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Thread: Disguised quadratic equations

  1. #1
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    Disguised quadratic equations

    I'm having troubles solving the following problem.

    x^6 + 5x^3 -24 = 0

    Please post solution.

    Thanks in advance

    Mouseman
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  2. #2
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    Quote Originally Posted by Mouseman View Post
    I'm having troubles solving the following problem.

    x^6 + 5x^3 -24 = 0

    Please post solution.

    Thanks in advance

    Mouseman
    (x^3 + 8)(x^3 - 3) = 0

    x = cubert(-8) or x = cubert(3)
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  3. #3
    A riddle wrapped in an enigma
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    $\displaystyle \sqrt[3]{-8}=-2$
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mouseman View Post
    I'm having troubles solving the following problem.

    x^6 + 5x^3 -24 = 0

    Please post solution.

    Thanks in advance

    Mouseman
    Rewrite this as

    $\displaystyle (x^3)^2+5x^3-24=0$

    Now let $\displaystyle u=x^3$

    giving us

    $\displaystyle u^2+5u-24=0$

    which implies that

    $\displaystyle u=\frac{-5\pm\sqrt{25-4(-24)(1)}}{2}=\frac{-5\pm{11}}{2}$

    So either

    $\displaystyle u=-8$

    or

    $\displaystyle u=3$

    So backsubbing we get

    $\displaystyle x^3=-8\Rightarrow{x=-2}$

    or

    $\displaystyle x^3=3\Rightarrow{x=\sqrt[3]{3}}$

    Note this is very useful trick...for example say you were given

    solve

    $\displaystyle \sin^2(x)+2\sin(3x)-8=0$

    Let $\displaystyle u=\sin(x)$

    do the same thign

    or

    $\displaystyle e^{2x}+10e^x-11=0$

    rewriting this as

    $\displaystyle (e^x)^2+10e^x-11=0$

    let $\displaystyle e^x=u$

    same trick

    Remember that when you solve the u formula with the quadratic formula that you must take the inverse to get the actual answer


    for example

    in the sin(x) one

    lets say after you sub you get

    $\displaystyle u=5$

    then by backsubbing you get

    $\displaystyle \sin(x)=5\Rightarrow{x=arcsin(5)}$

    This is a very useful tool
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