Question

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

• A. +ve
• B. $(ac-b^2)(ax^2 + 2bx + c)$
• C. -ve
• D. 0

Answer 1

Answer: (C)

-ve

Answer 2

We have $\begin{vmatrix}a&b&ax+b\\\ b &c&bx+c\\\ ax+b &bx+c&0\end{vmatrix}$ By $R_{3} \to R_{3} - \left(xR_{1}+ R_{2}\right);$ $= \begin{vmatrix}a&b&ax+b\\\ b&c&bx+c\\\ 0&0&-\left(ax^{2} + 2bx +c\right)\end{vmatrix}$ $= \left(ax^{2} +2bx+c\right)\left(b^{2}-ac\right) = \left(+\right)\left(-\right)=-ve$