A) real and distinct
B) imaginary
C) equal
D) rational and equal
Correct Answer: B
Solution :
Given, \[f(x)={{x}^{2}}+ax+b\]has imaginary roots. Discriminant, \[D<0\Rightarrow {{a}^{2}}-4b<0\] Now, \[f(x)=2x+a\] \[f(x)=2\] Also, \[f(x)+f(x)+f(x)=0\] ?(i) \[\Rightarrow \] \[{{x}^{2}}+ax+b+2x+a+2=0\] \[\Rightarrow \] \[{{x}^{2}}+(a+2)x+b+a+2=0\] \[\therefore \] \[x=\frac{-(a+2)\pm \sqrt{{{(a+2)}^{2}}-4(a+b+2)}}{2}\] \[=\frac{-(a+2)\pm \sqrt{{{a}^{2}}-4b-4}}{2}\] Since, \[{{a}^{2}}-4b<0\] \[\therefore \] \[{{a}^{2}}-4b-4<0\] Hence, Eq. (i) has imaginary roots.
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