Question

已知关于x的方程(1-2m)x²-2(√m+1)x-1=0有两个不相等的实数根,求m范围.

Answer 1

(1-2m)x²-2(√m+1)x-1=0

a=1-2m ,b=-2√(m+1) ,c=-1

△=b²-4ac

=[-2√(m+1)]²-4×(1-2m)×(-1)

=4(m+1)+4-8m

=4m+4+4-8m

=-4m+8

由题意,可知△﹥0 且 1-2m≠0

-4m+8﹥0 且 m≠1/2

m﹤2 且 m≠1/2