Question

$a{x}^2+bx+c=0$ has real and distinct roots $\alpha$ and $\beta$ $(\beta >\alpha )$. Further $a>0,b<0$ and $c<0$, then

• A. $0<\beta <|\alpha |$
• B. $0<|\alpha |<\beta$
• C. $\alpha +\beta <0$
• D. $|\alpha |+|\beta |=\left|\frac{b}{a}\right|$

Answer 1

The correct option is B $0<|\alpha |<\beta$

$a{x}^2+bx+c=0$ has real and distinct roots $\alpha$ and

$\beta (\beta >\alpha )$. Further $a>0,b<0$ and $c<0$.

$\Rightarrow \alpha +\beta =\frac{-b}{a}>0$ as $a>0,b<0$.

$\Rightarrow \alpha \beta =\frac{c}{a}<0$ as $a>0,c<0$.

Roots are of opposite sign and $\beta >0$ and $\alpha <0$.
and $|\alpha |<\beta$ as $\alpha +\beta >0$.

$\therefore 0<|\alpha |<\beta$.

Hence, option B.