How do you rationalize the denominator and simplify #2/(sqrt6-sqrt5)#?

1 Answer
Konstantinos Michailidis
Sep 24, 2015

Refer to explanation

Explanation:

When you have a formula like #sqrta-sqrtb# in the denominator you multiply with #sqrta+sqrtb# so

#2/(sqrt6-sqrt5)=2*(sqrt6+sqrt5)/((sqrt6-sqrt5)*(sqrt6+sqrt5))=2*(sqrt6+sqrt5)/((sqrt6)^2-(sqrt5)^2)=2*(sqrt6+sqrt5)/(6-5)=2*(sqrt6+sqrt5)#

Remarks

We used the identity #a^2-b^2=(a-b)*(a+b)#

Related questions
See all questions in Multiplication and Division of Radicals
Impact of this question
1663 views around the world
You can reuse this answer
Creative Commons License