Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
1. Simplifying Rational Expressions
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Exercise 57 Page 669

The area of the blue border is the difference between the area of the entire wall and the white rectangle.

1.4 feet

Practice makes perfect

We want to paint a wall by painting a blue border with width x around a white rectangle.

We are asked to find the approximate width x that makes the area of the blue border and the area of the white rectangle the same. To do so, we will first find the mentioned areas one by one and then set them equal. Let's begin by finding the area of the white rectangle.

Area of the White Rectangle

We can find the side lengths of the white rectangle by using the given figure.

From here we can write an expression for the area of the white rectangle.

Area of the Rectangle = (8-2x)(12-2x)
â–¼
Simplify right-hand side
Area of the Rectangle = 12(8-2x)-2x(8-2x)
Area of the Rectangle = 96-24x-2x(8-2x)
Area of the Rectangle = 96-24x-16x+4x^2
Area of the Rectangle = 96-40x+4x^2
Area of the Rectangle = 4x^2-40x+96

Next, we will find the area of the blue border.

Area of the Blue Border

To find the area of the border we will subtract the area of the white rectangle from the area of the entire wall. Area of the Blue Border = Area of the Wall - Area of the White Rectangle We found the area of the white rectangle in the previous subheading. We will find the area of the entire wall as well. Area of the Wall &= 8* 12 &= 96 Now we are ready to write an expression for the area of the blue border.

Area of the Blue Border = 96- (4x^2-40x+96)
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Simplify right-hand side
Area of the Blue Border = 96-4x^2+40x-96
Area of the Blue Border = 96-96-4x^2+40x
Area of the Blue Border = -4x^2+40x

As we found the required areas, we can find the approximate value of x.

Finding x

We want the areas of the blue border and the white rectangle to be equal. Therefore, let's equate them to find x.

Area of the Blue Border = Area of the White Rectangle
-4x^2+40x = 4x^2-40x+96
â–¼
Simplify
40x=8x^2-40x+96
0=8x^2-80x+96
0=8(x^2-10x+12)
0=x^2-10x+12
x^2-10x+12=0

To find the approximate value of x we will use the Quadratic Formula.

x=- b±sqrt(b^2-4ac)/2a
x=- ( -10)±sqrt(( - 10)^2-4( 1)( 12))/2( 1)
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Solve for x and Simplify
x=10±sqrt((- 10)^2-4(1)(12))/2(1)
x=10±sqrt((- 10)^2-4(12))/2
x=10±sqrt(100-4(12))/2
x=10±sqrt(100-48)/2
x=10±sqrt(52)/2
x=10±sqrt(4* 13)/2
x=10± sqrt(4)* sqrt(13)/2
x=10± 2 sqrt(13)/2
x=2(5± sqrt(13))/2
x=5± sqrt(13)
x_1 ≈ 8.6 x_2 ≈ 1.4

Since the width of the entire wall is 8 feet, the width of the border needs to be smaller than 8 feet. x_1 ≈ 8.6 x_2 ≈ 1.4 Therefore, x is approximately 1.4 feet.