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

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Answer 1 · 2016-05-30T08:22:17

#x=-1# or #x=-2#

First square both sides (noting that this is where any extraneous solutions may be introduced) to give:

#x+2 = (x+2)^2 = x^2+4x+4#

Subtract #x+2# from both ends to get:

#0 = x^2+3x+2 = (x+1)(x+2)#

So #x=-1# or #x=-2#

Checking each of these:

#sqrt((-1)+2) = sqrt(1) = 1 = (-1)+2#

#sqrt((-2)+2) = sqrt(0) = 0 = (-2)+2#

So both of the solutions of our derived quadratic equation are also solutions of the original equation.