42. a2(ax)2 = b2(b — x)2. 43. (ax) + (x —b)2 = (a — b)3. 44. (ax) (b— x) = 2(a - b)2. 45. (ax) (a − x) (x − b) + (x — b)2 = (a — b)3. 47. (a+b+c)x2 — (2a + b + c)x + a = 0. 51. x2-2 (a - b) x = (a + c − b) (b + c − a). 52. abx2-(a+b)x+1=0. 53. 4x2 - 4ax + a2 — b2 = 0. 59. (3x-5)2-8 (3x-5)+ 7 = 0. 60. (2x — a)2 = b (2x — a) + 2b3. 61. (3x-2a+b)2 + 2b (3x — 2a + b) = a2 -- b3. CHAPTER III EQUAL ROOTS AND IMAGINARY ROOTS 409. In the preceding chapter after having put the equation of the second degree in the form (2 ax + b)2 = b2 - 4ac, it was supposed that the second member was a positive quantity and then the formula was deduced which gave the two roots, or the two solutions of the equation. The first member is a perfect square and the unknown quantity 2 xb, whose square is 0, is itself 0; hence 2 a Here the equation is said to have two equal roots, each equal to b The reason for this is that, as the radical 1 bo 4 ac in (3) approaches zero, the quantity which is added to -b for one root and subtracted for the other becomes as small as is desired, the two roots differ by as small a quantity as we please, and are said to be equal in the limit. IMAGINARY ROOTS 410. Since the square of positive or negative quantities is always positive, it follows that it is impossible to take the square root of negative quantities (379). Hence, in the case of the equation of the second degree, if b2 - 4 ac is negative, it will be impossible to satisfy the equation of the second degree. For then a positive quantity equal to a negative quantity, which is impossible. In this case the formulae give fictitious values which have been introduced into mathematical analysis under the name of imaginary quantities. Consider the equation, x2 = + 4. Here the square root of +4 can be found and the equation will have the solutions, If, however, the second member is negative, and x2 = — 1, then there is no number, positive or negative, whose square is equal to 1, and it is impossible to satisfy the equation. If, however, Ꮴ 1 is represented by i and the letter i is introduced into algebraic calculations as though it represented a real quantity, with the convention that its square (i) is equal to - 1, the equation a2 = −1 can be satisfied by the imaginary values + i and i; for All imaginary values can be expressed in terms of the symbol i (88381, 382). For example, the equation, x2 = - 16 has the two imaginary solutions, 1 = +V-16, and x = If it is noticed that - 16 16 (-1), and the theorem concerning the square root of a product is applied (2356), it is possible to write Now apply the principle to the more general equation, 411. It is necessary to proceed in a similar manner in case of if the quantity b2-4ac is negative. Because, when this equation is put under the form (2 ax + b) = b2 4 ac it is impossible, as has been seen, to satisfy it with real values, but it can be satisfied by the imaginary values, 2 ax + b = ± √ − (b2 — 4 ac), with the convention, as has already been seen in the preceding examples, that the square of the symbol (2-4 ac) is always Hence, it will follow that the imaginary —4ac) equal to --(b2-4ac) 380. values of x are The roots x and x, may be written in the form A+ Bi, where 2 The ordinary rule of calculation in common Algebra has been extended to imaginary quantities as though the symbol i were a real number with the convention that i- -1 (Chap. VII, Book III). In view of the preceding consideration, the solution of the equation ax + bx + c = 0 presents itself under three aspects: 1. If 2 2. If b2 4ac0, both roots are real and different. 3. If b2 4 ac<0, both roots are imaginary and different.. SOLUTION OF THE EQUATION x2 + px + q = 0 412. This equation can be solved by comparing it with equation (1) 8406, thus: The solution of the equation x2+px+q=0 will present the same general cases found in equation (1) 411: 1. If p2 2. If p2 4q> 0, both roots are real and different. 4q = 0, both roots are real and equal. 3. If p2-4q<0, both roots are imaginary and different. EXAMPLES. 1. Solve the equation x2-7 x + 10 = 0. Here p7, q= 10, and on substituting in equation (5), -7±√49–40 −7 ± 3–5 or 2. X1, X2 2 2. Solve ab(x2 + 1) = (a2 + b2)x. 2 3. Find the value of k in order that the equation, b2 (a2 - x2) = a2 (mx + k)2, may have equal roots. - Develop and arrange with respect to x and x; then (a2 m2 + b2) x2 + 2a2 kmx + a2 k2 a b2 = 0. Py the condition for equal roots, (2a km) — 4 (a2 m2 + b2) (a2 k2 — a2 b2) = 0. Removing the parentheses and dividing by 4, a1 k2 m2 — a1 k2 m2 + a1 b2 m2 — a2 b2 k2 + a2 b1 = 0, Therefore, after dividing by a2b2, [2411, 2] k2 = b2 + a2 m2; k = ±√b2 + a2 m2. |