How do you divide #(2sqrt108)/sqrt(180y)#?

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Answer 1 · 2015-04-01T04:00:27

#(2sqrt(3))/sqrt(5y)# or #(2sqrt(15y))/(5y)#

#(2sqrt(108))/sqrt(180y)#

We can factor the numbers inside the square root into perfect squares:

#(2sqrt(36 * 3))/sqrt(36*5y)#

Since the square root of #36# is #6#:

#(2*6sqrt(3))/(6sqrt(5y))#

Cancel out #6# from both the numerator and denominator:

#(2sqrt(3))/sqrt(5y)#

It is technically "simplified" here. If you are asked to rationalize the denominator, multiply both the numerator and the denominator by #sqrt(5y)#:

#((2sqrt(3))/sqrt(5y))*(sqrt(5y))/(sqrt(5y))#

#(2sqrt(15y))/(5y)#