Mathematics - Quadratic Equation Question with Solution | TestHub

The given equation is a quadratic in $\tan x$. For real roots, the discriminant must be non-negative.

Let $y=\tan x$. The equation is $(1+K)y^2-4y-1+K=0$.

Discriminant $\Delta =(-4{)}^2-4(1+K)(-1+K)\ge 0$.

$16-4(K^2-1)\ge 0$

$16-4K^2+4\ge 0$

$20-4K^2\ge 0$

$4K^2\le 20$

$K^2\le 5$.