函数,尽快!设函数y=3ax^2+2bx+c,若a+b+c=0,c>0,3a+2b+c>0.求证：（1）函数和x轴有两不

(1)
判别式△=4b²-12ac=4(a+c)²-12ac
=4a²+8ac+4c²-12ac
=4a²-4ac+c²+3c²
=(2a-c)²+3c²
∵c＞0
∴△＞0
∴函数和x轴有两不同交点
(2)
3a+2b+c=a+2a+2b+c=a-2c+c=a-c＞0
∴a＞c＞0,b=-a-c＜0
a+b=-c＜0,b＜-a,b/a＜-1
3a+2b+c=3a+2b-a-b=2a+b＞0
b＞-2a,b/a＞-2
∴-2＜b/a＜-1
(3)
由韦达定理
x1+x2=-2b/3a,x1x2=c/3a
∴|x1-x2|=√[(x1+x2)²-4x1x2]
=√(4b²/9a²-4c/3a)
=√[(4b²-12ac)/9a²]
=√[(4b²+12a(a+b))/9a²]
=2/3√[(b²+3a²+3ab)/a²]
=2/3√(b²/a²+3b/a+3)
=2/3√[(b²/a²+b/a+9/4)+3/4]
=2/3√[(b/a+3/2)²+3/4]
∵-2＜b/a＜-1
∴-1/2＜b/a+3/2＜1/2
∴√3/3≤2/3√[(b/a+3/2)²+3/4]＜2/3
即√3/3 ≤|x1-x2|＜2/3

答案见图片: