Step-by-step Solution:

Given: \[ (k-2)x^2 + 8x + (k+4) = 0 \] Let: \[ a = k-2,\quad b = 8,\quad c = k+4 \] For real and distinct roots: \[ D = b^2 - 4ac > 0 \] \[ 64 - 4(k-2)(k+4) > 0 \] \[ 64 - 4(k^2 + 2k - 8) > 0 \] \[ 64 - 4k^2 - 8k + 32 > 0 \] \[ 96 - 4k^2 - 8k > 0 \] \[ k^2 + 2k - 24 < 0 \] \[ (k+6)(k-4) < 0 \Rightarrow -6 < k < 4 \] For both roots negative sum of roots < 0: \[ -\frac{b}{a} < 0 \Rightarrow \frac{-8}{k-2} < 0 \Rightarrow k > 2 \] Product of roots > 0: \[ \frac{c}{a} > 0 \Rightarrow \frac{k+4}{k-2} > 0 \Rightarrow k > 2 \] Combining conditions: \[ -6 < k < 4 \quad and \quad k > 2 \] \[ 2 < k < 4 \] Hence, \[ k = 3 \]