Question

If set of values of ' $a$ ' for which $f\left(x\right)=a{x}^2-(3+2a)x+6$ $a\ne 0$ is positive for exactly three distinct negative integral values of $x$ is $(c,d],$ then find the value of $(c^2+4|d|)$.

Answer 1

$f\left(x\right)=a{x}^2-(3+2a)x+6$

$=(ax-3)(x-2)$

Here, roots of the equation $f\left(x\right)=0$ are 2 and $\frac{3}{a},$ and $f\left(0\right)=6$.

$f\left(x\right)$ should be positive for exactly three negative integral values of $x$

Therefore, graph of $f\left(x\right)$ must be a downward parabola passing through $x=2$ and $x=3/a$ and $-4\le \frac{3}{a}<-3$

$\therefore a\in (-1,-\frac{3}{4}]$

$\therefore c=-1,d=-\frac{3}{4}$

$\Rightarrow c^2+4|d|=1+3=4$