How do you rationalize the denominator and simplify #1/(sqrt3 - sqrt5 - 2)#?

1 Answer
George C.
Apr 22, 2016

#1/(sqrt(3)-sqrt(5)-2) = -10/47-4/47sqrt(15)+7/47sqrt(3)-1/47sqrt(5)#

Explanation:

Multiply numerator and denominator by:

#(sqrt(3)+sqrt(5)-2)(sqrt(3)-sqrt(5)+2)(sqrt(3)+sqrt(5)+2)#

First note that:

#(sqrt(3)-sqrt(5)+2)(sqrt(3)+sqrt(5)+2)#

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

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

#=2+4sqrt(3)+4-5#

#= 1+4sqrt(3)#

Similarly:

#(sqrt(3)-sqrt(5)-2)(sqrt(3)+sqrt(5)-2)#

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

#=(sqrt(3)-2)^2-(sqrt(5))^2#

#=2-4sqrt(3)+4-5#

#= 1-4sqrt(3)#

So the denominator becomes:

#(1-4sqrt(3))(1+4sqrt(3))=1^2-(4sqrt(3))^2=1-48=-47#

Meanwhile, the numerator becomes:

#(sqrt(3)+sqrt(5)-2)(1+4sqrt(3))#

#=sqrt(3)+sqrt(5)-2 + 4sqrt(3)(sqrt(3)+sqrt(5)-2)#

#=sqrt(3)+sqrt(5)-2+12+4sqrt(15)-8sqrt(3)#

#=10+4sqrt(15)-7sqrt(3)+sqrt(5)#

So:

#1/(sqrt(3)-sqrt(5)-2) = (10+4sqrt(15)-7sqrt(3)+sqrt(5))/(-47)#

#=-10/47-4/47sqrt(15)+7/47sqrt(3)-1/47sqrt(5)#

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