THE CUBIC AND QUARTIC EQUATIONS IN INTERMEDIATE ALGEBRA COURSES
J Villanueva Florida Memorial College 15800 NW 42nd Ave Miami, FL 33054 [email protected]
I. Introduction A. The cubic and quartic in history B. The quadratic equation II. Solution to the cubic III. Solution to the quartic IV. Examples V. Conclusion
I. Introduction A. Historical background
The solution of the cubic and quartic equations is important in the history of mathematics for several reasons. First, it was the first major advance by modern man since the time of the ancient Greeks. It was the first mathematical formula unknown to the ancients. Second, it forced mathematicians to take both complex numbers and negative numbers seriously. And, more importantly, it led to the study of the theory of equations, culminating in the nineteenth century in the proof of the insolvability of the quintic.
There are a number of key figures in the triumph of the cubic and quartic formulas. Omar Khayyam (1048-1123) used intersections of conics to give geometric constructions of roots of cubics. Leonardo of Pisa (Fibonacci) (c1180-1245) had an approximation formula for certain forms of the cubic. About 1515, Scipione del Ferro (d1526) discovered a method for finding the roots of several forms of the cubic and shared these secrets with some of his students. Antonio Ma Flor, one of these students, challenged mathematicians, in 1535, to a problem-solving contest involving the cubic. Niccolo Fontana (Tartaglia, 'the stammerer') (c1500-1557) answered the call and found the general solution. Giralamo Cardano (1501-1576), published the secret formula in his treatise Ars Magna (The Great Art, or The Rules of Algebra), in 1535, with due credit to Tartaglia.
The quartic formula was discovered by Lodovici Ferrari (1522-1565) in 1540. A similar formula was found by Rene Descartes (1596-1650) around the same time.
For hundreds of years, mathematicians sought some generalization of the classical formulas that would give the roots of any polynomial. Finally, P. Ruffini (1765-1822), in 1799, and Niels Henrik Abel (1802-1829), in 1824, proved that no such formula exists for the general quintic. Evariste Galois (1811-1832) was able to determine precisely those polynomials whose roots can be found (and, in so doing, founded the theory of groups).
B. The quadratic equation
The solution of the quadratic equation: ax 2 +bx +c =0 is familiar to every student of intermediate algebra: −b± b2 −4ac x ,x = 1 2 2a
with the discriminant ∆ = {+, two real roots; -, two complex roots; 0, one double root} and with the properties: b ; c . x + x = − x •x = 1 2 a 1 2 a
II. Solution to the Cubic
The solution to the cubic: ax 3 +bx 2 +cx +d = 0 will be patterned after the quadratic formula. Assume: = The discriminant ∆ = − 2 − 2 − 2 = {+, three real a 1. (x1 x2 ) (x1 x3) (x2 x3) roots; -, two complex roots; 0, multiple roots}.
Step 1. Let b ⇒ 3 + + = (reduced equation). Then x = y − . y py q 0 3 b3 bc 2b3 p = c− q = d − + . 3 3 27
2 2 2 Step 2. Let p ⇒ 2 p ⇒ q p q y = z − (z3) +q(z3)− = 0 z3 = − ± + 3z 3 2 9 4 Choose: p then: = + =ω +ω 2 =ω 2 +ω z z = − , y1 z1 z2 y2 z1 z2 y3 z1 z2 1 2 3 1 3 ω3 = +ω +ω 2 = ω = +i , complex cube root of unity, with identities: 1; 1 0. 2 2 ⇒ y + y + y = 0 ; + + = ; y y y = −q ⇒ ∆ = −4p3 −27 q2. 1 2 3 y1 y2 y2 y3 y3 y1 p 1 2 3
Case 1. ∆ =0
q z = z = 3 − 1 2 2 3 q = + = − ; 2 3 ; = y1 z1 z2 4q y =ωz +ω z = ()−1 z = y3 y2. 2 1 2 1 2
2 2 q p q Case 2. ∆ < 0. z ,z = 3 − ± + 1 2 2 3 2 y = z + z ; = 1 + ± i − 1 1 2 y , y ()z1 z2 ()z1 z2 3 2 3 2 2
Case 3. ∆ > 0. ⇒ p < 0. Reduced equation: y3 + py +q =0 ⇒ p θ = 3θ − θ Let y = 2 − z, and using the identity: cos 3 4cos 3cos , 3 2/3 ⇒ 3 − = = θ q 3 4z 3z k z cos ; k = − − . 2 p
p 2nπ , n = 0, 1, 2. ∴y = 2 − cos θ + n 3 3
III. Solution to the Quartic x4 +bx 3 +cx 2 +dx +e = 0.
A. Ferrari b ⇒ 4 2 2 Let x = y − y + py + qy + r = ,0 or y = −py −qy −r. 4 z2 Add: y2z + to both sides: 4 2 2 (1*) z z2 = (my +k) ; m, k to be determined. 2 + = ()− 2 − + − y z p y qy r 2 4 ⇒ z y2 + = ±()my +k . 2 In (*1), RHS is quadratic in y 2. It is a perfect square if discriminant is zero, ie: z2 , or (*2) 3 − 2 − + ( − 2 )= 2 − ()− − = z pz 4rz 4pr q ,0 q 4 z p r 0 4 called the resolvent cubic.
