標題:
f.4 a math呀~~help
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發問:
if one root of the equation 6x^2-(8a-3)x + (2a^2-a)=0 is the reciprocal of the other,find the real values of a and the roots.
最佳解答:
Let r, 1 / r be the two roots. Then r + (1 / r) = (8a - 3) / 6............(1) r(1 / r) = (2a^2 - a) / 6...................(2) From (2), 1 = (2a^2 - a) / 6 2a^2 - a = 6 2a^2 - a - 6 = 0 (2a + 3)(a - 2) = 0 Thus a = -3 / 2 or 2. When a = -3 / 2, r + (1 / r) = (8(-3 / 2) - 3) / 6 = -5 / 2 (r^2 + 1) / r = -5 / 2 2(r^2 + 1) = -5r 2r^2 + 5r + 2 = 0 (r + 2)(2r + 1) = 0 r = -2 or -1 / 2 The roots are -2 or -1 / 2. When a = 2, r + (1 / r) = (8(2) - 3) / 6 = 13 / 6 (r^2 + 1) / r = 13 / 6 6(r^2 + 1) = 13r 6r^2 - 13r + 6 = 0 (3r - 2)(2r - 3) = 0 r = 2 / 3 or 3 / 2 The roots are 2 / 3 or 3 / 2.
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