| Exam Board | Edexcel |
|---|---|
| Module | CP AS (Core Pure AS) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Roots with special relationships |
| Difficulty | Challenging +1.2 This is a Core Pure AS question requiring use of Vieta's formulas with an unusual root relationship. Students must set up equations from sum and product of roots, then solve a system involving the constraint. While the algebraic manipulation is non-trivial and the root relationship is unconventional, the problem follows a standard template for this topic with clear signposting across two parts totaling moderate marks. |
| Spec | 4.02i Quadratic equations: with complex roots4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(\alpha + \beta + \left(\alpha + \frac{12}{\alpha} - \beta\right) = 8\) so \(2\alpha + \frac{12}{\alpha} = 8\) | M1 | Equates sum of roots to 8 and obtains equation in just \(\alpha\) |
| \(\Rightarrow 2\alpha^2 - 8\alpha + 12 = 0\) or \(\alpha^2 - 4\alpha + 6 = 0\) | A1 | Obtains a correct equation in \(\alpha\) |
| \(\Rightarrow \alpha = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(6)}}{2(1)}\) or \((\alpha-2)^2 - 4 + 6 = 0 \Rightarrow \alpha = \ldots\) | M1 | Forms three-term quadratic in \(\alpha\), attempts to solve by completing the square or quadratic formula |
| \(\Rightarrow \alpha = 2 \pm i\sqrt{2}\) are the two complex roots | A1 | \(\alpha = 2 \pm i\sqrt{2}\) |
| Third root via sum: \(= 8 - (2+i\sqrt{2}) - (2-i\sqrt{2}) = \ldots\) OR via product \(= \frac{24}{(2+i\sqrt{2})(2-i\sqrt{2})} = \ldots\) OR via \((z-\alpha)(z-\beta) = z^2 - 4z + 6 \Rightarrow f(z) = (z^2-4z+6)(z-\gamma) \Rightarrow \gamma = \ldots\) | M1 | Any correct method for finding remaining root |
| Roots of \(f(z) = 0\) are \(2 \pm i\sqrt{2}\) and \(4\) | A1 | Third root found with all three roots correct; \(\alpha\) and \(\beta\) need not be identified |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(f(4) = 0 \Rightarrow 4^3 - 8\times4^2 + 4p - 24 = 0 \Rightarrow p = \ldots\) OR \(p = (2+i\sqrt{2})(2-i\sqrt{2}) + 4(2+i\sqrt{2}) + 4(2-i\sqrt{2}) \Rightarrow p = \ldots\) OR \(f(z) = (z-4)(z^2-4z+6) \Rightarrow p = \ldots\) | M1 | Any correct method of finding \(p\); e.g. factor theorem, sum of pairs, or using roots to form \(f(z)\) |
| \(\Rightarrow p = 22\) | A1 | \(p = 22\) by correct solution only; can be found using complex roots from (a) |
# Question 7:
## Part (a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $\alpha + \beta + \left(\alpha + \frac{12}{\alpha} - \beta\right) = 8$ so $2\alpha + \frac{12}{\alpha} = 8$ | M1 | Equates sum of roots to 8 and obtains equation in just $\alpha$ |
| $\Rightarrow 2\alpha^2 - 8\alpha + 12 = 0$ or $\alpha^2 - 4\alpha + 6 = 0$ | A1 | Obtains a correct equation in $\alpha$ |
| $\Rightarrow \alpha = \frac{4 \pm \sqrt{(-4)^2 - 4(1)(6)}}{2(1)}$ or $(\alpha-2)^2 - 4 + 6 = 0 \Rightarrow \alpha = \ldots$ | M1 | Forms three-term quadratic in $\alpha$, attempts to solve by completing the square or quadratic formula |
| $\Rightarrow \alpha = 2 \pm i\sqrt{2}$ are the two complex roots | A1 | $\alpha = 2 \pm i\sqrt{2}$ |
| Third root via sum: $= 8 - (2+i\sqrt{2}) - (2-i\sqrt{2}) = \ldots$ OR via product $= \frac{24}{(2+i\sqrt{2})(2-i\sqrt{2})} = \ldots$ OR via $(z-\alpha)(z-\beta) = z^2 - 4z + 6 \Rightarrow f(z) = (z^2-4z+6)(z-\gamma) \Rightarrow \gamma = \ldots$ | M1 | Any correct method for finding remaining root |
| Roots of $f(z) = 0$ are $2 \pm i\sqrt{2}$ and $4$ | A1 | Third root found with all three roots correct; $\alpha$ and $\beta$ need not be identified |
## Part (b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $f(4) = 0 \Rightarrow 4^3 - 8\times4^2 + 4p - 24 = 0 \Rightarrow p = \ldots$ OR $p = (2+i\sqrt{2})(2-i\sqrt{2}) + 4(2+i\sqrt{2}) + 4(2-i\sqrt{2}) \Rightarrow p = \ldots$ OR $f(z) = (z-4)(z^2-4z+6) \Rightarrow p = \ldots$ | M1 | Any correct method of finding $p$; e.g. factor theorem, sum of pairs, or using roots to form $f(z)$ |
| $\Rightarrow p = 22$ | A1 | $p = 22$ by correct solution only; can be found using complex roots from (a) |
---7.
$$\mathrm { f } ( z ) = z ^ { 3 } - 8 z ^ { 2 } + p z - 24$$
where $p$ is a real constant.\\
Given that the equation $\mathrm { f } ( z ) = 0$ has distinct roots
$$\alpha , \beta \text { and } \left( \alpha + \frac { 12 } { \alpha } - \beta \right)$$
\begin{enumerate}[label=(\alph*)]
\item solve completely the equation $\mathrm { f } ( z ) = 0$
\item Hence find the value of $p$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel CP AS 2019 Q7 [8]}}