A problem on nested radicals Find the value of $x$ for all $a>b^2$ if:
$$\large x=\sqrt{a-b\sqrt{a+b\sqrt{a-b{\sqrt{a+b.......}}}}}$$
My attempt
$$\large x=\sqrt{a-b\sqrt{(a+b)x}}$$
$$\large x^4=(a-b)^2(a+b)x$$
$$\large x=((a-b)^2(a+b))^{1/3}$$ (real root)
Question: Is my solution correct??

Here's another one:
$$\text{Let} x=\sqrt{a-b\sqrt{a+b\sqrt{a-b\sqrt{a+b\cdots}}}}$$
$$\text{Let}y=\sqrt{a+b\sqrt{a-b\sqrt{a+b\sqrt{a-b\cdots}}}}$$
Now $$x^2=a-by---1$$ and $$y^2=a+bx----2$$ $$\implies x^2-y^2=-b(x+y)$$ $$\implies(x+y)(x-y+b)=0$$(Considering only positive solutions of $x$) $$y=x+b$$, putting it in eq $2$, $$x^2+bx+b^2-a=0$$, on Solving:
$$\large x=\frac{-b+\sqrt{4a-3b^2}}{2}$$
