How do you simplify $\sqrt{3} (4 + \sqrt{12})$ $sqrt3 (4 + sqrt12)$ ?

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How do you simplify $\sqrt{3} (4 + \sqrt{12})$ $sqrt3 (4 + sqrt12)$ ?

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Answer 1 · 2016-06-24T04:05:15

$4 \sqrt{3} + 6$ $4sqrt(3)+6$

Explanation:

$\sqrt{3} \cdot (4 + \sqrt{12}) =$ $sqrt(3)*(4+sqrt(12)) =$
$= 4 \sqrt{3} + \sqrt{3} \cdot \sqrt{12} =$ $= 4sqrt(3)+sqrt(3)*sqrt(12) =$
$= 4 \sqrt{3} + \sqrt{3 \cdot 12} =$ $= 4sqrt(3)+sqrt(3*12) =$
$= 4 \sqrt{3} + \sqrt{36} =$ $= 4sqrt(3)+sqrt(36) =$
$= 4 \sqrt{3} + 6$ $= 4sqrt(3)+6$

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How do you simplify $\sqrt{3} (4 + \sqrt{12})$ $sqrt3 (4 + sqrt12)$ ?

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