1. **Problem statement:** Find a quadratic equation whose roots are $\alpha + \frac{1}{\beta}$ and $\beta + \frac{1}{\alpha}$, given that $\alpha$ and $\beta$ are roots of the equation $x^2 - 6x + 7 = 0$. 2. **Recall the sum and product of roots:** For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, - Sum of roots: $\alpha + \beta = -\frac{b}{a}$ - Product of roots: $\alpha \beta = \frac{c}{a}$ 3. **Apply to given equation:** Given $x^2 - 6x + 7 = 0$, so $a=1$, $b=-6$, $c=7$. - Sum: $\alpha + \beta = 6$ - Product: $\alpha \beta = 7$ 4. **Find sum of new roots:** New roots are $\alpha + \frac{1}{\beta}$ and $\beta + \frac{1}{\alpha}$. Sum: $$ \left(\alpha + \frac{1}{\beta}\right) + \left(\beta + \frac{1}{\alpha}\right) = (\alpha + \beta) + \left(\frac{1}{\beta} + \frac{1}{\alpha}\right) $$ Calculate $\frac{1}{\beta} + \frac{1}{\alpha}$: $$ \frac{1}{\beta} + \frac{1}{\alpha} = \frac{\alpha + \beta}{\alpha \beta} = \frac{6}{7} $$ So sum of new roots: $$ 6 + \frac{6}{7} = \frac{42}{7} + \frac{6}{7} = \frac{48}{7} $$ 5. **Find product of new roots:** $$ \left(\alpha + \frac{1}{\beta}\right) \left(\beta + \frac{1}{\alpha}\right) = \alpha \beta + \alpha \cdot \frac{1}{\alpha} + \frac{1}{\beta} \cdot \beta + \frac{1}{\beta} \cdot \frac{1}{\alpha} $$ Simplify terms: $$ = \alpha \beta + 1 + 1 + \frac{1}{\alpha \beta} = 7 + 2 + \frac{1}{7} = 9 + \frac{1}{7} = \frac{64}{7} $$ 6. **Form the quadratic equation:** If roots are $r_1$ and $r_2$, quadratic equation is: $$ x^2 - (r_1 + r_2)x + r_1 r_2 = 0 $$ Substitute sum and product: $$ x^2 - \frac{48}{7}x + \frac{64}{7} = 0 $$ Multiply both sides by 7 to clear denominators: $$ 7x^2 - 48x + 64 = 0 $$ **Final answer:** $$ 7x^2 - 48x + 64 = 0 $$