A) \[(-\infty ,6)\]
B) \[(-\infty ,\infty )\]
C) \[(6,\infty )\]
D) \[(-100,100)\]
Correct Answer: B
Solution :
(b): Given in equation is \[{{x}^{2}}+6x+13>0\]. Here, factorization is not possible. Rewriting the given in equation we get, \[\left( {{x}^{2}}+6x+9 \right)+4>0\] \[\Rightarrow \] \[{{\left( x+3 \right)}^{2}}+4>0\] We know that \[{{\left( x+3 \right)}^{2}}\ge 0,\forall x\in R\] \[{{\left( x+3 \right)}^{2}}+4\ge 4>0,\forall x\in R\]. \[\therefore \] The required solution is the set of all real numbers, i.e., \[\left( -\infty ,\infty \right)\].
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