How do you simplify #sqrt(8) / sqrt(3)#?

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Answer 1 · 2016-09-09T18:28:14

#(2 sqrt(6)) / (3)#

We have: #(sqrt(8)) / (sqrt(3))#

Let's express the numerator as a product:

#= (sqrt(4) cdot sqrt(2)) / (sqrt(3))#

#= (2 sqrt(2)) / (sqrt(3))#

Then, let's rationalise the denominator:

#= (2 sqrt(2)) / (sqrt(3)) cdot (sqrt(3)) / (sqrt(3))#

#= (2 cdot sqrt(6)) / (3)#

#= (2 sqrt(6)) / (3)#