I have to show that (1-2k)x^2-5kx-2k=0 has real roots for all integer values of k. I think all you have to do is find the discriminant and then make it a square but I couldn't get it to simplify. Can anyone help?

## discriminant question in homework

### Untouchable

Posted 16 January 2007 - 12:27 AM

For (1-2k)x -5kx-2k=0 to have real roots the discriminant 0 (i.e. positive)

b - 4ac a = 1-2k b = -5k and c = -2k

(-5k) - (4 x (1-2k) x (-2k))

25k -(-8k + 16k )

25k + 8k - 16k

9k + 8k

9k will always be positive and 8k < 9k

9k + 8k will always be positive

discriminant 0

(1-2k)x 5kx-2k=0 has real roots for all integer values of k

b - 4ac a = 1-2k b = -5k and c = -2k

(-5k) - (4 x (1-2k) x (-2k))

25k -(-8k + 16k )

25k + 8k - 16k

9k + 8k

9k will always be positive and 8k < 9k

9k + 8k will always be positive

discriminant 0

(1-2k)x 5kx-2k=0 has real roots for all integer values of k