Question

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

• A. $(-2,1)$
• B. $(-1,0)$
• C. $(0,1)$
• D. $(-2,3)$

Answer 1

Answer: (A), (B), (D)

The correct options are
A $(-2,1)$
B $(-1,0)$
D $(-2,3)$

Let $x$ be a real root.
Then $x^3+(3+2i)x+(-1+ia)=0$
$\Rightarrow (x^3+3x-1)+i(a+2x)=0$
Comparing real and imaginary parts, we get
$a+2x=0$ and $x^3+3x-1=0$
$\Rightarrow {\left(\frac{-a}{2}\right)}^3+3\left(\frac{-a}{2}\right)-1=0$
$\Rightarrow a^3+12a+8=0$

Let $f\left(a\right)=a^3+12a+8$
Since, $f\left(a\right)$ is increasing functions.
So, check the sign of $f\left(a\right)$ at the end points of the given intervals.
$f(-2)=-2<0$
$f(-1)=-5<0$
$f\left(0\right)=8>0$
$f\left(1\right)=21>0$
$f\left(3\right)=71>0$

$\therefore$ By using Intermediate value theorem, possible intervals are $(-2,1),(-1,0),(-2,3)$