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Introduction Quadratic Equations Cubic Equations Quartic Equations
Solving Polynomial Equations
Bernd Schroder
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Solving Polynomial Equations
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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
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