _ ?i(n-l)(n-2)(7i-3) — (n - r + 1) 
 
 625. Theorem. The number of combinations of n things taken 
 r at a time is the same as the number of combinatioyis of n tilings 
 taken n — r at a time. 
 
 Dem. For each combination containing r things there is left a 
 combination containing the remaining n — r things ; hence, 
 
 fl — C 
 
 626. Sen. In numerical applications the formula of Art. 625 
 often involves much less labor than that of Art. 624. 
 
 Thus, in determining the number of combinations of 15 things taken 12 at 
 a time, the formula of Art. 624 gives 
 
 c _ 15 . 14 • 13 • 12 . 11 . 10 . 9 • 8 . 7 • 6 • 5 • 4 
 15 12 ~ 2.3.4.5.6.7.8.9.10.11.12 ' 
 
 while that of Art. 625 gives simply 
 
 „ 15-14-13 
 15 ° 3 = 2-3 ' 
 
 627. Theorem. If p specified things are to be included in each 
 combination, the number of combinations is the same as of n — p 
 things taken r — p at a time. 
 
 Dem. The only combinations possible are those arising from 
 the combinations of the n—p things, of which r — p are in each 
 combination. 
 
 628. Theorem. The total number of combinations of n things 
 taken any number at a time is 2 n — 1. 
 
 Dem. Each thing may be treated in two ways, as it may be 
 included or excluded. Either way of treating any one of the 
 
COMBINATIONS 385 
 
 things may be combined with either way of treating each of the 
 other things, giving as the number of such combinations 
 
 2 x 2 x 2 x • • • to n factors = 2 n . 
 
 But this includes the case in which all the things are excluded. 
 Hence the total number of combinations is 2 n — 1. 
 
 629. Cor. .Q + n C, + jC t -f «- n C r = 2" - 1. 
 
 630. Theorem. The total number of combinations of n things 
 taken any number at a time/ when p of one kind are alike, q of 
 another kind alike, r of another kind alike, and so on, is 
 
 ( P +i)( 9 + i)0- + i)... -i. 
 
 Dem. The p like things may be treated in p + 1 ways, as all 
 may be excluded, or 1, 2, 3, ••• ov p included. 
 
 Similarly the q like things may be treated in q + 1 ways, the 
 r like things in r -+- 1 ways, and so on. Hence the total , number 
 of ways of treating all the things is (p -f- 1) (q + 1) (r + 1) •••. 
 
 But this includes the case in which all the things are excluded. 
 Hence the total number of combinations is 
 
 0> + i)(<, + i)(>- + i)--i. 
 
 631. Cor. TJie total number of combinations of n things taken 
 any number at a time, when p of one kind are alike, q of another 
 kind alike, r of another kind alike, and so on, ivhile the remaining t 
 things are unlike, is 
 
 O> + i)(7 + l)0' + i)-"(2')-l. 
 
 This follows from Arts. 630 and 628. 
 
 632. Theorem. The number of combinations of p + q + r + ••• 
 things ivith p things in each combination of one set, q in each of 
 another, r in each of another, etc., is 
 
 \ p + q + r+ ♦•- 
 
 [p X [q X \r_ X 
 dowxey's alg. — 25 
 
386 HIGHER ALGEBRA 
 
 Dem. Multiplying numerator and denominator of the formula 
 of Art. 624 by \n—_r, it becomes 
 
 c = n(n- 1)Q - 2) ... (n - r + 1) (n-r) — 3 • 2 • 1 = |» 
 
 |r(*— r) •••3.2.1 [r x| w-/ 
 
 Since for each combination of p things there is left a combina- 
 tion of n—p things (Art. 625), the number of combinations of 
 p + q things, with p things in each combination of one set and q 
 in each of another, is, by the above formula, 
 
 \p + q 
 
 \px\q_ 
 
 If there are p + q -\-r things, the number of combinations, with 
 p things in each combination of one set and q + r in each of 
 another, is 
 
 \p + q + r 
 
 \px \q + r 
 
 But the number of combinations of p + q things, with p things 
 in each combination of one set and q in each of another, is 
 
 text 1 
 
 hence the whole number is 
 
 \ p + q + r [g + r \p + g + r 
 x 
 
 \p X \q + r \qx\r_ [p X \q X \r 
 
 As the same reasoning applies to any number of sets of com- 
 binations, the formula is as stated in the theorem. 
 
 633. Cor. If there are m sets of combinations, and p=q=r=---, 
 the above formula becomes 
 
 \mp \mp 
 
 either ~~~ or 
 
 (\p) m \m x (\p) m ' 
 
 according as the combinations of the different sets are treated as dis- 
 tinct or identical. 
 
COMBINATIONS 387 
 
 EXAMPLES CL 
 
 1. How many different couples can be selected from 6 people ? 
 
 2. How many different sums of 3 figures each can be formed 
 from the digits 1, 2, 3, 4, 5 ? 
 
 3. How many different products of 2 factors each can be formed 
 from the nine digits ? How many of 3 factors each ? Of 4 factors 
 each? Of 5 factors each? Of 6 factors each? Of 7 factors 
 each ? Of 8 factors each ? Of 9 factors each ? 
 
 4. How many sets of fours can be formed, at different times, 
 from 6 soldiers ? 
 
 5. How many different committees of 5 each can be formed 
 out of a corporation of 12 members ? 
 
 6. From a company of 50 soldiers, how many pickets of 6 men 
 each can be formed ? 
 
 7. In how many ways can 4 vacancies be filled from 10 
 applicants ? 
 
 8. How many span can be formed from 10 horses ? 
 
 9. A druggist accepts an offer of $ 5 for as many glasses of 
 soda water as can be flavored with any two of his 20 sirups. Does 
 he gain or lose by the transaction, and how much, the price being 
 5 cents per glass ? 
 
 10. In how many ways can a base-ball nine be selected from 
 15 players, the pitcher and the catcher being always the same 
 men? 
 
 11. How many different amounts can be weighed with 5 weights 
 of 1 pound, 2 pounds, 4 pounds, 8 pounds, 16 pounds, respectively ? 
 
 12. How many different products can be formed from 5 differ- 
 ent factors, taken any number at a time ? 
 
 Query. Why must 5 of the combinations be rejected ? 
 
 13. In how many different ways can 12 stops of an organ, any 
 number at a time, be opened ? . 
 
 14. How many different sums can be formed with a three-cent 
 piece, a five-cent piece, a dime, a quarter dollar, a half dollar, and 
 a dollar ? Would the answer be the same with any six pieces of 
 money of different values ? 
 
388 HIGHER ALGEBRA 
 
 15. In how many ways can a selection from 5 rubies, 4 dia- 
 monds, and 3 emeralds, taking at least one of each kind, be made ? 
 
 16. In how many ways can a selection from 5 pears and 6 
 apples, taking at least one of each kind, be made ? 
 
 17. With 5 dimes and 5 half dimes, how many different amounts 
 could be put into a contribution box ? 
 
 18. With 3 weights of one denomination, 5 of another, and 4 
 of another, how many different amounts can be weighed ? 
 
 19. From 5 apples, 4 apricots, an orange, a pear, a peach, and 
 a banana, how many choices of fruit may be made ? 
 
 20. From a committee of 10, how many subcommittees of 2, 3, 
 and 5 members respectively can be appointed ? 
 
 21. From 12 soldiers, how many different scouting parties of 
 2, 4, and 6, respectively, could be formed ? 
 
 22. In how many different ways can the 52 cards of a pack be 
 divided equally (a) among 4 players, (b) into 4 piles on the table ? 
 
 Sug. When divided among the players, the mere exchange of what they 
 hold would give different hands to the respective players ; but when divided 
 into piles on the table, the exchange of positions would not give different 
 piles. See Art. 633. 
 
 23. In how many different pairs can a railroad section boss 
 distribute his 10 men for work along the track ? 
 
 Query. In what way does this differ from example 1 ? 
 
ANSWERS 
 
 Examples I 
 
 2. 2 m 2 + 2 mn + 2 n\ 3. 6 a6 - a 2 x + 7 ax 2 + ax 3 . 4. x 4 . 
 5. 15 a 2 cx 2 + 2 a 2 6x 2 + 7 wixty*. 6. ab 2 - 2 a 2 6 + a 3 - 6«c - 4 ae 2 - c 3 . 
 7. a 2 - xy. 8. a 2 6 3 -f x 2 y. 9. 4 a 2 - 6 2 + a 2 6 + a6 2 + 2 6 3 - 3 a 3 . 
 
 10. 6w 2 + 2» 2 + 2fm 2 « + li»i» 2 . 11. 4x?-x£ + 5x 2 + 5x-*/-a6-x 3 -3. 
 
 12. 5c2 - Vx + 2V2. 
 
 Examples II 
 
 1. (a + 6c -36 - 2c + 4)x. 2. 2 6x +(c - 5 6)y. 
 
 3. (a + 4)x + (2 6 -3)y+(c-4)z. 4. (a + »)(* + */) + (6 - n)(x - y). 
 5. 2 a. 6. (2 + 4 a) Vx - y. 7. fv/o^x. 8 - JL + ~ m + Sb. 
 
 . vx y 
 
 9. 5 a Vm 2 - x 2 . 10. (a 4- 6 + c) Vx 2 - **, 
 
 Examples III 
 
 2. x 3 + x 2 -2x-8. 3. 4xy. 4. 6x+2x 3 . 5. 3x 3 - 13x 2 + 9x- 3. 
 
 6. 2a 3 - 2a 2 + 4a. 7. 4a 2 . 8. x 4 + 5 x 3 y - 2 x 2 ?/ 2 - xy\ 
 
 9. 3a + 36 + 3a' + 3c. 10. GVX + 2VX 3 . 11. 4x^yi 
 
 12. 3(x + y)+6(x-*/). 13. 2 6VT372. 14. | vaTx 2 . 
 
 15. 2(a-6 + c)Vx 2 + y 2 . 16. (a - 6) Vx + y + (a + 6) Vx - y. 
 
 17. 3x 2 + 7x-8. 18. 2x 3 -4x 2 y- 5x*/ 2 + 3?/ 3 . 19. 3a -66 + 4c. 
 20. 6 vx + y + (a - 6) Vx 4- y + (c - a*) (x - y). 
 
 Examples IV 
 1. 3x 2 -3x-5. 2. 6a -46 -2. 3. 3a + 2. 4. 4x 2 -4x + 5. 
 
 5. 6m + 2. 8. -(3-a)x 3 -(c-26 - 4)x 2 -(2d - 4)x. 
 
 6. 2z. 9. - (c - « 2 )x 3 - (6 - a + 5)x 2 . 
 
 7. x + y + z. 10. -(6-a)x 3 -(6 +2c)x 2 -(6+ c + d)x. 
 
 389 
 
390 ANSWERS 
 
 Examples V 
 
 5. 20« 5 6 5 . 6. 72ab~ 1 c 3 . 7. -42x 3 ^. 8. 160x^"i 9. 30a 6 6 7 c 3 . 
 
 10. 72aWy 5 - 11. -42xt*/ 2 . 12. - 216 a%c~ h 13. -180z 3 ™+»+y»+ 4 «+ 4 . 
 
 Examples VI 
 
 2. 6 a* - 23 a 3 ?) + 41 a 2 & 2 - 42 a& 3 4" 18 b*. 3. 3 x 4 - 7 x 3 - 4x 2 + 16x - 8. 
 
 4. 8x 5 - 6x 4 - 25x 3 + 13x 2 + 18 x - 8. 5. x°> + y\ 6. x 4 4 afy 8 4 */ 4 . 
 7. ra 6 - am 2 )^ 2 + »« +p 6 . 8. a 2 " 1 - 0"+* 4 a m+2 - a m+1 + a n+1 - a 3 . 
 9. 8 a 3 4 27 6 3 - c 3 4 18 abc. 10. a 3 + 6 3 -(- c 3 - 3 a&c. 
 
 12. 2x 7 4 10 x 6 - 9x 5 + 21x 4 -22x 3 -llx 2 + 24x- 12. 
 
 13. G x 6 - 3 x b y 4 xV 2 - 1 1 xy 5 + 15 ^ 6 . 
 
 14. ?^ 8 - 10 m 6 4 5 mfi 4 25 m 4 - 43 m 3 - 12 m 2 + 44 m - 40. 
 
