Question : If $(y^{2} + \frac{1}{y^{2}}) = 167$ and $y > 0$ , find the value of $(y + \frac{1}{y})$ .
Option 1: 13
Option 2: $- \sqrt{165}$
Option 3: $\sqrt{165}$
Option 4: –13

Question

Question : If $(y^{2} + \frac{1}{y^{2}}) = 74$ and $y > 1$ , then find the value of $(y - \frac{1}{y})$ .
Option 1: $6 \sqrt{2}$
Option 2: $- 2 \sqrt{19}$
Option 3: $2 \sqrt{19}$
Option 4: $- 6 \sqrt{2}$

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Answer 1

Correct Answer: 13

Solution : Given, $(y^{2} + \frac{1}{y^{2}}) = 167$
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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Question : If (y2+1y2)=167 and y>0, find the value of (y+1y).Option 1: 13Option 2: −165Option 3: 165Option 4: –13

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Question : If $(y^{2} + \frac{1}{y^{2}}) = 167$ and $y > 0$ , find the value of $(y + \frac{1}{y})$ .
Option 1: 13
Option 2: $- \sqrt{165}$
Option 3: $\sqrt{165}$
Option 4: –13