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If a1, a2, a3 ......an (n>= 2) are real...

If `a_1, a_2, a_3 ......a_n (n>= 2) ` are real and `(n-1) a_1^2 -2na_2 < 0` then prove that at least two roots of the equation ` x^n+a_1 x^(n-1) +a_2 x^(n-2) +......+a_n = 0 `are imaginary.

Text Solution

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Let `alpha_(1),alpha_(2),alpha_(3),..,alpha_(n)` are the roots of the given equation.
Then `sumalpha-(1)=alpha_(1)+alpha_(2)+alpha_(3)+………+alpha_(n)=-a_(1)`
and `sumalpha_(1)alpha_(2)=alpha_(1)alpha_(2)+alpha_(1)alpha_(3)+……..+alpha_(n-1)alpha_(n)=a_(2)`
Now `(n-1)a-(1)^(2)-2na_(2)=(n-1)(sumalpha_(1))^(2)-2nsumalpha_(1)alpha_(2)`
`=n{(sumalpha_(1))^(2)-2sumalpha_(1)alpha_(2)}-(sumalpha_(1))^(2)`
`=nsumalpha_(1)^(2)-(suma_(1))^(2)`
`=sum_(1leiltjlen)sum(alpha_(i)-alpha_(j))^(2)`
But given that `(n-1)a_(1)^(2)-2na_(2)lt0`
`impliessum_(1leiltjlen).sum(alpha_(i)-alpha_(j))^(2)lt0`
which is true only when atleast two roots are imaginary.
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