How do you solve #root3(7n-8)=3 # and find any extraneous solutions?

1 Answer
A08
Aug 18, 2016

#n=5# and there are no extraneous solutions.

Explanation:

#root3(7n-8)=3#

Taking cube of both sides we get
#(root3(7n-8))^3=3^3#
#=>7n-8=27#

Rearranging and solving for #n#
#7n=27+8#
#n=35/7=5#
Insert #n=5# in original equation. We see that the equation holds good. We have a solution and there are no extraneous solutions.

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