## Precalculus (6th Edition)

$x=1$
Let $u=x^{-1}$ $u^2=x^{-2}$ Using the representations above, the given equation becomes: $$-u^2+2u=1$$ Add $u^2$ and subtract $2$ to both sides of the equation to obtain: $$-u^2+2u+u^2-2u=1+u^2-2u \\0=u^2-2u+1$$ Factor the trinomial to obtain: $$0=(u-1)(u-1)$$ Use the Zero-Factor Property by equating each factor to zero to obtain: $$u-1=0 \text{ or } u-1=0$$ Solve each equation to obtain: $$u=1$$ Replacing $u$ with $x^{-1}$ gives: $$u=1 \\x^{-1} = 1$$ Use the rule $a^{-m} = \dfrac{1}{a^m}$, then cross-multiply to obtain: $$\dfrac{1}{x} = 1 \\x(1) = 1 \\x=1$$