A2 June 2019 Paper 2 Q12
12 Abel and Bonnie are trying to solve this mathematical problem:
\(z = 2 - 3\mathrm{i}\) is a root of the equation Find the value of \(m\) and the value of \(p\). |
Abel says he has solved the problem.
Bonnie says there is not enough information to solve the problem.
Since \(z = 2 - 3\mathrm{i}\) is a root of the equation, |
State one extra piece of information about \(m\) and \(p\) which could be added to the problem to make the beginning of Abel’s solution correct. [1 mark]
| Scheme | Marks | AO |
|---|---|---|
| States appropriate extra piece of information. Condone “\(m\) and \(p\) are integers” | R1 | 2.3 |
Typical solution
\(m\) and \(p\) are real numbers.
| Scheme | Marks | AO |
|---|---|---|
| Finds one possible pair of values for \(m\) and \(p\) or finds the root ±7/2 in Abel’s case or substitutes \(z = 2 - 3\mathrm{i}\) in the cubic and expands or uses the sum of roots or the sum of pairwise products of roots (not using \(z = 2 + 3\mathrm{i}\)) to form an equation (condone sign errors) | R1 | 2.1 |
| Uses product of roots = ±91/2 to find another possible pair of complex roots of the cubic or expresses \(m\) and \(p\) as complex numbers in their expansion or uses product of roots = ±91/2 to form another equation (not using \(z = 2 + 3\mathrm{i}\)) | M1 | 3.1a |
| Finds another possible correct pair of complex roots of the cubic (condone sign errors) or forms simultaneous equations for real and imaginary parts or reduces their system of equations to a system of linear equations | M1 | 2.1 |
| Completes a rigorous argument to prove that Bonnie is correct; for example, explains that they have shown that there is more than one possible set of values for \(m\) and \(p\) or explains why their simultaneous linear equations have no unique solution | R1 | 2.4 |
| (5 marks) |
Typical solution
Product of roots = − 91/2
One possible solution to the cubic is given by (for example)
\[z = 2 - 3\mathrm{i},\ z = 2 + 3\mathrm{i},\ z = -\frac{7}{2}\]since
\[(2 - 3\mathrm{i})(2 + 3\mathrm{i})\left(-\frac{7}{2}\right) = -\frac{91}{2}\]This gives \(m = -1\) and \(p = -2\)
Another solution to the cubic is (for example)
\[z = 2 - 3\mathrm{i},\ z = 4 + 6\mathrm{i},\ z = -\frac{7}{4}\]since
\[(2 - 3\mathrm{i})(4 + 6\mathrm{i})\left(-\frac{7}{4}\right) = -\frac{91}{2}\]This gives different values of \(m\) and \(p\).
Because there is more than one possible set of values of \(m\) and \(p\), there is not sufficient information to solve the problem, and Bonnie is right.