An algebra problem by Wildan Bagus Wicaksono

$\begin{aligned} x & = \frac {\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}} \\ & = \frac {\left(\sqrt{a+2b}+\sqrt{a-2b}\right)\left(\sqrt{a+2b}+ \sqrt{a-2b}\right)}{\left(\sqrt{a+2b}-\sqrt{a-2b}\right)\left(\sqrt{a+2b}+\sqrt{a-2b}\right)} \\ & = \frac {a+2b+2\sqrt{(a+2b)(a-2b)}+a-2b}{a+2b-a+2b} \\ & = \frac {2a+2\sqrt{a^2-4b^2}}{4b} \\ & = \frac {a+\sqrt{a^2-4b^2}}{2b} \end{aligned}$

We note that the roots of $bx^2-ax+b=0$ are $x = \dfrac {a \pm \sqrt{a^2-4b^2}}{2b}$ . This means that $x = \dfrac {a + \sqrt{a^2-4b^2}}{2b}$ is one of the roots, implying that $bx^2-ax+b=\boxed{0}$ .

Given $\large x = \frac {\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}}$ or, $\large \frac{1+ x}{1- x} = \frac {\sqrt{a+2b}}{\sqrt{a-2b}}$ Squaring both sides, $\large \frac{1+x^{2}+2x}{1+x^{2}-2x}=\frac{a+2b}{a-2b}$ or, $\large a-2b+ax^{2}-2bx^{2}+2ax-4bx=a+2b+ax^{2}+2bx^{2}-2ax-4bx$ or, $\large 4bx^{2}-4ax+4b=0$ or, $\large bx^{2}-ax+b=\boxed{0}$