AS October 2020 Paper 1 Q5
5 In this question you must show detailed reasoning.
The cubic equation \(5x^3 + 3x^2 - 4x + 7 = 0\) has roots \(\alpha\), \(\beta\) and \(\gamma\).
Find a cubic equation with integer coefficients whose roots are \(\alpha + \beta\), \(\beta + \gamma\) and \(\gamma + \alpha\). [7]
| Scheme | Marks | AO |
|---|---|---|
| \(\alpha + \beta + \gamma = -\dfrac{3}{5},\ \alpha\beta + \beta\gamma + \gamma\alpha = -\dfrac{4}{5},\ \alpha\beta\gamma = -\dfrac{7}{5}\) | B1 | 1.1a |
| \(A + B + \Gamma = 2(\alpha + \beta + \gamma) = -\dfrac{6}{5}\) | B1ft | 2.1 |
| \(AB + B\Gamma + \Gamma A = (\alpha + \beta)(\beta + \gamma) + (\beta + \gamma)(\gamma + \alpha) + (\gamma + \alpha)(\alpha + \beta) = 3(\alpha\beta + \beta\gamma + \gamma\alpha) + \alpha^2 + \beta^2 + \gamma^2\) | M1 | 2.1 |
| \(= \alpha\beta + \beta\gamma + \gamma\alpha + (\alpha + \beta + \gamma)^2 = -\dfrac{11}{25}\) | A1 | 1.1 |
| \(AB\Gamma = (\alpha + \beta)(\beta + \gamma)(\gamma + \alpha) = 2\alpha\beta\gamma + \alpha(\alpha\beta + \beta\gamma + \gamma\alpha) - \alpha\beta\gamma + \beta(\alpha\beta + \beta\gamma + \gamma\alpha) - \alpha\beta\gamma + \gamma(\alpha\beta + \beta\gamma + \gamma\alpha) - \alpha\beta\gamma = (\alpha + \beta + \gamma)(\alpha\beta + \beta\gamma + \gamma\alpha) - \alpha\beta\gamma\) | M1 | 2.1 |
| \(= -\dfrac{3}{5} \times -\dfrac{4}{5} - -\dfrac{7}{5} = \dfrac{12}{25} + \dfrac{35}{25} = \dfrac{47}{25}\) | A1 | 1.1 |
| \(a = 25 \Rightarrow 25x^3 + 30x^2 - 11x - 47 = 0\) | A1 | 1.1 |
| [7] |
Notes
B1: For at least 2 correct
B1ft: \(2 \times\) their \(\Sigma\alpha\)
\(A\), \(B\) and \(\Gamma\) are the roots of the new equation
M1: (1st) Attempting to find the new \(\Sigma\alpha\beta\) and expanding to a symmetrical form
M1: (2nd) Opening brackets convincingly and writing in symmetrical form
Other correct forms are possible eg \(\alpha^2\beta + \alpha^2\gamma + \beta^2\alpha + \beta^2\gamma + \gamma^2\alpha + \gamma^2\beta + 2\alpha\beta\gamma\)
A1: (3rd) Or any non-zero integer multiple
Must be integer coefficients
Alternative method
| Scheme | Marks |
|---|---|
| Substitution \(x = -\dfrac{3}{5} - u\) | B1* |
| \(\alpha + \beta + \gamma = -\dfrac{3}{5},\ \left[\alpha\beta + \beta\gamma + \gamma\alpha = -\dfrac{4}{5},\ \alpha\beta\gamma = -\dfrac{7}{5}\right]\) | B1dep* |
| When \(x = \alpha\), \(u = -\dfrac{3}{5} - \alpha\) \(= \alpha + \beta + \gamma - \alpha\) \(= \alpha + \beta\) | B1** |
| When \(x = \beta\) / \(\gamma\), \(u = -\dfrac{3}{5} - \beta\) / \(\gamma\) \(= \alpha + \beta + \gamma - \beta\) / \(\gamma\) \(= \alpha + \gamma\) / \(\beta\) | B1 dep** |
| \(5\left(-\dfrac{3}{5} - u\right)^3 + 3\left(-\dfrac{3}{5} - u\right)^2 - 4\left(-\dfrac{3}{5} - u\right) + 7 = 0\) \(-5\left(u^3 + 3u^2 \cdot \tfrac{3}{5} + 3u \cdot \left(\tfrac{3}{5}\right)^2 + \left(\tfrac{3}{5}\right)^3\right) + 3\left(u^2 + 2u \cdot \tfrac{3}{5} + \left(\tfrac{3}{5}\right)^2\right) + 4\left(u + \tfrac{3}{5}\right) + 7 = 0\) | M1 |
| \(-5u^3 - 6u^2 + \dfrac{11}{5}u + \dfrac{47}{5}\ [= 0]\) | A1 |
| \(25u^3 + 30u^2 - 11u - 47 = 0\) | A1 |
B1*: Used or stated
B1dep*: Might occur before the first B1 in candidates working
Note only the sum of roots needed here
B1**: Show that when \(x\) is one of the roots of the original equation, \(u\) is one of the roots of the new equation
B1 dep**: Show that \(x = -\dfrac{3}{5} - u\) gives the other two required roots
M1: Use substitution \(x = -\dfrac{3}{5} - u\) and expand \(\left(-\dfrac{3}{5} - u\right)^3\) and \(\left(-\dfrac{3}{5} - u\right)^2\)
\(\left(-\dfrac{3}{5} - u\right)^3 = -\left(u^3 + \dfrac{9}{5}u^2 + \dfrac{27}{25}u + \dfrac{27}{125}\right)\)
A1: (2nd) Or any non-zero integer multiple
Must be integer coefficients. Must have “=0”