With i in hand, every quadratic has exactly two roots — no exceptions.
For most of school, a quadratic could have two solutions, one, or — the awkward case — none. "No solution" meant a negative slipped under the square root, and the real line had nowhere to put it. But we have just built a place: i, with i2 = −1. Once √(−16) = 4i is a real, honest number, the quadratic formula never stalls. In this lesson the discriminant Δ = b2 − 4ac stops being a gatekeeper and becomes a classifier: two real roots, one repeated root, or a conjugate pair — and "how many roots does a quadratic have?" finally has one clean answer: always exactly two.
Recall the quadratic formula and its discriminant from the quadratic formula and the discriminant: for a x2 + b x + c = 0 (with a ≠ 0),
x = −b ± √Δ2a, Δ = b2 − 4ac.
The whole story lives under that root. When Δ ≥ 0 we take a real square root and read off real roots. When Δ < 0 a negative slips under the radical — and that is exactly the case we once labelled "no solution."
But "no real solution" is not the same as "no solution." Write Δ' = −Δ > 0 (a genuine positive). Using i2 = −1 from the imaginary unit and the rule √(−9) = √9·√(−1) = 3i,
√Δ = √(−Δ') = i√Δ'.
The square root is back — it just carries an i. So the formula keeps running.
Δ < 0 does not mean "no roots." It means no real roots — the roots live off the real line, on the complex plane.
The parabola makes this visible. From the shape of a parabola, the real roots are exactly where the curve crosses the x-axis. When Δ < 0 the parabola never touches that axis — it floats entirely above (or, for a < 0, entirely below). The roots are not there; the graph genuinely cannot show them, because they are not real numbers.
√(−16) = ±4 drops the i — that is the classic error. The honest statement is √(−16) = 4i, because (4i)2 = 16·i2 = −16. And √(−1)·√(−1) = √1 = 1 is also wrong: i·i = i2 = −1. The "√a·√b = √(ab)" shortcut fails for negatives.
With √Δ = i√(−Δ), the quadratic formula gives both roots directly:
x = −b ± i√(−Δ)2a (when Δ < 0).
Let us walk the signature example all the way out. Solve x2 − 2x + 5 = 0, so a = 1, b = −2, c = 5.
Δ = (−2)2 − 4·1·5 = 4 − 20 = −16 < 0.
√(−16) = 4i.
x = 2 ± 4i2 = 1 ± 2i.
So the two roots are 1 + 2i and 1 − 2i. Check the first by substituting back: (1 + 2i)2 = 1 + 4i + 4i2 = 1 + 4i − 4 = −3 + 4i, then −2(1 + 2i) = −2 − 4i, and adding 5:
(−3 + 4i) + (−2 − 4i) + 5 = 0. ✓
The imaginary parts cancel, the real parts cancel — it really is a root.
Solve x2 − 6x + 13 = 0. Δ = 36 − 52 = −16, so √(−16) = 4i and x = (6 ± 4i)/2 = 3 ± 2i. The two roots are 3 + 2i and 3 − 2i.
Look again at every example: 1 ± 2i, 3 ± 2i. The two complex roots are always conjugates of each other — same real part, opposite imaginary part. This is no accident, and it holds whenever the coefficients a, b, c are real.
Here is why, using the conjugate rules from conjugate and modulus. Suppose z is a root, so a z2 + b z + c = 0. Take the conjugate of both sides. Conjugation passes through sums and products — (zw)‾ = z̄·w̄, (z + w)‾ = z̄ + w̄ — and real numbers are their own conjugates (ā = a). So
(a z2 + b z + c)‾ = a z̄2 + b z̄ + c = 0̄ = 0.
That last line says z̄ satisfies the same equation: if z is a root, so is z̄. The roots arrive in mirror pairs.
Geometrically the conjugate is the reflection across the real axis (Stage 31.4), so a complex root pair sits symmetric above and below the real axis — the same height, opposite signs. The equal scale of the complex plane makes that symmetry exact in the picture.
The conjugate-pair law needs real coefficients. For an equation like x2 − i = 0 the coefficients are not all real, and the two roots are not conjugates. And never write 1 + 2i < 1 − 2i — ℂ has no order. You can compare their moduli (both are √5), but not the numbers themselves.
Now the three cases line up into one statement. The sign of Δ decides the kind of roots, but never the count:
| Discriminant | The graph | The roots | How many |
|---|---|---|---|
| Δ > 0 | crosses the x-axis twice | two distinct reals | 2 |
| Δ = 0 | touches the x-axis once | one repeated real | 2 (counted twice) |
| Δ < 0 | misses the x-axis | a conjugate pair a ± bi | 2 |
Counting multiplicity, a quadratic always has exactly two roots. Complex numbers are what make this uniform — they fill the gap that "no solution" used to leave. No quadratic is ever short of roots again.
