Quadratic Functions & Complex Numbers

Quadratics with Complex Roots

Solving quadratic equations whose solutions are complex numbers.

Advanced20 min lesson3 min readUpdated August 12, 2026Author not yet attributed

Prerequisites

The Discriminant

What a Parabola Looks Like When Its Roots Are Complex

A parabola that never touches the x-axis has no real x-intercepts. Before reading on, think about what that fact should tell you about the equation's solutions when set equal to 0 — and reconsider it now that 'no real solutions' doesn't mean 'no solutions at all.'

A parabola's x-intercepts are exactly the real solutions of setting the quadratic equal to 0. When the vertex sits entirely above the x-axis and the parabola opens upward (or entirely below and opens downward), it never crosses — meaning zero real solutions, which now translates directly to two complex solutions instead. The graph and the algebra tell the same story: no real crossing means the discriminant is negative.

Worked Example — Solving a Quadratic with Complex Roots

Solve x² + 4x + 13 = 0. Using the quadratic formula: x = (−4 ± √(16 − 52))/2 = (−4 ± √−36)/2 = (−4 ± 6i)/2 = −2 ± 3i. The solutions are −2 + 3i and −2 − 3i.

Worked Example — Connecting Complex Roots to the Graph

Without graphing, predict whether y = x² + 4x + 13 crosses the x-axis, using the solutions just found. Since both solutions are complex (not real), the parabola has no real x-intercepts — it never crosses the x-axis. Checking with vertex form (completing the square): y = (x + 2)² + 9, a vertex at (−2, 9), entirely above the x-axis and opening upward, confirming it.

Worked Example — Verifying a Complex Solution

Verify that x = −2 + 3i is a solution to x² + 4x + 13 = 0. Substitute: (−2 + 3i)² + 4(−2 + 3i) + 13. First, (−2 + 3i)² = 4 − 12i + 9i² = 4 − 12i − 9 = −5 − 12i. Then 4(−2 + 3i) = −8 + 12i. Adding everything: (−5 − 12i) + (−8 + 12i) + 13 = (−5 − 8 + 13) + (−12i + 12i) = 0 + 0i = 0. Confirmed.

Tip

Vertex form makes the real-vs-complex question visible at a glance: if the vertex's y-coordinate has the same sign as the leading coefficient's direction of opening (positive vertex, opens up; or negative vertex, opens down), the parabola never crosses the x-axis, matching complex roots.

Common Mistakes

Trying to plot complex solutions as x-intercepts on the real coordinate plane.
Complex solutions aren't points on the real x-axis at all — a parabola with complex roots simply has no x-intercepts to plot; the complex solutions live in the complex plane, a completely separate picture.
Making a sign error when substituting a complex number back in to verify a solution, especially when squaring it.
Square a complex number the same careful way as any binomial — (a + bi)² = a² + 2abi + b²i², remembering to replace i² with −1 in the last term.

Key Takeaways

A parabola with complex roots never crosses the x-axis, matching a negative discriminant.
Solving with complex roots uses the same quadratic formula process, simplified using i.
Vertex form can predict whether a quadratic's roots will be real or complex, based on the vertex's position relative to the x-axis.

Summary

Complex roots connect algebra and graphing into one consistent picture. The next unit shifts from quadratics specifically to polynomials of any degree, starting with how a polynomial behaves far from the origin.

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