Negative Exponents

The goal is to get the exponents of 2 and -2 into 1 and -1. A feasible way is to use identities relating to sums and differences of squares. Since the answer asks for $|x-\frac{1}{x}|$ , it could be nice to use $(a-b)^2=a^2-2ab+b^2$ . Plugging in $x$ for $a$ and $\frac{1}{x}$ for $b$ , we get $(x-\frac{1}{x})^2=x^2-2+\frac{1}{x^2}$ . This works out nicely because we have two of the terms stated in the original equation and a constant. Now we can rewrite that equation substituting in our previous work: $(x-\frac{1}{x})^2+2=3$ , so $(x-\frac{1}{x})^2=1$ . Taking the square root gives $x-\frac{1}{x}=\pm 1$ , so $|x-x^{-1}|=\boxed{1}$ .