Question

$x^7+3x^5-13x^3-15x=0$ and if $|\alpha_1| \ge |\alpha_2| \ge |\alpha_3| \ge |\alpha_4| \ge |\alpha_5| \ge |\alpha_6| \ge |\alpha_7|$ then $(\alpha_1\alpha_2 - \alpha_3\alpha_4 + \alpha_5\alpha_6) = ?

Answer 1

9

Answer 2

The roots of the polynomial $x^7+3x^5-13x^3-15x=0$ are found by factoring and solving the resulting cubic equation in $x^2$. The roots are $0, \pm\sqrt{3}, \pm i, \pm i\sqrt{5}$. Their absolute values are $0, \sqrt{3}, \sqrt{3}, 1, 1, \sqrt{5}, \sqrt{5}$. The condition $|\alpha_1| \ge \dots \ge |\alpha_7|$ groups roots with equal absolute values. Thus, $\{\alpha_1, \alpha_2\} = \{i\sqrt{5}, -i\sqrt{5}\}$, $\{\alpha_3, \alpha_4\} = \{\sqrt{3}, -\sqrt{3}\}$, and $\{\alpha_5, \alpha_6\} = \{i, -i\}$. Calculating the products $\alpha_1\alpha_2=5$, $\alpha_3\alpha_4=-3$, and $\alpha_5\alpha_6=1$, the desired expression is $5 - (-3) + 1 = 9$.