1. **Problem statement:** Find all roots of the biquadratic equation $x^4 - 5x^2 + 4 = 0$. 2. **Rewrite the equation:** Let $y = x^2$. Then the equation becomes a quadratic in $y$: $$y^2 - 5y + 4 = 0$$ 3. **Solve the quadratic:** Use the quadratic formula: $$y = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 1 \cdot 4}}{2 \cdot 1} = \frac{5 \pm \sqrt{25 - 16}}{2} = \frac{5 \pm 3}{2}$$ 4. **Find roots for $y$:** - $y_1 = \frac{5 + 3}{2} = 4$ - $y_2 = \frac{5 - 3}{2} = 1$ 5. **Back-substitute for $x$:** Since $y = x^2$, solve $x^2 = 4$ and $x^2 = 1$: - $x = \pm 2$ - $x = \pm 1$ 6. **Final roots:** $x = -2, -1, 1, 2$ This completes the solution for part (a).