If exactly one of the roots of the equation x² + (k + 3) x + k = 0 lies in [1, 3] then the minimum value of $\frac{1-{\mathrm{k}}^2}{\mathrm{k}}$ is

The given equation is $x^2+\left(k+3\right)x+k=0$

Given that there exist only one root between $1,3$

Hence, $f\left(1\right)·f\left(3\right)<0$

It implies that 

      $\begin{array}{rcl}\left[1+k+3+k\right]\left[9+3k+9+k\right]& <& 0\\ \left(k+2\right)\left(2k+9\right)& <& 0\\ k& \in & \left(-\frac{9}{2},-2\right)\end{array}$

The minimum value of $\frac{1-{\mathrm{k}}^2}{\mathrm{k}}$ obtained at $k=-2$

it implies that the required value is $\frac{1-4}{-2}=\frac{3}{2}=1.5$