<s>[INST] Solve the equation $\frac{\sqrt{4+x}}{2+\sqrt{4+x}}=\frac{\sqrt{4-x}}{2-\sqrt{4-x}}$ Solve the equation $$\dfrac{\sqrt{4+x}}{2+\sqrt{4+x}}=\dfrac{\sqrt{4-x}}{2-\sqrt{4-x}}$$ The domain is $4+x\ge0,4-x\ge0,2-\sqrt{4-x}\ne0$.
Note that the LHS is always positive, so the roots must also satisfy: $A:2-\sqrt{4-x}>0$.
Firstly, I decided to raise both sides of the equation to the power of $2$. I came at $$\dfrac{4+x}{8+x+4\sqrt{4+x}}=\dfrac{4-x}{8-x-4\sqrt{4-x}}$$
Another thing I tried is to let $\sqrt{4+x}=u\ge0$ and $\sqrt{4-x}=v\ge0$. Then $$\begin{cases}\dfrac{u}{2+u}=\dfrac{v}{2-v}\\4+x=u^2\\4-x=v^2\end{cases}$$ Adding the second to the third equation, we get $8=u^2+v^2$.
And the last thing: I cross-multiplied $$2\sqrt{4+x}-\sqrt{16-x^2}=2\sqrt{4-x}+\sqrt{16-x^2}\\\sqrt{4+x}=\sqrt{4-x}+\sqrt{16-x^2}$$ Raising to the power of 2 gives $$4+x=4-x+16-x^2+2\sqrt{(4-x)(16-x^2)}\\2\sqrt{(4-x)(16-x^2)}=x^2+2x-16$$ Is there something easier?
 [/INST] Your analysis is good.
Starting in the middle:

*

*$u = \sqrt{4+x}.$


*$v = \sqrt{4-x}.$


*$\dfrac{u}{2+u} = \dfrac{v}{2-v}.$
So,
$$2u - uv = 2v + uv \implies 2u - 2v = 2uv \implies $$
$$2\sqrt{4+x} - 2\sqrt{4-x} = 2\sqrt{16 - x^2} \implies $$
$$4(4 + x) + 4(4-x) - 8\sqrt{16 - x^2} = 4(16 - x^2) \implies $$
$$32 - 8\sqrt{16 - x^2} = 4(16 - x^2).$$
Let $t = \displaystyle \sqrt{16 - x^2} \implies t \geq 0.$
Then
$$32 - 8t = 4t^2 \implies t^2 + 2t - 8 = 0 \implies$$
$$(t - 2)(t + 4) = 0.$$
$(t + 4) = 0$ must be rejected, since $t \geq 0.$
Therefore,
$$\sqrt{16 - x^2} = t = 2 \implies 16 - x^2 = 4 \implies x = \pm 2\sqrt{3}.$$
These two roots of $x$ may be checked as follows:
$$\left(\sqrt{3} + 1\right)^2 = 4 + 2\sqrt{3} \implies \sqrt{4 + 2\sqrt{3}} = \sqrt{3} + 1.$$
$$\left(\sqrt{3} - 1\right)^2 = 4 - 2\sqrt{3} \implies \sqrt{4 - 2\sqrt{3}} = \sqrt{3} - 1.$$
Plugging $x = 2\sqrt{3}$ back into the original equation gives
$$\frac{\sqrt{3} + 1}{2 + [\sqrt{3} + 1]} = \frac{\sqrt{3} - 1}{2 - [\sqrt{3} - 1]}. \tag1 $$
Plugging $x = -2\sqrt{3}$ back into the original equation gives
$$\frac{\sqrt{3} - 1}{2 + [\sqrt{3} - 1]} = \frac{\sqrt{3} + 1}{2 - [\sqrt{3} + 1]}. \tag2 $$
The assertion in (1) above is easily determined to to be true and the assertion in (2) above is easily determined to be false.
Therefore, of the two candidate roots of
$~\displaystyle x = \pm 2\sqrt{3}~$, the first root of $~\displaystyle x = 2\sqrt{3}~$  is confirmed and the second root of $~\displaystyle x = -2\sqrt{3}~$ must be rejected.
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