A2 June 2022 Paper 2 Q6
6. The cubic equation
\[4x^3 + px^2 - 14x + q = 0\]where \(p\) and \(q\) are real positive constants, has roots \(\alpha\), \(\beta\) and \(\gamma\)
Given that \(\alpha^2 + \beta^2 + \gamma^2 = 16\)
Given that \(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma} = \dfrac{14}{3}\)
Without solving the cubic equation,
| Scheme | Marks | AO |
|---|---|---|
| \(4x^3 + px^2 - 14x + q = 0 \Rightarrow x^3 + \dfrac{p}{4}x^2 - \dfrac{14}{4}x + \dfrac{q}{4} = 0\) \(\alpha + \beta + \gamma = -\dfrac{p}{4}\quad \alpha\beta + \alpha\gamma + \beta\gamma = -\dfrac{14}{4}\) or \(-\dfrac{7}{2}\) | B1 | 3.1a |
| \((\alpha + \beta + \gamma)^2 = \alpha^2 + \beta^2 + \gamma^2 + 2(\alpha\beta + \alpha\gamma + \beta\gamma)\) \(\left(-\dfrac{p}{4}\right)^2 = 16 + 2\left(-\dfrac{7}{2}\right) \Rightarrow p = \ldots\) or \((\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \alpha\gamma + \beta\gamma) = \alpha^2 + \beta^2 + \gamma^2\) \(\left(-\dfrac{p}{4}\right)^2 - 2\left(-\dfrac{7}{2}\right) = 16 \Rightarrow p = \ldots\) | M1 | 3.1a |
| \(p = 12\) * cso | A1* | 1.1b |
| (3) |
Notes
B1: Identifies the correct values for the sum and pair sum. This may be implied by substituting into an equation, it must be clear
M1: Uses the correct identity and values of their sum and their pair sum to find a value of \(p\)
A1*: \(p = 12\) cso there is no need to see a reason
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma} = \dfrac{\beta\gamma + \alpha\gamma + \alpha\beta}{\alpha\beta\gamma}\) | M1 | 1.1b |
| \(\dfrac{\left(-\dfrac{7}{2}\right)}{\left(\dfrac{-q}{4}\right)} = \dfrac{14}{3} \Rightarrow q = \ldots\) | M1 | 1.1b |
| \(q = 3\) | A1 | 1.1b |
| (3) |
Notes
M1: Establishes a correct identity
M1: Uses their identity and their pair sum and their product of roots to find a value of \(q\). Condone a slip but the intention must be clear.
A1: \(q = 3\) Allow this mark from incorrect sign of both pair sum and product
Alternative
| Scheme | Marks | AO |
|---|---|---|
| \(4\left(\dfrac{1}{w}\right)^3 + 12\left(\dfrac{1}{w}\right)^2 - 14\left(\dfrac{1}{w}\right) + q\{= 0\}\) | M1 | 1.1b |
| \(qw^3 - 14w^2 + 12w + 4 = 0 \Rightarrow \dfrac{14}{3} = -\dfrac{-14}{q} \Rightarrow q = \ldots\) | M1 | 1.1b |
| \(q = 3\) | A1 | 1.1b |
| (3) |
M1: Uses \(x = \dfrac{1}{w}\) the substitution
M1: Simplifies to an quartic equation of the form \(aw^3 + bw^2 + cw + d = 0\) and uses \(\dfrac{14}{3} = -\dfrac{b}{a}\) to find a value for \(q\)
A1: \(q = 3\)
| Scheme | Marks | AO |
|---|---|---|
| \((\alpha - 1)(\beta - 1)(\gamma - 1) = \ldots\) \(= \alpha\beta\gamma - (\alpha\beta + \alpha\gamma + \beta\gamma) + (\alpha + \beta + \gamma) - 1\) | M1 A1 | 1.1a 1.1b |
| \(= \left(-\dfrac{\text{their } 3}{4}\right) - \left(-\dfrac{7}{2}\right) + \left(-\dfrac{12}{4}\right) - 1 = \ldots\) | dM1 | 1.1b |
| \(= -\dfrac{5}{4}\) | A1 | 1.1b |
| (4) | ||
| (10 marks) |
Notes
M1: Attempts to multiply out the three brackets.
A1: Correct expansion.
dM1: Dependent on previous method. Substitutes in the value of their sum, pair sum and the value of their product as appropriate. Condone a slip but the intention must be clear
A1: Correct value
Alternative
| Scheme | Marks | AO |
|---|---|---|
| \(4(x + 1)^3 + 12(x + 1)^2 - 14(x + 1) + \text{`}3\text{'}\{= 0\}\) or substitutes in 1 | M1 | 1.1a |
| \(= \ldots 4 + \ldots 12 + \ldots{-14} + \text{`}3\text{'} = 5\) or \(4x^3 + 24x^2 + 22x + 2 + \text{`their } q\text{'}\) | A1ft | 1.1b |
| \(= -\dfrac{\text{`their constant'}}{4}\) | dM1 | 1.1b |
| \(= -\dfrac{5}{4}\) | A1 | 1.1b |
M1: Substitutes \((x + 1)\) or \(x = 1\) into the cubic with their value of \(q\). Allow the use of different letters e.g. \((w + 1)\)
A1ft: Correct constant terms, follow through on their value of \(q\)
dM1: Applies \(-\dfrac{\text{`their constant'}}{4}\)
A1: Correct value