Foundation June 2024 Paper 2 Q27
27 In this question, all lengths are in centimetres.
Not to scale
The area of the rectangle is 70 cm\(^2\).
(a) Show that \(x^2 + 5x - 84 = 0\). [4]
(b)
(i) Solve by factorising.
\(x^2 + 5x - 84 = 0\) [3]
(ii) Find the length of the longer side of the rectangle. [1]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| \((x - 2)(x + 7)\) | B1 | B1 implied by \((x - 2)\) and \((x + 7)\) in a multiplication grid Condone missing final bracket e.g. \((x - 2)(x + 7\) | |
| \(x^2 - 2x + 7x - 14\) or better | M2 | M1 for 3 out of 4 terms correct | \(+5x\) is two terms |
| \(x^2 + 5x - 14 = 70\) or \(x^2 - 2x + 7x - 14 = 70\) | A1 | A1 dep on B1M2 With no errors leading to the answer | A1 alternatives: \(x^2 + 5x - 14 - 70 = 0\) or \(x^2 - 2x + 7x - 14 - 70 = 0\) |
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| (i) \((x + 12)(x - 7)\) [\(= 0\)] | M2 | M1 for \((x + a)(x + b)\) where \(ab = -84\) or \(a + b = 5\) or \(x(x + 12) - 7(x + 12)\) or \(x(x - 7) + 12(x - 7)\) | Condone \((x + 12)(x - 7) = y\) for 2 marks For M2 and M1 condone the omission of the final bracket. |
| \(-12\) and 7 | B1FT | correct or FT their linear factors | \((x - 12)(x + 7)\) then \(-12\) and 7 scores M1B0 If both correct after \(x(x + 12) - 7(x + 12)\) or \(x(x - 7) + 12(x - 7)\) allow M2B1 BOD |
| (ii) 14 | 1 | FT dep on 2 integer answers given in part (b)(i) their largest positive answer + 7 | |