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

1 Answer
Apr 3, 2015

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

The rationalization of this denominator is a two-set process.
First rationalize the #sqrt(5)#
and then, after some simplification
rationalize the #sqrt(3)#

Let #a=(1+sqrt(3))#
So our initial stage is to rationalize the denominator of
#1/(a-sqrt(5))#

As usual to do this we multiply the numerator and the denominator by the conjugate of the denominator:
#1/(a-sqrt(5)) * (a+sqrt(5))/(a+sqrt(5))#

#= (a+sqrt(5))/(a^2 - 5)#

Substituting #(1+sqrt(3))# back in to this expression in place of #a#.
we get
#= (1+sqrt(3)+sqrt(5))/((1+sqrt(3))^2-5)#

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

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

Rationalizing this denominator (using the conjugate of #2sqrt(3)-1#
results in
#((1+sqrt(3)+sqrt(5)))/((2sqrt(3)-1)) * ((2sqrt(3)+1))/((2sqrt(3)+1))#

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