While doing a certain sum I got stuck at a step.I am getting $(a+b+c)(a-b+c)>0$.I need to prove $(1+c/a+|b/a|)<0$.Is it possible?How?
a is not 0.
The original question : If $ax^2+bx+c=0$ has two real roots $\alpha$ and $\beta$ where $\alpha<-1$ and $\beta>1$ then show that $1+c/a+|b/a|<0$
The implication you wish to prove is incorrect:
$(a+b+c)(a-b+c)>0 \Rightarrow 1+\frac{c}{a}+|\frac{b}{a}|<0$ does not hold for $a=1$, $b=2$, $c=3$
In the context of the question it is correct, but then I may as well start from scratch:
$(x-\alpha)(x-\beta)=x^2+\frac{b}{a}x+\frac{c}{a}$
$\Rightarrow \alpha\beta=\frac{c}{a}<-1$ and $\alpha+\beta=-\frac{b}{a}$
Now let $\alpha=-1-\delta$ and $\beta=1+\epsilon$ where $\delta, \epsilon>0$
$1+\frac{c}{a}=1-(1+\delta)(1+\epsilon)=-(\delta+\epsilon+\delta\epsilon)$
and $|\frac{b}{a}|=-|\epsilon-\delta|=\min(\epsilon-\delta, \delta-\epsilon)$
Now it is obvious that $1+\frac{c}{a}+|\frac{b}{a}|<0$