Solving the Quartic Equation by Radicals (A Plain-Language Version)

四次方程的根式求解（通俗版） 苏剑林 · 2009-09-06 · Mathematics equation radical algebra

Some time ago I published General Solution of the Cubic Equation, which got linked to from Wikipedia and brought this site a surprising amount of traffic — clearly there's real demand for this kind of material. Back then I promised I would eventually write up the quartic formula as well, and now that I finally have some time, here it is. I hope it proves useful.

$$ax^4+bx^3+cx^2+dx+e=0(a!=0)$$

The same two complaints as before apply: among the resources available online, one problem is that many sites lack the tools to properly render mathematical formulas; the other is that the language tends to be too technical for a general audience (as with Wikipedia). If I'd started out reading the Wikipedia article, I'm fairly sure I still wouldn't understand it today. more

The story goes that after Cardano got the formula for solving cubic equations out of Tartaglia, he had another remarkable encounter. One day, a man named "Ferrari" became a servant in his household. But this "servant" turned out to have extraordinary mathematical talent, and so the relationship between Cardano and Ferrari quickly evolved from master-and-servant into teacher-and-student. Through Cardano's lessons, Ferrari learned Latin, Greek, and mathematics — and later went on to achieve something unprecedented: he discovered the "radical solution of the quartic equation"!

The derivation:

Following the same spirit as Cardano's solution of the cubic, Ferrari first took the general quartic equation
$ax^4+bx^3+cx^2+dx+e=0$,
and by substituting $y=x+{b}/{4a}$, turned it into a quartic equation in y:
$y^4+py^2+qy+r=0$ ———(A)
In this equation, he introduced an extra variable z, and made the following transformation:
$(y^2+p+z)^2=(p+2z)y^2-qy+(p^2-r+2pz+z^2)$ ———(B)
One can show that (A) and (B) are equivalent.
If we can also turn the right-hand side of (B) into a perfect square, then the original equation becomes a quadratic in y. So how exactly can we make the right-hand side of (B) into the perfect square we want? It depends on z.
We know that for a quadratic trinomial of the form $ax^2+bx+c(a!=0)$, a sufficient condition for it to be a perfect square is $b^2-4ac=0$. The right-hand side of (B) happens to be exactly such a trinomial, so we may as well set
$q^2-4(p+2z)(p^2-r+2pz+z^2)=0$ ———(C)
which achieves exactly what we want. (C) is a cubic equation in z, and hence solvable. Once we've found z this way, we can take the square root of both sides of (B), turning it into two quadratic equations in y. This gives us 4 values of y (in the general case), and from there, 4 values of x.

So the general procedure for solving a quartic equation by radicals is:

1. Substitute $y=x+{b}/{4a}$ to turn it into a quartic equation in y: $y^4+py^2+qy+r=0$.

1. Solve the equation $q^2-4(p+2z)(p^2-r+2pz+z^2)=0$.

1. Solve the equation $(y^2+p+z)^2=(p+2z)y^2-qy+(p^2-r+2pz+z^2)$

1. Solve the equation $y=x+{b}/{4a}$.

You can imagine just how complicated the general root formula for a quartic equation must be! It seems that, so far, no one capable enough has actually bothered to write the whole thing out explicitly (or perhaps no one has been willing to). So here it is, produced with the help of a computer instead.

The 4 roots of a general quartic equation:

(1)

(2)

(3)

(4)

For reference, see:

http://planetmath.org/encyclopedia/QuarticFormula.html

http://www.wolframalpha.com/input/?i=a+x%5E4%2Bb+x%5E3%2Bc+x%5E2%2Bd+x%2Be+%3D+0

**And so, looking back, it was Cardano — the "talent scout" — who discovered Ferrari, the "thousand-mile horse," giving him the chance to shine so brilliantly in mathematics!