The quartic equation $ax^4 + bx^3 + cx^2 + dx + e = 0$ has roots $\alpha, \beta, \gamma, \delta$. Given that $\alpha \beta = p$ find the value of $k$
So I have deduced that $\gamma \delta = \frac{e}{ap}$ using product of roots $=-\frac{e}{a}$ but I am not sure how to proceed from here.
I have written out Vieta's formulae, but can't seem to manipulate to get $k$.
Is there an efficient way to do this?
For simplicity, I denote the roots by $a,b,c,d$.
Observe that you only need $ac+ad+bc+bd = (a+b)(c+d)$ since you already have $ab$ and $cd$.
Since the coefficient of $x$ is zero, $$abc+abd+acd+bcd = 0 \iff ab(c+d) = -cd(a+b)$$From here, substitute $(a+b)+(c+d) = 3/2$ to get the required answer
By Vieta’s theorem we have $\alpha\beta\gamma\delta=-8$, so $\gamma\delta=-\frac85$. Also by Vieta’s theorem we have:
$$\alpha\beta\gamma + \alpha\beta\delta + \alpha\gamma\delta + \beta\gamma\delta = 0$$
$$\alpha\beta(\gamma +\delta )+ \gamma\delta(\alpha + \beta) = 0$$
$$5(\gamma +\delta)-\frac85(\alpha + \beta) = 0.\tag1$$
Again by Vieta’s theorem we have
$$\alpha+\beta+\gamma+\delta=\frac32.\tag2$$
Solving $(1)$ and $(2)$ together we get $\alpha+\beta=\frac{25}{22}$, $\gamma+\delta=\frac4{11}$.
Now, by the same theorem, we have
$$k=2(\alpha\beta+ \alpha\gamma+ \alpha\delta +\beta\gamma+ \beta\delta+\gamma\delta )=$$
$$=2(\alpha\beta+( \alpha+\beta)(\gamma+\delta)+\gamma\delta )=$$
$$=2(5+\frac{25}{22}\cdot\frac{4}{11}-\frac85)=$$
$$=\frac{4614}{605}.$$
HINT.-From Vieta's formulas we have $$\begin{cases}\alpha+\beta=\frac32-(\gamma+\delta)\\\alpha\beta=4\end{cases}\Rightarrow\begin{cases}\alpha=f_1(\gamma,\delta)\\\beta=f_2(\gamma,\delta)\end{cases}$$ It follows the three equations we need for find out the three unknowns $\gamma,\delta, k$ :$$\begin{cases}f_1(\gamma,\delta)+f_2(\gamma,\delta)+\gamma+\delta=\frac32\\ [f_1(\gamma,\delta)+f_2(\gamma,\delta)](\gamma+\delta)+\gamma\delta+5=\frac k2\\ \gamma\delta=\frac85\end{cases}$$
