A2 June 2019 Paper 1 Q13
13 The equation \(z^3 + kz^2 + 9 = 0\) has roots \(\alpha\), \(\beta\) and \(\gamma\).
(a)
(i) Show that\[\alpha^2 + \beta^2 + \gamma^2 = k^2\] [3 marks]
(ii) Show that\[\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = -18k\] [4 marks]
(b) The equation \(9z^3 - 40z^2 + rz + s = 0\) has roots \(\alpha\beta + \gamma\), \(\beta\gamma + \alpha\) and \(\gamma\alpha + \beta\).
(i) Show that\[k = -\frac{40}{9}\] [1 mark]
(ii) Without calculating the values of \(\alpha\), \(\beta\) and \(\gamma\), find the value of \(s\).
Show working to justify your answer. [6 marks]
| Scheme | Marks | AO |
|---|---|---|
| (i) Recognises that result can be obtained by expanding \((\alpha + \beta + \gamma)^2\) and completes correct expansion. | M1 | 3.1a |
| Correctly uses sum of roots and sum of pairs of roots, in their expansion. Condone \(\alpha + \beta + \gamma = k\) | B1 | 1.1a |
| Completes a rigorous argument to show the required result. Do not condone \(\alpha + \beta + \gamma = k\) | R1 | 2.1 |
| (ii) Expands \((\alpha\beta + \beta\gamma + \gamma\alpha)^2\) | M1 | 1.1a |
| Rearranges to express \(\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2\) in terms of sum, product and sum of pairs of roots | M1 | 3.1a |
| Correctly states product of roots – could be seen in (a)(i) | B1 | 1.1a |
| Completes a rigorous argument to show the required result. | R1 | 2.1 |
Typical solution
(i)
\[(\alpha + \beta + \gamma)^2 = \alpha^2 + \beta^2 + \gamma^2 + 2(\alpha\beta + \beta\gamma + \gamma\alpha)\]But \(\alpha + \beta + \gamma = -k\) and
\[\alpha\beta + \beta\gamma + \gamma\alpha = 0\]So
\[\alpha^2 + \beta^2 + \gamma^2 = k^2\](ii)
\[\begin{aligned}(\alpha\beta + \beta\gamma + \gamma\alpha)^2 &= \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 \\ &\quad + 2\alpha\beta^2\gamma + 2\alpha^2\beta\gamma + 2\alpha\beta\gamma^2 \\ &= \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 \\ &\quad + 2\alpha\beta\gamma(\alpha + \beta + \gamma)\end{aligned}\]But \(\alpha + \beta + \gamma = -k\)
And \(\alpha\beta + \beta\gamma + \gamma\alpha = 0\)
And \(\alpha\beta\gamma = -9\)
So \(\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = -18k\)
| Scheme | Marks | AO |
|---|---|---|
| (i) Deduces result correctly using both equations and previous working. | R1 | 2.2a |
| (ii) Forms correct equation from product of roots | B1 | 1.1b |
| Expands product of roots. Condone one or two errors | M1 | 1.1a |
| Identifies the degree 5 terms and fully factorises them in the form \(\alpha\beta\gamma\left(\alpha^2 + \beta^2 + \gamma^2\right)\) | M1 | 3.1a |
| Collects all other terms on RHS as \((\alpha\beta\gamma)^2 + \alpha\beta\gamma + \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2\) \((\alpha\beta\gamma)^2\) PI 81 or \((-9)^2\) | M1 | 3.1a |
| Correctly substitutes their values into their equation | M1 | 1.1a |
| Obtains the correct answer from correct reasoning. | R1 | 2.1 |
| (14 marks) |
Typical solution
(i)
\[\alpha\beta + \gamma + \beta\gamma + \alpha + \gamma\alpha + \beta = \frac{40}{9}\]But \(\alpha + \beta + \gamma = -k\) And \(\alpha\beta + \beta\gamma + \gamma\alpha = 0\)
so \(k = -\dfrac{40}{9}\)
(ii)
\[\begin{aligned}-\frac{s}{9} &= (\alpha\beta + \gamma)(\beta\gamma + \alpha)(\gamma\alpha + \beta) \\ &= \alpha^2\beta^2\gamma^2 + \alpha\beta^3\gamma + \alpha^3\beta\gamma + \alpha^2\beta^2 \\ &\quad + \alpha\beta\gamma^3 + \beta^2\gamma^2 + \gamma^2\alpha^2 + \alpha\beta\gamma \\ &= (\alpha\beta\gamma)^2 + \alpha\beta\gamma + \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 \\ &\quad + \alpha\beta\gamma\left(\alpha^2 + \beta^2 + \gamma^2\right) \\ &= 81 - 9 - 18k - 9k^2 \\ &= 81 - 9 + 80 - 9 \times \frac{1600}{81} \\ &= -\frac{232}{9}\end{aligned}\]So \(s = 232\)