McGraw Hill Integrated II, 2012
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McGraw Hill Integrated II, 2012 View details
6. Solving x^2+bx+c=0
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Exercise 32 Page 49

Practice makes perfect
a

The width of a high school soccer field is 45 yards shorter than its length as shown below.

We see above that the variable l represents the length of the field. Since the field has the shape of a rectangle, its area A can be calculated by multiplying the length by the width. A=l(l-45) ⇔ A=l^2-45l

b

Given that the area of the field is 9000 square yards, let's first substitute 9000 for A into the equation we wrote in Part A. Then, we will solve the equation by factoring and using the Zero Product Property.

A=l^2-45l
9000=l^2-45l
0=l^2-45l-9000
l^2-45l-9000=0
â–¼
Factor
l^2-120l+75l-9000=0
l(l-120)+75l-9000=0
l(l-120)+75(l-120)=0
(l-120)(l+75)=0
To solve this equation, we will apply the Zero Product Property.

(l-120)(l+75)=0
lcl-120=0 & (I) l+75=0 & (II)
ll=120 l+75=0
ll_1=120 l_2=-75

Since the length cannot be negative, we will ignore the negative value of l. Therefore, we can determine the dimensions of the field as shown below. Length:& 120 yd Width:& 120-45=75 yd