IB Maths AI HLComplex NumbersPaper 1 & 2Discriminant~8 min read
Complex Roots of Quadratics
A quadratic with a negative discriminant has no real solutions — but it always has two complex ones. When the coefficients are real, those two roots are a conjugate pair: identical except for the sign of the imaginary part.
📘 What you need to know
A quadratic az² + bz + c = 0 has complex roots when the discriminant b² − 4ac < 0.
Solve them with the quadratic formula — or by completing the square.
When a, b, c are real, the two roots are a complex conjugate pair: p ± qi.
If z = p + qi is one root, the other is automatically z* = p − qi.
This conjugate-pair rule fails if any coefficient is itself complex.
A quadratic with roots z1, z2 factorises as a(z − z1)(z − z2).
When does a quadratic have complex roots?
The number of real roots of az² + bz + c = 0 is decided by the discriminant, Δ = b² − 4ac. If Δ < 0 there are no real roots — instead the equation has two complex roots.
Graphically, Δ < 0 means the parabola never crosses the x-axis: it sits entirely above it (or entirely below). With no x-intercepts, the solutions cannot be real numbers.
The teal parabola has Δ < 0 and never crosses the x-axis — its roots are a complex conjugate pair. The grey parabola has Δ > 0 and cuts the axis twice, giving two real roots.
Solving a quadratic with complex roots
Once Δ < 0 is confirmed, solve exactly as for any quadratic — the quadratic formula still applies. The square root of the negative discriminant simply produces an imaginary term.
Discriminant & quadratic formula
Δ = b² − 4ac — complex roots when Δ < 0
z = −b ± √(b² − 4ac)2aa negative value under the root gives the imaginary part
Split the negative root as √−k = (√k)i, simplify the surd, then divide through by 2a. The two answers always come out as a conjugate pair.
Conjugate pairs and factorising
For a quadratic with real coefficients, the ± in the formula guarantees the roots differ only in the sign of their imaginary part. So if one root is known, the other is free — just take its conjugate. This also works in reverse: from a conjugate pair you can rebuild the quadratic.
Building a quadratic from its roots: with roots z1 and z2, the equation is z² − (z1 + z2)z + z1z2 = 0. For a conjugate pair the sum and product are both real, so the coefficients are real.
Because a quadratic factorises through its roots, once z1 and z2 are known the expression can be written as a(z − z1)(z − z2).
🧠Recipe — solving a quadratic with complex roots
Identify a, b, c and check the discriminant b² − 4ac is negative.
Substitute into the quadratic formula, leaving the negative value under the root.
Split the root: write √−k as (√k)i and simplify the surd fully.
Divide through by 2a and write each root in Cartesian form p + qi.
State both roots as the conjugate pair p + qi and p − qi.
Worked examples
WE 1
Testing the discriminant
Use the discriminant to decide whether each equation has complex roots: (a) 2x² − 5x + 4 = 0, (b) x² − 6x + 9 = 0.
(a) Δ = b² − 4ac(−5)² − 4(2)(4) = 25 − 32 = −7Δ < 0 ⇒ complex roots(b) Δ = (−6)² − 4(1)(9)= 36 − 36 = 0(a) complex roots · (b) one repeated real rootΔ = 0 is the boundary — equal real roots, not complex.
WE 2
Solving with the quadratic formula
Solve z² + 4z + 13 = 0, giving your answers in Cartesian form.
a = 1, b = 4, c = 13 — apply the formulaz = (−4 ± √(16 − 52)) ÷ 2= (−4 ± √−36) ÷ 2 = (−4 ± 6i) ÷ 2divide through by 2z = −2 ± 3ithe roots −2 + 3i and −2 − 3i are a conjugate pair.
WE 3
A leading coefficient that isn’t 1
Solve 2z² − 4z + 5 = 0, giving your answers in exact Cartesian form.
a = 2, b = −4, c = 5z = (4 ± √(16 − 40)) ÷ 4= (4 ± √−24) ÷ 4√−24 = √24 i = 2√6 iz = (4 ± 2√6 i) ÷ 4z = 1 ± (√6 / 2) idivide every term by 4 — both the real 4 and the 2√6.
WE 4
Building a quadratic from one root
A quadratic equation with real coefficients has z = 3 − 5i as one root. (a) Write down the other root. (b) Find a quadratic equation with these roots and integer coefficients.
(a) real coefficients ⇒ roots are conjugatesother root = 3 + 5i(b) sum = (3−5i) + (3+5i) = 6product = (3−5i)(3+5i) = 9 + 25 = 34z² − (sum)z + (product) = 0(a) 3 + 5i · (b) z² − 6z + 34 = 0sum and product of a conjugate pair are always real — so the coefficients are too.
WE 5
Factorising with complex roots
The expression z² + 2z + 10 has complex roots. Factorise it fully.
Consider 3z² − 6z + 7 = 0. (a) Show it has complex roots. (b) Solve it, giving roots in Cartesian form. (c) Hence write 3z² − 6z + 7 in factorised form.
(a) Δ = (−6)² − 4(3)(7)= 36 − 84 = −48 < 0 ⇒ complex roots(b) z = (6 ± √−48) ÷ 6, √−48 = 4√3 iz = (6 ± 4√3 i) ÷ 6(c) factorise as a(z − z₁)(z − z₂), a = 3(b) z = 1 ± (2√3 / 3) i · (c) 3(z − 1 − (2√3/3)i)(z − 1 + (2√3/3)i)keep the leading 3 in front of the factorised form — it is the coefficient a.
💡 Top tips
Always check the discriminant first — Δ < 0 confirms the roots are complex before you solve.
The quadratic formula still works — only the √ of a negative number is new.
With real coefficients, find one root and the other is just its conjugate — no extra work.
Divide every term by 2a, including the surd, when simplifying the roots.
Check a root by substituting back — a correct root gives 0 + 0i.
âš Common mistakes
Calling Δ = 0 complex — that gives one repeated real root; complex needs Δ < 0.
Stopping at “no solutions” — a negative discriminant means two complex roots, not none.
Forgetting to divide the surd by 2a — both parts of the numerator must be divided.
Sign slips when factorising — z − (−1 + 3i) becomes z + 1 − 3i.
Assuming conjugate roots when a coefficient is complex — the pairing rule needs real coefficients.
Next up: Modulus & Argument — every complex number also has a size and a direction. The modulus measures its distance from the origin, and the argument the angle it turns through.
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