Question:

**Step 1:** Identify properties of the given roots. Let $\alpha$ and $\beta$ be the roots of the equation $x^2 - 3x + 1 = 0$. By Vieta's formulas, the sum of the roots is $\alpha + \beta = -(-3)/1 = 3$, and the product of the roots is $\alpha\beta = 1/1 = 1$. **Step 2:** Define the new roots. Let the new roots be $y_1 = \frac{1}{\alpha + 2}$ and $y_2 = \frac{1}{\beta + 2}$. **Step 3:** Calculate the sum of the new roots. The sum of the new roots is: $$y_1 + y_2 = \frac{1}{\alpha + 2} + \frac{1}{\beta + 2}$$ $$y_1 + y_2 = \frac{(\beta + 2) + (\alpha + 2)}{(\alpha + 2)(\beta + 2)}$$ $$y_1 + y_2 = \frac{\alpha + \beta + 4}{\alpha\beta + 2\alpha + 2\beta + 4}$$ $$y_1 + y_2 = \frac{\alpha + \beta + 4}{\alpha\beta + 2(\alpha + \beta) + 4}$$ Substitute the values from Step 1: $$y_1 + y_2 = \frac{3 + 4}{1 + 2(3) + 4} = \frac{7}{1 + 6 + 4} = \frac{7}{11}$$ **Step 4:** Calculate the product of the new roots. The product of the new roots is: $$y_1 y_2 = \left(\frac{1}{\alpha + 2}\right) \left(\frac{1}{\beta + 2}\right)$$ $$y_1 y_2 = \frac{1}{(\alpha + 2)(\beta + 2)}$$ $$y_1 y_2 = \frac{1}{\alpha\beta + 2(\alpha + \beta) + 4}$$ Substitute the values from Step 1: $$y_1 y_2 = \frac{1}{1 + 2(3) + 4} = \frac{1}{1 + 6 + 4} = \frac{1}{11}$$ **Step 5:** Formulate the new quadratic equation. A quadratic equation with roots $y_1$ and $y_2$ is given by $y^2 - (y_1 + y_2)y + y_1 y_2 = 0$. Substitute the sum and product calculated in Steps 3 and 4: $$y^2 - \left(\frac{7}{11}\right)y + \left(\frac{1}{11}\right) = 0$$ Multiply the entire equation by 11 to clear the denominators: $$11y^2 - 7y + 1 = 0$$ Thus, the equation whose roots are $\frac{1}{\alpha + 2}$ and $\frac{1}{\beta + 2}$ is $11x^2 - 7x + 1 = 0$.