2016 SSC CGL 29 Aug Evening Shift

Question 59

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

Solution

Given : $$x^2+\frac{1}{x^2}=2$$ ----------(i)

Now, $$(x-\frac{1}{x})^2=x^2+\frac{1}{x^2}-2(x)(\frac{1}{x})$$

=> $$(x-\frac{1}{x})^2=2-2$$     [Using (i)]

=> $$x-\frac{1}{x}=0$$

=> Ans - (B)

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