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

1 Answer
Jul 6, 2015

2sqrt(-1/6) + sqrt(-4/3) = (isqrt3)/3(2+sqrt2)

Explanation:

2sqrt(-1/6) + sqrt(-4/3) = 2sqrt((-1))/sqrt6 + sqrt(-4)/sqrt3

= (2i)/sqrt6 + (2i)/sqrt3

= (2i)/sqrt(3×2) + (2i)/sqrt3

= (2i)/(sqrt3× sqrt2) + (2i)/sqrt3

= (2i)/sqrt3(1/sqrt2 + 1)

= (2isqrt3)/3(sqrt2/2+ 1)

= (cancel(2)isqrt3)/3((sqrt2+ 2)/cancel(2))

= (isqrt3)/3(2+sqrt2)

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