Browsing as Guest. Progress, bookmarks and attempts are disabled.
Log in to track your work.
FM June 2024 p11 q01
4157
The cubic equation \(2x^3 + x^2 - px - 5 = 0\), where \(p\) is a positive constant, has roots \(\alpha, \beta, \gamma\).
(a) State, in terms of \(p\), the value of \(\alpha\beta + \beta\gamma + \gamma\alpha\).
(b) Find the value of \(\alpha^2\beta\gamma + \alpha\beta^2\gamma + \alpha\beta\gamma^2\).
(c) Deduce a cubic equation whose roots are \(\alpha\beta, \beta\gamma, \alpha\gamma\).
(d) Given that \(\alpha^2 + \beta^2 + \gamma^2 = \frac{1}{3}\), find the value of \(p\).
Solution
(a) By Vieta's formulas, the sum \(\alpha\beta + \beta\gamma + \gamma\alpha = -\frac{b}{a} = -\frac{1}{2}p\).
(b) The expression \(\alpha^2\beta\gamma + \alpha\beta^2\gamma + \alpha\beta\gamma^2\) can be factorized as \(\alpha\beta\gamma(\alpha + \beta + \gamma)\). Using Vieta's formulas, \(\alpha + \beta + \gamma = -\frac{1}{2}\) and \(\alpha\beta\gamma = -\frac{5}{2}\). Thus, the value is \(-\frac{5}{2} \times -\frac{1}{2} = \frac{5}{4}\).
(c) The roots \(\alpha\beta, \beta\gamma, \alpha\gamma\) imply a new cubic equation. Using Vieta's formulas, the sum of the roots is \(\alpha\beta + \beta\gamma + \alpha\gamma = -\frac{1}{2}p\), the sum of the products of the roots taken two at a time is \(\alpha^2\beta\gamma + \alpha\beta^2\gamma + \alpha\beta\gamma^2 = \frac{5}{4}\), and the product of the roots is \((\alpha\beta)(\beta\gamma)(\alpha\gamma) = (\alpha\beta\gamma)^2 = \left(-\frac{5}{2}\right)^2 = \frac{25}{4}\). The cubic equation is \(4z^3 + 2pz^2 - 5z - 25 = 0\).
(d) Given \(\alpha^2 + \beta^2 + \gamma^2 = \frac{1}{3}\), we use the identity \(\alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha)\). Substituting the known values, \(\frac{1}{3} = \left(-\frac{1}{2}\right)^2 - 2\left(-\frac{1}{2}p\right)\). Solving for \(p\), we get \(\frac{1}{3} = \frac{1}{4} + p\), leading to \(p = \frac{1}{12}\).
