Quartic Equation Solver | ax⁴+bx³+cx²+dx+e=0
Solve quartic equations (ax⁴+bx³+cx²+dx+e=0) and find all four roots — real and complex — with step-by-step working, factored form, and root verification.
Quick Presets
What Is the Quartic Equation Solver | ax⁴+bx³+cx²+dx+e=0?
A quartic equation (degree 4) has four roots in the complex numbers, counted with multiplicity (Fundamental Theorem of Algebra). Unlike cubics, quartics can have 0, 2, or 4 real roots depending on the discriminant. Historically, Ferrari solved the general quartic in 1540 using a resolvent cubic, but the process is lengthy for manual computation.
This solver detects two important special cases automatically. If b = d = 0 (biquadratic), substituting u = x² reduces the equation to a quadratic — far simpler. If e = 0, x = 0 is an immediate root and a cubic handles the rest. For the general case, the Durand-Kerner method iteratively refines all four roots simultaneously in the complex plane, converging to full double-precision accuracy in under 200 iterations for virtually all inputs.
Formula
General form: ax⁴ + bx³ + cx² + dx + e = 0, a ≠ 0
Biquadratic case (b = d = 0):
Let u = x²: au² + cu + e = 0 → solve quadratic → x = ±√u
General case — Durand-Kerner method:
pⱼ_new = pⱼ − f(pⱼ) / ∏_{k≠j} (pⱼ − pₖ), iterate until convergence
Four initial guesses on a spiral in ℂ; converges to all roots simultaneously
Root count (real coefficients):
4 real roots | 2 real + 1 complex conjugate pair | 2 complex pairs
How to Use
- 1Enter coefficients: Type a, b, c, d, e for ax⁴+bx³+cx²+dx+e=0. The equation preview updates live.
- 2Use a preset: Try "Biquadratic" or "4 Real Roots" to see different root configurations.
- 3Solve: Click "Solve Quartic". Results are labeled Real or Complex with full verification.
- 4Check verification: |f(xᵢ)| < 1×10⁻⁶ means the root is accurate. Values near machine epsilon (10⁻¹⁵) are exact.
- 5Step-by-step: Toggle the working to see which special case was detected and how roots were found.
Example Calculation
Example 1 — Biquadratic: x⁴ − 5x² + 4 = 0
- ›a=1, b=0, c=−5, d=0, e=4 → biquadratic (b=d=0)
- ›Let u=x²: u² − 5u + 4 = 0 → (u−1)(u−4) = 0
- ›u₁=1, u₂=4 → x = ±√1 = ±1, x = ±√4 = ±2
- ›Four real roots: x = −2, −1, 1, 2
Example 2 — Four distinct real roots: x⁴ − 10x³ + 35x² − 50x + 24 = 0
- ›a=1, b=−10, c=35, d=−50, e=24
- ›Rational roots test: factors of 24 over 1 → try 1, 2, 3, 4
- ›Factored form: (x−1)(x−2)(x−3)(x−4) = 0
- ›Roots: x = 1, 2, 3, 4 (all real, all distinct)
Example 3 — Two complex pairs: x⁴ + 2x² + 1 = 0
- ›Biquadratic: u² + 2u + 1 = 0 → (u+1)² = 0 → u = −1 (double)
- ›x² = −1 → x = ±i (repeated twice)
- ›Roots: x = +i, +i, −i, −i (all complex, double pairs)
Understanding Quartic Equation | ax⁴+bx³+cx²+dx+e=0
Historical Context
Lodovico Ferrari solved the general quartic in 1540, just years after Cardano published the cubic solution. Ferrari's method reduces the quartic to a cubic “resolvent,” whose solution provides the key to factoring the quartic. This was the last polynomial equation for which a general algebraic formula was found — Évariste Galois proved in 1832 that no such formula can exist for degree ≥ 5.
Root Combinations
| Configuration | Real roots | Complex roots | Example |
|---|---|---|---|
| 4 distinct real | 4 | 0 | x⁴−10x³+35x²−50x+24 |
| 2 real + 1 conjugate pair | 2 | 2 | x⁴−1 (roots ±1, ±i) |
| 1 double real + 1 pair | 2 | 2 | (x−1)²(x²+1) |
| 2 double real roots | 4 | 0 | (x−1)²(x−2)² |
| 2 conjugate pairs | 0 | 4 | x⁴+5x²+6 |
| 1 double conjugate pair | 0 | 4 | (x²+1)² |
Frequently Asked Questions
How many real roots can a quartic have?
A quartic with real coefficients can have 0, 2, or 4 real roots (counting multiplicity). It cannot have exactly 1 or 3 real roots because complex roots of real polynomials come in conjugate pairs, a±bi always appear together.
What is a biquadratic equation?
A biquadratic is a quartic where b = d = 0: ax⁴ + cx² + e = 0. Setting u = x² reduces it to a quadratic in u. Each positive u gives two real roots (±√u); each negative u gives two complex roots. The "Biquadratic" preset demonstrates this.
What is the Durand-Kerner method?
Durand-Kerner simultaneously refines all root approximations each iteration: pⱼ ← pⱼ − f(pⱼ)/∏_{k≠j}(pⱼ−pₖ). Starting from spiral initial guesses in the complex plane, it converges quadratically to all four roots at once. It works for any polynomial degree and naturally finds complex roots.
Why does the verification show a very small but non-zero value?
IEEE 754 double-precision has ~15 digits of precision, so |f(root)| of 10⁻¹⁴ is machine epsilon — essentially perfect. Values below 10⁻⁸ are all practically exact. Only readings above 10⁻⁵ indicate a conditioning problem, such as two roots very close together.
Can I solve a polynomial of degree other than 4?
This tool handles degree-4 only. For degree 2: Quadratic Solver. For degree 3: Cubic Equation Solver. For degree ≥ 5: no general formula exists (Abel–Ruffini theorem) — use numerical methods or the Polynomial Evaluator for evaluation.
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