Question : If $\left(y^2+\frac{1}{y^2}\right)=167$ and $y>0$, find the value of $\left(y+\frac{1}{y}\right)$.
Option 1: 13
Option 2: $-\sqrt{165}$
Option 3: $\sqrt{165}$
Option 4: –13
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Correct Answer: 13
Solution :
Given, $\left(y^2+\frac{1}{y^2}\right)=167$
Adding 2 on both sides, we get
⇒ $y^2+\frac{1}{y^2}+2=167+2$
⇒ $y^2+\frac{1}{y^2}+2\times y\times\frac{1}{y}=169$
⇒ $(y+\frac{1}{y})^2=169$
⇒ $y+\frac{1}{y}=\sqrt{169}$
⇒ $y+\frac{1}{y}=13$ [As $y>0$]
Hence, the correct answer is 13.
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