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

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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!

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