AQA FP1 2006 January — Question 5 11 marks

Exam BoardAQA
ModuleFP1 (Further Pure Mathematics 1)
Year2006
SessionJanuary
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicComplex Numbers Arithmetic
TypeVerifying roots satisfy equations
DifficultyStandard +0.3 This is a straightforward Further Pure 1 question testing basic complex number operations and standard results about conjugate roots. Part (a) involves routine multiplication and direct substitution to verify a root. Part (b) applies the standard theorem that complex roots of real quadratics come in conjugate pairs, then uses sum/product of roots formulas—all textbook exercises requiring recall and careful arithmetic but no problem-solving insight.
Spec4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02g Conjugate pairs: real coefficient polynomials4.05a Roots and coefficients: symmetric functions

5
    1. Calculate \(( 2 + \mathrm { i } \sqrt { 5 } ) ( \sqrt { 5 } - \mathrm { i } )\).
    2. Hence verify that \(\sqrt { 5 } - \mathrm { i }\) is a root of the equation $$( 2 + \mathrm { i } \sqrt { 5 } ) z = 3 z ^ { * }$$ where \(z ^ { * }\) is the conjugate of \(z\).
  1. The quadratic equation $$x ^ { 2 } + p x + q = 0$$ in which the coefficients \(p\) and \(q\) are real, has a complex root \(\sqrt { 5 } - \mathrm { i }\).
    1. Write down the other root of the equation.
    2. Find the sum and product of the two roots of the equation.
    3. Hence state the values of \(p\) and \(q\).

AnswerMarks Guidance
(a)(i) Full expansion of productM1
Use of \(i^2 = -1\)m1
\((2 + \sqrt{5i})(\sqrt{5} - i) = 3\sqrt{5} + 3i\)A1 \(\sqrt{5} \cdot \sqrt{5} = 5\) must be used – Accept not fully simplified
(ii) \(z^* = x - iy = (\sqrt{5} + i)\)M1
Hence resultA1 Convincingly shown (AG)
Other root is \(\sqrt{5} + i\)B1
(b)(i) Sum of roots is \(2\sqrt{5}\)B1
(ii) Product is \(6\)M1A1
(iii) \(p = -2\sqrt{5}\), \(q = 6\)B1; B1√ ft wrong answers in (ii)
Total for Q5: 11 marks
**(a)(i)** Full expansion of product | M1
Use of $i^2 = -1$ | m1
$(2 + \sqrt{5i})(\sqrt{5} - i) = 3\sqrt{5} + 3i$ | A1 | $\sqrt{5} \cdot \sqrt{5} = 5$ must be used – Accept not fully simplified

**(ii)** $z^* = x - iy = (\sqrt{5} + i)$ | M1
Hence result | A1 | Convincingly shown (AG)
Other root is $\sqrt{5} + i$ | B1 |

**(b)(i)** Sum of roots is $2\sqrt{5}$ | B1

**(ii)** Product is $6$ | M1A1 | 

**(iii)** $p = -2\sqrt{5}$, $q = 6$ | B1; B1√ | ft wrong answers in (ii)

**Total for Q5: 11 marks**

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5
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Calculate $( 2 + \mathrm { i } \sqrt { 5 } ) ( \sqrt { 5 } - \mathrm { i } )$.
\item Hence verify that $\sqrt { 5 } - \mathrm { i }$ is a root of the equation

$$( 2 + \mathrm { i } \sqrt { 5 } ) z = 3 z ^ { * }$$

where $z ^ { * }$ is the conjugate of $z$.
\end{enumerate}\item The quadratic equation

$$x ^ { 2 } + p x + q = 0$$

in which the coefficients $p$ and $q$ are real, has a complex root $\sqrt { 5 } - \mathrm { i }$.
\begin{enumerate}[label=(\roman*)]
\item Write down the other root of the equation.
\item Find the sum and product of the two roots of the equation.
\item Hence state the values of $p$ and $q$.
\end{enumerate}\end{enumerate}

\hfill \mbox{\textit{AQA FP1 2006 Q5 [11]}}