Solution to equation (is it possible?)

After stumbling through some dense fog of algebra, I have come across the following problem - any help would be much appreciated!

I'm currently trying to solve the following equation, but my efforts to date have been inconsequential. I was wondering whether anyone had some suggestions on how to proceed with solving the following equation for $$x$$? Clearly, in the case where $$w_3 = 0$$, one can easily solve a quadratic equation, but, if possible, I'd ideally like a general method for solving the equation rather than thinking about this on a case by case basis (since the $$w_i$$ are themselves functions of many parameters which ideally shouldn't be restricted to special cases).

The equation I'm trying to solve for $$x$$ is: $$a_1x^2 + a_2x + a_3x^{1-\frac{1}{\alpha}} + a_4 = 0$$

where $$a_1, a_2, a_3, a_4 > 0$$, and $$\alpha > 0$$ is a fixed parameter.

Thank you!

• This isn't a polynomial equation if one of the terms is $w_3x^{1-1/\alpha}$ – Peter Foreman Jul 27 at 22:34
• For what are you wanting to solve? – William Elliot Jul 28 at 0:48
• Is $\alpha$ a rational number? – hardmath Jul 28 at 1:47
• @hardmath If we could solve it for the rational case, that'd be good, but in general, $\alpha$ is a positive real number. – AlwaysNeedHelp Jul 28 at 9:09
• If $a=1$, you have a quadratic equation. In the general case, it is a transcendental equation. Because it depends on $x$ and $e^{(1-\frac{1}{a})\ln(x)}$ which are algebraically independent, the equation cannot be solved by transforming it by elementary operations you can derive from the equation. – IV_ Jul 28 at 12:05

For arbitrary (irrational) $$\alpha$$ there is an obstacle to solving the equation beyond the mere fact that it is not a polynomial equation. The coefficients $$w_1,w_2,w_3,w_4$$ are assumed to be positive, so that there are no changes-in-sign in the equation and Descartes' rule (generalized) tells us there are no positive roots.
But on the other hand $$x^{1-\frac{1}{\alpha}}$$ is only well-defined for positive $$x$$ when $$\alpha \in (0,1)$$ is irrational. So putting these two observations together would say there is not even a possibility of numerical approximation of a root (no positive root exists and it is unclear what complex root might be meaningful).
• @IV_: The response given by OP to my asking about rationality of $\alpha$ suggests an interest in solutions for a continuous parameter rather than for an isolated rational value. So I did not pursue the idea of a related polynomial equation further. – hardmath Jul 29 at 0:10