AQA FP1 2008 January — Question 8 12 marks

Exam BoardAQA
ModuleFP1 (Further Pure Mathematics 1)
Year2008
SessionJanuary
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeQuadratic with transformed roots
DifficultyStandard +0.8 This FP1 question requires systematic application of root transformation theory (finding sum and product of α³ and β³ from those of α and β), complex number manipulation, and connecting abstract algebraic results to concrete calculations. While the individual techniques are standard for Further Maths, the multi-part structure requiring both symbolic manipulation without solving and verification through complex arithmetic makes it moderately challenging, above average difficulty but not requiring exceptional insight.
Spec1.02d Quadratic functions: graphs and discriminant conditions4.02a Complex numbers: real/imaginary parts, modulus, argument4.02e Arithmetic of complex numbers: add, subtract, multiply, divide4.02i Quadratic equations: with complex roots4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots

8
    1. It is given that \(\alpha\) and \(\beta\) are the roots of the equation $$x ^ { 2 } - 2 x + 4 = 0$$ Without solving this equation, show that \(\alpha ^ { 3 }\) and \(\beta ^ { 3 }\) are the roots of the equation $$x ^ { 2 } + 16 x + 64 = 0$$ (6 marks)
    2. State, giving a reason, whether the roots of the equation $$x ^ { 2 } + 16 x + 64 = 0$$ are real and equal, real and distinct, or non-real.
  1. Solve the equation $$x ^ { 2 } - 2 x + 4 = 0$$
  2. Use your answers to parts (a) and (b) to show that $$( 1 + \mathrm { i } \sqrt { 3 } ) ^ { 3 } = ( 1 - \mathrm { i } \sqrt { 3 } ) ^ { 3 }$$

Question 8(a)(i):
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(\alpha + \beta = 2,\; \alpha\beta = 4\)B1B1
\(\alpha^3 + \beta^3 = (2)^3 - 3(4)(2) = -16\)M1A1
\(\alpha^3\beta^3 = (4)^3 = 64\), hence resultM1A1 convincingly shown (AG)
Subtotal6
Question 8(a)(ii):
AnswerMarks Guidance
Working/AnswerMarks Guidance
Discriminant \(= 0\), so roots equalB1E1 or by factorisation
Subtotal2
Question 8(b):
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(x = \frac{2 \pm \sqrt{4-16}}{2}\)M1 or by completing the square
\(= 1 \pm \frac{1}{2}i\sqrt{12}\)A1
Subtotal2
Question 8(c):
AnswerMarks Guidance
Working/AnswerMarks Guidance
\(\alpha,\,\beta = 1 \pm i\sqrt{3}\) and \(\alpha^3 = \beta^3\), hence resultE2
Subtotal2
Total12
## Question 8(a)(i):

| Working/Answer | Marks | Guidance |
|---|---|---|
| $\alpha + \beta = 2,\; \alpha\beta = 4$ | B1B1 | |
| $\alpha^3 + \beta^3 = (2)^3 - 3(4)(2) = -16$ | M1A1 | |
| $\alpha^3\beta^3 = (4)^3 = 64$, hence result | M1A1 | convincingly shown (AG) |
| **Subtotal** | **6** | |

## Question 8(a)(ii):

| Working/Answer | Marks | Guidance |
|---|---|---|
| Discriminant $= 0$, so roots equal | B1E1 | or by factorisation |
| **Subtotal** | **2** | |

## Question 8(b):

| Working/Answer | Marks | Guidance |
|---|---|---|
| $x = \frac{2 \pm \sqrt{4-16}}{2}$ | M1 | or by completing the square |
| $= 1 \pm \frac{1}{2}i\sqrt{12}$ | A1 | |
| **Subtotal** | **2** | |

## Question 8(c):

| Working/Answer | Marks | Guidance |
|---|---|---|
| $\alpha,\,\beta = 1 \pm i\sqrt{3}$ and $\alpha^3 = \beta^3$, hence result | E2 | |
| **Subtotal** | **2** | |
| **Total** | **12** | |

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8
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item It is given that $\alpha$ and $\beta$ are the roots of the equation

$$x ^ { 2 } - 2 x + 4 = 0$$

Without solving this equation, show that $\alpha ^ { 3 }$ and $\beta ^ { 3 }$ are the roots of the equation

$$x ^ { 2 } + 16 x + 64 = 0$$

(6 marks)
\item State, giving a reason, whether the roots of the equation

$$x ^ { 2 } + 16 x + 64 = 0$$

are real and equal, real and distinct, or non-real.
\end{enumerate}\item Solve the equation

$$x ^ { 2 } - 2 x + 4 = 0$$
\item Use your answers to parts (a) and (b) to show that

$$( 1 + \mathrm { i } \sqrt { 3 } ) ^ { 3 } = ( 1 - \mathrm { i } \sqrt { 3 } ) ^ { 3 }$$
\end{enumerate}

\hfill \mbox{\textit{AQA FP1 2008 Q8 [12]}}