TITLE: How to solve for the roots of an equation of two variables ? For example $\frac{4L^6-19L^4x}{9}+\frac{8L^2x^2}{3}-x^3$.
QUESTION [5 upvotes]: I want to solve for the roots of an equation which has two variables in it. For example:
$$24L^2x^2+ 4L^2x-9x = 0$$
This is just an example which I made up. Mine is a bit more complicated. I am not used dealing with this sort of thing, so is there any method of solving for the roots of this equation?
For a real case I have complicated equations such as:
$$\frac{4L^6-19L^4x}{9}+\frac{8L^2x^2}{3}-x^3$$
Thanks in advance.

REPLY [5 votes]: Multiply by $9\;$:
$$4L^6-19L^4x+24L^2x^2-9x^3 = 0$$
If $L=0$ then the equation reduces to $x^3=0$, otherwise divide by $L^6\,$:
$$4-19\frac{x}{L^2}+24\frac{x^2}{L^4}-9\frac{x^3}{L^6} = 0$$
Let $\;\displaystyle y = \frac{x}{L^2}\,$, then the equation can be written as:
$$4-19 y + 24 y^2 - 9 y^3 = 0$$
The latter has the obvious root $y=1 \iff x = L^2$ (or see the rational root theorem if not obvious upfront), and factoring $y-1$ out leaves a quadratic in $y$ (which also happens to have "nice" roots).

REPLY [4 votes]: $$24L^2x^2 + (4L^2-9)x=0$$
$$x(24L^2x+4L^2-9)=0$$
$$x =0, x=\frac{9-4L^2}{24L^2} $$
Alternatively
$$4L^2(6x^2+x)=9x$$
$$L=\pm \sqrt{\frac{9x}{24x^2+4x}}$$