If k is a root of x^2 - 4x + 1 = 0 , then what is ^ -1 k + ^ -1...

The given equation is $x^2 - 4x + 1 = 0$.

The roots of the equation are given by $x = \dfrac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3}$.

Since $\sqrt{3} \approx 1.732$, both roots $2 + \sqrt{3}$ and $2 - \sqrt{3}$ are positive. Therefore, $k > 0$.

Using the standard inverse trigonometric identity for $x > 0$:

$\tan^{-1}x + \tan^{-1}\left(\dfrac{1}{x}\right) = \dfrac{\pi}{2}$

Substituting $x = k$, we get:

$\tan^{-1}k + \tan^{-1}\left(\dfrac{1}{k}\right) = \dfrac{\pi}{2}$

Answer: $\dfrac{\pi}{2}$