How do you simplify #sqrt(32/4)#?

1 Answer
Jan 29, 2016

Answer:

#2sqrt2#

Explanation:

There are 2 solutions :)

The first solution is:

Since #sqrt(a/b) = sqrt(a)/sqrt(b)#, where #b# is not equal to #0#.

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

#sqrt32 = sqrt16*sqrt2 #
#= 4sqrt2#

Since now we're done in the numerator, we're gonna simplify the denominator, since #4# is perfect square, #4 = 2 * 2#, the #sqrt4# is equal to #2#.

Plugging all the answers, we get:

#(4sqrt2)/2#

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

#2sqrt2#

this is the final answer :)

the 2nd solution is:

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

#sqrt(32/4) = sqrt(8)#

since, #32/4# = #8#, then we get #sqrt8#

#sqrt8 = sqrt4*sqrt2#

#= 2sqrt2#