How do you rationalize #(2sqrt3 - 3sqrt2)/ sqrt2#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Massimiliano May 8, 2015 In this way: #(2sqrt3 - 3sqrt2)/ sqrt2*sqrt2/sqrt2=(2sqrt3*sqrt2-3sqrt2*sqrt2)/2=# #=(2sqrt6-6)/2=(2(sqrt6-3))/2=sqrt6-3#. Answer link Related questions How do you simplify #\frac{2}{\sqrt{3}}#? How do you multiply and divide radicals? How do you rationalize the denominator? What is Multiplication and Division of Radicals? How do you simplify #7/(""^3sqrt(5)#? How do you multiply #(sqrt(a) +sqrt(b))(sqrt(a)-sqrt(b))#? How do you rationalize the denominator for #\frac{2x}{\sqrt{5}x}#? Do you always have to rationalize the denominator? How do you simplify #sqrt(5)sqrt(15)#? How do you simplify #(7sqrt(13) + 2sqrt(6))(2sqrt(3)+3sqrt(6))#? See all questions in Multiplication and Division of Radicals Impact of this question 1063 views around the world You can reuse this answer Creative Commons License