Question

已知关于x的方程x的平方-(m的平方+1)x+m-2=0的两实数跟x1,x2满足(1+x1)(1+x2)=2,求m的值

Answer 1

x^2-(m^2+1)x+m-2=0

x1+x2=m^2+1,x1x2=m-2

(1+x1)(1+x2)=2,

x1+x2+x1x2+1=2

m^2+1+m-2+1=2

m^2+m-2=0

m= -2,或m=1