Questions: x^-2-4 x^-1+3=0 x=-3, 3

x^-2-4 x^-1+3=0
x=-3, 3
Transcript text: \[ \begin{array}{l} x^{-2}-4 x^{-1}+3=0 \\ x=-3 \quad \times, 3 \quad \times \end{array} \]
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Solution

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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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