Higher November 2018 Paper 5 Q20
20
(a) Prove that \((2x + 1)(3x + 2) + x(3x + 5) + 2\) is a perfect square. [6]
(b) Gemma says
The equation \((2x + 1)(3x + 2) + x(3x + 5) + 2 = -12\) has no solutions.
Explain Gemma’s reasoning. [1]
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| [\((2x + 1)(3x + 2) =\)] \(6x^2 + 3x + 4x + 2\) or better | M2 | B1 for 3 out of 4 terms correct | \(6x^2 + 7x + 2\), \(7x\) counts as 2 terms For B1 accept terms on a grid |
| [\(x(3x + 5) =\)] \(3x^2 + 5x\) | M1 | ||
| \(9x^2 + 12x + 4\) | M1 | FT their expansions dep on 3 term quadratic | Condone if expression ‘= 0’ |
| \((3x + 2)^2\) | A1 | For A1 accept \((3x + 2)(3x + 2)\) | |
| which is a perfect square | A1 | ||
| Answer | Marks | Part marks and guidance | |
|---|---|---|---|
| Cannot square root a negative value oe or for \(9x^2 + 12x + 16\) [= 0] and \(b^2 - 4ac \lt 0\) with \(a\), \(b\), \(c\) substituted or values shown oe | 1 | Correction: the published mark scheme prints \(9x^2 + 12x + 4\) [= 0] here. Setting the expression equal to –12 gives \(9x^2 + 12x + 16 = 0\), for which \(b^2 - 4ac = 144 - 576 \lt 0\). | |