## Geometry: Common Core (15th Edition)

$2 \sqrt {3}$
First, we write both numbers under the same radical: $\frac{\sqrt {36}}{\sqrt {3}} = \sqrt \frac{36}{3}$ Evaluate what is under the radical sign: $\sqrt {12}$ Rewrite $12$ as a product of a perfect square and another factor. We can use $4$ as the perfect square factor: $\sqrt {4 • 3}$ Take the square root of $4$ to solve: $2 \sqrt {3}$