AS June 2020 Paper 1 Q9
9 The quadratic equation \(2x^2 + px + 3 = 0\) has two roots, \(\alpha\) and \(\beta\), where \(\alpha \gt \beta\).
(a)
(i) Write down the value of \(\alpha\beta\). [1 mark]
(ii) Express \(\alpha + \beta\) in terms of \(p\). [1 mark]
(b) Hence find \((\alpha - \beta)^2\) in terms of \(p\). [2 marks]
(c) Hence find, in terms of \(p\), a quadratic equation with roots \(\alpha - 1\) and \(\beta + 1\) [4 marks]
| Scheme | Marks | AO |
|---|---|---|
| (i) Obtains the correct value of \(\alpha\beta = \frac{3}{2}\). | B1 | 1.2 |
| (ii) Obtains the correct value of \(\alpha + \beta = -\frac{p}{2}\). | B1 | 1.2 |
Typical solution
(i)
\[\alpha\beta = \frac{3}{2}\](ii)
\[\alpha + \beta = -\frac{p}{2}\]| Scheme | Marks | AO |
|---|---|---|
| Expresses \((\alpha - \beta)^2\) in the form of \((\alpha + \beta)^2 + m\alpha\beta\). | M1 | 1.1a |
| Obtains the correct value of \((\alpha - \beta)^2 = \frac{p^2}{4} - 6\). (May be unsimplified.) FT their \(\alpha\beta\) and \(\alpha + \beta\). | A1F | 1.1b |
Typical solution
\[\begin{aligned}(\alpha - \beta)^2 &= \alpha^2 - 2\alpha\beta + \beta^2 \\ &= (\alpha + \beta)^2 - 2\alpha\beta - 2\alpha\beta \\ &= \left(-\frac{p}{2}\right)^2 - 4 \times \frac{3}{2} \\ &= \frac{p^2}{4} - 6\end{aligned}\]| Scheme | Marks | AO |
|---|---|---|
| Selects a method to find the quadratic equation with roots \(\alpha - 1\), \(\beta + 1\) by expressing the sum and product of roots in terms of \(\alpha\) and \(\beta\). | M1 | 3.1a |
| Obtains the sum of roots = their \(\alpha + \beta\). | B1F | 1.1b |
| Finds an expression for \(\alpha - \beta\) in terms of \(p\). FT their \((\alpha - \beta)^2\). | B1F | 1.1b |
| Obtains a correct quadratic equation with roots \(\alpha - 1\), \(\beta + 1\). FT their \((\alpha - \beta)^2\) and their \(\alpha + \beta\). Condone a quadratic expression. ISW after a correct equation or expression. | A1F | 1.1b |
| (8 marks) |
Typical solution
\[\text{New sum} = \alpha - 1 + \beta + 1 = \alpha + \beta = -\frac{p}{2}\]\[\alpha - \beta = \sqrt{\frac{p^2}{4} - 6}\]positive root only as \(\alpha \gt \beta\)
\[\begin{aligned}\text{New product} &= (\alpha - 1)(\beta + 1) = \alpha\beta + \alpha - \beta - 1 \\ &= \frac{3}{2} + \sqrt{\left(\frac{p^2}{4} - 6\right)} - 1 \\ &= \frac{1}{2} + \sqrt{\frac{p^2}{4} - 6}\end{aligned}\]\[x^2 - x\left(-\frac{p}{2}\right) + \frac{1}{2} + \sqrt{\left(\frac{p^2}{4} - 6\right)} = 0\]