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.

parmen Nov 12, 2023

#1**0 **

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)

bader Nov 12, 2023