How do you solve #sqrt(x+2) = x-4#?

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Answer 1 · 2016-05-10T23:01:49

#x=7#

First square both sides of the equation, noting that this may introduce spurious solutions:

#x+2 = (x-4)^2 = x^2-8x+16#

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

#0 = x^2-9x+14 = (x-2)(x-7)#

So #x=2# or #x=7#

If #x=2# then #sqrt(x+2) = sqrt(4) = 2 != -2 = 2 - 4 = x-4#

If #x=7# then #sqrt(x+2) = sqrt(9) = 3 = 7-4 = x-4#

So #x=7# is the only valid solution of the original equation.