To find the product of the roots of the quadratic equation \(x^2 - 3x + 2 = 0\),
we use Viète's formulas, which relate the coefficients of a polynomial to sums and products of its roots.
For a quadratic equation of the form \(ax^2 + bx + c = 0\),
the product of the roots (\(P\)) is given by:
\[P = \frac{c}{a}\]
Here, \(a = 1\), \(b = -3\), and \(c = 2\).
Substitute these values into the formula:
\[P = \frac{2}{1} = 2\]
Thus, the product of the roots is 2.
Let \( \alpha, \beta \) be the roots of the equation \( x^2 - ax - b = 0 \) with \( \text{Im}(\alpha) < \text{Im}(\beta) \). Let \( P_n = \alpha^n - \beta^n \). If \[ P_3 = -5\sqrt{7}, \quad P_4 = -3\sqrt{7}, \quad P_5 = 11\sqrt{7}, \quad P_6 = 45\sqrt{7}, \] then \( |\alpha^4 + \beta^4| \) is equal to: