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Math Help - Quadratic

  1. #1
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    Quadratic

    3x^1/3 -16x^1/3 -35=0



    I substituted x^1/3 for u and end up getting stuck at



    3u^2 -16u-35 What do I do from here?
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  2. #2
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    Re: Quadratic

    Quote Originally Posted by daword177 View Post
    3x^2/3 -16x^1/3 -35=0
    I substituted x^1/3 for u and end up getting stuck at
    3u^2 -16u-35 What do I do from here?
    3u^2-16u-35=(3u+5)(u-7)
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  3. #3
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    Re: Quadratic

    That's a quadratic equation (once you add the "= 0") so if you cannot factor, complete the square or use the quadratic formula.
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  4. #4
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    Re: Quadratic

    Quote Originally Posted by daword177 View Post
    3x^1/3 -16x^1/3 -35=0



    I substituted x^1/3 for u and end up getting stuck at



    3u^2 -16u-35 What do I do from here?
    Do you mean

    3x^2/3 -16x^1/3 -35=0

    If so: Your substitution is OK.

    Use the quadratice formula to solve the equation

    3u^2 -16u-35  = 0

    for u.

    Afterwards plug in the values of u into the equation

    \sqrt[3]{x} = u

    and solve for x.
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