Question

If $z^3+(3+2i)z+(-1+ia)=0$ has one real root, then the value of the $a$ lies in the interval $aϵR$

• A. $(-2,-1)$
• B. $(-1,0)$
• C. $(0,1)$
• D. $(1,2)$

Answer 1

The correct option is B $(-1,0)$

Let $z=\alpha$ is the real root of the equation
${\alpha }^3+(3+2i)\alpha +(-1+ia)=0$
$\Rightarrow ({\alpha }^3+3\alpha -1)+(2\alpha +a)i=0$
As $\alpha$ is a real root of the equation, the real & the imaginary part of the equation must be zero.
$\therefore a+2\alpha =0\Rightarrow \alpha =-\frac{a}{2}$
so using value of $\alpha$ we get, $-\frac{a^3}{8}-\frac{3a}{2}-1=0\Rightarrow a^3+12a+8=0$

Now, Let $f\left(x\right)=a^3+12a+8$
$f^{\prime }\left(a\right)=3a^2+12>0\forall a\in R$
$\therefore f$ is an increasing function$\forall x\in R$
$f(-2)=(-2{)}^3+12(-2)+8<0$
$f(-1)=(-1{)}^3+12(-1)+8<0$
$f\left(0\right)=(0{)}^3+12(0)+8>0$
$f\left(1\right)=(1{)}^3+12(1)+8>0$
$f\left(2\right)=(2{)}^3+12(2)+8>0$

$\therefore a$ lies in the interval $(-1,0)$