1. **Problem Statement:** Given the quadratic equation $ky^2 + ry + d = 0$ with roots $\alpha$ and $\beta$, show that $\alpha + \beta = -\frac{r}{k}$ and $\alpha \beta = \frac{d}{k}$. 2. **Formula Used:** For a quadratic equation $ay^2 + by + c = 0$, the sum of roots is $-\frac{b}{a}$ and the product of roots is $\frac{c}{a}$. 3. **Apply to Given Equation:** Here, $a = k$, $b = r$, and $c = d$. 4. **Sum of Roots:** $$\alpha + \beta = -\frac{b}{a} = -\frac{r}{k}$$ 5. **Product of Roots:** $$\alpha \beta = \frac{c}{a} = \frac{d}{k}$$ 6. **Explanation:** This follows from the factorization of the quadratic as $k(y - \alpha)(y - \beta) = 0$, expanding which gives $ky^2 - k(\alpha + \beta)y + k\alpha\beta = 0$. Comparing coefficients with $ky^2 + ry + d = 0$ yields the required relations. **Final Answer:** $$\alpha + \beta = -\frac{r}{k}, \quad \alpha \beta = \frac{d}{k}$$