If a >0 and discriminant of ax² + 2bx + c is negative, then $\mathrm{\Delta }=\left|\begin{array}{ccc}a& b& ax+b\\ b& c& bx+c\\ ax+b& bx+c& 0\end{array}\right|$ is

Here a > 0 and $4b^2-4ac<0\text{,i.e., }ac-b^2>0$
$\therefore a{x}^2+2bx+c>0,\mathrm{\forall }x\in R;$Now

$\mathrm{\Delta }=\left|\begin{array}{ccc}a& b& ax+b\\ b& c& bx+c\\ ax+b& bx+c& 0\end{array}\right|$

$\left[{\mathrm{R}}_3\to {\mathrm{R}}_3-{\mathrm{xR}}_1-{\mathrm{R}}_2\right]$
$\begin{array}{c}\mathrm{\Delta }=\left|\begin{array}{ccc}\mathrm{a}& \mathrm{b}& \mathrm{ax}+\mathrm{b}\\ \mathrm{b}& \mathrm{c}& \mathrm{bx}+\mathrm{c}\\ 0& 0& -\left({\mathrm{ax}}^2+2\mathrm{bx}+\mathrm{c}\right)\end{array}\right|\\ Expanding determinant along R_3\\ =-\left({\mathrm{ax}}^2+2\mathrm{bx}+\mathrm{c}\right)\left(\mathrm{ac}-{\mathrm{b}}^2\right)=-(+\mathrm{ve})(+\mathrm{ve})=-\mathrm{ve}\end{array}$