Question

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

Answer 1

$$\begin{gathered} \frac{{\displaystyle \sqrt{32}+\sqrt{48}}}{{\displaystyle \sqrt{8}+\sqrt{12}}} \\ =\frac{{\displaystyle \sqrt{16\times 2}+\sqrt{16\times 3}}}{{\displaystyle \sqrt{4\times 2}+\sqrt{4\times 3}}} \\ =\frac{{\displaystyle 4\sqrt{2}+4\sqrt{3}}}{{\displaystyle 2\sqrt{2}+2\sqrt{3}}}\left[\sqrt{ab}=\sqrt{a}\times \sqrt{b}\right] \end{gathered}$$
$$\begin{gathered} =\frac{{\displaystyle 4\left(\sqrt{2}+\sqrt{3}\right)}}{{\displaystyle 2\left(\sqrt{2}+\sqrt{3}\right)}} \\ =\frac{4}{2} \\ =2 \end{gathered}$$
Hence, the correct answer is option (b).