A2 October 2020 Paper 1 Q9
9 You are given that the cubic equation \(2x^3 + px^2 + qx - 3 = 0\), where \(p\) and \(q\) are real numbers, has a complex root \(\alpha = 1 + \mathrm{i}\sqrt{2}\).
(a) Write down a second complex root, \(\beta\). [1]
(b) Determine the third root, \(\gamma\). [2]
(c) Find the value of \(p\) and the value of \(q\). [2]
(d) Show that if \(n\) is an integer then \(\alpha^n + \beta^n + \gamma^n = 2 \times 3^{\frac{1}{2}n} \times \cos n\theta + \dfrac{1}{2^n}\) where \(\tan\theta = \sqrt{2}\). [4]
| Scheme | Marks | AO |
|---|---|---|
| \(\beta = 1 - \mathrm{i}\sqrt{2}\) oe | B1 | 2.2a |
| [1] |
| Scheme | Marks | AO |
|---|---|---|
| \(\alpha\beta\gamma = \dfrac{3}{2},\ \alpha\beta = \left(1 + \mathrm{i}\sqrt{2}\right)\left(1 - \mathrm{i}\sqrt{2}\right) = 3\) | M1 | 2.1 |
| \(\Rightarrow \gamma = \dfrac{1}{2}\) | A1 | 1.1 |
| [2] |
Notes
M1: Use of \(\alpha\beta\gamma = \dfrac{3}{2}\) to find the 3rd root.
Alternatively, find \(x^2 - 2x + 3\) and divide
| Scheme | Marks | AO |
|---|---|---|
| \(\left(x - \left(1 + \mathrm{i}\sqrt{2}\right)\right)\left(x - \left(1 - \mathrm{i}\sqrt{2}\right)\right)(2x - 1) = 0\) \(\Rightarrow \left(x^2 - 2x + 3\right)(2x - 1) = 0\) | M1 | 3.1a |
| \(\Rightarrow 2x^3 - 4x^2 + 6x - x^2 + 2x - 3 = 0\) \(\Rightarrow 2x^3 - 5x^2 + 8x - 3 = 0\) i.e. \(p = -5,\ q = 8\) | A1 | 1.1 |
| [2] |
Notes
M1: Multiply out (can be seen in (b))
Alternative method
| Scheme | Marks |
|---|---|
| \(\alpha + \beta + \gamma = -\dfrac{p}{2} = \dfrac{5}{2} \Rightarrow p = -5\) \(\alpha\beta + \beta\gamma + \gamma\alpha = \dfrac{q}{2}\) | M1 |
| \(= \dfrac{1}{2}\left(1 + \mathrm{i}\sqrt{2} + 1 - \mathrm{i}\sqrt{2}\right) + \left(1 + \mathrm{i}\sqrt{2}\right)\left(1 - \mathrm{i}\sqrt{2}\right)\) \(= 1 + 3 = 4 \Rightarrow q = 8\) | A1 |
| [2] |
M1: Use of symmetry forms for roots
(corrected from the printed mark scheme: the first line is printed as \(\alpha + \beta + \gamma = -\dfrac{a}{2}\); the coefficient is \(p\))
| Scheme | Marks | AO |
|---|---|---|
| \(\alpha = 1 + \mathrm{i}\sqrt{2} = \sqrt{3}\left(\dfrac{1}{\sqrt{3}} + \mathrm{i}\dfrac{\sqrt{2}}{\sqrt{3}}\right) = 3^{\frac{1}{2}}(\cos\theta + \mathrm{i}\sin\theta)\) where \(\cos\theta = \dfrac{1}{\sqrt{3}},\ \sin\theta = \dfrac{\sqrt{2}}{\sqrt{3}} \Rightarrow \tan\theta = \sqrt{2}\) | M1 | 2.1 |
| \(\beta = 1 - \mathrm{i}\sqrt{2} = \sqrt{3}\left(\dfrac{1}{\sqrt{3}} - \mathrm{i}\dfrac{\sqrt{2}}{\sqrt{3}}\right) = 3^{\frac{1}{2}}(\cos\theta - \mathrm{i}\sin\theta)\) oe | A1 | 1.1 |
| \(\Rightarrow \alpha^n = 3^{\frac{n}{2}}(\cos n\theta + \mathrm{i}\sin n\theta),\quad \beta^n = 3^{\frac{n}{2}}(\cos n\theta - \mathrm{i}\sin n\theta)\) \(\Rightarrow \alpha^n + \beta^n = 2 \times 3^{\frac{n}{2}} \times \cos n\theta\) | M1 | 2.1 |
| \(\Rightarrow \alpha^n + \beta^n + \gamma^n = 2 \times 3^{\frac{n}{2}} \times \cos n\theta + \dfrac{1}{2^n}\) AG | A1 | 2.1 |
| [4] |
Notes
M1: Either \(\alpha\) or \(\beta\) seen in mod/arg form
A1: For both of them – accept exponentials
M1: Derivation of \(\alpha^n\) or \(\beta^n\)