Question : If $x^2+\frac{1}{x^2}=2$, then the value of $x-\frac{1}{x}$ is:
Option 1: –2
Option 2: 0
Option 3: 1
Option 4: –1
Answer (1)
27th Jan, 2024
Correct Answer: 0
Solution :
Given: $x^2+\frac{1}{x^2}=2$
We know that the algebraic identity is $(x-\frac{1}{x})^2=x^2+\frac{1}{x^2}-2$
Substitute the value of $x^2+\frac{1}{x^2}=2$ in the above expression, we get,
$(x-\frac{1}{x})^2=2-2$
⇒ $(x-\frac{1}{x})^2=0$
⇒ $(x-\frac{1}{x})=0$
Hence, the correct answer is 0.
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