How do you rationalize the denominator and simplify #2/(sqrt3+sqrt2)#?

1 Answer
Jan 15, 2017

Answer:

#(2)/(sqrt3+sqrt2)=2(sqrt3-sqrt2)#

Explanation:

Simplify #(2)/(sqrt3+sqrt2)#

Rationalize the denominator using the difference of squares, #a^2-b^2=(a+b)(a-b)#, where #a=sqrt3# and #b=sqrt2#.

#(2)/((sqrt3+sqrt2))*((sqrt3-sqrt2))/((sqrt3-sqrt2))=#

#(2(sqrt3-sqrt2))/((sqrt3+sqrt2)(sqrt3-sqrt2))=#

#(2(sqrt3-sqrt2))/(sqrt3^2-sqrt2^2)=#

Apply the rule #sqrt(a)^2=a#

#(2(sqrt3-sqrt2))/(3-2)=#

#(2(sqrt3-sqrt2))/1=#

#2(sqrt3-sqrt2)#