If the quadratic equation $x^2+(2-\tan \theta )x-(1+\tan \theta )=0$ has two integral roots, Then sum of all possible values of $\theta$ in interval $(0,2\pi )$ is $k\pi .$ Then the value of k is.

$\alpha +\beta =\tan \theta -\mathrm{2................}\left(1\right)$

$\alpha \beta =-\tan \theta -1\mathrm{................}\left(2\right)$

$\left(1\right)+\left(2\right)\Rightarrow \alpha +\beta +\alpha \beta =-3\Rightarrow (\alpha +1)(\beta +1)=-2$

Hence either $\alpha +1=-2and\beta +1=1$ or $\alpha +1=-1and\beta +1=2$

i.e., either $\alpha =-3and\beta =0$ or $\alpha =-2and\beta =1$

if $\alpha =-3and\beta =0\Rightarrow \tan \theta =-1\Rightarrow \theta =\frac{3\pi }{4},\frac{7\pi }{4}$

if $\alpha =-2and\beta =1\Rightarrow \tan \theta =1\Rightarrow \theta =\frac{\pi }{4},\frac{5\pi }{4}$

Hence sum of all possible values of $\theta$ is 0 $\frac{\pi }{4}+\frac{3\pi }{4}+\frac{5\pi }{4}+\frac{7\pi }{4}=\frac{16\pi }{4}=4\pi$

Hence k = 4 Ans.]