From the image, the question states:

• The area of the vegetable garden is 80 m².
• A fence with a 1 m wide iron net was built around the green land.
• The shape in the diagram suggests a rectangular garden.
• The dimensions of the garden are marked as $x + 2$ and $x$ meters.

The goal is to find how many meters of wire (presumably the perimeter) are needed for the fence.

Step-by-step solution:

1. Area of the garden: The area of the rectangle is given by the formula for the area of a rectangle: $\text{Area} = \text{length} \times \text{width}$ Here, the length is $x + 2$ and the width is $x$. Therefore, we have: $(x + 2) \times x = 80$ Expanding this equation: $x^2 + 2x = 80$ Rearranging: $x^2 + 2x - 80 = 0$ Now we solve this quadratic equation using the quadratic formula: $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a = 1$, $b = 2$, and $c = -80$.

   Plugging in the values: $x = \frac{-2 \pm \sqrt{2^2 - 4(1)(-80)}}{2(1)}$ $x = \frac{-2 \pm \sqrt{4 + 320}}{2}$ $x = \frac{-2 \pm \sqrt{324}}{2}$ $x = \frac{-2 \pm 18}{2}$ So, we have two possible solutions: $x = \frac{-2 + 18}{2} = 8 \quad \text{or} \quad x = \frac{-2 - 18}{2} = -10$

   Since the width cannot be negative, we take $x = 8$.

2. Dimensions of the garden:

   • Width: $x = 8$ meters
   • Length: $x + 2 = 10$ meters

3. Perimeter: The perimeter of a rectangle is given by: $\text{Perimeter} = 2(\text{length} + \text{width}) = 2(10 + 8) = 2(18) = 36 \, \text{meters}$

Thus, 36 meters of wire are needed to fence the garden.

Do you have any questions or would you like more details?

Related questions:

1. How do you solve more complex quadratic equations?
2. What are some other methods to calculate the perimeter of irregular shapes?
3. How does the quadratic formula work for non-integer solutions?
4. Could this problem be solved using a different approach?
5. What would happen if the garden had a different shape, like a circle or triangle?

Tip:

Always ensure that the dimensions of any physical object make sense within the context of the problem. Negative values often indicate extraneous solutions.