1、a+b+c=-a/2 => -b=3/2a+c => b^2-4ac=(3/2a-1/3c)^2+8/9c^2>0
2、c=-3/2a-b>0 => b<-3/2a
c=-3/2a-b => b^2-4ac=(2a+b)^2+2a^2 => |x1-x2|=√[(b^2-4ac)/(a^2)]=√[(2+b/a)^2+2]
当a>0时,b/a<-3/2,2+b/a<1/2,|x1-x2|>=√2
3、f(0)=c>0,f(2)=4a+2b+c=5/2a+b
当a>0时,b/a<-3/2：当-4<=b/a<-3/2时,对称轴x=-b/(2a)∈(3/4,2],且f(0)>0,开口向上,必有一根∈(0,2)；当b/a<=-4时,b<=-4a,f(2)=5/2a+b<=5/2a-4a=-3/2a<0,且f(0)>0,必有一根∈(0,2)