To find the width of the border that Hassan wants to add to his photo, we can start by calculating the area of the picture itself and then relate it to the area of the entire framed picture.
The area of the picture (length = 10 inches, width = 6 inches) is given by: \[ \text{Area of the picture} = \text{length} \times \text{width} = 10 \times 6 = 60 \text{ square inches} \]
The final area of the entire framed picture is 96 square inches.
Next, we need to determine the area occupied by the border: \[ \text{Area of the border} = \text{Final area} - \text{Area of the picture} = 96 - 60 = 36 \text{ square inches} \]
Let \( x \) be the width of the border. When we add the border around the picture, the new dimensions of the framed picture become:
- Length: \( 10 + 2x \)
- Width: \( 6 + 2x \)
Now, we can express the area of the entire framed picture in terms of \( x \): \[ \text{Area of the framed picture} = (10 + 2x)(6 + 2x) \]
Setting this equal to 96 square inches: \[ (10 + 2x)(6 + 2x) = 96 \]
Expanding this equation: \[ 60 + 20x + 12x + 4x^2 = 96 \] \[ 4x^2 + 32x + 60 = 96 \]
Now, simplify the equation: \[ 4x^2 + 32x + 60 - 96 = 0 \] \[ 4x^2 + 32x - 36 = 0 \]
Next, we can divide the entire equation by 4: \[ x^2 + 8x - 9 = 0 \]
Now, we will factor this quadratic equation: \[ (x + 9)(x - 1) = 0 \]
Setting each factor to zero gives: \[ x + 9 = 0 \quad \Rightarrow \quad x = -9 \quad (\text{not feasible since width can't be negative}) \] \[ x - 1 = 0 \quad \Rightarrow \quad x = 1 \]
Thus, the width of the border is: \[ \boxed{1 \text{ inch}} \]