· logo1 Introduction Quadratic Equations Cubic Equations Quartic Equations Theorem. The quadratic formula. Let (F,+,·) be a ﬁeld and let a,b,c∈F with a6= 0. Then the number

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Introduction Quadratic Equations Cubic Equations Quartic Equations

Solving Polynomial Equations

Bernd Schroder

Bernd Schroder Louisiana Tech University, College of Engineering and Science


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Which Polynomial Equations Can Be SolvedExactly?

1. x2 + c = 0 (Done.)2. ax2 +bx+ c = 0 (This presentation.)3. ax3 +bx2 + cx+d = 0 (This presentation.)4. x4 +ax3 +bx2 + cx+d = 0 (This presentation.)5. For higher order there can be no formula. (Separate set of

presentations.)



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1. x2 + c = 0

(Done.)2. ax2 +bx+ c = 0 (This presentation.)3. ax3 +bx2 + cx+d = 0 (This presentation.)4. x4 +ax3 +bx2 + cx+d = 0 (This presentation.)5. For higher order there can be no formula. (Separate set of

presentations.)



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1. x2 + c = 0 (Done.)

2. ax2 +bx+ c = 0 (This presentation.)3. ax3 +bx2 + cx+d = 0 (This presentation.)4. x4 +ax3 +bx2 + cx+d = 0 (This presentation.)5. For higher order there can be no formula. (Separate set of

presentations.)



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1. x2 + c = 0 (Done.)2. ax2 +bx+ c = 0

(This presentation.)3. ax3 +bx2 + cx+d = 0 (This presentation.)4. x4 +ax3 +bx2 + cx+d = 0 (This presentation.)5. For higher order there can be no formula. (Separate set of

presentations.)



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1. x2 + c = 0 (Done.)2. ax2 +bx+ c = 0 (This presentation.)

3. ax3 +bx2 + cx+d = 0 (This presentation.)4. x4 +ax3 +bx2 + cx+d = 0 (This presentation.)5. For higher order there can be no formula. (Separate set of

presentations.)



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1. x2 + c = 0 (Done.)2. ax2 +bx+ c = 0 (This presentation.)3. ax3 +bx2 + cx+d = 0

(This presentation.)4. x4 +ax3 +bx2 + cx+d = 0 (This presentation.)5. For higher order there can be no formula. (Separate set of

presentations.)



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1. x2 + c = 0 (Done.)2. ax2 +bx+ c = 0 (This presentation.)3. ax3 +bx2 + cx+d = 0 (This presentation.)

4. x4 +ax3 +bx2 + cx+d = 0 (This presentation.)5. For higher order there can be no formula. (Separate set of

presentations.)



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1. x2 + c = 0 (Done.)2. ax2 +bx+ c = 0 (This presentation.)3. ax3 +bx2 + cx+d = 0 (This presentation.)4. x4 +ax3 +bx2 + cx+d = 0

(This presentation.)5. For higher order there can be no formula. (Separate set of

presentations.)



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1. x2 + c = 0 (Done.)2. ax2 +bx+ c = 0 (This presentation.)3. ax3 +bx2 + cx+d = 0 (This presentation.)4. x4 +ax3 +bx2 + cx+d = 0 (This presentation.)

5. For higher order there can be no formula. (Separate set ofpresentations.)



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1. x2 + c = 0 (Done.)2. ax2 +bx+ c = 0 (This presentation.)3. ax3 +bx2 + cx+d = 0 (This presentation.)4. x4 +ax3 +bx2 + cx+d = 0 (This presentation.)5. For higher order there can be no formula.

(Separate set ofpresentations.)



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1. x2 + c = 0 (Done.)2. ax2 +bx+ c = 0 (This presentation.)3. ax3 +bx2 + cx+d = 0 (This presentation.)4. x4 +ax3 +bx2 + cx+d = 0 (This presentation.)5. For higher order there can be no formula. (Separate set of

presentations.)



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Definition.

Let (F,+, ·) be a field, let a ∈ F and let n ∈ N\{1}.A number r ∈ F so that rn = a is called an nth root of a.Notation: r = n

√a or r = a

1n . Second roots will be called square

roots and they will be denoted√

a.



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Definition. Let (F,+, ·) be a field, let a ∈ F and let n ∈ N\{1}.

A number r ∈ F so that rn = a is called an nth root of a.Notation: r = n

√a or r = a



a.



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Definition. Let (F,+, ·) be a field, let a ∈ F and let n ∈ N\{1}.A number r ∈ F so that rn = a is called an nth root of a.

Notation: r = n√

a or r = a1n . Second roots will be called square


a.



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Definition. Let (F,+, ·) be a field, let a ∈ F and let n ∈ N\{1}.A number r ∈ F so that rn = a is called an nth root of a.Notation:

r = n√

a or r = a1n . Second roots will be called square


a.



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Definition. Let (F,+, ·) be a field, let a ∈ F and let n ∈ N\{1}.A number r ∈ F so that rn = a is called an nth root of a.Notation: r = n

√a

or r = a1n . Second roots will be called square


a.



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√a or r = a

1n .

Second roots will be called squareroots and they will be denoted

√a.



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√a or r = a


roots

and they will be denoted√

a.



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√a or r = a



a.



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Theorem.

The quadratic formula. Let (F,+, ·) be a field andlet a,b,c ∈ F with a 6= 0. Then the number of solutions in F ofthe equation ax2 +bx+ c = 0 equals the number of square rootsof b2−4ac in F. Moreover, if b2−4ac has at least one squareroot in F, then the solutions of ax2 +bx+ c = 0 are

x1,2 =−b±

√b2−4ac

2a.



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Theorem. The quadratic formula.

