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For what real values of k does the quadratic 12x^2 + kx + 27 = 0 have nonreal roots? Enter your answer as an interval.

 Jun 20, 2020
 #1
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If a quadratic has nonreal roots, then the discriminant \(b^2-4ac<0\).

Plug in the values for b, a, and c into this equation to get \(k^2-4*12*27<0\)

Rearrange to get \(k^2<4*12*27\)

Now factorize 4*12*27 and solve to get \(-36 < k < 36\)

 Jun 20, 2020
edited by thelizzybeth  Jun 20, 2020
edited by thelizzybeth  Jun 20, 2020
edited by thelizzybeth  Jun 20, 2020
edited by thelizzybeth  Jun 20, 2020
 #2
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Deleted

 Jun 20, 2020
edited by thelizzybeth  Jun 20, 2020
edited by thelizzybeth  Jun 20, 2020

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