the sum of the series, in five several cases, as below, are given, to find the number 22. (1) There are two numbers whose sum, sum of their squares, and their product, are all equal; and (2) two others whose sum, product, and difference of their squares, are all equal. What are these pairs of numbers? Ans. (1) impossible; (2) (3 + √5) and (15). 23. A gentleman bought a horse for a certain sum, and having re-sold it for 1197. found that he had gained as much per cent. by the transaction as the horse cost him; what was the prime cost of the horse? Ans. 70%. or 1707. The latter is incongruous. 24. The arithmetical mean of two numbers exceeds the geometrical mean by 13, and the geometrical exceeds the harmonical mean by 12. What are those numbers? 25. A traveller sets out for a certain place, and travels one mile the first day, two miles the second, three the third, and so on: and five days afterwards another sets out and travels 12 miles a day. Show how far he must travel before he overtakes the first, and explain the other answer. 26. A wine merchant sold 7 dozen of sherry and 12 dozen of claret for 501. and finds that he sold 3 dozen more of sherry for 10l. than of claret for 61. What was the price of each? Explain the double answer. 27. A parcel contained 24 coins, valued 18s, part of them silver and the other copper. Each silver coin is worth as many pence as there are copper coins, and each copper coin is worth as many pence as there are silver coins. How many more were there of copper than of silver? 28. Find four numbers which exceed one another by unity, such that their continued product may be 120. 29. There is a number consisting of two digits, which, when divided by the sum of its digits, gives a quotient greater by 2 than the first digit; but if the digits be inverted, and the resulting number be divided by a number greater by unity than the sum of the digits, the quotient is greater by 2 than the preceding quotient. Find the congruous answer. 30. "Some bees were sitting on a tree; at one time the square root of half their number flew away. Again, eight-ninths of the whole flew away the second How many were there * ?" time; two bees remained. 31. D sets out from F towards G, and travels 8 miles a day; after he had gone 27 miles, E sets out from G towards F, and goes every day of the whole journey; and, after he had travelled as many days as he goes miles in one day, he met D. What is the distance of the two places? 32. There is a number consisting of three digits, of which the first is to the second as the second to the third; the number itself is to the sum of its digits as 124 to 7; and if 594 be added to it, the digits will be inverted. Required the number. * From Strachey's translation of the Bija Ganita, the work from which the rule for the solution of Quadratic Equations, given at p. 197, was taken. The scientific world is indebted for the publication of this very curious and interesting work to my lamented friend, the late Professor Leybourn, of the Royal Military College, Sandhurst. 33. Bacchus having caught Silenus asleep by the side of a full cask, seized the opportunity of drinking, which he continued for two-thirds of the time that Silenus would have taken to empty the whole cask. After that, Silenus awakes, and drinks what Bacchus had left. Had they both drunk together it would have been emptied two hours sooner, and Bacchus would have drunk only half of what he left for Silenus. Required the time in which each would have emptied the cask separately. Ans. Bacchus in 6, and Silenus in 3 hours. 34. A and B travelled on the same road, and at the same rate, from H to L. At the 50th milestone L, A overtook a drove of geese, which were proceeding at the rate of 3 miles in 2 hours, and 2 hours afterwards met a stage-waggon which was moving at the rate of 9 miles in hours. B overtook the same drove of geese at the 45th mile-stone, and met the same stage-waggon exactly 40 minutes before he came to the 31st milestone. Where was B when A reached L? Ans, at the 25th milestone. THE SOLUTION OF CUBIC AND BIQUADRATIC EQUATIONS. ALTHOUGH Horner's general method of approximating to the roots of numerical equations of all degrees supersedes the special methods adapted