The correct option is option (d) $\frac{-6}{5}$
Since 2 is a zero of $k x^{2}+3 x+k$, we have:
$\mathrm{k} \times(2)^{2}+3(2)+\mathrm{k}=0$
$\Rightarrow 4 \mathrm{k}+\mathrm{k}+6=0$
$\Rightarrow 5 \mathrm{k}=-6$
$\Rightarrow \mathrm{k}=\frac{-6}{5}$
If one zero of the quadratic polynomial $\mathrm{kx}^{2}+3 \mathrm{x}+\mathrm{k}$ is 2 , then the value of $\mathrm{k}$ is (a) $\frac{5}{6}$ (b) $\frac{-5}{6}$ (c) $\frac{6}{5}$ (d) $\frac{-6}{5}$
If one zero of the quadratic polynomial $\mathrm{kx}^{2}+3 \mathrm{x}+\mathrm{k}$ is 2 , then the value of $\mathrm{k}$ is (a) $\frac{5}{6}$ (b) $\frac{-5}{6}$ (c) $\frac{6}{5}$ (d) $\frac{-6}{5}$