Box 5.3b: Solving a Quartic (4th degree) Polynomial© 1999 by Karl Hahn |
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How Ferrari Did It
| Four centuries before European mathematicians solved the problems of the cubic and quartic equations, Indian mathematician and astronomer, Bhaskara, had solved these problems. You can see a biography of Bhaskara by clicking here. |
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Only once in my life have I actually found use of the quartic formula. But seeing how to develop a solution to the general quartic is much more interesting than the formula itself, which is what I will do here. Once again, this is optional material, placed here for those who have both a curious mind and the extra time needed to spend on topics that won't be on the exam.
To be honest, I don't think I could have come up with a solution to the quartic on my own even if I had studied the problem for years. What I will do here is reconstruct, as best I can, Ferrari's thinking when he discovered the solution.
The general quartic equation is
x4 + px3 + qx2 + rx + s = 0 eq. 5.3b-1Just as with the cubic, it behooves us to get rid of the second-highest term (in this case the cubed term) by making a substitution of variables. Ferrari probably thought to do this because he already knew the solution to the cubic, which requires a similar simplification, when he solved the quartic. The substitution you make is
p
x = u - eq. 5.3b-2
4
when you multiply it out you will get something in the form of
u4 + au2 + bu + c = 0 eq. 5.3b-3I will let you multiply it out for yourself to see how a, b, and c relate to p, q, r, and s.
The next thing that is likely to have crossed Ferrari's mind is, "If only I could factor this into two quadratics, then I'd have it nailed." For convenience he made the squared term of each quadratic be u2 by itself (i.e. no coefficient), since you can always divide out the leading coefficient of any polynomial. So let the product of the two quadratics be
(u2 + αu + β)(u2 + γu + δ) = 0 eq. 5.3b-4The next thing Ferrari had to do was contrive a way so that equation 5.3b-4 would be the same as equation 5.3b-3. This means finding expressions for α, β, γ, and δ so that when you multiply out equation 5.3b-4 you get equation 5.3b-3.
One thing is perfectly clear. When you multiply out equation 5.3b-4, all the cubed terms must cancel. Why? Because equation 5.3b-3 has no cubed terms. When you multiply 5.3b-4 out, the only cubed terms are αu3 and γu3 (multiply it out for yourself if you can't see this). To get those two to cancel requires that γ = -α. This way Ferrari was able to eliminate one variable from the product:
(u2 + αu + β)(u2 - αu + δ) = 0 eq. 5.3b-4aThe next thing Ferrari probably focused on was how to get all the u2 terms in the product to add up to au2 (which is the u2 term in equation 5.3b-3). When you gather all the u2 terms in the product, this requirement becomes:
-α2u2 + βu2 + δu2 = au2 eq. 5.3b-5a -α2 + β + δ = a eq. 5.3b-5bMaking this happen is a little tricky. It seems that Ferrari had these thoughts about it:
Let β and δ each contribute half of the a that shows on the right (of equation 5.3b-5b). Let them each cancel half of the -α2 term as well. Then it will work out just right. But that would make β equal to δ, unless I add some quantity to one and subtract that same quantity from the other. If I named that quantity, (1/2)ρ, it would would mean
1
β = (a + α2 + ρ) eq. 5.3b-6a
2
1
δ = (a + α2 - ρ) eq. 5.3b-6b
2
If you substitute this β and δ back into
equation 5.3b-5b, you will see that it works no matter what value you choose for ρ.
So with the substitutions, the product of quadratics is now:
1 1
(u2 + αu + (a + α2 + ρ))(u2 - αu + (a + α2 - ρ)) = 0 eq. 5.3b-7
2 2
Now comes the next hurdle Ferrari had to get over. That is making all the linear u terms (that is terms with no exponent on the u) come out right. When you gather them up from the product given in equation 5.3b-7, they have to add up to bu in order to match equation 5.3b-3. That means that
1 1When you take the cancellations and solve what's left for ρ you getαu(a + α2 - ρ) -αu(a + α2 + ρ) = bu eq. 5.3b-8 2 2
b
ρ = - eq. 5.3b-9
α
which you substitute back into equation 5.3b-8.
The final hurdle for Ferrari was to find the α that makes the constant term come out right. To match the constant term in equation 5.3b-3, the constant term from the product in equation 5.3-8 has to be equal to c. That means that
1If you multiply through by 4 and then multiply out the difference of squares, you get(a + α2 + ρ)(a + α2 - ρ) = c eq. 5.3b-10 4
a2 + 2aα2 + α4 - ρ2 = 4c eq. 5.3b-11Now plug in the solution you got for ρ in equation 5.3b-9.
b2
a2 + 2aα2 + α4 - - 4c = 0 eq. 5.3b-12
α2
Multiply through by α2
a2α2 + 2aα4 + α6 - b2 - 4cα2 = 0 eq. 5.3b-13We have to solve for α in order to find the all the coefficients of the two quadratics. And here α is expressed as the solution to a 6th degree polynomial. But things aren't as bad as they look. All the exponents of α are even. So substitute z = α2, and now you have the cubic polynomial:
a2z + 2az2 + z3 - b2 - 4cz = 0 eq. 5.3b-13a z3 + 2az2 + (a2-4c)z - b2 = 0 eq. 5.3b-13bNow it's all downhill. Use the cubic formula to solve for z. Take the square root of z to get α. Then use α to find the coefficients of the two quadratics. Then use the quadratic formula on each of them to come up with solutions for u. Then use equation 5.3b-2 on your u's to find your solutions for x.
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