A2 June 2020 Paper 1 Q8
8 The three roots of the equation
\[4x^3 - 12x^2 - 13x + k = 0\]where \(k\) is a constant, form an arithmetic sequence.
Find the roots of the equation. [6 marks]
| Scheme | Marks | AO |
|---|---|---|
| Defines three roots in arithmetic progression. PI by a correct use of arithmetic series sum in terms of \(\alpha\) and \(\beta\) | M1 | 3.1a |
| Uses one of Vieta’s laws: \(\alpha + \beta + \gamma = 3\) \(\alpha\beta + \beta\gamma + \gamma\alpha = \dfrac{-13}{4}\) \(\alpha\beta\gamma = \dfrac{-k}{4}\) | M1 | 3.1a |
| Obtains 1 as a root of the equation (this might be \(a + D = 1\) without clear realisation that 1 is therefore a root) | A1 | 1.1b |
| Substitutes their root of 1 in the cubic equation to find \(k\) or uses another of Vieta’s laws with their root of 1 substituted. | M1 | 3.1a |
| Obtains the value of \(k\) or solves their equation to find \(D\) or the other roots. Allow one slip. | A1F | 1.1b |
| Correctly obtains all three roots: \(-1.5\), \(1\), \(3.5\) | A1 | 1.1b |
| (6 marks) |
Typical solution
Let the roots be \(\alpha - D\), \(\alpha\) and \(\alpha + D\)
\(\Sigma\alpha\): \(\quad \alpha - D + \alpha + \alpha + D = -\left(\dfrac{-12}{4}\right)\)
\[3\alpha = 3\]\[\alpha = 1\]\(\Sigma\alpha\beta\):
\[(1 - D)(1) + (1)(1 + D) + (1 + D)(1 - D) = \frac{-13}{4}\]\[1 - D + 1 + D + 1 - D^2 = \frac{-13}{4}\]\[3 + \frac{13}{4} = D^2 \Rightarrow D = \pm\frac{5}{2}\]\[x = -1.5,\ 1,\ 3.5\]