Question

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

• A. $<0$
• B. $>0$
• C. $=0$
• D. Cannot say

Answer 1

Answer: (A)

$<0$

Since $\alpha <-1$ and $\beta >1$
$\alpha +\lambda =-1$ and $\beta =1+\mu \{\lambda ,\mu >0\}$
Now, $1+\frac{c}{a}+\left|\frac{b}{a}\right|=1+\alpha \beta +|\alpha +\beta |$
$=1+(-1-\lambda )(1+\mu )+|-1-\lambda +1+\mu |=1-1-\mu -\lambda -\lambda \mu +|\mu -\lambda |=-\mu -\lambda -\lambda \mu +\mu -\lambda (if\mu >\lambda )\&=-\mu -\lambda -\lambda \mu +\lambda -\mu (if\lambda >\mu )$
$\therefore 1+\frac{c}{a}+\left|\frac{b}{a}\right|=-2\lambda -\lambda \mu or2\mu -\lambda \mu$
in the both the cases $1+\frac{c}{a}+\left|\frac{b}{a}\right|<0$ $(\because \lambda ,\mu >0)$