A) be
B) \[{{b}^{2}}c\]
C) \[-{{b}^{2}}c\]
D) \[\frac{1}{2}{{b}^{2}}c\]
Correct Answer: B
Solution :
Let i\[\alpha \], where \[\alpha \] is purely real be one of the imaginary roots. Then, \[-a{{\alpha }^{2}}+ib\alpha +c+i=0\] \[\Rightarrow \]\[c=a{{\alpha }^{2}},1+b\alpha =0\] \[\Rightarrow \]\[{{\alpha }^{2}}=\frac{c}{a},\alpha =\frac{-1}{b}\] \[\Rightarrow \]\[\frac{1}{{{b}^{2}}}=\frac{c}{a}\Rightarrow a={{b}^{2}}c\]\[\Rightarrow \]
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