An organic compound (X) with molecular formula $\mathrm{C}_{3} \mathrm{H}_{6} \mathrm{O}$ is not readily oxidised. On reduction it gives $\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}(\mathrm{Y})\right.$ which reacts with HBr to give a bromide (Z) which is converted to Grignard reagent. This Grignard reagent on reaction with (X) followed by hydrolysis give 2,3-dimethylbutan-2-ol. Compounds (X), (Y) and (Z) respectively are:
1. Compound (X): - $\mathrm{CH}_{3} \mathrm{COCH}_{3}$ (Acetone)
2. Reduction to (Y): - $\mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}$ (Isopropyl alcohol)
3. Reaction with HBr to form (Z): - $\mathrm{CH}_{3} \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{3}$ (2-Bromopropane)
4. Grignard reagent and reaction with (X): - The Grignard reagent formed from (Z) reacts with acetone to form 2,3-dimethylbutan-2-ol after hydrolysis.
Therefore, the correct answer is (2) $\mathrm{CH}_{3} \mathrm{COCH}_{3}, \mathrm{CH}_{3} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{3}, \mathrm{CH}_{3} \mathrm{CH}(\mathrm{Br}) \mathrm{CH}_{3}$.
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: