Imaginary & Complex Roots: When a Parabola Misses the X-Axis
Find the complex roots of polynomials — using the discriminant and the quadratic formula — and understand why complex roots come in conjugate pairs, for the ACT.
The Short Version
- If a quadratic's discriminant (b² − 4ac) is negative, its roots are complex, not real.
- Find them with the quadratic formula — the negative under the radical becomes an i.
- Complex roots always come in conjugate pairs: a + bi and a − bi.
- A degree-n polynomial has exactly n roots (counting complex and repeated). An ACT and Algebra II topic.
A parabola can cross the x-axis twice, touch it once, or miss it entirely. When it misses, the equation still has solutions — they're just not real numbers. They're complex, built from the imaginary unit i. The discriminant tells you in advance which case you're in, and the quadratic formula produces the complex roots directly, always in matched conjugate pairs.
This guide connects the graph, the discriminant, and the algebra of complex roots, with worked and practice problems matched to real test difficulty at Northside Tutoring.
Why Complex Roots Matter
The ACT tests complex roots more than the SAT does, often by asking for the solutions of a quadratic with a negative discriminant. The topic ties together complex numbers and the quadratic formula you already know. It's Algebra II material beyond the SSAT.
Quick Recap: i and the Discriminant
Recall i = √−1, so √−16 = 4i. And the discriminant of ax² + bx + c is the part under the radical in the quadratic formula: b² − 4ac. Its sign decides the nature of the roots.
When Roots Are Complex
| Discriminant b² − 4ac | Roots | Graph |
|---|---|---|
| Positive | 2 real | crosses x-axis twice |
| Zero | 1 real (repeated) | touches x-axis once |
| Negative | 2 complex | misses the x-axis |
Finding Complex Roots
When the discriminant is negative, use the quadratic formula and pull out the i. For x² − 2x + 5 = 0: a = 1, b = −2, c = 5, discriminant = 4 − 20 = −16.
Conjugate Pairs
Notice the roots above are 1 + 2i and 1 − 2i — identical except for the sign of the imaginary part. That's always true for polynomials with real coefficients: complex roots come in conjugate pairs.
One complex root gives you the other
If a real-coefficient polynomial has 3 + i as a root, then 3 − i must also be a root. The ACT sometimes hands you one complex root and expects you to know its conjugate is the partner.
Counting a Polynomial's Roots
The Fundamental Theorem of Algebra says a polynomial of degree n has exactly n roots, as long as you count complex roots and repeated roots. So a cubic (degree 3) has 3 roots; if one is real and two are complex, those two are a conjugate pair.
Where You'll See This — Test by Test
No reference sheet provides the quadratic formula or the conjugate rule. The ACT tests finding complex roots and recognizing conjugate pairs; the SAT touches complex numbers more lightly. It's beyond the SSAT.
ACT
Tests solving quadratics with negative discriminants and identifying complex conjugate roots.
Explore ACT Tutoring → College AdmissionsSAT
The SAT covers complex-number arithmetic but tests complex roots less than the ACT.
Explore SAT Tutoring → K-12 CurriculumAlgebra II
A core Algebra II topic combining the discriminant, the quadratic formula, and complex numbers.
Explore Algebra Tutoring → K-12 CurriculumPre-Calculus
Pre-Calculus extends this to higher-degree polynomials and the Fundamental Theorem of Algebra.
Explore Math Tutoring →Watch the Lesson
Sometimes a diagram needs a voice. In the short video below, one of our Northside tutors walks through the core idea and works through test-style problems in real time.
Complex Roots — In Plain English
A live walkthrough from our tutoring team.
— Featuring a Northside Tutoring instructor
Worked Example Problems
These problems are calibrated to the difficulty you'll actually see on test day. Try each one before opening the solution.
Solve x² + 9 = 0.
Show solution
x² = −9, so x = ±√−9 = ±3i.
How many real solutions does x² + x + 1 = 0 have?
Show solution
Discriminant = 1 − 4 = −3 < 0, so no real solutions (two complex).
Solve x² − 4x + 13 = 0.
Show solution
Discriminant = 16 − 52 = −36. x = (4 ± √−36)/2 = (4 ± 6i)/2 = 2 ± 3i.
A polynomial with real coefficients has 5 − 2i as a root. Name another root.
Show solution
Complex roots come in conjugate pairs, so 5 + 2i is also a root.
How many roots (real and complex) does a degree-4 polynomial have?
Show solution
Exactly 4, counting complex and repeated roots (Fundamental Theorem of Algebra).
Common Mistakes to Avoid
Three traps that catch students every year
- Saying "no solutions" for a negative discriminant. There are no real solutions, but there are two complex ones.
- Forgetting the conjugate partner. Real-coefficient polynomials have complex roots in a + bi / a − bi pairs.
- Dropping the i when simplifying. √−16 = 4i, not 4 — the negative under the radical becomes an imaginary unit.
Practice Problems — You Try
Three problems below. Work each before checking the solution.
Solve x² + 25 = 0.
Show solution
x = ±√−25 = ±5i.
Solve x² − 2x + 2 = 0.
Show solution
Discriminant = 4 − 8 = −4. x = (2 ± 2i)/2 = 1 ± i.
A cubic with real coefficients has roots 2 and 1 + 3i. What is the third root?
Show solution
Complex roots pair with their conjugates, so the conjugate 1 − 3i must be the third root.
The Northside Method — How We Teach This 1-on-1
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Every Northside student works through a four-step framework:
- Assessment. We diagnose which specific skills are slowing your student down — not just whether they "get it" in the abstract.
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