高悬赏！急！！！高一数学求解

1.f(0)=c>0
f(1)=3a+2b+c>0

3a+2b+c-2(a+b+c)>0
a-c>0
a>c>0

3a+2b+c-(a+b+c)>0
2a+b>0
b>-2a
b/a>-2

a+b+c=0
b=-a-c<-a
b/a<-1

所以，:a>0且－2<b/a<－1

2.(x1-x2)^2=(x1+x2)^2-4x1x2=(-2b/(3a))^2-4(c/(3a))
=(-2b/(3a))^2-4(-a-b)/(3a) t=2b/(3a)
-4/3<t<-2/3 =t^2+2t+4/3=(t+1)^2+1/3
1/3<=(x1-x2)^2<4/9
这道题我算了很久才算出来的,要是可以的话就把分给我吧,我急用

正确解答！

(1)a+b+c=0,得c=-a-b，
由f(0)*f(1)>0
得c*(3a+2b+c)>0,
(-a-b)*(2a+b)>0,
(a+b)*(2a+b)<0
2a^2+3ab+b^2<0 ,
又a!=0,得（b/a)^2+3(b/a)+2<0
得-2<b/a<-1.

(2)(x1-x2)^2=(x1+x2)^2-4x1x2=(-2b/(3a))^2-4(c/(3a))
=(-2b/(3a))^2-4(-a-b)/(3a) t=2b/(3a)
-4/3<t<-2/3 =t^2+2t+4/3=(t+1)^2+1/3
1/3<=(x1-x2)^2<4/9

则根号3/3小于等于（x1-x2)的绝对值<2/3

1.f(0)=c>0
f(1)=3a+2b+c>0