AS October 2020 Paper 1 Q7
7 In the quartic equation \(2x^4 - 20x^3 + ax^2 + bx + 250 = 0\), the coefficients \(a\) and \(b\) are real. One root of the equation is \(2 + \mathrm{i}\).
Find the other roots. [7]
| Scheme | Marks | AO |
|---|---|---|
| Another root is \(2 - \mathrm{i}\) | B1 | 1.2 |
| suppose other roots are \(\gamma\), \(\delta\) (say) | 3.1a | |
| \(4 + \gamma + \delta = 10\) | M1 | 1.1a |
| \(\Rightarrow \gamma + \delta = 6\) | A1 | 1.1 |
| \((2 + \mathrm{i})(2 - \mathrm{i})\gamma\delta = 125\) | M1 | 1.1 |
| \(5\gamma\delta = 125\) | A1 | |
| \(\gamma = 3 + 4\mathrm{i},\ \delta = 3 - 4\mathrm{i}\) | M1 | 1.1 |
| So other 2 roots are \(3 + 4\mathrm{i}\) and \(3 - 4\mathrm{i}\) | A1 | 3.2a |
| [7] |
Notes
Or \(c + d\mathrm{i}\) and \(c - d\mathrm{i}\). M1
\(4 + 2c = 10\) M1
\(\Rightarrow c = 3\) A1
\((2 + \mathrm{i})(2 - \mathrm{i})(3 + d\mathrm{i})(3 - d\mathrm{i}) = 125\)
\(\Rightarrow 5(9 + d^2) = 125\) M1A1
\(\Rightarrow d = 4\) A1
M1: (2nd) product of roots used
M1: (3rd) eliminating and solving quad
Alternative solution
| Scheme | Marks |
|---|---|
| Another root is \(2 - \mathrm{i}\) | B1 |
| \((x - 2 - \mathrm{i})(x - 2 + \mathrm{i}) = x^2 - 4x + 5\) | M1 A1 |
| \((x^2 - 4x + 5)(2x^2 + kx + 50) = 2x^4 - 20x^3 + ax^2 + bx + 250\) \(\Rightarrow k - 8 = -20\), so \(k = -12\) | M1 |
| Other factor is \(2x^2 - 12x + 50\) | A1 |
| \(x^2 - 6x + 25 = 0\) gives \(x = 3 \pm 4\mathrm{i}\) | M1 |
| Roots are \(2 + \mathrm{i}\), \(2 - \mathrm{i}\), \(3 + 4\mathrm{i}\) and \(3 - 4\mathrm{i}\) | A1 |
M1: (1st) Attempt to multiply factors
\(x^2 - 4x + 5\)
condone sign errors
M1: (2nd) Attempt to find other quad factor
\(2x^2 - 12x + 50\)
or long division