Question

若方程2(m+1)x2+4mx+3m-2=0的两根同号，则m的取值范围

Answer 1

首先m+1<>0, 即m<>-1
其次delta=16m^2-8(m+1)(3m-2)=-8(m^2+m-2)=-8(m+2)(m-1)>=0, 即-2=<m<=1
再次两根积为正数，即(3m-2)/[2(m+1)]>0, 即 m>2/3 or m<-1
综合得：2/3<m<=1 or -2=<m<-1

Answer 2

2(m+1)x2+4mx+3m-2=0的两根同号
则：x1*x2>0
△>0
根据韦达定理得：
x1x2=(3m-2)/[2(m+1)]>0
解得:m<-1或m>2/3
△=16m²-8(m+1)(3m-2)>0
m²+m-2<0
(m+2)(m-1)<0
-2<m<1
综合得：-2<m<-1或2/3<m<1