How do you solve and check for extraneous solutions in #root3(x-3) =1#?

1 Answer
Aug 7, 2015

Answer:

#x=4# with no extraneous solutions

Explanation:

Given #root(3)(x-3) = 1#

Cube both sides:
#color(white)("XXXX")##x-3 = 1^3 = 1#

Add 3 to both sides
#color(white)("XXXX")##x = 4#

Verify by substituting #4# for #x# in the original expression
#color(white)("XXXX")##root(3)(4-3) = root(3)(1) = 1#
so this solution is not extraneous