Let (F,+, ·) be a field andlet a,b,c ∈ F with a 6= 0. Then the number of solutions in F ofthe equation ax2 +bx+ c = 0 equals the number of square rootsof b2−4ac in F. Moreover, if b2−4ac has at least one squareroot in F, then the solutions of ax2 +bx+ c = 0 are

x1,2 =−b±

√b2−4ac

2a.



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Theorem. The quadratic formula. Let (F,+, ·) be a field andlet a,b,c ∈ F with a 6= 0.

Then the number of solutions in F ofthe equation ax2 +bx+ c = 0 equals the number of square rootsof b2−4ac in F. Moreover, if b2−4ac has at least one squareroot in F, then the solutions of ax2 +bx+ c = 0 are

x1,2 =−b±

√b2−4ac

2a.



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Theorem. The quadratic formula. Let (F,+, ·) be a field andlet a,b,c ∈ F with a 6= 0. Then the number of solutions in F ofthe equation ax2 +bx+ c = 0 equals the number of square rootsof b2−4ac in F.

Moreover, if b2−4ac has at least one squareroot in F, then the solutions of ax2 +bx+ c = 0 are

x1,2 =−b±

√b2−4ac

2a.



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Theorem. The quadratic formula. Let (F,+, ·) be a field andlet a,b,c ∈ F with a 6= 0. Then the number of solutions in F ofthe equation ax2 +bx+ c = 0 equals the number of square rootsof b2−4ac in F. Moreover, if b2−4ac has at least one squareroot in F, then the solutions of ax2 +bx+ c = 0 are

x1,2 =−b±

√b2−4ac

2a.



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Proof (number of solutions and formula).

Let x ∈ F be asolution.

0 = ax2 +bx+ c

0 = a(

x2 +ba

x)

+ c

0 = a

(x2 +

ba

x+(

b2a

)2

−(

b2a

)2)

+ c

0 = a(

x+b

2a

)2

−a(

b2a

)2

+ c

a(

x+b

2a

)2

= a(

b2a

)2

− c(x+

b2a

)2

=(

b2a

)2

− ca



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Proof (number of solutions and formula). Let x ∈ F be asolution.

0 = ax2 +bx+ c

0 = a(

x2 +ba

x)

+ c

0 = a

(x2 +

ba

x+(

b2a

)2

−(

b2a

)2)

+ c

0 = a(

x+b

2a

)2

−a(

b2a

)2

+ c

a(

x+b

2a

)2

= a(

b2a

)2

− c(x+

b2a

)2

=(

b2a

)2

− ca



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0 = ax2 +bx+ c

0 = a(

x2 +ba

x)

+ c

0 = a

(x2 +

ba

x+(

b2a

)2

−(

b2a

)2)

+ c

0 = a(

x+b

2a

)2

−a(

b2a

)2

+ c

a(

x+b

2a

)2

= a(

b2a

)2

− c(x+

b2a

)2

=(

b2a

)2

− ca



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0 = ax2 +bx+ c

0 = a(

x2 +ba

x)

+ c

0 = a

(x2 +

ba

x+(

b2a

)2

−(

b2a

)2)

+ c

0 = a(

x+b

2a

)2

−a(

b2a

)2

+ c

a(

x+b

2a

)2

= a(

b2a

)2

− c(x+

b2a

)2

=(

b2a

)2

− ca



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0 = ax2 +bx+ c

0 = a(

x2 +ba

x)

+ c

0 = a

(x2 +

ba

x+(

b2a

)2

−(

b2a

)2)

+ c

0 = a(

x+b

2a

)2

−a(

b2a

)2

+ c

a(

x+b

2a

)2

= a(

b2a

)2

− c(x+

b2a

)2

=(

b2a

)2

− ca



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0 = ax2 +bx+ c

0 = a(

x2 +ba

x)

+ c

0 = a

(x2 +

ba

x+(

b2a

)2

−(

b2a

)2)

+ c

0 = a(

x+b

2a

)2

−a(

b2a

)2

+ c

a(

x+b

2a

)2

= a(

b2a

)2

− c(x+

b2a

)2

=(

b2a

)2

− ca



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0 = ax2 +bx+ c

0 = a(

x2 +ba

x)

+ c

0 = a

(x2 +

ba

x+(

b2a

)2

−(

b2a

)2)

+ c

0 = a(

x+b

2a

)2

−a(

b2a

)2

+ c

a(

x+b

2a

)2

= a(

b2a

)2

− c

(x+

b2a

)2

=(

b2a

)2

− ca



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0 = ax2 +bx+ c

0 = a(

x2 +ba

x)

+ c

0 = a

(x2 +

ba

x+(

b2a

)2

−(

b2a

)2)

+ c

0 = a(

x+b

2a

)2

−a(

b2a

)2

+ c

a(

x+b

2a

)2

= a(

b2a

)2

− c(x+

b2a

)2

=(

b2a

)2

− ca



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Proof (number of solutions and formula, concl.).

