In mathematics, a quartic equation is the result of setting a quartic function equal to zero. An example of a quartic equation is the equation

2x⁴ + 4x³ - 26x² - 28x + 48 = 0;

the general form is

As the fundamental theorem of algebra tells us, a quartic equation always has four solutions (roots). They may be complex and there may be duplicate solutions.

Solving the quartic equation

Naturally, much effort has been turned to finding these roots. As with other polynomials, it is sometimes possible to factor a quartic equation directly; but more often such a feat is herculean, especially when the roots are irrational or complex. Hence it would be useful to have a general formula or algorithm (such as the quadratic equation which solves all quadratics). After much effort, such a formula was indeed found for quartics — but since then it has been proven (by Evariste Galois) that such an approach dead-ends with quartics; they are the highest-degree polynomial equations whose roots can be expressed in a formula using a finite number of arithmetic operators and n-th roots. From quintics on up, one can only try an ad-hoc approach.

Frankly, however, given the complexity of the quartic formulae (see below), they are not often used. If only the real rational roots are needed, they can be found (as is true for polynomials of any degree) via trial and error, using Ruffini's rule (so long as all the polynomial coefficients are rational). In the modern age of computers, furthermore, good numerical approximations for the roots are rapidly obtainable via Newton's method. But if the quartic must be solved entirely and precisely, the procedures are outlined below.

Special cases

Quartics in name only

If a₀ = 0, then one of the roots is x = 0, and the other roots can be found by dividing by x, and solving the resulting cubic equation,

a₄x³ + a₃x² + a₂x + a₁ = 0.

Biquadratic equations

A quartic equation where a₃ and a₁ are equal to 0 takes the form

and thus is a biquadratic equation, very easy to solve. Let z = x², so our equation turns to

which is a simple quadratic equation, whose solutions are easily found using the quadratic formula:

When we've solved it (i.e. found these two "z" values), we can extract x from them

If any of the z solutions were negative or complex numbers, some of the x solutions are complex numbers.

The General Case

To begin, the quartic must first be converted to a "depressed quartic."

Converting to a depressed quartic

Let

Ax⁴ + Bx³ + Cx² + Dx + E = 0

be the general quartic equation which it is desired to solve. Divide both sides by A,

The first step should be to eliminate the x³ term. To do this, change variables from x to u, such that

.

Then

Expanding the powers of the binomials produces

Collecting the same powers of u yields

Now rename the coefficients of u. Let

The resulting equation is

which is a depressed quartic equation.

Ferrari's solution

The depressed quartic can be solved by means of a method discovered by Ferrari. Once the depressed quartic has been obtained, the next step is to add the valid equation

(u² + α)² - u⁴ - 2αu² = α²

to equation (1), yielding

The effect has been to fold up the u⁴ term into a perfect square: (u² + α)². The second term, α u² did not disappear, but its sign has changed and it has been moved to the right side.

The next step is to insert a variable y into the perfect square on the left side of equation (2), and a corresponding 2y into the coefficient of u² in the right side. To accomplish these insertions, the following valid formulas will be added to equation (2),

and

0 = (α + 2y)u² - 2yu² - αu²

These two formulas, added together, produce

which added to equation (2) produces

(u² + α + y)² + βu + γ = (α + 2y)u² + (2yα + y² + α²).

This is equivalent to

The objective now is to choose a value for y such that the right side of equation (3) becomes a perfect square. This can be done by letting the discriminant of the quadratic function become zero. To explain this, first expand a perfect square so that it equals a quadratic function:

(ax + b)² = a²x² + 2abx + b².

The quadratic function on the right side has three coefficients. It can be verified that squaring the second coefficient and then subtracting four times the product of the first and third coefficients yields zero:

(2ab)² - 4a²b² = 0.

Therefore to make the right side of equation (3) into a perfect square, the following equation must be solved:

β² - 4(2y + α)(y² + 2yα + α² - γ) = 0.

Multiply the binomial with the polynomial,

β² - 4(2y³ + 5αy² + (4α² - 2γ)y + (α³ - αγ)) = 0

Multiply both sides by −1, then add β² to both sides, divide both sides by 4, then subtract β²/4 from both sides,

This is a cubic equation for y. Divide both sides by 2,

Conversion of the nested cubic into a depressed cubic

Equation (4) is a cubic equation nested within the quartic equation. It must be solved in order to solve the quartic. To solve the cubic, first transform it into a depressed cubic by means of the substitution

Equation (4) becomes

Expand the powers of the binomials,

Distribute, collect like powers of v, and cancel out the pair of v² terms,

This is a depressed cubic equation.

Relabel its coefficients,

The depressed cubic now is

Solving the nested depressed cubic

The solution of equation (5) is

therefore the solution of the original nested cubic is

Folding the second perfect square

With the value for y given by equation (6), it is now known that the right side of equation (3) is a perfect square, so that it can be folded:

Therefore equation (3) becomes

Equation (7) has a pair of folded perfect squares, one on each side of the equation. The two perfect squares balance each other. Note: if β equals zero then the ratio β/|β| (see sign function) will be indeterminate, but in such a case equation (1) becomes equivalent to a quadratic equation so that it is possible to use the quadratic formula to solve for u² directly.

If two squares are equal, then the sides of the two squares are also equal, viz.

Collecting like powers of u produces

Equation (8) is a quadratic equation for u. Its solution is

This is the solution of the depressed quartic, therefore the solution of the original quartic equation is

Summary of Ferarri's method

Given the quartic equation

Ax⁴ + Bx³ + Cx² + Dx + E = 0,

its solution can be found by means of the following calculations:

Quod Erat Faciendum.

There are other methods of solving the quartic equations, perhaps more optimal. Ferrari was the first to discover one of these labyrinthine solutions. The equation which he solved was

x⁴ + 6x² - 60x + 36 = 0

which was already in depressed form. It has a pair of solutions which can be found with the set of formulas shown above.

Obtaining alternative solutions

It could happen that the solution obtained through the seven formulae above is complex. It may also be the case that one is only looking for a real solution. Let x₁ denote the complex solution. If all the original coefficients A, B, C, D and E are real -- which should be the case when one desires only real solutions -- then there is another complex solution x₂ which is the complex conjugate of x₁. If the other two roots are denoted as x₃ and x₄ then the quartic equation can be expressed as

(x - x₁)(x - x₂)(x - x₃)(x - x₄) = 0,

but this quartic equation is equivalent to the product of two quadratic equations:

and

Since

then

Let

b = [Re(x₁)]² + [Im(x₁)]²

so that equation (9) becomes

Also let there be (unknown) variables w and v such that equation (10) becomes

Multiplying equations (11) and (12) produces

Comparing equation (13) to the original quartic equation, it can be seen that

and

Therefore

Equation (12) can be solved for x yielding

One of these two solutions should be the desired real solution.

• Ferrari's achievement