And the bridge between roots and coefficients survives intact. From Vieta's formulas, the sum of the roots is −b/a and the product is c/a — and these hold for the complex pair too, because the imaginary parts cancel:
For 1 ± 2i: sum = (1 + 2i) + (1 − 2i) = 2 = −(−2)/1 ✓; product = (1 + 2i)(1 − 2i) = 12 + 22 = 5 = 5/1 ✓.
Notice the product is just z·z̄ = |z|2 — a real number, exactly as Stage 31.4 promised. A conjugate pair always has a real sum (2·Re z) and a real product (|z|2), which is precisely why a real-coefficient quadratic can have them.
Build a quadratic whose roots are 2 ± 3i. Sum = 4, product = 22 + 32 = 13. With a = 1, Vieta gives b = −(sum) = −4 and c = product = 13: x2 − 4x + 13 = 0. Check: Δ = 16 − 52 = −36, x = (4 ± 6i)/2 = 2 ± 3i ✓.
Once no roots ever go missing, a new question becomes natural: how many of each kind? A quadratic gives exactly two; a cubic gives three; a degree-n polynomial gives exactly n roots in the complex numbers (counting multiplicity). That last fact is the Fundamental Theorem of Algebra — the complex numbers are "complete" for polynomials in a way the reals never were. We met its first instance here: the doorway from "maybe no roots" to "always exactly the right number."
Counting roots — and counting in general, arranging and selecting from finite collections — opens the next strand of the curriculum: Stage 32 · Counting (permutations and combinations). Complex numbers gave us a world where nothing is ever missing; counting is how we keep track of everything in it.
| Idea | The statement |
|---|---|
| Negative Δ | √Δ = i√(−Δ) — no longer "no solution," just no real solution |
| The roots | x = (−b ± i√(−Δ)) / (2a) when Δ < 0 |
| Conjugate pairs | real coefficients ⇒ if a + bi is a root, so is a − bi (mirror across the real axis) |
| The count | Δ > 0, = 0, < 0 → two reals / one repeated / a conjugate pair — always exactly two |
| Vieta still holds | sum = −b/a, product = c/a = |z|2 for the complex pair |
Next: with no roots ever missing, we turn to counting — how many ways can things be arranged and chosen (Stage 32).
Solve x2 + 4 = 0. Are the roots a conjugate pair?
x2 = −4, so x = ±√(−4) = ±2i. The roots are 2i and −2i = 0 ± 2i — yes, a conjugate pair (real part 0, opposite imaginary parts). Here a = 1, b = 0, c = 4, Δ = −16.
For x2 − 4x + 7 = 0: compute Δ, classify the roots, then solve.
Δ = (−4)2 − 4·1·7 = 16 − 28 = −12 < 0 → a conjugate pair. √(−12) = √12·i = 2√3 i, so x = (4 ± 2√3 i)/2 = 2 ± √3 i.
One root of a real-coefficient quadratic is −1 + 3i. What is the other root, and what is the quadratic (with a = 1)?
Real coefficients force the conjugate: the other root is −1 − 3i. Sum = −2, product = (−1)2 + 32 = 10. By Vieta with a = 1: b = −(sum) = 2, c = product = 10, so x2 + 2x + 10 = 0.
True or false: "Δ < 0 means the equation has no solutions." Explain.
False. Δ < 0 means no real solutions — the parabola misses the x-axis. There are still exactly two solutions; they are a complex conjugate pair off the real line. The roots exist; the graph just can't display them.
Verify Vieta's formulas for the roots 3 ± 2i of x2 − 6x + 13 = 0.
Sum = (3 + 2i) + (3 − 2i) = 6 = −b/a = −(−6)/1 ✓. Product = (3 + 2i)(3 − 2i) = 32 + 22 = 13 = c/a = 13/1 ✓. (The product is |3 + 2i|2, a real number.)
A student writes "√(−25) = ±5." Find and fix the two things wrong with this.
(1) The i is missing: √(−25) = 5i, since (5i)2 = −25. (2) The principal square root is a single value, 5i, not "±5"; the ± appears only when it sits in the quadratic formula's "−b ± √Δ." Correct statement: √(−25) = 5i.
Six questions to lock it in. Tap the answer you think is right.
This lesson closes Stage 31 by reuniting complex numbers with the quadratic the students already know. It addresses the Common Core standard CCSS HSN-CN.C.7 (solve quadratic equations with real coefficients that have complex solutions) and builds toward HSN-CN.C.8–9 (extend polynomial identities to the complex numbers; the Fundamental Theorem of Algebra), drawing on the conjugate and modulus ideas of HSN-CN.A.3. The big shift to reinforce: a negative discriminant never meant "no solution," only "no real solution," and complex numbers make "how many roots?" perfectly uniform — always exactly two for a quadratic, counting multiplicity. A natural at-home check: pick any b and c making b2 < 4c, and confirm together that the two roots are conjugates whose sum is −b and product is c.