If the quadratic equation {x^2} + left( {2 - | vedclass

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Question

Let $\alpha$ and $\beta$ be the roots of

$x^{2}+(2-	an 	heta) x-$ $(1+	an 	heta)=0$

$\alpha+\beta=	an 	heta-2 $

Vedclass · Accepted Answer

$4$

Vedclass · Answer

Let $\alpha$ and $\beta$ be the roots of

$x^{2}+(2-	an 	heta) x-$ $(1+	an 	heta)=0$

$\alpha+\beta=	an 	heta-2 $           .........$(1)$

$\alpha \beta=-	an 	heta-1$           ........$(2)$

$(1)+(2) \Rightarrow \alpha+\beta+\alpha \beta=-3 \Rightarrow(\alpha+1)(\beta+1)=-2$

Hence either $\alpha+1=-2 $ and $ \beta+1=1$ or

$\alpha+1=-1 $ and $ \beta+1=2$

i.e. either $\alpha=-3 $ and $ \beta=0$ or $\alpha=-2 $ and $ \beta=1$

i.e. either $	an 	heta=-1$ or $	an 	heta=1$

i.e. either $	heta=\frac{3 \pi}{4}, \frac{7 \pi}{4}$ or $	heta=\frac{\pi}{4}, \frac{5 \pi}{4}$

Hence sum of all possible values of $	heta$ is

$\frac{\pi}{4}+\frac{3 \pi}{4}+\frac{5 \pi}{4}+\frac{7 \pi}{4}=\frac{16 \pi}{4}=4 \pi$

If the quadratic equation ${x^2} + \left( {2 - \tan \theta } \right)x - \left( {1 + \tan \theta } \right) = 0$ has $2$ integral roots, then sum of all possible values of $\theta $ in interval $(0, 2\pi )$ is $k\pi $, then $k$ equals

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