| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2010 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Quadratic with transformed roots |
| Difficulty | Standard +0.8 This is a multi-step Further Maths question requiring manipulation of roots and their reciprocals. Part (a) is routine recall, part (b) is standard algebraic manipulation, but part (c) requires finding sum and product of transformed roots involving both α and β terms, demanding careful algebraic work with complex expressions and multiple substitutions—significantly above average difficulty but still a recognizable FP1 exercise type. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| \(\alpha + \beta = 4\), \(\alpha\beta = 10\) | B1, B1 | One mark each |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{1}{\alpha}+\frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{4}{10} = \frac{2}{5}\) | M1, A1 | M1 for correct method using Vieta's, A1 completion |
| Answer | Marks | Guidance |
|---|---|---|
| Sum of new roots: \(\alpha+\frac{2}{\beta}+\beta+\frac{2}{\alpha} = (\alpha+\beta)+2\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) = 4 + 2\cdot\frac{2}{5} = \frac{24}{5}\) | M1, A1 | |
| Product of new roots: \(\left(\alpha+\frac{2}{\beta}\right)\left(\beta+\frac{2}{\alpha}\right) = \alpha\beta + 2 + 2 + \frac{4}{\alpha\beta} = 10+4+\frac{4}{10} = \frac{142}{10} = \frac{71}{5}\) | M1, A1 | |
| Equation: \(x^2 - \frac{24}{5}x + \frac{71}{5}=0\); \(5x^2 - 24x + 71 = 0\) | A1, A1 | A1 correct equation with integer coefficients |
# Question 8:
**(a)**
| $\alpha + \beta = 4$, $\alpha\beta = 10$ | B1, B1 | One mark each |
**(b)**
| $\frac{1}{\alpha}+\frac{1}{\beta} = \frac{\alpha+\beta}{\alpha\beta} = \frac{4}{10} = \frac{2}{5}$ | M1, A1 | M1 for correct method using Vieta's, A1 completion |
**(c)**
| Sum of new roots: $\alpha+\frac{2}{\beta}+\beta+\frac{2}{\alpha} = (\alpha+\beta)+2\left(\frac{1}{\alpha}+\frac{1}{\beta}\right) = 4 + 2\cdot\frac{2}{5} = \frac{24}{5}$ | M1, A1 | |
| Product of new roots: $\left(\alpha+\frac{2}{\beta}\right)\left(\beta+\frac{2}{\alpha}\right) = \alpha\beta + 2 + 2 + \frac{4}{\alpha\beta} = 10+4+\frac{4}{10} = \frac{142}{10} = \frac{71}{5}$ | M1, A1 | |
| Equation: $x^2 - \frac{24}{5}x + \frac{71}{5}=0$; $5x^2 - 24x + 71 = 0$ | A1, A1 | A1 correct equation with integer coefficients |
---8 The quadratic equation
$$x ^ { 2 } - 4 x + 10 = 0$$
has roots $\alpha$ and $\beta$.
\begin{enumerate}[label=(\alph*)]
\item Write down the values of $\alpha + \beta$ and $\alpha \beta$.
\item Show that $\frac { 1 } { \alpha } + \frac { 1 } { \beta } = \frac { 2 } { 5 }$.
\item Find a quadratic equation, with integer coefficients, which has roots $\alpha + \frac { 2 } { \beta }$ and $\beta + \frac { 2 } { \alpha }$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2010 Q8 [10]}}