## Precalculus (6th Edition) Blitzer

$19i\sqrt{2}$
RECALL: (1) $\sqrt{-1}=i$ (2) For any real number $a\gt0$, $\sqrt{-a} = i\sqrt{a}$ Use rule (2) above to obtain: $=5\cdot i\sqrt{8}+3 \cdot i\sqrt{18} \\=5i\sqrt{4(2)} + 3i\sqrt{9(2)} \\=5i(\sqrt{2^2(2}) + 3i(\sqrt{3^2(2})$ Simplify each radical to obtain: $=5i\cdot 2\sqrt{2} +3i \cdot 3 \sqrt{2} \\=10i\sqrt{2} + 9i\sqrt{2}$ Combine like terms using the rule $ac+bc=(a+b)c$ to obtain: $=(10i+9i)\sqrt{2} \\=19i\sqrt{2}$