1. **State the problem:** Solve the equation $$x^4 - 10x^2 + 9 = 0$$ for $x$. 2. **Use substitution:** Let $y = x^2$. Then the equation becomes $$y^2 - 10y + 9 = 0$$. 3. **Solve the quadratic in $y$:** Use the quadratic formula $$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a=1$, $b=-10$, and $c=9$. 4. Calculate the discriminant: $$\Delta = (-10)^2 - 4 \times 1 \times 9 = 100 - 36 = 64$$. 5. Find the roots for $y$: $$y = \frac{10 \pm \sqrt{64}}{2} = \frac{10 \pm 8}{2}$$ 6. So, $$y_1 = \frac{10 + 8}{2} = 9$$ $$y_2 = \frac{10 - 8}{2} = 1$$ 7. **Back-substitute for $x$:** Recall $y = x^2$, so $$x^2 = 9 \implies x = \pm 3$$ $$x^2 = 1 \implies x = \pm 1$$ 8. **Final solution:** The solutions to the original equation are $$x = -3, -1, 1, 3$$.