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Does $x^a + bx + c = 0$ have analytic solution given $a \in \mathbb{R}, a > 1$ and $x > 0$? The case that I am interested also has $b < 0$ and $c < 0$. Searches have led me only to the quadratic formula, and to the the form $a^x + bx + c$, so far.

Slightly off-topic but I landed there when trying to show the 80-20 property of Pareto distribution.

Ott Toomet
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  • I guess you mean $x$ will be an analytic function of $a$, $b$, and $c$, right? – Robert Lewis Oct 28 '17 at 04:35
  • If $a$ is rational then $x^a+bx +c$ can be written as a polynomial in $x^{a-1}$ and there will be usual polynomial results. – abiessu Oct 28 '17 at 04:48
  • $x^5 - x - 1 = 0$ has a single root, but this root cannot be expressed through radicals. There's no quintic formula. – Theo Bendit Oct 28 '17 at 04:50
  • If you mean a general formula to calculates the roots, the answer is no for n= or >5 due to Abel. But you can always use: $x^a +bx+c=(x-d)(x-e)(x-f) . . . to (a) terms$. Then you have to solve a system of Diophantine equations. – sirous Oct 28 '17 at 09:11
  • Thank you for all the comments :-) I forgot to mention that a is real (rational will do as well) but I don't think it changes the main message. – Ott Toomet Oct 29 '17 at 04:45

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