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

(a) $l= \frac{600}{w}$ , $l=75-2w$ (b) $w=11.6 ft, 25.9 ft$ (c) $l= 23.2$ and $l= 51.7$
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. $w^2-\frac{75}{2}w+\frac{5625}{16}=-300+\frac{5625}{16}$ $\implies (w-\frac{75}{4})^2=\frac{825}{16}$ $\implies (w-\frac{75}{4})=7.2$ $w=11.6 ft, 25.9 ft$ (c) $l= \frac{600}{w}$ and $l= \frac{600}{w}$ $l= \frac{600}{11.6}=23.2$ and $l= \frac{600}{25.9}=51.7$