Ama rents a rectangular pitch at the market. The pitch measures 18 m by 12 m. She lays out her display in a smaller rectangle in the middle of the pitch, le...

A walkway of the same width all the way round reduces each dimension of the pitch by \(x\) at both ends, so each one loses \(2x\), not \(x\). Everything else follows from that.

(a) Length and width of the display. [2]
The pitch is 18 m by 12 m and the walkway is \(x\) m wide on all four sides:

\[\text{length} = (18 - 2x)\ \text{m}, \qquad \text{width} = (12 - 2x)\ \text{m}\]

(b) Show that \(x^2 - 15x + 26 = 0\). [3]
The display has area 112 m²:

\[(18 - 2x)(12 - 2x) = 112\]

Expand carefully, watching the signs:

\[216 - 36x - 24x + 4x^2 = 112\] \[4x^2 - 60x + 216 = 112 \quad\Rightarrow\quad 4x^2 - 60x + 104 = 0\]

Every coefficient is divisible by 4:

\[x^2 - 15x + 26 = 0\]

Marks: one for forming the product, one for the expansion, one for dividing through to the printed form. Both middle terms are negative, so they combine to \(-60x\); making one of them positive is the commonest error.

(c) Solve \(x^2 - 15x + 26 = 0\). [2]
Two numbers multiply to \(+26\) and add to \(-15\): those are \(-2\) and \(-13\).

\[(x - 2)(x - 13) = 0 \quad\Rightarrow\quad x = 2 \ \text{ or }\ x = 13\]

(d) Rejecting a root. [1]
Both roots are positive, so "widths cannot be negative" is not enough on its own here. Substitute into the display dimensions instead. If \(x = 13\) then the display width is \(12 - 2(13) = -14\) m, which is impossible, and the walkway would in any case be wider than the pitch itself. So \(x = 2\).

(e) Area of the walkway. [2]
The walkway is what is left of the pitch after the display is removed:

\[\text{pitch area} = 18 \times 12 = 216\ \text{m}^2\] \[\text{walkway area} = 216 - 112 = 104\ \text{m}^2\]

Check with \(x = 2\): the display measures \(18 - 4 = 14\) m by \(12 - 4 = 8\) m, and \(14 \times 8 = 112\) m², as stated.

Examination point: when both roots of a border problem are positive, test them in the reduced dimensions rather than quoting the usual "a length cannot be negative". The impossible root is the one that makes the inner rectangle vanish or turn negative.