(x+

b2a

)2

=(

b2a

)2

− ca(

x+b

2a

)2

=b2−4ac(2a)2

x1,2 +b

2a= ±

√b2−4ac(2a)2

x1,2 = − b2a±√

b2−4ac2a

x1,2 =−b±

√b2−4ac

2a



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Proof (number of solutions and formula, concl.).(x+

b2a

)2

=(

b2a

)2

− ca

(x+

b2a

)2

=b2−4ac(2a)2

x1,2 +b

2a= ±

√b2−4ac(2a)2

x1,2 = − b2a±√

b2−4ac2a

x1,2 =−b±

√b2−4ac

2a



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b2a

)2

=(

b2a

)2

− ca(

x+b

2a

)2

=b2−4ac(2a)2

x1,2 +b

2a= ±

√b2−4ac(2a)2

x1,2 = − b2a±√

b2−4ac2a

x1,2 =−b±

√b2−4ac

2a



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b2a

)2

=(

b2a

)2

− ca(

x+b

2a

)2

=b2−4ac(2a)2

x1,2 +b

2a= ±

√b2−4ac(2a)2

x1,2 = − b2a±√

b2−4ac2a

x1,2 =−b±

√b2−4ac

2a



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b2a

)2

=(

b2a

)2

− ca(

x+b

2a

)2

=b2−4ac(2a)2

x1,2 +b

2a= ±

√b2−4ac(2a)2

x1,2 = − b2a±√

b2−4ac2a

x1,2 =−b±

√b2−4ac

2a



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b2a

)2

=(

b2a

)2

− ca(

x+b

2a

)2

=b2−4ac(2a)2

x1,2 +b

2a= ±

√b2−4ac(2a)2

x1,2 = − b2a±√

b2−4ac2a

x1,2 =−b±

√b2−4ac

2a



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Proof (expressions are solutions).

ax21 +bx1 + c

= a

(−b+

√b2−4ac

2a

)2

+b

(−b+

√b2−4ac

2a

)+ c

= ab2−2b

√b2−4ac+b2−4ac

4a2 +−b2 +b

√b2−4ac

2a+ c

=b2−2b


4a+−2b2 +2b

√b2−4ac

4a+ c

=−4ac

4a+ c

= 0.



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Proof (expressions are solutions).ax2

1 +bx1 + c

= a

(−b+

√b2−4ac

2a

)2

+b

(−b+

√b2−4ac

2a

)+ c

= ab2−2b


4a2 +−b2 +b

√b2−4ac

2a+ c

=b2−2b


4a+−2b2 +2b

√b2−4ac

4a+ c

=−4ac

4a+ c

= 0.



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1 +bx1 + c

= a

(−b+

√b2−4ac

2a

)2

+b

(−b+

√b2−4ac

2a

)+ c

= ab2−2b


4a2 +−b2 +b

√b2−4ac

2a+ c

=b2−2b


4a+−2b2 +2b

√b2−4ac

4a+ c

=−4ac

4a+ c

= 0.



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1 +bx1 + c

= a

(−b+

√b2−4ac

2a

)2

+b

(−b+

√b2−4ac

2a

)+ c

= ab2−2b


4a2 +−b2 +b

√b2−4ac

2a+ c

=b2−2b


4a+−2b2 +2b

√b2−4ac

4a+ c

=−4ac

4a+ c

= 0.



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1 +bx1 + c

= a

(−b+

√b2−4ac

2a

)2

+b

(−b+

√b2−4ac

2a

)+ c

= ab2−2b


4a2 +−b2 +b

√b2−4ac

2a+ c

=b2−2b


4a+−2b2 +2b

√b2−4ac

4a+ c

=−4ac

4a+ c

= 0.



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1 +bx1 + c

= a

(−b+

√b2−4ac

2a

)2

+b

(−b+

√b2−4ac

2a

)+ c

= ab2−2b


4a2 +−b2 +b

√b2−4ac

2a+ c

=b2−2b


4a+−2b2 +2b

√b2−4ac

4a+ c

=−4ac

4a+ c

= 0.



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1 +bx1 + c

= a

(−b+

√b2−4ac

2a

)2

+b

(−b+

√b2−4ac

2a

)+ c

= ab2−2b


4a2 +−b2 +b

√b2−4ac

2a+ c

=b2−2b


4a+−2b2 +2b

√b2−4ac

4a+ c

=−4ac

4a+ c

= 0.



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1 +bx1 + c

= a

(−b+

√b2−4ac

2a

)2

+b

(−b+

√b2−4ac

2a

)+ c

= ab2−2b


4a2 +−b2 +b

√b2−4ac

2a+ c

=b2−2b


4a+−2b2 +2b

√b2−4ac

4a+ c

=−4ac

4a+ c

= 0.



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Theorem.

Every nonzero complex number has exactly ndistinct complex roots.

Proof(?) For z = reiθ , the n roots are wk = n√

rei( θ

n +k 2π

n ), wherek ∈ {0, . . . ,n−1}.



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Theorem. Every nonzero complex number has exactly ndistinct complex roots.


rei( θ

n +k 2π

n ), wherek ∈ {0, . . . ,n−1}.



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Proof

(?) For z = reiθ , the n roots are wk = n√

rei( θ

n +k 2π

n ), wherek ∈ {0, . . . ,n−1}.



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Proof(?)

For z = reiθ , the n roots are wk = n√

rei( θ

n +k 2π

n ), wherek ∈ {0, . . . ,n−1}.



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Proof(?) For z = reiθ

, the n roots are wk = n√

rei( θ

n +k 2π

n ), wherek ∈ {0, . . . ,n−1}.



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r

ei( θ

n +k 2π

n ), wherek ∈ {0, . . . ,n−1}.



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rei( θ

n +k 2π

n )

, wherek ∈ {0, . . . ,n−1}.



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rei( θ

n +k 2π

n ), wherek ∈ {0, . . . ,n−1}.



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rei( θ

n +k 2π

n ), wherek ∈ {0, . . . ,n−1}.



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Theorem.

The cubic formula, step I: Canonical form. Let(F,+, ·) be a field and let a,b,c,d ∈ F with a 6= 0. Then x ∈ F

solves the equation ax3 +bx2 + cx+d = 0 iff y = x+b3a

solves

the equation y3 +py+q = 0, where p =− b2

3a2 +ca

and

q =2b3

27a3 −bc3a2 +

da

.



