Quadratics with Complex Solutions
Aligned to N-CN.C.7 — Common Core State Standards for Mathematics.
What this lesson teaches
When a quadratic with real coefficients has a negative discriminant, its solutions are complex. The quadratic formula still works — the square root of the negative number produces the imaginary part.
Worked examples
Example 1
Solve x^2 − 4x + 13 = 0. Using x = (−b ± √(b^2 − 4ac))/(2a): x = (4 ± √(16 − 52))/2 = (4 ± √(−36))/2. Since √(−36) = 6i, x = (4 ± 6i)/2 = 2 ± 3i.
Example 2
Solve x² + 9 = 0. Subtract: x² = −9, so x = ±√(−9) = ±3i. The two solutions 3i and −3i are conjugates, and the graph of y = x² + 9 never touches the x-axis — which is the visual meaning of having no real roots.
Example 3
Common mistake: concluding no solution when the discriminant is negative. For x² − 2x + 5 = 0, b² − 4ac = 4 − 20 = −16. That is not the end: √(−16) = 4i, so x = (2 ± 4i)/2 = 1 ± 2i. Over the real numbers there is no solution; over the complex numbers there are always exactly two. Say which number system you are answering in.
Practice questions
- Solve x^2 + 9 = 0.
- Solve x^2 − 2x + 5 = 0.
- Solve x^2 + 6x + 10 = 0.
Watch the lesson
Every lesson comes with a video taught in English and Spanish — the same video the QR code in the printed workbook opens.
▶ Watch this lessonEn español
Cuadráticas con Soluciones Complejas
Cuando una ecuación cuadrática con coeficientes reales tiene discriminante negativo, sus soluciones son complejas. La fórmula cuadrática sigue funcionando — la raíz cuadrada del número negativo produce la parte imaginaria.
Ejemplo 1: Resuelve x^2 − 4x + 13 = 0. Usando x = (−b ± √(b^2 − 4ac))/(2a): x = (4 ± √(16 − 52))/2 = (4 ± √(−36))/2. Como √(−36) = 6i, x = (4 ± 6i)/2 = 2 ± 3i.
Ejemplo 2: Resuelve x² + 9 = 0. Resta: x² = −9, así que x = ±√(−9) = ±3i. Las dos soluciones 3i y −3i son conjugadas, y la gráfica de y = x² + 9 nunca toca el eje x — ese es el significado visual de no tener raíces reales.
Ejemplo 3: Error común: concluir que no hay solución cuando el discriminante es negativo. Para x² − 2x + 5 = 0, b² − 4ac = 4 − 20 = −16. Eso no es el final: √(−16) = 4i, así que x = (2 ± 4i)/2 = 1 ± 2i. En los números reales no hay solución; en los complejos siempre hay exactamente dos. Di en qué sistema numérico estás respondiendo.
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