EQUATIONS OF DEGREE 3 AND 4.

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1 EQUATIONS OF DEGREE AND 4. IAN KIMING Consider the equation. Equations of degree. x + ax 2 + bx + c = 0, with a, b, c R. Substituting y := x + a, we find for y an equation of the form: ( ) y + py + 2q = 0, where for certain reasons to be explained below we have chosen to write the coefficients as p and 2q. We shall assume that: pq 0, as ( ) is readily solved when p = 0 or q = 0. Below we give the general solution of ( ) when pq 0. But let us start with the following heuristic considerations: Suppose that θ is a root of ( ), and suppose further that u and v are complex numbers with θ = u + v. Then: 0 = θ + pθ + 2q = u + v + (u + v)(p + uv) + 2q ; so if additionally we had uv = p, we would have whence 0 = u + v + 2q = u p u + 2q, u 6 + 2qu p = 0. Now, this equation is a quadratic equation in u ; we can solve this to obtain: u = q ± p + q 2. So it would appear that we can express the roots of ( ) via cube roots of the numbers q ± p + q 2. Let us make these considerations precise. Remarks on n th roots: Let n N and consider the complex number ζ n := e 2πi n = cos 2π n + i sin 2π n. Then we find that = ζ 0 n, ζ n,..., ζ n n are n distinct roots of the polynomial x n ; hence these complex numbers are precisely the roots of this polynomial. Now let z be a complex number 0. Then if θ and θ 2 are any 2 roots of x n z, we necessarily have (θ i 0 and) (θ 2 /θ ) n =, whence θ 2 = ζ a n θ

2 2 IAN KIMING for some a f0,... n g. One denotes any root of x n n z by the symbol n z. I.e., the symbol z is determined only up to a factor in f, ζ n,..., ζn n g. The only possible exception n is when z is a positive real number; in that case the symbol z often denotes the unique root of x n z which is also a positive real number. For the case n = we shall denote ζ by ε. That is, ε := e 2πi = cos 2π + i sin 2π = 2 + i 2. Theorem. Consider the equation ( ), and let u be a cube root of the number q + p + q 2 : u = q + p + q 2. Then there is a unique cube root v of q p + q 2 such that uv = p, i.e.: v = q p + q 2, uv = p, where it is understood that we have chosen the same square root of p + q 2 in both cases. The roots of ( ) are then: θ := u + v, θ 2 := εu + ε 2 v, θ := ε 2 u + εv. Proof. If u and w are cube roots of q + p + q 2 and q p + q 2 respectively, then (uw) = u w = ( q + p + q 2 )( q p + q 2 ) = q 2 (p +q 2 ) = p = ( p). So, uw must be one of the numbers p, p ε, p ε 2. Then p is one of the numbers u w, u εw, u ε 2 w. So there is in fact a unique cube root v of q p + q 2 such that uv = p. With such u and v we now have (remember that ε =, ε 2 +ε+ = 0, uv = p): (x (u + v))(x (εu + ε 2 v))(x (ε 2 u + εv)) = x (( + ε + ε 2 )u + ( + ε + ε 2 )v) x 2 +(( + ε + ε 2 )u + (ε + ε 2 )uv + ( + ε + ε 2 )v) x (u + ( + ε + ε 2 )u 2 v + ( + ε + ε 2 )uv 2 + v ) = x uv x (u + v ) = x + px + 2q. Hence the roots of ( ) are as stated.

3 EQUATIONS OF DEGREE AND 4. Remark: We can write the roots of ( ) as ( ) θ = q + p + q 2 + q p + q 2 where we are allowed to choose the values of the cube roots. However, we must then remember that the values of the 2 cube roots must be chosen so that their product is p. The formula ( ) for the roots is called Cardano s formula after Girolamo Cardano (50 576) who in 545 published the formula in his book Ars Magna. It is known however that he had obtained the solution from Niccolo Tartaglia ( ) under an oath to keep the solution secret. Hence the publication by Cardano inevitably lead to a huge public controversy. 2. Geometric solution of third degree equations. In this section we show how the equation ( ) can be solved using trigonometric or hyperbolic functions. We will split the discussion up in the following subcases: Case : p + q 2 = 0 Case 2 : p < 0 and p + q 2 < 0 Case : p < 0 and p + q 2 > 0 Case 4 : p > Case. If p + q 2 = 0 then equation ( ) is easily solved: The roots are just 2 p, p, p: (y + 2 p)(y p) 2 = y + py 2p p = y + py + 2q Case 2. If p < 0 and p + q 2 < 0 then < q p <. Consequently, we can p choose ψ ( R) such that cos ψ = q p p. As cos x = 4 cos x cos x we then find that 4 cos ψ cos ψ q p p = 0 whence ( 2 p cos ψ ) ( + p 2 p cos ψ ) + 2q = 0. So, 2 p cos ψ is a root of ( ). We find: y + py + 2q = (y 2 p cos ψ )(y2 + (2 p cos ψ )y + ( 4 cos2 ψ )p). Solving the quadratic equation, we find the roots of ( ): θ = 2 p cos ψ, θ 2 = 2 ( p 2 cos ψ ) 2 sin ψ θ = 2 ( p 2 cos ψ ) + 2 sin ψ = 2 p cos π + ψ = 2 p cos π ψ,.

