设y=f(x)=ax²+bx+c
任取-b/(2a)≤x10
∴ax1≥-b/2,ax2>-bx2
相加:
a(x1+x2)>-b
∴a(x1+x2)+b>0
又x1-x20
∴(x1-x2)[a(x1+x2)+b]