unraveling the complexities of the cubic formula · pdf filea little history •quadratic...
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Unraveling the Complexities of the Cubic Formula
Bruce Bordwell &Mark Omodt
Anoka-Ramsey Community CollegeCoon Rapids, Minnesota
A little history…
• Quadratic Equations
• Cubic Equations
• del Ferro (1465 – 1526)
• Tartaglia (1500 – 1557)
• Cardano (1501 – 1576)
• Bombelli (1526 – 1572)
• Ferrari (1522 – 1565)
• Abel (1802 – 1829)/Galois (1811 – 1832)
The Cubic Formula
3 2 0ax bx cx d 3ba
x t
3 23 3 3
( ) ( ) ( ) 0b b ba a a
a t b t c t d
3 0t pt q 2
23
3ac b
ap 3 2
32 9 27
27b abc a d
aq
3pw
t w 33 3
( ) ( ) 0p pw w
w p w q
36 327
0pw qw
36 327
0pw qw
342273
2
pq qw
323
23 2 3 2 3
3 3
422 9 27 2 9 27
2727 273
2
ac bab abc a d b abc a d
a aw
Why ? 3pw
t w
If 3 0t pt q , what happens if we substitute t w v ?
3
0w v p w v q
23 32 03 3w v w pvw vv qpw
Note that if we’re trying to eliminate terms,
2 2 2 23 3 3 3 3 3 3w v wv pw pv w v pw wv pv w wv p v wv p w v wv p
So if we choose 3
pv
w
, then those 4 terms are eliminated from the equation and we get
3 33 3 3 3
30
3 27
p pw v q w q w q
w w
.
Multiplying through by 3w gives us 3
6 3 027
pw qw .
A simple example
3 23 3 14 0x x x 3
3 3(1)1b
ax t t t
3 0t pt q
2
2
3
36ac b
ap
3 2
3
2 9 27
279b abc a d
aq
3 6 9 0t t 23
p
w wt w w
32 2( ) 6( ) 9 0w w
w w 6 39 8 0w w
𝑡 − 1 3 + 3 𝑡 − 1 2 − 3(𝑡 − 1) − 14 = 0
6 3 3 39 8 ( 1)( 8) 0w w w w
3 1 0w
312 2
cos(120 ) sin(120 ) 1,w k i k i
3 8 0w
2 cos(120 ) sin(120 ) 2, 1 3w k i k i
21 1 3 1 2
ww t w x t
22 2 3 1 2
ww t w x t
3 3 33 51 23 2 2 2 2 2 2
1w
w i t w i x t i
3 33 524 2 2 2 2
1 3 1w
w i t w i x t i
3 3 33 51 25 2 2 2 2 2 2
1w
w i t w i x t i
3 33 526 2 2 2 2
1 3 1w
w i t w i x t i
Another simple example
3 23 27 31 0x x x 31b
ax t t
3 0t pt q
2
2
3
324ac b
ap
3 2
3
2 9 27
2756b abc a d
aq
3 24 56 0t t 83
p
w wt w w
6 3 3 356 512 ( 64)( 8) 0w w w w
3 64 0w 4 cos(60 120 ) sin(60 120 ) 4, 2 2 3w k i k i
3 8 0w 2 cos(120 ) sin(120 ) 2, 1 3w k i k i
81 4 2 1 1
ww t w x t
82 2 2 1 1
ww t w x t
83 2 2 3 1 3 3 1 2 3 3
ww i t w i x t i
84 1 3 1 3 3 1 2 3 3
ww i t w i x t i
85 1 3 1 3 3 1 2 3 3
ww i t w i x t i
86 2 2 3 1 3 3 1 2 3 3
ww i t w i x t i
What can go “wrong”?
3 23 4 12 0x x x 31b
ax t t
3 7 6 0t t 73 3
p
w wt w w
6 3 34327
6 0w w
2 343 400
27 276 6 4( ) 63 10 3
2 2 93w i
3 10 3
93w i
3 1 110 3 10 37 21
9 27 27cos tan 360 sin tan 360w k i k
1 110 3 10 37 21 1 131 9 3 27 3 27
cos tan sin tanw i
1 1.5 0.288675134595w i
331 2 6
w i
33 71 2 6 3
3 1 2w
w i t w x t
2 32 3
1w i 73
2 1 3w
t w x t
5 313 2 6
w i 73
1 1 2w
t w x t
1 110 3 10 37 21 1 132 9 3 27 3 27
cos tan 120 sin tan 120w i
1 110 3 10 37 21 1 133 9 3 27 3 27
cos tan 240 sin tan 240w i
What can go really “wrong”?
