For quadratic $ax^2+bx+c>0$ for all $x$: $a>0$ → $k-2>0$ → $k>2$ Discriminant $<0$ $D=b^2-4ac=8^2-4(k-2)(k+4)$ Compute: $D=64-4(k^2+2k-8)=64-4k^2-8k+32$ $D=96-4k^2-8k<0$ Divide by $-4$: $k^2+2k-24>0$ $(k+?)(k+?)$ → roots $4$ and $-6$ So $k>4$ or $k<-6$ Combine with $k>2$ ⇒ $k>4$ Least integer = $5$