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

Question

anonymous · Answer

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).

callisto · Answer

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?

anonymous · Answer

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

anonymous · Answer

an what about me!!

callisto · Answer

Welcome, glad to help!

anonymous · Answer

Hahaha, sorry. Thanks, Eric! (: