How do you simplify #sqrt540#? Prealgebra Exponents, Radicals and Scientific Notation Square Root 1 Answer Shwetank Mauria May 3, 2016 #sqrt540=6sqrt15# Explanation: #sqrt540=sqrt(ul(2xx2xx3xx3)xx3xx5)# = #2xx3xxsqrt(3xx5)# = #6sqrt15# 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 3048 views around the world You can reuse this answer Creative Commons License