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

If f(x) = ax2 + bx + c has no real zeros and a + b + c 0C. c < 0D. None of these

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If f(x) = ax2 + bx + c has no real zeros and a + b + c 0C. c < 0D. None of these

1.	If f(x) = ax2 + bx + c has no real zeros and a + b + c < 0, thenA. c = 0B. c > 0C. c < 0D. None of these
Answer» 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 b ² > a² + 2ac + c² 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.

Answer»

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

b ² > a² + 2ac + c²

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.