Cubic Equation Solver (ax³ + bx² + cx + d = 0)

Understanding the cubic equation solver — concept and principle

A cubic equation has the form ax³ + bx² + cx + d = 0 with a ≠ 0. Every cubic with real coefficients has at least one real root, because its graph runs from −∞ to +∞; it has either three real roots or one real root and a pair of complex conjugate roots. Cubics appear in volume problems, equations of state and interpolation curves. This solver returns all three roots in closed form.

How it is calculated

1. Divide by a: x³ + Bx² + Cx + D = 0 with B = b/a, C = c/a, D = d/a.
2. Substitute x = t − B/3 to remove the squared term, giving the depressed cubic t³ + Pt + Q = 0 with P = C − B²/3 and Q = 2B³/27 − BC/3 + D.
3. Compute Δ = Q²/4 + P³/27.

• Δ > 0: one real root. Cardano's formula: u = ∛(−Q/2 + √Δ), v = ∛(−Q/2 − √Δ), t = u + v; the other two roots are complex.
• Δ = 0: repeated roots, t = 2u and −u with u = ∛(−Q/2).
• Δ < 0: three distinct real roots, found with the trigonometric form tₖ = 2√(−P/3)·cos(θ − 2πk/3), k = 0, 1, 2, where θ = ⅓·arccos((3Q/2P)·√(−3/P)).

1. Shift back: x = t − B/3. Real roots are sorted from smallest to largest.

The principle

The substitution x = t − B/3 centres the cubic on its point of inflection, which removes the x² term. Cardano's idea is to write t = u + v; then t³ = u³ + v³ + 3uv(u + v), and choosing 3uv = −P turns the cubic into a quadratic in u³. Δ is the discriminant of that quadratic, which is why its sign decides the root pattern. When Δ < 0 Cardano's square root is imaginary even though all roots are real (the "casus irreducibilis"), so the solver uses the identity cos 3θ = 4cos³θ − 3cosθ instead.

Worked example

1. Inputs: a = 1, b = −6, c = 11, d = −6.
2. B = −6, C = 11, D = −6; shift −B/3 = 2.
3. P = 11 − 36/3 = −1 and Q = 2(−216)/27 − (−66)/3 − 6 = −16 + 22 − 6 = 0.
4. Δ = 0 + (−1)/27 ≈ −0.037 < 0, so there are three real roots.
5. 2√(1/3) ≈ 1.1547 and θ = ⅓·arccos(0) = π/6, so t = 1.1547·cos 30° = 1, 1.1547·cos(−90°) = 0 and 1.1547·cos(−210°) = −1.
6. x = t + 2 gives 3, 2, 1. Result: 1, 2, 3. Check: (x − 1)(x − 2)(x − 3) = x³ − 6x² + 11x − 6.

How to read the result

Vieta's formulas let you check any answer: the roots add up to −b/a (here 6), their pairwise products add up to c/a (11) and their product is −d/a (6).

Common mistakes

• Entering coefficients in the wrong order or forgetting a zero coefficient (x³ − 1 is 1, 0, 0, −1).
• Treating a tiny imaginary part from rounding as a genuinely complex root.

Frequently asked questions

Does every cubic have a real root?

Yes, if the coefficients are real. The graph must cross the x-axis at least once.

Can I factor a cubic instead?

If it has a rational root p/q, p divides d and q divides a (rational root theorem). Divide that factor out and solve the remaining quadratic.

Who discovered the cubic formula?

Scipione del Ferro and Niccolò Tartaglia solved it in the early 16th century; Gerolamo Cardano published it in Ars Magna (1545).

Sources: M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions, NBS Applied Mathematics Series 55, §3.8.2 · W. H. Press et al., Numerical Recipes, Cambridge University Press, §5.6 Quadratic and Cubic Equations