Question

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

Answer 1

Given:- $x=\frac{\sqrt{a+3b}+\sqrt{a-3b}}{\sqrt{a+3b}-\sqrt{a-3b}}$

To prove:- $3b{x}^2-2ax+3b=0$

Proof:-

$x=\frac{\sqrt{a+3b}+\sqrt{a-3b}}{\sqrt{a+3b}-\sqrt{a-3b}}$

Multiply and divide the above equation by conjugate of its denominator, we have

$x=\frac{\sqrt{a+3b}+\sqrt{a-3b}}{\sqrt{a+3b}-\sqrt{a-3b}}\times \frac{\sqrt{a+3b}+\sqrt{a-3b}}{\sqrt{a+3b}+\sqrt{a-3b}}$

$x=\frac{{(\sqrt{a+3b}+\sqrt{a-3b})}^2}{{\left(\sqrt{a+3b}\right)}^2-{\left(\sqrt{a-3b}\right)}^2}$

$x=\frac{(a+3b)+(a-3b)+2\sqrt{(a+3b)(a-3b)}}{(a+3b)-(a-3b)}$

$x=\frac{a+3b+a-3b+2\sqrt{(a+3b)(a-3b)}}{a+3b-a+3b}$

$\Rightarrow x=\frac{2a+2\sqrt{a^2-{\left(3b\right)}^2}}{6b}$

$\Rightarrow 6bx=2a+2\sqrt{a^2-9b^2}$

$\Rightarrow 3bx-a=\sqrt{a^2-9b^2}$

Squring both sides, we have

${(3bx-a)}^2={\left(\sqrt{a^2-9b^2}\right)}^2$

$9b^2{x}^2+a^2-6abx=a^2-9b^2$

$9b^2{x}^2-6abx+9b^2=0$

$\Rightarrow 3b(3b{x}^2-2ax+3b)=0$

$\Rightarrow 3b{x}^2-2ax+3b=0$

Hence proved.