A2 October 2021 Paper 1 Q8
8 The equation \(4x^4 - 4x^3 + px^2 + qx - 9 = 0\), where \(p\) and \(q\) are constants, has roots \(\alpha, -\alpha, \beta\) and \(\dfrac{1}{\beta}\).
(a) Determine the exact roots of the equation. [5]
(b) Determine the values of \(p\) and \(q\). [4]
| Scheme | Marks | AO |
|---|---|---|
| \(\alpha \times (-\alpha) \times \beta \times \dfrac{1}{\beta} = -\dfrac{9}{4}\) | M1 | 2.1 |
| \(\alpha = (\pm)\dfrac{3}{2}\) | A1 | 2.2a |
| Sum of roots \(= \beta + \dfrac{1}{\beta} = 1\) | M1 | 2.1 |
| \(\Rightarrow \beta^2 - \beta + 1 = 0\) \(\Rightarrow \beta = \dfrac{1 \pm \mathrm{i}\sqrt{3}}{2},\ \dfrac{1}{\beta} = \dfrac{1 \mp \mathrm{i}\sqrt{3}}{2}\) | A1 | 2.1 |
| so roots are \(\pm\dfrac{3}{2}\) and \(\dfrac{1 \pm \mathrm{i}\sqrt{3}}{2}\) | A1 | 2.2a |
| [5] |
Notes
Alternative solution
| Scheme | Marks |
|---|---|
| \((x - \alpha)(x + \alpha)(x - \beta)\left(x - \dfrac{1}{\beta}\right)\) | M1 |
| \(= x^4 - \left(\beta + \dfrac{1}{\beta}\right)x^3 + \left(1 - \alpha^2\right)x^2 + \alpha^2\left(\beta + \dfrac{1}{\beta}\right)x - \alpha^2\) | A1 |
| \(\Rightarrow \alpha^2 = \dfrac{9}{4} \Rightarrow \alpha = \pm\dfrac{3}{2}\) | A1 |
| \(\beta + \dfrac{1}{\beta} = 1 \Rightarrow \beta^2 - \beta + 1 = 0\) \(\Rightarrow \beta = \dfrac{1 \pm \mathrm{i}\sqrt{3}}{2},\ \dfrac{1}{\beta} = \dfrac{1 \mp \mathrm{i}\sqrt{3}}{2}\) | M1 |
| so roots are \(\pm\dfrac{3}{2}\) and \(\dfrac{1 \pm \mathrm{i}\sqrt{3}}{2}\) | A1 |
| [5] |
(Corrected from the printed mark scheme: the expansion in the alternative solution is printed as \(x^4 + \left(\beta + \frac{1}{\beta}\right)x^3 + \left(1 - \alpha^2\right)x^2 - \alpha^2\left(\beta + \frac{1}{\beta}\right)x - \alpha^2\); the signs of the \(x^3\) and \(x\) terms are corrected above.)
| Scheme | Marks | AO |
|---|---|---|
| Sum of pairs \(= \alpha(-\alpha) + \alpha\beta + \dfrac{\alpha}{\beta} - \alpha\beta - \dfrac{\alpha}{\beta} + 1\) | M1 | 2.1 |
| \(= 1 - \alpha^2 = -\dfrac{5}{4} \Rightarrow p = -5\) | A1 | 2.2a |
| Sum of triples \(= -\alpha^2\left(\beta + \dfrac{1}{\beta}\right) = -\dfrac{9}{4} \times 1 = -\dfrac{9}{4}\) | M1 | 2.1 |
| so \(q = 9\) | A1 | 2.2a |
| [4] |
Notes
M1: Unsimplified
M1: simplified
Alternative solution
| Scheme | Marks |
|---|---|
| \(\left(x - \dfrac{3}{2}\right)\left(x + \dfrac{3}{2}\right)\left(x - \dfrac{1 + \mathrm{i}\sqrt{3}}{2}\right)\left(x - \dfrac{1 - \mathrm{i}\sqrt{3}}{2}\right)\) | M1 |
| \(= \left(x^2 - \dfrac{9}{4}\right)\left(x^2 - x + 1\right)\) | A1 |
| \(= x^4 - x^3 - \dfrac{5}{4}x^2 + \dfrac{9}{4}x - \dfrac{9}{4}\) | A1 |
| so \(p = -5\) and \(q = 9\) | A1 |
| [4] |
A1: Award for either \(x^2\) or \(x\) coefficient correct