Question

The number of positive integral values of $a$ for which the root(s) of the equation $(a-2)x^2+2ax+(a+3)=0$ lies in $(-2,1)$ is

Answer 1

$(a-2)x^2+2ax+(a+3)=0$
Case $1:a-2\ne 0\Rightarrow a\ne 2$
Now,
$x^2+\frac{2a}{(a-2)}x+\frac{a+3}{a-2}=0$
Conditions
$\left(i\right)D\ge 0\Rightarrow \frac{4a^2}{(a-2{)}^2}-\frac{4a+12}{a-2}\ge 0\Rightarrow \frac{4a^2-4a^2-12a+8a+24}{(a-2{)}^2}\ge 0\Rightarrow \frac{a-6}{(a-2{)}^2}\le 0\Rightarrow a\in (-\infty ,6]-\left\{2\right\}\cdots \left(1\right)\left(ii\right)f(-2)>0\Rightarrow 4-\frac{4a}{a-2}+\frac{a+3}{a-2}>0\Rightarrow \frac{4a-8-4a+a+3}{a-2}>0\Rightarrow \frac{a-5}{a-2}>0\Rightarrow a<2,a>5\cdots \left(2\right)\left(iii\right)f\left(1\right)>0\Rightarrow 1+\frac{2a}{a-2}+\frac{a+3}{a-2}>0\Rightarrow \frac{a-2+2a+a+3}{a-2}>0\Rightarrow \frac{4a+1}{a-2}>0\Rightarrow a<-\frac{1}{4},a>2\cdots \left(3\right)\left(iv\right)-2<-\frac{b}{2a}<1\Rightarrow -2<-\frac{a}{a-2}<1\Rightarrow 2>\frac{a}{a-2}>-12>\frac{a}{a-2}\Rightarrow \frac{a-4}{a-2}>0\Rightarrow a<2,a>4\cdots \left(4\right)\frac{a}{a-2}>-1\Rightarrow \frac{2(a-1)}{a-2}>0\Rightarrow a<1,a>2\cdots \left(5\right)$

$\left(4\right)\cap \left(5\right)$$\Rightarrow a\in (-\infty ,1)\cup (4,\infty )\cdots \left(6\right)$

Case $2:a-2=0\Rightarrow a=2$
$0+4x+5=0\Rightarrow x=-\frac{5}{4}$
Which lies in $(-2,1)$

Therefore, from equations $\left(1\right),\left(2\right),\left(3\right),\left(6\right)$ and case $2$
$a\in (-\infty ,-\frac{1}{4})\cup \left\{2\right\}\cup (5,6]$
Hence, the positive integral values of $a$ are $2,6$.