Question

Let $a,b,c$ be real. If $a{x}^2+bx+c=0$ has two real roots $\alpha$ and $\beta$ , where $\alpha <-1$ and $\beta >1$ , then show that $1+\frac{c}{a}+\left|\frac{b}{a}\right|<0$

Answer 1

From figure, it is clear that, if $a>0$, then $f(-1)<0$ and $f(1)<0$ and if $a<0,f(-1)>0$ and $f(1)>0$. In both cases, $af(-1)<0$ and $af(1)<0$.
$\Rightarrow a(a-b+c)<0$ and $a(a+b+c)<0$
On dividing by $a^2$, we get
$1-\frac{b}{a}+\frac{c}{a}<0$ and $1+\frac{b}{a}+\frac{c}{a}<0$
On combining both, we get
$1\pm \frac{b}{a}+\frac{c}{a}<0\Rightarrow 1+\left|\frac{b}{a}\right|+\frac{c}{a}<0$