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?
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.