ex1.4_3

$i$ $\quad \sqrt{-3} \cdot \sqrt{-12}$

\sqrt{-3} \cdot \sqrt{-12} = i \, \sqrt{3} \, \cdot \, i \, \sqrt{12}\\[0.25 in] = i \, \sqrt{3} \, \cdot \, 2i \, \sqrt{3}\\[0.25 in] = 2i^2(\sqrt{3})^2\\[0.25 in] = (-2)(3) = -6

Another possible approach goes like this....

\sqrt{-3} \cdot \sqrt{-12} = i \, \sqrt{3} \, \cdot \, i \, \sqrt{12}\\[0.25 in] = i^2 \, \sqrt{36}\\[0.25 in] =(-1)(6) = -6