Solving Equation of Degree n, where n is any value between 1 and 2 How does one solve an equation of the form:
$$ax^n + bx + c = 0$$
where n is a non integer value between 1 and 2.
Is there a formula to provide an analytic solution?
 A: I don't have a formal proof that it doesn't have any analytic solutions but here are excellent reasons to think so :
First you might want to think, is $n$ a rational or an irrational?
Case 1: $n\in \mathbb Q$ hence $n = \frac pq$, and your equation can be written as :
$$ax^{p/q} + bx + c = 0\Longrightarrow x^{p/q} = \frac{bx + c}{a} \Longrightarrow x^p = a^{-q}(bx+c)^q$$
Which is a $p$ degree polynomial ($\frac pq \ge 1$). For $p \ge 5$ this does not normally have an analytic solution although I admit it has a special structure which could allow for some solutions. Since you can take any rational, then you would need a formula which works for any $p$ and $q$.
So the rational case doesn't look very promising. What if it's irrational?
Case 2: $n\in \mathbb R / \mathbb Q$ this is almost ridiculous. Take $n = \frac \pi e$ then you need to solve the following equation :
$$x^\pi = a^{-e}(bx + c)^e, \qquad e = \exp(1)$$
I never heard of such a theory so I hardly think that there is any known analytic solution for this very specific case. You need to solve it for all possible irrationals.
In order to have a general analytic solution, you need it to satisfy all the possible cases, reason I strongly believe it does not exist.
So how do you solve it? Well I would say numerically is a good way!
A: $$ax^n+bx+c=0\iff x^n=-\frac ca-\frac bax\iff x=\sqrt[^n]{-\frac ca-\frac bax}=\sqrt[^n]{-C-Bx}\iff$$
$$x=\sqrt[^n]{-C-B~\sqrt[^n]{-C-B~\sqrt[^n]{\ldots}}}$$
A: Solving
$$
a x^r + b x + c = 0
$$
for $r \in (1, 2)$ is equivalent to finding a fixed point $x^*$ thus 
$$
f(x^*) = x^*
$$ 
of the function
$$
f(x) = a x^r + (b+1) x + c
$$
Depending on the coefficients $a, b, c$ IMHO that seems to allow zero, one, two or infinite many solutions (last case if $a = b = c = 0$) to me.
For arbitrary coefficients and choice of exponent $r$ I would be surprised if there was a general analytic solution,
