1. **Problem:** Solve the equation $$x^4 - 6x^2 + 8 = 0$$. 2. **Formula and approach:** This is a quartic equation but can be treated as a quadratic in terms of $y = x^2$. 3. **Rewrite the equation:** Substitute $y = x^2$, so the equation becomes: $$y^2 - 6y + 8 = 0$$ 4. **Solve the quadratic in $y$:** Use the quadratic formula: $$y = \frac{6 \pm \sqrt{(-6)^2 - 4 \times 1 \times 8}}{2 \times 1} = \frac{6 \pm \sqrt{36 - 32}}{2} = \frac{6 \pm \sqrt{4}}{2}$$ 5. **Calculate the roots for $y$:** $$y_1 = \frac{6 + 2}{2} = 4, \quad y_2 = \frac{6 - 2}{2} = 2$$ 6. **Back-substitute to find $x$:** Recall $y = x^2$, so: $$x^2 = 4 \implies x = \pm 2$$ $$x^2 = 2 \implies x = \pm \sqrt{2}$$ 7. **Final solution:** The four roots of the original equation are: $$x = 2, -2, \sqrt{2}, -\sqrt{2}$$ This method simplifies quartic equations that are quadratic in form by substitution, making them easier to solve.