Question

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

• A. $0$
• B. $1$
• C. $-2$
• D. $2$

Answer 1

Answer: (B)

$1$

Let $x$be the root where $x\ne 0$ and $xϵR$
$x^4-a_1{x}^{3}i-a_2{x}^2+a_{3}xi+a_4=0$
$\Rightarrow x^4-a_2{x}^2+a_4=0$ (1)
and $a_1{x}^3-a_{3}x=0$ (2)
From Eq. (2),
$a_1{x}^2-a_3=0$
$\Rightarrow x^2=a_3/a_1$ (as $x\ne 0)$
Putting the value of $x^2$ in Eq. (1), we get
$\frac{a_3^2}{a_1^2}-\frac{a_2{a}_3}{a_1}+a_4=0$
$\Rightarrow a_3^2+a_4{a}_1^2=a_1{a}_2{a}_3$
$\Rightarrow \frac{a_3}{a_1{a}_2}+\frac{a_1{a}_4}{a_2{a}_3}=1$ (dividing by $a_1{a}_2{a}_3)$