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Theorem. The cubic formula, step I: Canonical form.

Let(F,+, ·) be a field and let a,b,c,d ∈ F with a 6= 0. Then x ∈ F


solves


3a2 +ca

and

q =2b3

27a3 −bc3a2 +

da

.



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Theorem. The cubic formula, step I: Canonical form. Let(F,+, ·) be a field and let a,b,c,d ∈ F with a 6= 0.

Then x ∈ F


solves


3a2 +ca

and

q =2b3

27a3 −bc3a2 +

da

.



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Theorem. The cubic formula, step I: Canonical form. Let(F,+, ·) be a field and let a,b,c,d ∈ F with a 6= 0. Then x ∈ F

solves the equation ax3 +bx2 + cx+d = 0 iff

y = x+b3a

solves


3a2 +ca

and

q =2b3

27a3 −bc3a2 +

da

.



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solves

the equation y3 +py+q = 0,

where p =− b2

3a2 +ca

and

q =2b3

27a3 −bc3a2 +

da

.



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solves


3a2 +ca

and

q =2b3

27a3 −bc3a2 +

da

.



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Proof (substitution).

x := y+α .

ax3 +bx2 + cx+d= a(y+α)3 +b(y+α)2 + c(y+α)+d= ay3 +3aαy2 +3aα

2y+aα3 +by2 +2bαy+bα

2 + cy+ cα +d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.

The quadratic term is zero for α =− b3a

.



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Proof (substitution). x := y+α .



2 + cy+ cα +d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.


.



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ax3 +bx2 + cx+d

= a(y+α)3 +b(y+α)2 + c(y+α)+d= ay3 +3aαy2 +3aα


2 + cy+ cα +d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.


.



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ax3 +bx2 + cx+d= a(y+α)3 +b(y+α)2 + c(y+α)+d

= ay3 +3aαy2 +3aα2y+aα

3 +by2 +2bαy+bα2 + cy+ cα +d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.


.



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2y+aα3

+by2 +2bαy+bα2 + cy+ cα +d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.


.



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2

+ cy+ cα +d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.


.



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2 + cy+ cα

+d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.


.



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2 + cy+ cα +d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.


.



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2 + cy+ cα +d

= ay3

+(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.


.



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2 + cy+ cα +d

= ay3 +(3aα +b)y2

+(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.


.



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2 + cy+ cα +d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.


.



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2 + cy+ cα +d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.


.



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2 + cy+ cα +d

= ay3 +(3aα +b)y2 +(

3aα2 +2bα + c

)y

+(

aα3 +bα

2 + cα +d)

.


.



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Proof.

With α =− b3a

, x solves ax3 +bx2 + cx+d = 0 iff

y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.

Divide by a and substitute α to obtain that this is the case iff

y = x+b

3asolves y3 +py+q = 0, where

p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and

q =aα3 +bα2 + cα +d

a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.



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Proof. With α =− b3a


y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.


y = x+b


p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and


a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.



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y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.


y = x+b


p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and


a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.



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y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.


y = x+b


p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and


a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.



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y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.


y = x+b


p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and


a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.



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y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.


y = x+b


p =3aα2 +2bα + c

a

=3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and


a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.



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y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.


y = x+b


p =3aα2 +2bα + c

a=

3b2

9a2

− 2b2

3a2 +ca

=− b2

3a2 +ca

and


a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.



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y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.


y = x+b


p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2

+ca

=− b2

3a2 +ca

and


a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.



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y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.


y = x+b


p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and


a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.



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y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.


y = x+b


p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2

+ca

and


a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.



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y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.


y = x+b


p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and


a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.



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y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.


y = x+b


p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and


a

=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.



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y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.


y = x+b


p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and


a=− b3

27a3

+b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.



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y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.


y = x+b


p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and


a=− b3

27a3 +b3

9a3

− bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.



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y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.


y = x+b


p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and


a=− b3

27a3 +b3

9a3 −bc3a2

+da

=2b3

27a3 −bc3a2 +

da.



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y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.


y = x+b


p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and


a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.



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y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.


y = x+b


p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and


a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3

− bc3a2 +

da.



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y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.


y = x+b


p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and


a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2

+da.



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y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.


y = x+b


p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and


a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.



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y = x+b

3asolves

ay3 +(

3aα2 +2bα + c

)y+(

aα3 +bα

2 + cα +d)

= 0.


y = x+b


p =3aα2 +2bα + c

a=

3b2

9a2 −2b2

3a2 +ca

=− b2

3a2 +ca

and


a=− b3

27a3 +b3

9a3 −bc3a2 +

da

=2b3

27a3 −bc3a2 +

da.



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Theorem.

The cubic formula, step 2: Cardano’s formula. Letp,q ∈ C\{0} and consider the equation x3 +px+q = 0. Let

α1,2,3 :=

(−q

2+

√q2

4+

p3

27

) 13

be the three cube roots of

−q2

+

√q2

4+

p3

27, where it does not matter which of the two

numbers whose square isq2

4+

p3

27is chosen as the square root.

Then the solutions of x3 +px+q = 0 arex1,2,3 = α1,2,3−

p3α1,2,3

.



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Theorem. The cubic formula, step 2: Cardano’s formula.

Letp,q ∈ C\{0} and consider the equation x3 +px+q = 0. Let

α1,2,3 :=

(−q

2+

√q2

4+

p3

27

) 13


−q2

+

√q2

4+

p3



4+

p3



p3α1,2,3

.



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Theorem. The cubic formula, step 2: Cardano’s formula. Letp,q ∈ C\{0} and consider the equation x3 +px+q = 0.

