Question

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

Answer 1

C. 0

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