# What is 2* square root of (-1/6) + square root of (-4/3)?

Jul 6, 2015

$2 \sqrt{- \frac{1}{6}} + \sqrt{- \frac{4}{3}} = \frac{i \sqrt{3}}{3} \left(2 + \sqrt{2}\right)$

#### Explanation:

$2 \sqrt{- \frac{1}{6}} + \sqrt{- \frac{4}{3}} = 2 \frac{\sqrt{\left(- 1\right)}}{\sqrt{6}} + \frac{\sqrt{- 4}}{\sqrt{3}}$

$= \frac{2 i}{\sqrt{6}} + \frac{2 i}{\sqrt{3}}$

 = (2i)/sqrt(3×2) + (2i)/sqrt3

= (2i)/(sqrt3× sqrt2) + (2i)/sqrt3

$= \frac{2 i}{\sqrt{3}} \left(\frac{1}{\sqrt{2}} + 1\right)$

$= \frac{2 i \sqrt{3}}{3} \left(\frac{\sqrt{2}}{2} + 1\right)$

$= \frac{\cancel{2} i \sqrt{3}}{3} \left(\frac{\sqrt{2} + 2}{\cancel{2}}\right)$

$= \frac{i \sqrt{3}}{3} \left(2 + \sqrt{2}\right)$

This does not look much simpler than the original equation, but it has the radicals in the numerator and reduced to their lowest terms.