How do you solve then check for extraneous solutions: #(x+5)^(2/3)=4#?

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Answer 1 · 2018-03-25T13:19:36

#x=3#

Raise both sides to the power of #3/2#

#((x+5)^(2/3))^(3/2) = 4^(3/2)#

#x+5 = (+-sqrt4)^3#

There are two possible solutions for #sqrt4#

#x+5 = (+2)^3" "or" "x+5 = (-2)^3#

#" "x+5 = 8" "or" "x+5 = -8#

#x = 8-5" "or" "x = -8-5#

#x = 3 or x=-13#

Check both in the original equation.

#(3+5)^(2/3) = 8^(2/3) = 4#

#(13+5)^(2/3) = 13^(2/3) =1.768" "larr# this does not give #4#