Edexcel F1 2023 June — Question 5 10 marks

Exam BoardEdexcel
ModuleF1 (Further Pure Mathematics 1)
Year2023
SessionJune
Marks10
PaperDownload PDF ↗
TopicRoots of polynomials
TypeQuadratic with transformed roots
DifficultyStandard +0.8 This is a Further Maths question requiring manipulation of symmetric functions of roots beyond basic sum/product. Part (a) needs expansion and substitution using Vieta's formulas. Part (b) requires forming a new equation from complex transformed roots (α/(α²+1)), which demands systematic application of sum and product of transformed roots—a non-trivial multi-step process typical of F1 but more demanding than standard root transformation questions.
Spec4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots

5. $$f ( x ) = x ^ { 2 } - 6 x + 3$$ The equation \(\mathrm { f } ( x ) = 0\) has roots \(\alpha\) and \(\beta\) Without solving the equation,
  1. determine the value of $$\left( \alpha ^ { 2 } + 1 \right) \left( \beta ^ { 2 } + 1 \right)$$
  2. find a quadratic equation which has roots $$\frac { \alpha } { \left( \alpha ^ { 2 } + 1 \right) } \text { and } \frac { \beta } { \left( \beta ^ { 2 } + 1 \right) }$$ giving your answer in the form \(p x ^ { 2 } + q x + r = 0\) where \(p , q\) and \(r\) are integers to be determined.

5.

$$f ( x ) = x ^ { 2 } - 6 x + 3$$

The equation $\mathrm { f } ( x ) = 0$ has roots $\alpha$ and $\beta$\\
Without solving the equation,
\begin{enumerate}[label=(\alph*)]
\item determine the value of

$$\left( \alpha ^ { 2 } + 1 \right) \left( \beta ^ { 2 } + 1 \right)$$
\item find a quadratic equation which has roots

$$\frac { \alpha } { \left( \alpha ^ { 2 } + 1 \right) } \text { and } \frac { \beta } { \left( \beta ^ { 2 } + 1 \right) }$$

giving your answer in the form $p x ^ { 2 } + q x + r = 0$ where $p , q$ and $r$ are integers to be determined.
\end{enumerate}

\hfill \mbox{\textit{Edexcel F1 2023 Q5 [10]}}