# How do you evaluate \frac { 1} { - \sqrt { 48} }?

Oct 22, 2017

$\frac{1}{- \sqrt{48}} = - \frac{\sqrt{3}}{12} \approx - \frac{195}{1351} \approx - 0.1443375$

#### Explanation:

Note that:

$48 = {4}^{2} \cdot 3$

$\sqrt{a b} = \sqrt{a} \sqrt{b} \text{ }$ if $a , b \ge 0$

So we find:

$\frac{1}{- \sqrt{48}} = - \frac{1}{\sqrt{{4}^{2} \cdot 3}}$

$\textcolor{w h i t e}{\frac{1}{- \sqrt{48}}} = - \frac{1}{\sqrt{{4}^{2}} \cdot \sqrt{3}}$

$\textcolor{w h i t e}{\frac{1}{- \sqrt{48}}} = - \frac{1}{4 \sqrt{3}}$

$\textcolor{w h i t e}{\frac{1}{- \sqrt{48}}} = - \frac{\sqrt{3}}{4 \sqrt{3} \sqrt{3}}$

$\textcolor{w h i t e}{\frac{1}{- \sqrt{48}}} = - \frac{\sqrt{3}}{12}$

That is the simplest form.

This is an irrational number. To calculate rational approximations we can use:

$\sqrt{{a}^{2} + b} = a + \frac{b}{2 a + \frac{b}{2 a + \frac{b}{2 a + \frac{b}{2 a + \ldots}}}}$

with $a = 7$ and $b = - 1$

as follows:

$\sqrt{48} = \sqrt{{7}^{2} - 1} = 7 - \frac{1}{14 - \frac{1}{14 - \frac{1}{14 - \frac{1}{14 - \frac{1}{14 - \ldots}}}}}$

So:

1/(-sqrt(48)) = -1/(7-1/(14-1/(14-1/(14-1/(14-1/(14-...)))))

$\textcolor{w h i t e}{\frac{1}{- \sqrt{48}}} \approx - \frac{1}{7 - \frac{1}{14 - \frac{1}{14}}} = - \frac{1}{7 - \frac{14}{195}} = - \frac{195}{1351} \approx - 0.1443375$