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I am a student of physics, stuck with the general solution of an algebraic equation of the form: $$x^{n+2} + x^2 =f$$ I asked Mathematica, she couldn't solve it.

Is there any way to get a general solution for this type of equation (may be in terms some special function) in Mathematics literature?

Please forgive me, if the question is too naive!

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  • $\begingroup$ You can't solve it in general. If $n$ is odd, then let $y=x^2$, which simplifies it. Else, resort to numerical methods. $\endgroup$ Apr 24, 2017 at 20:10
  • $\begingroup$ @SimplyBeautifulArt Shouldn't that be when $n$ is even? Still, that won't give a general solution. $\endgroup$
    – wythagoras
    Apr 24, 2017 at 20:12
  • $\begingroup$ @wythagoras Oops, yes, $n$ should read as even. $\endgroup$ Apr 24, 2017 at 20:16
  • $\begingroup$ But, $y=x^2$, doesn't seem to do much help. or maybe I am not seeing it! $\endgroup$
    – Archimedes
    Apr 24, 2017 at 20:17
  • $\begingroup$ Well, it might simplify the problem down to a solvable one, though usually not. $\endgroup$ Apr 24, 2017 at 20:22

2 Answers 2

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If $n$ is small, you can use the usual polynomial tricks. If $n,x$ are large enough $x^{n+2}$ will be much larger than $x^2$. A good approximation to the solution will then be $x=f^{1/(n+2)}$. You can then do fixed point iteration using $x_{i+1}=(f-x_i^2)^{1/(n+2)}$ and iterate to convergence.

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If you factor out $x^2$, then if $n$ is odd, $x=0$,$x=-1$ and if $n$ is even $x=0$, $x=i$.

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    $\begingroup$ assuming you want f=0 $\endgroup$
    – Jack
    Apr 24, 2017 at 20:32

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