To determine the range of \( f(x) = 3\sqrt{x-2} + \sqrt{4-x} \), we analyze the domain of \( f(x) \) by finding values of \( x \) for which both square roots are real.
Therefore, \( x \) is restricted to the interval \([2, 4]\).
Next, we evaluate \( f(x) \) at the endpoints to determine the minimum and maximum values.
Thus, the minimum value \( \alpha \) is \( \sqrt{2} \), and the maximum value \( \beta \) is \( 3\sqrt{2} \).
Now, we calculate \( \alpha^2 + 2\beta^2 \):
\[ \alpha^2 + 2\beta^2 = (\sqrt{2})^2 + 2(3\sqrt{2})^2 = 2 + 2 \times 18 = 2 + 36 = 42. \]
Therefore, \( \alpha^2 + 2\beta^2 \) is equal to 42.
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: