Let f (x) = ax & # 178; + BX + C (a > 0), and f (1) = - A / 2 (1) Prove that the function f (x) has two zeros. (2) let X1 and X2 be the two zeros of the function f (x), and find the absolute value range of x1-x2. (3) prove that the function f (x) has at least one zeros in (0,2) Is to find the value range of the absolute value of x1-x2

Let f (x) = ax & # 178; + BX + C (a > 0), and f (1) = - A / 2 (1) Prove that the function f (x) has two zeros. (2) let X1 and X2 be the two zeros of the function f (x), and find the absolute value range of x1-x2. (3) prove that the function f (x) has at least one zeros in (0,2) Is to find the value range of the absolute value of x1-x2

1) Because a > 0, i.e. the opening is upward, and f (1) = - A / 2 = 2
So | x1-x2 | > = √ 2
3)f(0)=c
f(2)=4a+2b+c=4a+2(-3a/2-c)=a-2c
If C > 0, then f (0) > 0, f (1)

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