Quick answer: For ax² + bx + c = 0 with real coefficients and a ≠ 0, D = b² − 4ac. If D < 0 there are no real roots. Complex roots can still be calculated using i² = −1.
Worked example
Solve x² + 2x + 5 = 0 over the complex numbers.
- D = 2² − 4(1)(5) = −16.
- Apply the formula: x = (−2 ± √(−16))/2.
- Since √(−16) = 4i, x = −1 ± 2i.
Check the result
For x = −1 + 2i, x² = −3 − 4i; adding 2x + 5 gives zero. The conjugate also works.
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Common mistake to avoid
Do not write √(−16) = −4. In a real-number-only question, the correct conclusion is no real roots, not no roots in every number system.
Try it yourself
1. Classify x² + 1 = 0. 2. Classify x² − 4x + 4 = 0.
Answers and reasoning
1. D = −4: no real roots; complex roots ±i. 2. D = 0: one repeated real root x = 2.
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