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Question

If a1,a2,a3,,an;(n2) are real and (n1)a212na2 <0, then atleast roots of the equation xn + a1xn1 + a2xn2 + +an=0, are imaginary.

A
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1
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Solution

The correct option is A 2
Given: a1,a2,a3,,an;(n2);
(n1)a212na2 <0;
xn + a1xn1 + a2xn2 + +an=0
Let α1,α2,α3,,αn are the n imaginary roots of the given equation.

α1=α1+α2+α3++αn=a1
and α1.α2=α1.α2+α2.α3+α3.α4++αn1.αn=a2

Now, (n1)a212na2=(n1)(α1)22nα1.α2

(n1)a212na2=n[(α1)22α1.α2](α1)2

(n1)a212na2=nα12(α1)2

(n1)a212na2=n1i<jn(αiαj)2

But we have given that (n1)a212na2<0

1i<jn(αiαj)2<0

This is true only when atleast two roots are imaginary.

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