We have the area of the photograph given by the expression \( x(x + 2) \), where \( x \) is the width and \( x + 2 \) is the length. We are told that the area is 48 square inches. Therefore, we can set up the equation:
\[ x(x + 2) = 48 \]
Expanding the left side, we get:
\[ x^2 + 2x = 48 \]
Next, we can rearrange this equation to set it to 0:
\[ x^2 + 2x - 48 = 0 \]
Now, we can factor this quadratic equation. We need to find two numbers that multiply to -48 and add up to 2. The correct factors are 8 and -6, as \( 8 \times -6 = -48 \) and \( 8 + (-6) = 2 \). Thus, we can factor the equation as:
\[ (x + 8)(x - 6) = 0 \]
Setting each factor equal to zero gives us:
- \( x + 8 = 0 \) \( \Rightarrow x = -8 \) (not applicable since width cannot be negative)
- \( x - 6 = 0 \) \( \Rightarrow x = 6 \)
This means the width of the photograph is:
\[ \boxed{6} \text{ inches} \]
To confirm, the length would be \( x + 2 = 6 + 2 = 8 \) inches. The area is \( 6 \times 8 = 48 \) square inches, which checks out.
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