to particular classes, yet there are many occasions in the higher departments of mathematical inquiry in which it would be of great advantage to possess the symbolical values of the roots in terms of the literal coefficients of any given equation. Beyond those of the third and fourth degrees the labours of mathematicians have been altogether unsuccessful in assigning these symbolical values: and, indeed, there are strong reasons for believing that this want of success arises from circumstances that render the solution of the problem altogether impossible. Were it, however, otherwise, it is evident from a comparison of the rapid increase in the complexity of the expressions of the roots of equations of the first, second, third, and fourth degrees, that the roots of an equation of the fifth degree in terms of the coefficients would be so unwieldy as to preclude the possibility of substituting them in any other expression, and effecting a sufficient degree of reduction in the result to be of the slightest mathematical utility. I. CARDAN'S SOLUTION OF THE CUBIC EQUATION. 1. To prepare the equation, it is necessary to transform it into another whose second term has the coefficient 0. Let ax + bx2 + cx + d = =0 be the given equation. Assume two unknown quantities u and z, such that u + z = x. Substitute this value of x in the given equation; then it reduces to az3 + (3au + b) z2 + (3au2 + 2bu + c) z + au3 + bu2 + cu + d=0. Now that the coefficient of z2 may be 0, we must have 3au + b = 0, or u = b 3a Put this value for u, and reduce the expression; then we have finally for the transformed equation 27 a31⁄23 — (9ab2 — 27a2c) z + 2b3 — 9abc + 27aad = 0. an integer, the expression takes a much simpler form. When the equation is of any higher degree, as the nth, the same process will b give for the value of u which will remove the second term u = — As an example of the cubic, let the equation a3 transformed into one wanting the second term. Here a = 1, b =― - 6, and hence u — — - 3 6 - 6x2+10x na 8 = 0 be =2, and x z + 2; which 21x2 +146x b 3 1200, where 3 2 = and x z + Substitute na 2' - 34·5z2 + 56z + 36·5625 = 0. 1. x4 423 EXAMPLES. Remove the second terms from each of the following equations: 1 x + 3 = 0; x2 + 18 x + 3 (∙1)3 — 0. 5. Transform y3+12y-2—64y-1+15y1=0, and 2-102+1522—182°—0 †, into equations deficient of the second term. 2. To resolve the transformed cubic equation. Let the transformed equation be z3 + 3ez 2f0, where e and ƒ denote the coefficients of the transformed equation, divided by the coefficient of z3. Assume the two unknowns x and y to fulfil the equations x + y = z, and xy=- - e. Substitute these in the given equation, we have 23+ y3 = 2f; or squaring, x + 2x3y3 + y = 4f2. Whence by subtraction, a6 2x3y3 + y 203 = ƒ± √ƒ2 + e3, and y3 = ƒ + √ƒ2 + e3. Hence extracting the cube roots, and putting the values of x and y in x + y =2, we have z = √/ƒ± √√ƒ2 + e3 + √ ƒ + √ƒ2 + e3, which are both contained in the single expression z = √ƒ + √ƒ2 + e3 + V ƒ − √ƒ2 + e3. And 2+ u gives the value of the root of the original or untransformed equation, u being determined according to the process already explained. * As a process for numerical work, it may be stated, that a much shorter one may be given than that of actual substitution; but as it will be detailed further on, (in Horner's method of solution,) it will be unnecessary to do more than refer to it here. + In the first of these two equations, it will be necessary to reduce the equations to the form of integer indices before the application of the method; and in the second, to consider z3 as the quantity according to which the arrangement is made. This last, however, may be put in a still different form, though not a more advantageous form. When e is negative, and e3 is greater than ƒ2, the radical √ƒ2 + e3 becomes imaginary, and hence the numerical calculation of the root becomes impossible without some further contrivance. Many have been proposed; as the expan sion of {ƒ + k}} + {ƒ — k} into series, by which the odd powers of k or √ƒ2 + e3 mutually cancel, and leave only terms which are real; also by means of trigonometrical tables, and by special tables devoted to the purpose. All these methods are now so completely superseded by Horner's process as to need no remark here. It will only be requisite to state, that the case now supposed is called The Irreducible Case of Cardan's Cubic; and that it is known from other considerations, that in this case all the three roots of the equation are real, whilst in that to which Cardan's formula applies, two of the roots are imaginary, and the real one is given by that formula. 