Question

$\frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}=?$
(a) $\sqrt{2}$
(b) 2
(c) 4
(d 8

Answer 1

(b) 2

We have:
$$\begin{gathered} \frac{\sqrt{32}+\sqrt{48}}{\sqrt{8}+\sqrt{12}} \\ =\frac{\sqrt{16\times 2}+\sqrt{16\times 3}}{\sqrt{4\times 2}+\sqrt{4\times 3}} \\ =\frac{4\sqrt{2}+4\sqrt{3}}{2\sqrt{2}+2\sqrt{3}} \\ =\frac{4\left(\sqrt{2}+\sqrt{3}\right)}{2\left(\sqrt{2}+\sqrt{3}\right)} \\ =2 \end{gathered}$$