Aspire Faculty ID #10207 · Topic: NIMCET 2015 · Just now
NIMCET 2015

The value of k for which the equation $(k-2)x^{2}+8x+k+4=0$ has both real, distinct and negative roots is

Solution

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: $k \in (2, 4) \quad \text{or} \quad 2 < k < 4$
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