Let \(\alpha\), \(\beta\), \(\gamma\) and \(\delta\) be the roots of the
quartic equation
\[
x^4 +px^3 +qx^2 +r x +s =0
\,.
\]
You are given that, for any such equation, \(\,\alpha \beta + \gamma\delta\,\), \(\alpha\gamma+\beta\delta\,\) and \(\,\alpha \delta + \beta\gamma\,\) satisfy a cubic equation of the form
\[
y^3+Ay^2+
(pr-4s)y+ (4qs-p^2s -r^2)
=0
\,.
\]
Determine \(A\).
Now consider the quartic equation given by \(p=0\,\), \(q= 3\,\), \(r=-6\,\) and \(s=10\,\).
Find the value of \(\alpha\beta + \gamma \delta\), given that it is the largest root of the corresponding cubic equation.
Hence, using the values of \(q\) and \(s\), find the value of \((\alpha +\beta)(\gamma+\delta)\,\)
and the value of \(\alpha\beta\) given that \(\alpha\beta >\gamma\delta\,\).
Using these results, and the values of \(p\) and \(r\), solve the quartic equation.
\begin{align*}(\alpha+\beta)(\gamma+\delta) &= \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta \\
&= 3 -(\alpha\beta + \gamma\delta) \\
&=3-7 = -4
\end{align*}
Let \(\alpha\beta\) and \(\gamma\delta\) be the roots of a quadratic; then the quadratic will be \(t^2-7t+10 = 0 \Rightarrow t = 2,5\) so \(\alpha\beta = 5\)
\(\alpha\beta = 5, \gamma\delta = 2\)
Consider the quadratic with roots \(\alpha+\beta\) and \(\gamma+\delta\), then
\(t^2-4 = 0 \Rightarrow t = \pm 2\).
Suppose \(\alpha+\beta = 2, \gamma+\delta=-2\) then
\(\alpha, \beta = 1 \pm 2i, \gamma,\delta = -1 \pm i\)
\(\alpha \beta \gamma + \beta\gamma\delta + \gamma\delta\alpha + \delta\alpha\beta = 5\gamma + 2\beta + 2\alpha + 5\delta = -6 \neq 6\)
Suppose \(\alpha+\beta = -2, \gamma+\delta=2\) then
\(\alpha, \beta = -1 \pm 2i, \gamma,\delta = 1 \pm i\)
\(\alpha \beta \gamma + \beta\gamma\delta + \gamma\delta\alpha + \delta\alpha\beta = 5\gamma + 2\beta + 2\alpha + 5\delta = 6\), therefore these are there roots. (In some order):
\(1 \pm i, -1 \pm 2i\)