The value of k for which (k−2)x2+8x+k+4=0 has both roots real, distinct and negative, is
Solutions · 1
Official solution
D>0: 64−4(k−2)(k+4)>0⟹(k+6)(k−4)<0⟹−6<k<4.
Both roots negative requires k−2k+4>0 (product positive) and sum negative. Checking candidates within (−6,4), only k=3 satisfies all conditions.