Bot GPT 4-o mini To solve the problem, let's define the width as \( w \) feet. According to the problem, the length of the frame is 6 feet longer than the width, so we can express the length as:
\[ l = w + 6 \]
Part A: Write an equation for the area of the frame
The area \( A \) of a rectangle is given by the formula:
\[ A = \text{length} \times \text{width} \]
Substituting our expressions for length and width into the area formula, we have:
\[ 96 = (w + 6) \times w \]
This expands to:
\[ 96 = w^2 + 6w \]
Rearranging gives us the quadratic equation:
\[ w^2 + 6w - 96 = 0 \]
Part B: Determine the dimensions of the frame
To find the width \( w \), we can use the quadratic formula, which is:
\[ w = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
In our equation \( w^2 + 6w - 96 = 0 \), we have:
- \( a = 1 \)
- \( b = 6 \)
- \( c = -96 \)
Let's calculate the discriminant:
\[ b^2 - 4ac = 6^2 - 4(1)(-96) = 36 + 384 = 420 \]
Now applying the quadratic formula:
\[ w = \frac{-6 \pm \sqrt{420}}{2(1)} = \frac{-6 \pm \sqrt{420}}{2} \]
Calculating \( \sqrt{420} \):
\[ \sqrt{420} \approx 20.49 \]
Now substituting back into the formula:
\[ w = \frac{-6 \pm 20.49}{2} \]
Calculating both possible values of \( w \):
\[ w = \frac{14.49}{2} \approx 7.245 \quad (\text{this is a valid width}) \] 2. \[ w = \frac{-26.49}{2} \approx -13.245 \quad (\text{not valid since width cannot be negative}) \]
Thus, we accept \( w \approx 7.245 \).
Now, we can find the length:
\[ l = w + 6 \approx 7.245 + 6 \approx 13.245 \]
Final Dimensions
- Width \( \approx 7.25 \) feet (approximately)
- Length \( \approx 13.25 \) feet (approximately)
So, the dimensions of the frame are approximately:
- Length: \( 13.25 \) feet
- Width: \( 7.25 \) feet
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