How do you simplify #sqrt12 / sqrt5#?

2 Answers
Jul 15, 2017

Answer:

See a solution process below:

Explanation:

First, we can rationalize the denominator by multiplying the the fraction by the appropriate form of #1#:

#sqrt(5)/sqrt(5) * sqrt(12)/sqrt(5) =>#

#(sqrt(5) * sqrt(12))/(sqrt(5) * sqrt(5)) =>#

#sqrt(5 * 12)/5 =>#

#sqrt(60)/5#

Next, we can simplify the numerator as follows:

#sqrt(60)/5 =>#

#sqrt(4 * 15)/5 =>#

#(sqrt(4) * sqrt(15))/5 =>#

#(2sqrt(15))/5#

Jul 15, 2017

Answer:

#(2sqrt15)/5#

Explanation:

First, multiply the numerator and denominator by #sqrt5#, which will get rid of the radical in the denominator. This is essentially the same as multiplying the expression by #1#.

#sqrt12/sqrt5 * color(blue)(sqrt5/sqrt5)#

#sqrt5 * sqrt 5 = sqrt25 = 5#, so the denominator becomes #5#. Similarly, #sqrt12 * sqrt 5 = sqrt60#.

#=sqrt60/5#

Now, we can split #sqrt60# into #sqrt4*sqrt15#.

#=(sqrt4*sqrt15)/5#

#=(2sqrt15)/5#