10
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Out of curiosity, what was the genesis of this equation, where did you run across it? $\endgroup$ – Steven Stadnicki Dec 14 '14 at 17:08
19
$\begingroup$

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

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

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

$\endgroup$
  • 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. $\endgroup$ – Kieren MacMillan Jun 27 '15 at 16:09
3
$\begingroup$

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

$\endgroup$
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.