# How do you rationalize the denominator and simplify sqrt(3)/(sqrt(7)-2)?

Mar 2, 2018

$\frac{\sqrt{21} + 2 \sqrt{3}}{3}$

#### Explanation:

To rationalize the denominator, the best move is to multiply the numerator and denominator by the reciprocal of the denominator (which is the denominator but with the middle sign reversed).

$\frac{\sqrt{3}}{\sqrt{7} - 2} \cdot \textcolor{b l u e}{\frac{\sqrt{7} + 2}{\sqrt{7} + 2}}$

(sqrt(3)(sqrt(7)+2))/((sqrt(7)-2)(sqrt(7)+2)

From here, we can simplify the numerator and denominator.

$\sqrt{\alpha} \sqrt{\beta} = \sqrt{\alpha \beta} \to \sqrt{3} \sqrt{7} = \sqrt{21}$

$\sqrt{3} \left(\sqrt{7} + 2\right) = \textcolor{red}{\sqrt{21} + 2 \sqrt{3}}$

$\left(\sqrt{7} - 2\right) \left(\sqrt{7} + 2\right) = 7 + \cancel{2 \sqrt{7} - 2 \sqrt{7}} - 4 = \textcolor{red}{3}$

$\frac{\sqrt{3}}{\sqrt{7} - 2} = \textcolor{red}{\frac{\sqrt{21} + 2 \sqrt{3}}{3}}$