高手来看下，写个过程．

证 :
(1)若a=0,则b+c=0,b≠0,方程f(x)=0有实根 若a≠0,则(2b)^2-4*3a*c=4(a+c)^2-12ac =4(a^2-ac+c^2)=4>0,方程f(x)=0有实根.

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

(3) m+n=-2b/(3a), m*n=c/(3a) a+b+c=0,c=-(a+b) |m-n|^2=(m+n)^2-4mn=^2-4c/(3a) =(4/9)(b/a)^2+(4/3)(b/a)+4/3 =(4/9)^2+1/3 由-2<b/a<-1得到,-1/2<b/a+3/2<1/2, 0<=^2<1/4 1/3<=|m-n|^2<1/3+1/4=7/12<4/9,得到(√3)/3≤｜m-n｜＜2/3

我才初一,太难了!