If $$\frac{{{x^2} - x + 1}}{{{x^2} + x + 1}} = \frac{3}{2}{\text{,}}$$ then the value of $$\left( {x + \frac{1}{x}} \right){\text{is?}}$$
A. 4
B. -5
C. 6
D. 8
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & \frac{{{x^2} - x + 1}}{{{x^2} + x + 1}} = \frac{3}{2}\left( {{\text{Given}}} \right) \cr & \Rightarrow \frac{{x\left\{ {\left( {x + \frac{1}{x}} \right) - 1} \right\}}}{{x\left\{ {\left( {x + \frac{1}{x}} \right) + 1} \right\}}} = \frac{3}{2} \cr & \Rightarrow \frac{{\left( {x + \frac{1}{x}} \right) - 1}}{{\left( {x + \frac{1}{x}} \right) + 1}} = \frac{3}{2} \cr & \,\,\,\,\,\,\,\,\,\,{\text{Let }}\left( {x + \frac{1}{x} = y} \right) \cr & \Rightarrow \frac{{y - 1}}{{y + 1}} = \frac{3}{2} \cr & \Rightarrow 2\left( {y - 1} \right) = 3\left( {y + 1} \right) \cr & \Rightarrow 2y - 2 = 3y + 3 \cr & \Rightarrow y = - 2 - 3 \cr & \Rightarrow y = - 5 \cr & \therefore x + \frac{1}{x} = - 5 \cr & {\text{ }} \cr} $$Related Questions on Algebra
If $$p \times q = p + q + \frac{p}{q}{\text{,}}$$ then the value of 8 × 2 is?
A. 6
B. 10
C. 14
D. 16
A. $$1 + \frac{1}{{x + 4}}$$
B. x + 4
C. $$\frac{1}{x}$$
D. $$\frac{{x + 4}}{x}$$
A. $$\frac{{20}}{{27}}$$
B. $$\frac{{27}}{{20}}$$
C. $$\frac{6}{8}$$
D. $$\frac{8}{6}$$
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