How do you simplify $\sqrt{3} (\sqrt{12} - \sqrt{6})$ $sqrt3(sqrt12-sqrt6)$ ?

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How do you simplify $\sqrt{3} (\sqrt{12} - \sqrt{6})$ $sqrt3(sqrt12-sqrt6)$ ?

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Answer 1 · 2015-11-09T08:43:02

First, expand the product: multiply $\sqrt{3}$ $sqrt(3)$ with $\sqrt{12}$ $sqrt(12)$ and with $\sqrt{6}$ $sqrt(6)$ , respectively:

$\sqrt{3} (\sqrt{12} - \sqrt{6}) = \sqrt{3} \cdot \sqrt{12} - \sqrt{3} \cdot \sqrt{6}$ $sqrt(3)(sqrt(12) - sqrt(6)) = sqrt(3) * sqrt(12) - sqrt(3) * sqrt(6)$

Now, use the rule $\sqrt{a \cdot b} = \sqrt{a} \cdot \sqrt{b}$ $sqrt(a*b) = sqrt(a) * sqrt(b)$ :

$\sqrt{3} \cdot \sqrt{12} - \sqrt{3} \cdot \sqrt{6}$ $sqrt(3) * sqrt(12) - sqrt(3) * sqrt(6)$
$= \sqrt{3 \cdot 12} - \sqrt{3 \cdot 6}$ $= sqrt(3 * 12) - sqrt(3 * 6)$
$= \sqrt{36} - \sqrt{18}$ $= sqrt(36) - sqrt(18)$
$= 6 - \sqrt{18}$ $= 6 - sqrt(18)$

The last one, $\sqrt{18}$ $sqrt(18)$ , can be simplifyed further since $18 = 2 \cdot 9$ $18 = 2 * 9$ and $9 = 3^{2}$ $9 = 3^2$ :

$6 - \sqrt{18}$ $6 - sqrt(18)$
$= 6 - \sqrt{2 \cdot 9}$ $= 6 - sqrt(2*9)$
$= 6 - \sqrt{2} \cdot \sqrt{9}$ $= 6 - sqrt(2)*sqrt(9)$
$= 6 - \sqrt{2} \cdot 3$ $= 6 - sqrt(2)*3$
$= 6 - 3 \sqrt{2}$ $= 6 - 3sqrt(2)$

I hope that this helped!

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How do you simplify $\sqrt{3} (\sqrt{12} - \sqrt{6})$ $sqrt3(sqrt12-sqrt6)$ ?

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