已知二次不等式ax2+bx+c>0的解集为{-1/3<x<2},求关于x的不等式cx2-bx+a
人气:387 ℃ 时间:2020-04-18 20:57:13
解答
ax2+bx+c>0的解集为{-1/3<x<2}
所以a<0
且-1/3和2是ax2+bx+c=0的跟
则-1/3+2=-b/a
-1/3*2=c/a
b=-5a/3
c=-2a/3
所以是-2a/3*x²+5ax/3+a<0
两边乘3/a<0
不等号改向
2x²-5x-3>0
(2x+1)(x-3)>0
{x|x<-1/2,x>3}
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