求证:不论k为何值时,方程(x-1)(x-k)=4有两个不相等的实数根

(x-1)(x-k)=4
x²-kx-x+k-4=0
x²-(k+1)x+(k-4)=0
b²-4ac=[-(k+1)]²-4×1×(k-4)
=k²+2k+1-4k+16
=k²-2k+1+16
=(k-1)²+16
≥16＞0
∴不论k为何值时,方程(x-1)(x-k)=4有两个不相等的实数根