Question

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

Answer 1

Given equation can be rearranged as
$x\left(x^6+3 x^4-13 x^2-15\right)=0$
clearly $x=0$ is one of the root and other part can be observed by replacing $x^2= t$ from which we have $t^3+3 t^2-13 t-15=0$
$\Rightarrow (t-3)\left(t^2+6 t+5\right)=0$
So, $ t=3, t=-1, t=-5$
Now we are getting $x^2=3, x^2=-1, x^2=-5$
$\Rightarrow x=\pm \sqrt{3}, x=\pm i , x=\pm \sqrt{5} i$
From the given condition $\left|\alpha_1\right| \geq\left|\alpha_2\right| \geq \ldots \geq\left|\alpha_7\right|$
We can clearly say that $\left|\alpha_\tau\right|=0$ and
and $\left|\alpha_6\right|=\sqrt{5}=\left|\alpha_5\right|$
and
$\left|\alpha_4\right|=\sqrt{3}=\left|\alpha_3\right|$ and $\left|\alpha_2\right|=1=\left|\alpha_1\right|$
So we can have, $\alpha_1=\sqrt{5} i , \alpha_2=-\sqrt{5} i , \alpha_3=\sqrt{3} i$,
$\alpha_4=-\sqrt{3}, \alpha_5= i , \alpha_6=- i$
Hence
$\alpha_1 \alpha_2-\alpha_3 \alpha_4 +\alpha_5 \alpha_6$
$ =1-(-3)+5=9$