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I have a very naive question. Is there a closed (analytic) way to write the solution (over $\mathbb{R})$ to the equation: $$Ax^a+Bx+C=0$$ Where $a\in \langle \frac{1}{2},1\rangle$. and where $A,B,C\in\mathbb{R}$. Or at least does anyone know any restrictions for which there exists a solution in closed form (by this I mean that can be found analytically).

Also, I've done some numerics and it also seems to me that this solution should be unique.
Can one prove that (if not what the explicit form of the solution is) that it is at least unique proved it exists and is real?

(I don't know much about this subject but it seems to me that there might be an algebraic geometry-ish kind of solution to this problem (if any).)

El Ruño
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    For $a = 5$ see Bring radical, and for a better known more general situation, google trinomial equation (non-school level hits, such as this AND this AND this). – Dave L. Renfro Mar 13 '22 at 17:40
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    $a\in (\frac{1}{2},1)$ so I don't think the case $a=5$ really helps :/ (or I don't understand how it helps at least). – El Ruño Mar 13 '22 at 17:46
  • Apparently I googled and commented too quickly to notice this! – Dave L. Renfro Mar 13 '22 at 18:37
  • If $x^a$ is the principal value of the power function, there are no negative solutions for $A \neq 0$. We can show that there cannot be more than two non-negative roots by considering the roots of the derivative of $x \mapsto A x^a + B x + C$. A closed form for the smaller root of $x^a - b x - c$ (provided a real root exists) with positive $b$ and $c$ is $$\frac c b \hspace {1.5 px} {1 ! \Psi{\hspace {-1.5 px} 1}} {\left( b c^{1/a - 1} \middle | {(1, \frac 1 a) \atop (2, \frac 1 a - 1)} \right)} - \frac c b.$$ – Maxim Mar 19 '22 at 01:16
  • see e.g. https://math.stackexchange.com/questions/629687/how-to-find-the-zeros-of-an-equation-of-nth-degree/630480#630480 – IV_ Mar 19 '22 at 13:49
  • For $A=0$ or $B=0$, the solution is simple. For $C=0$, we have $x={{\rm e}^{-\ln \left( -{\frac {A}{B}} \right) \left( a-1 \right) ^{-1}}}$. – IV_ Mar 19 '22 at 13:51

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If $a$ is rational (like $a=\frac{3}{4}$) then you can make a substitution (like $y=x^{1/4}$) and the equation becomes a polynomial equation (like $By^4+Ay^3+C=0$).

This polynomial equation will often not admit radical solutions. For example, if $a=\frac45$, you have $By^5+Ay^4+C=0$ which is famously not solvable by radicals for most $A,B,C$.

As for solutions being unique, this is not always the case. For example that last equation with $A=-1$, $B=1$ and $C=0$ is one of the rare cases with radical solutions, and there are two solutions for $y$: $y=0,1$. So two solutions for $x$: $x=0,1$.

2'5 9'2
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