<s>[INST] Real solution of the equation $\sqrt{a+\sqrt{a-x}} = x\;,$ If $a>0$ 
For a real number $a>0\;,$ How many real solution of the equation $\sqrt{a+\sqrt{a-x}} = x$

$\bf{My\; Try::}$ We can Write $\sqrt{a+\sqrt{a-x}} = x$ as $a+\sqrt{a-x}=x^2$
So we get $(x^2-a)=\sqrt{a-x}\Rightarrow (x^2-a)^2 = a-x\;,$ Where $x<a$
So we get $x^4+a^2-2ax^2=a-x\Rightarrow x^4-2ax^2+x+a^2-a=0$
Now How can i solve it after that, Help me
Thanks
 [/INST] Firstly, let's say $f(x) = \sqrt{a+\sqrt{a-x}} - x$.  If $f$ has real solutions, then it must hold that $x \in [0,a]$. Now, taking the derivative of $f$, we have:
$$f'(x) = -1-\dfrac{1}{4\sqrt{a+\sqrt{a-x}}\cdot (\sqrt{a-x})}.$$
Clearly, $f'(x)<0, \forall x \in [0,a).$ Thus, $f$ is strictly decreasing in $[0,a]$.
Also, we have that $f(0) = \sqrt{a+\sqrt{a}}>0, \, \forall a>0$. Moreover, $f(a) = \sqrt{a} - a$. Thus, everything depends on the value of $a.$ 
Thus, there will be either zero roots or one root of $f$.
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