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
1 answer:
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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