Quartic Equation Solver (4th-Degree Polynomial Roots)
Solve ax⁴ + bx³ + cx² + dx + e = 0 for all four roots, real and complex, with multiplicities, the real-root count and a Vieta check of the sum and product of the roots.
How it works
ax⁴ + bx³ + cx² + dx + e = 0 · x₁ + x₂ + x₃ + x₄ = −b/a · x₁x₂x₃x₄ = e/a · biquadratic (b = d = 0): u = x², au² + cu + e = 0
By the fundamental theorem of algebra a quartic has exactly four roots counted with multiplicity; non-real ones come in conjugate pairs, so there are 0, 2 or 4 real roots. Ferrari’s closed-form formula exists but is long, so the solver finds all four roots simultaneously with the Durand–Kerner iteration, polishes them with Newton’s method and merges repeated roots. Vieta’s formulas (sum −b/a, product e/a) let you check the result.
Worked example
x⁴ − 2x³ − 13x² + 14x + 24 = 0 factors as (x + 3)(x + 1)(x − 2)(x − 4), so the roots are x = −3, −1, 2, 4. Check: their sum is 2 = −(−2)/1 and their product is 24 = 24/1.
Sources: Wolfram MathWorld — Quartic Equation · OpenStax Precalculus 2e — Zeros of Polynomial Functions
Frequently asked questions
How many solutions does a quartic equation have?
Four, counting multiplicity and complex roots. The number of real roots is 4, 2 or 0, because complex roots come in conjugate pairs a ± bi.
How do I solve a biquadratic equation like x⁴ − 5x² + 4 = 0?
Substitute u = x²: u² − 5u + 4 = 0 gives u = 1 or 4, so x = ±1 and ±2. The solver marks this case when b = d = 0.
Is there a quartic formula?
Yes — Ferrari’s method (1540) reduces the quartic to a resolvent cubic, but the general formula is very long. Numerical methods give the same roots far more simply.
What does (×2) next to a root mean?
It is a double root: the graph touches the x-axis there without crossing it, and the factor (x − r) appears squared.
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