· logo1 introduction quadratic equations cubic equations quartic equations theorem. the quadratic...
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![Page 1: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/1.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
Solving Polynomial Equations
Bernd Schroder
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 2: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/2.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 3: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/3.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 4: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/4.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 5: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/5.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 6: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/6.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 7: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/7.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 8: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/8.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 9: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/9.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 10: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/10.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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 ofpresentations.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 11: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/11.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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 ofpresentations.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 12: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/12.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 13: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/13.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 14: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/14.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 15: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/15.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
roots and they will be denoted√
a.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 16: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/16.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
roots and they will be denoted√
a.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 17: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/17.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
roots and they will be denoted√
a.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 18: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/18.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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 squareroots and they will be denoted
√a.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 19: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/19.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 20: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/20.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 21: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/21.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 22: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/22.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 23: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/23.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 24: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/24.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 25: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/25.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 26: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/26.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 27: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/27.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 28: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/28.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 29: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/29.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 30: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/30.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 31: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/31.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 32: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/32.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 33: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/33.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 34: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/34.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 35: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/35.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 36: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/36.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 37: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/37.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 38: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/38.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 39: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/39.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 40: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/40.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
√b2−4ac+b2−4ac
4a+−2b2 +2b
√b2−4ac
4a+ c
=−4ac
4a+ c
= 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 41: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/41.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
Proof (expressions are solutions).ax2
1 +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
√b2−4ac+b2−4ac
4a+−2b2 +2b
√b2−4ac
4a+ c
=−4ac
4a+ c
= 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 42: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/42.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
Proof (expressions are solutions).ax2
1 +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
√b2−4ac+b2−4ac
4a+−2b2 +2b
√b2−4ac
4a+ c
=−4ac
4a+ c
= 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 43: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/43.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
Proof (expressions are solutions).ax2
1 +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
√b2−4ac+b2−4ac
4a+−2b2 +2b
√b2−4ac
4a+ c
=−4ac
4a+ c
= 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 44: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/44.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
Proof (expressions are solutions).ax2
1 +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
√b2−4ac+b2−4ac
4a+−2b2 +2b
√b2−4ac
4a+ c
=−4ac
4a+ c
= 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 45: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/45.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
Proof (expressions are solutions).ax2
1 +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
√b2−4ac+b2−4ac
4a+−2b2 +2b
√b2−4ac
4a+ c
=−4ac
4a+ c
= 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 46: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/46.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
Proof (expressions are solutions).ax2
1 +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
√b2−4ac+b2−4ac
4a+−2b2 +2b
√b2−4ac
4a+ c
=−4ac
4a+ c
= 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 47: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/47.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
Proof (expressions are solutions).ax2
1 +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
√b2−4ac+b2−4ac
4a+−2b2 +2b
√b2−4ac
4a+ c
=−4ac
4a+ c
= 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 48: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/48.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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}.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 49: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/49.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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}.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 50: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/50.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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}.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 51: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/51.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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}.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 52: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/52.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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}.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 53: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/53.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
Theorem. Every nonzero complex number has exactly ndistinct complex roots.
Proof(?) For z = reiθ , the n roots are wk = n√
r
ei( θ
n +k 2π
n ), wherek ∈ {0, . . . ,n−1}.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 54: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/54.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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}.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 55: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/55.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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}.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 56: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/56.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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}.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 57: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/57.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 58: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/58.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 59: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/59.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 60: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/60.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 61: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/61.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 62: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/62.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 63: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/63.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 64: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/64.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 65: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/65.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 66: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/66.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
Proof (substitution). x := y+α .
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)
.
The quadratic term is zero for α =− b3a
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 67: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/67.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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+bα2 + cy+ cα +d
= ay3 +(3aα +b)y2 +(
3aα2 +2bα + c
)y
+(
aα3 +bα
2 + cα +d)
.
The quadratic term is zero for α =− b3a
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 68: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/68.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 69: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/69.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 70: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/70.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 71: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/71.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 72: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/72.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 73: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/73.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 74: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/74.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 75: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/75.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 76: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/76.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 77: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/77.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 78: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/78.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 79: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/79.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 80: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/80.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 81: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/81.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 82: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/82.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 83: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/83.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 84: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/84.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 85: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/85.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 86: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/86.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 87: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/87.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 88: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/88.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 89: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/89.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 90: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/90.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 91: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/91.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 92: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/92.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 93: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/93.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 94: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/94.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 95: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/95.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 96: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/96.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 97: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/97.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 98: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/98.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 99: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/99.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 100: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/100.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 101: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/101.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 102: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/102.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 103: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/103.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 104: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/104.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 105: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/105.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 106: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/106.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
Proof. Let x be a solution. Substitution x = α +β .
0 = x3 +px+q= (α +β )3 +p(α +β )+q
= α3 +3α
2β +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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 107: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/107.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 108: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/108.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 109: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/109.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
= α3 +β
3 +q
= α3− p3
27α3 +q.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 110: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/110.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 111: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/111.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 112: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/112.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 113: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/113.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 114: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/114.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 115: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/115.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 116: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/116.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 117: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/117.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 118: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/118.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 119: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/119.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 120: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/120.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 121: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/121.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 122: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/122.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 123: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/123.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 124: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/124.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 125: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/125.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 126: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/126.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 127: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/127.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 128: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/128.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 129: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/129.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 130: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/130.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 131: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/131.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 132: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/132.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 133: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/133.jpg)
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Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 134: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/134.jpg)
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Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 135: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/135.jpg)
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Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 136: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/136.jpg)
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Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 137: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/137.jpg)
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Introduction Quadratic Equations Cubic Equations Quartic Equations
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.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 138: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/138.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 139: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/139.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 140: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/140.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 141: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/141.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 142: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/142.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 143: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/143.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 144: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/144.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 145: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/145.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 146: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/146.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 147: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/147.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 148: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/148.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 149: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/149.jpg)
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Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 150: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/150.jpg)
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Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 151: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/151.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 152: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/152.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 153: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/153.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 154: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/154.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 155: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/155.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 156: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/156.jpg)
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Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 157: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/157.jpg)
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Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 158: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/158.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 159: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/159.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
Proof (“⇐” only).
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)]
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 160: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/160.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 161: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/161.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 162: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/162.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 163: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/163.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 164: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/164.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 165: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/165.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
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.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 166: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/166.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
Proof (“⇐” concl.).
(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.
=(
x2 +p2
+α0
)2−2α0(x− z0)2
= 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 167: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/167.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
Proof (“⇐” concl.).(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.
=(
x2 +p2
+α0
)2−2α0(x− z0)2
= 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 168: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/168.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
Proof (“⇐” concl.).(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.
=(
x2 +p2
+α0
)2−2α0(x− z0)2
= 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 169: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/169.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
Proof (“⇐” concl.).(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.
=(
x2 +p2
+α0
)2−2α0(x− z0)2
= 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 170: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/170.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
Proof (“⇐” concl.).(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.
=(
x2 +p2
+α0
)2−2α0(x− z0)2
= 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
![Page 171: · 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](https://reader033.vdocuments.net/reader033/viewer/2022042621/5f67ea07e243e26f424dfe60/html5/thumbnails/171.jpg)
logo1
Introduction Quadratic Equations Cubic Equations Quartic Equations
Proof (“⇐” concl.).(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.
=(
x2 +p2
+α0
)2−2α0(x− z0)2
= 0
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations