Quadratic Equations

We are given the polynomial equation $x^4-x^3-8 x^2+2 x+12=0$ with roots $\alpha, \beta, \gamma$, and $\delta$. It is known that the sum of two of the roots, $\alpha$ and $\beta$, is zero, i.e., $\alpha+\beta=0$.From Vieta's formulas, the sum of all roots is:$S_1=\alpha+\beta+\gamma+\delta=1$Given $\alpha+\beta=0$, we find:$\gamma+\delta=1$Similarly, from Vieta's formulas, the sum of the products of the roots taken two at a time is:$S_2=\alpha\beta+\alpha\gamma+\alpha\delta+\beta\gamma+\beta\delta+\gamma\delta=-8$Using $\alpha+\beta=0$, simplify to:$\alpha\beta+\gamma\delta=-8$From the sum of the products of the roots taken three at a time, we have:$S_3=\alpha\beta\gamma+\alpha\beta\delta+\alpha\gamma\delta+\beta\gamma\delta=-2$Substitute $(\alpha+\beta)\gamma\delta+\alpha\beta(\gamma+\delta)=-2$ and rearrange:$\alpha\beta=-2$From Vieta's formulas, the product of the roots is given by:$S_4=\alpha\beta\gamma\delta=12$Use $\alpha\beta=-2$ to find:$\gamma\delta=-6 \; \; \ldots (v)$Next, solve the system of equations obtained from (ii) and (v):$\gamma+\delta=1$$\gamma\delta=-6$These are the equations for a quadratic:$t^2-(\gamma+\delta)t+\gamma\delta=0 \Rightarrow t^2-t-6=0$Solving, we find the roots to be:$t = \frac{1 \pm \sqrt{1+24}}{2} = \frac{1 \pm 5}{2}$Thus, $t=3$ or $t=-2$. Since $\gamma > \delta$, we assign $\gamma = 3$ and $\delta = -2$.Finally, compute:$3\gamma+2\delta=3 imes 3+2 imes (-2)=9-4=5$