Quadratic Equation and Inequalities Part 2

Given equation can be rearranged as
x(x6+3x4−13x2−15)=0
clearly x=0 is one of the root and other part can be observed by replacing x2=t from which we have ‌‌t3+3t2−13t−15=0
⇒‌‌(t−3)(t2+6t+5)=0
So, ‌‌t=3,t=−1,t=−5
Now we are getting x2=3,x2=−1,x2=−5
⇒x=±√3,x=±i,x=±√5i
From the given condition |α1|≥|α2|≥....≥|α7|
We can clearly say that |α7|=0 and
and|α6|=√5=|α5|
and |α4|=√3=|α3| and |α2|=1=|α1|
So we can have, α1=√5i,α2=−√5i,α3=√3i,
α4=−√3,α5=i,α6=−i
Hence
α1α2−α3α4‌+α5α6
‌=1−(−3)+5=9