How do you simplify #sqrt 8/ sqrt3#?

2 Answers
Mar 28, 2018

Answer:

To answer is this: #sqrt(8)/(sqrt3)*sqrt(3)/(sqrt(3))=(sqrt8*sqrt3)/3=(sqrt24)/3#

Explanation:

In this question, since you do not want square root functions as the divisors, you multiply the top and bottom by the square root divisor at the bottom.

In this case, #sqrt(3)#. After multiplying the top and bottom by #sqrt(3)#, you will remove the square root term from the bottom and as such getting you only #3#, but the top will be #sqrt 8 * sqrt 3#, which is #sqrt 24#.

In your final answer, it will then become #sqrt(24)/ 3#.

Mar 28, 2018

Answer:

#(2sqrt6)/3#

Explanation:

rationalise the denominator, by multiplying by #sqrt3:#

#(sqrt3 * sqrt8)/(sqrt3 * sqrt3)#

#= (sqrt24)/3#

#sqrt24 = sqrt4 * sqrt6#

#= 2 * sqrt6 = 2sqrt6#

#(sqrt24)/3 = (2sqrt6)/3#