OCR FP1 — Question 8 11 marks

Exam BoardOCR
ModuleFP1 (Further Pure Mathematics 1)
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeQuadratic with transformed roots
DifficultyModerate -0.3 This is a standard FP1 question on roots of polynomials using Vieta's formulas and symmetric functions. Part (a) involves routine application of sum/product of roots and algebraic manipulation to find a new equation. Part (b) uses Vieta's formulas with a constraint on the roots. All techniques are textbook exercises with no novel insight required, making it slightly easier than average A-level difficulty despite being Further Maths content.
Spec4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots

  1. The quadratic equation \(x^2 - 2x + 4 = 0\) has roots \(\alpha\) and \(\beta\).
    1. Write down the values of \(\alpha + \beta\) and \(\alpha\beta\). [2]
    2. Show that \(\alpha^2 + \beta^2 = -4\). [2]
    3. Hence find a quadratic equation which has roots \(\alpha^2\) and \(\beta^2\). [3]
  2. The cubic equation \(x^3 - 12x^2 + ax - 48 = 0\) has roots \(p\), \(2p\) and \(3p\).
    1. Find the value of \(p\). [2]
    2. Hence find the value of \(a\). [2]

\begin{enumerate}[label=(\alph*)]
\item The quadratic equation $x^2 - 2x + 4 = 0$ has roots $\alpha$ and $\beta$.
\begin{enumerate}[label=(\roman*)]
\item Write down the values of $\alpha + \beta$ and $\alpha\beta$. [2]
\item Show that $\alpha^2 + \beta^2 = -4$. [2]
\item Hence find a quadratic equation which has roots $\alpha^2$ and $\beta^2$. [3]
\end{enumerate}

\item The cubic equation $x^3 - 12x^2 + ax - 48 = 0$ has roots $p$, $2p$ and $3p$.
\begin{enumerate}[label=(\roman*)]
\item Find the value of $p$. [2]
\item Hence find the value of $a$. [2]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{OCR FP1  Q8 [11]}}