It is known that the quadratic function f (x) = AX2 + BX + C satisfies a > b > C, f (1) = 0. The function g (x) = f (x) + BX (1) proves that the function y = g (x) must have two different zeros (2) Let the two zeros of function y = g (x) be x1, X2, and find the absolute value range of x1-x2

It is known that the quadratic function f (x) = AX2 + BX + C satisfies a > b > C, f (1) = 0. The function g (x) = f (x) + BX (1) proves that the function y = g (x) must have two different zeros (2) Let the two zeros of function y = g (x) be x1, X2, and find the absolute value range of x1-x2

(1) G (x) = f (x) + BX = AX2 + 2bx + C, the discriminant = 4 (b ^ 2-ac) has f (1) = 0, has a + B + C = 0, and a > b > C, there must be a > 0, C0 is constant. Therefore, the function y = g (x) must have two different zeros. (2) the relationship between root and coefficient X1 + x2 = - 2b of a, X1 * x2 = C of a, the absolute value of x1-x2 = under the root sign [(x1 + x2) ^ 2-4, X1 * x2] = substitute B = -- a-c into [4 / 2 of a ^ 2 (b ^ 2-A * c)] under the root sign to get [C / 2 of a + 1 / 2 ^ 2 + 3 / 4] under the root sign