Solve the equation: $x=\sqrt{a-\sqrt{a+x}},(a\geq 1)$ 
Solve the equation: $x=\sqrt{a-\sqrt{a+x}},(a\geq 1)$ 

We have: $x^2=a-\sqrt{a+x}$ , if we take both sides to the power of 2 we will get a 4th order equation which I don't know of!
Please help.

We certainly have $x>0$. Then by squaring, the original equation can be written
$$a-x^2=\sqrt{x+a}.$$
The two members describe the parabolas of equations
$$\begin{cases}y=a-x^2,\\x=y^2-a\end{cases}$$ where $y>0$.
Adding these two equalities we get a pair of straight lines
$$y+x=y^2-x^2,$$ i.e. $$(y+x)(y-x-1)=0.$$

Hence, the requested solution lies in the first quadrant and is the positive root of
$$\begin{cases}y-x-1=0\\y=a-x^2\end{cases},$$
$$x^2+x+1-a=0,$$
$$x=\frac{-1+\sqrt{4a-3}}{2}.$$
