AS June 2018 Paper 1 Q8
8 \(2 - 3\mathrm{i}\) is one root of the equation
\[z^3 + mz + 52 = 0\]where \(m\) is real.
(a) Find the other roots. [3 marks]
(b) Determine the value of \(m\). [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Correctly identifies the complex conjugate as a root. | B1 | 1.1b |
| Forms one of the following equations (or their equivalents) \(\alpha + \beta + 2 - 3\mathrm{i} = 0\) or \(\alpha\beta(2 - 3\mathrm{i}) = \pm 52\) or \(\alpha\beta + \beta(2 - 3\mathrm{i}) + (2 - 3\mathrm{i})\beta = \pm m\) with an equation to find \(m\). Or forms an equation to find \(m\) and then solves the cubic for their value of \(m\). | M1 | 1.1a |
| Finds correct third root | A1 | 1.1b |
Typical solution
2nd complex root is \(2 + 3\mathrm{i}\)
sum of roots \(= -\dfrac{b}{a}\)
\[\therefore \alpha + 2 + 3\mathrm{i} + 2 - 3\mathrm{i} = 0\]\[\alpha + 4 = 0\]\[\alpha = -4\]Alternative
2nd complex root is \(2 + 3\mathrm{i}\)
product of roots \(= -\dfrac{d}{a}\)
\[\therefore \alpha(2 + 3\mathrm{i})(2 - 3\mathrm{i}) = -52\]\[\alpha(4 - 9\mathrm{i}^2) = -52\]\[\alpha = \frac{-52}{13}\]\[\alpha = -4\]| Scheme | Marks | AO |
|---|---|---|
| Forms a correct equation to find \(m\). May be seen in part (a). | M1 | 1.1a |
| Finds correct value of \(m\). | A1 | 1.1b |
| (5 marks) |
Typical solution
\[\therefore (-4)^3 + m(-4) + 52 = 0\]\[-64 - 4m + 52 = 0\]\[-12 = 4m\]\[m = -3\]Alternative
\[\sum\alpha\beta = \frac{c}{a}\]\[\therefore (2 - 3\mathrm{i})(2 + 3\mathrm{i}) + (-4)(2 - 3\mathrm{i}) + (-4)(2 + 3\mathrm{i}) = m\]\[4 + 9 - 8 + 12\mathrm{i} - 8 - 12\mathrm{i} = m\]\[m = -3\]