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Math Help - 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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    \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

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

    Now let u=x^3

    giving us

    u^2+5u-24=0

    which implies that

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

    So either

    u=-8

    or

    u=3

    So backsubbing we get

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

    or

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

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

    solve

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

    Let u=\sin(x)

    do the same thign

    or

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

    rewriting this as

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

    let 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

    u=5

    then by backsubbing you get

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

    This is a very useful tool
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