Simplify #2sqrt27+3sqrt12#? Prealgebra Exponents, Radicals and Scientific Notation Square Root 1 Answer Shwetank Mauria Jun 22, 2017 #2sqrt27+3sqrt12=12sqrt3# Explanation: #2# square root of #27# plus #3# square root of #12# can be written as #2sqrt27+3sqrt12# = #2sqrt(3xx3xx3)+3sqrt(2xx2xx3)# = #2sqrt(ul(3xx3)xx3)+3sqrt(ul(2xx2)xx3)# = #2xx3xxsqrt3+3xx2xxsqrt3# = #6sqrt3+6sqrt3# = #(6+6)sqrt3# = #12sqrt3# Answer link Related questions How do you simplify #(2sqrt2 + 2sqrt24) * sqrt3#? 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#? See all questions in Square Root Impact of this question 1521 views around the world You can reuse this answer Creative Commons License