Question:
$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=?$
(a) $\sqrt{2}$
(b) 2
(c) 4
(d) 8
Solution:
$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$
$=\frac{\sqrt{16 \times 2}+\sqrt{16 \times 3}}{\sqrt{4 \times 2}+\sqrt{4 \times 3}}$
$=\frac{4 \sqrt{2}+4 \sqrt{3}}{2 \sqrt{2}+2 \sqrt{3}} \quad[\sqrt{a b}=\sqrt{a} \times \sqrt{b}]$
$=\frac{4(\sqrt{2}+\sqrt{3})}{2(\sqrt{2}+\sqrt{3})}$
$=\frac{4}{2}$
$=2$
Hence, the correct answer is option (b).