Ex. 1. y4 +3y2 −2y +3= .0 (*2) becomes: 2 z3 −3z2 −12 z +32 = ,0 which has root: z = 4. Thus: (y 2 + 2) = y 2 + 2y +1 = ()y +1 2 + = + + ⇒ 1 3 y 2 (y 1) y = ±i 2 2 2 ⇒ 1 11 y + 2 = −(y+1) y = ±i . 2 2
B. Descartes: reduced quartic y4 + py 2 + qy + r = .0
Factor: (*3) (y2 +ky +m)(y2 −ky +n)= y4 + (m+n−k 2 )y2 +(kn −km )y + mn ⇒ m+ n−k 2 = p m+ n = p+ k 2 k(n−m)= q mn = r ⇒ 2n = p+k 2 +q/k 2m = p+ k 2 −q/k 3 2 q q 2 + 2 + 2 − 2 − 2 = 4r = 2n2m = p+k 2 + p+k 2 − , or: (k ) 2p(k ) (p 4r)k q .0 (*4) k k Any root of (*4) gives a factorization (*3).
3 2 Ex. 2. y4 −3y2 +6y −2 = 0 ⇒ (k 2 ) −6(k 2 ) +17 k 2 −36 =0 with root k 2 = 4. Thus: y4 −3y2 +6y −2 = (y2 +2y−1)(y2 +2y+2) ∴ = − + ; = − − ; = + ; = − . y1 1 2 y2 1 2 y3 1 i y4 1 i
IV. Examples
Ex. 1. x3 +7x2 +11 x+5= 0 = = = ⇒ b2 16 bc 2b3 128 b 7,c 11 ,d 5 p = c− = ; q = d − + = 3 3 3 27 27 ⇒ ∆ = −4p3 −27 q2 = 0 ⇒ multiple roots:
= 3 − = −q 3 q 4 ∴ = − b = − , = = − y 4q y = y = = x y x1 5 x2 x3 1 1 3 2 3 2 3 3
Ex. 2. x3 −6x−9 =0 x = y ⇒ p = −6;q = −9 ⇒ ∆ = −1323 ⇒ two complex roots ∴x = z + z =3; = 1 + ± i − = 1 − ± 1 1 2 x ,x ()z1 z2 ()z1 z2 3 ( 3 i 3) 2 3 2 2 2
Ex. 3. 2x3 −5x2 − x+6= 0 ⇒ 5 1 ⇒ 5 1 ⇒ 31 77 x3 − x2 − x +3= 0 b =− ,c =− ,d =3 p = − ,q = 2 2 2 2 12 54 ⇒ ∆ = 225 ⇒ three real roots 16
2/3 p 2nπ 1 −1 q 3 = − θ + θ = cos − − = .0 89102 yn 2 cos 2 p 3 3 3 = 2 b = − 3 7 ; 11 ; y3 ; ∴ = − = x 1; x = . y = y =− 3 x1 y1 ;2 2 3 1 6 2 6 3 2
Ex. 4. ( Curio): x3 + x2 −2=0 ⇒ = = = − ⇒ b2 1 bc 2b3 52 b 1,c 0,d 2 p = c− = − ; q =d − + = − 3 3 3 27 27 ⇒ 212 ∆ = −4p3 −27 q2 = − 27 3 2 q p q 13 ; y = z + z 13 3 ; z ,z = 3 − ± + = 26 ±15 3 1 1 2 = 26 ±15 3 + 26 −15 3 1 2 2 27 27 3 3 = −1 + ± i 3 − = −13 + +3 − ± i 33 + −3 − y , y ()z1 z2 (z1 z2 ) 26 15 3 26 15 3 26 15 3 26 15 3 2 3 2 2 6 6 1; 1 ∴x = y − x = 3 26 +15 3 +3 26 −15 3 −1 3 1 3 1 i 3 x ,x = 3 26 +15 3 +3 26 −15 3 +2± 3 26 +15 3 −3 26 −15 3 2 3 6 6 But the original equation has the root: x = 1. ⇒ 3 3 and . 26 +15 3 + 26 −15 3 = 4 3 26 +15 3 −3 26 −15 3 = 2 3
V. Conclusion
Any polynomial equation of degree ≤ 4, with real coefficients, is solvable by radicals or complex numbers. When the solution is impractical, we use other means, like the Rational Root Theorem, Descartes’ Rules of Signs, or approximation methods, like Newton's approximation method.
Bibliography:
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