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Complex Solutions of Quadratics

A free Algebra II lesson from the “Complex Numbers” unit, with a worked example and practice problems including step-by-step solutions.

When a quadratic's discriminant is negative, it has no real roots but two complex roots. Complex roots for real-coefficient quadratics come in conjugate pairs.

What you'll learn

Why it matters: When a parabola never touches the x-axis, its zeros are a complex conjugate pair. Engineering control systems, oscillation modeling, and signal filters all run into this case, and the quadratic formula still gives the answer — it just produces complex roots.

Worked example

Problem. Solve x^2 + 9 = 0.

  1. Subtract 9 to get x^2 = -9.
  2. Take square roots.
  3. x = +/- sqrt(-9) = +/- 3i.

Answer: x = 3i and x = -3i

Lesson preview: this page shows 5 practice items. The interactive activity contains 28 practice items and 15 quiz items (multiple choice and typed response). Counts include repeated prompts and variants.

Practice problems

1. Solve x^2 + 16 = 0.

Choices: x = 4i and x = -4i · x = 4 and x = -4 · x = 16i · No solutions

Show solution
  1. x^2 = -16.

Answer: x = 4i and x = -4i

2. Find the discriminant of x^2 + 4x + 8 = 0.

Show solution
  1. b^2 - 4ac = 16 - 32.

Answer: -16

3. A discriminant of -20 means the solutions are...

Choices: Complex · Two real · One repeated real · Linear

Show solution
  1. Negative discriminants give complex roots.

Answer: Complex

4. If 2 + 5i is a root of a real quadratic, the other root is...

Choices: 2 - 5i · -2 + 5i · -2 - 5i · 5 + 2i

Show solution
  1. Complex roots come in conjugate pairs.

Answer: 2 - 5i

5. Find the discriminant of x^2 - 6x + 13 = 0.

Show solution
  1. Use b^2 - 4ac.
  2. (-6)^2 - 4(1)(13) = 36 - 52.
  3. The discriminant is -16, so the roots are complex.

Answer: -16

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