How do you simplify #6/sqrt3#?

2 Answers
Mar 3, 2018

Answer:

#2sqrt3#

Explanation:

#"to simplify we "color(blue)"rationalise the denominator"#

#"that is eliminate the radical from the denominator"#

#"to do this multiply the numerator/denominator by "sqrt3#

#["noting in general that "sqrtaxxsqrta=a]#

#rArr6/sqrt3#

#=6/sqrt3xxsqrt3/sqrt3#

#=(6sqrt3)/3=2sqrt3#

Mar 3, 2018

Answer:

2#sqrt3#

Explanation:

A lot of times in mathematics, you don't want a square root in the denominator.

In order to get it out, multiply the fraction by another fraction, and that other fraction is the square root over itself.
In this example, what I'm talking about is this:
#(6/sqrt3)xx(sqrt3/sqrt3)#

By doing this, you actually are not changing the value of the original fraction, since #sqrt3/sqrt3# equals 1.

But when you multiply these two fractions (multiply across), you get: #6/sqrt3xxsqrt3/sqrt3=(6sqrt3)/(sqrt3sqrt3)=(6sqrt3)/3=6/3xxsqrt3=2xxsqrt3=2sqrt3#

It's simpler than the steps I included, I only made it long to try to run through the steps in a more understandable way. Hope this helped.