1. **State the problem:** We have two quadratic equations: - Equation 1: $2x^2 - mx + 8 = 0$ with roots $\alpha$ and $\beta$. - Equation 2: $5x^2 - 10x + 5n = 0$ with roots $\frac{1}{\alpha}$ and $\frac{1}{\beta}$. We need to find the product $mn$. 2. **Use sum and product of roots for the first equation:** For quadratic equation $ax^2 + bx + c = 0$ with roots $r_1$ and $r_2$, sum of roots $= -\frac{b}{a}$ and product of roots $= \frac{c}{a}$. From the first equation: Sum: $\alpha + \beta = \frac{m}{2}$ (since $-(-m)/2 = m/2$) Product: $\alpha \beta = \frac{8}{2} = 4$ 3. **Use sum and product of roots for the second equation:** Equation: $5x^2 - 10x + 5n = 0$. Sum of roots $= -\frac{-10}{5} = 2$. Product of roots $= \frac{5n}{5} = n$. Given roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$: Sum: $\frac{1}{\alpha} + \frac{1}{\beta} = 2$ Product: $\frac{1}{\alpha} \cdot \frac{1}{\beta} = n$ 4. **Express sum and product of reciprocals in terms of $\alpha$ and $\beta$:** Sum of reciprocals: $$\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{m/2}{4} = \frac{m}{8}$$ Given this sum equals 2: $$\frac{m}{8} = 2 \Rightarrow m = 16$$ 5. **Find product of reciprocals:** $$n = \frac{1}{\alpha \beta} = \frac{1}{4}$$ 6. **Calculate $mn$:** $$mn = 16 \times \frac{1}{4} = 4$$ **Final answer:** $\boxed{4}$