The value of frac{sqrt{32}+sqrt{48}}{sqrt{8}+ | vedclass

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Question

$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=\frac{\sqrt{16 	imes 2}+\sqrt{16 	imes 3}}{\sqrt{4 	imes 2}+\sqrt{4 	imes 3}}$

$=\frac{4 \sqrt{

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=\frac{\sqrt{16 	imes 2}+\sqrt{16 	imes 3}}{\sqrt{4 	imes 2}+\sqrt{4 	imes 3}}$

$=\frac{4 \sqrt{2}+4 \sqrt{3}}{2 \sqrt{2}+2 \sqrt{3}}=\frac{4(\sqrt{2}+\sqrt{3})}{(2 \sqrt{2}+\sqrt{3})}=\frac{4}{2}=2$

Hence, $(d)$ is the correct answer.

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

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