Question

Consider the equation $a{z}^2+z+1=0$ having purely imaginary root where $a=\cos \theta +i\sin \theta ,i=\sqrt{-1}$ and function $f\left(x\right)=x^3-3x^2+3(1+\cos \theta )x+5,$ then answer the following questions.

Which of the following is true?

• A. $f\left(x\right)=0$ has three real distinct roots
• B. $f\left(x\right)=0$ has one positive real root
• C. $f\left(x\right)=0$ has one negative real root
• D. $f\left(x\right)=0$ has three but not distinct roots

Answer 1

The correct option is C $f\left(x\right)=0$ has one negative real root

$a{z}^2+z+1=0$ where, $a=\cos \theta +i\sin \theta$
Let $z_1$ & $z_2$ be the roots of the above equation.
Since, $z_1$ & $z_2$ are both purely imaginary.
${\therefore z}_1=-\overline{z_1}\&z_2=-\overline{z_2}$
Sum of the roots of the equation= $z_1+z_2=\frac{-1}{a}$ ...(1)
conjugating equation (1)
$\overline{z_1}+\overline{z_2}=\frac{-1}{\overline{a}}$
$\Rightarrow \frac{1}{a}+\frac{1}{\overline{a}}=0$ ..{${\because z}_1=-\overline{z_1}\&z_2=-\overline{z_2}$}
$\Rightarrow \cos \theta =0\Rightarrow \theta =\frac{\Pi }{2}$

for $f\left(x\right)=0$
${\Rightarrow x}^3-3x^2+3x+5=0$
$\Rightarrow {(x-1)}^3+6=0$
$\Rightarrow x=1-6^{\frac{1}{3}}$
Therefore f(x) has one negative real root.
Ans: C