Answer :

Given;

f(x) = ax^{2} + bx + c has no real zeroes, and a + b + c<0

Suppose a = – 1,

b = 1,

c = – 1

Then a + b + c = – 1,

b^{2} – 4ac = – 3

Therefore it is possible that c is less tha zero.

Suppose c = 0

Then b^{2} – 4ac = b^{2} ≥ 0

So,

f(x) has at least one zero.

Therefore c cannot equal zero.

Suppose c > 0.

It must also be true that b^{2} ≥0

Then,

b^{2} – 4ac < 0 only if a > 0.

Therefore,

a + b + c < 0.

– b > a + c > 0

b^{2} > (a + c)^{2}

b^{2} > a^{2} + 2ac + c^{2}

b^{2} – 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.

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