Question:

Step 1: Find the roots of $x^2 - 3x - 2 = 0$. $$x = \frac{3 \pm \sqrt{9+8}}{2} = \frac{3 \pm \sqrt{17}}{2}$$ So $\tan\alpha = \frac{3-\sqrt{17}}{2}$ and $\tan\beta = \frac{3+\sqrt{17}}{2}$ (since $\beta > \alpha$). Step 2: Compute $\tan(\beta - \alpha)$ using the formula. $$\tan(\beta-\alpha) = \frac{\tan\beta - \tan\alpha}{1 + \tan\alpha\tan\beta}$$ $$\tan\beta - \tan\alpha = \sqrt{17}, \quad \tan\alpha\tan\beta = -2$$ $$\tan(\beta-\alpha) = \frac{\sqrt{17}}{1+(-2)} = \frac{\sqrt{17}}{-1} = -\sqrt{17}$$ Step 3: Determine the quadrant of $\beta - \alpha$. Since $\tan(\beta-\alpha) = -\sqrt{17} < 0$ and $\beta - \alpha > 0$ (as $\beta > \alpha$), we have $\beta - \alpha \in \left(\frac{\pi}{2}, \pi\right)$. Option (2) is correct. Step 4: Compute $\tan 2\alpha$. $$\tan 2\alpha = \frac{2\tan\alpha}{1-\tan^2\alpha}$$ $$\tan\alpha = \frac{3-\sqrt{17}}{2}, \quad \tan^2\alpha = \frac{(3-\sqrt{17})^2}{4} = \frac{26-6\sqrt{17}}{4} = \frac{13-3\sqrt{17}}{2}$$ $$1 - \tan^2\alpha = 1 - \frac{13-3\sqrt{17}}{2} = \frac{2-13+3\sqrt{17}}{2} = \frac{-11+3\sqrt{17}}{2}$$ $$\tan 2\alpha = \frac{2 \cdot \frac{3-\sqrt{17}}{2}}{\frac{-11+3\sqrt{17}}{2}} = \frac{(3-\sqrt{17}) \cdot 2}{-11+3\sqrt{17}} = \frac{2(3-\sqrt{17})}{3\sqrt{17}-11}$$ Rationalizing: multiply numerator and denominator by $(3\sqrt{17}+11)$: $$= \frac{2(3-\sqrt{17})(3\sqrt{17}+11)}{(3\sqrt{17})^2 - 11^2} = \frac{2(9\sqrt{17}+33-3\cdot17-11\sqrt{17})}{153-121} = \frac{2(-2\sqrt{17}-18)}{32} = \frac{-4(\sqrt{17}+9)}{32} = \frac{-(\sqrt{17}+9)}{8}$$ Alternatively, simplifying gives $\tan 2\alpha = \frac{1+\sqrt{17}}{1-\sqrt{17}}$. Option (3) is correct.