4 4 IAN KIMING Remark: Let us now notice that all roots of ( ) are real in this case. But p + q 2 < 0, so if we use Cardano s formula ( ) to find the roots, we are forced to work with square roots of negative numbers, and hence with complex numbers. This is in fact how complex numbers were first encountered in mathematics, and Cardano is generally regarded as their discoverer. Because of these circumstances, this case is classically called Casus irreducibilis. 2.. Case. Suppose that p < 0 and p + q 2 > 0. Then we have p >, and p we can choose ψ R such that As cosh x = 4 cosh x cosh x we find: hence ( 2 q So, 2 q p cosh ψ cosh ψ = p p. 4 cosh ψ cosh ψ + p p = 0, ) ( ψ p cosh + p 2 q is a root of ( ). We find: ) ψ p cosh + 2q = 0. y +py+2q = (y+2 q ψ p cosh )(y2 (2 q ψ p cosh )y+( 4 ψ cosh2 )p). Solving the quadratic equation, we find the roots of ( ): θ = 2 q ψ p cosh, θ 2 = q ψ p cosh + i p sinh ψ, θ = q ψ p cosh i p sinh ψ Case 4. Suppose then finally that p > 0, and let ψ R be determined by: sinh ψ = q p p. As sinh x = 4 sinh x + sinh x we find: hence So, 2 p sinh ψ 4 sinh ψ + sinh ψ + q p p = 0, ( 2 p sinh ψ ) ( + p 2 p sinh ψ ) + 2q = 0. is a root of ( ). We find: y + py + 2q = (y 2 p sinh ψ )(y2 + (2 p sinh ψ )y + ( + 4 sinh2 ψ )p).

5 EQUATIONS OF DEGREE AND 4. 5 Solving the quadratic equation, we find the roots of ( ): θ = 2 p sinh ψ, θ 2 = p sinh ψ + i p cosh ψ, θ = p sinh ψ i p cosh ψ. Remark: If we allow ourselves the use of trigonometric functions of a complex variable, the above cases can be dealt with simultaneously: Choose ψ C such that cos ψ = q p p ; then by reasoning similar to the one used in case 2 above, we find that the roots of ( ) are: θ = 2 p cos ψ, θ 2 = 2 p cos 2π + ψ θ = 2 p cos 2π ψ However, if one wants to actually use these formulas in practice, they ultimately boil down to the various cases above.,.. Equations of degree 4. Using a convenient change of variables of form x x + η for some constant η, we quickly realize that we can solve general equations (with real coefficients) of degree 4 in x if we can solve equations of the form: ( ) x 4 + αx 2 + βx + γ = 0 for α, β, γ R (do this as an exercise; compare with what we did with cubic equations). Theorem 2. Let θ, θ 2, θ be the roots of the polynomial: so that in particular θ θ 2 θ = β 2. y + 2αy 2 + (α 2 4γ)y β 2, If we choose the signs of the square roots θ i of the θ i in such a way that: θ θ2 θ = β,

