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

Question

2035 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-09-02T16:00:05

#{7}# is the solution set.

Square both sides.

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

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

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

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

#x = 7 and 2#

Checking in the original equation:

#sqrt(7 + 2) =^? 7 - 4#

#sqrt(9) = 3#

#sqrt(2 + 2) =^? 2 - 4#

#2 != -2#

Hence, the only actual solution is #x = 7#.

Hopefully this helps!