Question

If a > 0 and discriminant of ax² + 2bx + c is negative, then
\[∆ = \begin{vmatrix}a & b & ax + b \\ b & c & bx + c \\ ax + b & bx + c & 0\end{vmatrix} is\]

पर्याय

• positive

• \[\left( ac - b^2 \right) \left( a x^2 + 2bx + c \right)\]

• negative

• 0

Answer 1

\[\text{ Discriminant D of }{ax}^2 + 2bx + c = \left( 2b \right)^2 - 4ac < 0 \left[\text{ Given }\right]\]
\[ \Rightarrow 4 b^2 - 4ac < 0 \]
\[ \Rightarrow b^2 - ac < 0,\text{ where }a > 0 \ldots(1)\]
\[\Delta = \begin{vmatrix} a & b & ax + b\\ b & c & bx + c\\ax + b & bx + c & 0 \end{vmatrix}\]
\[ = \begin{vmatrix} ax & bx & {ax}^2 + bx\\ b & c & bx + c\\ax + b & bx + c & 0 \end{vmatrix} \left[\text{ Applying }R_1 \to x R_1 \right]\]
\[ = \frac{1}{x}\begin{vmatrix} ax + b 7 bx + c & {ax}^2 + bx + bx + c\\ b & c & bx + c\\ax + b & bx + c & 0 \end{vmatrix} \left[\text{ Applying }R_1 \to R_1 + R_2 \right]\]
\[ = \frac{1}{x}\begin{vmatrix} 0 & 0 & {ax}^2 + 2bx + c\\ b & c & bx + c\\ax + b & bx + c & 0 \end{vmatrix} \left[\text{ Applying }R_1 \to R_1 - R_3 \right]\]
\[ = \frac{1}{x}\left\{ {ax}^2 + 2bx + c \begin{vmatrix}b & c \\ ax + b & bx + c\end{vmatrix} \right\} \left[\text{ Expanding along }R_1 \right]\]
\[ = \frac{1}{x}\left( {ax}^2 + 2bx + c \right)\left( b^2 x + bc - acx - bc \right)\]
\[ = \frac{1}{x}\left( {ax}^2 + 2bx + c \right) x \left( b^2 - ac \right) \]
\[ = \left( {ax}^2 + 2bx + c \right)\left( b^2 - ac \right) < 0 \left[\text{ From eq . }(1) \right]\]
\[ \Rightarrow \Delta < 0 \]