Box 5.3b: Solving a Quartic (4th degree) Polynomial© 1999 by Karl Hahn |
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
x^{4} + px^{3} + qx^{2} + 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 -when you multiply it out you will get something in the form ofeq. 5.3b-2 4
u^{4} + au^{2} + 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 u^{2} 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
(u^{2} + αu + β)(u^{2} + 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 αu^{3} and γu^{3} (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:
(u^{2} + αu + β)(u^{2} - αu + δ) = 0 eq. 5.3b-4aThe next thing Ferrari probably focused on was how to get all the u^{2} terms in the product to add up to au^{2} (which is the u^{2} term in equation 5.3b-3). When you gather all the u^{2} terms in the product, this requirement becomes:
-α^{2}u^{2} + βu^{2} + δu^{2} = au^{2} eq. 5.3b-5a -α^{2} + 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 -α^{2} 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 =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:(a + α^{2} + ρ) eq. 5.3b-6a 2 1 δ =(a + α^{2} - ρ) eq. 5.3b-6b 2
1^{ } 1 (u^{2} + αu +(a + α^{2} + ρ))(u^{2} - αu +(a + α^{2} - 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 + α^{2} - ρ) -αu(a + α^{2} + ρ) = bu eq. 5.3b-8 2^{ } 2
b ρ = -which you substitute back into equation 5.3b-8.eq. 5.3b-9 α
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 + α^{2} + ρ)(a + α^{2} - ρ) = c eq. 5.3b-10 4
a^{2} + 2aα^{2} + α^{4} - ρ^{2} = 4c eq. 5.3b-11Now plug in the solution you got for ρ in equation 5.3b-9.
b^{2} a^{2} + 2aα^{2} + α^{4} -Multiply through by α^{2}- 4c = 0 eq. 5.3b-12 α^{2}
a^{2}α^{2} + 2aα^{4} + α^{6} - b^{2} - 4cα^{2} = 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 = α^{2}, and now you have the cubic polynomial:
a^{2}z + 2az^{2} + z^{3} - b^{2} - 4cz = 0 eq. 5.3b-13a z^{3} + 2az^{2} + (a^{2}-4c)z - b^{2} = 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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