Core Connections Algebra 2, 2013
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Core Connections Algebra 2, 2013 View details
2. Section 3.2
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Exercise 123 Page 159

Let the width of the walkway be x. What do the outer dimensions become then given the gardens width and length?

5* 8 meters

Practice makes perfect

Let's draw a diagram of Phana's garden. It's 2 meters wide and 5 meters long.

The total area of her inner garden is the product of its width and length. A= w l ⇒ 2 (5)=10 m^2

Let's call the width of the new walkway x meters. This means we can write the following algebraic expressions for the walkway's outer dimensions.

The area of the walkway can be written as the difference of the gardens area including the walkway, subtracted by the area of the inner garden. (5+2x)(2+2x)- 10 We know that the area of the walkway is 30 m^2, so we can set this expression equal to 30.
(5+2x)(2+2x)-10=30
Simplify
10+10x+4x+4x^2-10=30
4x^2+14x=30
4x^2+14x-30=0
2x^2+7x-15=0
This is a quadratic equation. Let's solve it with the Quadratic Formula. Notice that x represents the sides of a polygon which means we cannot have negative solutions.
2x^2+7x-15=0
x=- 7±sqrt(7^2-4( 2)( -15))/2( 2)
Evaluate right-hand side
x=- 7±sqrt(49-4(2)(-15))/2(2)
x=- 7±sqrt(49+120)/4
x=- 7±sqrt(169)/4
x=-7±13/4

x > 0

x=-7+13/4
x=6/4
x=1.5
The sidewalk is x= 1.5 meters wide. We use this value to determine the width and length of walkways outer dimensions. w&= 2+2x ⇒ 2+2( 1.5) = 5 l &= 5+2x ⇒ 5+2( 1.5) = 8 The outer dimensions of the walkway are 5* 8 meters.