A) a
B) 0
C) c
D) \[{{b}^{2}}-4ac\]
Correct Answer: C
Solution :
\[a{{x}^{2}}+bx+c=a\,{{\left( -\frac{b}{a} \right)}^{2}}\,+b\left( -\frac{b}{a} \right)+c\] \[=\frac{a{{b}^{2}}}{{{a}^{2}}}-\frac{{{b}^{2}}}{a}+c\] \[=\frac{{{b}^{2}}}{a}\,-\frac{{{b}^{2}}}{a}+c=c\]
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