Answer
$$x_{1} = 2 + \sqrt 3 $$
$$x_{2} = 2 - \sqrt 3 $$
Work Step by Step
We have $$\frac{1}{x^2} - \frac{4}{x} + 1 = 0$$
As we see we have some rational terms so in order to eliminate all of these denominators we are going to multiply all the equation by the minimum common multiple of them. The minimum common multiple of $x^2$ and $x$ is $x^2$. Multiplying each side of the equation by $x^2$ $$x^2(\frac{1}{x^2} - \frac{4}{x} + 1) = x^2(0)$$ $$\frac{x^2*(1)}{x^2} - \frac{x^2(4)}{x} + x^2(1) = x^2(0)$$
Simplifying each rational expression and organizing the equation $$1 - x(4) + x^2(1) = x^2(0)$$ $$1 - 4x + x^2 = 0$$ $$x^2 - 4x + 1 = 0$$
We got a quadratic equation so we are going to solve it with the quadratic formula. In this case $a = 1$, $b = -4$ and $c = 1$ $$x_{1,2} = \frac{-b ± \sqrt {b^2 - 4ac}}{2a} =x_{1,2} = \frac{-(-4) ± \sqrt {(-4)^2 - 4(1)(1)}}{2(1)} $$
Doing each operation taking in count the signs law $$x_{1,2} = \frac{4 ± \sqrt {16 - 4}}{2} $$ $$x_{1,2} = \frac{4 ± \sqrt {12}}{2} $$
As we know $12$ does not has an exact square root, but we can distribute it like $4*3$ and 4 has exact square root, so $\sqrt {12} = \sqrt {4*3} = 2\sqrt {3}$. Then we got $$x_{1,2} = \frac{4 ± 2\sqrt {3}}{2} $$
For the first value of $x$ we have $$x_{1} = \frac{4 + 2\sqrt {3}}{2} = 2 + \sqrt 3$$
And for the second value we have $$x_{2} = \frac{4 - 2\sqrt {3}}{2} = 2 - \sqrt 3$$