3 24 5 6 0x x x
43 3ba
x t t
3 31 1463 27
0t t
313 9pw w
t w w
6 3146 2979127 729
0w w
2146 146 2979127 27 729
4( )3 30273
2 27 3w i
3 1 1302 31 31 9 302 9 3027327 3 27 73 73
cos tan 360 sin tan 360w i k i k
1 131 31 9 302 9 3021 1327 3 73 3 73
cos tan 120 sin tan 120w k i k
1 131 9 302 9 3021 13 3 73 3 73
cos tan 120 sin tan 120w k i k
1 131 9 302 9 3021 11 3 3 73 3 73
cos tan sin tan 1.7248775 0.6850125w i i
1 1
31 41 1 1 1 13 9 3
3.4497549 4.7830883pw w
t w w x t
1 131 9 302 9 3021 12 3 3 73 3 73
cos tan 120 sin tan 120 1.4556770 1.1512814w i i
2 2
31 42 2 2 2 23 9 3
2.9113540 1.5780206pw w
t w w x t
1 131 9 302 9 3021 13 3 3 73 3 73
cos tan 240 sin tan 240 0.2692005 1.8362940w i i
3 3
31 43 3 3 3 33 9 3
0.5384010 0.7949324pw w
t w w x t
Tartaglia’s Formula,but Cardano’s Method
𝑇ℎ𝑒 solutions to 𝑥3 + 𝑝𝑥 + 𝑞 = 0 are
𝐴 + 𝐵, −𝐴 + 𝐵
2+(𝐴 − 𝐵)
2𝑖 3, −
𝐴 + 𝐵
2−(𝐴 − 𝐵)
2𝑖 3
where 𝐴 =
3
−𝑞
2+
𝑞2
4+𝑝3
27and 𝐵 =
3
−𝑞
2−
𝑞2
4+𝑝3
27
3
3 2 2 3
3 3
3 3
Let then substituting,
( ) ( ) 0
3 3 0 Factor by Grouping
3 ( ) ( ) 0
(3 )( ) 0
For this to equal 0, then If Note: 0,
A B x
A B p A B q
A A B AB B pA pB q
A B AB A B p A B q
A B AB a A B q
A B
3 3 3
33 3 3 3 3 3
2 3 3
then 0 and
3 0 and or 0 has trivial solutions.
and ( )( ) 027
(
*
Note:
)
q
AB p A B q x px
pA B A B q z A z B
z z A B
product sum3 3
3 3
32
0
and are solutions to the quadratic equation
z 027
A B
A B
pqz
3 32 2
3 3
2 3 2 33 3
2 3 2 3
3 3
4 4
27 27 and and are interchangable
2 2
and 2 4 27 2 4 27
and Cardano assumed that the solution is real.2
Note:
4 27 2 4 27
The solu
p pq q q q
A B A B
q q p q q pA B
q q p q q pA B
2 3 2 3
3 3
tion is
2 4 27 2 4 27
q q p q q px A B
To find the other two solutions, we simply solve the resulting quadratic equation when the original equation is divided by the linear factor. Using synthetic division,
The solutions of
are…..
2 3
2
3 2 2
( ) 1 0
( ) ( ) ( ) ( )
1 ( ) ( ) 0
( ) ( ) ( )
A B p q
A B A B p A B A B
A B p A B
x px q x A B x A B x p A B
2 2( ) ( ) 0x A B x p A B
2 2
2
22
2
2
( ) ( ) 4 ( )
2
( ) 3( ) 4
2
( ) 3 ( ) 4 ( )
2 2
( ) 3 where
2
4( )
2 3
( ) ( )3
Note: 3
( ) *
2 2
A B A B A B px
A B A B px
A B ix A B AB A B
A B i px AB
pA B
A B
A B A Bx i
2 2
3
2 3 2 3
3 3
4 or ( ) ( )
3
Therefore, all solutions to 0 are
( ) ( ) ( ) ( ) , 3, 3
2 2 2 2
where and 2 4 27 2 4 27
pA B A B
x px q
A B A B A B A BA B i i
q q p q q pA B
The first simple example
3 23 3 14 0x x x 1x t
3Solving 6 9 0 by substituting 6 and 9t t p q
3 3
3
2
1
3
3 3
1
9 81 216 9 81 216 and
2 4 27 2 4 27
9 818
2 4
9 49
2 4
9 7 9 7 and
2 2 2 2
2 and 1
The solutions are 23
2
A B
A
A
A B
A B
t
B B
xA B
A At
2
3 3
3 33
2 2 2
3 33
2 2 2 2
5 32 2
5 32 2
i i
A B A B
x i
x it i i
What can go “wrong”?
