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

##### 2 Answers

Attempt to make the three term denominator two terms..

#### Explanation:

Multiply the fraction by

The denominator you should get after multiplying the two fractions is

This can then be made is DOPS and written as :

Simplify that to get:

I'll leave it here :).

#### Explanation:

This involves two stages of rationalisation to get rid of terms in

#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#