Question
A rectangular piece of carpet covers 180 yd squared. The width is 2yd less than the length. Find the length and width.
Round your answers to the nearest tenth of a yard.
Round your answers to the nearest tenth of a yard.
Answers
GPT-4o mini
Let's denote the length of the carpet as \( L \) yards and the width as \( W \) yards. According to the problem, we have the following two relationships:
1. The area of the carpet:
\[
L \times W = 180 \quad \text{(1)}
\]
2. The width is 2 yards less than the length:
\[
W = L - 2 \quad \text{(2)}
\]
Now, we can substitute equation (2) into equation (1):
\[
L \times (L - 2) = 180
\]
Expanding this equation gives:
\[
L^2 - 2L = 180
\]
Rearranging this into standard quadratic form results in:
\[
L^2 - 2L - 180 = 0
\]
Next, we can solve this quadratic equation using the quadratic formula:
\[
L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = -2 \), and \( c = -180 \). Plugging these values into the formula, we have:
\[
L = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-180)}}{2 \cdot 1}
\]
Calculating the components:
1. \( b^2 = 4 \)
2. \( 4ac = 4 \cdot 1 \cdot (-180) = -720 \)
3. Therefore, \( b^2 - 4ac = 4 + 720 = 724 \)
Substituting these values in:
\[
L = \frac{2 \pm \sqrt{724}}{2}
\]
Now calculating \( \sqrt{724} \):
\[
\sqrt{724} \approx 26.9 \quad \text{(rounded to one decimal place)}
\]
Now substituting back to find \( L \):
\[
L = \frac{2 \pm 26.9}{2}
\]
Calculating the two possible solutions for \( L \):
1. \( L = \frac{28.9}{2} \approx 14.45 \)
2. \( L = \frac{-24.9}{2} \) (This will yield a negative length, which is not valid.)
So we take:
\[
L \approx 14.45
\]
Now substituting that back into (2) to find \( W \):
\[
W = L - 2 \approx 14.45 - 2 \approx 12.45
\]
Finally, rounding both to the nearest tenth, we have:
- Length \( L \approx 14.5 \) yards
- Width \( W \approx 12.5 \) yards
Thus, the dimensions of the carpet are:
- Length: **14.5 yards**
- Width: **12.5 yards**
1. The area of the carpet:
\[
L \times W = 180 \quad \text{(1)}
\]
2. The width is 2 yards less than the length:
\[
W = L - 2 \quad \text{(2)}
\]
Now, we can substitute equation (2) into equation (1):
\[
L \times (L - 2) = 180
\]
Expanding this equation gives:
\[
L^2 - 2L = 180
\]
Rearranging this into standard quadratic form results in:
\[
L^2 - 2L - 180 = 0
\]
Next, we can solve this quadratic equation using the quadratic formula:
\[
L = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
Here, \( a = 1 \), \( b = -2 \), and \( c = -180 \). Plugging these values into the formula, we have:
\[
L = \frac{-(-2) \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot (-180)}}{2 \cdot 1}
\]
Calculating the components:
1. \( b^2 = 4 \)
2. \( 4ac = 4 \cdot 1 \cdot (-180) = -720 \)
3. Therefore, \( b^2 - 4ac = 4 + 720 = 724 \)
Substituting these values in:
\[
L = \frac{2 \pm \sqrt{724}}{2}
\]
Now calculating \( \sqrt{724} \):
\[
\sqrt{724} \approx 26.9 \quad \text{(rounded to one decimal place)}
\]
Now substituting back to find \( L \):
\[
L = \frac{2 \pm 26.9}{2}
\]
Calculating the two possible solutions for \( L \):
1. \( L = \frac{28.9}{2} \approx 14.45 \)
2. \( L = \frac{-24.9}{2} \) (This will yield a negative length, which is not valid.)
So we take:
\[
L \approx 14.45
\]
Now substituting that back into (2) to find \( W \):
\[
W = L - 2 \approx 14.45 - 2 \approx 12.45
\]
Finally, rounding both to the nearest tenth, we have:
- Length \( L \approx 14.5 \) yards
- Width \( W \approx 12.5 \) yards
Thus, the dimensions of the carpet are:
- Length: **14.5 yards**
- Width: **12.5 yards**