| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2016 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Quadratic with transformed roots |
| Difficulty | Moderate -0.3 Part (a) is direct recall of Vieta's formulas (2 marks). Part (b) requires finding sum and product of the new roots using algebraic manipulation of α+β and αβ, then constructing the equation - a standard FP1 technique but requires careful algebra across multiple steps. Overall slightly easier than average as it's a textbook-style roots transformation question with no novel insight needed. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\alpha + \beta = 6\); \(\alpha\beta = 14\) | B1; B1 | If LHS is missing look for later evidence before awarding the B1s |
| (b) \(P = \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = 1\) | B1 | \(P = 1\) seen or used |
| \(S = \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta}\) | M1 | \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta\) OE seen or used |
| \(\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 36 - 28 = 8\) | M1 | |
| \(S = \frac{8}{14}\) | A1 | A correct value for \(S\) seen, or used in quadratic. Ft on wrong sign for \(\alpha + \beta\) |
| \(x^2 - Sx + P = 0\) | M1 | Using correct general form of LHS of eqn with ft substitution of \(c\)'s \(S\) and \(P\) non-zero values |
| \(x^2 - \frac{8}{14}x + 1 = 0\) | ||
| (Quadratic eqn is) \(7x^2 - 4x + 7 = 0\) | A1 | CSO. ACF of the equation, but must have integer coefficients |
**(a)** $\alpha + \beta = 6$; $\alpha\beta = 14$ | B1; B1 | If LHS is missing look for later evidence before awarding the B1s
**(b)** $P = \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = 1$ | B1 | $P = 1$ seen or used
$S = \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta}$ | M1 | $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta$ OE seen or used
$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = 36 - 28 = 8$ | M1 |
$S = \frac{8}{14}$ | A1 | A correct value for $S$ seen, or used in quadratic. Ft on wrong sign for $\alpha + \beta$
$x^2 - Sx + P = 0$ | M1 | Using correct general form of LHS of eqn with ft substitution of $c$'s $S$ and $P$ non-zero values
$x^2 - \frac{8}{14}x + 1 = 0$ | |
(Quadratic eqn is) $7x^2 - 4x + 7 = 0$ | A1 | CSO. ACF of the equation, but must have integer coefficients
**Total: 7 marks**
---The quadratic equation $x^2 - 6x + 14 = 0$ has roots $\alpha$ and $\beta$.
\begin{enumerate}[label=(\alph*)]
\item Write down the value of $\alpha + \beta$ and the value of $\alpha\beta$.
[2 marks]
\item Find a quadratic equation, with integer coefficients, which has roots $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$.
[5 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2016 Q1 [7]}}