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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.

this guy123
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    You can think of $L$ as constant and solve for $x$ in terms of $L$. Then any value you give $L$ will determine $x$. Or the reverse: solve for $L$ in terms of $x$. In more complicated examples you may not be able to solve explicitly. – Ethan Bolker Sep 12 '17 at 01:34
  • @EthanBolker I see. That makes sense because I want to solve for x in terms of L. Which would give me expressions with L in it. – this guy123 Sep 12 '17 at 01:37
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    This is just an example Then maybe you should post a real case, since this example has the obvious root $x=0,$, and after removing that one you are left with a linear equation in $x$. – dxiv Sep 12 '17 at 01:37
  • @dxiv I will edit my post and add a real example. – this guy123 Sep 12 '17 at 01:38

2 Answers2

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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).

dxiv
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$$24L^2x^2 + (4L^2-9)x=0$$ $$x(24L^2x+4L^2-9)=0$$ $$x =0, x=\frac{9-4L^2}{24L^2} $$

Alternatively

$$4L^2(6x^2+x)=9x$$ $$L=\pm \sqrt{\frac{9x}{24x^2+4x}}$$

John Lou
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