Question

The equation kx^2 + 2kx - 4 + k = 0 has no real roots. Find the range of values of k.

Answer 1

For exactly two real roots, k > sqrt(8).

The equation to be solved is x = (-k +- sqrt(k^2 - 4*2))/2

2 real roots occur when the discriminant (sqrt(k^2 - 4*2)) is larger than 0, thus k^2 > 8, so, k > sqrt(8).

Answer 2

Since it has no real roots
Δ<0
(b^2 -4ac) <0
Put b = 2k, a= k, and c = -4+k into the inequality
(2k)^2 - 4(k)(-4+k) <0
4k^2 + 16k - 4K^2 <0
Can you solve it from here?

Answer 3

Thank you, guys! (: Yes, thanks, Callisto!

Answer 4

an what about me!!

Answer 5

Welcome, glad to help!

Answer 6

Hahaha, sorry. Thanks, Eric! (: