How do you solve #sqrt(33+sqrtx)=6#?
1 Answer
Nov 26, 2016
Explanation:
Square both sides:
#(sqrt(33 + sqrt(x)))^2 = 6^2#
#33 + sqrt(x) = 36#
#sqrt(x) = 3#
#(sqrt(x))^2 = 3^2#
#x = 9#
Hopefully this helps!
