While designing the school year book, a teacher asked the student that the length and width of a particular photo is increased by x units each to double the area of the photo. The original photo is 18 cm long and 12 cm wide.
Based on the above information, answer the following questions.
Generated by claude-sonnet-4-6 · 2026-06-15 10:32 · grounding stimulus
Model Answer
(I) Original area = 18 × 12 = 216 cm²
New dimensions: length = (18 + x) cm, width = (12 + x) cm
Condition: new area = double the original area
$$( 18 + x)(12 + x) = 2 \times 216 = 432$$
(II) Expanding: $216 + 18x + 12x + x^2 = 432$
$$x^2 + 30x + 216 - 432 = 0$$
$$x^2 + 30x - 216 = 0$$
(III) Factorising: $x^2 + 30x - 216 = 0$
$$x^2 + 36x - 6x - 216 = 0$$
$$x(x + 36) - 6(x + 36) = 0$$
$$(x - 6)(x + 36) = 0$$
$$x = 6 \quad \text{or} \quad x = -36$$
Since x cannot be negative, x = 6 cm.
New length = 18 + 6 = 24 cm; New width = 12 + 6 = 18 cm
Source: Chapter 4 – Quadratic Equations
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Explanation
- (I) requires only setting up the equation — one line is enough.
- (II) must be in standard form $ax^2 + bx + c = 0$; expand and simplify.
- (III) carries 2 marks: 1 for finding x = 6 (with rejection of negative value) and 1 for stating both new dimensions. Always reject negative values for physical lengths.