How do you rationalize the denominator and simplify #1/{1+sqrt(3)-sqrt(5)}#?

2 Answers
Apr 2, 2018

Answer:

Attempt to make the three term denominator two terms..

Explanation:

Multiply the fraction by #(1+ sqrt3 + sqrt5)/(1+sqrt3+sqrt5)# . This is the same as multiplying by 1. This is two eliminate two surds.
The denominator you should get after multiplying the two fractions is
#(1+sqrt3+sqrt5) * (1+sqrt3-sqrt5)# . We can write this as #((1+sqrt3)-sqrt5)(1+sqrt3)+sqrt5)#.
This can then be made is DOPS and written as : #(1+sqrt3)^2-(sqrt5)^2#.
Simplify that to get: #1+3+2sqrt3-5#
I'll leave it here :).

Apr 2, 2018

Answer:

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

Explanation:

This involves two stages of rationalisation to get rid of terms in #sqrt(3)# and #sqrt(5)#. Both steps use the difference of squares identity:

#A^2-B^2=(A-B)(A+B)#

So:

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

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

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

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

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

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

#color(white)(1/(1+sqrt(3)-sqrt(5))) = ((1+sqrt(3)+sqrt(5))(2sqrt(3)+1))/(12-1)#

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

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

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

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