Chapter Test
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The area of a rectangle is length times width.
The equation must be in standard form before the Quadratic Formula can be applied.
Width: 7
Length: 8
Width: 7.26
Length: 8.26
We know that a rectangle has an area of 60 square feet, a width of x, and a length of x+1. Recall that the area of a rectangle is length times width.
| Width ⇔ x | Length ⇔ x+1 | Area ⇔ (x+1)(x) |
|---|---|---|
| 1 | 2 | 2 |
| 2 | 3 | 6 |
| 3 | 4 | 12 |
| 4 | 5 | 20 |
| 5 | 6 | 30 |
| 6 | 7 | 42 |
| 7 | 8 | 56 |
| 8 | 9 | 72 |
We want to use the values that get the area as close to 60 as possible. We can see that when the width is 7 feet and the length is 8 feet, the area is closest to 60 feet.
We will use the equation we set up in Part A.
60=(x+1)(x)
We can solve this equation for the width x and then use that value to find the length. Let's start to solve this equation by putting it in standard form.
Now we can apply the Quadratic Formula. 0= ax^2+ bx+ c ⇔ x=- b± sqrt(b^2-4 a c)/2 a We first need to identify the values of a, b, and c. 0=x^2+x-60 ⇔ 0= 1x^2+ 1x+( - 60)=0 We see that a= 1, b= 1, and c= - 60. Let's substitute these values into the Quadratic Formula.
Substitute values
Calculate power
a * 1=a
- a(- b)=a* b
Add terms
Calculate root
The solutions for this equation are x=-1± 15.52/2. Let's separate them into the positive and negative cases.
| x=-1± 15.52/2 | |
|---|---|
| x_1=-1 + 15.52/2 | x_2=-1 - 15.52/2 |
| x_1=14.52/2 | x_2=-16.52/2 |
| x_1=7.26 | x_2=-8.26 |
Using the Quadratic Formula, we found that the solutions of the given equation are x_1=7.26 and x_2=-8.26. Since x represents width, it wouldn't make sense to have a negative x-value. Therefore, x=7.26 feet is the only correct answer. We can use this to find the length. l=x+1 ⇒ l=7.26+1=8.26 We have found that the length is 8.26 feet and the width is 7.26 feet. These are rounded to the hundredths place.