Construct a quadratic equation whose roots are (1) \( \begin{array}{l}\text { (2) } \alpha+\frac{1}{\beta}, \beta+\frac{1}{\alpha}\end{array} \) (3) \( \alpha^{3} \) and \( \beta^{3} \) Assignment

To construct a quadratic equation whose roots are given as \( r_1 = \alpha + \frac{1}{\beta} \), \( r_2 = \beta + \frac{1}{\alpha} \), we can use Vieta's formulas. According to Vieta's, for a quadratic equation of the form \( x^2 - (r_1 + r_2)x + r_1r_2 = 0 \), the sum of the roots \( r_1 + r_2 \) and the product of the roots \( r_1r_2 \) can be used to form the equation. ### Step 1: Calculate the sum of the roots \( r_1 + r_2 \) \[ r_1 + r_2 = \left( \alpha + \frac{1}{\beta} \right) + \left( \beta + \frac{1}{\alpha} \right) = \alpha + \beta + \frac{1}{\beta} + \frac{1}{\alpha} \] ### Step 2: Calculate the product of the roots \( r_1r_2 \) \[ r_1r_2 = \left( \alpha + \frac{1}{\beta} \right) \left( \beta + \frac{1}{\alpha} \right) \] Expanding this product: \[ r_1r_2 = \alpha\beta + \frac{\alpha}{\alpha} + \frac{\beta}{\beta} + \frac{1}{\alpha\beta} = \alpha\beta + 1 + 1 + \frac{1}{\alpha\beta} = \alpha\beta + 2 + \frac{1}{\alpha\beta} \] ### Step 3: Form the quadratic equation Now we can substitute the sum and product into the quadratic equation: \[ x^2 - (r_1 + r_2)x + r_1r_2 = 0 \] Substituting the values we calculated: \[ x^2 - \left( \alpha + \beta + \frac{1}{\beta} + \frac{1}{\alpha} \right)x + \left( \alpha\beta + 2 + \frac{1}{\alpha\beta} \right) = 0 \] ### Final Quadratic Equation Thus, the quadratic equation whose roots are \( \alpha + \frac{1}{\beta} \) and \( \beta + \frac{1}{\alpha} \) is: \[ x^2 - \left( \alpha + \beta + \frac{1}{\beta} + \frac{1}{\alpha} \right)x + \left( \alpha\beta + 2 + \frac{1}{\alpha\beta} \right) = 0 \] This is the required quadratic equation.