Question

如果方程x2+(m-1)x+m2-2=0两个实数根一个小于-1,另一个大于1,那么实数m的取值范围是多少

Answer 1

x1+x2=1-m,x1x2=m�0�5-2

x1>1,x2>1
x1-1>0,x2-1>0
所以x1-1+x2-1>0
(x1-1)(x2-1)>0

x1-1+x2-1>0
x1+x2-2>0
1-m-2>0
m<-1

(x1-1)(x2-1)>0
x1x2-(x1+x2)+1>0
m�0�5-(1-m)+1>0
m�0�5+m>0
m<-1,m>0

有两个不等实根
(m-1)�0�5-4(m�0�5-2)>0
m�0�5-2m+1-4m�0�5+8>0
3m�0�5+2m-9<0
(1-2√7)/3<m<(1+2√7)/3

所以(1-2√7)/3<m<-1

Answer 2

用韦达定理x1+x2=-b/ax1*x2=c/a再把x1<1 x2>1代换进去就OK了