令x1
=ax1²+bx1+c-ax2²-bx2-c
=a(x1+x2)(x1-x2)+b(x1-x2)
=(x1-x2)[a(x1+x2)+b]
x1
所以x1+x2<-b/a
a>0,两边乘a
a(x1+x2)<-b
所以a(x1+x2)+b<0
所以(x1-x2)[a(x1+x2)+b]>0
即x1
所以x<-b/2a是减函数
令-b/2a
=(x1-x2)[a(x1+x2)+b]
x1
所以x1+x2>-b/a
a>0,两边乘a
a(x1+x2)>-b
所以a(x1+x2)+b>0
所以(x1-x2)[a(x1+x2)+b]<0
即-b/2a