Question 68

If $$x(3-\frac{2}{x})=\frac{3}{x}$$, then value of $$x^2+\frac{1}{x^2}$$ is

Solution

Given : $$x(3-\frac{2}{x})=\frac{3}{x}$$

=> $$3x-2=\frac{3}{x}$$

=> $$3x-\frac{3}{x}=2$$

=> $$3(x-\frac{1}{x})=2$$

=> $$x-\frac{1}{x}=\frac{2}{3}$$

Squaring both sides,

=> $$(x-\frac{1}{x})^2=(\frac{2}{3})^2$$

=> $$x^2+\frac{1}{x^2}-2(x)(\frac{1}{x})=\frac{4}{9}$$

=> $$x^2+\frac{1}{x^2}-2=\frac{4}{9}$$

=> $$x^2+\frac{1}{x^2}=2+\frac{4}{9}=2\frac{4}{9}$$

=> Ans - (D)

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