<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] Here is a simplification, which leads to a parametrized solution.
Write the original equation as two equations
$$
\sqrt{a +\sqrt{a-x}} = x +q\\
\sqrt{a-\sqrt{a+x}}= x -q
$$
where $q$ is a function of $a$. 
Double squaring both equations gives
$$
(x+q)^4-2a(x+q)^2+x+a^2-a = 0 \\ 
(x-q)^4-2a(x-q)^2-x+a^2-a = 0
$$ 
Subtracting and adding gives the two equations
$$
x^2+q^2-a -\frac{1}{4q}= 0 \\ 
x^4+ 6 q^2x^2 + q^4-2a(x^2+q^2)+a^2-a = 0
$$
The first equation gives the result 
$$
x = \sqrt{-q^2+a +\frac{1}{4q}}= \sqrt{a + f(q)}
$$ 
with the offset $f(q)$. In here, $q$ has to be inserted, where $q = q(a)$ is obtained by replacing all terms with $x^2$ in  the second equation with the first equation, giving
$$
(-q^2+a +\frac{1}{4q})^2+ 6 q^2(-q^2+a +\frac{1}{4q}) + q^4-2a(a +\frac{1}{4q})+a^2-a = 0
$$ 
Expanding this expression gives an implicit formula for $q(a)$ as follows:
$$
- 64 q^6 + 64 a q^4 + 16 q^3 - 16 a q^2 + 1 = 0
$$
I wasn't able to solve this further for $q$. However, the equations directly lead to a parametrized solution. Let $q$ be a free parameter. The range where $q$ is to be chosen from is discussed below. Then for each $q$, we have $a$ and $x$ according to
$$
a  = \frac{64 q^6 - 16 q^3 - 1}{64 q^4 - 16 q^2} \\
x = \sqrt{-q^2+a +\frac{1}{4q}}
$$
The following observations can be made:
For large $a$, the equation for $a$ gives $q^2 \to \frac{1}{4}$. This can also be observed by regarding the very first set of two equations and treating $q$ as a small disturbance, which can in first order be neglected if all other terms grow exceedingly large. We then have 
$$
\sqrt{a +\sqrt{a-x_1}} = x_1\\
\sqrt{a-\sqrt{a+x_2}}= x_2
$$
which are solved (see Solve the equation: $x=\sqrt{a-\sqrt{a+x}},(a\geq 1)$) by
$$
x_1 = \frac12 + \sqrt{a -3/4}\\
x_2 = -\frac12 + \sqrt{a -3/4}
$$
So indeed, the $x_{1,2}$ grow exceedingly large, and $q = -\frac12$ gives both the correct offset term $-\frac34$ under the previous root $
x = \sqrt{-q^2+a +\frac{1}{4q}}
$, as well as the two deviations $x - x_{1,2}$. For small $a$, correction terms  must be added to $q(a)$ (in the sense of an expansion). 

Addendum:
1. As a side observation, one can use the two solutions $x_{1,2}$ to establish that the modified equation
$$
\sqrt{a +\sqrt{a-x -\frac12}}+\sqrt{a-\sqrt{a+x-\frac12}}=2x
$$
is solved by $x = \sqrt{a -3/4}$.


*Note that in the general solution $x =\sqrt{a + f(q)}$, the offset $f(q) = -q^2 +\frac{1}{4q}$ actually has a maximum at $q = -0.5$. This can be seen by
$f'(q)  =  \frac{-1 - 8 q^3}{4 q^2}$ which is zero at $q = -0.5$. Hence, the offset has its maximum $f(q= -0.5) = -0.75$ for $\alpha \to \infty$. It will be lower for other $q$. The behavior of $q(a)$ is given in the following plot. For low $a=1.5$, for example, we have $q\simeq -0.81$ and $f(q) \simeq -0.965$. The plot $q(a)$, together with the offset behavior $f(q)$, establish that the offset is monotonously rising with $a$ in a small range of $f$-values.



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