A) \[{{b}^{2}}-4c>0\]
B) \[{{b}^{2}}-4c<0\]
C) \[{{c}^{2}}<b\]
D) \[{{b}^{2}}<c\]
Correct Answer: D
Solution :
\[{{x}^{2}}+2bx+c={{(x+b)}^{2}}+c-{{b}^{2}}\] \[\because \] \[{{(x+b)}^{2}}\]is a perfect square, therefore the given expression is positive if \[c-{{b}^{2}}>0\]or \[{{b}^{2}}<c\].
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