How do you solve #root3(3x - 8) = 1 # and find any extraneous solutions?

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Answer 1 · 2018-06-21T22:26:02

#x=3#

There are no extraneous solutions.

#root3(3x - 8) = 1#

#(root3(3x - 8))^3 = 1^3#

#3x-8=1#

#3x=9#

#x=3#

test for "extraneous solutions":

#root3(3x - 8) = 1#

#root3(3(3) - 8) = 1#

#root3(9- 8) = 1#

#root3(1) = 1#

#1=1# Checks.