Unraveling the Complexities of the Cubic Formula. Bruce Bordwell & Mark Omodt Anoka-Ramsey Community College Coon Rapids, Minnesota

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1 Unraveling the Complexities of the Cubic Formula Bruce Bordell & Mark Omodt Anoka-Ramsey Community College Coon Rapids, Minnesota

2 A little history Quadratic Equations Cubic Equations del Ferro ( ) Tartaglia ( ) Cardano ( ) Bombelli ( ) Ferrari ( ) Abel ( )/Galois ( )

3 The Cubic Formula

4 axbxcx d 0 x t b a b b b a a a a ( t ) b ( t ) c ( t ) d 0 t pt q 0 t p p acb a q 9 7 p p b abc a d 7a ( ) p ( ) q 0 6 p 0 7 q

5 6 p 0 7 q q q 4 p 7 4 acb a b 9abc 7a d b 9abc 7a d 7a 7a 7

6 t p Why? If 0, hat happens if e substitute t v? t pt q v p v q 0 v v v p pv q 0 Note that if e re trying to eliminate terms, v v p pv v p v pv v p v v p v v p So if e choose p v, then those 4 terms are eliminated from the equation and e get p p 0 v q q q 7. Multiplying through by gives us p q

7 A simple example

8 x x x 14 0 x t t t b a (1) 1 t 1 + t 1 (t 1) 14 = 0 t pt q 0 p acb a 6 b 9abc 7a d 7a q 9 t p 6t 9 0 t ( ) 6( )

9 6 9 8 ( 1)( 8) cos(10 k) isin(10 k) 1, i 80 cos(10 k) isin(10 k), 1 i

10 1 t x t 1 1 t x t 1 i t i x t 1 i i t i x t 1 i 5 4 i t i x t 1 i i t i x t 1 i 5 6

11 Another simple example

12 x x 7x 1 0 x t b t 1 a t pt q 0 p acb a 4 b 9abc 7a d 7a q 56 t p 4t 56 0 t ( 64)( 8) 0

13 cos(6010 k) i sin(6010 k) 4, i 80 cos(10 k) isin(10 k), 1 i t 1 x t 1 1 t x t 1 1 i t i x t i i t i x t i i t i x t i i t i x t i

14 What can go rong?

15 x x 4x 1 0 x t b t 1 a t p 7t 6 0 t ( ) i

16 10 9 i cos tan 60k i sin tan 60k i 10 cos tan sin tan i i 7 i t x t 1 1 6

17 i 10 cos tan 10 sin tan i t x t i 10 cos tan 40 sin tan i 7 t 1 x t 1

18 What can go really rong?

19 x t t 4 x x x b a t t 1 1t p ( ) i cos tan sin tan i k i k cos tan 7 10 sin tan 7 10 k i k 1 cos 1tan sin 1tan k i k

20 cos tan isin tan i p t x t cos tan 10 isin tan i p t x t cos tan 40 isin tan i p t x t

21 Tartaglia s Formula, but Cardano s Method The solutions to x + px + q = 0 are A + B, A + B + (A B) i, A + B (A B) i here A = q + q 4 + p 7 and B = q q 4 + p 7

22 Let A B x then substituting, ( A B) p( A B) q 0 A A B AB B pa pb q 0 Factor by Grouping A B AB A B p A B q ( ) ( ) 0 A B AB a A B q ( )( ) 0 For this to equal 0, then Note: If AB0, then q 0 and * AB p 0 and A B q or x px 0 has trivial solutions. p A B and A B q Note: ( z A )( z B ) 0 7 product sum z z( A B ) AB 0 A and B are solutions to the quadratic equation p 7 z qz 0

23 4p 4p q q q q 7 and 7 and are interchangable A B Note: A B A q q p q q p and B q q p q q p A and B Cardano assumed that the solution is real. The solution is q q p q q p x A B

24 To find the other to solutions, e simply solve the resulting quadratic equation hen the original equation is divided by the linear factor. Using synthetic division, ( A B) 1 0 p q ( A B) ( A B) p( A B) ( A B) 1 ( A B) p ( A B) 0 ( ) ( ) ( ) x px q x A B x A B x p A B The solutions of x ( A B) x p ( A B) 0 are..

25 x x ( ) ( ) 4 ( ) ( A B) ( A B) 4 p A B A B A B p ( A B) i 4 p x ( AB) Note: ( A B) 4 AB ( A B) ( A B) i p x ( A B) here * AB ( A B) ( A B) 4 p x i or ( A B) ( A B) Therefore, all solutions to x px q 0 are ( A B) ( A B) ( A B) ( A B) A B, i, i here q q p q q p A and B

26 The first simple example

27 x x x 14 0 xt1 Solving t 6t 9 0 by substituting p 6 and q 9 A A and B A A and B A and B1 The solutions are t A B x 1 AB AB 5 t i i x i AB AB 5 t i i x i 1

28 What can go rong?

29 x x x xt1 Solving t 7t 6 0 by substituting p 7 and q and 9 9 A i B i AB i i 9 9 Then, =? Ho can e use the formula ith such a cumbersome expression? Can e algebraically rerite A and Bas simplified complex numbers? AB That's hat the calculator said! Really! Solve: i 4p A B Since, ( A B) ( A B) Solving the system, produces A and B in complex form. A i and B i 6 6

30 x x x Solving t 7t 6 0 by substituting p 7 and q and 9 9 A i B i or A i and B i 6 6 The solutions are: t A B x t t xt1 1 A B A B 1 i x A B A B 1 i 1 x

31 A geometric representation of the solutions to a general cubic equation f ( x) ax bx cx d 0

32 Cardano s approach involves solving the depressed cubic (no squared term): g t t pt q ( ) 0 Since, every cubic can be depressed by substituting x b t in f( x) a Why this substitution? Cardano mentions that it becomes much easier to solve the depressed cubic than to deal ith the original.

