If \( \alpha, \beta \) are the roots of the equation \( 3 x^{2}+5 x-1=0 \) Construct a quadratic equation whose roots are (1) \( 5 \alpha, 5 \beta \) \( \begin{array}{lll}\text { (2) } \alpha+\frac{1}{\beta}, \beta+\frac{1}{\alpha} & \text { (3) } \alpha^{3} \text { and } \beta^{3} \text { Assignment }\end{array} \)

To construct the quadratic equations for the given roots, we will follow these steps: 1. **Find the roots \( \alpha \) and \( \beta \) of the original equation \( 3x^2 + 5x - 1 = 0 \)**. 2. **Use the relationships between the roots and coefficients to find the new roots for each case**. 3. **Construct the quadratic equations based on the new roots**. ### Step 1: Find the roots \( \alpha \) and \( \beta \) The roots of the quadratic equation \( ax^2 + bx + c = 0 \) can be found using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] For our equation \( 3x^2 + 5x - 1 = 0 \), we have \( a = 3 \), \( b = 5 \), and \( c = -1 \). Let's calculate the roots \( \alpha \) and \( \beta \): \[ \alpha, \beta = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 3 \cdot (-1)}}{2 \cdot 3} \] Calculating the discriminant: \[ 5^2 - 4 \cdot 3 \cdot (-1) = 25 + 12 = 37 \] Now substituting back into the formula: \[ \alpha, \beta = \frac{-5 \pm \sqrt{37}}{6} \] ### Step 2: Construct the quadratic equations for the new roots #### (1) Roots: \( 5\alpha, 5\beta \) The sum and product of the roots \( 5\alpha \) and \( 5\beta \) are: \[ \text{Sum} = 5\alpha + 5\beta = 5(\alpha + \beta) \] \[ \text{Product} = 5\alpha \cdot 5\beta = 25\alpha\beta \] From Vieta's formulas for the original equation: \[ \alpha + \beta = -\frac{b}{a} = -\frac{5}{3} \] \[ \alpha\beta = \frac{c}{a} = -\frac{1}{3} \] Now substituting these values: \[ \text{Sum} = 5\left(-\frac{5}{3}\right) = -\frac{25}{3} \] \[ \text{Product} = 25\left(-\frac{1}{3}\right) = -\frac{25}{3} \] The quadratic equation with roots \( 5\alpha \) and \( 5\beta \) is: \[ x^2 - \left(-\frac{25}{3}\right)x + \left(-\frac{25}{3}\right) = 0 \] Multiplying through by 3 to eliminate the fraction: \[ 3x^2 + 25x + 25 = 0 \] #### (2) Roots: \( \alpha + \frac{1}{\beta}, \beta + \frac{1}{\alpha} \) Calculating the sum and product of these roots: \[ \text{Sum} = \left(\alpha + \frac{1}{\beta}\right) + \left(\beta + \frac{1}{\alpha}\right) = (\alpha + \beta) + \left(\frac{1}{\beta} + \frac{1}{\alpha}\right) \] Using \( \frac{1}{\beta} + \frac{1}{\alpha} = \frac{\alpha + \beta}{\alpha\beta} \): \[ \text{Sum} = -\frac{5}{3} + \frac{-\frac{5}{3}}{-\frac{1}{3}} = -\frac{5}{3} + 5 = \frac{10}{3} \] Now for the product: \[ \text{Product} = \left(\alpha + \frac{1}{\beta}\right)\left(\beta + \frac{1}{\alpha}\right) = \alpha\beta + \alpha\cdot\frac{1}{\alpha} + \beta\cdot\frac{1}{\beta} + \frac{1}{\alpha\beta} \] This simplifies to: \[ \text{Product} = -\frac{1}{3} + 1 + 1 - 3 = \frac{5}{3} \] The quadratic equation is: \[ x^2 - \frac{10}{3}x + \frac{5}{3} = 0 \] Multiplying through by 3: \[ 3x^2 - 10x + 5 = 0 \] #### (3) Roots: \( \alpha^3, \beta^3 \) Using the identity \( \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \): First, we find \( \alpha^2 + \beta^2 \): \[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta = \left(-\frac{5}{3}\right)^2 - 2\left(-\frac{1}{3}\right) = \frac{25}{9} + \frac{2}{3} = \frac{25}{9} + \frac{6}{9} = \frac{31}{9} \] Now substituting into the identity: \[ \alpha^3 + \beta^3 = \left(-\frac{5}{3}\right)\left(\frac{31}{9} + \frac{1}{3}\right) = -\frac{5}{3}\left(\frac{31 + 3}{9}\right) = -\frac{5}{3}\left(\frac{34}{9}\right) = -\frac{170}{27} \] For the product \( \alpha^3 \beta^3 = (\alpha\beta)^3 = \left(-\frac{1}{3}\right)^3 = -\frac{1}{27} \). The quadratic equation is: \[ x^2 - \left(-\frac{170}{27}\right)x - \left(-\frac{1}{27}\right) = 0 \] Multiplying through by 27: \[ 27x^2 + 170x + 1 = 0 \] ### Summary of Quadratic Equations 1. For roots \( 5\alpha, 5\beta \): \( 3x^2 + 25x + 25 = 0 \) 2. For roots \( \alpha + \frac{1}{\