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Pearson Algebra 1 Common Core, 2011

PA

Pearson Algebra 1 Common Core, 2011 View details

5. Completing the Square

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Exercise 33 Page 580

Practice makes perfect

a A school plans to enclose the playground using fencing on three sides. Let w be the width of the playground.

We know that with the allocated budget, we can enclose 75ft of fencing. Then, we can write the equation below. 75=2w+l By subtracting 2w, we will get an expression for the length of the playground in terms of w. 75-2w=l

b We know the area of the playground, 600ft^2. The area of a rectangle is equal to its width times its length. Then, we have the following equation.

w * l = 600 Now we will substitute the expression we found in Part A and then solve for w.

w * l =600

l= 75-2w

w * ( 75-2w) =600

Distribute w

75w-2w^2=600

CommutativePropAdd

Commutative Property of Addition

- 2w^2+75w=600

We get a quadratic equation. We will solve it by completing the square. Let's divide each side by - 2 so the coefficient of x^2 will be 1.

- 2w^2+75w=600

.LHS /(- 2).=.RHS /(- 2).

- 2w^2+75w/- 2=600/- 2

▼

Simplify left-hand side

Write as a sum of fractions

- 2w^2/- 2+75w/- 2=600/- 2

MovePartNumRight

a* b/c=a/c* b

- 2/- 2w^2+75/- 2w=600/- 2

MoveNegDenomToFrac

Put minus sign in front of fraction

- 2/- 2w^2-75/2w=- 600/2

Calculate quotient

w^2-75/2w=- 300

In a quadratic expression, b is the linear coefficient. For the equation above, we have that b= 752. Let's now calculate ( b2 )^2.

( b/2 )^2

b= 75/2

( 75/2/2 )^2

▼

Simplify

Rewrite 75/2/2 as 75/2÷ 2

( 75/2 ÷ 2 )^2

Write as a fraction

( 75/2 ÷ 2/1 )^2

DivFracByFracDiv

a/b÷c/d=a/b*d/c

( 75/2 * 1/2 )^2

Multiply fractions

( 75/4 )^2

(a/b)^m=a^m/b^m

5625/16

Next, we will add ( b2 )^2= 562516 to both sides of our equation. Then, we will factor the trinomial on the left-hand side, and solve the equation.

w^2-75/2w=- 300

LHS+5625/16=RHS+5625/16

w^2-75/2w+ 5625/16=- 300+ 5625/16

FacPosPerfectSquare

a^2+2ab+b^2=(a+b)^2

(w-75/4)^2=- 300+5625/16

▼

Solve for w

Write as a fraction

(w-75/4)^2=- 4800/16+5625/16

Add fractions

(w-75/4)^2=825/16

sqrt(LHS)=sqrt(RHS)

sqrt((w-75/4)^2)=sqrt(825/16)

Calculate root

w-75/4=± sqrt(825/16)

sqrt(a/b)=sqrt(a)/sqrt(b)

w-75/4=± sqrt(825)/4

LHS+75/4=RHS+75/4

w=75/4± sqrt(825)/4

The solutions for this equation are w= 754± sqrt(825)4. Let's separate them into the positive and negative cases.

w=75/4± sqrt(825)/4
w_1=75/4+ sqrt(825)/4	w_2=75/4- sqrt(825)/4
w_1=75/4+ 5sqrt(33)/4	w_2=75/4- 5sqrt(33)/4
w_1 ≈ 25.9	w_2≈ 11.6

We have two solutions to the equation. The width of the playground is 25.9 or 11.6.

c To find the length, we will use the equation in Part A.

w	75-2w	l= 75-2w
w_1= 25.9	75-2* 25.9	l=23.2
w_2= 11.6	75- 2* 11.6	l=51.8

The length of the playground should be 23.2ft or 51.8ft.

Subchapter links

Lesson Check p.579

Practice and Problem-Solving Exercises p.579-581

Standardized Test Prep p.581

Mixed Review p.581