本帖最後由 Ivanhy92 於 31/10/2010 09:15 PM 編輯 Let a/k, a, and ak be the roots of the equation x^3+px^2+qx+r=0, where k≠0 From the relations of roots and coefficients, a/k + a + ak = -p => a(1/k + 1 + k) = -p --------(1) a^2/k + a^2 + (a^2)k = q => (a^2)(1/k + 1 + k) = q -----(2) a^3 = -r => a = (-r)^(1/3) -------------(3) (2)/(1) : a = q/(-p) Put (3) into above, (-r)^(1/3) = q/(-p) -r = (q^3)/(-p)^3 ∴(p^3)r = (q^3) For 3x^3 - 26x^2 + 52x - 24=0, x^3 - (26/3)x^2 + (52/3)x - 8 = 0 ∵[-(26/3)]^3 x 8 = -140608/27 = (-52/3)^3 ∴ By (3), a = 8^(1/3) = 2 Put a=2 into (1), 2(1/k + 1 + k) = 26/3 k = …… |