Solve for $x$

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.

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

Consider the identity


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


Now, the equation becomes,


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



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

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

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

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


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

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

  • 3
    $\begingroup$ That's just amazing!! $\endgroup$
    – user196761
    Dec 13 '14 at 16:10
  • 4
    $\begingroup$ Wow! How do you think of such things!? $\endgroup$
    – Henry
    Dec 13 '14 at 16:15
  • 5
    $\begingroup$ Well, he is a Math God. . . $\endgroup$
    – HDE 226868
    Dec 13 '14 at 16:15
  • $\begingroup$ @KierenMacMillan Can you tell me what solution does maxima give? $\endgroup$
    – MathGod
    Jun 27 '15 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 '15 at 16:09

you will get $$(x-1)(x^2+x+1)(x^4-4x-1)=0$$ and you can solve your problem

  • 5
    $\begingroup$ 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 '14 at 16:12

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