<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] My solution:
Let us consider two positive reals $\alpha,\beta$ defined as$$\alpha=\sqrt{4+x} \quad \beta=\sqrt{4-x} \Rightarrow \alpha^2+\beta^2=8 \Rightarrow (\alpha-\beta)^2+2\alpha\beta=8$$
Now the equation becomes
$$\frac{2+\alpha}{\alpha}=\frac{2-\beta}{\beta} \Rightarrow \frac{\alpha}{\alpha}+\frac{\beta}{\beta}=2\left(\frac{1}{\beta}-\frac{1}{\alpha}\right)$$
So we get
$$\alpha-\beta=\alpha\beta \Rightarrow \alpha >\beta \Rightarrow x>0$$
Finally $$(\alpha\beta)^2+2\alpha\beta-8=0 \Rightarrow \alpha\beta=2\Rightarrow \sqrt{16-x^2}=2 \Rightarrow x=2\sqrt{3}$$
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