3. Given the equation x2 — x + 301⁄2 = 523. Transposing and cancelling the denominators gives 3x2 2x 133. Hence the special rule in the note on the Hindû method gives 9x2-6x+1 = 400, or 3x − 1 = +20. x=7, and x = — - 61. Whence 4. Given x1 2ax2b, to find x. The Italian method can be used here. Complete the square, then a1 2ax2+ a2a2 + b ; Extract the roots, then x2 - a = ± √a2 +b; Resolve pure quadratic x = ± √a±√a2 + b. That is, a has the four values, +✔a+ √a2+b‚−√ a+ √a2+b, +√a √a2+b, and ✔a-√a2+b. complete the square by the Hindû method, and we get 6. Given √2-4 {√x + 13} * + 7 = x − 2√x — 9, to find x. Add 10 to both sides: then we get both sides squares, viz. : √x + 13 − 4√ √x + 13 + 4 = x − 2√ + 1, and extracting, To continue the solution, take the results + and separately. (1.) Take + : then √x − 1 − √√√x + 13 = − √x + 13 √√x+13=12, which is again of the quadratic form. Complete by the Hindû method, and extract, which gives 4 { √ x + 13 } − 4 √√√x + 13 + 1 = 49, or √ √ + 13 = 1+7 = or - 3. Squaring √ √ Squaring √ √ x = 16. + 13 = 4, we get √x + 13 = 16, or √x=3, and μ = 9. + 13 = − 3, we get + 139, or x 4, and (2.) Take : then √✅ +13 + √ √ x + 13 = 16; and completing the square (Hindû method) as before, extracting and reducing, we obtain ultimately two other values of x. Collecting the four together, we have x = 9, x = 16, and 57+7/65 2 x= This example exhibits a class of contrivances for completing the square of very frequent use; but no general rule can be laid down respecting it. The only general remark that can be made is, to endeavour to render the parts without the radical, the square of the radical itself; and then, if on completing the square of the side so transformed, the other side is also a square, the method will be effective. m x2); to find x. 7. Given / (1 + x)2 — TM/ (1 − x)2 = "√ (1 This is evidently in the form of a quadratic; by resolving which, we get, 2 + 1. 16 13. Solve +2 = 7.56; 32-2.5x+592 = 0; and a-√/a2-x2 = b a+√a2+x2 3 14. Resolve 2x 3x 3x = 20, and 2x2 + 3x−5√/2x2 + 3x + 9 = − -3. 16. Resolve (x + 1) (x2 + 1) (x3 + 1) = 30x3, and x1 SIMULTANEOUS EQUATIONS OF THE SECOND AND HIGHER DEGREES. THE same conditions that had place in the case of simple equations also hold with respect to those of the higher degrees. The elimination of any number of them, and the actual determination of the values of any one of them, becomes, however, much more difficult. Even when the elimination of all but one of the quantities is effected, the equation in which it is involved is generally of a high degree, and the trouble of actually determining its numerical value becomes considerable but except in very particular cases, when more than one of the simultaneous equations is above the first degree, the equation which results from the elimination takes a form which does not admit of any solution in terms of the literal quantities involved in the given ones, or at most by means of expressions of extreme complexity. In the case of the simultaneous equations involving two of the second degree, and the remaining ones simple, the solution is theoretically possible in the terms of the given symbols; but still the transformations to be made in the forms of the equations to effect it are too numerous to render it capable of practical application. Into the general reasonings concerning the principles of elimination, the limits of this Course forbid our entering. Nevertheless, as cases of frequent occurrence, in almost every branch of mathematics, involve this problem under more or less confined conditions, it has been considered necessary to devote a page or two to the simpler uses of it, and to give a few exercises, as a praxis for the student, at the close of the chapter. I. All the operations described in the section on Simultaneous Equations of the first degree are to be applied to others of a higher order, in any case that admits of it. II. If all the terms of an equation contain the same number of unknown factors (in which case it is called a homogeneous equation), we may put one of the factors equal to the other, multiplied by a new unknown, assumed for the purpose. As for instance, in 3xy + 2y2 = 4x2, we may put y = vx, which gives (3v + 2v2)x2 = 4x2, or from which may be obtained. The same substitution being made in another equation simultaneously given, but with the value of v, instead of v itself, gives an equation