How do you simplify #(sqrt2-sqrt3)/(sqrt2+sqrt3)#?

1 Answer
Nov 6, 2017

Answer:

See a solution process below:

Explanation:

To eliminate the radicals from the denominator (or to rationalize the denominator) we can multiply the fraction by the appropriate form of #1#. In this case;

#(sqrt(2) - sqrt(3))/(sqrt(2) - sqrt(3))#

#(sqrt(2) - sqrt(3))/(sqrt(2) - sqrt(3)) xx (sqrt(2) - sqrt(3))/(sqrt(2) + sqrt(3)) =>#

#((sqrt(2))^2 - sqrt(2)sqrt(3) - sqrt(3)sqrt(2) + (sqrt(3))^2)/((sqrt(2))^2 + sqrt(2)sqrt(3) - sqrt(3)sqrt(2) - (sqrt(3))^2) =>#

#(2 - sqrt(6) - sqrt(6) + 3)/(2 + 0 - 3) =>#

#(5 - 2sqrt(6))/(-1) =>#

#-5 + 2sqrt(6)#

Or

#2sqrt(6) - 5#