Question

If $a_1,a_2,a_3,\cdots ,a_n;(n\ge 2)$ are real and $(n-1)a_1^2-2n{a}_2<0,$ then atleast roots of the equation $x^n+a_1{x}^{n-1}+a_2{x}^{n-2}+\cdots +a_n=0$, are imaginary.

• A. $2$
• B. $4$
• C. $5$
• D. $1$

Answer 1

The correct option is A $2$

Given: $a_1,a_2,a_3,\cdots ,a_n;(n\ge 2);$
$(n-1)a_1^2-2n{a}_2<0;$
$x^n+a_1{x}^{n-1}+a_2{x}^{n-2}+\cdots +a_n=0$
Let ${\alpha }_1,{\alpha }_2,{\alpha }_3,\cdots ,{\alpha }_n$ are the $n$ imaginary roots of the given equation.

$\therefore \sum {\alpha }_1={\alpha }_1+{\alpha }_2+{\alpha }_3+\cdots +{\alpha }_n=-a_1$
$\text{and}\sum {\alpha }_1.{\alpha }_2={\alpha }_1.{\alpha }_2+{\alpha }_2.{\alpha }_3+{\alpha }_3.{\alpha }_4+\cdots +{\alpha }_{n-1}.{\alpha }_n=a_2$

$\text{Now},(n-1)a_1^2-2n{a}_2=(n-1)(\sum {\alpha }_1{)}^2-2n\sum {\alpha }_1.{\alpha }_2$

$\Rightarrow (n-1)a_1^2-2n{a}_2=n[(\sum {\alpha }_1{)}^2-2\sum {\alpha }_1.{\alpha }_2]-(\sum {\alpha }_1{)}^2$

$\Rightarrow (n-1)a_1^2-2n{a}_2=n\sum {{\alpha }_1}^2-(\sum {\alpha }_1{)}^2$

$\Rightarrow (n-1)a_1^2-2n{a}_2=n\sum _{1\le i<j\le n}\sum {({\alpha }_i-{\alpha }_j)}^2$

$\text{But we have given that}(n-1)a_1^2-2n{a}_2<0$

$\Rightarrow \sum _{1\le i<j\le n}\sum {({\alpha }_i-{\alpha }_j)}^2<0$

This is true only when atleast two roots are imaginary.