AS June 2018 Paper 1 Q18
18 \(\alpha\), \(\beta\) and \(\gamma\) are the real roots of the cubic equation
\[x^3 + mx^2 + nx + 2 = 0\]By considering \((\alpha - \beta)^2 + (\gamma - \alpha)^2 + (\beta - \gamma)^2\), prove that
\[m^2 \geqslant 3n\][4 marks]
| Scheme | Marks | AO |
|---|---|---|
| Writes the expression in terms of \(\sum\alpha\) and \(\sum\alpha\beta\) Award for correct expansion followed by use of \(\sum\alpha^2 = \left(\sum\alpha\right)^2 - 2\sum\alpha\beta\) | M1 | 3.1a |
| Substitutes \(\pm m\) for \(\sum\alpha\) and \(\pm n\) for \(\sum\alpha\beta\) | M1 | 1.1a |
| Gives a reason for expression \(\geqslant 0\) Condone lack of reference to roots being real. | E1 | 2.4 |
| Completes fully correct proof to reach the required result. This mark is only available if all previous marks have been awarded. Lose this mark for sight of \(\sum\alpha = m\) | R1 | 2.1 |
| (4 marks) |
Typical solution
\[\begin{aligned}&(\alpha - \beta)^2 + (\gamma - \alpha)^2 + (\beta - \gamma)^2 \\ &\quad = \alpha^2 - 2\alpha\beta + \beta^2 + \gamma^2 - 2\gamma\alpha + \alpha^2 + \beta^2 - 2\beta\gamma + \gamma^2 \\ &= 2\alpha^2 + 2\beta^2 + 2\gamma^2 - 2\alpha\beta - 2\gamma\alpha - 2\beta\gamma \\ &= 2\sum\alpha^2 - 2\sum\alpha\beta \\ &= 2\left(\left(\sum\alpha\right)^2 - 2\sum\alpha\beta\right) - 2\sum\alpha\beta \\ &= 2\left(\sum\alpha\right)^2 - 6\sum\alpha\beta \\ &= 2(-m)^2 - 6 \times n = 2m^2 - 6n\end{aligned}\]But as \(\alpha\), \(\beta\) and \(\gamma\) are real then each of \((\alpha - \beta)^2\), \((\gamma - \alpha)^2\) and \((\beta - \gamma)^2\) must be non-negative.
\[\therefore (\alpha - \beta)^2 + (\gamma - \alpha)^2 + (\beta - \gamma)^2 \geqslant 0\]\[\therefore 2m^2 - 6n \geqslant 0\]\[2m^2 \geqslant 6n\]\[m^2 \geqslant 3n\]AG