How do you rationalize the denominator: #3/sqrt3#?

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Answer 1 · 2018-03-07T19:36:51

#sqrt3#

This means that you do not want an irrational number (surd) in the denominator

#3/sqrt3 xx sqrt3/sqrt3" "larr sqrt3/sqrt3 =1#

#=(3sqrt3)/(sqrt3^2)#

#=(cancel3sqrt3)/cancel3#

#=sqrt3#

Answer 2 · 2018-03-07T19:41:12

#3/sqrt3=color(blue)((3sqrt3)/3#

Given:

#sqrt3#

Rationalize the denominator by multiplying the numerator and denominator by #sqrt3#.

#(3*sqrt3)/(sqrt3sqrt3)#

Apply the rule #sqrtasqrta=a#.

#(3sqrt3)/3#

Simplify.

#(color(red)cancel(color(black)(3))^1sqrt3)/color(red)cancel(color(black)(3))^1#

#sqrt3# #larr# answer