How do you rationalize #2/(5- sqrt 3)#?

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Answer 1 · 2015-05-11T20:04:19

Try multiplying the numerator (top) and denominator (bottom) by the conjugate, #(5+sqrt(3))# :

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

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

#= (10+2sqrt(3))/(25-3)#

#= (10+2sqrt(3))/22#

#= (5+sqrt(3))/11#

This is based on the equation:

#(a-b)(a+b) = a^2 - b^2#