Solving Third Degree Polynomial Equations · 2. An algorithm to beautify the solutions 3. Solving a...

Solving Third Degree Polynomial EquationsMichel Beaudin, [email protected]

Frédérick Henri, [email protected]

École de technologie supérieure, Montréal, Québec, Canada

TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd 2017-01-17 1

Abstract

Starting with some concrete examples, we use TI-NspireCX CAS to solve third degree polynomial equations withreal coefficients. Then we compare the answers obtainedto the ones provided by other CAS.

Using Cardano’s method or François Viète’s formulas, wecreate a homemade function allowing Nspire to produceclear and compact answers for the solutions of a thirddegree polynomial equation: "Everything should be madeas simple as possible, but not simpler".

TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd 2017-01-17 2

Table of contents

1. Two examples showing CAS imperfections

2. An algorithm to beautify the solutions

3. Solving a third degree polynomial equation with 1 realroot

4. Solving a third degree polynomial equation with 3 realroots

5. Implementation of a new TI-Nspire function

6. Conclusion

2017-01-17 3TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd

Two examples showing CAS imperfections –first exampleConsider the following third degree polynomial function

Since :a. the derivative of the function is strictly positive ;

b. and ;

we infer that the function has a single real root locatedbetween and .

Let’s look at the graph of the function to confirm this.2017-01-17 4TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd

Two examples showing CAS imperfections –first example


Two examples showing CAS imperfections –first exampleLet’s compare the solutions we obtain when we solve theequation

using 4 different CAS:

MapleTM,

MathematicaTM,

TI-NspireTM and

DeriveTM.2017-01-17 6TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd

Two examples showing CAS imperfections –first exampleWith MapleTM: 12 4 4 5 24 4 514 4 4 5 14 4 5 12 I 3 12 4 4 5 24 4 514 4 4 5 14 4 5 12 I 3 12 4 4 5 24 4 5


Two examples showing CAS imperfections –first exampleWith MathematicaTM:

21 5 12 1 512 1 i 3 12 1 5 1 i 32 1 512 1 i 3 12 1 5 1 i 32 1 5


Two examples showing CAS imperfections –first exampleWith TI-NspireTM:

2 5 32 5 3 2 5 3 2 · 25 15 1 · 2 2 5 1 2 · 2 2 5 1 · 5 1 · 2 2 · 4 · 316 i5 15 1 · 2 2 5 1 2 · 2 2 5 1 · 5 1 · 2 2 · 4 · 316 i


Two examples showing CAS imperfections –first exampleWith DeriveTM (notice the simplicity of these solutions):

4 5 42 4 5 424 5 44 4 5 44 i 3 4 5 44 3 4 5 444 5 44 4 5 44 i 3 4 5 44 3 4 5 44


Two examples showing CAS imperfections –second exampleConsider now this other simple third degree polynomialfunction

Having 2 critical points, we may expect this function tohave 3 real roots.

Let’s look at the graph of the function to confirm this.


Two examples showing CAS imperfections –second example


Two examples showing CAS imperfections –second exampleAgain, let’s compare the solutions we obtain when wesolve the equation

using 4 different CAS:

MapleTM,

MathematicaTM,

TI-NspireTM and

DeriveTM.2017-01-17 13TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd

Two examples showing CAS imperfections –second exampleWith MapleTM (it’s hard to believe these are real values)12 4 4I 3 24 4I 3

14 4 4I 3 14 4I 3 12 I 3 12 4 4I 3 24 4I 314 4 4I 3 14 4I 3 12 I 3 12 4 4I 3 24 4I 3


Two examples showing CAS imperfections –second exampleWith MathematicaTM:

112 1 i 3 12 1 i 31 i 32 1 i 3 12 12 1 i 3 1 i 3

12 1 i 3 12 1 i 3 1 i 32 1 i 32017-01-17 15TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd

Two examples showing CAS imperfections –second exampleWith TI-NspireTM :

cos 2 ·9 sin 2 ·9 · 31cos 2 ·9 sin 2 ·9 · 3 1

cos 2 ·9 sin 2 ·9 · 3 1cos 2 ·9 sin 2 ·9 · 32017-01-17 16TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd

Two examples showing CAS imperfections –second exampleWith DeriveTM (again, so much simpler):

2 · cos 292 · cos 92 · sin 18


An algorithm to beautify the solutions

Our goal is to create a new function for TI-Nspire thatfinds the roots of a third degree polynomial functionwhile:

a. avoiding the presence of radicals in thedenominators of the roots;

b. avoiding needless occurrence of in the real rootsand using, instead, trig expressions for the roots;

The next slide presents the algorithm we will implement.


Transform into a 6th degree polynomial equation

An algorithm to beautify the solutions


Choose a 3rd degree function

Get rid of the 2nd degree term

Apply a trigonometric substitution

Ascertain the number of real roots

Transform into a 2nd degree polynomial equation

Select one solution of the previous equation

Compute the 3 cubic roots of this solution

Compute the roots of the original function

1 real root 3 real roots

Solve the equation

Compute the roots of the original function

Let’s use our algorithm to solve a third degree polynomialequation that possesses a single real root.

To solve the equation, we will need to accomplish 8 steps,as seen in the algorithm’s organigram.

Solving a third degree polynomial equation with 1 real root


START 2 43 5 6 7 8

Solving a third degree polynomial equation with 1 real rootStep #1 – Pick a 3rd degree polynomial function

SELECT A 3RD DEGREE POLYNOMIAL FUNCTION IN THE FORM OF.

