A) a is real
B) \[a=1/2+i\]
C) \[a=1/2-i\]
D) the other root is also common
Correct Answer: C
Solution :
Let\[\alpha \]be the common root. Then, \[{{\alpha }^{2}}+\alpha i+a=0,\] \[{{\alpha }^{2}}-2\alpha +ia=0\] \[\Rightarrow \] \[\frac{{{\alpha }^{2}}}{a}=\frac{\alpha }{a(1-i)}=\frac{1}{-2-i}\] \[\Rightarrow \] \[\alpha =\frac{1}{1-i}\]\[\therefore \]\[a=\frac{-2-i}{{{(1-i)}^{2}}}\] \[\Rightarrow \] \[a=\frac{1}{2}-i\]
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