## Precalculus (6th Edition) Blitzer

$5$ inches
The original side is $x$ inches (uknown). The sides are now $(x+3)$ inches each, and Area = $(x+3)^{2}$ $64=x^{2}+6x+9$ $0=x^{2}+6x-55$ Factor the RHS by finding factors of $-55$ with sum $+6$. We find $+11$ and $-5$ $0=(x+11)(x-5)$ $w=-11$ is discarded as a negative side length, which makes no sense. $w=5$ (inches) is the original side length.