Define variables for the dimensions of the rectangular plot.

You're on a roll — a) Form quadratic equation to represent this situation. Step 1: Define variables for the dimensions of the rectangular plot. Let $w$ be the width of the plot in meters. Let $l$ be the length of the plot in meters. Step 2: Express the length in terms of the width. The problem states that the length of the plot is 3 meters more than its width. $$l = w + 3$$ Step 3: Use the given area to form an equation. The area of a rectangle is given by $A = l \times w$. The total area is $180 \text{ m}^2$. Substitute the expressions for $l$ and $A$: $$180 = (w+3)w$$ Step 4: Expand and rearrange the equation into standard quadratic form $ax^2 + bx + c = 0$. $$180 = w^2 + 3w$$ Subtract 180 from both sides: $$w^2 + 3w - 180 = 0$$ The quadratic equation is: $$\boxed{w^2 + 3w - 180 = 0}$$ b) Solve the equation by factorisation method to find the dimensions of the plot. Step 1: Factor the quadratic equation $w^2 + 3w - 180 = 0$. We need to find two numbers that multiply to -180 and add up to 3. These numbers are 15 and -12. $$15 \times (-12) = -180$$ $$15 + (-12) = 3$$ Step 2: Rewrite the middle term using these two numbers. $$w^2 + 15w - 12w - 180 = 0$$ Step 3: Factor by grouping. $$w(w + 15) - 12(w + 15) = 0$$ $$(w - 12)(w + 15) = 0$$ Step 4: Solve for $w$. Set each factor equal to zero: $$w - 12 = 0 \implies w = 12$$ $$w + 15 = 0 \implies w = -15$$ Step 5: Determine the valid width. Since width cannot be negative, we discard $w = -15$. Therefore, the width of the plot is $w = 12 \text{ m}$. Step 6: Calculate the length of the plot. Using the relationship $l = w + 3$: $$l = 12 + 3$$ $$l = 15 \text{ m}$$ The dimensions of the plot are: $$\boxed{\text{Width} = 12 \text{ m}, \text{Length} = 15 \text{ m}}$$ Got more? Send 'em!