The adjoining graph of $\mathrm{y}={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}$ shows that 

It is clear from graph that the equation $\mathrm{y}={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}=0$ has two real and district roots. Therefore,  
                     ${\mathrm{b}}^2-4\mathrm{ac}>0 ....\left(\mathrm{i}\right)$
$\because$Parabola open downwards.
$\therefore \mathrm{a}<0$
$\text{ and }\mathrm{y}={\mathrm{ax}}^2+\mathrm{bx}+\mathrm{c}\text{ cuts-oft }\mathrm{Y}\text{-axis at, }\mathrm{x}=0$
$\begin{array}{cc}\therefore & \mathrm{y}=\mathrm{c}<0\\ \Rightarrow & \mathrm{c}<0\end{array}$
$\text{ and }\mathrm{x}\text{-coordinate of vertex }>0$
$\begin{array}{l}\Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}-\frac{\mathrm{b}}{2\mathrm{a}}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}>0\Rightarrow \frac{\mathrm{b}}{\mathrm{a}}<0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{b}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}>0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em} \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}[\because \mathrm{a}<0]\end{array}$
It is clear that a and b are of opposite signs.