Let

α1,2,3 :=

(−q

2+

√q2

4+

p3

27

) 13


−q2

+

√q2

4+

p3



4+

p3



p3α1,2,3

.



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Theorem. The cubic formula, step 2: Cardano’s formula. Letp,q ∈ C\{0} and consider the equation x3 +px+q = 0. Let

α1,2,3 :=

(−q

2+

√q2

4+

p3

27

) 13


−q2

+

√q2

4+

p3

27

, where it does not matter which of the two


4+

p3



p3α1,2,3

.



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α1,2,3 :=

(−q

2+

√q2

4+

p3

27

) 13


−q2

+

√q2

4+

p3



4+

p3



p3α1,2,3

.



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α1,2,3 :=

(−q

2+

√q2

4+

p3

27

) 13


−q2

+

√q2

4+

p3



4+

p3



p3α1,2,3

.



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Proof.

Let x be a solution. Substitution x = α +β .

0 = x3 +px+q= (α +β )3 +p(α +β )+q= α

3 +3α2β +3αβ

2 +β3 +pα +pβ +q

Choose α and β so that 3αβ =−p (and α +β = x).

= α3−pα−pβ +β

3 +pα +pβ +q= α

3 +β3 +q

= α3− p3

27α3 +q.



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Proof. Let x be a solution.

Substitution x = α +β .

0 = x3 +px+q= (α +β )3 +p(α +β )+q= α

3 +3α2β +3αβ

2 +β3 +pα +pβ +q



3 +pα +pβ +q= α

3 +β3 +q

= α3− p3

27α3 +q.



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Proof. Let x be a solution. Substitution x = α +β .

0 = x3 +px+q= (α +β )3 +p(α +β )+q= α

3 +3α2β +3αβ

2 +β3 +pα +pβ +q



3 +pα +pβ +q= α

3 +β3 +q

= α3− p3

27α3 +q.



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0 = x3 +px+q

= (α +β )3 +p(α +β )+q= α

3 +3α2β +3αβ

2 +β3 +pα +pβ +q



3 +pα +pβ +q= α

3 +β3 +q

= α3− p3

27α3 +q.



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0 = x3 +px+q= (α +β )3 +p(α +β )+q

= α3 +3α

2β +3αβ

2 +β3 +pα +pβ +q



3 +pα +pβ +q= α

3 +β3 +q

= α3− p3

27α3 +q.



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0 = x3 +px+q= (α +β )3 +p(α +β )+q= α

3 +3α2β +3αβ

2 +β3 +pα +pβ +q



3 +pα +pβ +q= α

3 +β3 +q

= α3− p3

27α3 +q.



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0 = x3 +px+q= (α +β )3 +p(α +β )+q= α

3 +3α2β +3αβ

2 +β3 +pα +pβ +q



3 +pα +pβ +q= α

3 +β3 +q

= α3− p3

27α3 +q.



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0 = x3 +px+q= (α +β )3 +p(α +β )+q= α

3 +3α2β +3αβ

2 +β3 +pα +pβ +q



3 +pα +pβ +q

= α3 +β

3 +q

= α3− p3

27α3 +q.



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0 = x3 +px+q= (α +β )3 +p(α +β )+q= α

3 +3α2β +3αβ

2 +β3 +pα +pβ +q



3 +pα +pβ +q= α

3 +β3 +q

= α3− p3

27α3 +q.



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0 = x3 +px+q= (α +β )3 +p(α +β )+q= α

3 +3α2β +3αβ

2 +β3 +pα +pβ +q



3 +pα +pβ +q= α

3 +β3 +q

= α3− p3

27α3 +q.



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Proof.

Substitution u := α3.

α3− p3

27α3 +q = 0

α6 +qα

3− p3

27= 0

u2 +qu− p3

27= 0

u1,2 =−q±

√q2 +4 p3

27

2

= −q2±√

q2

4+

p3

27.

Therefore α must be one of the three third roots of

u1 =−q2

+

√q2

4+

p3

27or of u2 =−q

2−√

q2

4+

p3

27.



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Proof. Substitution u := α3.

α3− p3

27α3 +q = 0

α6 +qα

3− p3

27= 0

u2 +qu− p3

27= 0

u1,2 =−q±

√q2 +4 p3

27

2

= −q2±√

q2

4+

p3

27.


u1 =−q2

+

√q2

4+

p3

27or of u2 =−q

2−√

q2

4+

p3

27.



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α3− p3

27α3 +q = 0

α6 +qα

3− p3

27= 0

u2 +qu− p3

27= 0

u1,2 =−q±

√q2 +4 p3

27

2

= −q2±√

q2

4+

p3

27.


u1 =−q2

+

√q2

4+

p3

27or of u2 =−q

2−√

q2

4+

p3

27.



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α3− p3

27α3 +q = 0

α6 +qα

3− p3

27= 0

u2 +qu− p3

27= 0

u1,2 =−q±

√q2 +4 p3

27

2

= −q2±√

q2

4+

p3

27.


u1 =−q2

+

√q2

4+

p3

27or of u2 =−q

2−√

q2

4+

p3

27.



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α3− p3

27α3 +q = 0

α6 +qα

3− p3

27= 0

u2 +qu− p3

27= 0

u1,2 =−q±

√q2 +4 p3

27

2

= −q2±√

q2

4+

p3

27.


u1 =−q2

+

√q2

4+

p3

27or of u2 =−q

2−√

q2

4+

p3

27.



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α3− p3

27α3 +q = 0

α6 +qα

3− p3

27= 0

u2 +qu− p3

27= 0

u1,2 =−q±

√q2 +4 p3

27

2

= −q2±√

q2

4+

p3

27.


u1 =−q2

+

√q2

4+

p3

27or of u2 =−q

2−√

q2

4+

p3

27.