containing y only, and which may be resolved by the usual methods and known quantities. If there be three or more unknowns, so many independent but simultaneous substitutions must be assumed for them as there are of quantities besides the one selected above. III. Sometimes we can effect the reduction by substituting for one of the unknowns the sum, and for the other the difference of two other quantities, of course unknown too. This method applies to the case of two unknowns, and obviously is confined to it. Methods in some degree analogous to this have, however, been devised for the case of three or more unknowns. IV. It often happens that by raising one equation to some power, several of its terms will be identical with those of some other power of another equation, and in this case the equations are simplified by subtracting one from the other. The same is true of multiples of one, and powers of another equation. Sometimes too it happens that adding some quantities to one side of an equation, to render it a complete square, cube, or higher power of a binomial, the other side, so increased, becomes also a square, cube, and so on. The roots then being taken, the equation is reduced to lower dimensions, and the ultimate elimination more easily effected. Other rules and remarks will occur to the intelligent student as he proceeds; and the teacher will often, in the actual solution of individual problems, be able to enforce a rule, and to point out the circumstances under which it can be applied, which could scarcely be rendered intelligible in print without extreme prolixity. Ex. 1. Given x + y = a, and xy = b2, to find x and y. By squaring (1) we have x2 + 2xy + y2 = a2, and multiplying (2) by 4, 4xy = a2 = : 462; 4b2, - y = + √ a2 — 4b2. and extracting, æ 3(x + y) a3 Then by means of this and (1) proceed to find x and y as in the last example. Or thus. Put x = u + v and y = u — v: then we have x + y = 2u = a, or a u = 2 Insert the assumed values of u and v in x3 + y3 = b3; then we get 2u3 + 6uv2 = b3, or putting in this the value found for u, we have v2 = Putting in the equations x = u + v and y = u — v these values, we have the result required. Or, thus again. Put y = ux: then the equations become x(1 + u) = a, and x3(1 + u3) = b3. Divide the cube of the first by the second of these: then Hence u becomes known by the solution of the quadratic equation. Also, inserting the values of u thus found in x(1 + u) = a, we get x= and from this, again, y = ux will be obtained. 1 1 5 Ex. 3. Given x2y + y2x =30, and + = to find x and y. y 6' Break (1) into factors, and cancel denominators in (2), then xy(x + y) : = 30, and 6(x + y) = 5xy. Multiply the second of these by xy and the first by 6, and subtract: then x2y2 = 36, or xy = ± 6. As there is subsequent work to perform, it will be advisable to work with xy = 6 and xy = - 6 separately *. First, take ay 6: then inserting this in (1) we get x + y = 5; and resolving this as in Ex. 1, we find x − y = + 1; and this again combined with (1) by addition and subtraction gives x = 3 or 2, and y = 2 or 3. Secondly. Take xy=— 6: then, proceeding as before, we have x = 6 or and y = - 1 or 6. The latter pair of results not fulfilling the equation, do not come properly under the denomination of answers. They come into the work from the ambiguous root of x2y2: = 36; but it might have been inferred from this being (+ xy). (+ xy) = 36, that only xy + 6 was admissible. Or thus. Put xu + v and y = u- v. Then substituting in the given equations and reducing, we find u(u2 v2) = 15, and 12u = 5(u2 — v2). Divide the first of these equations by the second: then 4u2 (see remark on last solution). 5 = 25, or u = Substitute this in either of the last equations: then we get v = ± Or, thus again. Put y = ux. Then the equations reduce to uy3 (1 + u) = 30, and 6(1 + u) = 5uy. Equating the values of y3 derived from these we obtain 36 (1 + u)1 = 625u2, or 6(1 + u)2 = ± 25u: that is, 6u2 and 6u2+37u + 6 = 0. 3 6 The former gives u = and and the latter u = — and 3' 13u60, 30, and x = uy = 180. 6, and x = uy = 1. The two last results are, as in the other case, the consequence of the ambiguous sign in 6(1 + u)2 = ± 25, and their inapplicability might have been inferred at the outset, as in the former solutions. In questions of this kind, these separations should, for the sake of secure working, be always employed. |