Problem: The equation $x^4 - 6x^3 - 73 x^2 + kx + m = 0$ has two positive roots $\alpha ,\beta $ and two negative roots $\gamma , \delta $. It is given that $\alpha \beta = \gamma \delta = 4$.
Find the values of the constant $k$ and $m$.
Show that $(\alpha +\beta )(\gamma + \delta ) = -81$.
Find the quadratic equation which has roots $\alpha + \beta $ and $\gamma + \delta $.
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Solution: The given equation is
\begin{equation}\label{eq:05Apr2025-1}
x^4 - 6x^3 - 73 x^2 + kx + m = 0.
\end{equation}
Since $\alpha ,\beta ,\gamma ,\delta $ are roots of the equation \eqref{eq:05Apr2025-1}, we have
\begin{gather*}
\alpha + \beta + \gamma + \delta = 6, \\
\alpha \beta + \alpha \gamma + \alpha \delta + \beta \gamma + \beta \delta + \gamma \delta = -73,\\
\alpha \beta (\gamma +\delta ) + \gamma \delta(\alpha +\beta ) = -k,\\
\alpha \beta \gamma \delta = m.
\end{gather*}
Since $\alpha \beta = 4 = \gamma \delta $, so
\begin{align*}
\alpha \beta \gamma \delta = 16 \implies m = 16.
\end{align*}
Now we have
\begin{align*}
& \alpha \beta (\gamma +\delta ) + \gamma \delta(\alpha +\beta ) = - k \\
\implies & 4(\gamma +\delta ) + 4(\alpha +\beta ) = -k \\
\implies & 4\left( \alpha +\beta +\gamma +\delta \right) = -k \\
\implies & k = -24.
\end{align*}
We have
\begin{align*}
& (\alpha + \beta )(\gamma + \delta ) + \alpha \beta + \gamma \delta = -73 \\
\implies & (\alpha + \beta)(\gamma + \delta) + 8 = -73 \\
\implies & (\alpha + \beta)(\gamma + \delta) = -81
\end{align*}
The quadratic equation with roots $\alpha + \beta $ and $\gamma + \delta $ is given by
\[
x^2 - (\alpha + \beta + \gamma + \delta )x + (\alpha + \beta)(\gamma + \delta ) = 0.
\]
Since $\alpha + \beta + \gamma + \delta = 6 $ and $(\alpha + \beta)(\gamma + \delta) = -81$. Thus the quadratic equation is
\[
x^2 - 6x - 81 = 0.
\]