Let the width of the path be denoted as \( x \). The rectangular piece of land has dimensions of 12 m by 5 m. When the path is added along one side and one end, the dimensions of the new area will change as follows:
- The length is extended by the width of the path on one end: \[ \text{New length} = 12 + x \]
- The width is extended by the width of the path on one side: \[ \text{New width} = 5 + x \]
The total area of the plot (land + path) can be expressed as the product of the new dimensions: \[ \text{Total area} = (\text{New length}) \times (\text{New width}) = (12 + x)(5 + x) \]
According to the problem, the total area is given as 120 m²: \[ (12 + x)(5 + x) = 120 \]
Now, we will expand the left side of the equation: \[ 12 \cdot 5 + 12x + 5x + x^2 = 120 \] \[ 60 + 17x + x^2 = 120 \]
Next, we will rearrange the equation to bring all terms to one side: \[ x^2 + 17x + 60 - 120 = 0 \] \[ x^2 + 17x - 60 = 0 \]
To solve the quadratic equation \( x^2 + 17x - 60 = 0 \), we can use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] where \( a = 1, b = 17, c = -60 \).
First, we calculate the discriminant: \[ b^2 - 4ac = 17^2 - 4 \cdot 1 \cdot (-60) = 289 + 240 = 529 \]
Now we compute \( x \): \[ x = \frac{-17 \pm \sqrt{529}}{2 \cdot 1} \] \[ \sqrt{529} = 23 \] \[ x = \frac{-17 \pm 23}{2} \]
This results in two potential solutions: \[ x = \frac{6}{2} = 3 \quad \text{(since \( x \) cannot be negative)} \] and \[ x = \frac{-40}{2} = -20 \quad \text{(not feasible)} \]
Thus, the width of the path is: \[ \boxed{3 \text{ m}} \]
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