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Math Help - Solving an equation

  1. #1
    Junior Member
    Joined
    Jul 2007
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    Solving an equation

    I'm not getting this one.

    3z^4 - 5z^2 +1 = 0

    If I substitute variable u in place of z^2, I get

    3u^2 - 5u + 1 = 0

    solving with the Quadratic Equation gives me

    u= 5 \pm \frac{\sqrt13}{6}

    and since u = z^2, the solutions should be

    z = \pm \sqrt{5 \pm \frac{\sqrt13}{6}}

    But, the book says the solution is

    z = \pm \frac{ \sqrt{30 \pm 6\sqrt13}} {6}

    What have I done wrong?
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  2. #2
    MHF Contributor red_dog's Avatar
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    Medgidia, Romania
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    You made a mistake.
    u=\frac{5\pm\sqrt{13}}{6}.
    Then z=\pm\sqrt{\frac{5\pm\sqrt{13}}{6}}=\pm\frac{\sqrt  {30\pm 6\sqrt{13}}}{6}
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  3. #3
    Junior Member
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    D'oh! It's amazing, you think you're getting it and then go and do a stupid thing like that.
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