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Import work from year 2014-2015

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Benjamin Bertrand
2017-06-16 09:48:07 +03:00
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\documentclass[a4paper,10pt, table]{/media/documents/Cours/Prof/Enseignements/2014-2015/Archive/2014-2015/tools/style/classCours}
\usepackage{/media/documents/Cours/Prof/Enseignements/2014-2015/Archive/2014-2015/2014_2015}
% Title Page
\titre{Factorisation des polynômes du second degré}
% \seconde \premiereS \PSTMG \TSTMG
\classe{\premiereS}
\date{Janvier 2015}
\begin{document}
\maketitle
Dans le chapitre sur la dérivation, nous étions bloqués pour l'étude des polynômes de degré 3 car nous ne savions pas comment analyser le signe du polynôme dérivé.
\textit{Avec un exemple}.
\section{Solutions de l'équation $ax^2 + bx + c = 0$}
\textit{ On reprend ce qu'on sais déjà du chapitre sur la forme canonique et avec le tableu de variation, on met en valeur le rôle de $b^2 - 4ac$ dans la determination du nombre de solutions}
\begin{Prop}
Soit $ax^2 + bx + c = 0$ une équation du 2nd degré.
On définit le discriminant: $\Delta = b^2 - 4ac$
Le signe de $\Delta$ va determiner le nombre de solution à cette équation
\begin{itemize}
\item Si $\Delta > 0$ alors il y a 2 solutions
\begin{eqnarray*}
x_1 = \frac{-b + \sqrt{\Delta}}{2a} & \mbox{ et } & x_2 = \frac{-b - \sqrt{\Delta}}{2a}
\end{eqnarray*}
\item Si $\Delta = 0$ alors il y a 1 solution
\begin{eqnarray*}
x_1 & = & \frac{-b}{2a}
\end{eqnarray*}
\item Si $\Delta < 0$ il n'y a pas de solution
\end{itemize}
\end{Prop}
\section{Tableau de signe}
\textit{On fait les differents cas en fonction de $\Delta$}
\section{Factorisation des polynômes du seconde degré}
\section{Algorithme pour résoudre $ax^2 + bx + c= 0$}
\end{document}
%%% Local Variables:
%%% mode: latex
%%% TeX-master: "master"
%%% End:

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Notes sur Cours sur la factorisation pour les 1S
################################################
:date: 2015-07-01
:modified: 2015-07-01
:tags: Cours,Analyse
:category: 1S
:authors: Benjamin Bertrand
:summary: Pas de résumé, note créée automatiquement parce que je ne l'avais pas bien fait...
`Lien vers Facto2ndDeg.tex <Facto2ndDeg.tex>`_
`Lien vers Facto2ndDeg.pdf <Facto2ndDeg.pdf>`_

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# Import de la racine carrée
from math import sqrt
# Saisir a
a = int(input("a?"))
# Saisir b
b = int(input("b?"))
# Saisir c
c = int(input("c?"))
# On calcul le discriminant
# delta prend la valeur b^2 - 4ac
delta = b**2 - 4*a*c
# On différencie 3 cas
# Si le discriminant est positif
if delta > 0:
# On affiche "2 solution:"
print("2 solutions:")
# On calcul x1
x1 = (-b - sqrt(delta))/(2*a)
# On calcul x2
x2 = (-b + sqrt(delta))/(2*a)
# On affiche x1
print("x1 = ", x1)
# On affiche x1
print("x2 = ", x2)
# Si le discriminant est nul
elif delta == 0:
# On affiche "Une solution: "
print("Une solution:")
# On calcule x1
x1 = -b / (2*a)
# On affiche x1
print("x1 = ", x1)
# Dernier cas, le discriminant est alors négatif
else:
# On affiche "Pas de solution"
print("Pas de solution")

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Notes sur act
#############
:date: 2015-07-01
:modified: 2015-07-01
:tags: Analyse
:category: 1S
:authors: Benjamin Bertrand
:summary: Pas de résumé, note créée automatiquement parce que je ne l'avais pas bien fait...
`Lien vers boite.pdf <boite.pdf>`_