Given the quadratic function f (x) = AX2 + BX + C (C ≠ 0) (1) if A.B.C and f (1) = 0, it is proved that the image of F (x) has two intersections with the X axis; (2) if Changshu x1 If x 2 ∈ R, and x 1, x 2, f (x 1) ≠ f (x 2), we prove that the equation f (x) = 1 / 2 [f (x 1) + F (x 2)] must have a root belonging to (x 1, x 2)

Given the quadratic function f (x) = AX2 + BX + C (C ≠ 0) (1) if A.B.C and f (1) = 0, it is proved that the image of F (x) has two intersections with the X axis; (2) if Changshu x1 If x 2 ∈ R, and x 1, x 2, f (x 1) ≠ f (x 2), we prove that the equation f (x) = 1 / 2 [f (x 1) + F (x 2)] must have a root belonging to (x 1, x 2)

(1) From F (1) = 0, we can know that a + B + C = 0
The discriminant B & sup2; - 4ac = (a + C) & sup2; - 4ac = (A-C) & sup2; > = 0
So the image of F (x) has two intersections with X axis;
Note: if the discriminant is equal to 0, there are two identical intersections
(2) The equation f (x) = 1 / 2 [f (x1) + F (x2)], and the deformation can be known
Let f (x) = [f (x) - f (x1)] + [f (x) - f (x2)] = 0
It is easy to know that f (x1) = [f (x1) - f (x1)] + [f (x1) - f (x2)] = f (x1) - f (x2),
F(x2)=[f(x2)-f(x1)]+[f(x2)-f(x2)]=f(x2)-f(x1),
So f (x1) f (x2)