<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] It factors!
$x^4-2ax^2+x+(a^2-a)=(x^2+x-a)(x^2-x+(1-a))$
How do I get that?  Well, there will be an extraneous root with the wrong signs for the inner square root, thus:
$x=\sqrt{a-\sqrt{a-x}}$
$-x=-\sqrt{a-\sqrt{a+(-x)}}$
The second extraneous equation is consistent with the simpler recursion
$y=-\sqrt{a+y}$
with $y=-x$.  So upon squaring the last equation we find that $y^2-y-a=x^2+x-a=0$ meaning the quadratic expression must be a factor of the quartic.
But perforce it is the wrong factor.  The intended roots must be in the complementary quadratic factor
$(x^2-x+(1-a))=0$.
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