Question
If $$a > 0$$ and discriminant of $$a{x^2} + 2bx + c$$ is $$- ve,$$ then \[\left| \begin{array}{l}
\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,ax + b\\
\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\,\,\,\,\,\,\,\,\,\,\,\,bx + c\,\,\\
ax + b\,\,\,\,\,\,bx + c\,\,\,\,\,\,\,\,\,\,\,0
\end{array} \right|\] is equal to
A.
$$+ ve$$
B.
$$\left( {ac - {b^2}} \right)\left( {a{x^2} + 2bx + c} \right)$$
C.
$$- ve$$
D.
0
Answer :
$$\left( {ac - {b^2}} \right)\left( {a{x^2} + 2bx + c} \right)$$
Solution :
\[{\rm{We}}\,{\rm{have}}\,\,\left| \begin{array}{l}
\,\,\,\,\,\,\,\,a\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,ax + b\\
\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,bx + c\\
ax + b\,\,\,\,\,\,bx + c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0
\end{array} \right|\]
$$By\,{R_3} \to {R_3} - \left( {x{R_1} + {R_2}} \right);$$
\[ = \left| \begin{array}{l}
a\,\,\,\,\,\,\,\,b\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,ax + b\\
b\,\,\,\,\,\,\,\,c\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,bx + c\\
0\,\,\,\,\,\,\,\,0\,\,\,\,\,\,\,\, - \left( {a{x^2} + 2bx + C} \right)
\end{array} \right|\]
$$ = \left( {a{x^2} + 2bx + c} \right)\left( {{b^2} - ac} \right) = \left( + \right)\left( - \right) = - ve{\text{.}}$$