How do you solve #sqrt(3x-8) = 2# and find any extraneous solutions?

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Answer 1 · 2016-05-29T02:04:32

#x = 4#

As we have #sqrt(3x-8)# in the equation, we will require that #3x-8 >= 0#, that is, #x >= 8/3#.

#sqrt(3x-8) = 2#

#=>(sqrt(3x-8))^2 = 2^2#

#=> 3x-8 = 4#

(Note that typically #(sqrt(a))^2 = sqrt(a^2) = |a|#, but we already required #3x-8>=0#, meaning #|3x-8| = 3x-8#)

#=> 3x = 12#

#:. x = 4#

Checking our answer, we find that

#sqrt(3(4)-8) = sqrt(4) = 2#

as desired.