已知方程组kx平方-x-y+1/2=0 y=k(2x-1)有两个不同的实数解

kx^2-x-y+1/2 = 0 (1)
y=k(2x-1) (2)
Sub (2) into (1)
kx^2-x-k(2x-1) +1/2 = 0
kx^2 -(2k+1)x + (k+1/2) =0
△>0
=> (2k+1)^2 - 4k(k+1/2) > 0
4k^2+4k+1 - 4k^2 - 2k>0
2k+1>0
k> -1/2