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.

Note that $$\sqrt{49-x^2}+\sqrt{25-x^2}=\frac{(49-x^2)-(25-x^2)}{\sqrt{49-x^2}-\sqrt{25-x^2}}=\frac{24}{3}=8\,.$$
Together with $\sqrt{49-x^2}-\sqrt{25-x^2}=3$, we conclude that
$$\sqrt{49-x^2}=\frac{3+8}{2}=\frac{11}{2}\,.$$

Therefore, $x^2=\dfrac{75}{4}$, or $x=\pm\dfrac{5\sqrt{3}}{2}$.

P.S.:  Oopsie, nextpuzzle already made a comment, giving the same solution.
