1. **State the problem:** Solve the equation $$4x^{-4} - 5x^{-2} + 1 = 0$$ for $x$. 2. **Rewrite the equation using substitution:** Let $y = x^{-2}$. Then $x^{-4} = (x^{-2})^2 = y^2$. The equation becomes: $$4y^2 - 5y + 1 = 0$$ 3. **Solve the quadratic equation:** Use the quadratic formula: $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=4$, $b=-5$, and $c=1$. Calculate the discriminant: $$\Delta = (-5)^2 - 4 \times 4 \times 1 = 25 - 16 = 9$$ Calculate the roots: $$y = \frac{5 \pm 3}{8}$$ So, - $$y_1 = \frac{5 + 3}{8} = 1$$ - $$y_2 = \frac{5 - 3}{8} = \frac{2}{8} = \frac{1}{4}$$ 4. **Back-substitute for $x$:** Recall $y = x^{-2} = \frac{1}{x^2}$. For $y_1 = 1$: $$\frac{1}{x^2} = 1 \implies x^2 = 1 \implies x = \pm 1$$ For $y_2 = \frac{1}{4}$: $$\frac{1}{x^2} = \frac{1}{4} \implies x^2 = 4 \implies x = \pm 2$$ 5. **Final answer:** $$x = -2, -1, 1, 2$$