How do you rationalize the denominator and simplify #5/(sqrt3-1)#?

1 Answer
MathFact-orials.blogspot.com
Jul 5, 2016

#(5sqrt3+5)/2=(5/2)(sqrt3+1)#

Explanation:

We can simplify this fraction by using a clever use of the number 1 and also by keeping in mind that something in the form of:

#(a+b)(a-b)=a^2-b^2#, so that will allow us to rationalize the denominator without getting into messy square root issues.

Starting with the original:

#5/(sqrt3-1)#

We can now multiply by 1 and not change the value of the fraction:

#5/(sqrt3-1)(1)#

and we can now pick a value of 1 that will suit our purposes. We have something in the form of #(a-b)# already in the denominator, so let's multiply by #(a+b)#:

#5/(sqrt3-1)((sqrt3+1)/(sqrt3+1))#

which gets us:

#(5(sqrt3+1))/((sqrt3-1)(sqrt3+1))#

simplifying:

#(5sqrt3+5)/(3-1)=(5sqrt3+5)/2=(5/2)(sqrt3+1)#

Related questions
See all questions in Multiplication and Division of Radicals
Impact of this question
1854 views around the world
You can reuse this answer
Creative Commons License