Question

j 136b

1. if the roots of (x^2 - x + 1 = 0) are (alpha) and (\beta), create a quadratic equation that has the roots (\frac{1}{alpha + 1}) and (\frac{1}{\beta + 1}). (note that the coefficient of (x^2) is 1.)

Explanation:

Step1: Find sum and product of original roots

$$\alpha + \beta = 1, \quad \alpha\beta = 1$$

Step2: Calculate sum of new roots

$$\frac{1}{\alpha+1} + \frac{1}{\beta+1} = \frac{\alpha+\beta+2}{\alpha\beta+\alpha+\beta+1} = \frac{1+2}{1+1+1} = 1$$

Step3: Calculate product of new roots

$$\frac{1}{\alpha+1} \cdot \frac{1}{\beta+1} = \frac{1}{\alpha\beta+\alpha+\beta+1} = \frac{1}{1+1+1} = \frac{1}{3}$$

Step4: Formulate the new quadratic equation

$$x^2 - x + \frac{1}{3} = 0 \implies 3x^2 - 3x + 1 = 0$$

Answer:

\(3x^2 - 3x + 1 = 0\)