How do you evaluate #\frac { 7} { 2+ 2\sqrt { 3} }#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Shwetank Mauria Dec 28, 2016 #7/(2+2sqrt3)=(7sqrt3-7)/4# Explanation: #7/(2+2sqrt3)=7/(2sqrt3+2)# = #7/(2sqrt3+2)xx(2sqrt3-2)/(2sqrt3-2)# and using #(a+b)(a-b)=(a^2-b^2)#, we get #(7(2sqrt3-2))/((2sqrt3)^2-2^2)# = #(14sqrt3-14)/(12-4)# = #(14sqrt3-14)/8# and dividing numerator and denominator by #2# = #(7sqrt3-7)/4# 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 996 views around the world You can reuse this answer Creative Commons License