## Precalculus (6th Edition) Blitzer

$i\sqrt{20}$; $i\sqrt{4\cdot5}$; $2i\sqrt{5}$
RECALL: (1) $\sqrt{ab} = \sqrt{a} \cdot \sqrt{b}$ (2) $\sqrt{-1} = i$ Use rule (1) above to obtain: $\sqrt{-20} \\= \sqrt{(-1)(20)} \\= \sqrt{-1} \cdot \sqrt{20}$ Use rule (2) above to obtain: $=i\sqrt{20}$ Use rule (1) again to obtain: $=i\sqrt{4\cdot5} \\=i\sqrt{4} \sqrt{5} \\=i\cdot 2 \sqrt{5} \\=2i\sqrt{5}$ Therefore, the missing expressions in the given statement, in order, are: $i\sqrt{20}$; $i\sqrt{4\cdot5}$; $2i\sqrt{5}$