1. **State the problem:** We are given the quadratic equation $$3x^2 - 6x - 4 = 0$$ with roots $$\alpha$$ and $$\beta$$. 2. **Find the sum and product of the roots:** For a quadratic equation $$ax^2 + bx + c = 0$$, the sum of roots is $$-\frac{b}{a}$$ and the product is $$\frac{c}{a}$$. - Here, $$a=3$$, $$b=-6$$, $$c=-4$$. - Sum: $$\alpha + \beta = -\frac{-6}{3} = \frac{6}{3} = 2$$. - Product: $$\alpha \beta = \frac{-4}{3} = -\frac{4}{3}$$. 3. **Calculate** $$\frac{1}{\alpha} + \frac{1}{\beta}$$: - Use the identity $$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta}$$. - Substitute values: $$\frac{2}{-\frac{4}{3}} = 2 \times -\frac{3}{4} = -\frac{6}{4} = -\frac{3}{2}$$. 4. **Evaluate** $$\frac{1}{\alpha \beta}$$: - From above, $$\alpha \beta = -\frac{4}{3}$$. - Therefore, $$\frac{1}{\alpha \beta} = \frac{1}{-\frac{4}{3}} = -\frac{3}{4}$$. **Final answers:** - Sum of roots: $$2$$ - Product of roots: $$-\frac{4}{3}$$ - $$\frac{1}{\alpha} + \frac{1}{\beta} = -\frac{3}{2}$$ - $$\frac{1}{\alpha \beta} = -\frac{3}{4}$$