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Thread: Solving a cubic equation.

  1. #1
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    Solving a cubic equation.

    I have $\displaystyle f(x)= x^4 + 2x^2$

    and am trying to solve

    $\displaystyle f'(x)= 1$, that is $\displaystyle 4x^3 + 4x= 1$
    Last edited by mr fantastic; Sep 23rd 2010 at 03:55 PM. Reason: Heavily edited for clarity. Re-titled.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    The common factor is 4x. Can you try to factor it?

    Either that, or I'm not understanding what you need to do.

    Do you need to find the values of x for which 4x^3 + 4x = 1 ?
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  3. #3
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    Quote Originally Posted by Unknown008 View Post
    The common factor is 4x. Can you try to factor it?

    Either that, or I'm not understanding what you need to do.

    Do you need to find the values of x for which 4x^3 + 4x = 1 ?
    Yeah, I need to find x.
    But I think that pulling 4x out is wrong.
    $\displaystyle 4x(x^2 + 1) = 1$
    $\displaystyle 4x=1 \quad x^2 + 1 = 1$<------ isn't that illegal or something?
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  4. #4
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    e^(i*pi)'s Avatar
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    Indeed, you can't do that

    You get $\displaystyle 4x^3+4x-1 = 0$

    There aren't any rational factors though

    From wolfram: http://www.wolframalpha.com/input/?i=4x^3%2B4x-1+%3D+0
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  5. #5
    MHF Contributor Unknown008's Avatar
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    Ok, no, that's not possible.

    Bring all the terms on the same side.

    $\displaystyle 4x^3 + 4x= 1$

    $\displaystyle 4x^3 + 4x- 1=0$

    However, this doesn't seem to have three real roots.

    So, we'll go by iteration.

    Set one x as subject of formula.

    $\displaystyle 4x = 1-4x^3$

    $\displaystyle x = \dfrac{1-4x^3}{4}$

    $\displaystyle x_{n+1} = \dfrac{1-4x_n\ ^3}{4}$

    Lets take... x = 0.5

    $\displaystyle x_1 = 0.125$

    $\displaystyle x_2 =0.248$

    $\displaystyle x_3 = 0.245$
    .
    .
    .
    $\displaystyle x \approx 0.236$
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