AS June 2024 Paper 1 Q7
7 In this question you must show detailed reasoning.
The roots of the equation \(2x^3 - 3x^2 - 3x + 5 = 0\) are \(\alpha\), \(\beta\) and \(\gamma\).
By considering \((\alpha + \beta + \gamma)^2\) and \((\alpha\beta + \beta\gamma + \gamma\alpha)^2\), determine a cubic equation with integer coefficients whose roots are \(\dfrac{\alpha\beta}{\gamma}\), \(\dfrac{\beta\gamma}{\alpha}\) and \(\dfrac{\gamma\alpha}{\beta}\). [6]
| Scheme | Marks | AO |
|---|---|---|
| DR \(\alpha\beta\gamma = -\dfrac{5}{2}, \sum\alpha\beta = -\dfrac{3}{2}, \sum\alpha = \dfrac{3}{2}\) | M1 | 1.1 |
| \(\alpha'\beta'\gamma' = \dfrac{\alpha\beta}{\gamma} \times \dfrac{\beta\gamma}{\alpha} \times \dfrac{\gamma\alpha}{\beta} = \alpha\beta\gamma\left(= -\dfrac{5}{2}\right)\) | B1 | 1.1 |
| \((\alpha + \beta + \gamma)^2 = \alpha^2 + \beta^2 + \gamma^2 + 2(\alpha\beta + \beta\gamma + \gamma\alpha)\) \((\alpha\beta + \beta\gamma + \gamma\alpha)^2 =\) \(\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 + 2(\alpha\beta^2\gamma + \beta\gamma^2\alpha + \gamma\alpha^2\beta)\) \(= \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 + 2\alpha\beta\gamma(\alpha + \beta + \gamma)\) | M1 | 1.1 |
| \(\sum\alpha'\beta' = \dfrac{\alpha\beta}{\gamma}\dfrac{\beta\gamma}{\alpha} + \dfrac{\beta\gamma}{\alpha}\dfrac{\gamma\alpha}{\beta} + \dfrac{\gamma\alpha}{\beta}\dfrac{\alpha\beta}{\gamma}\) \(= \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\) \(= \left(\dfrac{3}{2}\right)^2 - 2 \times \left(-\dfrac{3}{2}\right) = \dfrac{9}{4} + \dfrac{12}{4} = \dfrac{21}{4}\) | B1 | 1.1 |
| \(\sum\alpha' = \dfrac{\alpha\beta}{\gamma} + \dfrac{\beta\gamma}{\alpha} + \dfrac{\gamma\alpha}{\beta} = \dfrac{\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2}{\alpha\beta\gamma}\) \(= \dfrac{(\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma)}{\alpha\beta\gamma}\) \(= \dfrac{(\alpha\beta + \beta\gamma + \gamma\alpha)^2}{\alpha\beta\gamma} - 2(\alpha + \beta + \gamma)\) \(= \dfrac{\left(-\frac{3}{2}\right)^2}{\left(-\frac{5}{2}\right)} - 2 \times \left(\dfrac{3}{2}\right) = -\dfrac{9}{4} \times \dfrac{2}{5} - \dfrac{30}{10} = -\dfrac{39}{10}\) | B1 | 1.1 |
| Choosing \(a' = 20\) gives \(20x^3 + 78x^2 + 105x + 50 = 0\) | A1 | 1.1 |
| [6] |
Notes
M1: Attempt to find the three Vieta's formulae for the original equation.
This mark can be awarded if 2 of the 3 are correct and the 3rd has been attempted or if all are correct but for signs.
B1: Showing that the new product of roots is the same as the original.
M1: Both considered and at least one correctly written in desired form using Vieta's expressions
May be embedded in expressions for \(\sum\alpha'\beta'\) or \(\sum\alpha'\)
B1: For correct \(\sum\alpha'\beta'\) in terms of Vieta's expressions i.e. \(\sum\alpha'\beta' = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\)
Can be awarded before numerical value found
Correct numerical value implies correct \(\sum\alpha'\beta'\) (but must be identified as \(\sum\alpha'\beta'\))
B1: For correct \(\sum\alpha'\) in terms of Vieta's expressions i.e. \(\sum\alpha' = \dfrac{(\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma)}{\alpha\beta\gamma}\)
Can be awarded before numerical value found
Correct numerical value implies correct \(\sum\alpha'\) (but must be identified as \(\sum\alpha'\))
A1: Any non-zero integer multiple. Must be a cubic equation (i.e. it must have “=0”) with integer coefficients but can be in any unknown.
Correct answer from finding roots on calculator only scores 0/6.