## Algebra 1: Common Core (15th Edition)

$n=3$ is extraneous.
$n=\sqrt 12-n$ With $n=-4$ then $-4=\sqrt 12-(-4)$ $-4\ne4$ With $n=3$ then $3=\sqrt 12-(3)$ $3=\sqrt9$ $3=3$ Hence, $n=-4$ is extraneous.