Algebra 101 - Unit 6

Lesson 39: Solving Quadratic Equations with Complex Roots

Some quadratic equations have no real solutions because the discriminant is negative. In these cases, the solutions are complex numbers. We use the quadratic formula to find these solutions.

Quadratic Formula Review

For ax² + bx + c = 0:

x = [−b ± √(b² − 4ac)] / (2a)

If b² − 4ac < 0, we write the square root as √(−d) = i√d, where i = √−1.

Examples

Example 1: Solve x² + 4 = 0

1. x² = −4
2. x = ±√(−4) = ±2i

Example 2: Solve x² + 2x + 5 = 0

1. Identify coefficients: a = 1, b = 2, c = 5
2. Discriminant: Δ = b² − 4ac = 4 − 20 = −16
3. x = [−2 ± √(−16)] / 2 = [−2 ± 4i]/2
4. Solutions: x = −1 ± 2i

Example 3: Solve 2x² + 3x + 7 = 0

1. Δ = 3² − 4(2)(7) = 9 − 56 = −47
2. x = [−3 ± √(−47)] / 4 = (−3 ± i√47)/4

Practice Problems

1. Solve x² + 9 = 0
2. Solve x² − 6x + 13 = 0
3. Solve 3x² + 2x + 5 = 0
4. Solve x² + 4x + 8 = 0
5. Solve 2x² − 3x + 7 = 0