6x2+10x As an example, let us take the equation a3 80, whose second term we had eliminated by our previous transformation, giving 23 2z40. Here e = 2 3 and ƒ 2. Hence 2=2+ √√/4 − + √2 −√√⁄4 — § = 3/2 + 1 √/⁄3 + 3/2 — √/3 = (1 + } √/3) + (1 − } √√/3) = 2. Hence x = z + 2 = 4, which is the real value of x in the given equation. Other Examples for Practice. Ans. 7x2 2. Find the value of x in the equations 3 +14x20. 3. Find Cardan's roots of the equations x3 + 6x = 20, and æ3 = 7. 6x2+18x=22, and ∞3 2.32748, and x = 5 respectively. 5. Given yo + 18y6 + 216y3 = 3392, to find Cardan's value of y. 6. Find Cardan's root of y2+ ·24y2 = 245. 3/9. Ans. 2*. 8. Solve those of the equations given at p. 199 for transformation, that come under the form adapted to Cardan's solution. II. SIMPSON'S SOLUTION OF THE BIQUADRATIC EQUATION †. LET the given equation be x2 + 2ax3 + bx2 + cx + d = 0.......... (1), and assume * In all cases where there are higher powers of the unknown, if they be in the ratio of 1, 2, 3, the problem is treated as a cubic. That here referred to is one in which y is not directly found by the formula, but y3, which is = 8, and then from this again y = 2. The same is true of any inferior powers, where the indices are related in the same way, as in the next question. This method of solution was invented by Mr. Thomas Simpson, F.R.S. Professor of Mathematics in the Royal Military Academy from 1743 to 1761. It has often been erroneously ascribed to Dr. Waring, Lucasian Professor of Mathematics in the University of Cambridge, and even called by his name. it equal to (x2 + ax + A)2 (Bx + C)2 = 0.... (2), where A, B, C are unknown. Expand (2) and equate the coefficients of the powers of x in the result with those of (1). This gives, Also multiplying four times (3) and (5) together, and equating the product to the square of (4) we have 8A3 46A2 + 4 (ac 2d) A = 4a2d 4bd + c2 Suppose A to be found from this cubic equation; then from (3) we have B = ± √2A + a2 — b .... (7) 2aA -c x2+(a + √2A + a2 — b) x + AF 2/2A + a2—b .... =0 (10) in which the doubtful sign is to be taken the same in the second and third terms. Note I. Whenever, by taking away the second term of a biquadratic, after the manner described at page 199, the fourth term also vanishes, the roots may immediately be obtained by the solution of a quadratic only. NOTE II. A biquadratic may also be solved independently of cubics, in the following cases :— 1. When the difference between the coefficient of the third term, and the square of half that of the second term, is equal to the coefficient of the fourth term, divided by half that of the second. Then if p be the coefficient of the second term, the equation will be reduced to a quadratic by dividing it by x2 +1px. 2. When the last term is negative, and equal to the square of the coefficient of the fourth term divided by 4 times that of the third term, minus the square of that of the second: then to complete the square, subtract the terms of the proposed biquadratic from (x2 + px)2, and add the remainder to both its sides. 3. When the coefficient of the fourth term divided by that of the second term gives for a quotient the square root of the last term; then, to complete the square, add the square of half the coefficient of the second term to twice the square root of the last term, multiply the sum by a2, from the product take the third term, and add the remainder to both sides of the biquadratic. 4. The fourth term will be made to go out by the usual operation for taking away the second term, when the difference between the cube of half the coefficient of the second term and half the product of the coefficients of the second and third terms, is equal to the coefficient of the fourth term. EXAMPLES. Ex. 1. Given a + 12x = 17 to find a by Simpson's method. |