First slide

Theory of equations

Question

If $a x^{2} + 2 b x - 3 c = 0$ has no real roots and $c < \frac{4}{3} (a + b)$ then range of c is

Moderate

A

$(0, b)$
B

$(- 1, b)$
C

$(- \infty, - b^{2} / 3 a)$
D

$(- \infty, - b / 12 a)$

Solution

As $f (x) = a x^{2} + 2 b x - 3 c = 0$ has no real roots, $f (x) > 0 \forall x \in R$ or $f (x) < 0 \forall x \in R$
Since $4 a + 4 b - 3 c > 0, f (2) > 0$ .
$\begin{array}{ll} ∴ f (x) = a x^{2} + 2 b x - 3 c > 0 \\ \Rightarrow f (0) = - 3 c > 0 \Rightarrow c < 0 \end{array}$
Also, $a > 0$ and $b^{2} + 3 c a < 0$
$\Rightarrow$ $c < - b^{2} / 3 a \Rightarrow c \in (- \infty, - b^{2} / 3 a)$

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