🎥 Video solution / Text Solution of this question is given below:
For the quadratic equation $(k-2)x^2 + 8x + (k+4) = 0$ to have real, distinct, and negative roots:
1. Real & distinct roots ($D > 0$):
$D = 8^2 - 4(k-2)(k+4) > 0$
$64 - 4(k^2 + 2k - 8) > 0 \implies k^2 + 2k - 24 < 0$
$(k+6)(k-4) < 0 \implies -6 < k < 4$
2. Both roots negative (Sum < 0 & Product > 0):
Sum$ = -\frac{8}{k-2} < 0 \implies k - 2 > 0 \implies k > 2$
* Product $ = \frac{k+4}{k-2} > 0 \implies k \in (-\infty, -4) \cup (2, \infty)$
Taking the intersection of all conditions:
$k \in (2, 4) \quad \text{or} \quad 2 < k < 4$