1. **Problem statement:** Given the quadratic equation $x^2 - bx + c = 0$ with roots $\alpha$ and $\beta$, we need to form a new quadratic equation whose roots are $\alpha + \frac{1}{\beta}$ and $\beta + \frac{1}{\alpha}$. 2. **Recall the relationships:** For the original equation, by Vieta's formulas: $$\alpha + \beta = b$$ $$\alpha \beta = c$$ 3. **Find the sum of the new roots:** $$\left(\alpha + \frac{1}{\beta}\right) + \left(\beta + \frac{1}{\alpha}\right) = (\alpha + \beta) + \left(\frac{1}{\beta} + \frac{1}{\alpha}\right) = b + \frac{\alpha + \beta}{\alpha \beta} = b + \frac{b}{c} = b\left(1 + \frac{1}{c}\right) = b\frac{c+1}{c}$$ 4. **Find the product of the new roots:** $$\left(\alpha + \frac{1}{\beta}\right)\left(\beta + \frac{1}{\alpha}\right) = \alpha \beta + \alpha \cdot \frac{1}{\alpha} + \frac{1}{\beta} \cdot \beta + \frac{1}{\beta} \cdot \frac{1}{\alpha} = c + 1 + 1 + \frac{1}{\alpha \beta} = c + 2 + \frac{1}{c} = \frac{c^2 + 2c + 1}{c} = \frac{(c+1)^2}{c}$$ 5. **Form the new quadratic equation:** If the roots are $r_1$ and $r_2$, the quadratic equation is: $$x^2 - (r_1 + r_2)x + r_1 r_2 = 0$$ Substitute the sum and product: $$x^2 - b\frac{c+1}{c}x + \frac{(c+1)^2}{c} = 0$$ 6. **Final answer:** $$\boxed{x^2 - \frac{b(c+1)}{c}x + \frac{(c+1)^2}{c} = 0}$$ This is the quadratic equation whose roots are $\alpha + \frac{1}{\beta}$ and $\beta + \frac{1}{\alpha}$.