Some time ago, I published General Solution of Cubic Equations, which was linked here via Wikipedia. Unexpectedly, it brought in a lot of traffic, and I realized that everyone still has a significant need for materials in this area. Previously, I promised to write out the root-finding formula for quartic equations as well. Now that I finally have some time, I am writing this post, hoping it will be helpful to everyone.
ax^4+bx^3+cx^2+dx+e=0 \quad (a \neq 0)
The same two points still apply: First, online resources often lack the necessary programs to describe professional mathematical formulas (this is true for many websites); second, the language is often too professional and not accessible enough for the general public (such as Wikipedia). If I had looked at Wikipedia from the start, I guarantee I still wouldn’t understand it now.
It is said that after Cardano obtained the root-finding formula for cubic equations from Tartaglia, he had another extraordinary encounter. One day, a man named “Ferrari” became a servant in his household. However, this “servant” demonstrated superb mathematical abilities. Consequently, the relationship between Cardano and Ferrari quickly transformed from master-servant to teacher-student. By listening to Cardano’s lectures, Ferrari learned Latin, Greek, and mathematics. Later, he achieved an unprecedented feat: discovering the “radical solution method for quartic equations”!
Analysis Process:
Similar to Cardano’s method for cubic equations, Ferrari first takes the general quartic equation
ax^4+bx^3+cx^2+dx+e=0,
and by setting y=x+\frac{b}{4a}, transforms it into a quartic equation in terms of y:
y^4+py^2+qy+r=0 —(A)In this equation, an additional variable z is introduced, and the following transformation is made:
(y^2+p+z)^2=(p+2z)y^2-qy+(p^2-r+2pz+z^2) —(B)It can be proven that (A) and (B) are equivalent.
If we can also turn the right side of (B) into a perfect square form, then the original equation can be transformed into a quadratic equation in terms of y. How exactly can we turn the right side of (B) into the perfect square form we desire? This depends on z.
We know that for a quadratic trinomial ax^2+bx+c \ (a \neq 0), a sufficient condition for it to be a perfect square is b^2-4ac=0. Since the right side of (B) is exactly such a quadratic trinomial, we might as well set
q^2-4(p+2z)(p^2-r+2pz+z^2)=0 —(C)This achieves our goal, and (C) is a cubic equation in terms of z, which is solvable. Thus, after solving for z, we can take the square root of both sides of (B), turning it into two quadratic equations in terms of y. From there, we can find four values for y (in the general case), and subsequently find four values for x.
Therefore, the general steps for the radical solution of a quartic equation are:
By setting y=x+\frac{b}{4a}, transform the equation into a quartic equation in y: y^4+py^2+qy+r=0.
Solve the equation q^2-4(p+2z)(p^2-r+2pz+z^2)=0.
Solve the equation (y^2+p+z)^2=(p+2z)y^2-qy+(p^2-r+2pz+z^2).
Solve the equation y=x+\frac{b}{4a}.
One can imagine how complex the general root-finding formula for a quartic equation would be! Therefore, it seems no capable person has written this formula out yet (or rather, no one is willing to write it)! It has only been written out with the help of computers.
The four roots of a univariate quartic equation:
(1)
(2)
(3)
(4)
For reference, see:
Finally, looking back, Cardano, acting as a “Bo Le” (talent scout), discovered the “Thoroughbred” Ferrari, allowing him to blossom in the field of mathematics!
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