How do you solve #sqrt(x+2)=x+2 # and find any extraneous solutions?
1 Answer
May 30, 2016
Explanation:
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
#0 = x^2+3x+2 = (x+1)(x+2)#
So
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.