What is the value of #2sqrt2sqrt2sqrt2sqrt2#? Prealgebra Exponents, Radicals and Scientific Notation Square Root 1 Answer Shwetank Mauria Sep 9, 2016 #2sqrt2sqrt2sqrt2sqrt2=8# Explanation: #2sqrt2sqrt2sqrt2sqrt2# = #2xxsqrt2xxsqrt2xxsqrt2xxsqrt2# = #2xxsqrt(2xx2xx2xx2)# = #2xxsqrt(ul(2xx2)xxul(2xx2))# = #2xx2xx2# = #8# 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 1799 views around the world You can reuse this answer Creative Commons License