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(B) Given the quadratic equation $x^2-8kx+16(k^2-k+1)=0$.For the roots to be real and distinct,the discriminant $D > 0$:$D = (-8k)^2 - 4(1)(16(k^2-k+1

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$2$

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(B) Given the quadratic equation $x^2-8kx+16(k^2-k+1)=0$.For the roots to be real and distinct,the discriminant $D > 0$:$D = (-8k)^2 - 4(1)(16(k^2-k+1)) > 0$$64k^2 - 64(k^2-k+1) > 0$$64k^2 - 64k^2 + 64k - 64 > 0$$64k - 64 > 0 \Rightarrow k > 1 \cdots (1)$For both roots to be at least $4$,the vertex of the parabola $-\frac{b}{2a} \geq 4$:$-\frac{-8k}{2(1)} \geq 4$ $\Rightarrow 4k \geq 4$ $\Rightarrow k \geq 1 \cdots (2)$Additionally,$f(4) \geq 0$:$f(4) = (4)^2 - 8k(4) + 16(k^2-k+1) \geq 0$$16 - 32k + 16k^2 - 16k + 16 \geq 0$$16k^2 - 48k + 32 \geq 0$Dividing by $16$:$k^2 - 3k + 2 \geq 0$$(k-1)(k-2) \geq 0 \Rightarrow k \leq 1 	ext{ or } k \geq 2 \cdots (3)$Combining $(1)$,$(2)$,and $(3)$:$k > 1$ $AND$ $k \geq 1$ $AND$ $(k \leq 1 	ext{ or } k \geq 2)$This results in $k \geq 2$.Therefore,the smallest value of $k$ is $2$.

The smallest value of $k$,for which both the roots of the equation $x^2-8kx+16(k^2-k+1)=0$ are real,distinct and have values at least $4$,is

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