How do you simplify the square root #sqrt(169/196)#? Prealgebra Exponents, Radicals and Scientific Notation Square Root 1 Answer Shwetank Mauria Nov 3, 2016 #sqrt(169/196)=13/14# Explanation: #sqrt(169/196# = #sqrt((13xx13)/(2xx2xx7xx7))# = #sqrt(13xx13)/sqrt(2xx2xx7xx7)# = #sqrt(ul(13xx13))/sqrt(ul(2xx2)xxul(7xx7))# = #13/(2xx7)# = #13/14# 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 9871 views around the world You can reuse this answer Creative Commons License