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Use the discriminant to determine how many real number solutions exist for the quadratic equation –4j2 + 3j – 28 = 0. A. 3 B. 2 C. 0 D. 1

Respuesta :

wegnerkolmp2741o

Answer:

C. 0

Step-by-step explanation:

–4j^2 + 3j – 28 = 0

The discriminant is b^2-4ac  if >0 we have 2 real solutions

                                                   =0  we have 1 real solutions

                                                   <0 we have 2 imaginary solutions

a = -4, b =3  c = -28

b^2 -4ac

(3)^2 - 4(-4)*(-28)

9 - 16(28)

9 -448

This will be negative so we have two imaginary solutions.

Therefore we have 0 real solutions

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