A2 June 2022 Paper 1 Q5
5 It is given that \(z = -\dfrac{3}{2} + \mathrm{i}\dfrac{\sqrt{11}}{2}\) is a root of the equation
\[z^4 - 3z^3 - 5z^2 + kz + 40 = 0\]where \(k\) is a real number.
(a) Find the other three roots. [5 marks]
(b) Given that \(x \in \mathbb{R}\), solve\[x^4 - 3x^3 - 5x^2 + kx + 40 \lt 0\] [1 mark]
| Scheme | Marks | AO |
|---|---|---|
| States complex conjugate | B1 | 1.1b |
| Obtains a quadratic factor or Obtains sum and product of their two complex roots or Substitutes given root or their conjugate into the quartic | M1 | 1.1a |
| Obtains \(z^2 + 3z + 5\) or Obtains \(k = -6\) | A1 | 1.1b |
| Forms a second quadratic or Solves the quartic equation with their value of \(k\) | M1 | 1.1a |
| Obtains 2 and 4 | A1 | 1.1b |
| (5) |
Typical solution
\(z = -\dfrac{3}{2} - \mathrm{i}\dfrac{\sqrt{11}}{2}\) is another root
Sum of complex roots = −3
Product of complex roots = 5
So \(z^2 + 3z + 5\) is a factor of \(z^4 - 3z^3 - 5z^2 + kz + 40\)
The other quadratic factor is
\[z^2 - 6z + 8 = (z - 2)(z - 4)\]The other roots are
\[2,\ 4,\ -\frac{3}{2} - \mathrm{i}\frac{\sqrt{11}}{2}\]| Scheme | Marks | AO |
|---|---|---|
| Deduces correct solution (ft from their exactly two distinct real roots) | B1F | 2.2a |
| (1) | ||
| (6 marks) |