A2 October 2020 Paper 2 Q2
2 In this question you must show detailed reasoning.
The roots of the equation \(3x^3 - 2x^2 - 5x - 4 = 0\) are \(\alpha\), \(\beta\) and \(\gamma\).
| Scheme | Marks | AO |
|---|---|---|
| DR \(u = x^2\) | B1 | 3.1a |
| \(3\left(\sqrt{u}\right)^3 - 2\left(\sqrt{u}\right)^2 - 5\sqrt{u} - 4\ (= 0)\) | M1 | 1.1 |
| \(3u\sqrt{u} - 5\sqrt{u} = 2u + 4 \Rightarrow u(3u - 5)^2 = (2u + 4)^2\) | M1 | 1.1 |
| \(u(9u^2 - 30u + 25) = 4u^2 + 16u + 16 \Rightarrow\) \(9u^3 - 34u^2 + 9u - 16 = 0\) | A1 | 3.2a |
| [4] |
Notes
B1: Correct substitution chosen
M1: Oe. Attempting to make substitution
or preparation for substitution by removing odd powers.
eg \(x^2(3x^2 - 5)^2 = (2x^2 + 4)^2\)…
M1: Rearranging and squaring bs to remove the square root(s)
…and then substituting \(u(3u - 5)^2 = (2u + 4)^2\)
A1: Rearranging to answer. Equation can be in \(x\)
Alternative method
| Scheme | Marks |
|---|---|
| DR \(\alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = \left(-\dfrac{-4}{3}\right)^2 = \dfrac{16}{9}\) | B1 |
| \(\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2\) \(= (\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma)\) | M1 |
| \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\) | M1 |
| \(u^3 - \left(\left(\dfrac{2}{3}\right)^2 - 2 \times \dfrac{-5}{3}\right)u^2 + \left(\left(\dfrac{-5}{3}\right)^2 - 2 \times \dfrac{4}{3} \times \dfrac{2}{3}\right)u - \dfrac{16}{9}\) \(= u^3 - \dfrac{34}{9}u^2 + u - \dfrac{16}{9} = 0 \Rightarrow 9u^3 - 34u^2 + 9u - 16 = 0\) | A1 |
| [4] |
B1: Must include one intermediate step
M1: Writing the expression in terms of standard symmetrical forms
NB \(\sum\alpha = \frac{2}{3},\ \sum\alpha\beta = -\frac{5}{3},\ \alpha\beta\gamma = \frac{4}{3}\)
M1: Writing the expression in terms of standard symmetrical forms
Condone without factorisation of “2”
A1: Substituting in and rearranging to answer
NB \(\sum\alpha^2 = \frac{34}{9},\ \sum\alpha^2\beta^2 = 1\)
| Scheme | Marks | AO |
|---|---|---|
| DR \(\dfrac{\sum\alpha^2\beta^2}{\alpha\beta\gamma} = \dfrac{\left(\frac{9}{9}\right)}{\left(\frac{4}{3}\right)}\) or \(\dfrac{1}{\left(\frac{4}{3}\right)}\) | M1 | 3.1a |
| \(= \dfrac{3}{4}\) | A1 | 1.1 |
| [2] |
Notes
M1: Their \(\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2\) from part (a) over \(\pm\dfrac{4}{3}\). Strict ft