A) equal to 1
B) less than 2 but not equal to 1
C) greater than 2
D) equal to 2
Correct Answer: C
Solution :
Given quadratic equation is \[p{{x}^{2}}+qx+r=0\] ?(1) \[D={{q}^{2}}-4pr\] Since \[\alpha \] and \[\beta \] are two complex root \[\therefore \]\[\beta =\overline{\alpha }\Rightarrow |\beta |=|\overline{\alpha }|\Rightarrow |\beta |=|\alpha |\]\[(\because |\overline{\alpha }|=||\alpha |)\] Consider \[|\alpha |+|\beta |=|\alpha |+|\alpha |\] \[(\because |\beta |=|\alpha |)\] \[=2|\alpha |>2.1=2\] \[(\because |\alpha |>1)\] Hence,\[|\alpha |+|\beta |\]is greater than 2.
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