AS October 2021 Paper 1 Q8
8 In this question you must show detailed reasoning.
The equation \(x^3 + kx^2 + 15x - 25 = 0\) has roots \(\alpha\), \(\beta\) and \(\dfrac{\alpha}{\beta}\). Given that \(\alpha \gt 0\), find, in any order,
- the roots of the equation,
- the value of \(k\). [7]
| Scheme | Marks | AO |
|---|---|---|
| DR Product of roots \(= \alpha^2 = 25\) | M1 | 3.1a |
| \(\Rightarrow \alpha = 5\) | A1 | 1.1 |
| \(\alpha\beta + \beta \times \dfrac{\alpha}{\beta} + \dfrac{\alpha}{\beta} \times \alpha = 15 \Rightarrow 5\beta + 5 + \dfrac{25}{\beta} = 15\) | M1 | 1.1 |
| \(\Rightarrow \beta^2 - 2\beta + 5 = 0\) | A1 | 1.1 |
| \(\Rightarrow (\beta - 1)^2 + 4 = 0\) | M1 | 1.1 |
| \(\Rightarrow \beta = 1 \pm 2\mathrm{i}\) so roots are \(5,\ 1 + 2\mathrm{i},\ 1 - 2\mathrm{i}\) | A1 | 2.2a |
| Sum of roots: \(-k = 5 + 1 + 2\mathrm{i} + 1 - 2\mathrm{i} = 7 \Rightarrow k = -7\) | B1 | 2.2a |
| [7] |
Notes
A1: (2nd) or \(5\beta^2 - 10\beta + 25 = 0\)
M1: (3rd) Or equivalent use of formula
A1: (3rd) SCB1 if ow
Alternative solution
| Scheme | Marks |
|---|---|
| Product of roots \(= \alpha^2 = 25\) | M1 |
| \(\Rightarrow \alpha = 5\) | A1 |
| Substituting \(x = 5\): \(125 + 25k + 75 - 25 = 0\) | M1 |
| \(\Rightarrow k = -7\) | A1 |
| \(x^3 - 7x^2 + 15x - 25 = (x - 5)(x^2 - 2x + 5)\) | M1 |
| Other roots are given by \(x^2 - 2x + 5 = 0\) | M1 |
| \(\Rightarrow x = 1 \pm 2\mathrm{i}\) | A1 |
M1: (3rd) Factorising
M1: (4th) Solving the resulting quadratic