## Precalculus (6th Edition) Blitzer

$3\sqrt{3}$
Factor each radicand so that one of the factors is a perfect square to obtain: $=3\sqrt{4(3)} - \sqrt{9(3)} \\=3\sqrt{2^2(3)} - \sqrt{3^2(3)}$ Simplify each radical to obtain: $=3\cdot 2\sqrt{3} - 3\sqrt{3} \\=6\sqrt{3} - 3\sqrt{3}$ Factor out $\sqrt{3}$ to obtain: $=\sqrt{3}(6-3) \\=\sqrt{3} \cdot 3 \\=3\sqrt{3}$