How to solve $\displaystyle x^4+x^3+x^2+x+1 = 0$

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- Nov 10th 2010, 02:31 PM #1

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- Nov 10th 2010, 02:40 PM #2

- Nov 10th 2010, 02:50 PM #3
Note that $\displaystyle \displaystyle 1+\cdots+x^4=\frac{1-x^5}{1-x}$ for $\displaystyle x\ne 1$ (but $\displaystyle x=1$ doesn't satisfy you're equation anyways). But, this is zero if and only if $\displaystyle 1-x^5=0\implies x^5=1\implies x=1$ but we are determined that wasn't the case.

- Nov 10th 2010, 02:52 PM #4

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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

- Nov 10th 2010, 02:55 PM #5

- Nov 10th 2010, 02:59 PM #6
Then you should have specified that you wanted complex solutions.

To follow what Drexel was saying,

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

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

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

$\displaystyle \displaystyle 1 - x^5 = 0$

$\displaystyle \displaystyle x^5 = 1$

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

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

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

So the solutions are

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