Find the set of values of k for which this equation has two real roots.
9x^2 + 2kx + k = 0
\(\frac{-k-\sqrt{k^2-9k}}{9}\)
Using dududsu2's idea....
If we have real roots, the discriminant must be ≥ 0
So
(2k)^2 - 4 (9)k ≥ 0
4k^2 - 36k ≥ 0
k^2 - 9k ≥ 0
k ( k - 9) ≥ 0
The solutions for k are (-inf, 0] and [ 9, inf )