 15. x 7 + x 6 + x 5 + 16x 4 - 24x 2 + 41 x - 15. 
 
 16. 3x 9 + 10 x 7 - 4 x 5 - 9x l + 16x 3 + 6x 2 - 12. 
 
 17. 8 x 4 *> 4 6 x^yi + 3 x 2 ^ 2 ? 4 22 xPy* q 4 6 y*. 
 
 Examples VII 
 
 3. 9x 4 + 3x 3 -2x 2 + 6x-4. 4. x 5 - 5x 4 4 10x 3 - 10x 2 4- 5x - 1. 
 
 5. 6 « 4 - 10 a 3 x - 22 « 2 x 2 + 46 ax 3 - 20 x 4 . 6. 81 x 4 - yK 
 7. 6 as 9 + 10 x 5 - 38 x 4 + 11 x 3 + 46 x 2 - 17 x - 15. 8. m 5 4 4 mrc 4 . 
 9. x"> - 50 x 4 4 769x 2 -3600. 10. x 8 4 x 4 + 1. 
 
 11. x 5 - 5 ax 4 + 10 a 2 x 3 - 13 a 3 x 2 + 13 a 4 x - 6 a 5 . 
 
 12. x 6 4 3 x h y - 3 x 4 y 2 - 11 x 3 y 3 + 6 xV + 12 xif - 8 jfi. 
 
 13. 2x 7 - 11 x 6 + 14x 5 + 17 x 4 - 50 x 3 + 23x 2 + 38x - 30. 
 
 14. x 7 - 7 x 6 + 21 x 5 - 17 x 4 - 25 x 3 + x 2 - 2 x - 4. 
 
 Examples VIII 
 
 2. x 4 - x 3 .y - 13 xV - 40 x?/ 3 4 25 y 4 . 
 
 3. 2x 5 4 7x*y + 3x 3 y 2 + 8x 2 ?/ 3 - 5x?/ 4 4 300 y*>. 
 
 4. 4 m 6 4 12 m b n - 24 m 4 n 2 4- 6 m 3 n 3 - 72 ra% 4 4 36 ran 5 - 20 n 6 . 
 
 6. 5 x 7 - 15 xhj - 6 x 5 ?/ 2 4 18 x 4 */ 3 - 4 x 3 */ 4 4 12 x 2 y° + 7 xy* - 21 y 7 . 
 
 7. x 5 4 3 x 4 */ 4 28 x 3 ?/ 2 4 3 x 2 ?/ 3 4 12 x?/ 4 - 63 ?/ 5 . 
 
 8. a 5 4 8 « 4 & - 43 « 3 6 2 4 67 a 2 ?; 3 4 72 a& 4 - 144 6 5 . 
 10. 2 x 5 4 13 x 4 4 19 x 3 4 25 x 2 4 31 x 4 30. 
 
ANSWERS 391 
 
 11. 4 x 5 - 10^4- 10x3 + 7x 2 4-3x + 4. 
 
 12. x 6 + 4 x 5 - 16 x 4 4- 37 x 3 + 8 x 2 - 50 x - 56. 
 
 13. x 6 4-6x 6 -8x 3 -48x 2 + 5x + 30. 14. Sx? - 27 x 4 -f 7x - 63. 
 15. 5x' i + 60x s -5x-60. 16. x 7 - 10 x 6 + 3x* + 30 x 3 + 6 x - 30. 
 
 Examples IX 
 
 2. x 5 -3x*- 6x3 + 12x 2 -40x + 96. 3. x 4 - 9x 3 -f 16 x 2 - 9x - 35. 
 
 4. x 4 + 4 x 3 - 19 x 2 - 46 x + 120. 
 
 5. 2 x 6 + 3 x 5 - 25 x 4 - 13 x 3 + 11 x 2 - 26 x + 168. 
 
 6. x 5 - 9X 4 + x 3 + 105 x 2 - 74 x - 168. 
 
 7. 3 x 6 - 7 x 5 - 51 x 4 + 41 x 3 4- 96x 2 - 142x + 60. 
 
 8. 9 & - 18 x 5 - 193 x 4 + 206 x 3 + 444 x 2 - 88 x - 160. 
 
 9. x 7 - 2 x 6 - 10 x 5 + 28 x 4 + 5 x 3 - 74x 2 + 76 x - 24. 
 
 5. x 2 + 4 x - 60. 
 
 8. m 2 + 19 m + 48. 
 
 11. y 2 + 80|/-900. 
 15. x 4 -5x 2 -84. 
 20. m'W - 10 ?nx 3 - 39. 
 
 23. a 6 x« -f 20 a 3 x 3 - 69. 
 27. 9 z 6 - 18 z 8 - 72. 28. 25 z 4 - 100 z 2 4- 75. 30. x 2 - 16 xy 4- 60 y*. 
 31. x 2 4- 16 xy 4- 48 ?/ 2 . 32. x 2 - 16 xy 4- 63 y 2 . 33. ^4-4 xhj - 32 y 2 . 
 34. 4z 4 4-8z 2 y-45y 2 . 35. ^ _ 4 &tf _ 32 ^6. 36. ^ - 4 z 3 */ 2 - 77 y*. 
 37. 9x 6 4-15xV- 14 y 4 . 39. x 2 -(8y - 3 z)x - 24 yz. 
 40. x* + (11 y -4 z)x* -Uyz. 41. 9X 6 4 3(7 y- 6 z)x* - 35 yz. 
 
 42. 25 a 4 4- 5(6 wi 8 - 3 n^a 2 - 18 m% 2 . 
 
 Examples XI 
 
 1. 1 4- 2 x + x 2 . 2. x 2 4- 4 x 4- 4. 3. x 4 4- 2 x 2 y 4- y 2 . 6. x 2w 4- 4 x rt 4- 4. 
 
 7 . £*,+£ 8. ^-l*-*-. 9. xt4-4 + ^-. 
 
 2/ 2 * 2 16x1 Syl 36 ^ 
 
 12. a% - a// 3 4- }a* 66. 13. x* n - 2 x^y n 4- y 2n . 16. (16 + 56 x n 4- 49x 2 «)x 2 . 
 
 17. 49x 3 -42xV^ + 9v" 5 . 18. i^4 : 2 4-^- 19. x 4 -4y 2 . 
 
 9 o 2n \a lm 
 
 20. 9 ?>i 4 - 25 n c >. 21. 1 - $ a 2 . 22. 25 x 6 - 9 y 4 * 2 . 23. 4 xf - 9 y. 
 
 24. (81 ax - l)ax. 
 
 
 Examples X 
 
 
 
 6. 
 
 X 2 _ 4 x _ 96. 
 
 7. 
 
 m 2 4- 16 m - 80. 
 
 9. 
 
 x 2 - 18x + 77. 
 
 10. 
 
 y*-l8y-88. 
 
 12. 
 
 x 2 + 25 x - 1250. 
 
 14. 
 
 x* _ 14 x 2 4- 48. 
 
 16. 
 
 x 6 + 9 x 3 - 70. 
 
 17. 
 
 z s _ 25 z 4 - 150. 
 
 21. 
 
 ro 4 z 8 4- 9 m 2 z 4 - 90. 
 
 22. 
 
 a 6 x 4 - 16 « 8 x 2 - 
 
 25. 
 
 4 z'2 - 4 x - 63. 
 
 26. 
 
 16 x 4 - 12 x 2 - 8 
 
392 ANSWERS 
 
 Examples XII 
 
 1. a 2 4- b 2 f c 2 - 2 ab - 2 ac + 2 5c. 2. a 4 + £> 4 + c 2 - 2a 2 6 2 + 2a 2 c - 26 2 c. 
 
 3. m 2 4- n 2 + p 2 + q 2 — 2 row — 2 mp + 2 rog 4 2 wp — 2 nq - 2pq. 
 
 4. x 4 + 4 y 2 + « 2 + 9 v 2 + 4 x 2 y - 2 x 2 !^- 6 x 2 v - 4 yit - 12 yv + 6 uv. 
 
 5. x 6 4- 2 x 5 4- 3 x 4 4- 4 x 3 4- 3 x 2 + 2 x 4- 1. 
 
 7. -x 4 - 2 x 3 */ 4- 3 x 2 y 2 - 2 x*/ 3 4- ?/ 4 . 
 
 8. x 6 4- 2 x 5 y 4- 3 xV + 4 x 3 y 3 + 3 x 2 ?/ 4 + 2 x?/ 5 4- y 6 . 
 
 9. 9x 4 4-25y 2 4- 16^4-4w 2 -30x 2 ?/4-24x% 3 -12x 2 w-40^ 3 4-20yM-16 2; 3 M. 
 10. a 8 4- 4 6 6 4- 9c 4 + 16 d 2 - 4 a 4 6 3 - 6 a 4 c 2 4- 8 a*d + 12 6 3 c 2 - 16 bH - 24 c 2 d. 
 
 Examples XIII 
 
 2. x 2 - 6 xy 4- 9 y 2 - 2 r dz 2 . 3. x 4 4- x 2 y 2 + */ 4 . 4. 4 x 2 - 4 xz 4- * 2 - 16 y 2 . 
 5. 16 x 8 + 23 x l y 4 4- 9 y*. 7. x 4 - 2 x 3 + x 2 - 81. 8. x 4 - x 2 4- 14 x - 49. 
 9. 9x 4 4- 14 x 2 4- 25. 10. a 2 + lab + 4 b 2 - 9 c 2 - 6 cd - d 2 . 
 
 Examples XIV 
 1. 2 a 2 . 2. 5x 2 . 3. 4x%" 2 . 4. 3x m -». 5. 6x w + n . 6. a 2 xi 
 
 m—n , 
 
 7. a^x - " - . 8. (ab) Sm . 9. 6x" 5 . 10. ro 2 x3. 11. a 2 (a - x) 2 . 
 
 12. 4ro(a 2 -3x) 5 . 13. 3 3 (x 2 - y 2 ). 14. a 2 (2x 4- 4x 3 )". 
 
 15 <W T 16 3« 3 d 3 x 5 17 4x"+ w 
 
 .'. &3y3 # ' 5fc 2 c ' ' 7y m + n zP 
 
 Examples XV 
 
 1. 2 ab 2 4- 5 a 3 6 - 4 ab. 2. 4x 2 y - 12y 2 + 9. 3. - 3x 2 4- 4ax - a 2 . 
 
 4. 5 a 2 - 3 b 2 - 4 c 2 .. 5. 3 x m + 2 x 2 "» - 4 x 3 " 1 . 6. 2 a™ & s ~ 2n - 3 «"*+ 2n b*. 
 
 7. 7 a 20 6 4 - & 8 + 13 a V2 b*. 8. x* 4- 3 xV _ 2. 9. a 1 - 2 "- ^-"-a + a^". 
 
 10. 2{x-y) -3(x-*/) 2 4-4(x-?/) 3 . 
 
 Examples XVI 
 
 2: 5x 2 -4x4-3. 3. 2 x 2 - 7 x - 8. 4. 2 a% 4- 3 a 2 fc 2 - a& 3 4- 4 6 4 . 
 
 Examples XVII 
 
 1. 5 a 2 -6 ax - 2 x 2 . 2. x 2 - 5 «x 4- 4 a 2 . 3. 2 ?/ 4 - 8 y 3 4- 12 y 2 - 8 ?/ + 2. 
 4. x 2 - 3 X 4- 5. 5. 2 a 3 & 4- 3 a 2 b 2 - a& 3 . 6. a 3 4- 3 a 2 x + 3 ax 2 4- x 3 . 
 
 7. 9x4-6?/4-3z. 8. 2y 3 -4x«/ 2 4-x 2 y4-3x 3 . 9. 3 x 3 4- x 2 y - 4 xy 2 4- 2 y 3 . 
 
ANSWERS 393 
 
 10. l - x + x s - x 4 + 2 x 5 . 11. 4 x 2 + 6 x + 9. 12. - x 2 - 2 x - 4. 
 
 13. 2a 2 -a6 + 26 2 . fc ? 2 5 2 2 7 
 
 16. x + y. 2x 2 + 5xy + 7y 2 
 
 17. a 4 - a 8 6 + a 2 6 2 - a& 3 + 6 4 . • 18. x 4 + Zxhj + 8x 2 y 2 - 8y*. 
 19. | a 2_^ a _j. 20. 6x-iy-|. 21. ax 2 -2x + 6. 23. 3x 2 +2x + 1. 
 24. x 2 -3x + 7. 25. 2 a* - 6 a 2 6 + 18 ab' 2 - 27 6 3 . 26. 2x3-x + l. 
 27. 2 a 3 + 4 a' 2 + 8 a + 16. 28. a + b. 29. m w + an an . 
 
 30. 2a 3 -3a 2 H a& 2 + 4 6 3 . 
 
 31. 2 x 4 — 3 x 3 y + 4 x 2 y3 — 5 xy 3 + 6 y 4 , with remainder 4 y 6 . 
 
 Examples XVIII 
 
 2. 7 x 2 + 11 xy + 20 y 2 . 3. x 3 + G x 2 y - xy 2 - 30 y 3 . 
 
 4. 2 x 3 + 3 x 2 y - 3 xy 2 + 4 y 3 . 5. 12 x 3 + 11 x 2 y 2 - «/ 3 , with remainder 3 y 4 . 
 
 6. x 4 + x*y - 5 x 2 y 2 - 3 xy 3 + 3 y\ 7. 3x 4 + Sxhj - 5x 2 ?/ 2 - 24xy 3 - 24 y*. 
 
 8. 3x 4 + 2 x*y + 3 x 2 y 2 - 4 xy 3 + 2 y*. 
 
 9. 4 x 5 - 4 xV + 3 x 3 y 2 + x 2 */ 3 + 6 xy 4 + 5 y b . 
 
 10. x 5 - 8 x*y + S x 3 ?/ 2 + 4 x 2 ?/ 3 + 2 y 5 . 12. x 3 + 8 x 2 + 6x + 2. 
 13. 3 x 4 - 9 x 3 + 27 x 2 + 4 x - 15. 14. 8 x 2 - 40 x + 100. 
 
 15. 5x 3 -6x 2 + 7. 16. x 4 + 5x 3 +5x 2 +5x + 5. 
 
 17. X 4 - x 3 + 3x 2 + 6 x - 3. 18. 2x 3 4- 5 x 2 + 5x. 
 
 19. x 5 + 5x 4 + 3x 3 -4x 2 -2x + 5. 20. x 4 - 2x 3 + x 2 + 2 x + 3. 
 
 22. x 3 + 3 x 2 + 9 x + 27. 23. x 4 - x 3 y + x 2 y 2 - xf + y*. 
 
 24. x 5 + 3 x*y + 9 x 3 ?/ 2 + 27 x 2 y 3 + 81 xy 4 + 243 y*. 
 
 Examples XIX 
 2. x 2 -3x+2. 3.X+7. 4. x 3 -3x 2 + 2x-3. 5. 1. 7.x 2 -6x-2. 
 
 Examples XXI 
 
 2. x 2 +3x + 2. 3. x 2 + 3x + 2. 4. x 4 + x 2 - 4x - 4. 
 
 5. x 4 + 4x 3 + 6x 2 + 4x + l. 6. x 3 -2x 2 + 3x-5. 7. 3x 3 + 4x 2 + 5x + 7. 
 
 Examples XXII to XXVIII 
 
 The student would derive little benefit from these examples if the answers 
 were given. 
 
 Examples XXIX 
 
 2. 12a 2 & 2 . 3. 3x 2 ?/ 2 . 4. 3 w 2 (x - y)\ 5. x 2 y 2 (2x + y). 6. x + 2. 
 
 7. x - 2. . 8. x - 1. 9. x + 1. 10. 2x + 1. 11. a + b. 12. x 2 - x. 
 13. x 2 + 2 xy. 14. x + 0. 15. x - 3. 16. x 2 + 4 x + 3. 17. x + 2. 
 
394 ANSWERS 
 
 18. x 2 -2x + 4. 19. x-1. 20. x 2 -3x-4. 21. x 2 + 3x-4. 
 
 22. x 2 - 6 x + 8. 23. x 2 - x + 1. 24. 4(x 2 - 2 x?/ + */ 2 ). 25. a - 2 6. 
 
 26. a 2 -3a&-46 2 . 27. 3x-4. 29. x 3 - 6x 2 + llx - 6. 
 
 30. x 3 + 5x 2 -2x-24. 31. x 4 + 4x 3 - 2x 2 - 12x + 9. 
 
 32. 3x 4 + 4x 3 -7x 2 -4x + 4. 33. 2x 3 - 3x 2 + 4x - 5. 
 
 Examples XXX 
 
 2. x 2 + x + 1. 3. 3 x - 4. 4. x 2 + xy + 3 y 2 . 
 
 5. x 4 — x 3 y + x 2 y 2 — xy 3 + y 4 . 
 
 Examples XXXI 
 
 2. 3x-3. 3. 7x + l. 4. 2x 2 -3x + 3. 5. 2x 2 -3x+l. 
 
 Examples XXXII 
 1. x + 6. 2. x 2 -8x+15. 3. 2(x + 2/). 
 
 Examples XXXIII 
 2. x 2 -4. 3. (* + y}*(x - y)*. 4. x 3 + 7x 2 - 9x-63. 
 
 5. x 3 - 9z 2 -x + 105. 6. x 4 - 10x 3 -39x 2 + 460x-700. 
 
 7. x 4 + 9 x 3 + 11 x 2 - 81 x - 180. 8. x 3 - 2 x 2 - 5 x + 6. 
 
 9. (6x 2 - 7x-5)(x-4)(2x + 4). 10. a 5 + 2a 4 - 10a 3 -20a 2 + 9a + 18. 
 11. x*-16y*. 12. x 4 + 5x 3 + 5x 2 - 5x-6. 13. a 4 - 1. 
 
 14. x 5 + 7 x 4 - 10 x 3 - 70 x 2 + 9 x + 63. 
 
 15. (2x-3*/ + 4)(3x + 2?/-5)(4x-y + 3). 
 
 Examples XXXIV 
 
 1 ^. 2 2ac " 3 ?•**& , 4 * 5 1 6 3_a 
 
 7 c '36 llc 2n "a* ' x 2 + y 2 ' 
 
 7. *±1 8 . ^L3. 9. _J_. 10 . ErJ. 
 
 x — 1 x + 5 x + 1 x + 3 
 
 12 a 2 - ax + x 2 ]3 x 2 + 3x + 9 H x-5 
 
 a + x x+1 ' x — 3 
 
 16. *-**. 17. EH*. 18. a + 6 ~ c . 19. 
 
 1— 4a y a + 6 + c 
 
 ^ 2x + 1 21 a + b + c + g 22 x + 3 | 
 
 ' 2 x + 3 " 6 — a — c — d " x — 3 
 
 24 2g + 6 25 x ~ 3 . 26. x -ii- 27. 2 g + 3 - 
 
 7x-5 x(3x-7) x-5 3x-l 4x - 3 
 
 f 
 
 46 
 
 11 1 + 
 
 X 
 
 (1- 
 
 X) 2 
 
 15. l ~y- 
 
 a + 6 — c 
 
 -a" 
 
 a — 6 + c 
 
 -d 
 
 23. 5 
 
 x ■ 
 
 -3 
 -5 
 
 28. * x 
 
 -5 
 
AN S WEBS 395 
 
 Examples XXXV 
 
 5. 2x-4-— . 3. 3x 2 -2x + 2+ ' 
 
 3x x-2 
 
 4. 2x 2 4-2xy4- 10 y 2 + — £~ 5. m + ft 
 
 x — 3 y wi + w 
 
 6. 3x-2- 5a; ~ 1 . 7. 16 x 4 -24x 3 + 36x 2 - 54x + 81. 
 
 4 x 2 + 3 
 
 8. 2 x - 1 + X ~ 6 — 9. 4x 2 4-6x-2 
 
 x 1 - x - 1 2 x 2 4- x - 3 
 
 10. 5x-14 4--^ L - 11. x + y. 
 
 x + 2 
 
 
 Examples XXXVI 
 
 
 
 
 6 3 18 x 2 
 a* 3x-4 
 
 3. 4 " X . 4. 
 
 a — x 
 
 X 3 
 X 
 
 -y 8 
 
 4-y 
 
 5. ai+ » 
 a — b 
 
 6 (a + 64c)(Hc 
 2 6c 
 
 -a) 7 x 2 4- 3x4-1. 
 x4- 5 
 
 8. 
 
 a(a 4- b) 
 a — b 
 
 
 Examples XXXVII 
 
 
 
 
 x(a-b) y(a + b) 
 
 * . o ** 
 
 
 6 
 
 4 4- 4x 
 
 a a _ 7>2 a 2 _ fta a _ 6 2 6 (x _ x 2) 6(1 - x 2 ) 6(1 - x 2 ) 
 
 3 x* - x 3 y 4- x 2 y 2 - gy 8 x 4 4- x 2 y 2 x 3 
 
 x 4 - y 4 x 4 — y 4 x 4 — y 4 
 
 4 x(a 2 -6 2 ) ) y(q-fe) 2 ) g(q 4- ft) 2 
 (a 2 - 6 2 ) 2 ' (a 2 - 6 2 ) 2 ' (a 2 - 6 2 ) 2 ' 
 
 6 2x(x4-y) ) 3y 4g(x 2 -xy + y 2 ) 
 
 x 3 4- y 3 x 3 4- y 3 x 3 4- y 3 
 
 6 (x - y) 2 x(x - y) x 2 
 
 (x - y) 3 ' (x - y) 3 ' (x - y) 8 ' 
 
 7 X4-2 2(x-2) 3(x 4- 3) 
 
 (x 4- 3) (x 2 - 4)' (x4-3)(x 2 -4)' (x4-3)(x 2 -4) 
 
 8 (x 4- 2) 2 X4-2 1 
 
 x 3 4- 6 x 2 4- 12x4- 4' x 3 4- 6 x 2 4- 12x4-4' x 3 4- Ox 2 4- 12 x 4- 4* 
 
 9 (x-2) 2 (x4-l) 2 (x 4- 2) 2 (x - l) 2 (x 4- 2) 2 (x 4- l) 2 t (x-2) 2 (x-l) 2 
 
 X 4_5 x 2 + 4 ' X*-5s 2 +4 ' x 4_53'2 + 4 ' x*-5« a + 4 , 
 
 Examples XXXVIII 
 
 2 *« 3 -2— 4 2b * . 5 _-J 6. 5£±i. 
 
 x 2 - y 2 x - x 3 6» - a 2 2 x - x 2 x 4 - 1 
 
 7 2x - a g x - 5 9 x + 4 1Q 1 
 
 x + a" ' (x-3)(x-4)' x-4 x 3 - 1 
 
396 
 
 n. 
 
 x + 2 
 
 (!-*}(* T 3) 
 
 12. 
 
 ANSWERS 
 1 
 
 1 + 2x 
 
 13. 0. 
 
 15. 0. 
 
 16. 
 
 3x 2 + 2x+ 13 
 x 3 + 27 
 
 17. 
 
 14. 3(x 2 + l) 
 x 4 + x 2 + 1 
 
 o 9qgft«y 
 4x3 
 
 (a 2 - 6 3 ) 5 
 
 11. J 
 
 16. 
 20. 
 
 12. 
 
 x — 3 y 
 x + 3y 
 
 x 2 + 6 x + 8 
 x 2 + 8 x + 15* 
 
 3. 
 
 a 
 
 x + y 
 
 8. 
 
 3x + 1 
 
 x — 5 
 
 1 - 
 
 y 
 
 Examples XXXIX 
 
 a z + 4 
 
 9. a 2 + 1 4- ^ 
 
 10. 
 
 -£-. 6 . ^- c . 
 
 x-2 a 3 
 
 2 x 4 + 6 x 3 - 36 x 2 
 
 17. 
 
 13. 1. 
 
 14. 
 
 18. 
 
 (x-6)(x 2 -6) 
 1 
 
 x(x 2 - ?/ 2 ) 
 x + 3 
 
 1 6 <*ft 2 r 
 
 12 xy 2 ' 
 6 x(x + 7) 
 
 10 2(a - 6) 2 
 
 3 6 2 (a + 6) 
 
 14. *« + * 15. 
 
 a + 2x 
 
 j ad/ft — bcfh 
 bdeh + bdfg 
 
 fi a + 3 
 
 2(x 2 + y 2 ) 
 21. 1. 
 
