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.
 A: Hint. You may just write
$$
{49-x^2}+{25-x^2}-9=2\sqrt{(49-x^2)(25-x^2)}
$$ simplify and square both sides once again.
A: Another solution
\begin{align*}
\sqrt {49 - x^2} - \sqrt {25 - x^2} &= 3\\
\sqrt {49-x^2} &= 3+ \sqrt {25-x^2}\\
49-x^2 &= 25 -x^2 + 9 + 6\sqrt {25-x^2}\\
15 &= 6 \sqrt {25 - x^2}\\
25 - x^2 &= \frac {25} 4\\
x^2 &= \frac {75}4\\
x &= \pm \frac 5 2 \sqrt 3.
\end{align*}
A: We have
$$\sqrt{49-x^2}-\sqrt{25-x^2}=3$$ and 
$$\left(\sqrt{49-x^2}-\sqrt{25-x^2}\right)\left(
\sqrt{49-x^2}+\sqrt{25-x^2}\right)=3\left(\sqrt{49-x^2}+\sqrt{25-x^2}\right),$$
which gives 
$$\sqrt{49-x^2}+\sqrt{25-x^2}=8$$ and from here
$$\sqrt{49-x^2}=5.5,$$ which gives the answer:
$$\left\{-\sqrt{18.75},\sqrt{18.75}\right\}.$$
A: First express one root: $$\sqrt{49-x^2}=3+\sqrt{25-x^2}$$
Now square:
$$ 49-x^2=9+6\sqrt{25-x^2}+ 25-x^2\implies \boxed{5=2\sqrt{25-x^2}}$$
and square again...
A: 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.
A: This is a good start. Now rearrange the equation so that only the square root remains on the left hand side. 
Take square again. Then you obtain a bi-quadratic equations, or "fake quartic" equation. Introduce a new variable $y=x^2$, and then you have a quadratic equation in $y$ that you can solve easily. 
Note that taking squares is not an equivalent transformation, so in the end, you have to check all the solutions. Some, or in fact, all of them can be extraneous. 
A: 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 $x=\pm \frac{5}{2}\sqrt 3.$