3
3 3
Solving 7 6 0 by substituting 7 and 6
10 10 3 3 and 3 3
9 9
Then, =
How can we use the formula with such a cumbersome expression?
Can we algeb ai
r
3 3 ?10 103 3 3 39 9
t t p q
A i B i
A B i i
2 2
cally rewrite and as simplified complex numbers?
3 That's what the calculator said! Really!
Solve: 3 4 Since, ( ) ( )
3 3
Solving the system, produces and
A B
A B
i pA B A B A B
A
in complex form.
3 3 3 3 and
2 6 2 6
B
A i B i
3 23 4 12 0x x x 1x t
1
2
3
3 3
1
2
3
Solving 7 6 0 by substituting 7 and 6
10 10 3 3 and 3 3
9 9
3 3 3 3 or and
2 6 2 6
The solutions are:
3
3 13 2
2
2
2 2 2
2 2
3
t t p q
A i B i
A i B i
t A B
A B A Bt i
A B A Bt
x
x
2
31 2
13
2 2i x
3 23 4 12 0x x x 1x t
A geometric representation of the solutions to a general
cubic equation
3 2( ) 0f x ax bx cx d
3( ) 0g t t pt q
3
bx t
a ( )f x
Cardano’s approach involves solving the depressed cubic (no squared term):
Since, every cubic can be depressed by substituting
in
Why this substitution? Cardano mentions that it becomes much easier to solve the depressed cubic than to deal with the original.
Creating a depressed cubic
3 2 0ax bx cx d
If x t e , what does e need to be in order to eliminate the 2x term?
3 2
0a t e b t e c t e d
3 2 2 3 2 23 3 2 0at aet ae t ae bt bet be ct ce d
If we want 2 2 23 3 0aet bt t ae b , then 3
be
a
.
Geometrically: The 𝑥 = 𝑡 −𝑏
3𝑎substitution defines a horizontal
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.
( )f x
( )f x( )g t
3
bx
a
Solving ''( ) 6 2 0 yields 3
bf x ax b x
a
( )g t
hq
3
The Geometry of the depressed cubic:
(t) tg pt q
2
2
2Note: is only used in
defining the parameters so no restriction on is necessa
Here and are defined by the local extrema of ( ).
Solving '( ) 3 0 yields .3
Therefore: r 3
o 3
p
h g t
pg t p t
p p
t
3 2
63 2
ry.
Defining :
(0) ( )
( ), substituting
2 or
Note: is the vertical trans
3
lati
4
on.
h
p
h
h g g
q p q
h
q
Interpreting Cardano’s Discriminant
3 3
2 3 2 3
2 3
2 3
and 2 2
If 0 then the solutions will be4 27
3 real where at least 2 are the same.
If 0 then the solutions will be4 27
3 distinct real.
4 27 4 27
q qA B
q
q p
p
p
p
q
q
2 3
If 0 then the solutions will be4 27
1 real and 2 complex.
q p
3
22 3 3
6
2 2
2
We can now define in terms of the geometric
variables , and :
44( )
4 27 4 4
Substituting we get the :
3
4
h q
p
h
p q
Cardano's discriminant
geometric discriminant
2 2
The geometric interpretation of the discriminant helps us
understand the nature of the roots by comparing the relative
size of
4
and .q
q h
h
q h
2 2
Scenario 1:
If 0 or , then 1 real and 2 complex roots.q h q h
2 3
Cardano Discriminant: 04 27
q p
2 2
Scenario 2:
If 0 or , then 3 real roots where one is a double root.q h q h
q h
2 3
Cardano Discriminant: 04 27
q p
2 2
Scenario 3:
If 0 or , then 3 real distinct roots.q h q h
q h
2 3
Cardano Discriminant: 04 27
q p
3
3 2
In Conclusion:
Instead of solving 0 and having no geometric understanding
of the solutions,
solve: where the parameters , and in the
formula have a direct correlation to th
03
t pt q
h qt t q
2
3 3
2 6
2
2 2
2 2
e geometry of the cubic.
