Question

For what real values of does the quadratic 12x^2+k*x+27=0 have nonreal roots

Answer 1

Answer:

If a quadratic has nonreal roots, then the discriminant .  b^2 - 4ac

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 < *12*27

Now factorize 4*12*27 and solve to get - 36 < k<26

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

Discriminant : k² - 4(12)(27) < 0

So k² - 1296 < 0, -36 < k < 36.