# How do you simplify (12+sqrt108)/6?

May 29, 2017

$= 2 + \sqrt{3}$

#### Explanation:

Each term in the numerator must be divided by $6 :$

$\frac{12 + \sqrt{108}}{6} = \frac{12}{6} + \frac{\sqrt{108}}{6} \text{ } \leftarrow$ prime factor 108

$= 2 + \frac{\sqrt{2 \times 2 \times 3 \times 3 \times 3}}{6} = 2 + \frac{\sqrt{{2}^{2} \times {3}^{2} \times 3}}{6}$

Find the roots where possible.

$2 + \frac{2 \cdot 3 \sqrt{3}}{6} = 2 + \frac{\cancel{6} \sqrt{3}}{\cancel{6}}$

$= 2 + \sqrt{3}$

May 29, 2017

$2 + \sqrt{3}$

#### Explanation:

$\text{simplifying " sqrt108" to begin with}$

$\text{using the "color(blue)"law of radicals}$

$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \sqrt{a} \times \sqrt{b} \Leftrightarrow \sqrt{a b}$

$\Rightarrow \sqrt{108} = \sqrt{{2}^{2} \times {3}^{3}} = 2 \times 3 \sqrt{3} = 6 \sqrt{3}$

$\Rightarrow \frac{12 + \sqrt{108}}{6}$

$= \frac{12 + 6 \sqrt{3}}{6}$

$= \frac{12}{6} + \frac{6 \sqrt{3}}{6}$

$= 2 + \sqrt{3}$