 Examples XL 
 
 x 2 — 3y „ x — ?/ 
 
 x + 4 y 3 «x 
 
 7 x 2 + xy + y 2 g 
 
 x 2 — xy + y 2 
 
 11. a 2 - 2 ac + c 2 - b 2 . 
 1. 16. 1. 17. 1. 
 
 Examples XLI 
 
 x-1 
 
 1 a n — b m 
 
 
 ab 
 
 
 x — a 
 
 
 x + a 
 
 
 12 ^ x ~ 
 
 1 
 
 2x- 
 
 o 
 
 18. * + 8 
 
 
 7. 
 
 2. a-b. 3 
 
 a g + ft2 + C 2 
 a 2 6 2 + a 2 c 2 + 6 2 c 2 ' 
 
 4. 
 
 15. 
 
 19. 
 
 x + y 
 
 3 x + 1 
 
 x + 3 " 
 
 5. 3(a + &). 
 
 x(x + 2y) 
 
 x+ 1 
 
 13. 
 
 x + 5 
 
 2a 
 
 2(w - ri) 
 (ro + w) 2 ' 
 
 5. 
 
 x-y 
 
 i ^ + x + z 
 
 yz + 1 
 4. 1. 
 
 Examples XLII 
 
 bdf+be + cf' 
 5. (a 3 -5 3 ) 2 . 
 
 x 2 + x - 1 
 
 3(x+l) 
 
ANSWERS 397 
 
 Examples XLIII 
 
 « xyc 4 2 ab * ~ aS . 3 *l2? 
 
 a 2 fc 3 ' ' &• + a*b ' (x - y) 3 (x 8 + y 8 ) 
 
 4 afrgg(q - 6) 5 ^(y 2 - x) 2 g (1 - x 2 )*> 
 
 y 8 (a 2 - aft + b 2 )' ' y(y + x 2 ) 8 ' ' x"»+ 2 p(x 2 - y)* 
 
 Examples XL IV 
 
 1. VtfW. 2. ViV$. 3. — * 4. b& 5. 
 
 \/(ax) 8 3 7</x 8 ^ 
 
 >/7 - 3Va"7 y/s« Q ^Iv'x^"' 
 
 3 
 
 a*x 3 
 
 io. *y. a. b*v». i2. is* is. «y. 
 
 yt *" 
 
 *+t *+t 
 
 15. «P*T. 16. £±$4 17. (a 2 +2&)*(x + */)l 18. M+3&. 
 
 Examples XLV 
 
 
 1. a 6 *) 4 . 2. a 4 6~ 4 c. 3. a*b M. 4. 3 6 xV 2 - 5. ^ 2 ^ 3 . 
 
 6 6W. 7 ^ 8 *te. 9 3W ^ 
 
 11 bahfi 12. *** 13. J^L 14 ** 15 8a3c ^ 
 
 3 6*„ 5 ^ * 66 
 
 16. 16x 4 ^. 17. |* 18. *3T. 
 
 Vy 
 
 Examples XL VI 
 
 2. x 4 + 4 xfy + 6 x 2 y 2 + 4 x?/ 3 + y 4 . 
 
 3. « 5 - 5 a 4 6 + 10 a 3 6 2 - 10 a 2 6 8 + 5 ab* - 6 5 . 
 
 4. 8&-12x*y + Gxy 2 -y 3 . 5. 1 - 4x 2 + Ox 4 - 4x« + x 8 . 
 
 6. a 6 + 6 « 5 6 + 15 a 4 b* + 20 a 3 6 3 + 15 a 2 b* + 6 ab b + & c . 
 
 7. x 8 - 4 x«y 2 + G xV - 4 x 2 ^ + y 8 . 
 
 8. x 2 + 8 x$y + 24 xy 2 + 32 x^ 3 + 10 y*. 
 
 9. 1 + 10x* + 40 x 4 + 80 x 6 + 80x 8 + 32x 10 . 
 
 10 a 2 , 8 A 2 24 aft 4 32 A 8 16 6 8 < 
 
 X X 2 X 8 X* 
 
398 
 
 ANSWERS 
 
 11. x 6 - 3x 5 + -^x 4 - | x 3 + ^x 2 - T \x + &. 
 
 12. a 2 + 8 aM + 28 A + 56 aM + 70 ab 2 + 56 a*^ + 28 ah* + 8 aM + 6 4 
 
 13 1 3«V 15 oV 20 flV 60 a»y" 06a*y« ■ 6i 1g 1<t 
 ' 64x 12 8x 10 4x 8 x« x 4 x 2 " * 
 
 14. x 4 - 4 x B y + 6 x 2 */ 2 - 4 x?/ 3 •+ ?/ 4 + 4 x*z - 12 x 2 ?/z + 12 xy 2 z - 4 -fz + x 2 z 2 
 - 12 xyz 2 + 6 */ 2 2 2 + 4 xz 3 - 4 */2 3 + z 4 . 
 
 15. 8x 2 - 4V(x + 1) 8 (« - 1)- 4V(x + l)(x - l) 3 - 4. 
 
 16. 32 x 6 + 6x 5 Vx 2 - 1 - 48 x 4 + 20x 3 (x 2 - 1)* + 18x 2 + 6x(x 2 - 1)*-1. 
 
 Examples XL VII 
 
 2. 8 x 6 - 36 x 5 + 114 x 4 - 207 x 3 + 285 x 2 - 225 x + 125. 
 
 3. 216 x 6 - 432 x b y + 504 x 4 ^ 2 - 64 x 3 */ 3 + 168 x 2 y* - 48 xy 5 + 8 if. 
 
 4. 64 x 9 - 96 x 8 + 192 x 7 + 88 x 6 - 96 x 5 + 366x 4 +147x 3 -15x 2 +225x+125. 
 
 5. x 12 - 6 x 11 + 21 x 10 - 47 x 9 + 81 x 8 - 108 x 7 + 126 x 6 - 117 x 5 
 
 + 99 x 4 - 61 x 3 + 42 x 2 - 12 x + 8. 
 
 Examples XL VIII 
 
 5. 5 x 2 + 2 y - z\ 9. 6 a 3 - 2 b 2 + 3 c - 5 & 
 
 Examples XLIX 
 
 3. 6x 3 -3x 2 -x. 10. 4 x 4 - 4 x 2 y 2 - 7 y 4 . 
 
 16. 4x 3 -5x 2 + 2x-3. 20. 11 x 3 - Sx 2 y + bxy 2 + 9y\ 
 
 28. 4 x 6 - 6 x 4 + 2 x 3 - 3 x 2 + 9. 32. 3x 6 -5x 5 -4x 2 + 2x-5. 
 
 33. 6 x 6 - 4 x 5 - 2 x 4 + x 3 + 5 x 2 - 3 x - 7. 
 
 34. 7 x 7 + 5x 6 - 3 x 5 - x 4 + 8 x 3 + 4 x 2 - 2 x + 6. 
 
 Examples LII 
 3. 2x 4 -3x 2 -l. 6. 4x 3 -2x 2 + 3x + 5. 9. 3 x 5 -2 x 4 +x 3 +4z 2 -3x--5. 
 
 Examples LIV 
 
 3. 5V3. 
 
 6. 15 VS. 
 
 9. Za 2 b 2 VTa. 
 
 12. 21 a&x 2 V5~6x. 
 
 15. 9xyVSy. 
 
 18. 20 a m b»+ 3 x / 2a™. 
 
 2. 4V2. 
 5. 14 V2. 
 
 8. iabVWb. 
 11. 24x^ 2 V3. 
 14. 2a6c 2 vT& 2 . 
 17. 12 a m+1 & 2n V3a^. 
 
 4. 6V2. 
 7. 24 y/E. 
 
 10. 10 6c 3 V5a&c. 
 13. 60a 2 rax 2 V2m. 
 16. YobcVbabtf. 
 19. - 21 a 2 6 2 c 3 ^4&. 
 

 
 
 ANSWERS 
 
 
 39! 
 
 20. 
 
 22 xy 2 VSx. 
 a\/a + b. 
 x(x + y) V2. 
 
 21. 
 24. 
 27. 
 
 36 a6- 2 c 2 V5l;. 
 
 22. 
 25. 
 28. 
 
 20 a£r 2 c 2 #7^. 
 
 23. 
 
 3xV2x-3y. 
 
 (x 2 - xy 2 ) V3x. 
 
 2 a 2 &V5 a - 8 a 2 6. 
 
 26. 
 
 2av / 3a-46. 
 
 Examples LV 
 V7. 3. VlS. 4. aVl56. 5. #7. 6. VB. 7. &VlO~a. 8. VS. 
 
 14. 
 
 1. 
 2. 
 
 4. 
 10. 
 
 14. 
 
 6. 
 
 9. 
 11. 
 13. 
 15. 
 18. 
 21. 
 
 a\/bb. 10. V4. 11. -a</9b*c- 12. V12 a m +*. 13. - V2 a. 
 y/UaW 2 . 15. 3v / 3aP. 16. V3a6. 17. 4Vx + 2y. 18. 2Vx-3y. 
 
 Examples LVI 
 
 \/9a^, V49 xfy*, Vi2hy, V4 x 2 - 12 x*/ 2 + 9 J/ 4 . 
 #64 x 6 ^, #125 <rx 9 , #-27 a- 6 6 3 c 6 , #216 x 3 */ 6 *" 9 . 
 VIO. 5. #2. 6. y/\. 7. #8~I. 8. -W 
 
 > x + y 
 
 Vx(a + 6). 
 
 11. #a-6. 
 
 12. 
 
 3. VT«. 
 
 9. #2x. 
 13. (a 2 -6 2 )*. 
 
 Jw + w. 15 (4a;a_4 y 2)l i 6 . (a* - a*)!. 17. Vx 2 - 1. 
 
 * in — n 
 
 Examples LVII 
 3. #32, #8. 4. #32, #9. 5. #256~^, WW¥. 
 
 #8, #9. 
 
 a-" 1 , V6 4m . 7. 
 
 #8T«*, #8~P, y/mV 2 . 
 
 IT 
 
 'ft*. 8. #a 6 , #M, #c 3 . 
 
 10. #8l, #86, #fl. 
 
 V04a 6 , v8a 3 , v4a 2 
 
 12- #a 3 +3a 2 6+3a& 2 +& 3 , #a 2 -2a& + 6 2 . 
 14. 3#125a 3 x^, 2v / x 4 y 2 , 4#12^ 3 . 
 5#100x«, 7#256xV, 6#343 xV 5 . 16. #7. 17. #3. 
 
 #16. 19. ML 20. #x* when x^l, #x»whenx<l. 
 
 Vx^ when x>l, v/x 3 when x<l. 
 
 Examples LVIII 
 f#3. 5. ^V5x. 6. \VQ. 7. &V42. 
 
 ay/x 8 + 2/ 
 
 8Vx 2 -2 
 
 «vr — y 8 
 b(x - y) ' 
 
 10. 
 
 b(x + y) 
 
 x 2 -2 
 11. f#l5. 12. 2V5I. 13. 4>/55x. 
 
 14. f#9. 15. f#36. 16. i-^iex 2 . 17. *#54. 18 
 
 4 a 3, 
 
 7 6x 
 
 4x 
 
 19. 
 24. 
 27. 
 
 9 6, 
 
 $#48. 20. 3#768. 21. bV$a\ 22. ~ #18xy . 23. 4#4x. 
 ^—-^#13^. 25. (2 + x)#(2-x) 8 . 26. (x 2 - xy + y 2 ) #xT7- 
 
 (x 2 -t, 2 )#(x 2 +y 2 )*. 29. 3+V2. 30. (^+^) 2 . 31. 14 + 5v ^ . 
 
 x-y 11 
 
400 
 
 ANSWERS 
 
 I. 5(VT+ »*+»). 
 
 33. 2 x 2 - 2 xVx 2 + 1 + 1. 
 
 35. x + Vx 2 
 
 36. Vx 2 + a. 37. x + vx 2 -l. 
 
 38 
 
 34. x + Vx 2 - 1 . 
 Va 2 +6 2 x 2 -a 
 6x 
 
 40. 2(2+V2-V6). 41 (2V2+V8-V5)(2V0-3) 
 
 y 30 
 
 Examples LIX 
 
 2. v/9 + v/6 + ^i. 3. 4-2^4 + 2^2. 
 
 4. \/32 + 4\/5 + 2\/2v / 5 + 2V5 + V2v / 25 + \/55. 
 
