# Roots of quartic equation - given product of two roots, find missing coefficient

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

$$k=2(\alpha\beta+ \alpha\gamma+ \alpha\delta +\beta\gamma+ \beta\delta+\gamma\delta )=$$

$$=2(\alpha\beta+( \alpha+\beta)(\gamma+\delta)+\gamma\delta )=$$

$$=2(5+\frac{25}{22}\cdot\frac{4}{11}-\frac85)=$$

$$=\frac{4614}{605}.$$

• Just to check (links to Wolfram Alpha)
– Aig
Feb 24 at 19:30

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 $$\gamma,\delta, k$$ :$$\begin{cases}f_1(\gamma,\delta)+f_2(\gamma,\delta)+\gamma+\delta=\frac32\\ [f_1(\gamma,\delta)+f_2(\gamma,\delta)](\gamma+\delta)+\gamma\delta+5=\frac k2\\ \gamma\delta=\frac85\end{cases}$$