How do you solve #sqrt (x+2) +4 = x# and find any extraneous solutions?

1 Answer
May 6, 2018

Answer:

#color(maroon)("Solutions : " x = 2, 7#

Explanation:

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

#sqrt(x + 2) = x - 4#

Squaring both sides, #x + 2 = (x -4)^2#

#x + 2 = x^2 - 8x + 16#

#x^2 -9x + 14 = 0#

#x^2 - 2x - 7x + 14 = 0#

#x(x - 2) - 7 (x - 2) = 0#

#(x - 2) * (x - 7) = 0#

#x = 2, 7#