Algebra 1

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

Chapter 9 - Quadratic Functions and Equations - 9-5 Completing the Square - Practice and Problem-Solving Exercises - Page 565: 33


(a) $l= \frac{600}{w}$ , $l=75-2w$ (b) $w=11.6 ft, 25.9 ft$ (c) $l= 23.2$ and $l= 51.9$

Work Step by Step

Length(l) x width(w) =600 $\implies l= \frac{600}{w}$ ...(1) (a) Total length of the playground$=2w+l$ Thus, $2w+l=75 ft$ so, $l=75-2w$ (b) From equation (1), we have $(\frac{600}{w})=75-2w$ $w^2-\frac{75}{2}w=-300$ Compare it with the standard form of quadratic equation $ax^2+bx+c$, we have $a=1, b=-\frac{75}{2}$ To complete the square, add $\dfrac{(-\frac{75}{2})^2}{4}=\frac{5625}{16}$ on both sides. $w^2-\frac{75}{2}w+\frac{5625}{16}=-300+\frac{5625}{16}$ $\implies (w-\frac{75}{4})^2=\frac{825}{16}$ Neglect negative sign, we have $\implies (w-\frac{75}{4})=7.2$ $w=11.6 ft, 25.9 ft$ (c) $l= \frac{600}{w}$ and $l= \frac{600}{w}$ $l= \frac{600}{11.6}=23.2$ and $l= \frac{600}{25.9}=51.9$
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