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. Positive
• B. $(ac-b^2)(a{x}^2+2bx+c)$
• C. Negative
• D. 0.0

Answer 1

The correct option is C Negative

Let$\Delta =\left|\begin{array}{ccc}a& b& ax+b\\ b& c& bx+c\\ ax+b& bx+c& 0\end{array}\right|$
Applying$R_3\to R_3-x{R}_1-R_2;$ we get
$\Delta =\left|\begin{array}{ccc}a& b& ax+b\\ b& c& bx+c\\ 0& 0& -(a{x}^2+2bx+c)\end{array}\right|$
$\Delta =(b^2-ac)(a{x}^2+2bx+c)$
Now,$b^2-ac<0anda>0$
$\Rightarrow$Discriminant of$a{x}^2+2bx+c$ is -ve and $a>0$
$\Rightarrow (a{x}^2+2bx+c)>0$for all $xϵR$
$\Rightarrow \Delta =(b^2-ac)(a{x}^2+2bx+c)<0,i.e.-ve.$