How to solve $y=2x^3+x$ for $x$

Question

I'm doing something that has throw out: $$y=nx^{2n-1}+x$$ and do make any progress with this problem I needed to make x the subject. Correct me if I'm wrong but this seems impossible to work with for any values other than $n=0,1,2$.

$n=2$ gives $y=2x^3+x$, and putting that into wolfram alpha gives a very messy:

But I have no idea how I'd have gotten with this without wolfram alpha, and can't seem to find anything helpful online. Any kind of help on this would be appreciated.

*edit: forget to mention I only care about real values of x and y

It's messy for what I'm doing with it afterwards. Also, not exactly in that form: $1/6^{2/3}$ & $1/6^{1/3}$, but I get your point — Banbadle, Jan 05 '16 at 21:10
It's the best you are going to get as a general expression. There may be better formulas depending on what you are doing with it afterwards. — Justpassingby, Jan 05 '16 at 21:24

score 3 · Answer 1 · answered Jan 05 '16 at 21:05

3

The general cubic equation has a "closed form" solution and this is probably what it looks like for your particular equation. Wikipedia gives the general solution in the article "cubic function".

Starting with the fifth degree ($n=3$) even that is no longer guaranteed and you are looking at numerical solution methods.

answered Jan 05 '16 at 21:05

Justpassingby

10,029

I know there's no guarantee of algebraic solutions to 5th powers and above, that's why i thought $n=2$ is a good stopping point.
The only things I could find online were the roots to a cubic, not a cubic equal to a variable. For the life of me I couldn't/cannot find a thing about it on wikipedia
– Banbadle Jan 05 '16 at 21:16
But the whole point of 'solving for $x$' is that you consider $y$ to be given, so it should be treated as a constant term in a polynomial equation. – Justpassingby Jan 05 '16 at 21:19
I phrased it poorly in the post title. I mean to make x the subject – Banbadle Jan 05 '16 at 21:21
I liked the first formulation better, but it comes down to the same thing: $y$ is given and $x$ has to be found. Either way you are looking for the zeroes of the polynomial (in $X$) $nX^{2n-1}+X-y.$ – Justpassingby Jan 05 '16 at 21:22
So it's the cubic formula where d=y? facepalm I'm so stupid sometimes. thanks so much – Banbadle Jan 05 '16 at 21:26

score 0 · Answer 2 · answered Jan 06 '16 at 02:30

The equation $2x^3+x-y=0$ shows only one real root. If you look here, you will notice that, in such a case, the is also an hyperbolic method; applied to your case $(p=\frac 12$, $q=-\frac y2)$, the solution write $$x=\sqrt{\frac{2}{3}} \sinh \left(\frac{1}{3} \sinh ^{-1}\left(3 \sqrt{\frac{3}{2}} y\right)\right)$$ which I personally find much nicer than the one Wolfram Alpha gave (it corresponds to the real root obtained using Cardano method).

How to solve $y=2x^3+x$ for $x$

2 Answers2