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α3− p3

27α3 +q = 0

α6 +qα

3− p3

27= 0

u2 +qu− p3

27= 0

u1,2 =−q±

√q2 +4 p3

27

2

= −q2±√

q2

4+

p3

27.


u1 =−q2

+

√q2

4+

p3

27or of u2 =−q

2−√

q2

4+

p3

27.



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α3− p3

27α3 +q = 0

α6 +qα

3− p3

27= 0

u2 +qu− p3

27= 0

u1,2 =−q±

√q2 +4 p3

27

2

= −q2±√

q2

4+

p3

27.


u1 =−q2

+

√q2

4+

p3

27or of u2 =−q

2−√

q2

4+

p3

27.



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Proof.

u1 =−q2

+

√q2

4+

p3

27, u2 =−q

2−√

q2

4+

p3

27. Our

overall goal is to find x = α +β . α and β must satisfy27α3β 3 =−p3. It is easy to check that 27u1u2 =−p3. Hence,if α3 = u1, then β 3 = u2 and vice versa. Thus x = α +β isalways the sum of a cube root of u1 and a cube root of u2.WLOG we can assume that α3 = u1. But then

α = 3√

u1 =

(−q

2+

√q2

4+

p3

27

) 13

and from 3αβ =−p we

obtain β =− p3α

.



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Proof. u1 =−q2

+

√q2

4+

p3

27, u2 =−q

2−√

q2

4+

p3

27.

Our


α = 3√

u1 =

(−q

2+

√q2

4+

p3

27

) 13


obtain β =− p3α

.



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Proof. u1 =−q2

+

√q2

4+

p3

27, u2 =−q

2−√

q2

4+

p3

27. Our

overall goal is to find x = α +β .

α and β must satisfy27α3β 3 =−p3. It is easy to check that 27u1u2 =−p3. Hence,if α3 = u1, then β 3 = u2 and vice versa. Thus x = α +β isalways the sum of a cube root of u1 and a cube root of u2.WLOG we can assume that α3 = u1. But then

α = 3√

u1 =

(−q

2+

√q2

4+

p3

27

) 13


obtain β =− p3α

.



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Proof. u1 =−q2

+

√q2

4+

p3

27, u2 =−q

2−√

q2

4+

p3

27. Our

overall goal is to find x = α +β . α and β must satisfy27α3β 3 =−p3.

It is easy to check that 27u1u2 =−p3. Hence,if α3 = u1, then β 3 = u2 and vice versa. Thus x = α +β isalways the sum of a cube root of u1 and a cube root of u2.WLOG we can assume that α3 = u1. But then

α = 3√

u1 =

(−q

2+

√q2

4+

p3

27

) 13


obtain β =− p3α

.



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Proof. u1 =−q2

+

√q2

4+

p3

27, u2 =−q

2−√

q2

4+

p3

27. Our

overall goal is to find x = α +β . α and β must satisfy27α3β 3 =−p3. It is easy to check that 27u1u2 =−p3.

Hence,if α3 = u1, then β 3 = u2 and vice versa. Thus x = α +β isalways the sum of a cube root of u1 and a cube root of u2.WLOG we can assume that α3 = u1. But then

α = 3√

u1 =

(−q

2+

√q2

4+

p3

27

) 13


obtain β =− p3α

.



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Proof. u1 =−q2

+

√q2

4+

p3

27, u2 =−q

2−√

q2

4+

p3

27. Our

overall goal is to find x = α +β . α and β must satisfy27α3β 3 =−p3. It is easy to check that 27u1u2 =−p3. Hence,if α3 = u1, then β 3 = u2 and vice versa.

Thus x = α +β isalways the sum of a cube root of u1 and a cube root of u2.WLOG we can assume that α3 = u1. But then

α = 3√

u1 =

(−q

2+

√q2

4+

p3

27

) 13


obtain β =− p3α

.



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Proof. u1 =−q2

+

√q2

4+

p3

27, u2 =−q

2−√

q2

4+

p3

27. Our

overall goal is to find x = α +β . α and β must satisfy27α3β 3 =−p3. It is easy to check that 27u1u2 =−p3. Hence,if α3 = u1, then β 3 = u2 and vice versa. Thus x = α +β isalways the sum of a cube root of u1 and a cube root of u2.

WLOG we can assume that α3 = u1. But then

α = 3√

u1 =

(−q

2+

√q2

4+

p3

27

) 13


obtain β =− p3α

.



logo1


Proof. u1 =−q2

+

√q2

4+

p3

27, u2 =−q

2−√

q2

4+

p3

27. Our

overall goal is to find x = α +β . α and β must satisfy27α3β 3 =−p3. It is easy to check that 27u1u2 =−p3. Hence,if α3 = u1, then β 3 = u2 and vice versa. Thus x = α +β isalways the sum of a cube root of u1 and a cube root of u2.WLOG we can assume that α3 = u1.

But then

α = 3√

u1 =

(−q

2+

√q2

4+

p3

27

) 13


obtain β =− p3α

.



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Proof. u1 =−q2

+

√q2

4+

p3

27, u2 =−q

2−√

q2

4+

p3

27. Our


α = 3√

u1 =

(−q

2+

√q2

4+

p3

27

) 13


obtain β =− p3α

.



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Proof. u1 =−q2

+

√q2

4+

p3

27, u2 =−q

2−√

q2

4+

p3

27. Our


α = 3√

u1 =

(−q

2+

√q2

4+

p3

27

) 13


obtain β =− p3α

.



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Proof (concl.).