 5. x z Vx + x 2 Vx 2 Vy» + x 2 y Vy 2 + x?/ 2 Vx Vy + ?/ 3 v'x 2 Vy + y 4 V^. 
 
 Examples LX 
 
 39V2. 3. 10V3 + 21V5. 4. 59V3. 5. 72V7. 6. 8a/6. 7. 26VTL 
 7V3. 9. SVl. 10. V0V2. 11. 9 VI. 12. 9abV2~aV 2 . 
 
 27x\/2x. 14. 2x^/Cx 2 . 15. 70a 2 V2. 16. (9 + 4x)v / 2x 2 . 
 
 13x(3 + 2x)V2. 19. 16V6. 20. 2VI5. 21. ^ 2 -Va. 22. 0. 
 VS. 24. VS. 25. ^VlO. 26. f\/U. 27. 20 VS. 28. 4\/9. 
 
 13. 
 17. 
 23. 
 
 29. 17a6V2a 2 -36 2 
 33. 88-4 V0. 
 
 30. Z>abVb + 2a. 
 
 31. 
 
 2 a 2 
 
 1. 
 
 8. 
 14. 
 20. 
 25. 
 31. 
 36. 
 42. 
 45. 
 
 Examples LXI 
 C. 2. 7V3. 3. 10. 4. 3. 5. 6x. 
 xy. 9. 48. 10. 14V6. 11. 30 V2. 
 13V21. 15. 77a/C. 16. 91 V2. 
 
 20^108. 21. 12 VS. 22. 2aVab. 
 Vim). 27. \Zl35. 28. 3^32. 
 
 6. 15V30. 
 12. 5V42. 
 17. iV5. 
 23. ^27. 
 29. \/2. 
 
 2_a 
 x 
 
 7. 72 V2. 
 
 13. 35V91. 
 
 18. iVlO. 
 
 24. 3x\/3x. 
 
 30. l^v^P. 
 
 V(l+x) 8 . 
 
 J2. V (a 2 -6 2 ) 2 . 34. 2 + 7V3. 35. 12x-6 + 16V2x. 
 3+V15 + VC. 37. 7. 38. 33. 39. 3. 40. a 3 . 41. 4-2VI0. 
 2+2VT5. 43. -48 + 54VG + 12Vl0+60\/l5. 44. 9a 2 +3a&+4 6 2 . 
 18 a -x. 46. x + y. 47. 3^0-12^3-^180+12. 
 
 Examples LXII 
 
 3. 4. 4. bVS. 5. 2. 6. \V2. 7. 2^2^. 8. 2a\/2, 
 
 9. Vx 2 -xy + y 2 . 10. Vx 2 -Sy. 11. fv35. 12. VT5. 13. VS6. 
 
 14. -VStf. 15. Vl. 16. V2. 17. $£ 18. - VWab. 19. ^V6. 
 a b 
 
ANSWERS 
 
 401 
 
 20. 
 
 28. 
 
 1. 21. y/E. 22. V5-3V3. 23. 8-8V3 + V5. 24. V7. 
 V3 - 1^243 + £\/27. 26. 2 + VB. 27. -^ + f VlO. 
 
 Va + V&+Vc. 29. ^ + f* 2 - SO. |V6. 
 
 x 4 + aW. 
 a 4 
 
 1. 75 a;. 
 
 Examples LXIII 
 2. 24 z. 3. 16x\/25^. 4. 
 
 54 «62. 
 
 8a 
 25 
 
 V5a. 
 
 6. 
 
 9Vl4a. 
 
 10. 
 
 aby/Tb*. 
 
 14. 
 
 12-2V35. 
 
 1. 
 
 6. 
 11. 
 
 3. 
 8. 
 
 4. 
 
 8. 
 
 12. 
 
 3. 
 
 9. 
 
 14. 
 
 18. 
 
 22, 
 30. 
 37. 
 
 7. abVWb. 8. 32aVa 2 6. 9. 2a 3 & 2 V2a 3 & 2 . 
 
 11. 5y/2xy> 12. -SabK 13. 2xy/y. 
 
 15. 15 + 6V6. 16. 9-6V2. 17. 3Vl8 - 3VI2 - 1. 
 18. v^ + 3v36-f 3v48 + 2. 
 
 Examples LXIV 
 y/ab. 2. 6v/5a^. 3. -3V40& 2 . 4. a?by/2aF*. 5. 2v / 7ax. 
 2\/xp. 7. Va. 8. aVl. 9. y/43xy*. 10. x(x - y) n . 
 
 y/1 - 5 a 2 . 12. 2Vo6. 
 
 Examples LXV 
 1 + VS. 4. 1 + y/2. 5. V5 - V2. 6. V6 - V2. 7. 5 f V6. 
 
 y/l- V3. 10. VT0 + 2V2. 11. V8-V3. 12. 2 + 3V5. 
 
 Examples LXVI 
 
 8a + 15&V^T. 5. 78V^T. 6. 28V^T. 7. 5xV^~I. 
 
 2V^T. 9. 18V3~V^T. 10. 14V^1. 11. 5aV2~V-I. 
 
 10+8V^3. 13. yfcX. 14. 6+V^a. 15. 2V^2. 16. vCl. 
 
 Examples LXVII 
 -xy. 4. -20Vl5. 5. 180. 6.-4 
 -2(14+V2). 10. 4(11 -V^T). 
 
 l&f/Sy/^lyY^l. 15. 16\^73\^T\^ 
 
 256 y/l. 
 
 23. 2. 
 
 19. 972V2V-1. 20. - 
 
 25. V3L 26, 27. 5. 28. 
 
 7. -42V3. 
 
 8. - 72. 
 
 11. 2. 
 
 12. 153. 
 
 [. 16. -50. 
 
 17. -24. 
 
 128V2V^T. 
 
 21. 2304 
 
 »+*£*. 29. 
 
 7 + 2Vtf 
 
 11 
 
 W-i. 
 
 31. 
 
 19 -6 VlO. 34. 12. 35. §V3. 
 
 5 
 V^l. 
 
 3V-T. 38. 7V4. 39. v^V^T. 41. V-l. 42. §±|£3. 
 
 21 
 
 DOWNEY'S ALG. 
 
 20 
 
402 
 
 ANSWERS 
 
 43. 
 
 i*-x + 2a- 
 
 a? + x 
 48. 3-4V^l. 
 52. 6-V^i. 
 
 57. 
 
 A 
 
 44. 1. 
 
 45 2 (<* 2 ~ *0 
 a' 2 + 6 2 
 
 8. 
 
 49. 4 + 3\/^l. 
 54. 7-4VC1. 
 
 6 + 2\/5x-2v/2x-6 
 
 47. 2 + 3' 
 
 1. 
 
 50. 5-2V-1. 
 55. 6 + 8.\Cl. 
 
 3. 
 
 -10. 
 
 4. -55. 
 
 10. 
 
 1. 
 
 11. -1. 
 
 17. 
 
 3. 
 
 ,« ac 
 18. -• 
 
 1. 
 
 81. 
 
 2. c 2 -2 6c. 
 
 7. 
 
 1. 
 
 8. 6. 9. 6. 
 
 Examples LXVIII 
 5. 23$. 6. |. 7. 7. 
 
 12. 4. 13. ^. 14. i 
 
 19. 0. 
 
 21. 0. 
 
 8. 4. 9. 4. 
 
 15. 4|. 16. 2. 
 
 22. -f. 
 
 Examples LXIX 
 3. 16. 4. 5. 
 
 5. 4(a-l). 
 
 10. 3. 
 
 11. 
 
 2aVb 
 
 b 4 r 
 
 12. 
 
 i4. («±^), i5., { i- (|^)n. 
 
 17. ±<*-l* 18- f 19. v« 2 -(^y 
 
 81 
 
 a 
 
 16. 
 
 6. 8. 
 13. 4. 
 
 4 m 2 
 
 (m + 1) 2 
 
 (a - I) 2 
 2a 
 
 21. 
 
 22. *£. 
 
 23. 1. 24. 4. 
 Examples LXX 
 
 3. 
 
 35. 4. an ? 
 nq — mq — 
 
 • 5. 112|. 6. an + hm . 7. ii. 8. 175. 
 np an + bn 
 
 9. 
 
 T \. 10. 40. 11 
 
 , 30. 12. ftf 13 - 6 - 14 - 77 - 15 - 40 - 
 
 16. 
 
 1. 7. $1200. 
 
 18. n ( c ~ b ), n(a-c\ 19 40) ga 
 a — b a — b 
 
 20. 
 
 6, 15, oo. 21. 5712. 22. 105. 23. 50. 24. A, 32 ; B, 25. 
 
 26. 
 
 f 3 hr. 16 T 4 T m. 
 t 3 hr. 32 T 8 r m. 
 
 27. 4 hr. 41 T 5 T m. 28. 4 hr., 54 mi., 60 mi. 
 
 29. 
 
 43. 30. 142,857. 
 
 31. 225 . 32. $1. 33. 30. 34. 300. 
 m -f n 
 
 35. 
 
 15, 90. 36. 10. 
 
 37. 8. 38. 10. 39. 140, 60, 45, 80. 
 
 40. 
 
 50, $1350, $1200. 
 
 41. 45, 30. 
 Examples LXXI 
 
 1. 
 
 10, 7. 2. 10, 3. 
 
 3. 4, 1. 4. 2, 2. 5. 2, - 3. 6. 12, 9. 
 
 7. 
 
 1, 2. 8. - 3, 5. 
 
 9. - 2, - 4. 10. - 2, 1. 11. 6, - 10. 
 

 
 ANSWERS 403 
 
 12. 
 
 4, - 3. 13. 1, - 2. 
 
 14. - 8, 5. 15. 18, 12. 16. - 2, - ft. 
 
 17. 
 
 be' — b'c a'c - ac f 
 a'b — ab' a'b — ab' 
 
 18. 36 , ?. 19. a,«. 20. «+ 26 , a " 26 . 
 2 2 2 ' 2 
 
 21. 
 
 4, -0. 22. -6, 
 a + 6, — i-r 
 
 -2. 23. -i 1 24. -\ -A 25. ! , J. 
 
 a 6 6 a 6 a 
 
 26. 
 
 
 a + 6 
 
 Examples LXXII 
 
 4. 18f, 31£. 5. 5, 60. 6. 30, 90. 7. *£- 8. 3, 9. 9. 24 
 
 10. 15, 22*. 11. 24, 30. 12. bm + a , an + b . 13. 48, 1G 
 
 mn — 1 mn — 1 
 
 14. 30, 20. 15. 42, 11, 7. 16. 50, 75. 17. 4£, 7$ 
 
 18 pm + qn - qmn pmn - qn - pm 19 g^ 2Q 2Q ^ g 
 mn — m — n ' mn — m — n 
 
 21. lO^, 17. 22. 24, 32. 23. 10, 24. 24. $10,000, $7200 
 
 25. 100, 12. 26. $3000, 4%. 27. Ill, 9|. 28. a ( b ~ c ) , P( c ~ a ) 
 
 b — a b — a 
 
 29. 50, 30. 30. 18, 15. 31. 48, 80; $63, $33. 33. 5%, 8%, 
 
 34. n ( r ~P\ <P ~ g) . 35. 11, 7. 
 
 r — q r — q 
 
 Examples LXXIII 
 
 1. 1, 2, 3. 2. 2, 3, 4. 3. 3, 2, 1. 4. 2, 5, - 1. 5. j£, - J, J. 
 
 6. 25, 55, 65. 7. 3, 4, 7, 1. 8. G, 3, 4, 5. 9. - 5, 4, - 3, - 2. 
 
 10. 2, 1, 1, 2. 11. I, 2, 1. 12. |, - J, ft. 13. 3f, 2|, 2^- 
 
 14. £,$,*. 15. 2 a, 2 6, 2 c. 16. |, f, f. 17. f, |, 2. 18. 1,2,3. 
 
 19. 3$, 2§, 2f. 20. -A&L, _JL«£., -JL^. 21. 4, 5, 6. 
 
 6+c a+c a + 6 
 
 22. 2, 1, 4, 3. 23. 1, - 2, 4, - 3. 24. 3, 2, - 5, 1. 25. 1, 2, 3, 4, 5. 
 
 26. 2, 1, -2, 3, -1, 4. 27. 2. 1, 3, 4. 28. 4, -2, 1, -3. 
 29. 1,3, 5, 7. 30." 3, 4, - G, 5. 31. 4, 6, 8, 10, 12. 32. 5, -3, 7, -2, 4. 
 
