# How do you rationalize the denominator and simplify (15+3sqrt3)/(7+sqrt8)?

Jun 18, 2017

See a solution process below:

#### Explanation:

To rationalize the denominator, multiply the fraction by $1$ in the form of: $\frac{7 - \sqrt{8}}{7 - \sqrt{8}}$:

$\frac{7 - \sqrt{8}}{7 - \sqrt{8}} \times \frac{15 + 3 \sqrt{3}}{7 + \sqrt{8}} \implies$

$\frac{\left(7 \cdot 15\right) + \left(7 \cdot 3 \sqrt{3}\right) - 15 \sqrt{8} - 3 \sqrt{8} \sqrt{3}}{\left(7 \cdot 7\right) + 7 \sqrt{8} - 7 \sqrt{8} - {\left(\sqrt{8}\right)}^{2}} \implies$

$\frac{105 + 21 \sqrt{3} - 15 \sqrt{8} - 3 \sqrt{8 \cdot 3}}{49 + \left(7 \sqrt{8} - 7 \sqrt{8}\right) - 8} \implies$

$\frac{105 + 21 \sqrt{3} - 15 \sqrt{8} - 3 \sqrt{24}}{41} \implies$

$\frac{105 + 21 \sqrt{3} - 15 \sqrt{8} - 3 \sqrt{4 \cdot 6}}{41} \implies$

$\frac{105 + 21 \sqrt{3} - 15 \sqrt{8} - 3 \sqrt{4} \sqrt{6}}{41} \implies$

$\frac{105 + 21 \sqrt{3} - 15 \sqrt{8} - \left(3 \cdot 2 \sqrt{6}\right)}{41} \implies$

$\frac{105 + 21 \sqrt{3} - 15 \sqrt{8} - 6 \sqrt{6}}{41}$