How do you simplify #(2sqrt2 + 2sqrt24) * sqrt3#? Prealgebra Exponents, Radicals and Scientific Notation Square Root 1 Answer Shwetank Mauria Jun 9, 2016 #(2sqrt2+2sqrt24)*sqrt3=2sqrt6+12sqrt2# Explanation: #(2sqrt2+2sqrt24)*sqrt3# = #2sqrt2*sqrt3+2sqrt24*sqrt3# and as #sqrtaxxsqrtb=sqrtab#, this is equal to #2sqrt6+2sqrt72# = #2sqrt(2xx3)+2sqrt(ul(2xx2)xx2xxul(3xx3))# = #2sqrt6+12sqrt2# Answer link Related questions How do you simplify #sqrt735/sqrt5#? How do you rationalize the denominator and simplify #1/sqrt11#? How do you multiply #sqrt[27b] * sqrt[3b^2L]#? How do you simplify #7sqrt3 + 8sqrt3 - 2sqrt2#? How do you simplify #sqrt468 #? How do you simplify #sqrt(48x^3) / sqrt(3xy^2)#? How do you simplify # sqrt ((4a^3 )/( 27b^3))#? How do you simplify #sqrt140#? How do you simplify #sqrt216#? How do you simplify #sqrt540#? See all questions in Square Root Impact of this question 3505 views around the world You can reuse this answer Creative Commons License