如果一元二次方程ax的平方+bx+c=0的两根之比为2：3.求证6b的平方=25AC

设 x1 和 x2 分别为方程的两个根,则有 x1 + x2 = -b/a (1) x1 * x2 = c/a (2) x1 / x2 = 2/3 ==> x1 = 2x2/3 (3) (3)代入(2) ==> 2 x2^2 / 3 = c/a ==> x2^2 = 3c/2a (4) 由方程 a x2^2 + b x2 + c = 0 ==> a x2^2 = - (b x2 + c) ==> x2^2 = - (bx2 + c ) / a (5) (4) - (5) ==> 3c/2a + (bx2 + c ) / a = 0 ==> 3c + 2(b x2 + c) = 0 ==> bx2 + c = -3c/2 ==> bx2 = -5c/2 ==> x2 = -5c / 2b (6) (6) 代入 (3) ==> x1 = 2 * x2 /3 = 2 * (-5c / 2b) / 3 = -5c/3b (7) (6)和(7) 代入 (1) ==> -5c/3b + (-5c / 2b) = -b/a ==> 5c/3b + 5c/2b = b/a ==> 10c/6b + 15c/6b = b/a ==> 25c / 6b = b / a ==> 6b^2 = 25ac