2015-02-08 16:04:05 +00:00
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\documentclass[a4paper,10pt]{article}
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\RequirePackage[utf8x]{inputenc}
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\RequirePackage{amssymb}
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\RequirePackage{amsmath}
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\RequirePackage{amsfonts}
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\RequirePackage{graphicx}
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2015-01-06 08:22:52 +00:00
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% Title Page
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2015-02-08 16:04:05 +00:00
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\title{Calcul littéral et statistiques}
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\date{\today}
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2015-01-06 08:22:52 +00:00
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\begin{document}
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\maketitle
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2015-02-08 16:04:05 +00:00
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\section{Polynômes}
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2015-01-06 08:22:52 +00:00
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2015-02-08 16:04:05 +00:00
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Résoudre l'équation suivante
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\begin{eqnarray*}
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2015-03-19 21:05:00 +00:00
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- 3 x^{ 2 } + 6 x - 3 & = & 0
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2015-02-08 16:04:05 +00:00
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\end{eqnarray*}
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Solution:
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2015-01-06 08:22:52 +00:00
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2015-02-08 16:04:05 +00:00
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On commence par calculer le discriminant
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\begin{eqnarray*}
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\Delta & = & b^2-4ac \\
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2015-03-19 21:05:00 +00:00
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\Delta & = & 6^{ 2 } - 4 \times ( -3 ) \times ( -3 ) \\
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\Delta & = & 36 - ( -12 ) \times ( -3 ) \\
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\Delta & = & 36 - 36 \\
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2015-02-08 16:04:05 +00:00
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\Delta & = & 0
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\end{eqnarray*}
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2015-01-06 08:22:52 +00:00
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2015-02-08 16:04:05 +00:00
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Alors $\Delta = 0 = 0$ donc il y a une solution
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2015-01-06 08:22:52 +00:00
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2015-02-08 16:04:05 +00:00
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2015-01-06 08:22:52 +00:00
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2015-02-08 16:04:05 +00:00
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\begin{eqnarray*}
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2015-03-19 21:05:00 +00:00
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x_1 = \frac{-b}{2a} = \frac{ -6 }{ 2 \times ( -3 ) } = \frac{ -6 }{ -6 } = \frac{ 6 }{ 6 } = 1 = \frac{ -6 }{ -6 }
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2015-02-08 16:04:05 +00:00
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\end{eqnarray*}
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2015-01-06 08:22:52 +00:00
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2015-03-19 21:05:00 +00:00
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Les solutions sont donc $\mathcal{S} = \left\{ \frac{ -6 }{ -6 }\right\}$
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2015-01-06 08:22:52 +00:00
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2015-02-08 16:04:05 +00:00
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\bigskip
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~\dotfill
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\bigskip
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2015-01-06 08:22:52 +00:00
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Résoudre l'équation suivante
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\begin{eqnarray*}
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2015-03-19 21:05:00 +00:00
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- 7 x^{ 2 } - 7 x + 9 & = & - 2 x^{ 2 } + x - 9
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2015-01-06 08:22:52 +00:00
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\end{eqnarray*}
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Solution:
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2015-02-08 16:04:05 +00:00
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On commence par se ramener à une équation de la forme $ax^2+bx+c = 0$.
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2015-01-06 08:22:52 +00:00
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\begin{eqnarray*}
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2015-03-19 21:05:00 +00:00
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- 7 x^{ 2 } - 7 x + 9 = - 2 x^{ 2 } + x - 9 & \Leftrightarrow & - 7 x^{ 2 } - 7 x + 9 - (- 2 x^{ 2 } + x - 9) = 0 \\
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& \Leftrightarrow & - 7 x^{ 2 } + 2 x^{ 2 } - 7 x - x + 9 + 9= 0 \\
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& \Leftrightarrow & ( ( -7 ) + 2 ) x^{ 2 } + ( ( -7 ) + ( -1 ) ) x + 9 + 9= 0 \\
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& \Leftrightarrow & - 5 x^{ 2 } - 8 x + 18= 0
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2015-01-06 08:22:52 +00:00
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\end{eqnarray*}
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2015-02-08 16:04:05 +00:00
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2015-03-19 21:05:00 +00:00
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On cherche maintenant à résoudre l'équation $- 5 x^{ 2 } - 8 x + 18 = 0$.
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2015-02-08 16:04:05 +00:00
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On commence par calculer le discriminant
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2015-01-06 08:22:52 +00:00
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\begin{eqnarray*}
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2015-02-08 16:04:05 +00:00
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\Delta & = & b^2-4ac \\
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2015-03-19 21:05:00 +00:00
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\Delta & = & ( -8 )^{ 2 } - 4 \times ( -5 ) \times 18 \\
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\Delta & = & 64 - ( -20 ) \times 18 \\
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\Delta & = & 64 - ( -360 ) \\
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\Delta & = & 424
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2015-02-08 16:04:05 +00:00
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\end{eqnarray*}
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2015-01-06 08:22:52 +00:00
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2015-03-19 21:05:00 +00:00
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Alors $\Delta = 424 > 0$ donc il y a deux solutions
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2015-02-08 16:04:05 +00:00
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2015-01-06 08:22:52 +00:00
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2015-02-08 16:04:05 +00:00
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\begin{eqnarray*}
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2015-03-19 21:05:00 +00:00
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x_1 & = & \frac{-b - \sqrt{\Delta}}{2a} = \frac{-8 - \sqrt{424}}{2 \times -5} = 1.26 \\
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x_2 & = & \frac{-b + \sqrt{\Delta}}{2a} = \frac{-8 + \sqrt{424}}{2 \times -5} = -2.86
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2015-02-08 16:04:05 +00:00
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\end{eqnarray*}
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2015-03-19 21:05:00 +00:00
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Les solutions sont donc $\mathcal{S} = \left\{ 1.26; -2.86 \right\}$
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2015-02-08 16:04:05 +00:00
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2015-01-06 08:22:52 +00:00
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\end{document}
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%%% Local Variables:
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%%% mode: latex
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%%% TeX-master: "master"
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%%% End:
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