How do you simplify #(5sqrt12+4sqrt147)/sqrt2#? Algebra Radicals and Geometry Connections Multiplication and Division of Radicals 1 Answer Binayaka C. Oct 23, 2016 #19sqrt6# Explanation: #(5sqrt12+4sqrt147)/sqrt2 = (5*sqrt(2*2*3)+ 4*sqrt(7*7*3))/sqrt2 = (5*2*sqrt3+4*7*sqrt3)/sqrt2 =(sqrt3*(10+28))/sqrt2=(38*sqrt3)/sqrt2=(19*sqrt2*cancelsqrt2*sqrt3)/cancelsqrt2=19*sqrt2*sqrt3=19sqrt6#[Ans] 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 895 views around the world You can reuse this answer Creative Commons License