Question

If the quadratic equation $x^2+(2-\tan \theta) x-(1+\tan \theta)=0$ has two integral roots and sum of all possible values of $\theta$ in the interval $(0,2 \pi)$ is $k \pi$, then the value of $\mathrm{k}$ is equal to

Answer 1

$ x^2+(2-\tan \theta) x-(1+\tan \theta)=0 $
$ \alpha+\beta=\tan \theta-2 $
$ \alpha \beta=-\tan \theta-1$
From equation (1) $+(2)$
$ \alpha+\beta+\alpha \beta=-3 $
$ (\alpha+1)(\beta+1)=-2$
Hence either $\alpha+1=-2$ and $\beta+1=1$
then $\alpha=-3, \beta=0$
or $\alpha+1=-1$ and $\beta+1=2$
$\alpha=-2$ and $\beta=1$
If $\alpha=-3, \beta=0$ then $\tan \theta=-1$
$\Rightarrow \alpha=\frac{3 \pi}{4}, \frac{7 \pi}{4}$
If $\alpha=-2, \beta=1$ then $\tan \theta=1$
$\Rightarrow \theta=\frac{\pi}{4}, \frac{5 \pi}{4}$
Sum $=\frac{3 \pi}{4}+\frac{7 \pi}{4}+\frac{\pi}{4}+\frac{5 \pi}{4}=4 \pi=4$