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.

This is a very intriguing problem. First of all,  $x\geq 0$ and $a\geq\sqrt{a+x}$ must hold for real $x$.
Squaring both sides of the equation we get $x^2=a-\sqrt{a+x}$.
$(x^2-a)^2=a+x$
$\Rightarrow x^4-2ax^2-x+a^2-a=0$
$\Rightarrow (x^2-x-a)(x^2+x+(1-a))=0$
$\therefore x= \frac12\pm\sqrt{a+\frac14},-\frac12\pm\sqrt{a-\frac34}$
All four of these solutions may not work because some of them could be extraneous roots due to squaring. Two of the solutions give negative values for $x$, so it must be
$x= \frac12+\sqrt{a+\frac14},-\frac12+\sqrt{a-\frac34}$
But the 1st one again does not satisfy the given equation.
Hence the unique solution to the equation  is $x=-\frac12+\sqrt{a-\frac34}$, and that solution only exists if $a\geq 1$.
