Can every quartic equation be solved with a formula?

Yes. Ferrari's method (1540) provides a closed-form solution for any quartic equation using radicals, analogous to the quadratic formula but far more complex. This makes the quartic the highest-degree polynomial for which a general radical formula exists; by the Abel–Ruffini theorem, no such formula exists for degree 5 or higher.

How many real roots can a quartic equation have?

A quartic equation always has exactly four roots in the complex numbers (counting multiplicity). The number of real roots can be 0, 2, or 4. It cannot have exactly 1 or 3 real roots if all coefficients are real, because complex roots of real-coefficient polynomials come in conjugate pairs.

What is the difference between a quartic equation and a quartic function?

A quartic function is $f(x) = ax^4 + bx^3 + cx^2 + dx + e$, which defines a curve you can graph. A quartic equation sets that expression equal to zero: $f(x) = 0$. Solving the equation means finding the $x$-intercepts of the function's graph.

Quartic Equation — Definition, Formula & Examples

A quartic equation is a polynomial equation of degree 4, meaning the highest power of the variable is 4. Its general form is

ax^4 + bx^3 + cx^2 + dx + e = 0

where

a \neq 0

A quartic equation is an algebraic equation of the form $ax^4 + bx^3 + cx^2 + dx + e = 0$ , where $a, b, c, d, e$ are coefficients from a given field (typically $\mathbb{R}$ or $\mathbb{C}$ ) and $a \neq 0$ . By the Fundamental Theorem of Algebra, every quartic equation has exactly four roots in $\mathbb{C}$ , counted with multiplicity. A general closed-form solution exists (discovered by Lodovico Ferrari in 1540), though it is considerably more complex than the quadratic or cubic formulas.

Key Formula

ax^4 + bx^3 + cx^2 + dx + e = 0

Where:

$a$ = Leading coefficient (must not be zero)
$b$ = Coefficient of the cubic term
$c$ = Coefficient of the quadratic term
$d$ = Coefficient of the linear term
$e$ = Constant term
$x$ = The unknown variable

How It Works

Solving a quartic equation depends on its structure. If the equation is **biquadratic** (missing the

x^3

and

x

terms), you can substitute

u = x^2

to reduce it to a quadratic in

u

. For other special forms, factoring into two quadratics or using substitutions may work. The general method—Ferrari's method—reduces the quartic to a resolvent cubic equation, which is then solved to find the four roots. In practice, many quartic equations encountered in coursework are designed to factor nicely or have rational roots that can be found using the Rational Root Theorem. Once one root

r

is found, you can divide out the factor

(x - r)

and solve the resulting cubic equation.

Worked Example

Problem: Solve the biquadratic equation

x^4 - 13x^2 + 36 = 0

Step 1: Recognize the equation is biquadratic (no

x^3

x

terms). Substitute

u = x^2

to reduce the degree.

u^2 - 13u + 36 = 0

Step 2: Solve the resulting quadratic using factoring. Find two numbers that multiply to 36 and add to −13.

(u - 9)(u - 4) = 0 \implies u = 9 \text{ or } u = 4

Step 3: Substitute back

u = x^2

and solve each equation for

x

x^2 = 9 \implies x = \pm 3 \qquad x^2 = 4 \implies x = \pm 2

Step 4: List all four roots.

x \in \{-3,\, -2,\, 2,\, 3\}

Answer: The four solutions are

x = -3,\; -2,\; 2,\; 3

Another Example

Unlike the first example, this quartic is not biquadratic. It requires the Rational Root Theorem and repeated polynomial division—a more general technique applicable to any quartic with rational roots.

Problem: Solve

x^4 - 5x^3 + 5x^2 + 5x - 6 = 0

Step 1: Use the Rational Root Theorem. Possible rational roots are

\pm 1, \pm 2, \pm 3, \pm 6

. Test

x = 1

1 - 5 + 5 + 5 - 6 = 0 \; \checkmark

Step 2: Divide the polynomial by

(x - 1)

using synthetic division to obtain the cubic factor.

x^4 - 5x^3 + 5x^2 + 5x - 6 = (x - 1)(x^3 - 4x^2 + x + 6)

Step 3: Find another root of the cubic. Test

x = -1

-1 - 4 - 1 + 6 = 0

. Divide again.

x^3 - 4x^2 + x + 6 = (x + 1)(x^2 - 5x + 6)

Step 4: Factor the remaining quadratic.

x^2 - 5x + 6 = (x - 2)(x - 3)

Step 5: Combine all factors and read off the roots.

(x - 1)(x + 1)(x - 2)(x - 3) = 0 \implies x \in \{-1,\, 1,\, 2,\, 3\}

Answer: The four solutions are

x = -1,\; 1,\; 2,\; 3

Visualization

Why It Matters

Quartic equations arise naturally in optimization problems, intersection calculations in computer graphics, and physics (e.g., lens optics and orbital mechanics). Students encounter them in college algebra, abstract algebra (when studying solvability by radicals), and Galois theory. Engineers routinely meet quartic equations when modeling fourth-order systems or computing eigenvalues of 4×4 matrices.

Common Mistakes

Mistake: Forgetting that a biquadratic substitution requires solving

x^2 = u

for both the positive and negative square roots.

Correction: Each positive value of

u

yields two values of

x

x = +\sqrt{u}

and

x = -\sqrt{u}

. A negative value of

u

gives no real roots for

x

Mistake: Assuming a quartic with real coefficients can have exactly 1 or 3 real roots.

Correction: Complex roots of real-coefficient polynomials always appear in conjugate pairs (by the Conjugate Pair Theorem). So the count of real roots must be 0, 2, or 4.

Mistake: Dividing by

x

without checking whether

x = 0

is a root.

Correction: If the constant term

e = 0

, then

x = 0

is a root. Factor it out as

x(ax^3 + bx^2 + cx + d) = 0

and include

x = 0

in your solution set.

Check Your Understanding

Solve

x^4 - 10x^2 + 9 = 0

Hint: This is biquadratic. Substitute

u = x^2

to get

u^2 - 10u + 9 = 0

Answer:

x = -3, -1, 1, 3

How many real roots does

x^4 + 1 = 0

have?

Hint: Think about whether

x^4

can ever be negative.

Answer: Zero real roots. Since

x^4 \geq 0

for all real

x

, the expression

x^4 + 1

is always at least 1.

x = 2

is a root of

x^4 - 6x^3 + 11x^2 - 6x = 0

, find all four roots.

Hint: Factor out

x

first, then use the Rational Root Theorem on the cubic.

Answer:

x = 0, 1, 2, 3

Related Terms

Cubic Polynomial — One degree lower; resolvent cubic used in Ferrari's method
Degree of a Polynomial — A quartic has degree 4
Coefficient — The constants a, b, c, d, e in the quartic
Conjugate Pair Theorem — Explains why complex roots come in pairs
Conjugates — Complex conjugate roots of real quartics
Constant Polynomial — Degree 0; simplest polynomial type
Degree of a Term — The leading term has degree 4
Binomial Theorem — Used when expanding binomials raised to the 4th power
Binomial Coefficients — Appear in expansions like $(x+a)^4$
Binomial Coefficients in Pascal's Triangle — Row 4 gives coefficients for quartic binomial expansions