Question

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

• A. $+$ve
• B. $(ac-b{)}^2(a{x}^2+2bx+c)$
• C. $-$ve
• D. $0$

Answer 1

The correct option is C $-$ve

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

Now, $\Delta =\left|\begin{array}{ccc}a& b& ax+b\\ b& c& bx+c\\ 0& 0& -(a{x}^2+2bx+c)\end{array}\right|$
$[\text{Operating}{R}_3\to R_3-x{R}_1-R_2]$
$=-(a{x}^2+2bx+c)(ac-b^2)$
$=-(+ve)(+ve)=-ve$