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Math Help - I need help finding the zeros of a function.

  1. #1
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    I need help finding the zeros of a function.

    The function is:

    <br />
4x^4+17x^2+4 = 0<br />

    I have tried to do the following:
    <br />
x^2(4x^2+17) + 4 = 0<br />
    <br />
x^2(4x^2+17) = -4<br />
    <br />
x^2 = -4<br />
    <br />
x = 2i<br />

    <br />
4x^2+17 = -4<br />
    <br />
4x^2 = -21<br />
    <br />
x^2 = -21/4<br />
    <br />
x^2 = sqrt(-21)/2<br />


    But I know I do not have the right answer for there is not sqrt(-21)/2 as one of the answers in the back of the book. Any help would be appreciated.
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  2. #2
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    let x^2 = w then x^4 = w^2

    4w^2+17w+4=0
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  3. #3
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by thyrgle View Post
    The function is:

    <br />
4x^4+17x^2+4 = 0<br />

    I have tried to do the following:
    <br />
x^2(4x^2+17) + 4 = 0<br />
    <br />
x^2(4x^2+17) = -4<br />
    <br />
x^2 = -4<br />
    <br />
x = 2i<br />

    <br />
4x^2+17 = -4<br />
    <br />
4x^2 = -21<br />
    <br />
x^2 = -21/4<br />
    <br />
x^2 = sqrt(-21)/2<br />


    But I know I do not have the right answer for there is not sqrt(-21)/2 as one of the answers in the back of the book. Any help would be appreciated.
    just after the line x=2i how did you get 4x^2+17 = -4??
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  4. #4
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    \displaystyle 4w^2+17w+4=0\Rightarrow w=-4, \ -\frac{1}{4}\Rightarrow x^2=-4, \ -\frac{1}{4}\Rightarrow x=\pm 2i, \ \pm\frac{i}{2}
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  5. #5
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    I get it know:

    <br />
(w+4)(4w+1)<br />
    Then
    <br />
(x^2+4) = 0 or (4x^2+1 = 0)<br />
    <br />
-2i, 2i, -1/2i, 1/2i<br />

    Thanks dwsmith!
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  6. #6
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    It is true that if ab= 0, then either a= 0 or b= 0 but that is a property of 0 only!
    Saying that x^2(4x^2+17) = -4 does NOT imply that " x^2= -4" or that " 4x^2+ 17= -4".

    As dwsmith said, you original equation is a quadratic in w= x^2. By the way, I would be inclined to write those last two roots as (1/2)i and (-1/2)i or even -i/2, i/2. -1/2i and 1/2i are too likely to be interpreted as -1/(2i) and 1/(2i) (which, it suddenly occurs to me, are exactly (1/2)i and (-1/2)i!)
    Last edited by HallsofIvy; January 17th 2011 at 04:18 PM.
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  7. #7
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    Just following up on what HallsofIvy said, here is a specific counterexample:

    ab=-4 has a=2, b=-2 as a solution. Notice that neither a nor b is -4.

    Of course, there are infinitely many other solutions to this equation.
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