How do you simplify #sqrt(72/9)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Lovecraft Oct 3, 2015 #sqrt(72/9) = 2sqrt(2)# Explanation: We know #72 = 9*8# so we can rewrite that as #sqrt(72/9) = sqrt((9*8)/9) = sqrt(8)# We know that #8 = 4*2# and that #sqrt(4) = 2# thus, we can rewrite and solve this as #sqrt(72/9) = 2sqrt(2)# 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 3387 views around the world You can reuse this answer Creative Commons License