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Solve for $x$
$$x^7-5x^4-x^3+4x+1=0$$

This equation has been bugging me since the past few days. I have found, using the Rational Root Theorem that $x=1$ is a root of this equation. However, after dividing, I cannot solve the six degree equation thus generated. I have also tried factorizing the equation, but it's not working.

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  • $\begingroup$ Out of curiosity, what was the genesis of this equation, where did you run across it? $\endgroup$ Dec 14, 2014 at 17:08

3 Answers 3

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Consider the identity

$(x^4-4x-1)^2=x^8-8x^5-2x^4+16x^2+8x+1$

Differentiating both sides w.r.t. $x$, we get,

$x^7-5x^4-x^3+4x+1=(x^4-4x-1)(x^3-1)$

Now, the equation becomes,

$(x^4-4x-1)(x^3-1)=0$

$\implies x=1,\omega, \omega^2$ (where $\omega$ is a non real cube root of unity)

or

$x^4-4x-1=0$

$\implies x^{4}+2x^{2}+1=2x^{2}+4x+2$

$\implies (x^{2}+1)^{2}=2(x+1)^{2}$

$\implies \{x^{2}+1+\sqrt{2} (x+1)\}\cdot \{x^{2}+1-\sqrt{2} (x+1)\}=0$

$\implies x=\dfrac{-\sqrt{2} \pm i\sqrt{2+4\sqrt{2}}}{2}$

or

$x=\dfrac{\sqrt{2} \pm \sqrt{4\sqrt{2}-2}}{2}$

$\therefore x=1,\omega,\omega^2,\dfrac{-\sqrt{2} \pm i\sqrt{2+4\sqrt{2}}}{2},\dfrac{\sqrt{2} \pm \sqrt{4\sqrt{2}-2}}{2}$

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  • 3
    $\begingroup$ That's just amazing!! $\endgroup$
    – user196761
    Dec 13, 2014 at 16:10
  • 4
    $\begingroup$ Wow! How do you think of such things!? $\endgroup$
    – Henry
    Dec 13, 2014 at 16:15
  • 5
    $\begingroup$ Well, he is a Math God. . . $\endgroup$
    – HDE 226868
    Dec 13, 2014 at 16:15
  • $\begingroup$ @KierenMacMillan Can you tell me what solution does maxima give? $\endgroup$
    – MathGod
    Jun 27, 2015 at 15:09
  • $\begingroup$ Apologies. After doing some arithmetic, the maxima solution is the same as yours. My fault. I deleted my comment, and upvoted your answer. $\endgroup$ Jun 27, 2015 at 16:09
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you will get $$(x-1)(x^2+x+1)(x^4-4x-1)=0$$ and you can solve your problem

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    $\begingroup$ I believe it will be more helpful, if you show, how you get this form. (And no, I didn't down-vote.) $\endgroup$
    – Tacet
    Dec 13, 2014 at 16:12
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$x^4(x^3 + 1) + (x^3 + 1) = 0$

$\implies(x^3 + 1)(x^4 + 1) = 0$

$\implies(x + 1)(x^2 - x + 1)(x^4 + 2x^2 + 1 - 2x^2) = 0$

$\implies(x + 1)(x^2 - x + 1)((x^2 + 1)^2 - 2x^2) = 0$

$\implies(x + 1)(x^2 - x + 1)(x^2 + 1 + (\sqrt 2)(x))(x^2 + 1 - (\sqrt 2)(x)) = 0$

$ x^2 - x + 1 = 0$

$\implies x = \frac{1 \pm \sqrt{1 - 4\times1}}2 = \frac{1 \pm i \sqrt 3}2$

$ x^2 + \sqrt 2x + 1 = 0$

$\implies x = \frac{-\sqrt 2 \pm \sqrt{2 - 4 \times 1}}2 = \frac{-\sqrt 2 \pm i \sqrt 2}2$

$ x^2 - \sqrt 2x + 1 = 0$

$\implies 4x = \frac{\sqrt 2 \pm \sqrt{2 - 4 \times 1}}2 = \frac{\sqrt 2 \pm i sqrt 2}2$

The $7$ roots of $x$ are:

$x = -1, \frac{1 \pm i \sqrt 3}2, \frac{-\sqrt 2 \pm i \sqrt 2}2, \frac{\sqrt 2 \pm i \sqrt 2}2$

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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Nov 20, 2022 at 17:51

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