How do you simplify #5sqrt8-4sqrt72+3sqrt96#? Prealgebra Exponents, Radicals and Scientific Notation Square Root 1 Answer Shwetank Mauria Apr 30, 2016 #5sqrt8-4sqrt72+3sqrt96=12sqrt6-14sqrt2# Explanation: #5sqrt8-4sqrt72+3sqrt96# = #5sqrt(ul(2xx2)xx2)-4sqrt(ul(2xx2)xx2xxul(3xx3))+3sqrt(ul(2xx2xx2xx2)xx2xx3)# = #5xx2xxsqrt2-4xx2xx3xxsqrt2+3xx2xx2xxsqrt(2xx3)# = #10sqrt2-24sqrt2+12sqrt6# = #12sqrt6-14sqrt2# 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 2148 views around the world You can reuse this answer Creative Commons License