2017.3.3 Question 3

By Vieta’s Theorem, from the quartic equation in $x$, we have

and from the cubic equation in $y$, we have

Therefore, $A=-q$.

1. Since $(p,q,r,s)=(0,3,-6,10)$, the cubic equation is reduced to

   $$y^3-3y^2-10y+84=0,$$

   and therefore

   $$(y-2)(y-7)(y+6)=0.$$

   Therefore, $y_1=7,y_2=2,y_3=-6$, and $\mathit{\alpha \beta }+\mathit{\gamma \delta }=7$.

2. We have

   $$\begin{array}{llll}\hfill (\alpha +\beta )(\gamma +\delta )& =\mathit{\alpha \gamma }+\mathit{\alpha \delta }+\mathit{\beta \gamma }+\mathit{\beta \delta }& \hfill & \\ \hfill & =\left(\mathit{\alpha \beta }+\mathit{\alpha \gamma }+\mathit{\alpha \delta }+\mathit{\beta \gamma }+\mathit{\beta \delta }+\mathit{\gamma \delta }\right)-\left(\mathit{\alpha \beta }+\mathit{\gamma \delta }\right)& \hfill & \\ \hfill & =q-7& \hfill & \\ \hfill & =3-7& \hfill & \\ \hfill & =-4.& \hfill & \end{array}$$

   By Vieta’s Theorem, we have $\mathit{\alpha \beta \gamma \delta }=s=10$. Therefore, $\mathit{\alpha \beta }$ and $\mathit{\gamma \delta }$ must be roots to the equation

   $$x^2-7x+10=0.$$

   The two roots are $x=2$ and $x=5$, and therefore $\mathit{\alpha \beta }=5$.

3. We have from the other root that $\mathit{\gamma \delta }=2$.

   We notice that $(\alpha +\beta )+(\gamma +\delta )=-p=0$. Therefore, from part 2, $(\alpha +\beta )$ and $(\gamma +\delta )$ are roots to the equation

   $$x^2-4=0.$$

   This gives us $\alpha +\beta =\pm 2$ and $\gamma +\delta =\mp 2$.

   Using the value of $r$ and Vieta’s Theorem, we have

   $$\mathit{\alpha \beta \gamma }+\mathit{\alpha \beta \delta }+\mathit{\alpha \gamma \delta }+\mathit{\beta \gamma \delta }=-r=6.$$

   Plugging in $\mathit{\alpha \beta }=5$ and $\mathit{\gamma \delta }=2$, we have

   $$5(\gamma +\delta )+2(\alpha +\beta )=6.$$

   Therefore, it must be the case that $\alpha +\beta =-2$ and $\gamma +\delta =2$.

   Hence, using the values of $\mathit{\alpha \beta }$ and $\mathit{\gamma \delta }$, $\alpha$ and $\beta$ are solutions to the quadratic equation $x^2+2x+5=0$, and $\gamma$ and $\delta$ are solutions to the quadratic equation $x^2-2x+2=0$.

   Solving this gives us $\alpha ,\beta =-1\pm 2i$ and $\gamma ,\delta =1\pm i$. The solutions to the original quartic equation is

   $$x_{1,2}=-1\pm 2i,x_{3,4}=1\pm i.$$