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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