Question

Let ${\alpha }_1,{\alpha }_2,\dots ,{\alpha }_7$ be the roots of the equation $x^7+$ $3x^5-13x^3-15x=0$ and $\left|{\alpha }_1\right|\ge \left|{\alpha }_2\right|\ge \dots \ge \left|{\alpha }_7\right|$. Then ${\alpha }_1{\alpha }_2-{\alpha }_3{\alpha }_4+{\alpha }_5{\alpha }_6$ is equal to

Answer 1

Answer: 9

Given equation can be rearranged as
$x\left(x^6+3x^4-13x^2-15\right)=0$
clearly $x=0$ is one of the root and other part can be observed by replacing $x^2=t$ from which we have $t^3+3t^2-13t-15=0$
$\Rightarrow (t-3)\left(t^2+6t+5\right)=0$
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 $\left|{\alpha }_1\right|\ge \left|{\alpha }_2\right|\ge \dots \ge \left|{\alpha }_7\right|$
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
${\alpha }_1{\alpha }_2-{\alpha }_3{\alpha }_4+{\alpha }_5{\alpha }_6$
$=1-(-3)+5=9$