How do you rationalize the denominator and simplify #(10-sqrt3)/(6+sqrt6)#?

1 Answer
Mar 15, 2016

Answer:

#(60-10sqrt(6)-6sqrt(3)+3sqrt(2))/30#

Explanation:

#1#. Start by multiplying the numerator and denominator by the conjugate of the fraction's denominator, #6-sqrt(6)#.

#(10-sqrt(3))/(6+sqrt(6))#

#=(10-sqrt(3))/(6+sqrt(6))((6-sqrt(6))/(6-sqrt(6)))#

#2#. Simplify the numerator.

#=(60-10sqrt(6)-6sqrt(3)+sqrt(18))/(6+sqrt(6))(1/(6-sqrt(6)))#

#=(60-10sqrt(6)-6sqrt(3)+3sqrt(2))/(6+sqrt(6))(1/(6-sqrt(6)))#

#3#. Simplify the denominator. Note that it contains a difference of squares #(color(red)(a^2-b^2=(a+b)(a-b)))#.

#=(60-10sqrt(6)-6sqrt(3)+3sqrt(2))/(36-6)#

#=color(green)(|bar(ul(color(white)(a/a)(60-10sqrt(6)-6sqrt(3)+3sqrt(2))/30color(white)(a/a)|)))#