## Precalculus (6th Edition) Blitzer

$\sqrt{7x}$
Simplify each radical to obtain $\sqrt{9(7x)} - \sqrt{4(7x)} \\=\sqrt{3^2(7x)}-\sqrt{2^2(7x)} \\=3\sqrt{7x}-2\sqrt{7x}.$ RECALL: The distributive property states that for any real numbers a, b, and c, $(ac-bc)=(a-b)c.$ Use the rule above to obtain $(3-2)\sqrt{7x} \\=1\cdot \sqrt{7x} \\=\sqrt{7x}$