Question

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

Answer 1

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

$x=\frac{a+2b+a-2b+2\sqrt{(a^2-4b^2)}}{a+2b-a+2b}$

$4bx=2a+2\sqrt{a^2-4b^2}$

$\Rightarrow 2bx-a=\sqrt{a^2-4b^2}$ by squring both sides

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

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

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