How do you solve #sqrt(x+2) = x# and are there any extraneous solutions?

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Answer 1 · 2018-03-13T14:17:22

#x = 2#

#sqrt (x + 2) = x#

square both sides:

#x + 2 = x^2#

#x^2 - x - 2 = 0#

# (x - 2)(x + 1) = 0#

#x = -1, x = 2#

Check:

#sqrt (-1 + 2) = -1#
#sqrt 1 = -1 : " extraneous solution"#

#sqrt (2 + 2) = 2#
#sqrt 4 = 2 # : this works in the original equation.