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5 November, 06:27

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

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  1. 5 November, 10:07
    0
    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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