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

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Answer 1 · 2016-06-02T23:49:36

Square both sides of the equation to get rid of the square root.

#(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 back in the original equation we find that only #x = 7# works. Therefore, the solution set is #{x = 7}#.

Hopefully this helps!