## Precalculus (6th Edition)

$\color{blue}{12}$
RECALL: (1) For $a\gt0$ and $b\gt0$, $\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}$ (2) For $a\ge0$, $\sqrt{a^2} = a$ Use rule (1) above to obtain: $\sqrt{6} \cdot \sqrt{24} \\=\sqrt{6\cdot24} \\=\sqrt{144} \\=\sqrt{12^2}$ Use rule (2) above to obtain: $=\color{blue}{12}$