Solving $\sqrt{a-\sqrt{a+x}} = x$ Solve the equation
$$\sqrt{a-\sqrt{a+x}} = x$$
My approach:
Tried shifting the variables into different options, but couldn't get anything out of it. So, please help.
 A: $$a-\sqrt{a+x}=x^2$$
$$a-x^2=\sqrt{a+x}$$
$$a+x=x^4-2ax^2+a^2$$
Consider it in terms of a quadratic in $a$
$$a^2+a(-2x^2-1)+x^4-x=0$$
Now we see that $x^4-x=x(x^3-1)=x(x-1)(x^2+x+1)=(x^2-x)(x^2+x+1)$. And $(x^2+x+1)+(x^2-x)=2x^2+1$. By Vieta, its roots are $x^2-x$ and $x^2+x+1$. You could also do it using quadratic formula.
A: $$a-\sqrt{a+x}=x^2$$
$$a-x^2=\sqrt{a+x}$$
$$a+x=x^4-2ax^2+a^2$$
$$x^4-2ax^2-x+a^2-a=0$$
$$(x^2+x+1-a)(x^2-x-a)=0$$
You can factor the last expression by setting up $(x^2+px+q)(x^2+rx+s)=x^4-2ax^2-x+a^2-a$. And comparing the coefficient to find $p,q,r,s$.
A: Let $\sqrt{a+x}=y\ge0\implies x=y^2-a$
$$y^2-a=\sqrt{a-y}\ge0\implies  y^2\ge a$$
$$(y^2-a)^2=a-y\iff a^2-(2y^2+1)a+y^4-y=0$$
$$a=\dfrac{2y^2+1\pm\sqrt{(2y^2+1)^2-4(y^4-y)}}2=\dfrac{2y^2+1\pm(2y+1)}2$$
A: Let $t=x+a$ to rewrite the nested radical equation $\sqrt{a-\sqrt{a+x}} = x $ as
$$t-a- \sqrt{a-\sqrt t}=
0$$
Avoid extraneous solutions from squaring the equation and instead  factorize it as
$$(\sqrt{a-\sqrt t}+\sqrt t)(\sqrt{a-\sqrt t}+1-\sqrt t)=0$$
Note that $ \sqrt{a-\sqrt t}+\sqrt t>0$ and any solutions result from
$$ \sqrt{a-\sqrt t}+1-\sqrt t =0$$
which factorizes as
$$\left(\sqrt{a-\sqrt t}+\frac{1+\sqrt{4a-3}}2\right)
 \left(\sqrt{a-\sqrt t}+\frac{1-\sqrt{4a-3}}2\right)=0
$$
Again, the first factor is positive, permitting no solutions, and the second factor yields the sole solution
$$ x=t-a= \frac12(\sqrt{4a-3}-1)$$
(Note: the solution exists for $a\ge 1$.)
