AQA Further Paper 1 2020 June — Question 8 6 marks

Exam BoardAQA
ModuleFurther Paper 1 (Further Paper 1)
Year2020
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicRoots of polynomials
TypeRoots with special relationships
DifficultyStandard +0.8 This is a Further Maths question requiring students to use the arithmetic sequence property with Vieta's formulas. While it involves multiple steps (setting roots as a-d, a, a+d, applying sum and product of roots, solving the resulting system), the approach is relatively standard for Further Maths students who have practiced roots with special relationships. It's moderately challenging but follows a well-established technique.
Spec4.05a Roots and coefficients: symmetric functions4.05b Transform equations: substitution for new roots

8 The three roots of the equation $$4 x ^ { 3 } - 12 x ^ { 2 } - 13 x + k = 0$$ where \(k\) is a constant, form an arithmetic sequence. Find the roots of the equation.

Question 8:
AnswerMarks Guidance
Answer/WorkingMark Guidance
Let the roots be \(\alpha - D\), \(\alpha\) and \(\alpha + D\)M1 Defines three roots in arithmetic progression; PI by correct use of arithmetic series sum in terms of \(\alpha\) and \(\beta\)
\(\Sigma\alpha: \quad \alpha - D + \alpha + \alpha + D = -\left(\dfrac{-12}{4}\right)\) \(3\alpha = 3\), \(\alpha = 1\)M1 Uses one of Vieta's laws: \(\alpha + \beta + \gamma = 3\), \(\alpha\beta + \beta\gamma + \gamma\alpha = \dfrac{-13}{4}\), \(\alpha\beta\gamma = \dfrac{-k}{4}\)
\(\Sigma\alpha\beta\): \((1-D)(1) + (1)(1+D) + (1+D)(1-D) = \dfrac{-13}{4}\) \(1 - D + 1 + D + 1 - D^2 = \dfrac{-13}{4}\) \(3 + \dfrac{13}{4} = D^2 \Rightarrow D = \pm\dfrac{5}{2}\)A1 Obtains 1 as a root of the equation (this might be \(\alpha + D = 1\) without clear realisation that 1 is therefore a root)
\(x = -1.5, 1, 3.5\)M1 Substitutes their root of 1 in the cubic equation to find \(k\) or uses another of Vieta's laws with their root of 1 substituted
A1FObtains the value of \(k\) or solves their equation to find \(D\) or the other roots; allow one slip
\(x = -1.5, 1, 3.5\)A1 Correctly obtains all three roots: \(-1.5, 1, 3.5\)
# Question 8:

| Answer/Working | Mark | Guidance |
|---|---|---|
| Let the roots be $\alpha - D$, $\alpha$ and $\alpha + D$ | M1 | Defines three roots in arithmetic progression; PI by correct use of arithmetic series sum in terms of $\alpha$ and $\beta$ |
| $\Sigma\alpha: \quad \alpha - D + \alpha + \alpha + D = -\left(\dfrac{-12}{4}\right)$ $3\alpha = 3$, $\alpha = 1$ | M1 | Uses one of Vieta's laws: $\alpha + \beta + \gamma = 3$, $\alpha\beta + \beta\gamma + \gamma\alpha = \dfrac{-13}{4}$, $\alpha\beta\gamma = \dfrac{-k}{4}$ |
| $\Sigma\alpha\beta$: $(1-D)(1) + (1)(1+D) + (1+D)(1-D) = \dfrac{-13}{4}$ $1 - D + 1 + D + 1 - D^2 = \dfrac{-13}{4}$ $3 + \dfrac{13}{4} = D^2 \Rightarrow D = \pm\dfrac{5}{2}$ | A1 | Obtains 1 as a root of the equation (this might be $\alpha + D = 1$ without clear realisation that 1 is therefore a root) |
| $x = -1.5, 1, 3.5$ | M1 | Substitutes their root of 1 in the cubic equation to find $k$ or uses another of Vieta's laws with their root of 1 substituted |
| | A1F | Obtains the value of $k$ or solves their equation to find $D$ or the other roots; allow one slip |
| $x = -1.5, 1, 3.5$ | A1 | Correctly obtains all three roots: $-1.5, 1, 3.5$ |
8 The three roots of the equation

$$4 x ^ { 3 } - 12 x ^ { 2 } - 13 x + k = 0$$

where $k$ is a constant, form an arithmetic sequence.

Find the roots of the equation.\\

\hfill \mbox{\textit{AQA Further Paper 1 2020 Q8 [6]}}