2014/017) show that if x^3 + ax +b=0 has 3 real roots then a < 0

If a = 0 then we have x^3 + b = 0 has 1 real roots

If a > 0 then

f(x) changes sign zero times for b >= 0 and once for b < 0

So as per Descartes' Rule of Signs there is zero positive roots for b >=0 and one for b < 0

F(-x) = - x^3 – ax + b and

f(-x) changes sign zero times for b <= 0 and once for b> 0

So as per Descartes' Rule of Signs there is zero positive roots for b<=0 and one for b > 0

So combining the two there cannot be more than one root

So if it has 3 roots then a < 0