Let #sqrt(1 + sqrt(3 + sqrt(5+ sqrt(x)))) = 2# What x? Prealgebra Fractions Fractions with Different Denominators 1 Answer Shwetank Mauria Apr 26, 2016 #x=961# Explanation: #sqrt(1+sqrt(3+sqrt(5+sqrtx)))=2# or #1+sqrt(3+sqrt(5+sqrtx))=4# or #sqrt(3+sqrt(5+sqrtx))=3# or #3+sqrt(5+sqrtx)=9# or #sqrt(5+sqrtx)=6# or #5+sqrtx=36# or #sqrtx=31# or #x=961# Answer link Related questions What is #3/8 -: 2/5#? What is #4##3/10##- ##3# #3/5#? What is #1/2 -: 3/4#? What is #1 1/4 xx 2 4/5#? What is the sum of #3 1/6 + (-5.5)#, expressed as a mixed number in simplest form? Josie had 1 gallon of ice cream. She used #3/10# gallon to make a chocolate milkshake and #3/10#... What is #4/5 -: 6 2/3#? How do you simplify #35/6 + 7/12 - 1 1/3#? How do you add #1/2 + 4/6#? How do you add #3 1/3 + 4 + 2/5#? See all questions in Fractions with Different Denominators Impact of this question 1870 views around the world You can reuse this answer Creative Commons License