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Solving the Quartic

Date: 05/27/98 at 23:10:32
From: Brian C.
Subject: Solving quartic equations directly

Do you have any pointers to a page that shows how to solve Quartic 
equations directly? I didn't have much trouble with the Cubic 
equations, and I'm hoping to be able to use it to add Quartic Splines 
to my realtime raytracer (computationally, as for 3rd/2nd degree, 
direct methods should be faster than doing Newton Raphson right?).
Brian Cowan

Date: 05/28/98 at 13:40:13
From: Doctor Rob
Subject: Re: Solving quartic equations directly

For a full discussion of quartic equations, see the Dr. Math FAQ:


I like this method to solve quartic equations. Divide by the leading
coefficient, to make it of the form:

   x^4 + a*x^3 + b*x^2 + c*x + d = 0

Substitute x = y - a/4, expand and simplify, to get:

   y^4 + e*y^2 + f*y + g = 0


   e = b - 3*a^2/8
   f = c + a^3/8 - a*b/2
   g = d - 3*a^4/256 + a^2*b/16 - a*c/4

This must factor into two quadratics with real coefficients, since
any complex roots must occur in conjugate pairs. Write:

   y^4 + e*y^2 + f*y + g = (y^2 + h*y + j)*(y^2 - h*y + g/j)

two factors). Then, equating coefficients of y^2 and y:

   e = g/j + j - h^2
   f = h*(g/j - j)


   g/j + j = e + h^2
   g/j - j = f/h

Adding and subtracting these two equations:

   2*g/j = e + h^2 + f/h
   2*j = e + h^2 - f/h

Multiplying these together:

   4*g = e^2 + 2*e*h^2 + h^4 - f^2/h^2


   h^6 + 2*e*h^4 + (e^2-4*g)*h^2 - f^2 = 0

This is a cubic equation in h^2, with known coefficients, since we 
know e, f, and g. We can use the solution of the cubic equation to 
find a positive real value of h^2, whose existence is guaranteed by 
This value of h will give a value of j from 2*j = e - h^2 - f/h. Once 

Once you have factored the quartic into two quadratics, finishing the

EXAMPLE 1:  Find the roots of x^4 + 6*x^3 - 5*x^2 - 10*x - 3 = 0.

Set x = y - 3/2, substitute and expand:

   y^4 + (-37/2)*y^2 + 32*y + (-231/16)
   h^6 + (-13)*h^4 + 46*h^2 - 1 = 0

Solving this cubic in h^2, we find the only real root is:

   h^2 = [13-cbrt(31*[31-3*sqrt(93)]/2)-cbrt(31*[31+3*sqrt(93)]/2)]/3

and h is the positive square root of this quantity. Approximately:

   h^2 = 0.0218741248
   h = 0.1478990357
   j = (e + h^2 - f/h)/2
   j = 0.1416217140

The exact expressions for j and g/j are too complicated for me to want
to write down explicitly. Then the approximate factorization is:

   (y^2 + 0.1478990357*y + 0.1416217140) * 
                                 (y^2 - 0.1478990357*y - 6.6197475893)


   (x^2 + 3.1468990357*x + 2.6134702676)*
                                  (x^2 + 2.8521009643x - 4.5915961433)

From here the rest is easy:

   x = -4, x = 1.1478990357, and x = -1.5739495179 +/- 0.3689894075*i

are the four approximate roots of the original equation.

Incidentally, the exact expression for the irrational real root is:

   x = [-4+cbrt(4*[47-3*sqrt(93)])+cbrt(4*[47+3*sqrt(93)])]/6

EXAMPLE 2:  Find the roots of x^4 + 6*x^3 + 7*x^2 - 7*x - 12 = 0.


   y^4 + (-13/2)*y^2 + (-1)*y + (-15/16)
   h^6 + (-37)*h^4 + 400*h^2 - 1024 = 0

Solving this cubic in h^2, we find:

   h^2 = 16,  h^2 = (21 + sqrt[185])/2,  h^2 = (21 - sqrt[185])/2.

We only need one of these, so we pick h^2 = 16, so:

   h = 4
   j = -21/4

and one factorization is:

   (y^2 + 4*y - 21/4) * (y^2 - 4*y + 11/4)


   (x^2 + 7*x + 3) * (x^2 - x - 1)

From here the rest is easy, and:


are the four roots of the original equation.

-Doctor Rob,  The Math Forum

Date: 07/22/99 at 12:24:38
From: Dan Collins 
Subject: Quartics

Hi - 

Compliments on the Web site, I've found it to be of use to me recently.

Near the beginning of the above answer, when substituting to produce a 
"reduced quartic," there's a degenerate case when (f = 0). If this occurs, 
the quartic has nicely reduced itself to a quadratic in (y^2) which can be 
easily solved by the quadratic formula (followed by simple factoring, 
solving each factor for 2 roots, and then re-substituting for all four 
solutions). However, the rest of the solution shown is then inappropriate - 
most clearly when the whole thing reduces to (y^4), at which point the only 
solution to the associated cubic is (h^2 = 0), despite the necessity of a 
"positive real value of h^2" thereafter (i.e., you have to take the square 
root and divide by it near the end).

Realizing this actually abbreviates the process in a few ways; in
particular, after we pass the degenerate case and have to solve the
associated cubic, it looks as if the first solution of the cubic formula at


always suffices for (h^2) - that is, there's no need to generate all three 
solutions to the associated cubic.

Thanks for your attention,
Daniel R. Collins (M.A. Mathematics University of Maine 1995)
Associated Topics:
High School Basic Algebra

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