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?
Best Answer
My first instinct is to note that for $x \ge -4$ and $x \ne 0$, $$\frac{\sqrt{4+x}}{2+\sqrt{4+x}} = \frac{\sqrt{4+x}(\sqrt{4+x}-2)}{(4+x)-4} = \frac{4+x - 2\sqrt{4+x}}{x}, \tag{1}$$ and similarly, for $x \le 4$ and $x \ne 0$, $$\frac{\sqrt{4-x}}{2-\sqrt{4-x}} = \frac{\sqrt{4-x}(2+\sqrt{4-x})}{4-(4-x)} = \frac{2\sqrt{4-x} +4-x }{x}. \tag{2}$$ Therefore the original equation implies $$4+x - 2\sqrt{4+x} = 2\sqrt{4-x} + 4-x$$ when $0 < |x| \le 4$. This in turn yields $$\sqrt{4+x} + \sqrt{4-x} = x, \tag{3}$$ and now we can square both sides:
$$4+x + 4-x + 2\sqrt{16-x^2} = x^2, \tag{4} $$ or $$4(16-x^2) = (x^2 - 8)^2, \tag{5} $$ hence $$x^4 - 12x^2 = 0. \tag{6}$$ This gives $x = \pm 2 \sqrt{3}$ as the only solutions satisfying the requirement that $0 < |x| \le 4$; however, the negative root, when substituted back into the original equation, is extraneous. The reason for this has to do with the first squaring step from $(3)$ to $(4)$, since the RHS of $(3)$ cannot be negative, being the sum of two (nonnegative) square roots on the LHS. Therefore the unique real-valued solution is $x = 2\sqrt{3}$.