Determine a quartic equation I am working on a puzzle from Popular Science from the 1980's. This was a puzzle that existed before pocket calculators or programmable computers. It results in one needing to solve a seeming quartic equation! The solution gives the equation:
$$x - \frac{x}{\sqrt{16-x^2}} - \frac{x}{\sqrt{9-x^2}} = 0.$$
As far as I can tell, this equation has an irrational denominator, and the solution page claims that this is a quartic equation, with an $x^4$ term. How can one tell from this equation that it’s a quartic? How can I remove the denominator and expand it out to collect the terms so that it results in the form of a general quartic equation? This seems too complicated!
Can you help?
 A: First, observe that $x = 0$ is a solution. After which cancel it out, so you can get equivalent equation to find other roots. After some algebra it'd look like
$$
\frac 1{\sqrt{16-x^2}} + \frac 1{\sqrt{9-x^2}} = 1
$$
You have several ways to make it quartic, but probably least algebraic is if you make a substitution $t = \sqrt{9-x^2}$, so $\sqrt{16-x^2} = \sqrt{t^2+7}$.
\begin{align}
\frac 1{\sqrt{t^2+7}} + \frac 1t &= 1 \\
t + \sqrt{t^2+7} &= t \sqrt{t^2+7} \\
\sqrt{t^2+7} (t-1) &= t \\
(t^2+7)(t-1)^2 &= t^2\\
t^4 - 2 t^3 + 7 t^2 - 14 t + 7  &= 0
\end{align}
Solving it is probably a different question, and Mathematica suggests that roots aren't nice.
A: You can do $$1 - \frac{1}{\sqrt{16-x^2}} - \frac{1}{\sqrt{9-x^2}} = 0\\
1 - \frac{1}{\sqrt{16-x^2}} = \frac{1}{\sqrt{9-x^2}}\\
1-\frac{2}{\sqrt{16-x^2}}+\frac{1}{16-x^2}=\frac {1}{9-x^2}\\
\frac{(16-x^2)(9-x^2)-7}{(16-x^2)(9-x^2)}=\frac{2x}{\sqrt{16-x^2}}$$ and one more square, cross multiply, and canceling the $16-x^2$ gets you a quartic in $x^2$.  Don't forget the root at $x=0$  Alpha finds real roots at $x\approx \pm 2.6033$
