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 $f( - 4) \leq 0 \Rightarrow 16a + 12 + 8a + 6 \leq 0$
$$\begin{matrix} & a \leq - \frac{3}{4} \\ \Rightarrow \ & a \in \left( - 1, - \frac{3}{4} \right\rbrack \end{matrix}$$
Create a free account to view solution
View Solution FreeMore Quadratic Equation Questions
If x = then -...How many roots does the equation $x^{3} + ax + 2 = 0$ possess depending on ' $a$ '?...Given that ax2 + bx + c = 0 has no real roots and a + b + c < 0, then -...The possible set of real values of ' $a$ ' for which, for all $x$ not exceeding unity in absolute value, the inequality ...If α, β are the roots of ax2 + bx + c = 0; α + h, β + h are the roots of px2 + qx + r = 0, and D1, D...