(k−2)x2+8x+k+4=0 If real roots then, 82−4(k−2)(k+4)>0 ⇒k2+2k−8<16 ⇒k2+6k−4k−24<0 ⇒(k+6)(k−4)<0 ⇒−6<k<4 If both roots are negative then αβ is + ve ⇒
k+4
k−2
>0⇒k>−4 Also,
k−2
k+4
>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