For what real values of $k$ does the quadratic $12x^2 + kx + 27 = 6x^2 + 12x + 18$ have nonreal roots? Enter your answer as an interval.
The quadratic has nonreal roots if and only if its discriminant is negative.
The discriminant of the quadratic ax2+bx+c is b2−4ac.
The discriminant of the quadratic 12x2+kx+27 is k2−4⋅12⋅27=k2−1296.
Therefore, the quadratic has nonreal roots if and only if k2−1296<0.
This inequality is equivalent to (k−36)(k+36)<0.
Therefore, the quadratic has nonreal roots if and only if k<−36 or k>36.
In interval notation, the solution is:
k \in (-\infty,-36) \cup (36,\infty)