1. **State the problem:** Solve the equation $$x^8 - 10x^4 + 9 = 0$$ for $x$ in the set of complex numbers. 2. **Rewrite the equation using substitution:** Let $$y = x^4$$. Then the equation becomes $$y^2 - 10y + 9 = 0$$. 3. **Solve the quadratic equation:** 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. **Evaluate each root:** - $$y_1 = \frac{10 + 8}{2} = \frac{18}{2} = 9$$ - $$y_2 = \frac{10 - 8}{2} = \frac{2}{2} = 1$$ 7. **Recall substitution:** Since $$y = x^4$$, solve $$x^4 = 9$$ and $$x^4 = 1$$. 8. **Solve $$x^4 = 9$$:** Rewrite as $$x^4 = 3^2$$. The fourth roots of 9 are given by $$x = 3^{1/2} \left(\cos\frac{2k\pi}{4} + i \sin\frac{2k\pi}{4}\right)$$ for $k=0,1,2,3$. Calculate: - For $k=0$: $$x = \sqrt{3}$$ - For $k=1$: $$x = \sqrt{3}i$$ - For $k=2$: $$x = -\sqrt{3}$$ - For $k=3$: $$x = -\sqrt{3}i$$ 9. **Solve $$x^4 = 1$$:** The fourth roots of unity are $$x = \cos\frac{2k\pi}{4} + i \sin\frac{2k\pi}{4}$$ for $k=0,1,2,3$. Calculate: - For $k=0$: $$x=1$$ - For $k=1$: $$x=i$$ - For $k=2$: $$x=-1$$ - For $k=3$: $$x=-i$$ 10. **Final solution set:** $$\{ \sqrt{3}, -\sqrt{3}, i\sqrt{3}, -i\sqrt{3}, 1, -1, i, -i \}$$ These are all the complex solutions to the original equation.