Let alpha_1, alpha_2, ldots, alpha_7 be the r | vedclass

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(A) The given equation is $x^7+3x^5-13x^3-15x=0$.Factoring out $x$,we get $x(x^6+3x^4-13x^2-15)=0$.Let $t = x^2$. The equation becomes $t^3+3t^2-13t-1

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(A) The given equation is $x^7+3x^5-13x^3-15x=0$.Factoring out $x$,we get $x(x^6+3x^4-13x^2-15)=0$.Let $t = x^2$. The equation becomes $t^3+3t^2-13t-15=0$.By testing values,$t=-1$ is a root: $(-1)^3+3(-1)^2-13(-1)-15 = -1+3+13-15 = 0$.Dividing by $(t+1)$,we get $(t+1)(t^2+2t-15)=0$,which factors as $(t+1)(t+5)(t-3)=0$.Thus,$x^2 = -1, -5, 3$.The roots are $x = 0, \pm i, \pm i\sqrt{5}, \pm \sqrt{3}$.The magnitudes are $|0|=0, |\pm i|=1, |\pm i\sqrt{5}|=\sqrt{5}, |\pm \sqrt{3}|=\sqrt{3}$.Ordering by magnitude: $|\alpha_1| = |\alpha_2| = \sqrt{5}$,$|\alpha_3| = |\alpha_4| = \sqrt{3}$,$|\alpha_5| = |\alpha_6| = 1$,$|\alpha_7| = 0$.Let $\alpha_1 = i\sqrt{5}, \alpha_2 = -i\sqrt{5}, \alpha_3 = \sqrt{3}, \alpha_4 = -\sqrt{3}, \alpha_5 = i, \alpha_6 = -i$.Then $\alpha_1 \alpha_2 - \alpha_3 \alpha_4 + \alpha_5 \alpha_6 = (i\sqrt{5})(-i\sqrt{5}) - (\sqrt{3})(-\sqrt{3}) + (i)(-i) = 5 + 3 + 1 = 9$.

Let $\alpha_1, \alpha_2, \ldots, \alpha_7$ be the roots of the equation $x^7+3x^5-13x^3-15x=0$ and $|\alpha_1| \geq |\alpha_2| \geq \ldots \geq |\alpha_7|$. Then $\alpha_1 \alpha_2 - \alpha_3 \alpha_4 + \alpha_5 \alpha_6$ is equal to $..................$.

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