JEE Main 2015 — Complex Number Question with Solution

If $2+3i$ in one of the roots, then $2-3i$ would be other.
Since coefficients of the equation are real.

Let $\gamma$ be the third root, then product of roots $\to \alpha \beta \gamma =\frac{13}{2}$
$\left(2+3i\right) \left(2-3i\right) \cdot \gamma =\frac{13}{2}$
$\left(4+9\right)\cdot \gamma =\frac{13}{2}$
$\Rightarrow \gamma =\frac{1}{2}$

The value of k will come if we put $x=\frac{1}{2}$ in the equation
$2\cdot \frac{1}{8}-\frac{9}{4}+k\cdot \frac{1}{2}-13=0$
$\Rightarrow \frac{k}{2}=15$
$\Rightarrow k=30$
$\therefore$ Equation will become $2x^3-9x^2+30x-13=0$
$\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =\frac{30}{2}=15$
$\Rightarrow \left(2+3i\right)\frac{1}{2}+\left(2-3i\right)\frac{1}{2}+\left(2+3i\right) \left(2-3i\right)=15$
$\Rightarrow 1+\frac{i}{2}+1-\frac{i}{2}+13=15$
$\Rightarrow 15=15$
Hence roots exists and is equal to $\frac{1}{2}$ '