If $a > 0$ and discriminant of $ax^2 + 2bx + c$ is negative, then $Δ=\begin{vmatrix}a&b&ax+b\\b&c&bx+c\\ax+b&bx+c&0\end{vmatrix}$, is

negative

It is given that the discriminant of $ax^2 + 2bx + c$ is negative.

$∴4b^2-4ac <0⇒b^2-ac <0⇒ ac-b^2>0$  ...(i)

Also, a > 0 and discriminant is negative.

$∴ax^2+2bx + c > 0$ for all $x ∈ R$  ...(ii)

Now,

$Δ=\begin{vmatrix}a&b&ax+b\\b&c&bx+c\\ax+b&bx+c&0\end{vmatrix}$

$⇒Δ=\begin{vmatrix}a&b&0\\b&c&0\\ax+b&bx+c&-x(ax+b)-(bx+c)\end{vmatrix}$

$⇒Δ=\begin{vmatrix}a&b&0\\b&c&0\\ax+b&bx+c&-(ax^2+2bx+c)\end{vmatrix}$

$⇒Δ=-(ax^2+2bx + c) (ac-b^2) <0$  [Using (i) and (ii)]