(K−2)x2+8x+K+4=0 If real roots then, ⇒⇒⇒⇒82−4(K−2)(K+4)>0K2+2K−8<16K2+6K−4K−24<0(K+6)(K−4)<0−6<K<4 If both roots are negative then αβ is +ve ⇒K−2K+4>0⇒K>−4 Also, K+4K−2>0⇒K>2 Roots are real so, −6<K<4 So, 6 and 4 are not correct. Since, K>2, so 1 is also not correct value of K. ∴K=3