Quadratic EquationHard
Question
The complete set of real values of ' $a$ ' for which the inequality $ax^{2} - (3 + 2a)x + 6 > 0$, $a \neq 0$ holds for exactly three negative integral values of x is
Options
A.$\left\lbrack - \frac{5}{4}, - \frac{3}{4} \right)$
B.$\left\lbrack - 2, - \frac{5}{4} \right\rbrack$
C.$( - 1,0)$
D.$\left( - 1, - \frac{3}{4} \right\rbrack$
Solution
$(x) = ax^{2} - (3 + 2a)x + 6$
$$f( - 3) > 0\ \Rightarrow \ 9a + 9 + 6a + 6 > 0\ \Rightarrow a > - 1 $$and $f( - 4) \leq 0 \Rightarrow 16a + 12 + 8a + 6 \leq 0$
$$\begin{matrix} & a \leq - \frac{3}{4} \\ \Rightarrow \ & a \in \left( - 1, - \frac{3}{4} \right\rbrack \end{matrix}$$
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