## Precalculus (6th Edition) Blitzer

Fill the blank with: $9$
$x^{2}+6x$ is a part of $\quad$ $x^{2}+2\cdot x\cdot\fbox{$3$} +\fbox{$3^{2}$}=(x+3)^{2}$ so we add $9$ (that is, $3^{2}$) to both sides to complete the square on the LHS. We obtain $x^{2}+6x+9=7+9$ $(x+3)^{2}=16$, from which, by the square root property, follows: $x+3=\pm 4$ $x=-3\pm 4$ Solution set = $\{-7,+1\}$ Fill the blank with: $9$