117 lines
2.7 KiB
TeX
117 lines
2.7 KiB
TeX
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\documentclass[a4paper,10pt]{article}
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\RequirePackage[utf8x]{inputenc}
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\RequirePackage[francais]{babel}
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\RequirePackage{amssymb}
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\RequirePackage{amsmath}
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\RequirePackage{amsfonts}
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\RequirePackage{subfig}
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\RequirePackage{graphicx}
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\RequirePackage{color}
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% Title Page
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\title{Calcul littéral et statistiques}
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\date{\today}
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\begin{document}
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\maketitle
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\section{Polynômes}
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Résoudre l'équation suivante
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\begin{eqnarray*}
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3 x^{ 2 } + 6 x + 3 & = & 0
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\end{eqnarray*}
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Solution:
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On commence par calculer le discriminant de $P(x) = 3 x^{ 2 } + 6 x + 3$.
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\begin{eqnarray*}
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\Delta & = & b^2-4ac \\
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\Delta & = & 6^{ 2 } - 4 \times 3 \times 3 \\
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\Delta & = & 36 - 4 \times 9 \\
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\Delta & = & 36 - 36 \\
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\Delta & = & 0
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\end{eqnarray*}
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Comme $\Delta = 0$ donc $P$ a une racine
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\begin{eqnarray*}
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x_1 = \frac{-b}{2a} = \frac{-6}{2\times 3} = -1 \\
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\end{eqnarray*}
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La solution de $3 x^{ 2 } + 6 x + 3 = 0$ est donc $\mathcal{S} = \left\{ -1\right\}$
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\bigskip
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~\dotfill
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\bigskip
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Résoudre l'équation suivante
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\begin{eqnarray*}
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x^{ 2 } + 4 x + 2 & = & - 9 x^{ 2 } + 9 x + 5
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\end{eqnarray*}
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Solution:
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On commence par se ramener à une équation de la forme $ax^2+bx+c = 0$.
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\begin{align*}
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& & x^{ 2 } + 4 x + 2 = - 9 x^{ 2 } + 9 x + 5 \\
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& \Leftrightarrow & x^{ 2 } + 4 x + 2 - ( - 9 x^{ 2 } + 9 x + 5 )= 0 \\
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& \Leftrightarrow & x^{ 2 } + 4 x + 2 + 9 x^{ 2 } - 9 x - 5= 0 \\
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& \Leftrightarrow & ( 1 + 9 ) x^{ 2 } + ( 4 - 9 ) x + 2 - 5= 0 \\
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& \Leftrightarrow & 10 x^{ 2 } - 5 x - 3= 0
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\end{align*}
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On cherche maintenant à résoudre l'équation $10 x^{ 2 } - 5 x - 3 = 0$.
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On commence par calculer le discriminant de $P(x) = 10 x^{ 2 } - 5 x - 3$.
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\begin{eqnarray*}
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\Delta & = & b^2-4ac \\
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\Delta & = & -5^{ 2 } - 4 \times 10 \times ( -3 ) \\
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\Delta & = & 25 - 4 \times ( -30 ) \\
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\Delta & = & 25 - ( -120 ) \\
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\Delta & = & 145
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\end{eqnarray*}
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comme $\Delta = 145 > 0$ donc $P$ a deux racines
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\begin{eqnarray*}
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x_1 & = & \frac{-b - \sqrt{\Delta}}{2a} = \frac{-5 - \sqrt{145}}{2 \times 10} = - \frac{\sqrt{145}}{20} + \frac{1}{4} \\
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x_2 & = & \frac{-b + \sqrt{\Delta}}{2a} = \frac{-5 + \sqrt{145}}{2 \times 10} = \frac{1}{4} + \frac{\sqrt{145}}{20}
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\end{eqnarray*}
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Les solutions de l'équation $10 x^{ 2 } - 5 x - 3 = 0$ sont donc $\mathcal{S} = \left\{ - \frac{\sqrt{145}}{20} + \frac{1}{4}; \frac{1}{4} + \frac{\sqrt{145}}{20} \right\}$
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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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