<s>[INST] Solving $\sqrt{a +\sqrt{a-x}}+\sqrt{a-\sqrt{a+x}}=2x, \ a\in\Bbb{R}$ 
Given
  $$\sqrt{a +\sqrt{a-x}}+\sqrt{a-\sqrt{a+x}}=2x$$
  and $a\in\Bbb{R}$, express $x$ in terms of $a$.

I rationalised the above expression and then again rationalised which gave me :
$$\sqrt{a+\sqrt{a-x}}-\sqrt{a-\sqrt{a+x}}=\frac{1}{\sqrt{a+x}-\sqrt{a-x}}$$
Now, what should I do?
If didn't rationalised the original Equation, or if the squared both sides twice then too bi-quadratic or higher degree polynomial will be obtained.
What should be done?
 [/INST] Squaring both sides, $$a\require{\cancel}\cancel{+\sqrt{a-x}}+a\cancel{-\sqrt{a-x}}+2\sqrt{\left(a+\sqrt{a-x}\right)\left(a-\sqrt{a-x}\right)}=2\left(a+\sqrt{a^2-a+x}\right)=4x^2.$$ since for all $m,n$ one has that $(m+n)(m-n)=m^2-n^2$. Now, dividing both sides by $2$,$$\begin{align} a+\sqrt{a^2-a+x}&=2x^2 \\ \Leftrightarrow (2x^2-a)^2 &= 4x^4 + a^2 - 4ax^2 \\ &= a^2 - a + x.\end{align}$$ Cancel out the term $a^2$ from both sides and expanding,$$4x^2(x^2-a) = x-a.$$ Now, notice that $$\frac{x^2-a}{x-a} = \frac{x^2-a+x-x}{x-a} = \frac{x^2-x}{x-a}+1.$$ Therefore, $$\begin{align}4x^2\left(\frac{x^2-a}{x-a}\right)= 4x^2\left(\frac{x^2-x}{x-a}+1\right)&=1 \\ \Leftrightarrow \frac{4x^3(x-1)}{x-a}+4x^2&=1.\end{align}$$ Now, since $4x^2 = (2x)^2$ then converting it to that and subtracting $1$ from both sides, $$\begin{align} \frac{4x^3(x-1)}{x-a} + (2x+1)(2x-1)&=0 \\ \Leftrightarrow \frac{4x^3(x-1)}{x-a} + x(2x+1) + (x-1)(2x+1) &= 0 \\ \Leftrightarrow (x-1)\left(\frac{4x^3}{x-a} + (2x+1)(x+1)\right)&=0.\end{align}$$ Solve for $x$ and $a$. To start you off, $x=1$.
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