For what values of k, the inequality (k- 2)x² + 8x + k + 4 < 0 satisfies for all real values of x. ?

Clearly, k- 2 < 0 k<2 …(I)

Also, D<0

$\begin{array}{l}64-4(k-2)(k+4)<0\\ (k-2)(k+4)-16>0\\ k^2+2k-24>0\\ (k+6)(k-4)>0\\ k<-6\text{ and }k>4\end{array}$

From (i) and (ii), we get, k<-6