B Why is k not smaller than zero?

1. Jul 4, 2016

Clever Penguin

Question:

The equation kx2+kx+3-k = 0, where k is a constant, has no real roots

a) Show that 5k2-12k < 0
b) Find the set of possible values of k

Solution attempted:
a) For no real roots, b2-4ac < 0

a = k
b = k
c = 3 - k

k2 - 4k(3 - k) < 0
k2 -12k + 4k2 < 0
5k2 - 12k < 0
Hence shown

b) 5k2 - 12k < 0
k(5k-12) < 0
So k < 0 or (5k-12) < 0, k< 12/5, k<2.4

So k < 2.4

But the answer says 0 < k < 2.4

Why?

Last edited: Jul 4, 2016
2. Jul 4, 2016

micromass

Your $b$ should be squared.

3. Jul 4, 2016

Clever Penguin

Done.
I'm self-studying maths, and it doesn't say why in the book.

4. Jul 4, 2016

Staff: Mentor

If k<0, then k(5k-12) < 0 cannot be true because both factors are negative.

5. Jul 4, 2016

Clever Penguin

Because if k<0, (5k-12) would be positive
got it