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

1 Answer
George C.
Mar 14, 2016

#2/(sqrt(6)-sqrt(5))=2(sqrt(6)+sqrt(5))#

Explanation:

Note the difference of squares identity:

#a^2-b^2=(a-b)(a+b)#

We use this with #a=sqrt(6)# and #b=sqrt(5)# below...

Multiply both numerator and denominator by #sqrt(6)+sqrt(5)# ...

#2/(sqrt(6)-sqrt(5))=2/(sqrt(6)-sqrt(5))*(sqrt(6)+sqrt(5))/(sqrt(6)+sqrt(5))#

#=(2(sqrt(6)+sqrt(5)))/(6-5)#

#=2(sqrt(6)+sqrt(5))#

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