# Solving a non-integer polynomial where the exponent is greater than one

I'm trying to solve an equation of the form:

$ax + bx^{1+c} + d = 0$, where $0 < c < 1$, and the reciprocal of $c$ is not necessarily an integer either.

Mathematica protests to me that it is not up to the task of solving this, and I'd like a general solution rather than FindRoot at a particular value.

I've poked around and there are a couple work-arounds - usually involving some substitution - for solving a non-integer polynomials but (as far as I can tell - because I'm having trouble applying them to this problem) they all only really seem to work if the non-integer polynomial is less than one and this is definitely more than one.

I've come to the point of getting a derivative w.r.t. to one of the parameters of interest with the implicit function theorem, which is fine for my purposes, but obviously getting a general solution would be preferable.

Is there a general solution for this?

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Already when the reciprocal of $c$ is an integer there is no expression for the solution(s) in terms of elementary functions. –  André Nicolas Aug 24 '12 at 17:38
Nitpick: This is not a polynomial unless $c$ is an integer. –  Jyrki Lahtonen Aug 28 '12 at 17:38

If $|b/a|$ and $|d/a|$ are small, you can solve it in series. For convenience, take $a=1$. Then $$x = -d + \sum_{n=0}^\infty \frac{\Gamma(nc+n+1)}{n! \Gamma(nc+2)} (-d)^{nc+1} (-b)^n$$

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No, especially in radicals. First of all, if there was a general formula, then it would work for all $c\in\mathbb R$, not just those values in your interval. But if you take $c=\frac 1{n-1}$, you get $bx^\frac n{n-1}+ax+d=0$. Applying the transformation $x=z^{n-1}$ yields $bz^n+az^{n-1}+d=0$. For $n\geq 5$, there is no general solution because of the Abel-Ruffini Theorem.

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