Why is k not smaller than zero?

[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?

 

[micromass]

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

k - 4k(3 - k) < 0

Your $b$ should be squared.

 

[Clever Penguin]

Your $b$ should be squared.

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

 

[mfb]

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

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

 

[Clever Penguin]

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

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