If $\mathrm{\alpha }\text{ and }\mathrm{\beta }(\mathrm{\alpha }<\mathrm{\beta })$  are the roots of the equation ${\mathrm{x}}^2+\mathrm{bx}+\mathrm{c}=0,\mathrm{where} \mathrm{c}<0<\mathrm{b},\mathrm{then}$

Solution:

Since$\alpha , \beta$ are the roots of the equation ${\mathrm{x}}^2+\mathrm{bx}+\mathrm{c}=0.$Here,$\mathrm{D}={\mathrm{b}}^2-4\mathrm{c}>0 \mathrm{because} \mathrm{c}<0<\mathrm{b}.$so, roots are real and unequal. 
Now ,$\mathrm{\alpha }+\mathrm{\beta }=-\mathrm{b}<0\text{ and }\mathrm{\alpha \beta }=\mathrm{c}<0$ 
$\therefore$ One root is positive and the other is negative, then the negative root being numerically bigger. As,$\mathrm{\alpha }<\mathrm{\beta } , \mathrm{\alpha }$ is the negative root while $\beta$is the positive root. so$\left|\mathrm{\alpha }\right|>\mathrm{\beta }$
and  $\mathrm{\alpha }<0<\mathrm{\beta }$