## Precalculus (6th Edition) Blitzer

$\dfrac{\sqrt{10}}{5}$
RECALL: for any nonnegative real number $a$, $\sqrt{a} \cdot \sqrt{a} = a.$ Rationalize the denominator by multiplying $\sqrt{10}$ to the numerator and the denominator, then use the rule above to obtain $\dfrac{2}{\sqrt{10}} \cdot \dfrac{\sqrt{10}}{\sqrt{10}} \\=\dfrac{2\sqrt{10}}{10}.$ Simplify by canceling the common factor to obtain $\require{cancel} \dfrac{\cancel{2}\sqrt{10}}{\cancel{2}(5)} \\=\dfrac{\sqrt{10}}{5}.$