# 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?

• At the original, subtract the RHS from both sides. Combine the fractions by making them have the same denominator. See what it looks like and if there are good steps to move forward. Nov 1, 2022 at 16:14

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 $$x = 2\sqrt{3}$$.

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.

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 $$(\alpha\beta)^2+2\alpha\beta-8=0 \Rightarrow \alpha\beta=2\Rightarrow \sqrt{16-x^2}=2 \Rightarrow x=2\sqrt{3}$$