Given the quadratic function f (x) = ax ^ 2 + BX + C (1) if a > b > C and f (1) = 0, it is proved that the image of F (x) has two different intersections with the x-axis; (2) it is proved that for x1, X2 and x1-2

Given the quadratic function f (x) = ax ^ 2 + BX + C (1) if a > b > C and f (1) = 0, it is proved that the image of F (x) has two different intersections with the x-axis; (2) it is proved that for x1, X2 and x1-2

1. If f (1) = 0, then a + B + C = 0, then B ^ 2-4ac = (a + C) ^ 2 - 4ac = (A-C) ^ 2 > 0, suppose a > C, none = 0,
So f (x) = 0 has two different roots, that is, two different intersections (1)
2. The formula is changed to 2F (x) - f (x1) - f (x2) = 0. Substitute f (x) = a ^ 2 + BX + C to get an equation. The range of X is determined by the coefficient. If you don't write it, it's troublesome. You can't do it. If you do it, I'm a teacher and I'll doubt you
3. F (x) + a = a ^ 2 + BX + A + C = a ^ 2 + BX - B = 0 has a solution, that is, the discriminant is greater than 0
b^2 -4ab>0 b(b-4a)>0 -a x0x1(-a x0x1-4a )>0