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Math Help - x^4+x^3+x^2+x+1 = 0

  1. #1
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    x^4+x^3+x^2+x+1 = 0

    How to solve x^4+x^3+x^2+x+1 = 0
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  2. #2
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    There are not any real solutions. A simple graph of the function will show this.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Garas View Post
    How to solve x^4+x^3+x^2+x+1 = 0
    Note that \displaystyle 1+\cdots+x^4=\frac{1-x^5}{1-x} for x\ne 1 (but x=1 doesn't satisfy you're equation anyways). But, this is zero if and only if 1-x^5=0\implies x^5=1\implies x=1 but we are determined that wasn't the case.
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    I know that there is no real solutions, but that equation have 4 complex solutions x^4+x^3+x^2+x+1 = 0 - Wolfram|Alpha
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Garas View Post
    I know that there is no real solutions, but that equation have 4 complex solutions x^4+x^3+x^2+x+1 = 0 - Wolfram|Alpha
    Ok. Then use what I said, again. x^5-1=0\implies x^5=1\implies x^5=e^{2\pi i}\implies x=e^{\frac{2\pi i}{5}},\cdots,e^{2\pi i}
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  6. #6
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    Quote Originally Posted by Garas View Post
    I know that there is no real solutions, but that equation have 4 complex solutions x^4+x^3+x^2+x+1 = 0 - Wolfram|Alpha
    Then you should have specified that you wanted complex solutions.

    To follow what Drexel was saying,

    \displaystyle x^4 + x^3 + x^2 + x + 1 = \frac{1- x^5}{1 - x} for \displaystyle x \neq 1.

    Therefore, if \displaystyle x^4 + x^3 + x^2 + x + 1 = 0

    \displaystyle \frac{1 - x^5}{1 - x} = 0 for \displaystyle x \neq 1

    \displaystyle 1 - x^5 = 0

    \displaystyle x^5 = 1

    \displaystyle x^5 = e^{2\pi i}

    \displaystyle x = e^{\frac{2\pi i}{5}}.


    There will be 5 solutions, all evenly spaced about a circle, so they will differ by an angle of \displaystyle \frac{2\pi}{5}.

    So the solutions are

    \displaystyle \left\{e^{\frac{2\pi i}{5}}, e^{\frac{4\pi i}{5}}, e^{\frac{6\pi i}{5}}, e^{\frac{8\pi i}{5}}\right\} excluding \displaystyle e^{\frac{10\pi i}{5}} = e^{2\pi i} = 1.
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