How do you solve and check for extraneous solutions in $sqrt(x+2)=x-4$ ?

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How do you solve and check for extraneous solutions in $sqrt(x+2)=x-4$ ?

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Answer 1 · 2015-08-01T17:07:30

$color(red)(x = 7)$ is a solution.
$color(red)(x=2)$ is an extraneous solution.

Explanation:

SOLVE:

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

Square each side.

$x+2=(x-4)^2$

Remove parentheses.

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

Move all terms to the right hand side.

$0=x^2-8x+16-x-2$

Combine like terms.

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

Factor:

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

$x-7=0$ and $x-2=0$

$x = 7$ and $x=2$

CHECK FOR EXTRANEOUS SOLUTIONS

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

If $x=7$ ,

$sqrt(7+2) = 7-4$

$sqrt9 = 3$

$3=3$

$x=7$ is a solution.

If $x=2$ ,

$sqrt(2+2) = 2-4$

$sqrt4=-2$

$2 = -2$

This is impossible, so $x=2$ is an extraneous solution.

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How do you solve and check for extraneous solutions in $sqrt(x+2)=x-4$ ?

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