 Examples LXXIV 
 1. 426. 2. $2, 20cts., lOcts. 3. 37,45,52. 4. 480,400,560. 
 
 5. 3 7G0, 200, 200. 6. $ 1000, $- 500, $0. 7. 120,114,110. 
 8. 1 to 2, 6 min. 9. 21, 17, 23, 15. 10. 18&, 34?, 23/ r , 80. 
 
 Examples LXXV 
 4. The 1st. 7. Less. 9. The 1st. 11. The 1st. 12. Between 34 
 and 37. 13. Between 15 and 20. 14. Between $ 12 and $ 14. 15. 15, 12. 
 16. 7£. 17. 5. 18. 20 or 5. 
 
404 Answers 
 
 Examples LXXVI 
 
 1. z-5. 2. -^~ 3. x + y. 4. K«LzJ& 5 . 5^8 
 
 x-2 * 2(a + &) a:_l 
 
 6. £. 7. 11. 8. T \V 9. f. 10. Greater. 11. Greater. 
 
 12. Less. 13. 15, 21. 14. 14, 35. 15. 7. 16. 5:8. 
 
 17. „ m2 " 2 • 18. 4:1. 19. 3 : 2. 20. 120, 160, 200. 
 
 m 2 -f mn + w 2 
 
 21. 42, 12. 22. 8:9. 23. 3 : 4. 24. 242, 300, 358. 25. 16, 20. 
 26. B, 11 : 10. 28. (a) 72, (b) 84, (c) 75, (d) 77. 
 
 Examples LXXVII 
 1. f. 3. f 4. xy - — • 5. 3. 6. 3, 12. 13. 
 
 xy a _&_ c _l_d 
 
 18. - ~ ac — 20. 200, 300. 21. 8, 6. 22. 2 or lOf. 23. 13, 26, or 39. 
 
 24. Volte" °25. -!£-, -!£_. li'-'jL + Jtt JL-.±1 27.78,66. 
 
 8-o *4« 2 m 2 a 2 m 2 a 
 
 28. .0089. 29. & 31. 8:45 a.m. 32. 10. 33. 6. 34. 45,30. 
 
 35. 100. 
 
 Examples LXXVIII 
 
 2. zxy. 3. xk2*. 6. x = 8y. 7. x = — . 8. z = — . 10. 24. 
 
 11. 5. 12. 9. 13. \. 14. 4. 15. 6. 17. 48. 18. 579. 
 19. 67£. 20. 18. 21. 47. 22. 250. 23. 26|. 25. ysr-fs + ii. 
 
 2 Sit 
 
 26. x = 1 + 2 y + 3 y 2 . 27. s = $/**« 28. r(- V. 29. 27, 74. 
 
 30. .00611. **' 
 
 Examples LXXIX 
 
 1. 58,590. 2. 83,903. 3. -58, -828. 4. -3,0. 5. Z = 116, 
 £=1404. 6. a = l, £=540. 7. d = 6, £ = 473. 8. n = 12, £=582. 
 
 9. a = 32, Z = 84. 10. d = - 5, Z = - 95. 11. n = 18, d = & 
 
 12. a = | , d = - T V 13. 103, 133, 163. 14. d = 9. 15. - 46. 
 16. f. 17. 576. 18. 1512. 19. 78th and 79th. 20. (2 t - l)a, at 2 . 
 21. 14,475. 22. 10 sec., 4f min. 23. 19,600 ft. 24. 11. 
 
 Examples LXXX 
 
 2. 2048, 4095. 3. 19,683, 29,524. 4. 16,384, 21,845f£. 5. 3072, 2049. 
 6. a = 5, I = 640. 7. a = 4, £ = 39,364. 8. I = - ^, £ = - $ 1. 
 9. a = l, £=511. 10. r=-4, £=1638. 11. r=±3, £=2186 or -1094. 
 
 12. n = 5, £ = 121. 13. a = 5, « = 6. 14. 68, 272, 1088. 15. r = 3. 
 16.21. 17.13,120. 18.3072. 19. ± 2. 20.2. 21.6,24,96,384,1536. 
 
ANSWERS 405 
 
 22. 2. 23. |. 24. f. 25. ft*. 26. ft. 27. tff 28. \. 
 
 29. 150. 30. (a) $127,093,291,416.30; (6) $25,418,658,283.26. 
 
 (c) $169,457.72. 
 
 Examples LXXXI 
 
 1. TiV 2. T V,*V 3. 28. 4. 16|,21,28. 5. *,*.*• 
 
 Examples LXXXII 
 
 ±5. 3. ± 1. 4. ± 1. 5. ± 3. 6. ± Vl9 
 
 8. ±|. 9. ± v'2 aft - b 2 . 10. ± | a/3. 11. ± a/3. 
 
 1. 
 
 ±3. 
 
 2. 
 
 7. 
 
 
 6 
 - 1 
 
 2. 
 
 ± — 
 
 2 
 
 a/4 aft - P 
 
 Examples LXXXIII 
 
 1. 6 , 15. 2. mV * , wV * , P V * . 3. 6, 15. 
 
 Vra 2 + w 2 + p a Vm 2 + n 2 + .p 2 a/mi 2 + n 2 + p 2 
 
 4. 12 ft., 18 ft. 5. 4,16. 6. 40,36. 7. 4550. 8. 3. 9. 5.57 + , 18.12-. 
 
 10. 161124.3 + . 11. 1 _ from the weaker light. 
 
 a/7±V17 
 12. _E^L from the weaker bell. 13. 5. 14. 
 
 1 ± a/3 a/2 n — m a/2 n — m 
 
 15. 20, 10. 16. 150. 17. 48. 
 
 Examples LXXXIV 
 
 3. 5, 1. 4. -1, -5. 5. 11, -1. 6. -1, -9. 7. 11, 1. 8. 8, -6. 
 
 9. 8, - 2. 10. 6, 2. 11. 10, - 6. 12. 8, - 1. 13. 32, - 2. 
 
 14. 14, - 2. 15. 15, - 3. 16. 11, - 3. 17. 4, - \. 18. 2, - 1. 
 
 19. £ *L 20. 2, -tt. 21. *, -*. 22. «±h 23. U* |. 
 
 ac ac a b c 4 2 
 
 Examples LXXXV 
 
 3. 12, 36. 4. 14, 10. 5. 6. 6. 20, 3. 7. 12. 8. 70, 80. 9. 40. 
 
 10. 7. 11. 10,10.|. 12. 3. 13. 18, $20. 14. 15. 15. 576. 16. 12 or 8. 
 
 17. 9, 36. 18. 1 hr. 48 min., 2 hr. 15 min. 19. 3 or 4$. 20. 15. 
 
 Examples LXXXVI 
 
 3. 5, 1. 4. 8, - 3. 5. 3, 3. 6. 7, - 7. 7. 0, 7. 8. 6, 6. 
 
 9. z 2 -2x = 8. 10. z 2 -10x = -21. 11. x 2 + 3 x = 40. 
 
 12. x 2 - 15 x = - 54. 13. x 2 = 3. 14. x 2 - 4 a; = 1. 15. 6, 4. 
 
 16. 6,2. 17. 2^ 2 = 9r/. 18. ja-I. 
 
406 ANSWERS 
 
 Examples LXXXVII 
 1. 3, 0. 2. 5. (Reject - 2). 3. 6, 7. 4. 1, 8, - 5. 5. - 1, ± 4. 
 6. 1, J T - 7. 0, 4(a+ b). 8. 3, 2. 9. 4, \. 10. 8, - §. 11. 4, }. 
 12. ±V3. 13. 5,3. 14. ± |V3. 15.0,4,12. 16. ±5. 
 
 Examples LXXXVIII 
 
 1. ± 4. 2. 3. 3. 36. 4. ± 729. 5. ± 125. 6. 27. 7. J. 8. f. 9. 64. 
 
 Examples LXXXIX 
 
 2. 2. 3. 7, - 1. 4. ± 5. 5. ± 2. 6. - 3. 7. ± 2. 
 8. K5±V5). 9. -|(l±vl3). 10. 1,-5. 
 
 Examples XC 
 
 2. ± 2, ± 1. 3. ± V5, ± 1. 4. ± 3, ± V^T. 5. ± 3, ± 4. 
 
 6. ^2, -2. 7. 4, v^iO. 8. 16, ifffi. 9. 243, (-28)1 10. 8, iff. 
 
 11. ±|V30, ±1. 12. 64, (-33)1 13. — , — 14. v^, V^VS*. 
 
 15. 16, T V 16. 2, i. V^ ^=6 
 
 Examples XCI 
 2. 5, -2|. 3. 1, 1. 4. 2, -6|. 5. 9, -12. 6. ±5, ± 3V2. 
 
 7. 3, -\. 8. 1,0,-3,-4. 9. 4,69. 10. 8, - 1, |(7 ± V53). 
 11. 2, 1, |(7 ± V33). 12. 2, - |, K 3 ±V505). 13. 3, -.1, 1 ± |V5L 
 14. 2, -|, K-17±V305). 15. 1, K-3±V5). 16. 5, -2. 
 
 17. 3, -31, i(- 1*^^251). 18. K1±V5)- 19- f(l±Vfy 
 
 Examples XCIV 
 
 1 (3,-ff; 2 /±Vf; 3 {2; 4 (l,-lt; 5 (5,10; f a, a ; 
 
 * \_4,jy^. l2Wf- *-2. '1 2, -8§. I 10, 5. \b,b. 
 
 Examples XCV 
 
 f2,-5; 1.P.-1; 3 f-2,0; 4 U ± *V33 ;_ 
 
 * 14, -3. ' 10, -f. '12, 1. '\_i_tiV33. 
 
 Examples XCVI 
 . fl, -1|,8, -2; 2 f±6; 
 
 ' 1 3, — 3f, — 3, 2. ' I ± 3. 
 
 3 (3, -2£, Kl±v/105); 4 fl, - 1$, &(- 1 ± V409); 
 
 ' 11. -Jt, Kl±Vl06). " I 2, -2|, T V(-1±V409). 
 
ANSWERS 407 
 
 Examples XCVII 
 
 • fiVifVfj 8 f±2, ±$>/lO; 3 f±3,T8; 
 
 'I ±2, ±iV2. "li^TlVlO. '\±5. 
 
 4 /±1. ±¥VE?; gf±2, ±^V3_; r ± i 0| ±ttV=47 
 
 Examples XCVIII 
 
 3,-5 
 5,6. 
 
 6 f ± fV2l ; fi f 2, 10, - 2, - 10 ; - f 10, - 5 ; g f ± 3, ± 2 
 'l±j>/21. '11,-3,-1,3. '1 5, -10. \±2j±3. 
 
 Examples XCIX 
 4,2, £(_13±V377); 3 / 3, l f 8±v^j 
 
 2, 4, i(-13TV377). ' 1 1, 3, 2 =F V^. 
 
 4 (0,12,6; 5 (4,2, K~7±V^35); 
 
 ' 1 0,6, 12. ' 12, 4, K-7TV-35). 
 
 Examples C 
 
 , f±fVi4; fK3±VT3); f9; r ± 4, ± 14 : 
 
 *l±fVl4. 'lKl±Vl3). ' 1 6. 'i±l,>4 
 
 '■0: 
 
 .(_ 3 ± 2V- 15); 6 f ± 8, ± 2 V=l . < 3, 4 ; 
 
 1,^(_3 ± 2a^15). *\±2, T8V^1. ' U, 3. 
 
 /±fV6, ±3; - 12,3; 1Q f 3, 2, J(5 ± V- 151); n f 3, 1 ; 
 
 *<-±*V6, ±1. '13, 2. * 1 2, 3, i(5TV^lM). * 1 1, 3. 
 
 13 f±3, ±fV2; 14 f ±0, ±4; 15 / 3, - 1 
 
 'I ±2, ±$V2. 'l±4, ±9. 11,-3. 
 
 12 l*^*' 
 
 14 ±»- 
 
 Examples CI 
 1. 20,15. 2. 20,15. 3. 10, $125. 4. 4,5. 5. Dis., 15 ; rates, 3, 2£, 4. 
 6. 48, 42. 7. 6, \\. 8. Dis., 46$ or 30 ; rate, 4 or 3. 9. Rowing, 
 4i and 6 ; walking, \\. 10. 16, 5f 
 
 Examples CII 
 
 3. x = 3 renders /(a;) a max. /(x) at a max. = 9. 
 
