If $$x\left( {3 - \frac{2}{x}} \right) = \frac{3}{x}{\text{,}}$$ then the value of $${x^2}{\text{ + }}\frac{1}{{{x^2}}}$$ is?
A. $${\text{2}}\frac{1}{9}$$
B. $${\text{2}}\frac{4}{9}$$
C. $${\text{3}}\frac{1}{9}$$
D. $${\text{3}}\frac{4}{9}$$
Answer: Option B
Solution(By Examveda Team)
$$\eqalign{ & x\left( {3 - \frac{2}{x}} \right) = \frac{3}{x} \cr & \Rightarrow 3x - 2 = \frac{3}{x} \cr & \Rightarrow 3x - \frac{3}{x} = 2 \cr & \Rightarrow 3\left( {x - \frac{1}{x}} \right) = 2 \cr & \Rightarrow x - \frac{1}{x} = \frac{2}{3} \cr & \left( {{\text{Squaring both sides}}} \right) \cr & \Rightarrow {x^2}{\text{ + }}\frac{1}{{{x^2}}} - 2 = \frac{4}{9} \cr & \Rightarrow {x^2}{\text{ + }}\frac{1}{{{x^2}}} = \frac{4}{9} + 2 \cr & \Rightarrow {x^2}{\text{ + }}\frac{1}{{{x^2}}} = 2\frac{4}{9} \cr} $$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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