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

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Answer 1 · 2015-07-06T21:12:51

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

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

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

$= (2i)/sqrt(3×2) + (2i)/sqrt3$

$= (2i)/(sqrt3× sqrt2) + (2i)/sqrt3$

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

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

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

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

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