 4. x = 4 renders /(x) a min. /(x) at a min. = 3. 
 
408 ANSWERS 
 
 5. x = — 2 renders /(x) a min. /(x) at a min. = 0. 
 
 6. x = 2 renders /(x) a max. f(x) at a max. = 15. 
 
 7. x = 2 renders /(x) a min. /(x) at a min. = 0. 
 
 8. x = 4 renders /(x) a max. /(x) at a max. = 50. 
 
 9. x = renders /(x) a min. /(x) at a min. = 11. 
 10. x = — £ renders /(x) a max. /(x) at a max. = 7f. 
 
 Examples CIII 
 
 2. x = — 1 renders /(x) a min. 
 
 3. x = — | renders /(x) a max. 
 
 4. x = 4 renders /(x) a max. 
 
 5. x = 2 renders /(x) a min. 
 
 so 
 
 Examples CIV 
 
 8; /(x) at a min. = — £, when x = 3. /(x) at a max. = — £, when x = 1. 
 9. /(x) at a min. = b, when x = a. 
 
 10. /(x) at a max. = |, when x = 3. 
 
 11. /(x) at a max. = ^, when x = — 5. 
 
 12. /(x) at a min. = 0, when x = — 1. 
 
 13. /(x) at a max. = a, when x = a. 
 
 14. /(x) at a min. = 6, when x = ± 2. /(x) at a max. = 10, when x = 0. 
 
 15. /(x) at a min. ss — §, when x = 2. /(x) at a max. = 1, when x = 0. 
 
 16. /(x) at a max. = &, when x = a. 
 
 17. /(x) at a min. = —27, when x= ± V2. /(x) at a max. = — 15, when x=0. 
 
 /(x) at a min. 
 
 = h 
 
 /(x) at a max. 
 
 = 6. 
 
 /(x) at a max. 
 
 = - 2 m. 
 
 /(x) at a min. 
 
 _ n — 4 
 
 { 
 
 fy at a 
 lx at a 
 
 Examples CV 
 
 y at a max. = 6, when x = 1. y at a min. = — 2, when x = 1. 
 x has neither a max. nor a min. 
 
 max. = 4, when x = 2. y at a min.= — 2, when x = 2. 
 
 max. = 2 + V22, when y = \. x at a min.=2— V22, when y=l. 
 m f y at a max. = 5, when x = 0. y at a min. = — 3, when x = 0. 
 
 I x at a max. = 4, when y = 1. x at a min. = — 4, when y = 1. 
 
 Examples CVI 
 
 3. 6 and 6. Min. val. = 72. 4. A square, each side 40 rd. 5. 384 sq. yd. 
 7. 6 knots, 32 m. 8. Dimensions, 8 by 16. Area, 128 sq. rd. 
 
 9. The corners are at the middle points of the given square. Side of min. 
 
 square = — • 11. The length is twice the breadth. 
 
 V2 
 12. Each side is \y/2 times the hypotenuse. 13. 7 ft. by If ft. 
 
ANSWERS 409 
 
 Examples CVII 
 
 3. oo. 4. oo. 5. 0. 6. 0. 7. & 8. |. 9. f. 10. -. 11. f. 12. 1. 
 
 Examples CIX 
 
 2. dy = (3 x 2 + 10 as - 6)dz. 3. tZy =s - (3 x~ 4 + 8 x)dx. 
 
 4. cty= (28x 3 -f 6x-3x~*jdx. 5. dy = 10(as* + 2 x" 5 - x)dx. 
 7. dy = 120(2 + 3 x 2 ) 3 xdx. 8. dy = -120(1 - 5 x 2 ) 2 xdx. 
 9. dy = 5 6(o + bx*)%x 2 dx. 10. dy = 12(3 - x 3 )- 3 x 2 dx. 
 
 12. dw = 3 x 2 y*dx + $ x 3 y*dy. 13. dw = (y 2 - 3)dx + 2 xydy. 
 
 14. du = (y 3 + 2)dx + 3(x - 1) y 2 dy. 15. (x + a)$<te + f x(x + a) *dx. 
 
 17. dy= . 3<to . 18 du = 3x*ydx-2x*dy, 
 (1 + x) 4 y« 
 
 20. dv = — ^— . 21. dy= 4 V~ 3x ± dx. 
 
 2Vl + x V2x^"3z 2 
 
 Examples CX 
 
 2. ^=3x 2 -6x. 3. ^=9(x 3 -5) 2 x 2 . 4. /'(afl = — . 
 
 dx dx K J J w Vx 2 ^ 
 
 5. /'(x) = 2x+l-3x-*. 6. & = **-_. , m=s __Jj2i_. 
 
 V ,» dx (1 + x) 3 ' V ' (x + a)» 
 
 9. «J = 2. 10. ^ = 36 x 2 - 10. 11. /"(x) = 450(2 + 5 x). 
 
 dx 2 dx 2 
 
 12. 12(a + x)- 6 . 13. &y = _AH± — 14. 80(2 x 2 - 3) 2 (14 x 2 - 3). 
 
 dx 2 (a + x) 3 
 
 Examples CXI 
 1. l + 3x + 4x 2 + 7r 3 + Hx 4 + etc. 2. 2 - 5x + 3x 2 + 2x 8 - 5x* + etc. 
 
 3. l + 2x + x 2 -x 3 -2x 4 -x 5 + etc. 4. l-x-x 2 + 5x 3 -7x 4 -x 8 + etc. 
 5. 5 + 27 x + 130 x 2 + 623 x 3 + etc. 6. 1 + 2 x + 3 x 2 + 6 x 3 + 9 x 4 + etc. 
 7. 2 + 2 x - 3 x 2 - 5 x 3 - x 4 - etc. 8. 1 + x + x 2 - x 3 + etc. 
 
 9. } + fx + 3x 2 + Hz 3 + e tc 
 
 Examples CXII 
 
 1. fB'Hfar^+ff+Ws + Hf^+etc. 2. ar 1 +3+2 x-5x 2 -16x 3 +etc. 
 
 3. |x + $ x 3 + 2 V s 6 + ir^ 7 + etc - 4 - x -2 - aT 1 - 2 x + 2 x 2 - 4 x 3 + etc. 
 5. x-x 2 -2x 3 -5x 4 -12x 5 -etc. 6. f x- 3 -^- 2 - £x _1 + f| + $f x + etc. 
 
 Examples CXIII 
 
 2. 1 + \ x + f x 2 - T \ x 8 + T f f x 4 + etc. 3. 1 + ^ x - f x 2 + ^ x 8 + etc. 
 
 4. 3 + ix-tt$a* + flftz* + etc. 6. 1 + £x+ f x 2 - if x 8 ^^x 4 + etc. 
 
410 ANSWERS 
 
 Examples CXIV 
 
 3. -? I + __1 4 3 * 
 
 x 2(1 -x) 2(1+ a) 2(x-2) 2x 
 
 5.—* ? 6 _ 5 4 
 
 3(x-2) 3(x + l) x-4 a: -3 
 
 7 __2 5 1 8 2 23 
 
 (x-3) 3 (x-3) 2 x-3 x + 5 (x + 5) 2 
 
 9. -J 2 !_. 10. » 4 
 
 x+1 (x+1) 2 (x + 1) 3 5(5 x- 2) 5(5 x- 2)3 
 11. -2 + ^ + _^_. 12 . _J *_ + 9 
 
 x 3x+l 2x-5 2(x+l) x + 2 2(x + 3) 
 
 u. _2 + _^ + ^« u. I + -L+-L + . i 
 
 x x + 2 (x + 2) 2 x x - 1 x - 2 (x - 2) 
 
 Examples CXV 
 3 2x- 11 _ 2 4 1 2 x - 2 
 
 x 2 + 1 x - 4 x 2 + 1 (x 2 + 1) 
 
 x + 1 6 _ 1 x + 2 
 
 3(x-l) 3(x 2 +2) 2(x-l) 2(x 2 +x + 4) 
 
 - 7 5x - 3 8 __as 1 
 
 x - 1 x 2 + x + l" x 2 + 2 x 2 + x + 2 
 
 Examples CXVI 
 
 3. xH|x"^t/-| x~ V + T V x~ V - ili £~V + etc. 
 
 4. x~ 2 + 2 x" 3 */ + 3 x" 4 !/ 2 + 4 x~ 5 y z + 5 x~ 6 */ 4 + etc. 
 
 5. x~3 - | %% + f x~V - |f x~*V + ^ aT^V - etc. 
 
 1 _2 _5 
 
 1 r» 3n 1 v ~Sifi J §_ ~ 3 
 
 V, 
 
 6. x 3 + } x 3 y - ) x V + ^ x 3 ?/ 3 - J& x ~ 3 y 4 + etc. 
 
 7. x^ + f x^y + f x~V - T V x - ^/ 3 + T f -g x~V - etc. 
 
 Examples CXVII 
 
 2. 3 x 5 - 2 x 2 + (15 x 4 - 4 x)h + (30 x 3 - 2)fr 2 + 30 x 2 ft 3 + 15 x/i 4 + 3 h*. 
 
 3. 2 x 4 - 4 x 3 + x 2 - 5 + (8 x 3 - 12 x 2 + 2 x)h 
 
 + (12 x 2 - 12 x + l)/i 2 + (8 x - 4)/i 3 + 2 A 4 . 
 
 Examples CXVIII 
 
 2. 16 x 4 + 32 x s y 3 + 24 x 2 ?/ 6 + 8 xy 9 + ?/ 12 . 
 
 3. x 2 - 2 x 4 y + 4 x 6 ?/ 2 - 8 x 8 # 3 + etc. 
 
 4. x* - f x~V _ 4 ^.-1^4 _ 1 2 x -f y 6 _ e tc. 
 
ANSWERS 
 
 411 
 
 3. 
 
 4. 
 6. 
 
 7. 
 
 11. 
 15. 
 
 16. 
 17. 
 18. 
 19. 
 23. 
 
 Examples CXIX 
 a? - 6 cfib + 15 a 4 6 2 - 20 « 3 6 3 + 15 aW - 6 a& 6 + 6 6 . 
 x 7 - 7 x«y + 21 x 5 y 2 - 35 x*y* + 35 x 3 y 4 - 21 x 2 y 5 + 7 xy« - y 7 . 
 1 + 4 x + 6 x 2 + 4 x 3 + x 4 . 5. 1 - 5 y + 10 y 2 - 10 y 3 + 5 y 4 - y 6 . 
 
 x -2 - 2 x _3 y + 3 x~ 4 y 2 - 4 x~ 6 y 3 -f 5 x-*y* - etc. 
 
 X 3 X 4 X 5 X 6 X' 
 
 1 - $a 2 - | a 4 - T V« 6 - T f^a8 _ et c. 
 a 6 + 16 a "^ + 96 « J ^ + 256 a* + 256 at. 
 
 2 2 
 
 * 
 
 8 
 
 + etc. 
 
 1+ 3 x 2 + 6 x 4 + 10 x 6 + 15 x 8 + etc. 
 a^ - \ a~^x 2 - $ af^x 4 - ^ a^x 6 - etc. 
 
 1 x 2 2x 4 
 
 27 a 3 27 a 4 81 a* 
 
 10 x 3 
 
 352. 
 
 729 a 
 1 + 4 x +'2 x 2 - 8 x'' 
 
 + etc. 22. 2(x 2 + 6 x + 1). 
 
 5 x 4 + 8 x 5 -f 2 x 6 - 4 x 7 + x 8 . 
 
 Examples CXX 
 
 = | log x + i[log (1 + x) + log (1 - x)]. 
 
 3. log y = £[log s + log (s - a) - log 6 - log c]. 
 
 4. logy = £[loga + 21ogfi -f 41ogc — 51ogd]. 
 
 5. logy = 2 log a + K lo g(« + &)+ log(a - 6)- log(l + &)]. 
 
 6. logy = £[logx-f log(l-x)]-£logz. 
 
 7. logy = £[raloga + plogfo - «logc]. 
 
 8. logy = $[log(a+ b) + log (a - b)- log a]. 
 
 9. log y = i [log x + log (x - 4) - log (x + 1)]. 
 
 11. 
 15. 
 18. 
 21. 
 
