If $a_{1}, a_{2}, a_{3}, . . . . ., a_{n} (n \geq 2)$ are real and $(n - 1) a_{1}^{2} = 2 n a_{2} < 0$ , then show that atleast two roots of the equation $x^{n} + a_{1} x^{n - 1} + a_{2} x^{n - 2} + . . . . + a_{n} = 0$ are imaginary.

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Solution

Let $α_{1}, α_{2}, α_{3}, . . . . ., α_{n}$ are the roots of the given equation.
Then ${\sum α_{1} = α}_{1} + α_{2} + α_{3} + . . . . . + α_{n} = - a_{1}$
${\sum α_{1} α_{2} = α}_{1} α_{2} + α_{2} α_{3} + . . . . . + α_{n - 1} α_{n} = - a_{2} N o w (n - 1) a_{1}^{2} - 2 n a_{2} = (n - 1) {(\sum α_{1})}^{2} - 2 n \sum α_{1} α_{2} = n {{(\sum α_{1})}^{2} - 2 \sum α_{1} α_{2}} - {(\sum α_{1})}^{2} = n \sum α_{1}^{2} - {(\sum α_{1})}^{2} = \sum \sum {(α_{i} α_{j})}^{2} 1 \leq i < j \leq n$
But given that $(n - 1) a_{1}^{2} - 2 n a_{2} < 0$
$\sum \sum {(α_{i} α_{j})}^{2} < 0 1 \leq i < j \leq n$
which is true only when at least two roots are imaginary.
Ans: 1

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