What value of k will have no real solution in kx^2+8x+5=0?

5 Answers

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  • Como
    Lv 7
    3 weeks ago

    :-

    64 < 20 k

    k > 16 / 5

  • 3 weeks ago

    To have "no real solution" for x, you need

    b^2 < 4ac, so

    64 < 20k, or

    k > 3.2.

  • David
    Lv 7
    3 weeks ago

    No real solutions if k is greater than 3 as an integer because the value of the discriminant of b^2 -4ac will be less than 0.

  • Ian H
    Lv 7
    3 weeks ago

    No real if 64 < 4*k *5, so, k > 3.2

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  • 3 weeks ago

    No real solutions will occur when the discriminant is less than 0.

    b^2 - 4ac < 0

    (8)^2 - 4(k)(5) < 0

    64 - 20k < 0

    -20k < -64

    20k > 64

    k > 64/20

    k > 32/10

    k > 16/5 <== answer

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