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

Question

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

Dr. Abubacker · Accepted Answer

(a)
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 $af(1) < 0$
[image:5_image12.emf]
$\Rightarrow \ \ a(a - b + c) < 0$ and $a(a + b + c) < 0$
$\Rightarrow \ \ 1 - \frac{b}{a} + \frac{c}{a} < 0$ and $1 + \frac{b}{a} + \frac{c}{a} < 0$ [divide by $a^{2}$]
$\Rightarrow \ \ 1 \pm \frac{b}{a} + \frac{c}{a} < 0\ \ \Rightarrow \ \ 1 + \left| \frac{b}{a} \right| + \frac{c}{a} < 0$

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

Explanation

Reviewed by Dr. Abubacker

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