Hence the solutions of the equationx3 +px+q = 0 must be of the formx1,2,3 = α1,2,3 +β1,2,3 = α1,2,3−

p3α1,2,3

, where α1,2,3 denote

the three cube roots of u1 (or of u2).The verification that these numbers are indeed solutions of theequation x3 +px+q = 0 is quite an exercise.



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Proof (concl.). Hence the solutions of the equationx3 +px+q = 0 must be of the formx1,2,3 = α1,2,3 +β1,2,3

= α1,2,3−p

3α1,2,3, where α1,2,3 denote




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Proof (concl.). Hence the solutions of the equationx3 +px+q = 0 must be of the formx1,2,3 = α1,2,3 +β1,2,3 = α1,2,3−

p3α1,2,3





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p3α1,2,3


the three cube roots of u1

(or of u2).The verification that these numbers are indeed solutions of theequation x3 +px+q = 0 is quite an exercise.



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p3α1,2,3


the three cube roots of u1 (or of u2).

The verification that these numbers are indeed solutions of theequation x3 +px+q = 0 is quite an exercise.



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p3α1,2,3





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p3α1,2,3





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Theorem.

The quartic formula/Ferrari method, step I:Elimination of the cubic term. Let (F,+, ·) be a field and leta,b,c,d ∈ F. Then x ∈ F solves the equationx4 +ax3 +bx2 + cx+d = 0 iff y = x+

a4

solves the equation

y4 +py2 +qy+ r = 0 where p =−3a2

8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.

Proof. Exercise. (I’ve done that one.)



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Theorem. The quartic formula/Ferrari method, step I:Elimination of the cubic term.

Let (F,+, ·) be a field and leta,b,c,d ∈ F. Then x ∈ F solves the equationx4 +ax3 +bx2 + cx+d = 0 iff y = x+

a4

solves the equation


8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.




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Theorem. The quartic formula/Ferrari method, step I:Elimination of the cubic term. Let (F,+, ·) be a field and leta,b,c,d ∈ F.

Then x ∈ F solves the equationx4 +ax3 +bx2 + cx+d = 0 iff y = x+

a4

solves the equation


8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.




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Theorem. The quartic formula/Ferrari method, step I:Elimination of the cubic term. Let (F,+, ·) be a field and leta,b,c,d ∈ F. Then x ∈ F solves the equationx4 +ax3 +bx2 + cx+d = 0 iff

y = x+a4

solves the equation


8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.




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Theorem. The quartic formula/Ferrari method, step I:Elimination of the cubic term. Let (F,+, ·) be a field and leta,b,c,d ∈ F. Then x ∈ F solves the equationx4 +ax3 +bx2 + cx+d = 0 iff y = x+

a4

solves the equation

y4 +py2 +qy+ r = 0

where p =−3a2

8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.




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a4

solves the equation


8+b

, q =a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.




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a4

solves the equation


8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.




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a4

solves the equation


8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.




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a4

solves the equation


8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.

Proof.

Exercise. (I’ve done that one.)



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a4

solves the equation


8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.

Proof. Exercise.

(I’ve done that one.)



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a4

solves the equation


8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.




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a4

solves the equation


8+b, q =

a3

8− ab

2+ c

and r =−3a4

256+

a2b16

− ac4

+d.




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Theorem.

The quartic formula/Ferrari method, step 2:Splitting the quartic. Let p,q,r ∈ C. Then x ∈ C solves theequation x4 +px2 +qx+ r = 0 iff x solves the equation(

x2 +p2

+α0

)2−2α0(x− z0)2 = 0, where α0 is a solution of

the cubic equation q2−4 ·2α

(α

2 +pα +p2

4− r)

= 0 and

z0 =q

4α0.



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Theorem. The quartic formula/Ferrari method, step 2:Splitting the quartic.

Let p,q,r ∈ C. Then x ∈ C solves theequation x4 +px2 +qx+ r = 0 iff x solves the equation(

x2 +p2

+α0



(α

2 +pα +p2

4− r)

= 0 and

z0 =q

4α0.



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Theorem. The quartic formula/Ferrari method, step 2:Splitting the quartic. Let p,q,r ∈ C.

Then x ∈ C solves theequation x4 +px2 +qx+ r = 0 iff x solves the equation(

x2 +p2

+α0



(α

2 +pα +p2

4− r)

= 0 and

z0 =q

4α0.



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Theorem. The quartic formula/Ferrari method, step 2:Splitting the quartic. Let p,q,r ∈ C. Then x ∈ C solves theequation x4 +px2 +qx+ r = 0 iff

x solves the equation(x2 +

p2

+α0



(α

2 +pα +p2

4− r)

= 0 and

z0 =q

4α0.



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Theorem. The quartic formula/Ferrari method, step 2:Splitting the quartic. Let p,q,r ∈ C. Then x ∈ C solves theequation x4 +px2 +qx+ r = 0 iff x solves the equation(

x2 +p2

+α0

)2−2α0(x− z0)2 = 0

, where α0 is a solution of


(α

2 +pα +p2

4− r)

= 0 and

z0 =q

4α0.



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x2 +p2

+α0



(α

2 +pα +p2

4− r)

= 0

and

z0 =q

4α0.



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x2 +p2

+α0



(α

2 +pα +p2

4− r)

= 0 and

z0 =q

4α0.



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Proof (“⇐” only).

x4 +px2 +qx+ r x4 =((

x2 +p2

+α0

)−(p

2+α0

))2

=(

x2 +p2

+α0

)2−2x2

(p2

+α0

)−2(p

2+α0

)2+(p

2+α0

)2

+px2 +qx+ r

=(

x2 +p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]

Because q2−4 ·2α0

(α

20 +pα0 +

p2

4− r)

= 0, the

term in square brackets is a perfect square,

2α0

(x− q

4α0

)2

= 2α0(x− z0)2.



