To determine the width of the border for the photo, let's first find the dimensions of the framed picture that Hasn't wants to create.
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Calculate the area of the picture:
The length of the picture is 10 inches and the width is 6 inches: \[ \text{Area of the picture} = \text{length} \times \text{width} = 10 \times 6 = 60 \text{ square inches} \] -
Calculate the total area of the frame (including the border):
The total area of the framed picture is 96 square inches, as given. -
Set up the equation for the framed dimensions:
Let \( x \) be the width of the border. The dimensions of the entire frame will then be:- Length: \( 10 + 2x \) (adding border on both sides)
- Width: \( 6 + 2x \) (adding border on both sides)
The area of the frame can be expressed as: \[ (10 + 2x)(6 + 2x) = 96 \]
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Expand the equation: \[ 60 + 20x + 12x + 4x^2 = 96 \] Which simplifies to: \[ 4x^2 + 32x + 60 = 96 \]
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Move 96 to the left side of the equation: \[ 4x^2 + 32x + 60 - 96 = 0 \] \[ 4x^2 + 32x - 36 = 0 \]
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Divide the equation by 4: \[ x^2 + 8x - 9 = 0 \]
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Now use the quadratic formula to solve for \( x \): The quadratic formula is given by: \[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \] Where \( a = 1, b = 8, c = -9 \): \[ x = \frac{{-8 \pm \sqrt{{8^2 - 4 \cdot 1 \cdot (-9)}}}}{2 \cdot 1} \] \[ x = \frac{{-8 \pm \sqrt{{64 + 36}}}}{2} \] \[ x = \frac{{-8 \pm \sqrt{100}}}{2} \] \[ x = \frac{{-8 \pm 10}}{2} \]
This gives two potential solutions: \[ x = \frac{2}{2} = 1 \quad \text{or} \quad x = \frac{-18}{2} = -9 \]
Since a negative border width doesn't make sense, we take \( x = 1 \).
Therefore, the width of the border is 1 inch.
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