Let’s apply our algorithm to solve the equation

As we will see, this function only possess a single realroot.


START 2 43 5 6 7 8

Solving a third degree polynomial equation with 1 real rootStep #2 – Get rid of the 2nd degree term

DEFINE BY SUBSTITUTING FOR AND THENDIVIDING BY .

Since 5 , we obtain:


START 2 43 5 6 7 8

Solving a third degree polynomial equation with 1 real rootStep #3 – Ascertain the number of real roots

REWRITE IN THE FORM OF WHERE

AND .

Since , we find that

and .


START 2 43 5 6 7 8

Solving a third degree polynomial equation with 1 real rootStep #3 – Ascertain the number of real roots

IF , THERE IS A SINGLE REAL SOLUTION. OTHERWISE,THERE ARE 3 REAL SOLUTIONS.

Since and , .

We conclude that our equation only has a single realsolution.


START 2 43 5 6 7 8

Solving a third degree polynomial equation with 1 real rootStep #4 – Transform the equation into a 6th degreepolynomial equation

DEFINE BY SUBSTITUTING FORIN AND THEN MULTIPLYING BY .

Since , we find:


START 2 43 5 6 7 8

Solving a third degree polynomial equation with 1 real rootStep #5 – Transform the equation into a 2nd degreepolynomial equation

DEFINE BY SUBSTITUTING FOR IN.

Since , we find:


START 2 43 5 6 7 8

Solving a third degree polynomial equation with 1 real rootStep #6 – Select 1 of the 2 solutions of the 2nd degreepolynomial equation

SOLVE THE EQUATION AND SELECT

THE SOLUTION .

After solving , we select the

solution · .


START 2 43 5 6 7 8

Solving a third degree polynomial equation with 1 real rootStep #7 – Compute the cubic roots of the chosen solution

SINCE AND , SOLVE THE EQUATION

.

Solving · , we find 3 solutions that we note, and .


START 2 43 5 6 7 8

Solving a third degree polynomial equation with 1 real rootStep #7 – Compute the cubic roots of the chosen solution

The solutions of the equation · are:

9 · 401 191 · 269 · 401 191 · 212 9 · 401 191 · 3 · 2129 · 401 191 · 212 9 · 401 191 · 3 · 2122017-01-17 29TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd

START 2 43 5 6 7 8

Solving a third degree polynomial equation with 1 real rootStep #8 – Compute the roots of the original function

USING THE 3 ROOTS OF THE FUNCTION , COMPUTE THE 3ROOTS OF THE FUNCTION .

Since , we compute the 3 roots of :


START 2 43 5 6 7 8

Solving a third degree polynomial equation with 1 real rootStep #8 – Compute the roots of the original function


Since , we compute the 3 roots of :


START 2 43 5 6 7 8

Let’s use our algorithm to solve a third degree polynomialequation that possesses 3 real roots.

To solve the equation, we will need to accomplish only 6steps, as seen in the algorithm’s organigram.

Solving a third degree polynomial equation with 3 real roots


START 2 43 5 6

Solving a third degree polynomial equation with 3 real rootsStep #1 – Pick a 3rd degree polynomial function

SELECT A 3RD DEGREE POLYNOMIAL FUNCTION IN THE FORM OF.

Let’s apply our algorithm again to solve the equation

As we will see, this function only possess 3 real roots.


START 2 43 5 6

Solving a third degree polynomial equation with 3 real rootsStep #2 – Get rid of the 2nd degree term

DEFINE BY SUBSTITUTING FOR AND THENDIVIDING BY .

Since , we obtain:


START 2 43 5 6

Solving a third degree polynomial equation with 3 real rootsStep #3 – Find the number of real roots

REWRITE IN THE FORM OF WHERE

AND .

Since , we find that

and .


START 2 43 5 6

Solving a third degree polynomial equation with 3 real rootsStep #3 – Ascertain the number of real roots

IF , THERE IS SINGLE REAL SOLUTION. OTHERWISE,THERE ARE 3 REAL SOLUTIONS.

Since and , .

We conclude that our equation has 3 real solutions.


START 2 43 5 6

Solving a third degree polynomial equation with 3 real rootsStep #4 – Apply a substitution

DEFINE BY SUBSTITUTING

FOR IN , WHERE .

Since , we find:


START 2 43 5 6

Solving a third degree polynomial equation with 3 real rootsStep #5 – Solve the equation

COMPUTE THE 3 ROOTS OF :

, AND .

And so we find:


START 2 43 5 6

Solving a third degree polynomial equation with 3 real rootsStep #6 – Compute the roots of the original function


Since , we compute the 3 roots of:


START 2 43 5 6

Solving a third degree polynomial equation with 3 real rootsStep #6 – Compute the roots of the original function


Since , we compute the 3 roots of :4 3 2 · sin 4 34 3 2 · sin 4 34 3 2 · sin 4 32017-01-17 40TIME 2016, UNAM, Mexico City, Mexico, June 29th - July 2nd

START 2 43 5 6

This new function has been implemented by MichelBeaudin and can be downloaded. Go to

https://cours.etsmtl.ca/seg/mbeaudin/

And save the TI-Nspire CX CAS library “Kit_ETS_MB” intoMyLib. The name of the function is “compact_cubic”.

Live examples: on the TI-Nspire CX CAS file.

Implementation of a new TI-Nspire function


Teaching mathematics using CAS is, as far as we areconcerned, an excellent opportunity for mathematicsteachers to stay enthusiastic even though the curriculumhas been quite the same for the several past years!

Conclusion


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