A2 June 2023 Paper 1 Q10
10 The equation \(x^3 - 4x^2 + 7x + c = 0\), where \(c\) is a constant, has roots \(\alpha\), \(\beta\) and \(\alpha + \beta\).
(a) Determine the roots of the equation. [6]
(b) Find \(c\). [1]
| Scheme | Marks | AO |
|---|---|---|
| \(\alpha + \beta + \alpha + \beta = 4 \qquad [\Rightarrow \alpha + \beta = 2]\) | B1 | 3.1a |
| \(\alpha\beta + (\alpha + \beta)\alpha + (\alpha + \beta)\beta = 7\) \(\Rightarrow \alpha^2 + \beta^2 + 3\alpha\beta = 7\) | B1 | 1.1 |
| \(x + \frac{3}{x} = 2\) or \(3x(2 - x) + (2 - x)^2 + x^2 = 7\) | M1 | 3.1a |
| \(\Rightarrow x^2 - 2x + 3 = 0\) | A1 | 1.1 |
| \(\Rightarrow x = \dfrac{2 \pm \sqrt{-8}}{2} = 1 \pm \mathrm{i}\sqrt{2}\) | M1 | 1.1 |
| so roots are \(1 + \mathrm{i}\sqrt{2}\), \(1 - \mathrm{i}\sqrt{2}\) and \(2\) | A1 | 2.2a |
| [6] |
Notes
B1: (2nd) or \(\alpha\beta + 2\alpha + 2\beta = 7\) or \(\alpha\beta + 2(\alpha + \beta) = 7\)
M1: (1st) substitution which could lead to a quadratic in \(\alpha\) or \(\beta\) or \(x\)
A1: (1st) could be in \(\alpha\) or \(\beta\)
M1: (2nd) a method to solve their quadratic
| Scheme | Marks | AO |
|---|---|---|
| \([c =]\ -6\) | B1 | 2.2a |
| [1] |