we may take √(3x2 − 2 x + 4) as the unknown number, replacing it temporarily by y. We then have (1±√37). The numbers 3, satisfy the given equation, and are therefore roots of that equation. The numbers √(1±√37) do not satisfy the given equation. But if the value of the radical be not restricted to the principal root, the given equation comprises the two equations Ex. 2. Solve the equation (3x2 + 13) + √(3 x2 + 13) = 6. Assuming (3x2 + 13) as the unknown number, and representing it by y, we have y + y2 = 6. Equating (3x2 + 13) to each of these roots, we have √(3x2 + 13) = 2, whence x = ± 1, (3x2 + 13)=-3, whence x = ±√3=±}√51. 16+√16=6. The numbers ± 1 are roots of the given equation, since The numbers±√51 are evidently not roots of the given equation, but are found to be roots of the equation − √(3 x2 + 13) + √(3 x2 + 13) = 6. The preceding examples illustrate the following principle : If a radical equation contain one radical, and an expression which is equal to the radicand or which can be made to differ from the radicand (or a multiple of the radicand) by a constant term, it can be solved as a quadratic equation. The same is true if the equation contain two radicals, one the square of the other, and in addition only constant terms. In both cases, the radicand must, in general, be a linear or a quadratic expression. 7. Irrational equations containing cube and higher roots in general lead to rational, integral equations of a higher degree than the second, and therefore cannot be solved by means of quadratic equations. But in some cases their solutions can be effected by special devices. Ex. Solve the equation Cubing, (8x + 4) − √/ (8 x − 4) = 2. 8x+4-3[(8 x + 4)]2 (8x-4)+3(8x+4)[(8x-4)]2 − 8x + 4 = 8. Transferring and uniting terms, and dividing by - 3, 3 (1 [(8x+4)]2(8x-4) (8x+4)[/(8x-4)]2 = 0. (2) Factoring, 3 (8x + 4) (8 x − 4) [ √(8 x + 4) – (8x-4)]= 0. (3) · − † This equation is equivalent to the three equations Equation (6) is not satisfied by any finite value of x. The numbers — } and { are found to satisfy the given equation. EXERCISES. If a Solve each of the following equations, and check the results. result does not satisfy an equation as written, determine what signs the radical terms must have in order that the result may satisfy the equation. 40. x2-x+2√(x2−x−11)=14. 1 x + √(2− x2) 42. (2 x2 - 3 x + 5) + 39. √x+x=2x. 13 6 41. x2+24=2x+6√(2x2-4x+16). 2 x2 - 3 x = 1. 43. 2x√(4x2-27 x) = 5 x2 + 27 x + 9. 44. √(3x2+7x-1) - √(3x2 - 4x+5)= 8. (2x27x+7) + √(2x2+9x-1)=6. 45. 60. 63. a = √(a + x) a√(a2x2) a + √(a2x2) 59. √(1 − x + x2) + √(1 + x + x2) = m. a√(x-b)+b√(ax) CHAPTER XXIV. SIMULTANEOUS QUADRATIC AND HIGHER EQUATIONS. To obtain a definite solution of a system of two or more quadratic or higher equations, as many equations must be given as there are unknown numbers. As in linear systems, the given equations must be consistent and independent. The solution of a system of quadratic or higher equations in general involves the solution of an equation of higher degree than the second, and therefore cannot be effected by the methods for solving quadratic equations. But there are many special systems whose solutions can be made to depend upon the solutions of quadratic equations. § 1. SIMULTANEOUS QUADRATIC EQUATIONS. 1. Elimination by Substitution. When one equation of a system of two equations is of the first degree, the solution can be obtained by the method of substitution. Ex. Solve the system y+2x=5, (1) (2) (3) (4) x = 1, (5) Substituting 5 - 2 x for y in (2), x2-25 +20 x − 4 x2 — — 8. From this equation we obtain Substituting 1 for x in (3), The system (1), (2) is equivalent to the system (3), (4), which is equivalent to the two systems (3), (5) and (3), (6). Therefore the solutions of the given system are 1, 3; 53, – 63, the first number of each pair being the value of x, and the second the corresponding value of y. Had we substituted 1 for x in (2), we should have obtained y= ± 3. But the solution 1,3 does not satisfy equation (1). By the principle of equivalent equations, proved in Ch. XIII., § 2, Art. 2 (iii.), equation (3), obtained from (1) by solving for y, and equation (4), obtained by substituting this value for y in (2), form a system equivalent to the given system. This principle does not, however, prove that (2) and (4) are necessarily equivalent to the given system. In this example, since (2) and (4) give more solutions than (1) and (4), the system formed by (2) and (4) cannot be equivalent to the given system. Therefore, having obtained the values of one of the unknown numbers, we should obtain the values of the other by substituting in the equation of the first degree. This advice was unnecessary in solving systems of linear simultaneous equations, since then both equations were of the first degree. |