The smallest integral value of k for which both the roots of x² - 8kx +16(k² - k + 1)= 0 are real and distinct and have value at-least 6.

Given equation is x²-9kx+ 16(k²-k+ 1)=0 Since the roots are real and distinct, so Since the roots are real and distinct , so 64k² - 64(k² - k + 1) > 0

$$\begin{gathered} \Rightarrow k^2-(k^2-k+1)>0 \\ \Rightarrow k-1>0 \\ \Rightarrow k>1 \end{gathered}$$

Thus, k= 2, 3, 4, 5 Hence the smallest integral value of k is 2.