Quadratic Equations with Complex Roots

TipLearning Objectives

By the end of this lesson, you’ll be able to:

  • Recognize when a quadratic has complex (non-real) solutions.
  • Solve quadratics with negative discriminants.
  • Express solutions in the form \(a \pm bi\).
  • Interpret complex solutions in algebraic and graphical contexts.

Key Ideas

A quadratic equation has the form: \[ ax^2 + bx + c = 0 \]

The quadratic formula gives: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

The discriminant: \[ D = b^2 - 4ac \]

Tells us the type of solutions:

Discriminant \(D\) Type of Solutions
\(D > 0\) Two real, distinct
\(D = 0\) One real, repeated
\(D < 0\) Two complex conjugates

When \(D < 0\), the square root becomes imaginary: \[ \sqrt{-k} = i\sqrt{k} \]

So solutions take the form: \[ a \pm bi \]

Common Problem Types

Quadratics with Negative Discriminant

Use the quadratic formula, convert \(\sqrt{-k}\) into \(i\sqrt{k}\).

Example:
\(x^2 + 4x + 13 = 0\)
\(D = 16 - 52 = -36\)

Completing the Square with Complex Results

Completing the square sometimes reveals imaginary parts.

Example:
\(x^2 + 6x + 12 = 0\)
Leads to \((x + 3)^2 = -3\)

Graph Interpretation

A quadratic with no x-intercepts → complex roots.

Example:
Parabola entirely above the x-axis.

Expressions Written in \(a \pm bi\) Form

After solving, rewrite cleanly.

Example:
\(\frac{-4 \pm i\sqrt{7}}{6}\) → \(-\frac{2}{3} \pm \frac{\sqrt{7}}{6}i\)

Strategies

  • Compute the discriminant first to determine the root type.
  • Replace \(\sqrt{-k}\) with \(i\sqrt{k}\) right away.
  • Always simplify fractions and radicals.
  • Final answers must be written in \(a \pm bi\) form.
  • Graph the parabola mentally to predict whether roots are real or complex.

Worked Examples

Example 1 — Negative Discriminant

Solve: \[ x^2 + 2x + 5 = 0 \]

Compute \(D\): \[ D = 2^2 - 4(1)(5) = 4 - 20 = -16 \]

Quadratic formula: \[ x = \frac{-2 \pm \sqrt{-16}}{2} \]

Convert: \[ \sqrt{-16} = 4i \]

Final: \[ x = -1 \pm 2i \]

Example 2 — Completing the Square

Solve: \[ x^2 + 6x + 12 = 0 \]

Rewrite: \[ x^2 + 6x = -12 \]

Complete square: \[ (x + 3)^2 = -3 \]

Take square root: \[ x + 3 = \pm i\sqrt{3} \]

Final: \[ x = -3 \pm i\sqrt{3} \]

Example 3 — Graph Interpretation

A quadratic opens upward and its vertex is at \((2,5)\).
Since the vertex is above the x-axis, the graph never touches the axis → no real roots → complex solutions.

WarningCommon Mistakes
  • Forgetting to convert \(\sqrt{-k}\) into \(i\sqrt{k}\).
  • Using the quadratic formula incorrectly (sign or denominator errors).
  • Leaving solutions unsimplified (not in \(a \pm bi\) form).
  • Misinterpreting a graph and believing complex roots appear on the x-axis.
  • Forgetting that complex roots always come in conjugate pairs for quadratics.

Practice Problems

  1. Solve:
    \[ x^2 + 4x + 20 = 0 \]

  2. Solve using the quadratic formula:
    \[ 2x^2 + 3x + 5 = 0 \]

  3. Solve by completing the square:
    \[ x^2 + 10x + 34 = 0 \]

  4. A parabola has vertex \((1,4)\) and opens upward. How many real solutions does it have?

1.
\(D = 16 - 80 = -64\)
\(\sqrt{-64} = 8i\)
\[ x = \frac{-4 \pm 8i}{2} = -2 \pm 4i \]

2.
\(D = 9 - 40 = -31\)
\[ x = \frac{-3 \pm i\sqrt{31}}{4} \]

3.
Complete square:
\(x^2 + 10x = -34\)
\((x+5)^2 = -9\)
\[ x = -5 \pm 3i \]

4.
Vertex above the x-axis → no real solutions, only complex roots.

Summary

  • Quadratics with \(D < 0\) have two complex conjugate roots.
  • Use the quadratic formula or completing the square.
  • Always convert to \(a \pm bi\) form.
  • Graphs with no x-intercepts correspond to complex solutions.
  • Compute \(D\) first — it predicts everything.
  • Replace \(\sqrt{-k}\) with \(i\sqrt{k}\) instantly.
  • Complex roots come in conjugate pairs for quadratics.
  • Final answers must be written cleanly as \(a \pm bi\).