If 2+3i is one of the roots of the equation 2x^3-9x^2+kx-13=0, k∈ R, then the real root of this equation (where i^2=-1) :

If $2+3i$2+3i in one of the roots, then $2-3i$2−3i would be other.

Since coefficients of the equation are real.

Let $\gamma$γ be the third root, then product of roots $\rightarrow \alpha \beta \gamma =\frac{13}{2}$→αβγ=213​

$$(2+3i) (2-3i) \cdot \gamma =\frac{13}{2}$$(2+3i)(2−3i)⋅γ=213​

$$(4+9)\cdot \gamma =\frac{13}{2}$$(4+9)⋅γ=213​

$$\Rightarrow \gamma =\frac{1}{2}$$⇒γ=21​

The value of k will come if we put $x=\frac{1}{2}$x=21​ in the equation

$$2\cdot \frac{1}{8}-\frac{9}{4}+k\cdot \frac{1}{2}-13=0$$2⋅81​−49​+k⋅21​−13=0

$$\Rightarrow \frac{k}{2}=15$$⇒2k​=15

$$\Rightarrow k=30$$⇒k=30

$\therefore$∴ Equation will become $2{x}^{3}-9{x}^{2}+30x-13=0$2x3−9x2+30x−13=0

$$\Rightarrow \alpha \beta +\beta \gamma +\gamma \alpha =\frac{30}{2}=15$$⇒αβ+βγ+γα=230​=15

$$\Rightarrow (2+3i)\frac{1}{2}+(2-3i)\frac{1}{2}+(2+3i) (2-3i)=15$$⇒(2+3i)21​+(2−3i)21​+(2+3i)(2−3i)=15

$$\Rightarrow 1+\frac{i}{2}+1-\frac{i}{2}+13=15$$⇒1+2i​+1−2i​+13=15

$$\Rightarrow 15=15$$⇒15=15

Hence roots exists and is equal to $\frac{1}{2}$21​ '