+1768 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.
+129883 Simplify as
6x^2 + ( k - 12)x + 9 = 0
We have non-real roots when the discriminant > 0
So
(k -12)^2 - 4(6)(9) < 0
(k -12)^2 - 216 < 0
(k -12)^2 < 216 take both roots
k - 12 < sqrt (216) and k - 12 > -sqrt (216)
k < 12 + 6sqrt (6) k > 12 - 6 sqrt (6)
k on this interval → ( 12 -6sqrt(6) , 12 + 6sqrt (6) )
