A2 June 2025 Paper 2 Q6
6. The quartic equation
\[2x^4 + Ax^3 - Ax^2 - 5x + 6 = 0\]where \(A\) is a real constant, has roots \(\alpha\), \(\beta\), \(\gamma\) and \(\delta\)
Given that \(\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = -\dfrac{3}{4}\)
| Scheme | Marks | AO |
|---|---|---|
| \(2x^4 + Ax^3 - Ax^2 - 5x + 6 = 0 \Rightarrow x^4 + \dfrac{A}{2}x^3 - \dfrac{A}{2}x^2 - \dfrac{5}{2}x + \dfrac{6}{2} = 0\) \(\alpha\beta\gamma\delta = \dfrac{6}{2}\) \(\alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta = \dfrac{5}{2}\) | B1 | 3.1a |
| \(\dfrac{3}{\alpha} + \dfrac{3}{\beta} + \dfrac{3}{\gamma} + \dfrac{3}{\delta} = \dfrac{3(\alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta)}{\alpha\beta\gamma\delta} = \dfrac{3 \times \left(\frac{5}{2}\right)}{\left(\frac{6}{2}\right)} = \ldots\) | M1 | 1.1b |
| \(= \dfrac{5}{2}\) | A1 | 1.1b |
| (3) |
Notes
B1: Identifies the correct values for the product and triple pair sum. If seen these must be correct, but they may be implied if not explicitly extracted, e.g. \(\displaystyle\sum \dfrac{3}{\alpha_i} = 3\dfrac{\sum \alpha_i\alpha_j\alpha_k}{\alpha\beta\gamma\delta} = 3 \times \dfrac{5}{6} = \dfrac{5}{2}\) does not technically contain an incorrect step if one realises the 2’s will cancel. A good indication will be what happens in part (b) – if there they realise the 2’s are involved then allow b.o.d. for a calculation such as here given in (a).
Note \(\sum \alpha\beta\gamma\delta\) is a correct notation for the product of roots.
M1: Uses the correct identity and their values of the product and triple sum. (The values need not be correct for this mark.)
A1: Correct value following correct identity with no indication of incorrect values for triple sum and product (see comment on the B mark).
Alt (a)
| Scheme | Marks | AO |
|---|---|---|
| \(x = \dfrac{3}{w} \Rightarrow 2\left(\dfrac{3}{w}\right)^4 + \ldots - 5\left(\dfrac{3}{w}\right) + 6 = 0\) or \(x = \dfrac{1}{w} \Rightarrow 2\left(\dfrac{1}{w}\right)^4 + \ldots - 5\left(\dfrac{1}{w}\right) + 6 = 0\) | B1 | 3.1a |
| \(\Rightarrow \ldots + \ldots w + \ldots - 15w^3 + 6w^4 = 0 \Rightarrow \displaystyle\sum \dfrac{3}{\alpha_i} = -\left(\dfrac{-15}{6}\right)\) or \(\Rightarrow 2 + \ldots - 5w^3 + 6w^4 = 0 \Rightarrow 3\displaystyle\sum \dfrac{1}{\alpha_i} = 3 \times -\left(\dfrac{-5}{6}\right)\) | M1 | 1.1b |
| \(= \dfrac{5}{2}\) | A1 | 1.1b |
| (3) |
Alt (a): Some may use a transformation. These can be scored as
B1: Makes a correct substitution into the equation to solve the problem. This will probably be \(x = \dfrac{3}{w}\) but note that \(x = \dfrac{1}{w}\) can also be used.
M1: Multiplies through by \(w^4\) and extracts the correct sum of roots from the new equation. If using \(x = \dfrac{1}{w}\) they must also multiply through by the 3 to gain this mark.
A1: Correct answer from correct work on the relevant coefficients (the others need not be seen).
| Scheme | Marks | AO |
|---|---|---|
| \(\alpha + \beta + \gamma + \delta = -\dfrac{A}{2} \qquad \alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta = -\dfrac{A}{2}\) | B1 | 1.1b |
| \((\alpha + \beta + \gamma + \delta)^2 =\) \(\alpha^2 + \beta^2 + \gamma^2 + \delta^2 + 2(\alpha\beta + \alpha\gamma + \alpha\delta + \beta\gamma + \beta\delta + \gamma\delta)\) | M1 A1 | 3.1a 1.1b |
| \(\left(-\dfrac{A}{2}\right)^2 = -\dfrac{3}{4} + 2\left(-\dfrac{A}{2}\right) \Rightarrow A^2 + 4A + 3 = 0 \Rightarrow A = \ldots\) | dM1 | 1.1b |
| \(A = -1,\ -3\) | A1 | 1.1b |
| (5) | ||
| (8 marks) |
Notes
B1: Identifies the correct values for the sum and pair sum. Allow when first seen – some will list all these in part (a), which is fine for this mark.
M1: Attempts to find the identity for \((\alpha + \beta + \gamma + \delta)^2\) in terms of the sum of squares and pair sum (seen or implied). Allow attempts where the “2” is incorrect with no other method shown.
A1: Correct identity (seen or implied).
dM1: Dependent on previous method mark. Substitutes the sum and pair sum of the roots into their identity and forms and solves a 3TQ for \(A\) (usual rules).
A1: Correct values of \(A\)