Let $ax^2+bx+c$ = 0 be a quadratic equation and $\alpha$,$\beta$ are real roots. Condition for $\alpha < -1$ and $\beta > 1$. Show that $1 +\frac{c}{a}$ + $\left|\frac{b}{a}\right| < 0$.

I have tried but could not prove this inequality. I want to know how to solve this. Show that I can get an idea and solve further questions myself.

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    $\begingroup$ If you look at $f(x) = x^2 + \frac{b}{a}x + \frac{c}{a}$, what you want to show is that $f(1) < 0$ and $f(-1) < 0$. $\endgroup$ Jul 9, 2015 at 7:56
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    $\begingroup$ And then by writing $f(x) = (x-\alpha)(x-\beta)$, when $\alpha<x<\beta$, the two factors of $f(x)$ have different signs. $\endgroup$
    – peterwhy
    Jul 9, 2015 at 8:03

2 Answers 2


Note that $-b/a=\alpha+\beta$ and $c/a=\alpha\beta$.

Furthermore, $(\alpha+1)(\alpha-1)(\beta+1)(\beta-1)>0 \\\implies (\alpha\beta)^2-2\beta\sqrt{\alpha^2}+1>\alpha^2+2\alpha\beta+\beta^2 \\ \implies|\alpha\beta|>|\alpha+\beta|+1.$


Without loss of generality, let $a \gt 0$.

Now firstly, Since $\alpha$ is a root of $f(x)$, hence $a\alpha^2+b\alpha+c=0$. $\Rightarrow$ $\alpha = \frac {-b - \sqrt{b^2-4ac}}{2a} \lt -1$ $\Rightarrow$ $-b - \sqrt{b^2-4ac} \lt -2a$ $\Rightarrow$ $\sqrt{b^2-4ac} \gt 2a-b$ $\Rightarrow$ $b^2-4ac \gt 4a^2-4ab+b^2$ $\Rightarrow$ $a^2+ab-ac \lt 0$ $\Rightarrow$ $1-\frac ba+\frac ca \lt 0$.

Similarly, $\beta=\frac {-b+\sqrt {b^2-4ac}}{2a} \gt 1$ $\Rightarrow$ $\sqrt {b^2-4ac} \gt b+2a$ $\Rightarrow$ $b^2-4ac \gt b^2+4ab+4a^2$ $\Rightarrow$ $a^2+ab+ac \lt 0$ $\Rightarrow$ $ 1+\frac ba+\frac ca \lt 0$.

Hence, $1+\frac ca+|\frac ba| \lt 0$.


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