Question

The value of $\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$ is equal to $\frac{\sqrt{8}+\sqrt{12}}{\sqrt{m}+\sqrt{3}}$.Find $m$

Answer 1

Given:-

$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=\frac{\sqrt{8}+\sqrt{12}}{\sqrt{m}+\sqrt{3}}$

L.H.S.-

$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$

$=\frac{\sqrt{4\times 8}+\sqrt{4\times 12}}{\sqrt{4\times 2}+\sqrt{4\times 3}}$

$\Rightarrow =\frac{(\sqrt{4}\times \sqrt{8})+(\sqrt{4}\times \sqrt{12})}{(\sqrt{4}\times \sqrt{2})+(\sqrt{4}\times \sqrt{3})}$

$\Rightarrow =\frac{\sqrt{4}\times (\sqrt{8}+\sqrt{12})}{\sqrt{4}\times (\sqrt{2}+\sqrt{3})}$

$\Rightarrow =\frac{\sqrt{8}+\sqrt{12}}{\sqrt{2}+\sqrt{3}}$

R.H.S.-

$\frac{\sqrt{8}+\sqrt{12}}{\sqrt{m}+\sqrt{3}}$

On comparing value of L.H.S. and R.H.S., we have,

$m=2$