1. **State the problem:** Given the quadratic equation $x^2 - 2x - 1 = 0$ with roots $\alpha$ and $\beta$, find the equation whose roots are $2\alpha + \frac{1}{\beta}$ and $-2\alpha - \frac{1}{\beta}$.\n\n2. **Recall the sum and product of roots:** For a quadratic $x^2 + px + q = 0$, roots $r_1$ and $r_2$ satisfy:\n$$r_1 + r_2 = -p$$\n$$r_1 r_2 = q$$\nFor our equation $x^2 - 2x - 1 = 0$, we have:\n$$\alpha + \beta = 2$$\n$$\alpha \beta = -1$$\n\n3. **Find the sum of the new roots:** Let the new roots be\n$$r_1 = 2\alpha + \frac{1}{\beta}, \quad r_2 = -2\alpha - \frac{1}{\beta}$$\nSum:\n$$r_1 + r_2 = \left(2\alpha + \frac{1}{\beta}\right) + \left(-2\alpha - \frac{1}{\beta}\right) = 0$$\n\n4. **Find the product of the new roots:**\n$$r_1 r_2 = \left(2\alpha + \frac{1}{\beta}\right) \left(-2\alpha - \frac{1}{\beta}\right) = -\left(2\alpha + \frac{1}{\beta}\right)^2$$\nExpand the square:\n$$\left(2\alpha + \frac{1}{\beta}\right)^2 = 4\alpha^2 + 4\alpha \cdot \frac{1}{\beta} + \frac{1}{\beta^2}$$\nSo,\n$$r_1 r_2 = -\left(4\alpha^2 + \frac{4\alpha}{\beta} + \frac{1}{\beta^2}\right)$$\n\n5. **Express terms in known quantities:**\nSince $\alpha \beta = -1$, then $\beta = \frac{-1}{\alpha}$. Substitute:\n$$\frac{1}{\beta} = -\alpha$$\n$$\frac{1}{\beta^2} = \left(\frac{1}{\beta}\right)^2 = (-\alpha)^2 = \alpha^2$$\nAlso,\n$$\frac{4\alpha}{\beta} = 4\alpha \cdot (-\alpha) = -4\alpha^2$$\n\n6. **Simplify the product:**\n$$r_1 r_2 = -\left(4\alpha^2 - 4\alpha^2 + \alpha^2\right) = -\left(0 + \alpha^2\right) = -\alpha^2$$\n\n7. **Find $\alpha^2$ using the original equation:**\nFrom $x^2 - 2x - 1 = 0$, for root $\alpha$:\n$$\alpha^2 - 2\alpha - 1 = 0 \implies \alpha^2 = 2\alpha + 1$$\n\n8. **Substitute $\alpha^2$ back:**\n$$r_1 r_2 = -\alpha^2 = -(2\alpha + 1)$$\nSince $\alpha$ is a root, but we want a numeric value, use the fact that $\alpha + \beta = 2$ and $\alpha \beta = -1$. We can find $\alpha^2 + \beta^2$:\n$$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta = 2^2 - 2(-1) = 4 + 2 = 6$$\nBut we only need $\alpha^2$ for the product, so we keep it symbolic. However, since the sum of roots is zero, the quadratic with roots $r_1$ and $r_2$ is:\n$$x^2 - (r_1 + r_2)x + r_1 r_2 = 0 \implies x^2 + r_1 r_2 = 0$$\n\n9. **Final equation:**\n$$x^2 - (2\alpha + 1) = 0$$\nBut since $r_1 r_2 = -(2\alpha + 1)$, the equation is:\n$$x^2 + (2\alpha + 1) = 0$$\n\n10. **Replace $2\alpha + 1$ with a numeric value:**\nSince $\alpha$ satisfies $\alpha^2 - 2\alpha - 1=0$, solve for $\alpha$:\n$$\alpha = \frac{2 \pm \sqrt{4 + 4}}{2} = 1 \pm \sqrt{2}$$\nChoose $\alpha = 1 + \sqrt{2}$ (either root works, the equation will be the same):\n$$2\alpha + 1 = 2(1 + \sqrt{2}) + 1 = 2 + 2\sqrt{2} + 1 = 3 + 2\sqrt{2}$$\n\n11. **Therefore, the quadratic equation is:**\n$$x^2 + (3 + 2\sqrt{2}) = 0$$\nOr equivalently,\n$$x^2 + 3 + 2\sqrt{2} = 0$$