smc

Expanding, we have $2k{x}^2-8x-x^2+6=0$, or $(2k-1)x^2-8x+6=0$. For this quadratic not to have real roots, it must have a negative discriminant. Therefore, $(-8{)}^2-4(2k-1)(6)<0⟹64-48k+24<0⟹k>\frac{11}{6}$. From here, we can easily see that the smallest integral value of $k$ is $2,\boxed{2}$.