AIEEE 2002 (Offline)Question 1
If $a>0$ and discriminant of $ax^{2}+2bx+c$ is $-ve$, then $\begin{vmatrix}a&b&ax+b\\b&c&bx+c\\ax+b&bx+c&0\end{vmatrix}$ is equal to
Determinant
Solution
Expanding, $\Delta=-(ac-b^2)(ax^2+2bx+c)$. Since $a>0$ and $b^2-ac<0$, we have $ac-b^2>0$ and $ax^2+2bx+c>0$, so $\Delta<0$.