Question

If both the roots of the equation $x^2+ax+2=0$ lie in the interval $(0,3)$, then the number of possible integral values of $a$ is

Answer 1

Let $f\left(x\right)=x^2+ax+2$
$\Rightarrow \Delta \ge 0\Rightarrow a\in (-\infty ,-2\sqrt{2}]\cup [2\sqrt{2},\infty )...\left(1\right)$

$\Rightarrow 0<\frac{\text{sum of roots}}{2}<3\Rightarrow 0<-\frac{a}{2}<3\Rightarrow a\in (-6,0)...\left(2\right)$

$\Rightarrow f\left(0\right)>0\&f\left(3\right)>0\Rightarrow a\in (-\frac{11}{3},\infty )...\left(3\right)$

From $\left(1\right),\left(2\right),\&\left(3\right)$
$a\in (-\frac{11}{3},-2\sqrt{2}]$
$\Rightarrow a=-3$
Only one integral value of $a$.