7 A cubic equation has roots \(\alpha , \beta\) and \(\gamma\) such that
$$\begin{aligned}
\alpha + \beta + \gamma & = 4 \\
\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } & = 14 \\
\alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 } & = 34
\end{aligned}$$
Find the value of \(\alpha \beta + \beta \gamma + \gamma \alpha\).
Show that the cubic equation is
$$x ^ { 3 } - 4 x ^ { 2 } + x + 6 = 0$$
and solve this equation.
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Question 7:
Answer Marks
Guidance
Working/Answer Marks
Guidance
\((\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = \alpha^2+\beta^2+\gamma^2 \Rightarrow \sum\alpha\beta = 1\) M1A1
Either: Required equation is \(x^3 - 4x^2 + x + c = 0\)M1
\(\Rightarrow \sum\alpha^3 - 4\sum\alpha^2 + 4 + 3c = 0\) M1
\(\Rightarrow 3c = 56 - 34 - 4 = 18 \Rightarrow c = 6\) (AG) A1
Or: \(\alpha^3+\beta^3+\gamma^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha)\)(M1)
\((\alpha+\beta+\gamma)^3 = \alpha^3+\beta^3+\gamma^3+3(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha)-3\alpha\beta\gamma\) (M1A1)
\(\Rightarrow \ldots \Rightarrow \alpha\beta\gamma = -6\) \(\Rightarrow x^3 - 4x^2 + x + 6 = 0\) (AG) A1
\(\Rightarrow (x+1)(x-2)(x-3) = 0 \Rightarrow x = -1, 2, 3\) M1A1
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## Question 7:
| Working/Answer | Marks | Guidance |
|---|---|---|
| $(\alpha+\beta+\gamma)^2 - 2(\alpha\beta+\beta\gamma+\gamma\alpha) = \alpha^2+\beta^2+\gamma^2 \Rightarrow \sum\alpha\beta = 1$ | M1A1 | |
| **Either:** Required equation is $x^3 - 4x^2 + x + c = 0$ | M1 | |
| $\Rightarrow \sum\alpha^3 - 4\sum\alpha^2 + 4 + 3c = 0$ | M1 | |
| $\Rightarrow 3c = 56 - 34 - 4 = 18 \Rightarrow c = 6$ (AG) | A1 | |
| **Or:** $\alpha^3+\beta^3+\gamma^3 - 3\alpha\beta\gamma = (\alpha+\beta+\gamma)(\alpha^2+\beta^2+\gamma^2-\alpha\beta-\beta\gamma-\gamma\alpha)$ | (M1) | |
| $(\alpha+\beta+\gamma)^3 = \alpha^3+\beta^3+\gamma^3+3(\alpha+\beta+\gamma)(\alpha\beta+\beta\gamma+\gamma\alpha)-3\alpha\beta\gamma$ | (M1A1) | |
| $\Rightarrow \ldots \Rightarrow \alpha\beta\gamma = -6$ | | |
| $\Rightarrow x^3 - 4x^2 + x + 6 = 0$ (AG) | A1 | |
| $\Rightarrow (x+1)(x-2)(x-3) = 0 \Rightarrow x = -1, 2, 3$ | M1A1 | |
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7 A cubic equation has roots $\alpha , \beta$ and $\gamma$ such that
$$\begin{aligned}
\alpha + \beta + \gamma & = 4 \\
\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 } & = 14 \\
\alpha ^ { 3 } + \beta ^ { 3 } + \gamma ^ { 3 } & = 34
\end{aligned}$$
Find the value of $\alpha \beta + \beta \gamma + \gamma \alpha$.
Show that the cubic equation is
$$x ^ { 3 } - 4 x ^ { 2 } + x + 6 = 0$$
and solve this equation.
\hfill \mbox{\textit{CAIE FP1 2012 Q7 [8]}}