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Math Help - How to find the zeros?

  1. #1
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    How to find the zeros?

    Im having trouble finding the zeros of this polynomial:

    -x^2+2x+4

    I have tried factoring it and it dosent work i have tried the quadratic formula too but no luck. I am supposed to graph this but I need the zeros first. Anyone have any suggestions im really lost. Thanks for your help in advance.

    PS. The first term is read as "negative x squared" im not sure how to do the symbols...
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  2. #2
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    Quote Originally Posted by nascar77 View Post
    Im having trouble finding the zeros of this polynomial:

    -x^2+2x+4

    I have tried factoring it and it dosent work i have tried the quadratic formula too but no luck. I am supposed to graph this but I need the zeros first. Anyone have any suggestions im really lost. Thanks for your help in advance.

    PS. The first term is read as "negative x squared" im not sure how to do the symbols...
    -x^2+2x+4 = 0

    a = -1

    b = 2

    c = 4

    x = \frac{-2 \pm \sqrt{2^2 - 4(-1)(4)}}{2(-1)}

    now ... what exactly is your problem in evaluating the above expression that uses the quadratic formula?
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  3. #3
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    Because the quadratic formula produces zeros that have imaginary numbers. But im actually supposed to graph this on a coordinate plane. The original funcion was f(x)=-x^3+4x^2-8. From here i used synthetic division and found one zero to be two. The quadratic i posted was the quotient left from the division.
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  4. #4
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    Because the quadratic formula produces zeros that have imaginary numbers.
    is that so?

    x = \frac{-2 \pm \sqrt{2^2 - 4(-1)(4)}}{2(-1)}

    x = \frac{-2 \pm \sqrt{4 - (-16)}}{-2}

    x = \frac{-2 \pm \sqrt{20}}{-2}

    x = \frac{-2 \pm 2\sqrt{5}}{-2}

    x = 1 \mp \sqrt{5}

    I'd say that x has two real zeros.
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  5. #5
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    I feel stupid I must have gotten lost somewhere in the algebra. thanks so much for your help!

    -Kimberly
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