A) 6
B) 3
C) 4
D) 1
Correct Answer: B
Solution :
\[(K-2){{x}^{2}}+8x+K+4=0\]If real roots then, \[{{8}^{2}}-4(K-2)(K+4)>0\] \[\Rightarrow \]\[{{K}^{2}}+2K-8<16\] If both roots are negative then \[\alpha \beta \]is + ve \[\Rightarrow \]\[\frac{K+4}{K-2}>0\Rightarrow K>-4\] Also,\[\frac{K-2}{K+4}>0\Rightarrow K>2\] Roots are real so, \[-6<K<4\] So, 6 and 4 are not correct. Since, \[K>2,\] sol is also not correct value of K. \[\therefore \]K=3
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