The moment of inertia of a hollow sphere about its diameter axis is given by:
\[ I_{\text{sphere}} = \frac{2}{3} MR^2 = M k_1^2 \]
where \(k_1\) is the radius of gyration of the hollow sphere.
The moment of inertia of a solid cylinder about its diameter axis is:
\[ I_{\text{cylinder}} = \frac{1}{12} M(4R^2) + \frac{1}{4} MR^2 + M(2R)^2 = \frac{67}{12} MR^2 = M k_2^2 \]
Calculating the ratio of the radii of gyration:
\[ \frac{k_1}{k_2} = \sqrt{\frac{\frac{2}{3}}{\frac{67}{12}}} = \sqrt{\frac{8}{67}} \]
Hence, the value of \(x\) is 67.
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: