A) \[3\]
B) \[-3\]
C) \[\pm 3\]
D) \[4\]
Correct Answer: C
Solution :
\[{{\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)}^{2}}={{x}^{4}}+\frac{1}{{{x}^{4}}}+2{{x}^{2}}.\frac{1}{{{x}^{2}}}\] \[=47+2=49\] \[\Rightarrow \] \[{{\left( {{x}^{2}}+\frac{1}{{{x}^{2}}} \right)}^{2}}={{(7)}^{2}}\] \[\Rightarrow \] \[{{x}^{2}}+\frac{1}{{{x}^{2}}}=7\] because \[{{x}^{2}}=a\] positive quantity. Again, \[{{\left( x+\frac{1}{x} \right)}^{2}}={{x}^{2}}+2.x.\frac{1}{x}+\frac{1}{{{x}^{2}}}\] \[={{x}^{2}}+\frac{1}{{{x}^{2}}}+2\] \[\Rightarrow \] \[{{\left( x+\frac{1}{x} \right)}^{2}}=7+2=9\] \[\Rightarrow \] \[x+\frac{1}{x}=\pm 3\]
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