<s>[INST] How to solve $\sqrt{49-x^2}-\sqrt{25-x^2}=3$? I recognize the two difference of squares: $49-x^2$ and $25-x^2$.
I squared the equation to get:
${49-x^2}-2(\sqrt{(49-x^2)(25-x^2)})+{25-x^2}=9$
However, I can't quite figure out how to remove the root in the middle. Any help is appreciated.
 [/INST] Set $\;\;t^2=25-x^2,\;$ the equation becomes $$\begin{align}\sqrt{24+t^2}-\sqrt{t^2}&=3\\ \sqrt{24+t^2}&=|t|+3\\24+t^2&=t^2+6|t|+9\\|t|&=\frac{5}{2}
\end{align}$$
Then $x^2=25-\frac{25}{4}=3\cdot \frac{25}{4},$ from where $x=\pm \frac{5}{2}\sqrt 3.$
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