Then, Cardano's solutions would be rewritten as:
and B2 2
where 4 and the discriminant determines the
nature of the r
4
o
4
o ts.
q h q hq qA
h q h
3 23 3 14 0x x x 1x t
3 2 2Solving 6 9 0 by defining 2 and 32t t h
3 3
3
2 23
3 3
the geometric discriminant No q 0te:
9 81 32 9 81 32 and
2 4 2 4
9 818
2 4
9 49
2 4
9 7 9 7 and
2 2 2 2
2 and 1
The solu
A B
A
A h
A B
A B
1
2
1
2
33
tions are 3
3 33
2 2 2 2
3 33
2 2 2 2
2
5 3
2 2
5 3
2 2
t A B
A B A Bt i i
A B A Bt i
x
ii
x i
x
An example with 15°reference angles
3 12 8 2 0x x
3ba
x t t
3 0t pt q
2
23
312ac b
ap
3 2
32 9 27
278 2b abc a d
aq
3 12 8 2 0t t
43p
wwt w w
6 38 2 64 0w w
28 2 8 2 4(1)(64)
3 8 2 12822(1)
4 2 4 2w i
3 4 2 4 2 8 cos( 45 360 ) sin( 45 360 )w i k i k
2 cos( 15 120 ) sin( 15 120 )w k i k
41
2 cos(135 ) sin(135 ) 2 2 2 2ww i i t x w
6 2 6 2 42 2 2
2 cos(255 ) sin(255 ) 6 2ww i i t x w
6 2 6 2 43 2 2
2 cos(15 ) sin(15 ) 6 2ww i i t x w
6 2 6 2 44 2 2
2 cos(105 ) sin(105 ) 6 2ww i i t x w
45
2 cos(225 ) sin(225 ) 2 2 2 2ww i i t x w
6 2 6 2 46 2 2
2 cos(345 ) sin(345 ) 6 2ww i i t x w
A surprisingly nice example
3 26 11 6 0x x x
32b
ax t t
3 2( 2) 6( 2) 11( 2) 6 0t t t
3 0t pt q
2
23
31ac b
ap
3 2
32 9 27
270b abc a d
aq
3 0t t
13 3pw w
t w w
31 13 3
( ) ( ) 0w w
w w
6 127
0w
6 1 127 27
cos(180 360 ) sin(180 360 )w k i k
31627 3
cos(30 60 ) sin(30 60 ) cos(30 60 ) sin(30 60 )w k i k k i k
3 31 11 3 2 6 3
cos(30 ) sin(30 ) 1 2 3w
w i i t w x t
3 3 12 3 3 3
cos(90 ) sin(90 ) 0 2 2w
w i i t w x t
3 31 13 3 2 6 3
cos(150 ) sin(150 ) 1 2 1w
w i i t w x t
3 31 14 3 2 6 3
cos(210 ) sin(210 ) 1 2 1w
w i i t w x t
3 3 15 3 3 3
cos(270 ) sin(270 ) 0 2 2w
w i i t w x t
3 31 16 3 2 6 3
cos(330 ) sin(330 ) 1 2 3w
w i i t w x t
𝑀 =𝑥3 + 3𝑥2
2
An Interesting Expression:
If for any value of x, Let
and
Then 𝑥 =3𝑀 + 𝑁 +
3𝑀− 𝑁
Examples:
1 =3
2 + 5 +3
2 − 5
2 =3
10 + 108 +3
10 − 108
3 =3
27 + 756 +3
27 − 756
4 =3
56 + 3200 +3
56 − 3200
𝑁 = 𝑀2 + 𝑥3
If a solution to is to be of this form then,𝒙𝟑 + 𝒑𝒙 + 𝒒 = 𝟎
M = 𝑥3+3𝑥2
2= −
𝑞
2𝑁 =
𝑥3 + 3𝑥2
2
2
+ 𝑥3 =𝑞2
4+𝑝3
27and
By substitution we see that 𝒙 =𝒑
𝟑is a solution and −𝒒 =
𝒑𝟑
𝟐𝟕+𝒑𝟐
𝟑3
3 2 3 2
2 2
1
Solve 6 4 0, where 6 and 4
( 6) ( 6) + 4 +
27 3 27 36
2 is a solution.3 3
4Set and since, ( ) ( ) .
3 The solutions
Note:
2
are
2
i
x x p q
p pq
p
pA B A B A B A B
x
x
Example:
1
22
3 3
2
2 21 3
2 2
3 32 2 2 2
3 32
12
32 2
x
i x
i x
A B
A B A Bx i i
A B A Bx i i
Additional “nice” examples
3 29 21 18 0x x x 3 6 9 0t t 6 39 8 0w w 3 3
6,2 2
ix
3 212 24 40 0x x x 3 24 72 0t t 6 372 512 0w w 10, 1 3x i
3 26 32 0x x 3 12 16 0t t
6 316 64 0w w 2, 4, 4x
3 23 9 27 0x x x 3 12 16 0t t
6 316 64 0w w 3, 3, 3x
3 26 15 14 0x x x 3 3 0t t
6 1 0w 2, 2 3x i
3 29 15 9 0x x x 3 3 0t t
6 64 0w 3, 3 2 3x