1. Let's start by understanding what "Case 3" means in the context of solving quadratic equations using the quadratic formula. 2. The quadratic formula is given by: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ where $a$, $b$, and $c$ are coefficients from the quadratic equation $ax^2 + bx + c = 0$. 3. The term under the square root, $b^2 - 4ac$, is called the discriminant. It tells us about the nature of the roots: - If the discriminant is positive, there are two real and distinct roots. - If it is zero, there is exactly one real root (a repeated root). - If it is negative, there are no real roots; instead, there are two complex roots. 4. Case 3 refers to when the discriminant is negative, i.e., $$b^2 - 4ac < 0$$ 5. Since the square root of a negative number is not a real number, we use imaginary numbers to express the roots. 6. We rewrite the square root of the negative discriminant as: $$\sqrt{b^2 - 4ac} = \sqrt{-1 \times (4ac - b^2)} = i\sqrt{4ac - b^2}$$ where $i$ is the imaginary unit with the property $i^2 = -1$. 7. Therefore, the roots in Case 3 are: $$x = \frac{-b \pm i\sqrt{4ac - b^2}}{2a}$$ 8. This means the solutions are complex conjugates, having a real part $-\frac{b}{2a}$ and an imaginary part $\pm \frac{\sqrt{4ac - b^2}}{2a}$. 9. In simple terms, when the discriminant is negative, the parabola does not cross the x-axis, and the solutions involve imaginary numbers. 10. Remember, imaginary numbers are just an extension of the number system that helps us solve equations that have no real solutions. This explanation should help you understand Case 3 smoothly!