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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.