(1)
x=1,f(x)=-a/2代入函數(shù)方程：
a+b+c=-a/2
b=-3a/2-c
對于方程ax^2+bx+c=0,由韋達(dá)定理,得
x1+x2=-b/a
x1x2=c/a
(x1-x2)^2
=(x1+x2)^2-4x1x2
=(-b/a)^2-4c/a
=(b^2-4ac)/a^2
=9/4-c/a+(c/a)^2
=[(c/a)-1/2]^2+2≥2
|x1-x2|≥√2
(2)
a>0
f(0)=c
f(1)=-a/20,f(1)