The resistance of the heating element is:
\[R = \rho \frac{\ell}{A}.\]
The power is inversely proportional to the length:
\[P \propto \frac{1}{\ell}.\]
From the relation \(P_1 \times t_1 = P_2 \times t_2\):
\[\frac{P_1}{P_2} = \frac{t_2}{t_1} = \frac{15}{20}.\]
Substituting \(P \propto \frac{1}{\ell}\):
\[\frac{\ell_2}{\ell_1} = \frac{t_2}{t_1} = \frac{15}{20} = \frac{3}{4}.\]
Thus, the new length should be decreased to:
\[\ell_2 = \frac{3}{4} \ell_1.\]
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: