The value of
Question:

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

(a) $\sqrt{2}$

(b) 2

(c) 4

(d) 8

Solution:

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

$=\frac{4 \sqrt{2}+4 \sqrt{3}}{2 \sqrt{2}+2 \sqrt{3}}=\frac{4(\sqrt{2}+\sqrt{3})}{2(\sqrt{2}+\sqrt{3})}=2$