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x4 +px2 +qx+ r

x4 =((

x2 +p2

+α0

)−(p

2+α0

))2

=(

x2 +p2

+α0

)2−2x2

(p2

+α0

)−2(p

2+α0

)2+(p

2+α0

)2

+px2 +qx+ r

=(

x2 +p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]


(α

20 +pα0 +

p2

4− r)

= 0, the


2α0

(x− q

4α0

)2

= 2α0(x− z0)2.



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x4 +px2 +qx+ r x4 =((

x2 +p2

+α0

)−(p

2+α0

))2

=(

x2 +p2

+α0

)2−2x2

(p2

+α0

)−2(p

2+α0

)2+(p

2+α0

)2

+px2 +qx+ r

=(

x2 +p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]


(α

20 +pα0 +

p2

4− r)

= 0, the


2α0

(x− q

4α0

)2

= 2α0(x− z0)2.



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x4 +px2 +qx+ r x4 =((

x2 +p2

+α0

)−(p

2+α0

))2

=(

x2 +p2

+α0

)2

−2x2(p

2+α0

)−2(p

2+α0

)2+(p

2+α0

)2

+px2 +qx+ r

=(

x2 +p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]


(α

20 +pα0 +

p2

4− r)

= 0, the


2α0

(x− q

4α0

)2

= 2α0(x− z0)2.



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x4 +px2 +qx+ r x4 =((

x2 +p2

+α0

)−(p

2+α0

))2

=(

x2 +p2

+α0

)2−2x2

(p2

+α0

)−2(p

2+α0

)2

+(p

2+α0

)2

+px2 +qx+ r

=(

x2 +p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]


(α

20 +pα0 +

p2

4− r)

= 0, the


2α0

(x− q

4α0

)2

= 2α0(x− z0)2.



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x4 +px2 +qx+ r x4 =((

x2 +p2

+α0

)−(p

2+α0

))2

=(

x2 +p2

+α0

)2−2x2

(p2

+α0

)−2(p

2+α0

)2+(p

2+α0

)2

+px2 +qx+ r

=(

x2 +p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]


(α

20 +pα0 +

p2

4− r)

= 0, the


2α0

(x− q

4α0

)2

= 2α0(x− z0)2.



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x4 +px2 +qx+ r x4 =((

x2 +p2

+α0

)−(p

2+α0

))2

=(

x2 +p2

+α0

)2−2x2

(p2

+α0

)−2(p

2+α0

)2+(p

2+α0

)2

+px2 +qx+ r

=(

x2 +p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]


(α

20 +pα0 +

p2

4− r)

= 0, the


2α0

(x− q

4α0

)2

= 2α0(x− z0)2.



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x4 +px2 +qx+ r x4 =((

x2 +p2

+α0

)−(p

2+α0

))2

=(

x2 +p2

+α0

)2−2x2

(p2

+α0

)−2(p

2+α0

)2+(p

2+α0

)2

+px2 +qx+ r

=(

x2 +p2

+α0

)2

−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]


(α

20 +pα0 +

p2

4− r)

= 0, the


2α0

(x− q

4α0

)2

= 2α0(x− z0)2.



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x4 +px2 +qx+ r x4 =((

x2 +p2

+α0

)−(p

2+α0

))2

=(

x2 +p2

+α0

)2−2x2

(p2

+α0

)−2(p

2+α0

)2+(p

2+α0

)2

+px2 +qx+ r

=(

x2 +p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]


(α

20 +pα0 +

p2

4− r)

= 0, the


2α0

(x− q

4α0

)2

= 2α0(x− z0)2.



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x4 +px2 +qx+ r x4 =((

x2 +p2

+α0

)−(p

2+α0

))2

=(

x2 +p2

+α0

)2−2x2

(p2

+α0

)−2(p

2+α0

)2+(p

2+α0

)2

+px2 +qx+ r

=(

x2 +p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]


(α

20 +pα0 +

p2

4− r)

= 0, the


2α0

(x− q

4α0

)2

= 2α0(x− z0)2.



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Proof (“⇐” concl.).

(x2 +

p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]


(α

20 +pα0 +

p2

4− r)

= 0, the


2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

=(

x2 +p2

+α0

)2−2α0(x− z0)2

= 0



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Proof (“⇐” concl.).(x2 +

p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]


(α

20 +pα0 +

p2

4− r)

= 0, the


2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

=(

x2 +p2

+α0

)2−2α0(x− z0)2

= 0



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p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]


(α

20 +pα0 +

p2

4− r)

= 0, the


2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

=(

x2 +p2

+α0

)2−2α0(x− z0)2

= 0



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p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]


(α

20 +pα0 +

p2

4− r)

= 0, the


2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

=(

x2 +p2

+α0

)2−2α0(x− z0)2

= 0



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p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]


(α

20 +pα0 +

p2

4− r)

= 0, the


2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

=(

x2 +p2

+α0

)2−2α0(x− z0)2

= 0



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p2

+α0

)2−[

2α0x2−qx+(

α20 +pα0 +

p2

4− r)]


(α

20 +pα0 +

p2

4− r)

= 0, the


2α0

(x− q

4α0

)2

= 2α0(x− z0)2.

=(

x2 +p2

+α0

)2−2α0(x− z0)2

= 0



Documents

· logo1 Introduction Quadratic Equations Cubic Equations Quartic Equations Theorem. The quadratic formula. Let (F,+,·) be a ﬁeld and let a,b,c∈F with a6= 0. Then the number