Question

If $\frac{x}{1}=\frac{\sqrt{a+3b}+\sqrt{a-3b}}{\sqrt{a+3b}-\sqrt{a-3b}}$, then, show that $3b{x}^2+3b-2ax=0$.

Answer 1

$\frac{x}{1}=\frac{\sqrt{a+3b}+\sqrt{a-3b}}{\sqrt{a+3b}-\sqrt{a-3b}}$
Applying componendo and dividendo
$\frac{x+1}{x-1}=\frac{\sqrt{a+3b}+\sqrt{a-3b}+\sqrt{a+3b}-\sqrt{a-3b}}{\sqrt{a+3b}+\sqrt{a-3b}-\sqrt{a+3b}+\sqrt{a-3b}}$
= $\frac{2\sqrt{a+3b}}{2\sqrt{a-3b}}$
= $\frac{\sqrt{a+3b}}{\sqrt{a-3b}}$
Squaring on both sides, we get:
$\frac{(x+1{)}^2}{(x-1{)}^2}=\frac{(\sqrt{a+3b}{)}^2}{(\sqrt{a-3b}{)}^2}$
= $\frac{(x+1{)}^2}{(x-1{)}^2}=\frac{a+3b}{a-3b}$
Again, applying componendo and dividendo
$\frac{(x+1{)}^2+(x-1{)}^2}{(x+1{)}^2-(x-1{)}^2}=\frac{a+3b+a-3b}{a+3b-a+3b}$
$\frac{2(x^2+1)}{4x}=\frac{2a}{6b}$
=$\frac{(x^2+1)}{2x}=\frac{a}{3b}$
=$3b{x}^2+3b=2ax$
= $3b{x}^2+3b-2ax=0$