Let the width of the rectangle be \( w \) meters. According to the problem, the length \( l \) of the rectangle is given by:
\[ l = 2w - 4 \]
The area \( A \) of the rectangle is given by the formula:
\[ A = l \times w \]
Substituting the expression for the length into the area formula:
\[ 96 = (2w - 4) \times w \]
Expanding this equation:
\[ 96 = 2w^2 - 4w \]
Rearranging it into a standard quadratic equation format:
\[ 2w^2 - 4w - 96 = 0 \]
Dividing the entire equation by 2 for simplicity:
\[ w^2 - 2w - 48 = 0 \]
Now, we will solve this quadratic equation using the trial and error method. We are looking for two numbers that multiply to \(-48\) and add to \(-2\).
Trying some factors of \(-48\):
- \(6\) and \(-8\) multiply to \(-48\) and add to \(-2\).
Thus, we can factor the equation as follows:
\[ (w - 8)(w + 6) = 0 \]
Setting each factor to zero gives:
\[ w - 8 = 0 \quad \Rightarrow \quad w = 8 \] \[ w + 6 = 0 \quad \Rightarrow \quad w = -6 \quad \text{(not valid since width cannot be negative)} \]
So, the width \( w \) is \( 8 \) meters. Now we can find the length \( l \):
\[ l = 2w - 4 = 2(8) - 4 = 16 - 4 = 12 , \text{meters} \]
Therefore, the dimensions of the rectangle are:
Width: \( 8 \) meters
Length: \( 12 \) meters