If the equations x2+2(k+1)x + 9a-5 has only negative roots, then

Condition for negative root only is that, D≥0sum of roots0a∈(−1,∞)........(2)and third condition, 9a−5>0a∈(59,∞)........(3)Now intersection of 1, 2, 3 will be the final answer so we get, a∈[6,∞)0So here for x2+2(a+1)x+9a−5=04(a+1)2−4×(9a−5)≥04a2+4+8a−36a+20≥0a2−7a+6≥0(a−6)(a−1)≥0a∈(−∞,1]∪[6,∞).......(1)second condition, −2(a+1)