Simplify #sqrt300-sqrt48#? Prealgebra Exponents, Radicals and Scientific Notation Square Root 1 Answer Shwetank Mauria Oct 4, 2016 #sqrt300-sqrt48=6sqrt3# Explanation: #sqrt300# = #sqrt(2xx2xx3xx5xx5)# = #sqrt(ul(2xx2)xx3xxul(5xx5))# = #2xx5xxsqrt3# = #10sqrt3# Similarly #sqrt48# = #sqrt(2xx2xx2xx2xx3)# = #sqrt(ul(2xx2)xxul(2xx2)xx3)# = #2xx2xxsqrt3# = #4sqrt3# Hence #sqrt300-sqrt48# = #10sqrt3-4sqrt3# = #(10-4)sqrt3# = #6sqrt3# 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 2923 views around the world You can reuse this answer Creative Commons License