1. **Stating the problem:** Given the quadratic equation $$by^2 + xy + pd = 0,$$ where $$\alpha$$ and $$\beta$$ are its roots, show that $$\alpha + \beta = -\frac{r}{k}$$ and $$\alpha \beta = \frac{g}{p}$$. 2. **Understanding the standard form and formulas:** A quadratic equation in standard form is $$ax^2 + bx + c = 0$$. For roots $$\alpha$$ and $$\beta$$, the sum and product of roots are given by: $$\alpha + \beta = -\frac{b}{a}$$ $$\alpha \beta = \frac{c}{a}$$ 3. **Comparing given equation to standard form:** The given equation is $$by^2 + xy + pd = 0$$. Here, the coefficient of $$y^2$$ is $$b$$, coefficient of $$y$$ is $$x$$ (which is unusual, but we treat it as a constant coefficient), and constant term is $$pd$$. 4. **Applying the sum and product formulas:** Sum of roots: $$\alpha + \beta = -\frac{\text{coefficient of } y}{\text{coefficient of } y^2} = -\frac{x}{b}$$ Product of roots: $$\alpha \beta = \frac{pd}{b}$$ 5. **Relating to the expressions to prove:** The problem states to show: $$\alpha + \beta = -\frac{r}{k}$$ $$\alpha \beta = \frac{g}{p}$$ Since the original equation is given as $$by^2 + xy + pd = 0$$, the letters $$r, k, g, p$$ must correspond to the coefficients in some way. 6. **Conclusion:** If we identify $$r = x$$ and $$k = b$$, then $$\alpha + \beta = -\frac{r}{k} = -\frac{x}{b}$$. If we identify $$g = pd$$ and $$p = b$$, then $$\alpha \beta = \frac{g}{p} = \frac{pd}{b}$$. Thus, the relations hold by matching the coefficients accordingly. **Final answer:** $$\boxed{\alpha + \beta = -\frac{r}{k}, \quad \alpha \beta = \frac{g}{p}}$$