maths F.4

The question has missed something in it, I put an “n” to see what happens:
(x – k)[x – (k + n)] = 1
x2 – kx – (k + n)x + k(k + n) = 1
x2 – (2k + n)x + k2 + kn – 1 = 0
D = (2k + n)2 – 4(k2 + kn – 1)
= 4k2 + 4nk + n2 – 4k2 – 4nk + 4
= n2 + 4 >0
So no matter what real values of k and n are, the quadratic equation always has 2 distinct real roots.

2009-08-21 00:29:06 補充：
For (x-k)(x-(k+1))=1, the above calculation is still valid. Just put n = 1. The quadratic equation is x^2 - (2k + 1)x + k^2 +k + 1 = 0.
The discriminant will be(2k +1)^2 - 4(k^2 + k -1) = 5. So the equation always has 2 distinct real roots.