System Quadratic Roots 6A6713

1. **Problem:** Solve the first item only: the system $3x + 2y - z = 4$, $2x + y - 3z = 9$, $x + 3y - 2z = -1$; then find $\alpha + \beta$, $\alpha\beta$ for $3x^2 - 6x + 2 = 0$; and form the calibrated quadratic whose roots are $2\alpha + \frac{1}{\beta}$ and $2\beta + \frac{1}{\alpha}$.\n\n2. **Solve the system of equations.**\n Use elimination.\n $$\begin{aligned}3x + 2y - z &= 4 \\ 2x + y - 3z &= 9 \\ x + 3y - 2z &= -1\end{aligned}$$\n Multiply the second equation by $2$:\n $$4x + 2y - 6z = 18$$\n Subtract the first equation from this:\n $$\cancel{4x} + \cancel{2y} - 6z - (\cancel{3x} + \cancel{2y} - z) = 18 - 4$$\n $$x - 5z = 14$$\n So, $$x = 14 + 5z$$\n\n Multiply the third equation by $2$:\n $$2x + 6y - 4z = -2$$\n Subtract the second equation from this:\n $$\cancel{2x} + 6y - 4z - (\cancel{2x} + y - 3z) = -2 - 9$$\n $$5y - z = -11$$\n So, $$z = 5y + 11$$\n\n Substitute $x = 14 + 5z$ into the first equation:\n $$3(14 + 5z) + 2y - z = 4$$\n $$42 + 15z + 2y - z = 4$$\n $$2y + 14z = -38$$\n $$y + 7z = -19$$\n\n Now use $z = 5y + 11$ in $y + 7z = -19$:\n $$y + 7(5y + 11) = -19$$\n $$y + 35y + 77 = -19$$\n $$36y = -96$$\n $$y = -\frac{8}{3}$$\n\n Then\n $$z = 5\left(-\frac{8}{3}\right) + 11 = -\frac{40}{3} + \frac{33}{3} = -\frac{7}{3}$$\n\n And\n $$x = 14 + 5\left(-\frac{7}{3}\right) = \frac{42}{3} - \frac{35}{3} = \frac{7}{3}$$\n\n Therefore, $$x = \frac{7}{3},\ y = -\frac{8}{3},\ z = -\frac{7}{3}$$\n\n3. **Find $\alpha + \beta$ and $\alpha\beta$.**\n The quadratic is\n $$3x^2 - 6x + 2 = 0$$\n For $ax^2 + bx + c = 0$, the roots satisfy\n $$\alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a}$$\n Here, $a = 3$, $b = -6$, and $c = 2$.\n So,\n $$\alpha + \beta = -\frac{-6}{3} = 2$$\n $$\alpha\beta = \frac{2}{3}$$\n\n4. **Form the new quadratic equation.**\n The new roots are\n $$r_1 = 2\alpha + \frac{1}{\beta}, \quad r_2 = 2\beta + \frac{1}{\alpha}$$\n\n Find the sum of the new roots:\n $$r_1 + r_2 = 2(\alpha + \beta) + \left(\frac{1}{\alpha} + \frac{1}{\beta}\right)$$\n Since\n $$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$$\n we get\n $$r_1 + r_2 = 2(2) + \frac{2}{2/3} = 4 + 3 = 7$$\n\n Find the product of the new roots:\n $$r_1r_2 = \left(2\alpha + \frac{1}{\beta}\right)\left(2\beta + \frac{1}{\alpha}\right)$$\n Expand:\n $$r_1r_2 = 4\alpha\beta + 2\frac{\alpha}{\alpha} + 2\frac{\beta}{\beta} + \frac{1}{\alpha\beta}$$\n $$r_1r_2 = 4\alpha\beta + 2 + 2 + \frac{1}{\alpha\beta}$$\n Substitute $\alpha\beta = \frac{2}{3}$:\n $$r_1r_2 = 4\left(\frac{2}{3}\right) + 4 + \frac{1}{2/3}$$\n $$r_1r_2 = \frac{8}{3} + 4 + \frac{3}{2}$$\n $$r_1r_2 = \frac{16}{6} + \frac{24}{6} + \frac{9}{6} = \frac{49}{6}$$\n\n Therefore the required quadratic is\n $$x^2 - 7x + \frac{49}{6} = 0$$\n Multiply by $6$ to clear fractions:\n $$6x^2 - 42x + 49 = 0$$\n\n5. **Final answers.**\n $$x = \frac{7}{3},\ y = -\frac{8}{3},\ z = -\frac{7}{3}$$\n $$\alpha + \beta = 2$$\n $$\alpha\beta = \frac{2}{3}$$\n New calibrated quadratic equation:\n $$6x^2 - 42x + 49 = 0$$