How do you simplify $9 \sqrt{2} + \sqrt{32}$ $9sqrt2 + sqrt32$ ?

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How do you simplify $9 \sqrt{2} + \sqrt{32}$ $9sqrt2 + sqrt32$ ?

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Answer 1 · 2016-02-13T13:00:24

I found $13 \sqrt{2}$ $13sqrt(2)$

Explanation:

We can manipulate the second root and write:
$9 \sqrt{2} + \sqrt{2 \cdot 16} =$ $9sqrt(2)+sqrt(2*16)=$
$= 9 \sqrt{2} + [\sqrt{2} \cdot \sqrt{16}] =$ $=9sqrt(2)+[sqrt(2)*sqrt(16)]=$
$= 9 \sqrt{2} + 4 \sqrt{2} =$ $=9sqrt(2)+4sqrt(2)=$
add:
$= 13 \sqrt{2}$ $=13sqrt(2)$

Answer 2 · 2016-02-13T13:02:20

$13 \sqrt{2}$ $13sqrt2$

Explanation:

to begin , simplify $\sqrt{32}$ $sqrt32$

32 may be written as 16 $\times 2$ $xx 2$

when simplifying , look for factors which are 'squares' as 16 is.

hence $\sqrt{32} = \sqrt{16 \times 2} = \sqrt{16} \times \sqrt{2} = 4 \sqrt{2}$ $sqrt32 = sqrt(16 xx 2 ) = sqrt16 xxsqrt2 = 4sqrt2$

expression can now be written : $9 \sqrt{2} + 4 \sqrt{2} = 13 \sqrt{2}$ $9sqrt2 + 4sqrt2 = 13sqrt2$

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How do you simplify $9 \sqrt{2} + \sqrt{32}$ $9sqrt2 + sqrt32$ ?

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How do you simplify $9 \sqrt{2} + \sqrt{32}$ $9sqrt2 + sqrt32$ ?