AS June 2022 Paper 1 Q4
4. The roots of the quartic equation
\[3x^4 + 5x^3 - 7x + 6 = 0\]are \(\alpha\), \(\beta\), \(\gamma\) and \(\delta\)
Making your method clear and without solving the equation, determine the exact value of
| Scheme | Marks | AO |
|---|---|---|
| \(\sum\alpha_i = -\dfrac{5}{3}\) and \(\sum\alpha_i\alpha_j = 0\) This mark can be awarded if seen in part (ii) or part (iii) | B1 | 3.1a |
| So \(\alpha^2 + \beta^2 + \gamma^2 + \delta^2 = (\alpha + \beta + \gamma + \delta)^2 - 2\left(\sum\alpha_i\alpha_j\right) = \ldots\) | M1 | 1.1b |
| \(= \dfrac{25}{9} - 2 \times 0 = \dfrac{25}{9}\) | A1 | 1.1b |
| (3) |
Notes
B1: Correct sum and pair sum of roots seen or implied. Must realise the pair sum is zero.
Note: These values can be seen anywhere in the candidate’s solution
M1: Uses correct expression for the sum of squares.
A1: \(\dfrac{25}{9}\). Allow this mark from incorrect sign on sum of squares (but they will score B0 if the sign is incorrect).
| Scheme | Marks | AO |
|---|---|---|
| \(\sum\alpha_i\alpha_j\alpha_k = \dfrac{7}{3}\) and \(\prod\alpha_i = 2\) or for \(x = \dfrac{2}{w}\) used in equation This mark can be awarded if seen in part (i) or part (iii) | B1 | 2.2a |
| So \(2\left(\dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma} + \dfrac{1}{\delta}\right) = 2 \times \dfrac{\sum\alpha_i\alpha_j\alpha_k}{\alpha\beta\gamma\delta} = 2 \times \dfrac{\text{‘}\frac{7}{3}\text{’}}{\text{‘}\frac{6}{3}\text{’}}\) or for \(3\left(\dfrac{16}{w^4}\right) + 5\left(\dfrac{8}{w^3}\right) - 7\left(\dfrac{2}{w}\right) + 6 = 0 \Rightarrow 6w^4 - 14w^3 + \ldots = 0\) leading to \(\dfrac{14}{6}\) | M1 | 1.1b |
| \(\left(= 2 \times \dfrac{7/3}{2}\right)\left(= \dfrac{14}{6}\right) = \dfrac{7}{3}\) | A1 | 1.1b |
| (3) |
Notes
B1: Correct triple sum and product of roots seen or implied. May be stated in (i). Alternatively, this may be scored for sight of \(x = \dfrac{2}{w}\) used as a transformation in the equation.
Note: These values can be seen anywhere in the candidate’s solution
M1: Substitutes their values into \(2 \times \dfrac{\sum\alpha_i\alpha_j\alpha_k}{\alpha\beta\gamma\delta} = \ldots\) In the alternative it is for rearranging the equation to a quartic in \(w\) and uses to find the sum of the roots.
A1: \(\dfrac{7}{3}\) Allow this mark from incorrect sign of both triple sum and product (but they will score B0 if the sign is incorrect).
| Scheme | Marks | AO |
|---|---|---|
| \((3 - \alpha)(3 - \beta)(3 - \gamma)(3 - \delta) = \ldots\) expands all four brackets Or equation with these roots is \(3(3 - x)^4 + 5(3 - x)^3 - 7(3 - x) + 6 = 0\) | M1 | 3.1a |
| \(= 81 - 27\left(\sum\alpha_i\right) + 9\left(\sum\alpha_i\alpha_j\right) - 3\left(\sum\alpha_i\alpha_j\alpha_k\right) + \prod\alpha_i\) \(= 81 - 27\left(-\dfrac{5}{3}\right) + 9(0) - 3\left(\dfrac{7}{3}\right) + 2\) Or expands to fourth power and constant terms and attempts product of roots \(3x^4 + \ldots + 3 \times 3^4 + 5 \times 3^3 - 7 \times 3 + 6 \to \prod\alpha_i = \dfrac{\text{"}363\text{"}}{3}\) | dM1 | 1.1b |
| \(= 121\) | A1 | 1.1b |
| (3) | ||
| (9 marks) |
Notes
M1: A correct method to find the value used – may recognise structure as scheme, may expand the expression in stages, or may attempt to use a linear transformation \((3 - x)\) or e.g. \((3 - w)\) in original equation. Condone slips as long as the intention is clear.
dM1: Dependent on previous method mark. Uses at least 2 values of their sum of roots etc. in their expression. If using a linear shift this is for expanding to find the coefficient of \(x^4\) and constant term and attempts product of roots by dividing the constant term by the coefficient of \(x^4\).
A1: 121.