0
$\begingroup$

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!

$\endgroup$
  • 5
    $\begingroup$ This isn't a polynomial equation if one of the terms is $w_3x^{1-1/\alpha}$ $\endgroup$ – Peter Foreman Jul 27 at 22:34
  • $\begingroup$ For what are you wanting to solve? $\endgroup$ – William Elliot Jul 28 at 0:48
  • $\begingroup$ Is $\alpha$ a rational number? $\endgroup$ – hardmath Jul 28 at 1:47
  • $\begingroup$ @hardmath If we could solve it for the rational case, that'd be good, but in general, $\alpha$ is a positive real number. $\endgroup$ – AlwaysNeedHelp Jul 28 at 9:09
  • 1
    $\begingroup$ 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. $\endgroup$ – IV_ Jul 28 at 12:05
3
$\begingroup$

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).

$\endgroup$
  • $\begingroup$ @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. $\endgroup$ – hardmath Jul 29 at 0:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.