To determine the width of the border Hassan wants to add to the photo, we can start by calculating the area of the original picture and then find the area of the border.
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Calculate the area of the original picture: The area \( A \) of the picture is given by the formula: \[ A = \text{length} \times \text{width} \] Substituting the length \( 10 \) inches and width \( 6 \) inches: \[ A = 10 \times 6 = 60 \text{ square inches} \]
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Determine the area of the border: The area of the final framed picture is \( 96 \) square inches, so the area of the border added is: \[ \text{Area of border} = \text{Area of framed picture} - \text{Area of original picture} \] \[ \text{Area of border} = 96 - 60 = 36 \text{ square inches} \]
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Express the dimensions of the framed picture: Let \( x \) be the width of the border. The dimensions of the framed picture will then be:
- Length: \( 10 + 2x \)
- Width: \( 6 + 2x \)
Therefore, the area of the framed picture can be expressed as: \[ (10 + 2x)(6 + 2x) = 96 \]
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Expanding the equation: \[ (10 + 2x)(6 + 2x) = 60 + 20x + 12x + 4x^2 = 4x^2 + 32x + 60 \] Setting this equal to \( 96 \): \[ 4x^2 + 32x + 60 = 96 \]
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Rearranging the equation: \[ 4x^2 + 32x + 60 - 96 = 0 \] \[ 4x^2 + 32x - 36 = 0 \]
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Dividing the entire equation by 4: \[ x^2 + 8x - 9 = 0 \]
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Factoring the quadratic equation: \[ (x + 9)(x - 1) = 0 \]
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Finding the values of \( x \): \[ x + 9 = 0 \implies x = -9 \quad (\text{not valid as width cannot be negative}) \] \[ x - 1 = 0 \implies x = 1 \]
Thus, the width of the border is 1 inch.