Question #46f3e

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Question #46f3e

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Answer 1 · 2017-05-02T20:25:46

Let us assume a square root....

Explanation:

Assuming that we are simplifying a square root (rather than a cube root, fourth root, or other radical), we are looking for perfect squares that divide 108. It helps to factor 108 completely into primes.

$108 = 2 \cdot 54$ $108 = 2*54$

$108 = 2 \cdot 2 \cdot 27$ $108 = 2*2*27$

$108 = 2 \cdot 2 \cdot 3 \cdot 9$ $108 = 2*2*3*9$

FINALLY, we have:

$108 = 2 \cdot 2 \cdot 3 \cdot 3 \cdot 3$ $108 = 2*2*3*3*3$

Since we are searching for perfect squares, any pair of identical factors IS a perfect square. Perfect squares extract as a single factor of their type.

$2 \cdot 2$ $2*2$ extracts as a $2$ $2$ .

$3 \cdot 3$ $3*3$ extracts as a $3$ $3$ .

The remaining $3$ $3$ stays under the radical.

So

$\sqrt{108} = 2 \cdot 3 \cdot \sqrt{3}$ $sqrt(108) = 2*3*sqrt(3)$ .

Multiply the factors that are outside.

$\sqrt{108} = 6 \sqrt{3}$ $sqrt(108) = 6sqrt(3)$

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Question #46f3e

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