Why do some quartics have no real roots?

A real polynomial of even degree can stay entirely above or below the axis, as x⁴ + 1 does. The remaining roots exist only in the complex plane, in conjugate pairs.

How do I solve one by hand?

Try factoring first: x⁴ − 5x² + 4 is quadratic in x², so u = x² turns it into u² − 5u + 4 = 0 with u = 1 or 4.

Does the numerical method lose accuracy?

It converges fast for well-separated roots but slows near multiple roots, so an imaginary part around 1e−9 usually means a repeated real root.

🔢 Quartic Equation Calculator

Solve the quartic equation.

f(x) = ax⁴ + bx³ + cx² + dx + e

Value of a Value of b Value of c Value of d Value of e

What is a Quartic Equation?

A polynomial equation of degree 4, in the form ax⁴+bx³+cx²+dx+e=0 (a≠0). It has at most 4 real roots; including complex numbers, it always has exactly 4 roots.

Solution Method

Uses Ferrari's formula or the Durand-Kerner numerical method. This calculator uses the Durand-Kerner iteration.

Repeat until |Δrᵢ| < 10⁻¹²

Types of Roots

4 real roots: graph crosses x-axis 4 times

2 real + 2 complex: graph crosses x-axis twice

4 complex roots: no x-axis crossings

Double root: graph is tangent to x-axis

Vieta's Formulas

x₁+x₂+x₃+x₄ = −b/a

x₁x₂x₃x₄ = e/a

Worked examples

x⁴ − 5x² + 4 = 0 factors as (x² − 1)(x² − 4), so the four real roots are ±1 and ±2.

x⁴ + 1 = 0 has no real root at all, so all four roots are complex and appear as two conjugate pairs.

What the roots assume

A quartic always has exactly four roots counted with multiplicity, but it can show zero, two or four real ones. The calculation uses the Durand-Kerner iteration, so repeated or nearly repeated roots may leave tiny imaginary residues instead of exact zeros. Coefficients must be real and a must not be zero.

Frequently asked questions

Why do some quartics have no real roots?: A real polynomial of even degree can stay entirely above or below the axis, as x⁴ + 1 does. The remaining roots exist only in the complex plane, in conjugate pairs.
How do I solve one by hand?: Try factoring first: x⁴ − 5x² + 4 is quadratic in x², so u = x² turns it into u² − 5u + 4 = 0 with u = 1 or 4.
Does the numerical method lose accuracy?: It converges fast for well-separated roots but slows near multiple roots, so an imaginary part around 1e−9 usually means a repeated real root.