A2 June 2024 Paper 1 Q3
3 The equation \(2x^3 - 2x^2 + 8x - 15 = 0\) has roots \(\alpha\), \(\beta\) and \(\gamma\).
Determine the value of \(\alpha^2 + \beta^2 + \gamma^2\). [4]
| Scheme | Marks | AO |
|---|---|---|
| \(\alpha + \beta + \gamma = 1\) | B1 | 3.1a |
| \(\alpha\beta + \beta\gamma + \gamma\alpha = 4\) | B1 | 1.1 |
| \([\alpha^2 + \beta^2 + \gamma^2 =](\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\) | M1 | 3.1a |
| \(= -7\) | A1 | 1.1 |
| [4] |
Notes
B1: May be embedded
B1: May be embedded
M1: oe Accept numerical values in terms of their 1 and 4, i.e. \(1^2 - 2 \times 4\). No slips allowed.
Alternative method
| Scheme | Marks |
|---|---|
| Let \(z = x^2\) \(2\left(\sqrt{z}\right)^3 - 2\left(\sqrt{z}\right)^2 + 8\sqrt{z} - 15 = 0\) | M1 |
| \(\left(\sqrt{z}\right)^2(2z + 8)^2 = (15 + 2z)^2\) | M1 |
| \(4z^3 + 28z^2 + 4z - 225 = 0\) | A1 |
| \(\alpha^2 + \beta^2 + \gamma^2 = -7\) | A1 |
| [4] |
M1: Correct substitution chosen and attempted
M1: Rearranging and squaring both sides to remove square root
A1: Correct equation