A plot consists of a rectangular garden measuring 8 m by 10 m, surrounded by a path of constant width, as shown in the diagram - Junior Cycle Mathematics - Question 9 - 2011

Combining the areas from all sections gives:

$4x^2 + 36x + 80 = 143.$

Next, move 143 to the left side:

$4x^2 + 36x + 80 - 143 = 0$ Thus, the simplified equation is:

$4x^2 + 36x - 63 = 0.$

For Elaine's diagram, the overall width includes the width of the garden plus the width of the path on both sides. So the width is:

$Width = 8 + 2x$

Similarly, the overall length is:

$Length = 10 + 2x.$

The total area for Elaine's diagram is calculated as follows:

$(8 + 2x)(10 + 2x) = 143.$

Expanding this gives:

$80 + 16x + 20x + 4x^2 = 143.$

Collecting terms results in:

$4x^2 + 36x + 80 - 143 = 0$ Thus, we have:

$4x^2 + 36x - 63 = 0.$

Factoring the equation:

$4x^2 + 36x - 63 = 0$ We can divide through by 4, resulting in:

$x^2 + 9x - 15.75 = 0$ Applying the quadratic formula gives:

x = rac{-b \pm ext{sqrt}(b^2 - 4ac)}{2a} In our scenario, this results in potential widths of either:

x = rac{3}{2} ext{ or } x = -21 However, we take the positive width, where:

$x = 1.5 ext{ m}.$

Tony chooses to test values for $x$ such as:

• For $x = 1$: Area = $(8 + 2)(10 + 2) = 120 ext{ m}^2$.
• For $x = 2$: Area = $(8 + 4)(10 + 4) = 168 ext{ m}^2$.
• For $x = 3$: Area = $(8 + 6)(10 + 6) = 224 ext{ m}^2$.
• Finally, for $x = 1.5$: Area = $(8 + 3)(10 + 3) = 143 ext{ m}^2$. This leads him to find the correct area at $x = 1.5$.

Answer: Elaine's method is best.

Reason: Although all methods provide the correct answer, Elaine's method is the quickest and simplest. Tony's method of trial and error could result in possible errors, while Kevin's requires complex calculations to derive the equation.