Question: Given that $\alpha$ and $\beta$ are the roots of the equation $x^2 - 2x + 5 = 0$, Form the…

Given that $\alpha$ and $\beta$ are the roots of the equation $x^2 - 2x + 5 = 0$,

Form the equation whose roots are:

(i) $\alpha^2$, $\beta^2$

(ii) $\frac{1}{\alpha}$, $\frac{1}{\beta}$

Solution

Let’s solve the problem step by step. Given that \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 - 2x + 5 = 0 \). \[ \alpha + \beta = 2 \quad & \] \[ \alpha \beta = 5 \quad & \] We need to form a new equation whose roots are \( \frac{1}{\alpha} + \beta^2 \) and \( \frac{1}{\beta} + \alpha^2 \). First, express \( \alpha^2 \) and \( \beta^2 \) using the original equation: \[ \alpha^2 = 2\alpha - 5 \quad & \] \[ \beta^2 = 2\beta - 5 \quad & \] Now, the new roots are: \[ r_1 = \frac{1}{\alpha} + 2\beta - 5 \quad & \] \[ r_2 = \frac{1}{\beta} + 2\alpha - 5 \quad & \] Calculate the sum \( S \) of the new roots: \[ S = r_1 + r_2 = \frac{1}{\alpha} + \frac{1}{\beta} + 2\beta + 2\alpha - 10 \quad & \] Using \( \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta} = \frac{2}{5} \) and \( 2\alpha + 2\beta = 4 \): \[ S = \frac{2}{5} + 4 - 10 = -\frac{28}{5} \quad & \] Next, calculate the product \( P \) of the new roots: \[ P = \left( \frac{1}{\alpha} + 2\beta - 5 \right) \left( \frac{1}{\beta} + 2\alpha - 5 \right) \quad & \] Expanding and simplifying using \( \alpha + \beta = 2 \) and \( \alpha \beta = 5 \): \[ P = \frac{1}{5} + 4 \times 5 + 25 - 22 = \frac{136}{5} \quad & \] The new quadratic equation with roots \( r_1 \) and \( r_2 \) is: \[ x^2 - Sx + P = 0 \quad & \] \[ x^2 + \frac{28}{5}x + \frac{136}{5} = 0 \quad & \] Multiplying through by 5 to eliminate fractions: \[ 5x^2 + 28x + 136 = 0 \quad & \] Final Answer: \[ 5x^2 + 28x + 136 = 0 \]