Question on location of the solutions of $ax^2+bx+c=0$

Question

Let $a,b$ and $c$ be real numbers. The equation $f(x)=ax^2+bx+c=0$ has solutions $α$ and $β$ such that $α<-1$ and $β>1.$ Prove that $$1+\frac{c}{a}+\frac{|b|}{a} < 0.$$

What I noticed was $\frac{f(-1)}{a}<0$ and $\frac{f(1)}{a}<0.$ Furthermore, $\frac{f(-1)f(1)}{a^2}>0$ which implies $(1+\frac{c}{a})^2>\frac{b^2}{a^2}$ I don't know how to proceed.

Answer 1

Use the identities for the zeros of the quadratic equation \begin{align*} \alpha + \beta &= \frac{-b}{a},\\ \alpha\beta &= \frac{c}{a}. \end{align*} Then if $b<0$ $$ 1+\frac{c}{a}+\frac{|b|}{a} = \left(\alpha +1\right) + \beta\left(\alpha + 1\right) < 0, $$ or if $b>0$ $$ 1+\frac{c}{a}+\frac{|b|}{a} = \left(1-\beta\right) + \alpha\left(\beta- 1\right) < 0. $$

Answer 2

Notice that $f(-1)=a-b+c$ and $f(1)=a+b+c.$ The facts that you already observed, namely, $\frac{f(-1)}{a}<0$ and $\frac{f(1)}{a}<0,$ then imply that $$1+\frac{c}{a}-\frac{b}{a} < 0$$ and $$1+\frac{c}{a}+\frac{b}{a} < 0.$$ Therefore the greater of the two formulas on the left sides of these inequalities is negative.

You can finish by proving and applying the fact that $$ \frac{\lvert b\rvert}{a} \leq \left\lvert \frac{b}{a} \right\rvert \ = \max\left\{-\frac{b}{a}, \frac{b}{a} \right\}. $$