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Complex Solutions of Quadratics

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

Quadratics with negative discriminants have complex solutions. For real-coefficient quadratics, nonreal complex roots occur in conjugate pairs.

What you'll learn

Why it matters: When a parabola never crosses the x-axis, the equation still has solutions, but they are complex. This idea supports engineering, signal, and polynomial work where no real intercept does not mean no algebraic answer.

Worked example

Problem. Solve x^2 + 25 = 0.

  1. x^2 = -25.
  2. Take square roots.
  3. x = +/-5i.

Answer: x = 5i or x = -5i

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 or x = -4i · x = 4 or x = -4 · x = 16i · No solution

Show solution
  1. x^2 = -16.

Answer: x = 4i or x = -4i

2. Find the discriminant of x^2 + 2x + 10 = 0.

Show solution
  1. 4 - 40 = -36.

Answer: -36

3. If 6 - 2i is a root of a real quadratic, the other root is...

Choices: 6 + 2i · -6 - 2i · -6 + 2i · 2 - 6i

Show solution
  1. Use the conjugate.

Answer: 6 + 2i

4. Solve x^2 + 25 = 0.

Choices: x = 5i or x = -5i · x = 5 or x = -5 · x = 25i · No solution

Show solution
  1. x^2 = -25.

Answer: x = 5i or x = -5i

5. What is the discriminant of x^2 + 4x + 13?

Show solution
  1. b^2 - 4ac = 16 - 52.

Answer: -36

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