Question : If $\frac {x^2+3x+1}{x^2–3x+1}=\frac{1}{2 }$, then the value of $(x+\frac{1}{x})$ is:
Option 1: 9
Option 2: –9
Option 3: 1
Option 4: 2
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Correct Answer: –9
Solution : Given: $\frac {x^2+3x+1}{x^2–3x+1}=\frac{1}{2}$ Dividing both the numerator and the denominator by $x$ on the left-hand side, we get, $\frac{(x+3+\frac{1}{x})}{(x–3+\frac{1}{x})}=\frac{1}{2}$ ⇒ $\frac{(x+\frac{1}{x})+3}{(x+\frac{1}{x})–3}=\frac{1}{2}$ ⇒ $2(x+\frac{1}{x})+6={(x+\frac{1}{x})-3}$ ⇒ $(x+\frac{1}{x})=-3-6$ ⇒ $(x+\frac{1}{x})=-9$ Hence, the correct answer is –9.
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