2. If \( \alpha, \beta \) are the roots of \( 3 x^{2}+2 x+1=0 \), then find the equation whose roots are \( \frac{1-\alpha}{1+\alpha} \) and \( \frac{1-\beta}{1+\beta} \)

To find the equation with roots \(\frac{1-\alpha}{1+\alpha}\) and \(\frac{1-\beta}{1+\beta}\), we can use the given quadratic equation \(3x^2 + 2x + 1 = 0\). ### Step 1: Find \(\alpha + \beta\) and \(\alpha \beta\) For the quadratic \(ax^2 + bx + c = 0\), the roots \(\alpha\) and \(\beta\) satisfy: \[ \alpha + \beta = -\frac{b}{a} = -\frac{2}{3}, \quad \alpha \beta = \frac{c}{a} = \frac{1}{3}. \] ### Step 2: Find the new roots The transformation to the new roots can be represented as: \[ y = \frac{1-x}{1+x}. \] Solving for \(x\) gives: \[ x = \frac{1-y}{1+y}. \] Using Vieta's formulas, we express \(s = \frac{1-\alpha}{1+\alpha} + \frac{1-\beta}{1+\beta}\) and \(p = \frac{(1-\alpha)(1-\beta)}{(1+\alpha)(1+\beta)}\). #### Finding \(s\): \[ s = \frac{(1-\alpha)(1+\beta) + (1-\beta)(1+\alpha)}{(1+\alpha)(1+\beta)} = \frac{1 - \alpha + \beta - \alpha\beta + 1 - \beta + \alpha - \alpha\beta}{(1+\alpha)(1+\beta)} = \frac{2 - 2\alpha\beta}{(1+\alpha)(1+\beta)}. \] Substituting \(\alpha + \beta = -\frac{2}{3}\) and \(\alpha \beta = \frac{1}{3}\): \[ s = \frac{2 - 2\cdot\frac{1}{3}}{(1 - \frac{2}{3})(1 - \frac{2}{3})} = \frac{2 - \frac{2}{3}}{\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)} = \frac{\frac{4}{3}}{\frac{1}{9}} = 12. \] #### Finding \(p\): \[ p = \frac{(1-\alpha)(1-\beta)}{(1+\alpha)(1+\beta)} = \frac{1 - (\alpha+\beta) + \alpha\beta}{1 + (\alpha+\beta) + \alpha\beta}. \] Substituting the values gives: \[ p = \frac{1 + \frac{2}{3} + \frac{1}{3}}{1 - \frac{2}{3} + \frac{1}{3}} = \frac{1 + 1}{1 - \frac{1}{3}} = \frac{2}{\frac{2}{3}} = 3. \] ### Step 3: Form the quadratic equation The new quadratic equation whose roots are \(y_1\) and \(y_2\) is given by: \[ x^2 - sx + p = 0 \implies x^2 - 12x + 3 = 0. \] Thus, the equation with roots \(\frac{1-\alpha}{1+\alpha}\) and \(\frac{1-\beta}{1+\beta}\) is \[ \boxed{x^2 - 12x + 3 = 0}. \]