Let the width of the path be \( x \) meters. The path runs along one side and one end of the rectangular plot, which has dimensions of 12 m by 5 m.
The dimensions of the larger rectangle, which includes the path, would then be:
- Length: \( 12 + x \) (adding the width on one side)
- Width: \( 5 + x \) (adding the width on the other side)
The area of the larger rectangle (the plot plus the path) can be represented as:
\[ \text{Area} = (12 + x)(5 + x) \]
We know the area of the plot and the width of the path leads to a total area of 120 m²:
\[ (12 + x)(5 + x) = 120 \]
Expanding the left side, we get:
\[ 60 + 12x + 5x + x^2 = 120 \]
Combining like terms gives:
\[ x^2 + 17x + 60 = 120 \]
Subtracting 120 from both sides leads to:
\[ x^2 + 17x - 60 = 0 \]
Now, we can solve this quadratic equation using the quadratic formula, where \( a = 1 \), \( b = 17 \), and \( c = -60 \):
\[ x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} = \frac{{-17 \pm \sqrt{{17^2 - 4 \cdot 1 \cdot (-60)}}}}{2 \cdot 1} \]
Calculating the discriminant:
\[ 17^2 = 289 \] \[ 4 \cdot 1 \cdot 60 = 240 \]
Thus,
\[ b^2 - 4ac = 289 + 240 = 529 \]
Taking the square root:
\[ \sqrt{529} = 23 \]
Plugging this back into the formula gives:
\[ x = \frac{{-17 \pm 23}}{2} \]
Calculating the two possible values for \( x \):
- \( x = \frac{{-17 + 23}}{2} = \frac{6}{2} = 3 \)
- \( x = \frac{{-17 - 23}}{2} = \frac{-40}{2} = -20 \) (not a valid solution since width cannot be negative)
Therefore, the width of the path is:
\[ \boxed{3 \text{ meters}} \]
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