Q.
If the quadratic equation x2+(2−tanθ)x−(1+tanθ)=0 has two integral roots and sum of all possible values of θ in the interval (0,2π) is kπ, then the value of k is equal to
x2+(2−tanθ)x−(1+tanθ)=0 α+β=tanθ−2 αβ=−tanθ−1
From equation (1) +(2) α+β+αβ=−3 (α+1)(β+1)=−2
Hence either α+1=−2 and β+1=1
then α=−3,β=0
or α+1=−1 and β+1=2 α=−2 and β=1
If α=−3,β=0 then tanθ=−1 ⇒α=43π,47π
If α=−2,β=1 then tanθ=1 ⇒θ=4π,45π
Sum =43π+47π+4π+45π=4π=4