Question #46f3e

1 Answer
namralos · Stefan V.
May 2, 2017

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*54#

#108 = 2*2*27#

#108 = 2*2*3*9#

FINALLY, we have:

#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*2# extracts as a #2#.

#3*3# extracts as a #3#.

The remaining #3# stays under the radical.

So

#sqrt(108) = 2*3*sqrt(3)#.

Multiply the factors that are outside.

#sqrt(108) = 6sqrt(3)#

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