<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] You get a quadratic equation in terms of "$a$". Using the quadratic formula for $a$,
$$
a=\frac{2x^2+1\pm|2x-1|}{2}
$$
From this, we get $a=x^2+x$ or $a=x^2-x+1$, so
$$
x=\frac{-1\pm \sqrt{1+4a}}{2}\text{ or } x=\frac{1\pm\sqrt{-3+4a}}{2}
$$
Now we must get rid of possiblities:


*

*If $x=\frac{-1-\sqrt{1+4a}}{2}$, it violates $x\ge 0$.

*If $x=\frac{-1+\sqrt{1+4a}}{2}$, it must satisfy $\frac{-1+\sqrt{1+4a}}{2}\ge \sqrt{a}$, which has no roots.

*If $x=\frac{1-\sqrt{-3+4a}}{2}$, it must satisfy $\frac{1-\sqrt{-3+4a}}{2}\ge\sqrt{a}$, which has no roots.


The only remaining possibility is $x=\frac{1+\sqrt{-3+4a}}{2}$. Solve below inequality:
$$
\sqrt{a}\le \frac{1+\sqrt{-3+4a}}{2}\le a
$$
Solving left inequality, we get $a\ge 1$, and solving right inequality, we get $4(a-1)^2\ge 0$. Thus the root of given equality when $a\ge 1$ is
$$
x=\frac{1+\sqrt{-3+4a}}{2}\;(a\ge 1)
$$
If $0<a<1$, then there are no roots.

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