How do you simplify #1/4 * sqrt(72)#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Don't Memorise Apr 28, 2016 #= (3sqrt2)/2# Explanation: #1/4 * sqrt72# #sqrt72# can be simplified as : #sqrt 72 = sqrt ( 3 * 3 * 2 * 2 * 2) = sqrt ( 3^2 * 2^2 *2) = color(green)(6 sqrt2# #1/4 * sqrt72 = 1/4 * color(green)(6sqrt2# #= ( 6sqrt2) / 4# #= ( cancel6sqrt2) / cancel4# #= (3sqrt2)/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 1271 views around the world You can reuse this answer Creative Commons License