## Precalculus (6th Edition)

$\color{blue}{\left\{\frac{1}{5}-\frac{2}{5}i, \frac{1}{5}+\frac{2}{5}i\right\}}$
Multiply $x^2$ to both sides of the equation: $x^2(5-\frac{2}{x} + \frac{1}{x^2})=x^2(0) \\x^2(5)-x^2(\frac{2}{x}) + x^2(\frac{1}{x^2})=0 \\5x^2-2x+1=0$ The quadratic equation above has $a=5, b=-2$ and $c=1$. Solve using the quadratic formula to obtain: $x=\dfrac{-b \pm \sqrt{b^2-4ac}}{2a} \\x=\dfrac{-(-2)\pm \sqrt{(-2)^2-4(5)(1)}}{2(5)} \\x=\dfrac{2\pm\sqrt{4-20}}{10} \\x=\dfrac{2\pm\sqrt{-16}}{10} \\x=\dfrac{2\pm\sqrt{16(-1)}}{10} \\x=\dfrac{2\pm 4i}{10} \\x=\dfrac{2}{10} \pm \dfrac{4i}{10} \\x=\dfrac{1}{5} \pm \dfrac{2i}{5}$ Thus, the solution set is $\color{blue}{\left\{\frac{1}{5}-\frac{2}{5}i, \frac{1}{5}+\frac{2}{5}i\right\}}$.