| Exam Board | AQA |
|---|---|
| Module | FP1 (Further Pure Mathematics 1) |
| Year | 2015 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Roots of polynomials |
| Type | Quadratic with transformed roots |
| Difficulty | Standard +0.3 This is a standard Further Maths question on transformed roots requiring application of Vieta's formulas and forming new equations from sums and products of transformed roots. Part (a) is direct recall, part (b) requires computing (α²-1)+(β²-1) and (α²-1)(β²-1) using standard algebraic manipulation, and part (c) is straightforward solving. While it's Further Maths content, it follows a well-established template with no novel insight required, making it slightly easier than average overall. |
| Spec | 4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots |
| Answer | Marks | Guidance |
|---|---|---|
| \(\alpha + \beta = -3\) | B1 | |
| \(\alpha\beta = \frac{7}{2}\) | B1 | Both required for 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Sum of new roots: \((\alpha^2-1)+(\beta^2-1) = \alpha^2+\beta^2-2\) | M1 | |
| \(\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 9 - 7 = 2\) | A1 | |
| So sum \(= 2 - 2 = 0\) | A1 | |
| Product of new roots: \((\alpha^2-1)(\beta^2-1) = \alpha^2\beta^2 - \alpha^2 - \beta^2 + 1\) | M1 | |
| \(= \left(\frac{7}{2}\right)^2 - 2 + 1 = \frac{49}{4} - 1 = \frac{45}{4}\) | A1 | |
| Equation: \(4x^2 + 45 = 0\) (or \(x^2 + \frac{45}{4} = 0\)) | A1 | Must have integer coefficients |
| Answer | Marks | Guidance |
|---|---|---|
| From \(4x^2 + 45 = 0\), roots are \(\pm\frac{3\sqrt{5}}{2}i\), so \(\alpha^2 = \frac{-45}{4}\) and \(\beta^2 = \frac{-45}{4}\) | B1 ft | |
| Or: since sum \(= 0\) and product \(= \frac{45}{4}\), both roots equal \(-\frac{45}{4}\) | B1 | Both values required |
# Question 1:
## Part (a)
| $\alpha + \beta = -3$ | B1 | |
| $\alpha\beta = \frac{7}{2}$ | B1 | Both required for 2 marks |
## Part (b)
| Sum of new roots: $(\alpha^2-1)+(\beta^2-1) = \alpha^2+\beta^2-2$ | M1 | |
| $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta = 9 - 7 = 2$ | A1 | |
| So sum $= 2 - 2 = 0$ | A1 | |
| Product of new roots: $(\alpha^2-1)(\beta^2-1) = \alpha^2\beta^2 - \alpha^2 - \beta^2 + 1$ | M1 | |
| $= \left(\frac{7}{2}\right)^2 - 2 + 1 = \frac{49}{4} - 1 = \frac{45}{4}$ | A1 | |
| Equation: $4x^2 + 45 = 0$ (or $x^2 + \frac{45}{4} = 0$) | A1 | Must have integer coefficients |
## Part (c)
| From $4x^2 + 45 = 0$, roots are $\pm\frac{3\sqrt{5}}{2}i$, so $\alpha^2 = \frac{-45}{4}$ and $\beta^2 = \frac{-45}{4}$ | B1 ft | |
| Or: since sum $= 0$ and product $= \frac{45}{4}$, both roots equal $-\frac{45}{4}$ | B1 | Both values required |
---1 The quadratic equation $2 x ^ { 2 } + 6 x + 7 = 0$ has roots $\alpha$ and $\beta$.
\begin{enumerate}[label=(\alph*)]
\item Write down the value of $\alpha + \beta$ and the value of $\alpha \beta$.
\item Find a quadratic equation, with integer coefficients, which has roots $\alpha ^ { 2 } - 1$ and $\beta ^ { 2 } - 1$.
\item Hence find the values of $\alpha ^ { 2 }$ and $\beta ^ { 2 }$.
\end{enumerate}
\hfill \mbox{\textit{AQA FP1 2015 Q1 [9]}}