Solution to $x^\alpha + p x = q$?

Question

I was wondering if there was any tricks, similar in spirit to the Vieta's substitution, that would apply the equation $$ x^\alpha + p x = q, $$ where $p,q$ and $\alpha$ are real constants. In particular $\alpha$ is not necessarily an integer. The goal is to solve for $x$.

Thanks for your help!

Answer 1

There is no known closed form general solution for a random integer $\alpha$. However, rewriting the

equation as $\color{blue}x=\sqrt[\large\alpha]{q-p\color{blue}x}~$ yields the following formula: $x=\sqrt[\large\alpha]{q-p~\sqrt[\large\alpha]{q-p~\sqrt[\large\alpha]{q-\ldots}}}$

Answer 2

Let me choose a for alpha to make notation easier

x^a = -px+q

(a)LN( x) = LN(-px+q)

x = e^{LN(-px+q)}/a

hope that helps