A2 June 2023 Paper 1 Q2
2 In this question you must show detailed reasoning.
The equation \(z^4 + 4z^3 + 9z^2 + 10z + 6 = 0\) has roots \(\alpha\), \(\beta\), \(\gamma\) and \(\delta\).
| Scheme | Marks |
|---|---|
| DR \((w - 1)^4 + 4(w - 1)^3 + \ldots\ldots = 0\) | M1 |
| \(\Rightarrow w^4 - 4w^3 + 6w^2 - 4w + 1 + \ldots\ldots\) | M1 |
| \(\Rightarrow w^4 + 3w^2 + 2 = 0\) | A1 |
| [3] |
Notes
M1: Substitute \(z = w - 1\)
M1: Expanding with at least \((w - 1)^4\) seen
A1: Convincingly shown AG. Must include \(= 0\) on last line. Brackets must be fully expanded in working or evidence of collection of like terms
A0 if variable used is not \(w\).
Alternative method
| Scheme | Marks |
|---|---|
| \((z + 1)^4 + 3(z + 1)^2 + \ldots\ldots = 0\) | M1 |
| \(\Rightarrow z^4 + 4z^3 + \ldots\ldots\) \(\Rightarrow z^4 + 4z^3 + 9z^2 + 10z + 6 = 0 = 0\) | A1 |
M1: Substitute \(w = z + 1\) into end result
Alternative method using symmetry of roots
| Scheme | Marks |
|---|---|
| \(\sum\alpha = -4\), \(\sum\alpha\beta = 9\). \(\sum\alpha\beta\gamma = -10\), \(\alpha\beta\gamma\delta = 6\) \(\sum(\alpha + 1) = \sum\alpha + 4 = -4 + 4 = 0\) | B1 |
| \(\sum(\alpha + 1)(\beta + 1) = \sum\alpha\beta + 3\sum\alpha + 6 = 9 - 12 + 6 = 3\) | B1 |
| For \(\sum(\alpha + 1)(\beta + 1)(\gamma + 1) = 0\) and \((\alpha + 1)(\beta + 1)(\gamma + 1)(\delta + 1) = 2\) | B1 |
B1: For sums from original equation and finding the sum of the new roots
B1: For showing convincingly the sum of new roots in pairs
B1: For the last two
| Scheme | Marks |
|---|---|
| \(w^2 = -1, -2\) | M1 |
| \(\Rightarrow w = \pm\mathrm{i}, \pm\sqrt{2}\mathrm{i}\) | M1 |
| \(\Rightarrow z = \pm\mathrm{i} - 1, \pm\sqrt{2}\mathrm{i} - 1\) | A1 |
| [3] |
Notes
M1: Solving quadratic equation in \(w^2\) (or using their variable)
M1: Square rooting their \(w^2\), including \(\pm\), as long as their \(w^2\) not both non-negative and real.
A1: cao
Answers with no working is 0