6 6 IAN KIMING then the 4 roots ξ, ξ 2, ξ, ξ 4 of ( ) are given by: ξ = 2 ( θ + θ 2 + θ ) ξ 2 = 2 ( θ θ 2 θ ) ξ = 2 ( θ + θ 2 θ ) ξ 4 = 2 ( θ θ 2 + θ ) Proof. If ξ, ξ 2, ξ, ξ 4 are the 4 roots of ( ) we have (x ξ )(x ξ 2 )(x ξ )(x ξ 4 ) = x 4 + αx 2 + βx + γ, which implies: 0 = ξ + ξ 2 + ξ + ξ 4, α = ξ ξ 2 + ξ ξ + ξ ξ 4 + ξ 2 ξ + ξ 2 ξ 4 + ξ ξ 4, β = ξ ξ 2 ξ + ξ ξ 2 ξ 4 + ξ ξ ξ 4 + ξ 2 ξ ξ 4, γ = ξ ξ 2 ξ ξ 4. Now let us define the following numbers w, w 2, w : w := ξ ξ 2 + ξ ξ 4, w 2 := ξ ξ + ξ 2 ξ 4, w := ξ ξ 4 + ξ 2 ξ. Using the above, we then find that w + w 2 + w = α, w w 2 + w w + w 2 w = ξξ 2 2 ξ + ξ ξ2ξ ξ ξξ ξ 2 ξ ξ4 2 +ξξ 2 2 ξ 4 + ξ ξ2ξ 2 + ξ ξ ξ4 2 + ξ 2 ξξ 2 4 +ξξ 2 ξ 4 + ξ ξ 2 ξ 2 + ξ ξ 2 ξ4 2 + ξ2ξ 2 ξ 4 = (ξ ξ 2 ξ + ξ ξ 2 ξ 4 + ξ ξ ξ 4 + ξ 2 ξ ξ 4 )(ξ + ξ 2 + ξ + ξ 4 ) 4ξ ξ 2 ξ ξ 4 = 4γ, and w w 2 w = ξ ξ 2 ξ ξ 4 + ξ 2 ξ 2 2ξ ξ 2 ξ 2 ξ ξ ξ 2 ξ ξ 4 +ξ 2 ξ 2 2ξ 2 + ξ ξ 2ξ ξ 4 + ξ ξ 2 ξ ξ 4 + ξ 2 2ξ 2 ξ 2 4 = (ξ 2 + ξ ξ 2 + ξ 2 4)ξ ξ 2 ξ ξ 4 + (ξ ξ 2 ξ + ξ ξ 2 ξ 4 + ξ ξ ξ 4 + ξ 2 ξ ξ 4 ) 2 2(ξ 2 ξ 2 2ξ ξ 4 + ξ 2 ξ 2 ξ 2 ξ 4 + ξ 2 ξ 2 ξ ξ ξ ξ 2 2ξ 2 ξ 4 + ξ ξ 2 2ξ ξ ξ ξ 2 2ξ 2 ξ 2 4) = γ ( (ξ + ξ 2 + ξ + ξ 4 ) 2 2(ξ ξ 2 + ξ ξ + ξ ξ 4 + ξ 2 ξ + ξ 2 ξ 4 + ξ ξ 4 ) ) +β 2 2ξ ξ 2 ξ ξ 4 (ξ ξ 2 + ξ ξ + ξ ξ 4 + ξ 2 ξ + ξ 2 ξ 4 + ξ ξ 4 ) = 2αγ + β 2 2αγ = β 2 4αγ.

7 EQUATIONS OF DEGREE AND 4. 7 Consider then the polynomial f(y ) := (Y (ξ + ξ 2 )) (Y (ξ + ξ 4 )) (Y (ξ + ξ )) (Y (ξ 2 + ξ 4 )) (Y (ξ + ξ 4 )) (Y (ξ 2 + ξ )) = ( Y 2 + (w 2 + w ) ) ( Y 2 + (w + w ) ) ( Y 2 + (w + w 2 ) ) = Y 6 + 2(w + w 2 + w )Y 4 We compute that +(w 2 + w w 2 + w w 2 + w w + w 2 w )Y 2 +(w 2 w 2 + w 2 w + w w 2 + w w w 2 2w + w 2 w 2 + 2w w 2 w ). and that w 2 + w w 2 + w w 2 + w w + w 2 w = (w + w 2 + w ) 2 + (w w 2 + w w + w 2 w ) = α 2 4γ, w 2 w 2 + w 2 w + w w 2 + w w w 2 2w + w 2 w 2 + 2w w 2 w whence = (w w 2 + w w + w 2 w )(w + w 2 + w ) w w 2 w = 4αγ + (4αγ β 2 ) = β 2, f(y ) = Y 6 + 2αY 4 + (α 2 4γ)Y 2 β 2. The roots of f(y ) are ± θ, ± θ 2, ± θ, with θ, θ 2, θ the roots of the polynomial: g(y) := y + 2αy 2 + (α 2 4γ)y β 2. Since θ, θ 2, θ are the roots of g(y) we have θ θ 2 θ = β 2, so that for any choice of square roots of the θ i, we necessarily have θ θ2 θ = ±β; it follows that we can choose the signs of the square roots in such a way that ( ) θ θ2 θ = β. Now we first show that we can choose the numbering of the θ i and the signs of the square roots θ i in such a way that ( ) is satisfied and so that the stated formulas for the roots of ( ) are correct. Since g(y 2 ) = f(y ) = ( Y 2 + (w 2 + w ) ) ( Y 2 + (w + w ) ) ( Y 2 + (w + w 2 ) ), we see that the roots of g(y) are the numbers (w 2 + w ), (w + w ), and (w + w 2 ). Renumbering the θ i if necessary, we may thus assume: Now, θ = (w 2 + w ), θ 2 = (w + w ), θ = (w + w 2 ). (w 2 + w ) = (ξ ξ + ξ 2 ξ 4 + ξ ξ 4 + ξ 2 ξ 2 ) = (ξ + ξ 2 )(ξ + ξ 4 ) = (ξ + ξ 2 ) 2

