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Introduction

If C > 0 an...

Question

If $C > 0$ and the equation $3 a x^{2} + 4 b x + c = 0$ has no real root, then

2 a + c > b

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a + 2 c > b

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3 a + c > 4 b

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a + 3 c < b

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Solution

The correct option is D $3 a + c > 4 b$
$\begin{matrix} f (x) = 3 a x^{2} + 4 b x + c = 0 f (0) > 0 C l e a r l y, f (x) > 0 \forall x \in R f (- 1) > 0 3 a + c > 4 b H e n c e, o p t i o n C i s c o r r e c t a n s w e r . \end{matrix}$

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Similar questions

a x^{2} + b x + c = 0

3 a x^{2} + (2 a + 3 b) x + b + 3 c = 0

f (x) = 3 a x^{2} - 4 b x + c (a, b, c \in R, a \neq 0)

a, b, c

A . P

f (x) = 0

3 a x^{2} + 2 b x + c = 0

a, b, c

a + b + c = 0

3 a x^{2} + 2 b x + c = 0

2 a + b + 3 c = 1

> 0

a^{4} b^{2} c^{2}

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