Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
5. Completing the Square
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Exercise 32 Page 580

Draw a diagram to write an expression in terms of the width of the walkway.

3.1ft

Practice makes perfect

We will begin by drawing a diagram to model the situation. Then, we will use the diagram to write an expression in terms of the width of the walkway. From there, we will solve the expression by completing the square

Drawing a diagram

A reflecting pool will measure 42ft by 26ft. This reflecting pool is being surrounded by a concrete walkway of uniform width x. Let's draw a diagram to find the measures of the walkway.

We get a rectangle whose dimensions are (26+2x) and (42+2x).

Using the Diagram to Write an Expression

To find the area A_w of the walkway, we need to subtract the area of the reflecting pool from the area of the big rectangle. A_w =(26+2x)(42+2x)- 26* 42 Let's simplify the right-hand side of the equation by applying the Distributive Property.
A_w =(26+2x)(42+2x)- 26* 42
Simplify right-hand side
A_w=(26+2x)(42+2x)-1092
A_w=26(42+2x)+2x(42+2x)-1092
A_w=1092+52x+2x(42+2x)-1092
A_w=1092+52x+84x+4x^2-1092
A_w=4x^2+136x
This expression gives us the area of the walkway in terms of x.

Solving the Expression

Since the amount of concrete can cover 460ft^2 of the walkway, the area of the walkway should be equal to 460. Let's substitute it. A_w =4x^2+136x ⇒ 460=4x^2+136x We get a quadratic equation. We will solve it by completing the square. Now let's divide each side by 4 so the coefficient of x^2 will be 1.
460= 4x^2+136x
460/4= 4x^2+136x/4
Simplify right-hand side
460/4 =4x^2/4+136x/4
460/4 =4/4x^2+136/4x
115=x^2+34x
x^2+34x=115
In a quadratic expression, b is the linear coefficient. For the equation above, we have that b=34. Let's now calculate ( b2 )^2.
( b/2 )^2
( 34/2 )^2
(17)^2
289
Next, we will add ( b2 )^2=289 to both sides of our equation. Then, we will factor the trinomial on the left-hand side, and solve the equation.
x^2+34x=115
x^2+34x+ 289=115+ 289
(x+17)^2=115+289
Solve for x
(x+17)^2=115+289
(x+17)^2=404
sqrt((x+17)^2)=sqrt(404)
x+17=± sqrt(404)
x=- 17 ± sqrt(404)
The solutions for this equation are x=- 17 ± sqrt(404). Let's separate them into the positive and negative cases.
x=- 17 ± sqrt(404)
x_1=- 17 + sqrt(404) x_2=- 17 - sqrt(404)
x_1=- 17 + 2sqrt(101) x_2=- 17 - 2sqrt(101)
x_1 ≈ 3.1 x_2≈ - 37.1

The only reasonable solution is 3.1 as length cannot be negative. The value of x is 3.1ft.