A) Complex
B) Real and unequal
C) Real and equal
D) One real and one imaginary
Correct Answer: B
Solution :
Given equation is \[{{x}^{2}}+2kx+4=0\] Put\[k=-3\], \[{{x}^{2}}-6x+4=0\]\[\Rightarrow x=3+\sqrt{5},\,3-\sqrt{5}\] Put k = 3, \[{{x}^{2}}+6x+4=0\]\[\Rightarrow x=-3+\sqrt{5},\,-3-\sqrt{5}\] i.e., Roots are real and unequal.You need to login to perform this action.
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