1. **Problem statement:** Given a quadratic equation with roots $\alpha$ and $\beta$ satisfying $$x^2 + 3x + 7 = 0,$$ find the quadratic equation whose roots are $\alpha - 2\beta$ and $\beta - 2\alpha$. 2. **Find the sum and product of original roots:** From the given quadratic equation, $$x^2 + 3x + 7 = 0,$$ the sum of roots is $$\alpha + \beta = -3,$$ and the product is $$\alpha \beta = 7.$$ 3. **Express new roots in terms of $\alpha$ and $\beta$:** The new roots are $$r_1 = \alpha - 2\beta,$$ $$r_2 = \beta - 2\alpha.$$ 4. **Sum of the new roots:** Calculate $$r_1 + r_2 = (\alpha - 2\beta) + (\beta - 2\alpha) = (\alpha + \beta) - 2(\beta + \alpha) = (\alpha + \beta) - 2(\alpha + \beta) = - (\alpha + \beta).$$ Substitute the known sum: $$r_1 + r_2 = -(-3) = 3.$$ 5. **Product of the new roots:** Calculate $$r_1 r_2 = (\alpha - 2\beta)(\beta - 2\alpha).$$ Expand: $$r_1 r_2 = \alpha \beta - 2\alpha^2 - 2\beta^2 + 4\alpha \beta = 5\alpha \beta - 2(\alpha^2 + \beta^2).$$ 6. **Simplify $\alpha^2 + \beta^2$:** Recall that $$\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta = (-3)^2 - 2 \times 7 = 9 - 14 = -5.$$ 7. **Calculate the product using values:** Substitute $$r_1 r_2 = 5(7) - 2(-5) = 35 + 10 = 45.$$ 8. **Form the new quadratic equation:** Using sum and product of roots, $$x^2 - (r_1 + r_2)x + r_1 r_2 = 0,$$ thus $$x^2 - 3x + 45 = 0.$$ **Final answer:** The quadratic equation with roots $\alpha - 2\beta$ and $\beta - 2\alpha$ is $$x^2 - 3x + 45 = 0.$$