1997 solution of a quadratic function problem in mathematics college entrance examination

(1) Let f (x) = f (x)-x, where x 1 and x2 are two of the equations f (x)-x = 0.

There is f (x) = a (x-x 1) (x-x2).

When x∈(0, x 1) is from x 1≤x2, and A > 0, F (x) = A (x-x 1) (x-x2) > 0,

That is, f (x) = f (x)-x > 0, and f (x) > x.

And x1-f (x) = x1-[x+f (x)] = x1-x-a (x-x1) (x-x2) = (x/kloc-0)

Because 0 < x < x 1 < x2

So x 1-x>0-x > 0,1+a (x-x2) =1+ax-ax2 >1-ax2 > 0.

X 1 > f (x), so x < f (x) < x 1.

(2) X0 =-b/2a, because x 1 and x2 are the roots of F (x)-X = 0, that is, x 1 and x2 are equations.

The root of ax2+(b- 1) x+c = 0.

So x 1+x2 = 1-b/a,

x0 = a(x 1+x2)- 1/2a = ax 1+ax2- 1/2a =(x 1/2)+(ax2- 1)/2a

Because ax2 < 1, that is, ax2- 1 < 0, x0.