8 8 IAN KIMING (remember that ξ + ξ 2 + ξ + ξ 4 = 0). It follows that we can choose the sign of θ in such a way that θ = ξ + ξ 2. Similarly we find that we can choose the signs of θ 2 and θ such that θ 2 = ξ + ξ and θ = ξ + ξ 4. So, we fix the signs of the square roots so that: ( ) ξ + ξ 2 = θ, ξ + ξ = θ 2, ξ + ξ 4 = θ. We claim that ( ) is then satisfied; this is a consequence of the following computation: θ θ2 θ = (ξ + ξ 2 )(ξ + ξ )(ξ + ξ 4 ) = ξ + ξ 2 ξ 2 + ξ 2 ξ + ξ 2 ξ 4 + ξ ξ 2 ξ + ξ ξ 2 ξ 4 + ξ ξ ξ 4 + ξ 2 ξ ξ 4 = ξ 2 (ξ + ξ 2 + ξ + ξ 4 ) β = β. Now notice that ( ) implies: ( ) ξ + ξ 4 = θ, ξ 2 + ξ 4 = θ 2, ξ 2 + ξ = θ ; for we have ξ + ξ 4 = (ξ + ξ 2 ), and so on. The equalities in ( ) and ( ) give us a system of 6 linear equations for the ξ i that can be solved for the roots ξ i ; for instance we find: ξ = 2 (2ξ + (ξ + ξ 2 + ξ + ξ 4 )) = 2 ((ξ + ξ 2 ) + (ξ + ξ ) + (ξ + ξ 4 )) = 2 ( θ + θ 2 + θ ), and similar expressions for the other ξ i, more precisely: ( ) ξ ξ 2 ξ = ξ 4 2 ( θ + θ 2 + θ ) 2 ( θ θ 2 θ ) 2 ( θ + θ 2 θ ) 2 ( θ θ 2 + θ ). Now observe the following: If we renumber the θ i the only effect on the solution formula ( ) is a permutation of the roots ξ j. Similarly, if we change signs on some of the square roots θ, θ 2, θ in such a way that ( ) is still satisfied, we must have changed signs for precisely 2 of these square roots; again we check that the only effect this has on ( ) is a renumbering of the ξ j. So, the formula ( ) giving the 4 roots of ( ) is in fact independent of the various choices we made above (because we have obviously not fixed any numbering of the 4 roots). Remark: The solution of the general equation of degree 4 was discovered in 540 by Lodovico Ferrari ( ); have a look at the link for more on the history of this. Remark: In Galois theory one shows that there can not exist a general formula for the roots of a polynomial of degree 5, if one is only allowed to extract n th roots and use the usual arithmetic operations. For special polynomials however, this may of course be possible (consider x 5 2 for example). Galois theory gives a

9 EQUATIONS OF DEGREE AND 4. 9 deeper understanding of when the roots of a polynomial can be expressed via n th roots and the usual arithmetic operations. Galois theory also gives a much better understanding of what is really happening in the above proofs, and why a general solution via n th roots etc. actually is possible in degrees 2, and 4. kiming@math.ku.dk Dept. of math., Univ. of Copenhagen, Universitetsparken 5, 200 Copenhagen Ø, Denmark.