Solving Equation of Degree n, where n is any value between 1 and 2

Question

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?

Answer 1

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!

Answer 2

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,

Answer 3

$$ax^n+bx+c=0\iff x^n=-\frac ca-\frac bax\iff x=\sqrt[^n]{-\frac ca-\frac bax}=\sqrt[^n]{-C-Bx}\iff$$

$$x=\sqrt[^n]{-C-B~\sqrt[^n]{-C-B~\sqrt[^n]{\ldots}}}$$