How do you solve #3x^(4/3) + 5 = 53# and find any extraneous solutions?

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How do you solve #3x^(4/3) + 5 = 53# and find any extraneous solutions?

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Answer 1 · 2016-08-10T20:07:36

#x=8#

Explanation:

#3x^(4/3)+5=53#
or
#3x^(4/3)=53-5#
or
#3x^(4/3)=48#
or
#x^(4/3)=48/3#
or
#x^(4/3)=16#
or
#x=16^(3/4)#
or
#x=root4(16^3)#
or
#x=(root4(2times2times2times2))^3#
or
#x=2^3#
or
#x=8#

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How do you solve #3x^(4/3) + 5 = 53# and find any extraneous solutions?

1 Answer

Explanation:

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