If f(x) = ax² + bx + c has no real zeros and a + b + c < 0, then

Given;

f(x) = ax² + bx + c has no real zeroes, and a + b + c<0

Suppose a = – 1,

b = 1,

c = – 1

Then a + b + c = – 1,

b² – 4ac = – 3

Therefore it is possible that c is less tha zero.

Suppose c = 0

Then b² – 4ac = b² ≥ 0

So,

f(x) has at least one zero.

Therefore c cannot equal zero.

Suppose c > 0.

It must also be true that b² ≥0

Then,

b² – 4ac < 0 only if a > 0.

Therefore,

a + b + c < 0.

– b > a + c > 0

b² > (a + c)²

b² > a² + 2ac + c²

b² – 4ac > (a – c)² ≥ 0

As we know that the discriminant can’t be both greater than zero and less than zero,

So, C can’t be greater than zero.