1. **State the problem:** Solve the equation $$\frac{1}{x} + \frac{2}{x^2} = 3$$ for $x$. 2. **Identify the common denominator:** The denominators are $x$ and $x^2$. The least common denominator (LCD) is $x^2$. 3. **Multiply both sides by $x^2$ to clear denominators:** $$x^2 \times \left(\frac{1}{x} + \frac{2}{x^2}\right) = x^2 \times 3$$ 4. **Distribute multiplication:** $$x^2 \times \frac{1}{x} + x^2 \times \frac{2}{x^2} = 3x^2$$ 5. **Simplify each term:** $$\cancel{x^2} \times \frac{1}{\cancel{x}} + \cancel{x^2} \times \frac{2}{\cancel{x^2}} = 3x^2$$ which simplifies to $$x + 2 = 3x^2$$ 6. **Rewrite as a quadratic equation:** $$3x^2 - x - 2 = 0$$ 7. **Use the quadratic formula:** For $ax^2 + bx + c = 0$, $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=3$, $b=-1$, and $c=-2$. 8. **Calculate the discriminant:** $$\Delta = (-1)^2 - 4 \times 3 \times (-2) = 1 + 24 = 25$$ 9. **Find the roots:** $$x = \frac{-(-1) \pm \sqrt{25}}{2 \times 3} = \frac{1 \pm 5}{6}$$ 10. **Calculate each solution:** - $$x = \frac{1 + 5}{6} = \frac{6}{6} = 1$$ - $$x = \frac{1 - 5}{6} = \frac{-4}{6} = -\frac{2}{3}$$ 11. **Check for restrictions:** Since the original equation has denominators with $x$ and $x^2$, $x$ cannot be zero. Both solutions $1$ and $-\frac{2}{3}$ are valid. **Final answer:** $$x = 1 \text{ or } x = -\frac{2}{3}$$