If $$x(3-\frac{2}{x})=\frac{3}{x}$$, then value of $$x^2+\frac{1}{x^2}$$ is
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,Â
=> $$(x-\frac{1}{x})^2=(\frac{2}{3})^2$$
=> $$x^2+\frac{1}{x^2}-2(x)(\frac{1}{x})=\frac{4}{9}$$
=> $$x^2+\frac{1}{x^2}-2=\frac{4}{9}$$
=> $$x^2+\frac{1}{x^2}=2+\frac{4}{9}=2\frac{4}{9}$$
=> Ans - (D)
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