1. **State the problem:** Solve the equation $$x^4 - 5x^2 + 4 = 0$$ for $x$. 2. **Use substitution:** Let $y = x^2$. Then the equation becomes a quadratic in $y$: $$y^2 - 5y + 4 = 0$$ 3. **Solve the quadratic equation:** Use the quadratic formula: $$y = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 - 16}}{2} = \frac{5 \pm \sqrt{9}}{2}$$ 4. **Calculate the roots for $y$:** $$y = \frac{5 \pm 3}{2}$$ So, $$y_1 = \frac{5 + 3}{2} = \frac{8}{2} = 4$$ $$y_2 = \frac{5 - 3}{2} = \frac{2}{2} = 1$$ 5. **Back-substitute $y = x^2$:** $$x^2 = 4 \quad \text{or} \quad x^2 = 1$$ 6. **Solve for $x$:** $$x = \pm \sqrt{4} = \pm 2$$ $$x = \pm \sqrt{1} = \pm 1$$ 7. **Final answer:** $$x = -2, -1, 1, 2$$ These are the four real solutions to the original equation.