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I have the equation $f(x) = \frac{1}{3 a^2 b^2} × x^9 - \frac{1}{2} × x^5 + \frac{1}{12} (a^2 + b^2 - 24 c d) × x^3 - a b c d (c + d - e)$, where $a$, $b$, $c$, $d$, and $e$ are non-negative real numbers. I want to find the smallest non-negative real solution to $f(x) = 0$.

I know that in general, polynomials of degree five and higher are not solvable algebraically. However, I also know that there are specific cases where high-degree polynomials are solvable. My reading suggests that identifying whether or not my equation is solvable involves Galois theory, but explanations of Galois theory that I've found online are going way over my head. WolframAlpha has not been helpful either.

Can my specific equation be solved?

Lawton
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    s̶o̶l̶v̶a̶b̶l̶e̶ ̶a̶l̶g̶e̶b̶r̶a̶i̶c̶a̶l̶l̶y̶ solvable in radicals. – NDB Aug 02 '23 at 16:15
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    @Lawton: Even if you combine constants and rewrite it as $$f x^9 -\dfrac{x^5}{2}+g x^3-h=0$$ There is no easily representable solution. – Moo Aug 02 '23 at 17:27
  • Galois theory is beautiful, but it (usually) does not give simple answers to this type of questions about user prescribed families of polynomials. You may find this disappointing, but them's the breaks. I could try and build examples within your family (you really should specify the coefficients in the way Moo did above), but let's go with quintics instead as examples are easier to locate. For example here it is explained why $x^5-5x+12=0$ is solvable. OTOH we can show that $x^5-5x+13=0$ is not solvable in this precise sense. – Jyrki Lahtonen Aug 08 '23 at 05:35
  • (cont'd) So an answer might read that for choices of parameters $f,g,h$ in this or that family yield a solvable polynomial whereas other choices make it unsolvable. I dare not speculate whether such families exist in your case. – Jyrki Lahtonen Aug 08 '23 at 05:37
  • May be this gives a better ide of the type of results you can expect from Galois theory? Still about quintics. – Jyrki Lahtonen Aug 08 '23 at 05:44
  • @JyrkiLahtonen There is a broad family when it is solvable in radicals, namely $c+d-e=0$ which reduces the nonic to a sextic with even powers, though I don't know if this is useful to the OP. – Tito Piezas III Aug 16 '23 at 05:20

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Irreducible polynomials with degree $n\geq5$ that are solvable in radicals are relatively rare. If they are also sparse polynomials (with few terms) but non-binomial, then they are even rarer. You have a nonic quadrinomial of form,

$$c_1x^9+c_2x^5+c_3x^3+c_4 = 0$$

Specifically,

$$\tfrac{1}{3 a^2 b^2}x^9 - \tfrac{1}{2}x^5 + \tfrac{1}{12} (a^2 + b^2 - 24 c d)x^3 - a b c d (c + d - e)=0$$

Its discriminant is very messy and from it alone, the chances that it is solvable for general $(a,b,c,d,e)$ is virtually zero. However, there is a broad family when $cd(c+d-e) = 0$ and your nonic reduces to a sextic with only even exponents, hence is solvable. (Whether there is a second family is unknown.)

P.S. There is a solvable quadrinomial family of form,

$$x^9-3uvw\,x^\color{red}4+u^3w\,x^3+v^3w^2 = 0$$

On the slim chance you made a typo (that your second exponent is a 4 and not a 5), then there are infinitely many $(a,b,c,d,e)$, but not all, such that the equation is solvable.

Tito Piezas III
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