(1)/(4p)(x-h)^(2)+k=0 Multiply the equation by 4p. Explain how different values of k affect the number of zeros of the polynomial. Con

Question

(1)/(4p)(x-h)^(2)+k=0
Multiply the equation by 4p. Explain how different values of k affect the number of zeros of the polynomial. Consider k > 0, k = 0, and k < 0. Assume p > 0.​

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Đan Thu 1 year 2021-09-04T01:09:51+00:00 2 Answers 34 views 0

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    0
    2021-09-04T01:11:10+00:00

    Answer:

    Step-by-step explanation:

    Given p>0, multiply the equation by 4p:  (1/4p)*(x-h)^(2)+k=0

    (x-h)^2+4kp = 0

    k>0

    4kp>0

    (x-h)^2 = -4kp

    So x has imaginary roots only. There is no real zeros of the polynomial.

    k=0

    4kp=0

    (x-h)^2 = 0

    x=h

    So x has one real root and the polynomial has one zero

    k<0

    4kp<0

    (x-h)^2 = -4kp

    So x has two real roots and the polynomial has two real zeros.

    0
    2021-09-04T01:11:19+00:00

    Answer:

    Step-by-step explanation:

    Assume that  and multiply the equation  by 4p. Then you obtain the equation (x-h)^2+4pk=0.

    1) If k>0, then 4pk>0 and the equation does not have real solutions and there is no zero.

    2) If k=0, then 4pk=0 and . There is one solution x=h and there is one zero.

    2) If k<0, then 4pk<0 and the equation  has two different solutions and there are two zeros.

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