If \[x-\frac{1}{x}=4,\]then \[\left( x+\frac{1}{x} \right)\]is equal to [SSC (CGL) 2013] |
A) \[5\sqrt{2}\]
B) \[2\sqrt{5}\]
C) (c)\[4\sqrt{2}\]
D) \[4\sqrt{5}\]
Correct Answer: B
Solution :
\[x-\frac{1}{x}=4\] [given] |
On squaring both sides, we get |
\[{{\left( x-\frac{1}{x} \right)}^{2}}={{4}^{2}}\] |
\[\Rightarrow \]\[{{x}^{2}}+\frac{1}{{{x}^{2}}}-2\times x\times \frac{1}{x}=16\] |
\[\Rightarrow \] \[{{x}^{2}}+\frac{1}{{{x}^{2}}}=16+2=18\] |
On adding 2 both sides, we get |
\[{{x}^{2}}+\frac{1}{{{x}^{2}}}+2=18+2\] |
\[\Rightarrow \] \[{{\left( x+\frac{1}{x} \right)}^{2}}=20\] |
\[\therefore \] \[x+\frac{1}{x}=\sqrt{20}=2\sqrt{5}\] |
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