A problem on nested radicals

Question

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??

Answer 1

Your approach is not right unless your original equation was: $$x=\sqrt{(a-b)\sqrt{(a+b)\sqrt{(a-b)\sqrt{(a+b)\cdots}}}}$$

However, it's almost there.

$\displaystyle\begin{align} x & = \sqrt{a-b\sqrt{a+b\sqrt{a-b\sqrt{a+b\cdots}}}} \\ x & =\sqrt{a-b\sqrt{a+bx}} \\x^2 & = a-b\sqrt{a+bx} \\(a-x^2)^2 & = b^2(a+bx) \\ a^2-2ax^2+x^4 & = ab^2 + b^3x \\ 0 & = x^4-2ax^2-b^3 x +a^2-ab^2 \\ 0 & = (x^2 +bx-a+b^2)(x^2-bx-a) \\ \therefore x & \in \left\{\frac {-b-\sqrt{4a-3b^2}}2, \frac{-b+\sqrt{4a-3b^2}}2, \frac{b-\sqrt{4a+b^2}}2, \frac{b+\sqrt{4a+b^2}}2 \right\} \end{align}$

Now apply the criteria $a>b^2$ to select among them for the satisfactory answer.

Answer 2

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:

$$\large x=\frac{-b+\sqrt{4a-3b^2}}{2}$$