Question

已知方程X2+(2+a)x+1+a+b=0的两根为x1,x2.并且0<x1<1<x2,则b/a的取值范围是?

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Answer 1

已知方程X2+(2+a)x+1+a+b=0的两根为x1,x2.并且0<x1<1<x2,则b/a的取值范围是?
因0<x1<1<x2可知
1+a+b>0,2+a<0
则a<-2,=>0<a+b+1<-2+b+1
=>b>1
则b/a+1+1/a<0(因a<-2)
1/a<-1-b/a
令y=x-1
则(y+1)²+(2+a)(y+1)+a+b+1=0
得y²+(4+a)y+2a+b+4=0
由于x1<1<x2则关于y的二次方程两个根一正一负
则2a+b+4<0
则2+b/a+4/a>0
0<2+b/a+4/a<2+b/a+4(-1-b/a)(已证得1/a<-1-b/a)
=>b/a<-2/3
2+b/a+4/a>0
=>b/a>-2-4/a(因a<-2,-4/a>0)
则b/a>-2
综上b/a∈(-2,-2/3)