How do you simplify $\sqrt{\frac{32}{4}}$ $sqrt(32/4)$ ?

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How do you simplify $\sqrt{\frac{32}{4}}$ $sqrt(32/4)$ ?

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Answer 1 · 2016-01-29T17:02:55

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

Explanation:

There are 2 solutions :)

The first solution is:

Since $\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$ $sqrt(a/b) = sqrt(a)/sqrt(b)$ , where $b$ $b$ is not equal to $0$ $0$ .

First we simplify the numerator, since there is no exact value of $\sqrt{32}$ $sqrt32$ we take its perfect squares. $16$ $16$ is a perfect square since $4 \cdot 4 = 16$ $4*4 = 16$ . Dividing $32$ $32$ by $16$ $16$ , we get $16 \cdot 2 = 32$ $16 * 2 = 32$ , therefore:

$\sqrt{32} = \sqrt{16} \cdot \sqrt{2}$ $sqrt32 = sqrt16*sqrt2$
$= 4 \sqrt{2}$ $= 4sqrt2$

Since now we're done in the numerator, we're gonna simplify the denominator, since $4$ $4$ is perfect square, $4 = 2 \cdot 2$ $4 = 2 * 2$ , the $\sqrt{4}$ $sqrt4$ is equal to $2$ $2$ .

Plugging all the answers, we get:

$\frac{4 \sqrt{2}}{2}$ $(4sqrt2)/2$

since $4$ $4$ and $2$ $2$ is a whole number, we can divide these 2 whole numbers, we get:

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

this is the final answer :)

the 2nd solution is:

First we simply evaluate the fraction inside the radical sign (square root)

$\sqrt{\frac{32}{4}} = \sqrt{8}$ $sqrt(32/4) = sqrt(8)$

since, $\frac{32}{4}$ $32/4$ = $8$ $8$ , then we get $\sqrt{8}$ $sqrt8$

$\sqrt{8} = \sqrt{4} \cdot \sqrt{2}$ $sqrt8 = sqrt4*sqrt2$

$= 2 \sqrt{2}$ $= 2sqrt2$

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