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

Let w be the width of the table. The length is 2w−1. The area is w(2w−1). Given that the area is 10, solve for w. $w(2w−1)=10$ $2w^2−w=10$ $2w^2−w−10=10−10$ $2w^2−w−10=0$ $(2w−5)(w+2)=0$ Use the zero product property to solve for w. 2w−5=0 OR w+2=0 2w−5+5=0+5 OR w+2−2=0−2 2w=5 OR w=−2 2w÷2=5÷2 OR w=−2 w=2.5 OR w=−2 Since the width cannot be negative, the width is 2.5 ft. Find the length. l=2(2.5)−1=5−1=4 ft