Algebra 1

Published by Prentice Hall
ISBN 10: 0133500403
ISBN 13: 978-0-13350-040-0

Chapter 9 - Quadratic Functions and Equations - 9-4 Factoring to Solve Quadratic Equations - Lesson Check - Page 557: 4


width = 2.5 ft length = 4 ft

Work Step by Step

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 $w=-2$ $2w\div2=5\div2$ 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
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