$x^4 + 4x^3 - 2x^2 - 12x + k$ has 4 real roots. Find the condition on k. The question is: $f(x) = x^4 + 4x^3 - 2x^2 - 12x + k$ has 4 real roots. What values can k take? Please drop a hint!

Hint:
Lets call vieta for the rescue.
$f(x) = x^4 + 4x^3 - 2x^2 - 12x + k=0$, let $x_1, x_2,x_3$ and $x_4$are the roots.
$\sum x_i=-4$
$\sum x_ix_j=-2$
$\sum x_ix_jx_k=+12$
$x_1x_2x_3x_4=k$
$\sum (x_i)^2=(\sum x_i)^2-2\sum x_ix_j=16-(-4)=12$
$\dfrac{x_1^2+x_2^2+x_3^2+x_4^2}{4} \ge (x_1x_2x_3x_4)^\frac{1}{2}$
$3 \ge k^{\frac{1}{2}} \implies 9 \ge k$. Now check the lower bound using the discriminant formula.
HAIL VIETA!, like I always say.
