1. **State the problem:** We have two quadratic equations with roots related as follows: - 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 $\frac{1}{\beta}$. 2. **Express sums and products of roots for Equation 1:** Using Vieta's formulas for $ax^2 + bx + c = 0$, sum of roots is $-\frac{b}{a}$ and product is $\frac{c}{a}$. - Sum: $\alpha + \beta = \frac{m}{2}$. - Product: $\alpha \beta = \frac{8}{2} = 4$. 3. **Express sums and products of roots for Equation 2:** Roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$. - Sum: $\frac{1}{\alpha} + \frac{1}{\beta}$. - Product: $\frac{1}{\alpha} \cdot \frac{1}{\beta} = \frac{1}{\alpha \beta} = \frac{1}{4}$. 4. **Apply Vieta's formulas to Equation 2:** Equation: $5x^2 - 10x + 5n = 0$. - Sum of roots: $\frac{10}{5} = 2$. - Product of roots: $\frac{5n}{5} = n$. 5. **Equate the sums and products from step 3 and step 4:** - Sum: $\frac{1}{\alpha} + \frac{1}{\beta} = 2$. But $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{m/2}{4} = \frac{m}{8}$. So, $\frac{m}{8} = 2 \implies m = 16$. - Product: $n = \frac{1}{4}$. 6. **Calculate $mn$:** $$mn = 16 \times \frac{1}{4} = 4.$$