How do you rationalize the denominator and simplify #1 / (1+sqrt3-sqrt5)#?

1 Answer
George C.
May 10, 2016

#1/(1+sqrt(3)-sqrt(5))=(7 + 3sqrt(3)+sqrt(5)+2sqrt(15))/11#

Explanation:

We will use the difference of squares identity twice:

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

Let us first multiply numerator and denominator by:

#1+sqrt(3)+sqrt(5)#

as follows:

#1/(1+sqrt(3)-sqrt(5))#

#= (1+sqrt(3)+sqrt(5))/((1+sqrt(3)-sqrt(5))(1+sqrt(3)+sqrt(5)))#

#= (1+sqrt(3)+sqrt(5))/(((1+sqrt(3))-sqrt(5))((1+sqrt(3))+sqrt(5)))#

#=(1+sqrt(3)+sqrt(5))/((1+sqrt(3))^2-(sqrt(5))^2)#

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

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

Then multiply numerator and denominator by #(2sqrt(3)+1)# as follows:

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

#=((1+sqrt(3)+sqrt(5))(2sqrt(3)+1))/((2sqrt(3))^2-1^2)#

#=1/11 (1+sqrt(3)+sqrt(5))(2sqrt(3)+1)#

#=1/11 (2sqrt(3)(1+sqrt(3)+sqrt(5)) + (1+sqrt(3)+sqrt(5)))#

#=1/11 ((2sqrt(3)+6+2sqrt(15)) + (1+sqrt(3)+sqrt(5)))#

#=1/11 (7 + 3sqrt(3)+sqrt(5)+2sqrt(15))#

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