| Exam Board | WJEC |
|---|---|
| Module | Further Unit 1 (Further Unit 1) |
| Year | 2019 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Equation with nonlinearly transformed roots |
| Difficulty | Challenging +1.2 This is a Further Maths question on transformed roots requiring systematic application of Vieta's formulas and algebraic manipulation. Part (a) involves finding a cubic from given roots using sum and product relationships, while part (b) requires comparing coefficients between two quadratics. Both parts are methodical rather than requiring novel insight, making this moderately above average difficulty but standard for Further Maths content. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| From equation 1: \(\alpha\beta = \frac{r}{p}\), \(\alpha + \beta = -\frac{q}{p}\). Sum of roots = \(2\alpha + 2\beta (= 2(\alpha + \beta))\). Sum of pairs = \(\alpha\beta + \alpha(\alpha + \beta) + \beta(\alpha + \beta) (= \alpha\beta + (\alpha + \beta)^2)\). Triples = \(\alpha\beta(\alpha + \beta)\) | B1, B2 | Both correct; B1 for 2 correct |
| \(\therefore \frac{-b}{a} = 2 \times \left(-\frac{q}{p}\right) = -\frac{2q}{p}\), \(\frac{c}{a} = \frac{r}{p} + \left(-\frac{q}{p}\right)^2 = \frac{r}{p} + \frac{q^2}{p^2}\), \(\frac{-d}{a} = \frac{r}{p} \times \left(-\frac{q}{p}\right) = -\frac{qr}{p^2}\). The equation is \(x^3 + \frac{2qx^2}{p} + \left(\frac{r}{p} + \frac{q^2}{p^2}\right)x + \frac{qr}{p^2} = 0\) oe | B2, B1 | B1 for 2 correct; FT 2nd B2 above |
| Answer | Marks | Guidance |
|---|---|---|
| From equation 2: \(2\alpha\gamma = -\frac{r}{p}\), \(\alpha = \frac{r}{p\beta} = -\frac{r}{2p\gamma}\), \(\therefore \frac{1}{\beta} = -\frac{1}{2\gamma}\), \(\therefore \beta = -2\gamma\) | B1, M1, A1 | Convincing |
**Part (a):**
From equation 1: $\alpha\beta = \frac{r}{p}$, $\alpha + \beta = -\frac{q}{p}$. Sum of roots = $2\alpha + 2\beta (= 2(\alpha + \beta))$. Sum of pairs = $\alpha\beta + \alpha(\alpha + \beta) + \beta(\alpha + \beta) (= \alpha\beta + (\alpha + \beta)^2)$. Triples = $\alpha\beta(\alpha + \beta)$ | B1, B2 | Both correct; B1 for 2 correct
$\therefore \frac{-b}{a} = 2 \times \left(-\frac{q}{p}\right) = -\frac{2q}{p}$, $\frac{c}{a} = \frac{r}{p} + \left(-\frac{q}{p}\right)^2 = \frac{r}{p} + \frac{q^2}{p^2}$, $\frac{-d}{a} = \frac{r}{p} \times \left(-\frac{q}{p}\right) = -\frac{qr}{p^2}$. The equation is $x^3 + \frac{2qx^2}{p} + \left(\frac{r}{p} + \frac{q^2}{p^2}\right)x + \frac{qr}{p^2} = 0$ oe | B2, B1 | B1 for 2 correct; FT 2nd B2 above
**Part (b):**
From equation 2: $2\alpha\gamma = -\frac{r}{p}$, $\alpha = \frac{r}{p\beta} = -\frac{r}{2p\gamma}$, $\therefore \frac{1}{\beta} = -\frac{1}{2\gamma}$, $\therefore \beta = -2\gamma$ | B1, M1, A1 | Convincing
---10. The quadratic equation $p x ^ { 2 } + q x + r = 0$ has roots $\alpha$ and $\beta$, where $p , q , r$ are non-zero constants.
\begin{enumerate}[label=(\alph*)]
\item A cubic equation is formed with roots $\alpha , \beta , \alpha + \beta$.
Find the cubic equation with coefficients expressed in terms of $p , q , r$.
\item Another quadratic equation $p x ^ { 2 } - q x - r = 0$ has roots $2 \alpha$ and $\gamma$.
Show that $\beta = - 2 \gamma$.
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 1 2019 Q10 [9]}}