Smallest Value of Parameter for Roots of Quadratic Equation

To find the smallest value of $k$ for which both roots of the quadratic equation
$f(x) = x^2 - 8kx + 16(k^2 - k + 1) = 0$
are real, distinct, and have values at least $4$, we use the conditions for the location of roots of a quadratic equation.

Let the roots of the equation be $\alpha$ and $\beta$, with $\alpha \neq \beta$ and $\alpha, \beta \ge 4$.

For the quadratic function $f(x) = ax^2 + bx + c$ with $a = 1 > 0$:

1. Discriminant condition ($D > 0$):
   The roots must be real and distinct, so the discriminant must be strictly positive:
   $D = (-8k)^2 - 4(1)\left[16(k^2 - k + 1)\right] > 0$
   $64k^2 - 64(k^2 - k + 1) > 0$
   $64(k - 1) > 0 \implies k > 1$

2. Vertex position condition ($-\frac{b}{2a} > 4$):
   Since both roots are at least $4$ and are distinct, the $x$-coordinate of the vertex of the parabola (which is the average of the two roots) must be strictly greater than $4$:
   $-\frac{b}{2a} = \frac{8k}{2} = 4k > 4 \implies k > 1$

3. Function value condition ($f(4) \ge 0$):
   Since both roots lie in $[4, \infty)$ and the parabola opens upwards ($a = 1 > 0$), the value of the function at $x = 4$ must be non-negative:
   $f(4) = 4^2 - 8k(4) + 16(k^2 - k + 1) \ge 0$
   $16 - 32k + 16k^2 - 16k + 16 \ge 0$
   $16k^2 - 48k + 32 \ge 0$
   Dividing through by $16$:
   $k^2 - 3k + 2 \ge 0$
   $(k - 1)(k - 2) \ge 0$
   This inequality yields:
   $k \le 1 \quad \text{or} \quad k \ge 2$

Intersection of the conditions:
Taking the intersection of all conditions:

1. $k > 1$
2. $k \le 1 \text{ or } k \ge 2$

We get:
$k \in [2, \infty)$

Therefore, the smallest value of $k$ is $2$.