Question

已知x1,x2是一元二次方程ax2+x+1=0的两个实根，若(x1-1)(x2-1)<0,求a的范围

Answer 1

(x1-1)(x2-1)=x1*x2-(x1+x2)+1
因为x1+x2=-1/a
x1*x2=1/a
代入x1*x2-(x1+x2)+1整理得
2/a +1<0
解得-2<a<0
又有△=1-4a>0得a<1/4
综上可得-2<a<0

Answer 2

x1+x2=-1/a
x1*x2=1/a

1）(x1-1)(x2-1)<0
x1*x2-(x1+x2)+1<0
1/a-(-1/a)+1<0
2/a +1<0
2/a<-1
-2<a<0

2）△=1-4a≥0
a≤1/4
所以a的范围为(-2,0)