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
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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