α&nbsp;and&nbsp;β&nbsp;are the roots of the quadratic equation.ω≠1&nbsp;and&nbsp;ω13=1If&nbsp;α=ω+ω3+ω4+ω−4+ω−3+ω−1β=ω2+ω5+ω6+ω−6+ω−5+ω−2,then quadratic equation, whose roots are&nbsp;α&nbsp;and&nbsp;β&nbsp;is:I tried finding the sum and product of roots and placing in the equation&nbsp;x2-(α+β)x+(α×β)=0&nbsp;but I was not able to solve for it.

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α&amp;nbsp;and&amp;nbsp;β&amp;nbsp;are the roots of the quadratic equation.ω≠1&amp;nbsp;and&amp;nbsp;ω13=1If&amp;nbsp;α=ω+ω3+ω4+ω−4+ω−3+ω−1β=ω2+ω5+ω6+ω−6+ω−5+ω−2,then quadratic equation, whose roots are&amp;nbsp;α&amp;nbsp;and&amp;nbsp;β&amp;nbsp;is:I tried finding the sum and product of roots and placing in the equation&amp;nbsp;x2-(α+β)x+(α×β)=0&amp;nbsp;but I was not able to solve for it.

Lana Hamilton · Accepted Answer

Step 1By the formula for a finite geometric series,α+β=∑i=−6−1ωi+∑i=16ωi=∑i=712ωi+∑i=16ωi=(1−ω13)1−ω−1=&amp;nbsp;-1&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;(ω13=1)−1=−1For the product&amp;nbsp;αβ, note that&amp;nbsp;ωn=ω-nTaking&amp;nbsp;ω&amp;nbsp;as&amp;nbsp;e2iπ13=cos⁡(2π13)+isin⁡(2π13)&amp;nbsp;for example,&amp;nbsp;α=2(cos⁡(2π13)+cos⁡(6π13)+cos⁡(8π13))&amp;nbsp;and you can do the same for&amp;nbsp;β. Then you can use the product-to-sum formulacos⁡Acos⁡B=12(cos⁡(A−B)+cos⁡(A+B))by expanding the brackets, there are nine terms in this form.

Tristatex9tw · Accepted Answer

Step 1We have1+α+β=1+ω+ω2+⋯+ω12=0and so α+β=−1Therefore, αβ=−(α+α2)=−3Unfortunately, I can&#039;t see shortcut to compute α+α2 except by expanding it.

\(\displaystyle\alpha\) and \(\displaystyle\beta\) are the roots

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