A2 June 2022 Paper 2 Q3
3 In this question you must show detailed reasoning.
The roots of the equation \(4x^3 + 6x^2 - 3x + 9 = 0\) are \(\alpha\), \(\beta\) and \(\gamma\).
Find a cubic equation with integer coefficients whose roots are \(\alpha + \beta\), \(\beta + \gamma\) and \(\gamma + \alpha\). [6]
| Scheme | Marks | AO |
|---|---|---|
| DR \(\sum\alpha = -\dfrac{6}{4} = -\dfrac{3}{2},\ \sum\alpha\beta = \dfrac{-3}{4},\ \alpha\beta\gamma = -\dfrac{9}{4}\) | B1 | 1.1 |
| \(\sum\alpha' = \alpha + \beta + \beta + \gamma + \gamma + \alpha = 2(\alpha + \beta + \gamma) = -3\) | B1 | 1.1 |
| \(\sum\alpha'\beta' =\) \((\alpha + \beta)(\beta + \gamma) + (\beta + \gamma)(\gamma + \alpha) + (\gamma + \alpha)(\alpha + \beta)\) \(= \alpha\beta + \alpha\gamma + \beta\gamma + \beta^2 + \beta\gamma + \beta\alpha\) \(\quad + \gamma\alpha + \gamma^2 + \gamma\alpha + \gamma\beta + \alpha\beta + \alpha^2\) \(= \alpha^2 + \beta^2 + \gamma^2 + 3(\alpha\beta + \beta\gamma + \gamma\alpha)\) \(= \alpha^2 + \beta^2 + \gamma^2 + 2(\alpha\beta + \beta\gamma + \gamma\alpha) + (\alpha\beta + \beta\gamma + \gamma\alpha)\) \(= (\alpha + \beta + \gamma)^2 + \alpha\beta + \beta\gamma + \gamma\alpha\) or \(\alpha'\beta'\gamma' = (\alpha + \beta)(\beta + \gamma)(\gamma + \alpha)\) \(= (\alpha\beta + \alpha\gamma + \beta^2 + \beta\gamma)(\gamma + \alpha)\) \(= \left(\sum\alpha\beta + \beta^2\right)(\gamma + \alpha)\) \(= (\gamma + \alpha)\sum\alpha\beta + \beta^2(\gamma + \alpha)\) \(= (\alpha + \beta + \gamma)\sum\alpha\beta - \beta\sum\alpha\beta + \beta(\alpha\beta + \beta\gamma)\) \(= \sum\alpha\sum\alpha\beta - \beta\sum\alpha\beta + \beta(\alpha\beta + \beta\gamma + \gamma\alpha) - \alpha\beta\gamma\) \(= \sum\alpha\sum\alpha\beta - \alpha\beta\gamma\) | M1 | 1.1 |
| \(\therefore \sum\alpha'\beta' = \left(-\dfrac{3}{2}\right)^2 + \dfrac{-3}{4} = \dfrac{6}{4} = \dfrac{3}{2}\) | A1 | 1.1 |
| & \(\alpha'\beta'\gamma' = -\dfrac{3}{2} \times \dfrac{-3}{4} - \dfrac{-9}{4} = \dfrac{9}{8} + \dfrac{18}{8} = \dfrac{27}{8}\) | A1 | 1.1 |
| \(8u^3 + 24u^2 + 12u - 27 = 0\) | A1 | 2.5 |
| [6] |
Notes
2nd B1: (ie sum of new roots \(= -3\)) or \(\dfrac{b}{a} = 3\)
M1: Correctly writing either \(\alpha'\beta'\gamma'\) or \(\Sigma\alpha'\beta'\) in a form involving the basic symmetrical forms
A1: Must be an equation with integer coefficients. Allow use of any unknown.
Alternative method
| Scheme | Marks |
|---|---|
| \(u = \sum\alpha - x\) | B1 |
| \(\begin{aligned} u &= -\dfrac{3}{2} - x \\[4pt] x &= -\dfrac{3}{2} - u \end{aligned}\) | B1 |
| \(4\left(-\dfrac{3}{2} - u\right)^3 + 6\left(-\dfrac{3}{2} - u\right)^2 - 3\left(-\dfrac{3}{2} - u\right) + 9 = 0\) | M1 |
| \(-\dfrac{27}{2} - 27u - 18u^2 - 4u^3 + \dfrac{27}{2} + 18u + 6u^2 + \dfrac{27}{2} + 3u = 0\) \(8u^3 + 24u^2 + 12u - 27 = 0\) | A3 |
| [6] |
B1: Can be implied by 2nd B1.
M1: Rearrange and substitute into original equation
A3: for fully correct equation in the correct form.
Award A1 only for coefficients of \(u^3\) and one other correct in any form
Award A2 only for coefficients of \(u^3\) and two others correct in any form