Solve the equation \( x^{-2} - 4x^{-1} + 3 = 0 \).
Rewrite the equation in terms of \( y \).
Let \( y = x^{-1} \). Then, the equation becomes \( y^2 - 4y + 3 = 0 \).
Factor the quadratic equation.
The factorized form of the polynomial is \( (y - 3)(y - 1) = 0 \).
Solve for \( y \).
Setting each factor to zero gives \( y = 3 \) and \( y = 1 \).
Convert back to \( x \).
Using \( y = x^{-1} \), we find \( x = \frac{1}{3} \) and \( x = 1 \).
The solutions for \( x \) are \( \boxed{x = 1} \) and \( \boxed{x = \frac{1}{3}} \).
The solutions for the equation are \( \boxed{x = 1} \) and \( \boxed{x = \frac{1}{3}} \).
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