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

1 Answer
George C.
Apr 10, 2017

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

Explanation:

The difference of squares identity can be written:

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

Using this with #a=sqrt(6)# and #b=sqrt(5)#, we can multiply both numerator and denominator by #sqrt(6)+sqrt(5)# to eliminate the square roots from the denominator...

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

#color(white)(2/(sqrt(6)-sqrt(5))) = (2sqrt(6)+2sqrt(5))/((sqrt(6))^2-(sqrt(5))^2)#

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

#color(white)(2/(sqrt(6)-sqrt(5))) = (2sqrt(6)+2sqrt(5))/1#

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

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