33 Creating a depressed cubic ax bx cx d 0 If x t e, hat does e need to be in order to eliminate the x term? a t e b t e c t e d 0 at aet ae t ae bt bet be ct ce d 0 If e ant aet bt t ae b 0, then b e. a

34 Geometrically: The x = t b a substitution defines a horizontal f( x) translation of the inflection point of to the y-axis. The inflection point of any cubic is also the point of symmetry of its graph. Solving f ''( x) 6ax b 0 yields x b a gt () f( x) x b a

35 The Geometry of the depressed cubic: g(t) t pt q gt () q h

36 Here and h are defined by the local extrema of g( t). Solving g '( t) t p 0 yields t. p Therefore: o r p Note: defining the parameters so no restriction on p is necessary. Defining h : h g(0) g( ) q ( p q), substituting h or h 6 4 p p is only used in Note: q is the vertical translation.

37 Interpreting Cardano s Discriminant A q q p q q p and B q p If 0 then the solutions ill be 4 7 real here at least are the same. q p If 0 then the solutions ill be 4 7 distinct real. q p If 0 then the solutions ill be real and complex.

38 We can no define Cardano's discriminant in terms of the geometric variables, h and q: q 4 p q p q 4( ) Substituting h 4 e get the geometric discriminant: q h 4 The geometric interpretation of the discriminant helps us understand the nature of the roots by comparing the relative size of q and h.

39 Scenario 1: If q h 0 or q h, then 1 real and complex roots. q h q p Cardano Discriminant: 0 4 7

40 Scenario : If q h 0 or q h, then real roots here one is a double root. q h q p Cardano Discriminant: 0 4 7

41 Scenario : If q h 0 or q h, then real distinct roots. q h q p Cardano Discriminant: 0 4 7

42 In Conclusion: Instead of solving of the solutions, t tq t pt q 0 0 and having no geometric understanding solve: here the parameters, h and q in the formula have a direct correlation to the geometry of the cubic. Then, Cardano's solutions ould be reritten as: q q h q q h A and B here h 4 and the discriminant q h determines the nature of the roots.

43 x x x 14 0 xt1 Solving t 6t 9 0 by defining and h A A A and B te: the geometric discriminant h No q A and B A and B1 The solutions are t A B x 1 AB AB t i i x AB AB t i i x 1 5 i 5 i

44 An example ith 15 reference angles

45 x 1x8 0 x t b t a t pt q 0 p acb a 1 q b 9abc7a d 7a 8 p t 1t8 0 t (1)(64) i (1) 4 4 i 8 cos( k ) i sin( k ) cos( 1510 k) isin( 1510 k) 4 cos(15 ) isin(15 ) i t x cos(55 ) isin(55 ) i t x cos(15 ) isin(15 ) i t x cos(105 ) isin(105 ) i t x cos(5 ) isin(5 ) i t x cos(45 ) isin(45 ) i t x 6 6

46 A surprisingly nice example

47 x t t ( t ) 6( t ) 11( t ) 6 0 x x x b a t pt q 0 p acb a 1 q b 9abc7a d 7a 0 t t 1 ( 1 ) ( 1 ) 0 t p cos(18060 k ) i sin(18060 k ) cos(0 60 ) sin(0 60 ) cos(0 60 ) sin(0 60 ) k i k k i k cos(0 ) isin(0 ) i t 1 x t 1 cos(90 ) isin(90 ) i t 0 x t cos(150 ) isin(150 ) i t 1 x t cos(10 ) isin(10 ) i t 1 x t cos(70 ) isin(70 ) i t 0 x t cos(0 ) isin(0 ) i t 1 x t

48 An Interesting Expression: If for any value of x, Let M = x + x and N = M + x Then x = M + N + M N Examples: 1 = = = =

49 If a solution to M = x +x = q x + px + q = 0 and is to be of this form then, N = x + x + x = q 4 + p 7 By substitution e see that x = p is a solution and q = p 7 + p Example: p p ( 6) ( 6) Note: q p x 6 is a solution. 4p The solutions are x ABx 1 1 x AB AB i i i x 1 x AB AB i i i x 1 Solve x 6x4 0, here p 6 and q 4 Set A B and A B i since, ( A B) ( A B).

50 Questions?

51 Additional nice examples x x x t 6t i x 6, x x x x x 0 x x x x x x x x x t t t t t 4t 7 0 1t16 0 1t16 0 t 0 t x10, 1 i x, 4, x,, x, i x,

Appendix 1. Cardano s Method

Appendix 1. Cardano s Method Appendix 1 Cardano s Method A1.1. Introduction This appendix gives the mathematical method to solve the roots of a polynomial of degree three, called a cubic equation. Some results in this section can