AS October 2021 Paper 1 Q2
2 The equation \(3x^2 - 4x + 2 = 0\) has roots \(\alpha\) and \(\beta\).
Find an equation with integer coefficients whose roots are \(3 - 2\alpha\) and \(3 - 2\beta\). [3]
| Scheme | Marks | AO |
|---|---|---|
| Let \(y = 3 - 2x \Rightarrow x = \dfrac{3 - y}{2}\) | B1 | 1.1a |
| \(\Rightarrow \dfrac{3(3 - y)^2}{4} - \dfrac{4(3 - y)}{2} + 2 = 0\) | M1 | 1.1 |
| \(\Rightarrow 3y^2 - 10y + 11\ (= 0)\) | A1 | 1.1 |
| [3] |
Notes
B1: soi
M1: substituting their \((3 - y)/2\) for \(x\)
Alternative solution 1
| Scheme | Marks |
|---|---|
| \(\alpha + \beta = \tfrac{4}{3},\ \alpha\beta = \tfrac{2}{3}\) | B1 |
| \(\Rightarrow 3 - 2\alpha + 3 - 2\beta = 6 - 2(\alpha + \beta) = \tfrac{10}{3}\) and \((3 - 2\alpha)(3 - 2\beta) = 9 - 6(\alpha + \beta) + 4\alpha\beta = \tfrac{11}{3}\) | M1 |
| \(\Rightarrow 3y^2 - 10y + 11\ (= 0)\) | A1 |
M1: Attempt to find sum and product of new roots
Alternative solution 2
| Scheme | Marks |
|---|---|
| \(\alpha, \beta\) are \(\dfrac{4 \pm \sqrt{16 - 4 \times 3 \times 2}}{2 \times 3} = \dfrac{4 \pm \sqrt{8}\,\mathrm{i}}{6} = \dfrac{2 \pm \sqrt{2}\,\mathrm{i}}{3}\) new roots are \(\dfrac{5 \pm 2\sqrt{2}\,\mathrm{i}}{3}\) | B1 |
| sum \(= \tfrac{10}{3}\), product \(= \dfrac{(5 + 2\sqrt{2}\,\mathrm{i})(5 - 2\sqrt{2}\,\mathrm{i})}{9} = \tfrac{11}{3}\) | M1 |
| \(\Rightarrow 3y^2 - 10y + 11\ (= 0)\) | A1 |
B1: Any exact form