Answer
$7 (\sqrt 5 + 2)$ or $ 7 \sqrt5 + 14 $
Work Step by Step
$ \frac{7}{\sqrt 5 - 2}$
The conjugate of the denominator is $ \sqrt 5 + 2$. Multiply the denominator and numerator by $ \sqrt 5 + 2 $, so the simplified denominator will not contain a radical. Therefore, multiply by 1, choosing $\frac{ \sqrt 5 + 2}{ \sqrt 5 + 2}$ for 1.
$ \frac{7}{\sqrt 5 - 2}$ = $ \frac{7}{\sqrt 5 - 2} \times \frac{ \sqrt 5 + 2}{ \sqrt 5 + 2}$
= $\frac{ 7(\sqrt 5 + 2)}{( \sqrt 5 - 2)( \sqrt 5 + 2)}$
$( \sqrt a - \sqrt b)( \sqrt a + \sqrt b)$ = $ (\sqrt a)^{2}$ - $ (\sqrt b)^{2}$. Therefore,
$( \sqrt 5 - 2)( \sqrt 5 + 2)$ = $ (\sqrt 5)^{2}$ - $ (2)^{2}$.
= $\frac{ 7(\sqrt 5 + 2)}{ (\sqrt 5)^{2} - (2)^{2}}$
= $\frac{ 7(\sqrt 5 + 2)}{ 5 - 4}$
= $\frac{ 7(\sqrt 5 + 2)}{ 1}$
= $7 (\sqrt 5 + 2)$ or $ 7 \sqrt5 + 14 $
