FP1 June 2017 Q6
6. Given that 4 and \(2\mathrm{i} - 3\) are roots of the equation \[x^3 + ax^2 + bx - 52 = 0\] where \(a\) and \(b\) are real constants,
| Scheme | Marks |
|---|---|
| \(x^3 + ax^2 + bx - 52 = 0,\ a, b \in \mathbb{R},\) 4 and \(2\mathrm{i} - 3\) are roots | |
| \(-2\mathrm{i} - 3\) | B1 |
| (1) |
Notes
B1: \(-2\mathrm{i} - 3\) seen anywhere in solution for Q6.
| Scheme | Marks |
|---|---|
| Way 1 \(\left(x - (2\mathrm{i} - 3)\right)\left(x - \text{"}(-2\mathrm{i} - 3)\text{"}\right);\ = x^2 + 6x + 13\) or \(x = -3 \pm 2\mathrm{i} \Rightarrow (x + 3)^2 = -4;\ = x^2 + 6x + 13 (= 0)\) \((x - 4)\left(x - (2\mathrm{i} - 3)\right);\ = x^2 - (1 + 2\mathrm{i})x + 4(2\mathrm{i} - 3)\) \((x - 4)\left(x - \text{"}(-2\mathrm{i} - 3)\text{"}\right);\ = x^2 - (1 - 2\mathrm{i})x + 4(-2\mathrm{i} - 3)\) | M1; A1 |
| \((x - 4)(x^2 + 6x + 13)\ \left\{= x^3 + ax^2 + bx - 52\right\}\) | M1 |
| \(a = 2,\ b = -11\) or \(x^3 + 2x^2 - 11x - 52\) | A1 A1 |
| (5) | |
| (6 marks) |
Notes
M1; A1: Must follow from their part (a). Any incorrect signs for their part (a) in initial statement award M0; accept any equivalent expanded expression for A1.
M1: \(\left(x - 3^{\text{rd}}\text{ root}\right)\left(\text{their quadratic}\right)\). Could be found by comparing coefficients from long division.
A1: At least one of \(a = 2\) or \(b = -11\)
A1: Both \(a = 2\) and \(b = -11\)
(b) Way 2
| Scheme | Marks |
|---|---|
| Sum \(= (2\mathrm{i} - 3) + \text{"}(-2\mathrm{i} - 3)\text{"} = -6\) Product \(= (2\mathrm{i} - 3) \times \text{"}(-2\mathrm{i} - 3)\text{"} = 13\) So quadratic is \(x^2 + 6x + 13\) | M1 A1 |
| \((x - 4)(x^2 + 6x + 13)\ \left\{= x^3 + ax^2 + bx - 52\right\}\) | M1 |
| \(a = 2,\ b = -11\) or \(x^3 + 2x^2 - 11x - 52\) | A1 A1 |
| (5) |
M1: Attempts to apply either \(x^2 - (\text{sum roots})x + (\text{product roots}) = 0\) or \(x^2 - 2\,\mathrm{Re}(\alpha)x + \left|\alpha^2\right| = 0\)
A1: \(x^2 + 6x + 13\)
M1: \(\left(x - 3^{\text{rd}}\text{ root}\right)\left(\text{their quadratic}\right)\)
A1: At least one of \(a = 2\) or \(b = -11\)
A1: Both \(a = 2\) and \(b = -11\)
(b) Way 3
| Scheme | Marks |
|---|---|
| \((2\mathrm{i} - 3)^3 + a(2\mathrm{i} - 3)^2 + b(2\mathrm{i} - 3) - 52 = 0\) | |
| \(5a - 3b = 43\) (real parts) and \(6a - b = 23\) (imaginary parts) or uses f(4) = 0 and f(a complex root) = 0 to form equations in \(a\) and \(b\). | M1 A1 |
| So \(a = 2,\ b = -11\) or \(x^3 + 2x^2 - 11x - 52\) | M1 A1 A1 |
| (5) |
M1: Substitutes \(2\mathrm{i} - 3\) into the displayed equation and equates both real and imaginary parts.
A1: \(5a - 3b = 43\) and \(6a - b = 23\) or \(16a + 4b = -12\) and \((2i - 3)^3 + a(2i - 3)^2 + b(2i - 3) - 52 = 0\) / \((-2i - 3)^3 + a(-2i - 3)^2 + b(-2i - 3) - 52 = 0\)
M1: Solves these equations simultaneously to find at least one of either \(a = \ldots\) or \(b = \ldots\)
A1: At least one of \(a = 2\) or \(b = -11\)
A1: Both \(a = 2\) and \(b = -11\)
(b) Way 4
| Scheme | Marks |
|---|---|
| \(b = \text{sum of product pairs}\) \(= 4(2\mathrm{i} - 3) + 4\text{"}(-2\mathrm{i} - 3)\text{"} + (2\mathrm{i} - 3)\text{"}(-2\mathrm{i} - 3)\text{"}\) | M1 A1 |
| \(a = -(\text{sum of 3 roots}) = -(4 + 2\mathrm{i} - 3\text{"} - 2\mathrm{i} - 3\text{"})\) | M1 |
| \(a = 2,\ b = -11\) or \(x^3 + 2x^2 - 11x - 52\) | A1 A1 |
| (5) |
M1: Attempts sum of product pairs.
A1: All pairs correct o.e.
M1: Adds up all 3 roots
A1: At least one of \(a = 2\) or \(b = -11\)
A1: Both \(a = 2\) and \(b = -11\)
(b) Way 5
| Scheme | Marks |
|---|---|
| Uses f(4) = 0 | M1 |
| \(16a + 4b = -12\) | A1 |
| \(a = -(\text{sum of 3 roots}) = -(4 + 2\mathrm{i} - 3\text{"} - 2\mathrm{i} - 3\text{"})\) | M1 |
| \(a = 2,\ b = -11\) or \(x^3 + 2x^2 - 11x - 52\) | A1 A1 |
| (5) |
M1: Adds up all 3 roots
A1: At least one of \(a = 2\) or \(b = -11\)
A1: Both \(a = 2\) and \(b = -11\)