How do you simplify $2 \cdot \sqrt{- \frac{1}{6}} + \sqrt{- \frac{4}{3}}$ $2* sqrt(-1/6) + sqrt(-4/3)$ ?

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How do you simplify $2 \cdot \sqrt{- \frac{1}{6}} + \sqrt{- \frac{4}{3}}$ $2* sqrt(-1/6) + sqrt(-4/3)$ ?

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Answer 1 · 2016-03-25T18:29:56

$i \frac{\sqrt{3}}{3} (\sqrt{2} + 4)$ $isqrt(3)/3(sqrt(2)+4 )$

Explanation:

Simplify: $2 \cdot \sqrt{- \frac{1}{6}} + \sqrt{- \frac{4}{3}}$ $2*sqrt(-1/6)+sqrt(-4/3)$
$2 \cdot i \sqrt{\frac{1}{6}} + i \sqrt{\frac{4}{3}}$ $2*isqrt(1/6)+isqrt(4/3)$
$\frac{2}{\sqrt{2}} \cdot i \sqrt{\frac{1}{3}} + i \sqrt{\frac{4}{3}}$ $2/sqrt(2)*isqrt(1/3)+isqrt(4/3)$
$(2 i) [\frac{1}{\sqrt{2} \sqrt{3}} + \frac{\sqrt{4}}{\sqrt{3}}] = 2 i \frac{1 + 2 \sqrt{2}}{\sqrt{6}}$ $(2i)[1/(sqrt(2)sqrt(3))+sqrt(4)/sqrt(3)] = 2i(1+2sqrt(2))/(sqrt(6)$
Multiply top and bottom by $\sqrt{6}$ $sqrt(6)$
$2 i \frac{1 + 2 \sqrt{2}}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = 2 i \frac{\sqrt{6} + 2 \sqrt{3} \sqrt{2} \cdot \sqrt{2}}{\sqrt{6} \cdot \sqrt{6}}$ $2i(1+2sqrt(2))/(sqrt(6))*sqrt(6)/sqrt(6)= 2i(sqrt(6)+2sqrt(3)sqrt(2)*sqrt(2))/(sqrt(6)*sqrt(6)$

$2 i \frac{\sqrt{2} \sqrt{3} + 4 \sqrt{3}}{6} = i \frac{\sqrt{3}}{3} (\sqrt{2} + 4)$ $2i(sqrt(2)sqrt(3)+4sqrt(3))/(6)=isqrt(3)/3(sqrt(2)+4 )$

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How do you simplify $2 \cdot \sqrt{- \frac{1}{6}} + \sqrt{- \frac{4}{3}}$ $2* sqrt(-1/6) + sqrt(-4/3)$ ?

1 Answer

Explanation:

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