1. The problem is to find the roots of the equation $x^2 + 4 = 0$ in the form $a + bi$. 2. The general formula for solving quadratic equations $ax^2 + bx + c = 0$ is given by the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ 3. In this equation, $a = 1$, $b = 0$, and $c = 4$. Substitute these values into the formula: $$x = \frac{-0 \pm \sqrt{0^2 - 4 \times 1 \times 4}}{2 \times 1} = \frac{\pm \sqrt{-16}}{2}$$ 4. Since the discriminant ($b^2 - 4ac$) is negative, the roots are complex numbers. Recall that $\sqrt{-1} = i$. 5. Simplify the square root: $$\sqrt{-16} = \sqrt{16} \times \sqrt{-1} = 4i$$ 6. Substitute back: $$x = \frac{\pm 4i}{2} = \pm 2i$$ 7. Therefore, the roots in simplest $a + bi$ form are: $$x = 0 + 2i \quad \text{and} \quad x = 0 - 2i$$