Question

Find all the value of ${}^{\prime }{a}^{\prime }$ for which both the roots of the equation $(a-2)x^2+2ax+(a+3)=0$ lie in the interval $(-2,1).$

Answer 1

Case I
When $a-2>0$
$\Rightarrow$ $a>2$
Condition I: $f(-2)>0$
$\Rightarrow$ $(a-2)4-4a+a+3>0$
$\Rightarrow$ $a-5>0$
$\Rightarrow$ $a>5$
Condition II: $f\left(1\right)>0$
$\Rightarrow$ $4a+1>0$
$\Rightarrow$ $a>-\frac{1}{4}$
Condition III: $D\ge 0$
$\Rightarrow$ $4a^2-4(a+3)(a-2)\ge 0$
$\Rightarrow$ $a\le 6$
Condition IV: $-2<-\frac{b}{2a}<1$
$\Rightarrow$ $aϵ(-\infty ,1)\cup (4,\infty )$
intersection gives $aϵ(5,6]$
Case II
When $a-2<0$
$a<2$
Condition I: $f(-2)<0$
$\Rightarrow$ $a<5$
Condition II: $f\left(1\right)<0$
$\Rightarrow$ $a<-\frac{1}{4}$
Condition III: $-2<-\frac{b}{2a}<1$
$\Rightarrow$ $aϵ(-\infty ,1)\cup (4,\infty )$
Condition IV: $D\ge 0$
$\Rightarrow$ $a\le 6$
intersection gives $aϵ(-\infty ,-\frac{1}{4}]$
$\therefore$ Complete solution is $aϵ(-\infty ,-\frac{1}{4})\cup (5,6]$