AS June 2022 Paper 1 Q3
3 In this question you must show detailed reasoning.
The roots of the equation \(5x^3 - 3x^2 - 2x + 9 = 0\) 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 \(\Sigma\alpha' = \alpha\beta + \beta\gamma + \gamma\alpha = -2/5\) | B1 | 1.1 |
| \(\alpha'\beta'\gamma' = (\alpha\beta)(\beta\gamma)(\gamma\alpha) = (\alpha\beta\gamma)^2\ldots\) | M1 | 1.1 |
| \(\ldots = (-9/5)^2 = 81/25\) | A1 | 1.1 |
| \(\Sigma\alpha'\beta' = (\alpha\beta)(\beta\gamma) + (\beta\gamma)(\gamma\alpha) + (\gamma\alpha)(\alpha\beta)\) \(= \alpha\beta\gamma(\alpha + \beta + \gamma)\ldots\) | M1 | 1.1 |
| \(= (-9/5)(3/5) = -27/25\) | A1 | 1.1 |
| \(a = 25 \Rightarrow 25x^3 + 10x^2 - 27x - 81 = 0\) | A1 | 1.1 |
| [6] |
Notes
B1: Quantity must be either identified as, or used as, sum of new roots
NB for reference: \(\sum\alpha = \dfrac{3}{5},\ \sum\alpha\beta = -\dfrac{2}{5},\ \alpha\beta\gamma = -\dfrac{9}{5}\)
M1: (1st) For expressing product of new roots in terms of old roots
A1: (1st) For expressing product of new roots in terms of old roots
Condone \((9/5)^2\) if seen. Do not condone \(-9/5^2\) or \(-(9/5)^2\) unless recovered
M1: (2nd) Finding sum of products and rewriting into symmetric form
A1: (3rd) Or any non-zero integer multiple
Needs to be an equation.
Alternative method
| Scheme | Marks |
|---|---|
| \(\alpha\beta\gamma = -9/5\) | B1 |
| \(u = \alpha\beta\gamma/x = -9/(5x)\) | B1 |
| When \(x = \alpha\), \(u = \beta\gamma\), and similar for other roots | B1 |
| \(5\left(\dfrac{-9}{5u}\right)^3 - 3\left(\dfrac{-9}{5u}\right)^2 - 2\left(\dfrac{-9}{5u}\right) + 9 = 0\) | M1 |
| \(\dfrac{-729}{25u^3} - \dfrac{243}{25u^2} + \dfrac{18}{5u} + 9 = 0\) | M1 |
| \(25x^3 + 10x^2 - 27x - 81 = 0\) | A1 |
B1: (2nd) SOI