Question

If the equation $z^4+a_1{z}^3+a_2{z}^2+a_{3}z+a_4=0$ where $a_1,a_2,a_3,a_4$ are real coefficients different from zero, has a pure imaginary root, then the expression $\frac{a_3}{a_1{a}_2}+\frac{a_1{a}_4}{a_2{a}_3}$ has the value equal to

• A. $0$
• B. $1$
• C. $-2$
• D. $2$

Answer 1

The correct option is B $1$

Let $ix$ be the root where $x\ne 0$ and $x\in R$ (as if $x=0$ satisfies, then $a_4=0$ which contradicts)

$x^4-a_1{x}^{3}i-a_2{x}^2+a_{3}xi+a_4=0\Rightarrow x^4-a_2{x}^2+a_4-i(a_1{x}^3-a_{3}x)=0\therefore x^4-a_2{x}^2+a_4=0\dots \left(1\right)$
and $a_1{x}^3-a_{3}x=0\dots \left(2\right)$

From equation $\left(2\right),$ $a_1{x}^3-a_{3}x=0$
$\Rightarrow x^2=\frac{a_3}{a_1}$ (as $x\ne 0$)

Putting the value of $x^2$ in equation $\left(1\right),$ we get
$\frac{a_3^2}{a_1^2}-\frac{a_2{a}_3}{a_1}+a_4=0$ $\Rightarrow a_3^2+a_4{a}_1^2=a_1{a}_2{a}_3$
$\Rightarrow \frac{a_3}{a_1{a}_2}+\frac{a_1{a}_4}{a_2{a}_3}=1$