Question

If $x=\frac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}}$, prove that $b{x}^2-ax+b=0$.

Answer 1

$x=\frac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}}$

$=\frac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}-\sqrt{a-2b}}\times \frac{\sqrt{a+2b}+\sqrt{a-2b}}{\sqrt{a+2b}+\sqrt{a-2b}}$

$=\frac{{(\sqrt{a+2b}+\sqrt{a-2b})}^2}{(a+2b)-(a-2b)}$

$=\frac{a+\sqrt{a^2-4b^2}}{2b}$ (Using identities and then simplifying)
$\Rightarrow 2bx=a+\sqrt{a^2-4b^2}$
$\Rightarrow 2bx-a=\sqrt{a^2-4b^2}$

Squaring both sides, we get,
$(2bx-a{)}^2=a^2-4b^2$

$\Rightarrow 4b^2{x}^2+a^2-4abx=a^2-4b^2$

$\Rightarrow 4b^2{x}^2-4abx+4b^2=0$

$\Rightarrow b{x}^2-ax+b=0$.

Therefore, hence proved.