 ,„ , mdx 
 
 ay = 
 
 x 
 
 dy = 
 
 dy 
 
 x = — 
 
 _ 2 mxdx 
 a 2 - x 2 ' 
 log ft 
 log a 
 
 x = 2£, y = |. 
 
 2.902003. 
 1.909930. 
 2.888089. 
 
 mdx 
 1-x 
 
 16.' dy = 
 19. x 
 
 3 mdx 
 
 13. 
 
 x 
 
 4 mxdx 
 1 + x 2 ' 
 log a — log b 
 clog a 
 
 14. dy = 
 17. x 
 
 mdx 
 
 x 
 logc 
 
 2 V log a logbl 
 
 Examples CXXI 
 
 2. 0.935507. 3. 3.891203. 
 
 6. 3.930236. 7. 4.896273. 
 
 10. 3.938648. 11. 1.895266. 
 
 b log a 
 x = log 7. 
 
 4. 1.915716. 
 
 8. 1.929092. 
 
 12. 3.934229. 
 
412 
 
 ANSWERS 
 
 Examples CXXII 
 
 1. 8313. 2. 792.75. 3. 8.1743. 4. 86764.6. 
 
 5. 79.8252. 6. .83752. 7. .0773742. 8. .0085167. 
 
 9. 860900. 10. 820000. 11. .008. 12. 819554. 
 
 13. 8636.32. 14. 85.8052. 15. 790.3972. 16. 7.75987. 
 17. 5.4283. 18. 5.4641. 
 
 Examples CXXIII 
 
 5. 1, 1. 6. 7, 1 ; 2, 4. 7. 5, 1 ; 19, 6 ; 33, 11 ; etc. 8. 1,2; 3, 1. 
 
 9. Impossible. 10. 20, 3 ; 39, 7 ; etc. 11. 17, 1 ; 10, 4 ; 3, 7. 
 
 12. 1, 53 ; 3, 40 ; 5, 27 ; 7, 14 ; 9, 1. 13. 3, 7 ; 8, 21 ; 13, 35 ; etc. 
 
 14. Impossible. 15. 2, 1, 3. 16. 4, 2, 7. 
 
 Examples CXXIV 
 
 1. 16, 2 ; 12, 5 ; 8, 8 ; 4, 11. 2. 11, 15. 3. 4 ways. 
 
 4. 2, 74; 15, 55; 28, 36; 41, 17. 5. Impossible. 48 ways. 6. f, £, f, etc. 
 
 7. 6,3. 8. 117, 19; 77, 59; 37, 99. 9. 5,1. 10. A gives £6 and 
 
 receives $28. 11. 1, 6, 8, or 3, 3, 9. 12. 5, 10, 15. 
 
 3. y* + 3y* + 486 = 0. 
 
 Examples CXXV 
 
 6. y 5 -2y*-2y s + 45y*- 12*/ + 81 = 0. 
 
 2. 19. 
 
 Examples CXXVI 
 3. 1. 5. 7. 6. 12. 
 
 7. 0. 
 
 Examples CXXVII 
 
 2. 1 pos., 1 neg., 2 imag. 3. No pos., 1 neg., 4 imag. 
 
 4. 1 pos., no neg., 4 imag. 
 
 5. Not more than 2 pos., not more than 2 neg., at least 2 imag. 
 
 6. Not more than 3 pos. , not more than 1 neg. , at least 2 imag. 
 
 7. 1 pos., not more than 3 neg., at least 2 imag. 
 
 8. Not more than 3 pos., 1 neg., at least 2 imag. 
 
 9. 1 is the only real root. 10. 1 and - 1 are the only real roots. 
 11. — 1 is the only real root. 12. No real root. 
 
 2. 2, 3, - 4. 
 
 14. 2, ±V3, ±VS. 
 
 Examples CXXVIII 
 
 5. 7, ±V5. 9 
 
 17. 1, 1, - 1, |, f. 20 
 
 1,1, -1, i(_3±V5). 
 1, 1, 1, 2, 2, - 2, - 3. 
 
AN S WEBS 413 
 
 Examples CXXIX 
 
 4. ± V2, ± V2, ± VS. 8. ± 2, ± V£, ± V6. 
 
 12. ± V2, ± V6, ± V^T, ± a/^2. 13. 0, ± V2, ± V3, ± V6, ± V^2. 
 14. +_ V^l, ± \^~2, ± V^3, ± V^5. 17. 3, ± V2, ± V3, ± V5. 
 
 Examples CXXX 
 
 7. 4x 5 + 24 x 4 + 15x 3 - 80 x 2 - 39 x + 36 = 0. 
 9. « 6 -G^ + 8x* + 18x 3 - 63s 2 + 24* + 20 = 0. 
 
 Examples CXXXI 
 
 4. Between and 1, and —1,-5 and — 6. 
 
 7. Between 3 and 4. Two imag. 14. Between and 1, 1 and 2, 2 and 3. 
 18. Between 3.2 and 3.3, 3.4 and 3.5, - 1 and -2,-3 and - 4. 
 22. Between 1 and 2, and —1,-1 and — 2. Two imag. 
 
 Examples CXXXII 
 
 2. 2 + V3 and 2 — V3 are each double roots. 4. 3, 1 ± VE 1 1 ± V5. 
 •. ± V§, ± V2 f 1 ± VS. 9. ± V2, ± V2, ± V2, ± V3. 
 
 10. 1 + VS and (1 — V3) are each quadruple roots. 
 
 Examples CXXXIII 
 
 3. x 3 + 20.630 x 2 - . 343893 x + .000043597 = 0. 
 
 4. x 4 + 11.4x 3 + 36.735 x 2 + 36.1965 x - .29499375 = 0. 
 
 Examples CXXXIV 
 
 3. 2.7824, 4.166, - .9489. 4. 4.117. Two roots imag. 
 
 5. 4|, 2 ± VS ; or 4.125, .26795, 3.73205. 6. - 2.2134. Two roots imag. 
 7. 9.9666. Two roots imag. 8. 2.25, 3.6055, - 3.6055. 
 
 9. -2.1768. Two roots imag. 10. 2.3569,2.6920,-2.0489. 
 
 11. 2.4641, - 4.4641. Two roots imag. 
 
 12. 4.3166, -2.3166. Two roots imag. 13. .337, 1.636, - 1.691, - 4.283. 
 14. Two roots imag. 15. 4.5814. Four roots imag. 
 16. 4.5195. Four roots imag. 17. 5.874. 18. .4148. 19. 106.474. 
 
 Examples CXXXVI 
 
 4. Between and —1,-2 and — 3, two between 2 and 3. 
 
 6. Between — 1 and — 2, 4 and 5, three between 1 and 2. 
 
 7. Between — 1 and — 2, 4 and 5, two between 1 and 2, two imag. 
 
414 
 
 ANSWERS 
 
 11. Double root between -1 and -2, double root between 1 and 2, two imag. 
 13. Between — 1 and — 2, 3 and 4, double root between — 1 and — 2, double 
 root between 1 and 2. 
 
 Examples CXXXVII 
 2. 1, 1, |(1 ± V^~3). 3. 2 ± V3, 1(1 ± V="3). 
 5. ±1, K-7±3V5). 
 7. ±1,1(1±VU72). 8. 1,3,1,-2,-1. 
 
 10. 2, i, 2, |, K1±V^3). 
 
 4. ±1, i(l±V^3). 
 6. |(3±VS), ^(-7±3V5). 
 
 9. 1, ±V-1, i(-l±V-35) 
 2. 1, i(-l±V^3). 
 
 5. ±V±v^T. 
 
 7. 3 times the roots of ex. 1. 
 
 9. ± 
 
 Examples CXXXVin 
 
 3. -1, i(l±v^3). 4. ±1, ±V=1. 
 
 6. - 1, i(l ±V5±V- 10±2V5). 
 
 8. ±1, ±Vi(-l±V-3). 
 
 1, ±Vki±V-3). 
 1. Convergent. 
 
 Examples CXXXIX 
 3. Convergent. 
 
 4. Divergent. 
 
 2. Convergent. 
 6. Convergent. 
 
 Examples CXL 
 
 1. Convergent when x < 1, divergent when x>l. 
 
 3. Convergent when x <1. 4. Convergent. 
 
 7. Divergent. 8. Convergent for x > 1, divergent for x< 1. 
 
 9. Divergent. 10. Convergent. 
 
 11. Convergent for x< 1, divergent for x > 1. 14. Convergent. 
 
 15. Convergent when X is numerically equal to or less than 1, divergent 
 
 when x is numerically greater than 1. 
 
 16. Convergent. 
 
 2. 1,6; 408 x 5 . 
 5. 1, 1, -1; 4x 7 . 
 
 2- 0. 
 
 17. Convergent. 
 
 Examples CXLI 
 3. 2, 1 ; 99 x 5 . 
 6. - 1, -1 ; 2x 6 . 
 
 Examples CXLII 
 3. 0, 0. 
 
 18. Divergent. 
 
 4. - 2, 3 ; 489 x 5 . 
 7. 2, 1, -2; 173 x\ 
 
 4. 8, 32. 
 
ANSWERS 415 
 
 Examples CXL.III 
 2. 78. 3. 225. 4. 6396 a;". 
 
 Examples CXLIV 
 
 10 &= 5(§-(. + iK» + 2)) i s « = i 
 
 11. «, = ___________; «. = -• 
 
 i o o 3 2 , 1 t n o 
 
 13 An_ I rc + 2 + 2(N + l)(rc + 2)' ^ _ 4* 
 
 n ~6 7i + 3~ f "(tt + 3)0 + 4)' °°~6 
 
 15. 
 
 n "2V l-3.5...(2« + l)j' ac ~2* 
 Examples CXLV 
 
 J? 
 
 1 + 11 4 
 
 3 
 
 1 + a; 4 1 + 3 3 - 1+3 
 1 - X _ 3* (1 _ a;) 2 '1-23-2 X* 
 
 
 1 + 2 a; - 3 3 2 
 
 
 6. 
 
 2 + 5a; 
 
 
 7 1+3 g 3-3- 6 3 2 
 
 (1 - 3) 2 1 - 2 3 - 3 2 + 2 3 3 
 
 1 + 5 x + 4 x 2 
 
 9. 
 
 1 + 2 36 - 3 X 2 
 1 - * + X' 2 + X s 
 
 
 10- 1 "- 6x + 8a! ' ;p = 2, g = 3, r = 4. 
 1-23-3 3 2 -4 3 3 ' ^ y 
 
 Examples CXLVI 
 2. 400. 3. 2548. 4. 1275. 5. 278,256. 6. 147. 
 
 Examples CXLVII 
 
 1. 680. 2. 1540. 3. 1505. 4. 624. 5. 1240. 
 6. 2730 7. 7490. 8. 3880. 9. 36,256. 
 
 Examples CXLVIII 
 
 2. 10 hr. 19 min. 47.45 sec. 3. 10 hr. 6 rain. 39.1 sec. 
 4. 3.97905. 5. 44° 58 40'- 
 
416 . ANSWERS 
 
 Examples CXLIX 
 
 1. 336. 2. 60. 3. 72, 504, 3024, 15,120, 60,480, 181,440, 362,880, 
 
 362,880. 4. 720. 5. 210. 6. 360. 7. 840. 8. 40,320. 
 
 9. 360. 10. 2520, 4,989,600. 11. 64. 12. 623,529. 13. 7,257,600. 
 
 14. (a) 5040, (6) [7 x |8 x [5 x |3. 15. 1260. 16. 5040. 
 
 17. (a) 576, (6) 144. 18. 576. 19. 120. 20. 720. 21. 120. 
 
 22. 2880. 23. 181,440. 24. 408,408. 25. 4. 26. 6. 27. 8. 
 28. 15. 29. 7. 30. 3. 31. 13. 
 
 Examples CL 
 
 1. 15. 2. 10. 3. 36, 84, 126, 126, 84, 36, 9, 1. 4. 15. 5. 792. 
 
 6. 15,890,700. 7. 210. 8. 45. 9. Loses, $4.90. 10. 1716. 
 
 11. 31. 12. 26. 13. 4095. 14. 63 ; yes. 15. 3255. 16. 322. 
 
 17. 35. 18. 129. 19. 479. 20. 2520. 21. 1980. 
 
 152 152 
 22. (a) -*=- ; (b) — 23. 945. 
 
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