2x2+kx+5=0 has no real roots so D<0 k2−40<0 (k−√40)(k+√40)<0 k∈(−√40,√40) x2+(k−5)x+1=0 has two distinct real roots so D>0 (k−5)2−4>0 k2−10k+21>0 (k−3)(k−7)>0 k∈(−∞,3)∪(7,∞) Therefore possible value of k are −6,−5,−4,−3,−2,−1,0,1,2 In 9 total 9 integer values of k are possible.