Question
The area of a rectangle is 1,056 square inches. Its length is 4 inches longer than 2 times its width.
Which equation can you use to find the width of the rectangle, w?
What is the width of the rectangle?
inches
Which equation can you use to find the width of the rectangle, w?
What is the width of the rectangle?
inches
Answers
GPT-4o mini
Let's define the width of the rectangle as \( w \). According to the problem, the length \( l \) of the rectangle is given by:
\[
l = 2w + 4
\]
The area \( A \) of a rectangle can be calculated using the formula:
\[
A = l \cdot w
\]
Given that the area is 1,056 square inches, we can set up the equation:
\[
w(2w + 4) = 1056
\]
Now, we can expand this equation:
\[
2w^2 + 4w = 1056
\]
To form a standard quadratic equation, we move 1,056 to the left side:
\[
2w^2 + 4w - 1056 = 0
\]
Now we can simplify this equation by dividing everything by 2:
\[
w^2 + 2w - 528 = 0
\]
Now, we can use the quadratic formula to solve for \( w \):
The quadratic formula is given by:
\[
w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In our case, \( a = 1 \), \( b = 2 \), and \( c = -528 \). Plugging in these values:
\[
w = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-528)}}{2 \cdot 1}
\]
\[
w = \frac{-2 \pm \sqrt{4 + 2112}}{2}
\]
\[
w = \frac{-2 \pm \sqrt{2116}}{2}
\]
Calculating the square root:
\[
\sqrt{2116} = 46
\]
Now substituting back into the equation:
\[
w = \frac{-2 \pm 46}{2}
\]
This gives us two potential solutions:
\[
w = \frac{44}{2} = 22 \quad \text{(positive width)}
\]
\[
w = \frac{-48}{2} = -24 \quad \text{(not possible, as width cannot be negative)}
\]
Therefore, the width of the rectangle is:
\[
\boxed{22} \text{ inches}
\]
\[
l = 2w + 4
\]
The area \( A \) of a rectangle can be calculated using the formula:
\[
A = l \cdot w
\]
Given that the area is 1,056 square inches, we can set up the equation:
\[
w(2w + 4) = 1056
\]
Now, we can expand this equation:
\[
2w^2 + 4w = 1056
\]
To form a standard quadratic equation, we move 1,056 to the left side:
\[
2w^2 + 4w - 1056 = 0
\]
Now we can simplify this equation by dividing everything by 2:
\[
w^2 + 2w - 528 = 0
\]
Now, we can use the quadratic formula to solve for \( w \):
The quadratic formula is given by:
\[
w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
In our case, \( a = 1 \), \( b = 2 \), and \( c = -528 \). Plugging in these values:
\[
w = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 1 \cdot (-528)}}{2 \cdot 1}
\]
\[
w = \frac{-2 \pm \sqrt{4 + 2112}}{2}
\]
\[
w = \frac{-2 \pm \sqrt{2116}}{2}
\]
Calculating the square root:
\[
\sqrt{2116} = 46
\]
Now substituting back into the equation:
\[
w = \frac{-2 \pm 46}{2}
\]
This gives us two potential solutions:
\[
w = \frac{44}{2} = 22 \quad \text{(positive width)}
\]
\[
w = \frac{-48}{2} = -24 \quad \text{(not possible, as width cannot be negative)}
\]
Therefore, the width of the rectangle is:
\[
\boxed{22} \text{ inches}
\]