Question

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

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

Answer 1

The correct option is C - ve

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

here $a>0;4b^2-4ac<0,ac-b^2>0$

$\therefore a{x}^2+2bx+c>0,\forall x\in R$

Applying $(R_3\to R_3-x{R}_1-R_2)$

$\Delta =\left|\begin{array}{ccc}a& b& ax+b\\ b& c& bx+c\\ 0& 0& -(a{x}^2+2bx+c)\end{array}\right|$

$=-(a{x}^2+2bx+c)\left)\right(ac-b^2)=(+ve\left)\right(+ve)=-ve$