I missed the lectures for this topic, so I don't have the notes, so I was wondering if anyone could give me the idea behind how to solve quartics in radicals. I know its long and messy, so just the basic idea would do. For example: x^{4} + 2x³ + 3x² + 4x + 5 = 0 I recall something about getting rid of the cubic term, so maybe I should substitute x = (u - 1/2), giving: u^{4} - 2u³ + 3/4u² - 1/2u + 1/16 + 2(u³ - 3/2u² + 3/4u - 1/8) + 3(u² - u + 1/4) + 4(u - 1/2) + 5 = 0 u^{4} + 3/4u² - 1/2u + 1/16 - 3u² + 3/2u - 1/4 + 3u² - 3u + 3/4 + 4u - 2 + 5 = 0 u^{4} + 3/4u² + 57/16 = 0 16u^{4} + 12u² + 57 = 0 Okay, this one happened to work out nicely, with the degree-1 term going away as well. Now I just have a quadratic in u². So perhaps a different example would be more enlightening. Also, when asked to "solve in radicals" does that mean that the correct answer to the above problem should be given as: [tex]x = \frac{1}{2} \pm \sqrt{\frac{-12 \pm \sqrt{-3504}}{32}}}[/tex] So: [tex]x = \frac{1 \pm \sqrt{\frac{-3 \pm \sqrt{-219}}{2}}}{2}[/tex]
Okay, so first I get rid of the x³ term with a change of variables. Then I end up with: u^{4} + Au² + Bu + C = 0 and I want to find a, b, and c such that: (u² + au + b)(u² - au + c) = u^{4} + (b-a²+c)x² + (ac-ab)x + (cb) = u^{4} + Au² + Bu + C, so I have three equations: A = b+c-a² B = a(c-b) C = cb So the third equations gives: b = C/c Subbing this into the second equation gives: a = B/(c-b) = B/(c-C/c) = Bc/(c²-C) Subbing these two into the first gives: A = C/c + c - B²c²/(c²-C)² A = (c²+C)/c - (Bc)²/(c²-C)² Ac(c²-C)² = (c²+C)(c²-C)² - B²c³ I have a feeling this is going to give me a quartic in c to solve, which doesn't help the matter. I remember my professor doing something strange, like subbing in x = u+v, (might not be exactly that substitution, and it might have been for solving the cubic, not the quartic: can't remember) and then factoring the resulting polynomial, and setting one of the factors to zero, and solving for u and v that way, and then in turn solving for x. I guess I should ask: How do you solve the cubic in radicals?
If you do it right, to find a, b, and c only requires solving a cubic... though it will probably be a cubic in a² or something like that. Incidentally, I guess it will be a polynomial in a² due to the symmetry of the factored form: we can swap (a, b, c) with (-a, c, b). Anyways, the way I remember for solving the cubic is to make the substitution: x = u + v and later impose the constraint uv=D for a clever choice of D.
Find yourself a galois theory book, and use the solvability of S_4, I suppose, is one way of constructing the formula; I remember it was how I derived the formula, albeit a long time ago, so I can't be any more help than that.
http://planetmath.org/encyclopedia/GaloisTheoreticDerivationOfTheQuarticFormula.html might be a bit of a spoiler (and was the first hit in google under galois theory quartic formula, by the way).
this is well explained in van der waerden, and copied on page 29 my algebra notes 844-2 on my webpage at UGA. briefly, S(4) has a normal tower consisting of {e}, V, S(4) where V is the klein 4 group. the fixed field of V is some field F with galois group S(3). if we consider the splitting field K of our quartic, and the 4 roots a,b,c,d in there, the 3 sums of pairwise products, X = ab+cd, Y = ac+bd, Z = ad+bc, are V - invariant hence lie in F, and satisfy an explicit "resolvant" cubic polynomial over the base field. since we can solve cubics we can find X,Y,Z, and it is then easy to express a,b,c,d in etrms of them explicitly. teh same trick can be worked with X = (a+b)(c+d), Y = (a+c)(b+d), Z = (a+d)(b+c), in which case the roots a,b,c,d, are just sums and differences of the square roots of X,Y,Z.
Binary Quartics How about the theory of binary quartics [tex]Q(x,y)=A_1x^4+Q_2x^3y+Q_3x^2y^2+Q_4xy^3+Q_5y^4[tex] where can we find any sources for the theory of the +ve definiteness of quartics?