| Exam Board | OCR |
|---|---|
| Module | Further Pure Core 2 (Further Pure Core 2) |
| Year | 2018 |
| Session | September |
| Marks | 9 |
| Topic | Roots of polynomials |
| Type | Quadratic with transformed roots |
| Difficulty | Challenging +1.2 This is a Further Maths question on root transformations requiring systematic algebraic manipulation. Part (i) uses standard Vieta's formulas (α+β=-b/a, αβ=c/a) to construct a new equation—methodical but routine for FM students. Parts (ii) and (iii) require discriminant analysis with some algebraic insight, but follow predictable patterns once the setup is established. The multi-part structure and proof elements elevate it above average A-level difficulty, but it remains a standard FM exercise without requiring novel problem-solving approaches. |
| Spec | 4.05a Roots and coefficients: symmetric functions |
| Answer | Marks | Guidance |
|---|---|---|
| \(\alpha + \beta = -\frac{b}{a}\), \(\alpha\beta = \frac{c}{a}\) used | B1 | |
| \(\left(x - \frac{b}{a}\right)\left(x - \frac{c}{a}\right) = 0\) | M1 | |
| \(x^2 + \frac{b-c}{a}x - \frac{bc}{a} = 0\) | A1 | |
| \(a^2x^2 + a(b-c)x - bc = 0\) | A1 | Or any non-zero integer multiple (ie correctly multiplying by \(a^2\) (or \(ka^2\) or \(a\) etc) (could be done earlier)) |
| Answer | Marks | Guidance |
|---|---|---|
| Both having repeated roots \(\Rightarrow \alpha = \beta\) and \(\alpha\beta = \alpha + \beta \Rightarrow \alpha^2 = 2\alpha\) | M1 | Assume repeated roots and set up equations. |
| \(E1\) | 2.1 | For establishing contradiction |
| so either \(\alpha = \beta = 0 \Rightarrow b = c = 0\) or \(\alpha = \beta = 2 \Rightarrow b < 0\) (or \(a < 0\) & \(c < 0\)). But \(a, b, c > 0\) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\Delta = (a(b-c))^2 - 4a^2(-bc)\) | M1 | Correct substitution into "\(b^2 - 4ac\)" |
| \(\Delta = a^2(b^2 - 2bc + c^2 + 4bc) = a^2(b+c)^2\) which is positive since \(a, b\) and \(c\) are real and both \(a \neq 0\) and \(b + c \neq 0\) | A1 | |
| \(a^2 > 0\) and \((b+c)^2 > 0\) alone is insufficient | E1 |
## (i)
$\alpha + \beta = -\frac{b}{a}$, $\alpha\beta = \frac{c}{a}$ used | B1 |
$\left(x - \frac{b}{a}\right)\left(x - \frac{c}{a}\right) = 0$ | M1 |
$x^2 + \frac{b-c}{a}x - \frac{bc}{a} = 0$ | A1 |
$a^2x^2 + a(b-c)x - bc = 0$ | A1 | Or any non-zero integer multiple (ie correctly multiplying by $a^2$ (or $ka^2$ or $a$ etc) (could be done earlier))
[4]
Or $\alpha' + \beta' = -\frac{b}{a} - \frac{c}{a}$, $\alpha'\beta' = \frac{b}{a} \cdot \frac{c}{a}$ Correctly expanding brackets and collecting terms
## (ii)
Both having repeated roots $\Rightarrow \alpha = \beta$ and $\alpha\beta = \alpha + \beta \Rightarrow \alpha^2 = 2\alpha$ | M1 | Assume repeated roots and set up equations.
$E1$ | 2.1 | For establishing contradiction
so either $\alpha = \beta = 0 \Rightarrow b = c = 0$ or $\alpha = \beta = 2 \Rightarrow b < 0$ (or $a < 0$ & $c < 0$). But $a, b, c > 0$ | E1 |
[2]
## (iii)
$\Delta = (a(b-c))^2 - 4a^2(-bc)$ | M1 | Correct substitution into "$b^2 - 4ac$"
$\Delta = a^2(b^2 - 2bc + c^2 + 4bc) = a^2(b+c)^2$ which is positive since $a, b$ and $c$ are real and both $a \neq 0$ and $b + c \neq 0$ | A1 |
$a^2 > 0$ and $(b+c)^2 > 0$ alone is insufficient | E1 |
[3]
---The roots of the equation $ax^2 + bx + c = 0$, where $a$, $b$ and $c$ are positive integers, are $\alpha$ and $\beta$.
\begin{enumerate}[label=(\roman*)]
\item Find a quadratic equation with integer coefficients whose roots are $\alpha + \beta$ and $\alpha\beta$. [4]
\item Show that it is not possible for the original equation and the equation found in part (i) both to have repeated roots. [2]
\item Show that the discriminant of the equation found in part (i) is always positive. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR Further Pure Core 2 2018 Q7 [9]}}