The Quadratic Formula
Auteur:
Amanda
Last Updated:
il y a 9 ans
License:
Creative Commons CC BY 4.0
Résumé:
This paper will state and prove the quadratic formula.
\begin
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This paper will state and prove the quadratic formula.
\begin
Discover why 18 million people worldwide trust Overleaf with their work.
\documentclass[a4paper]{article}
\usepackage{amsmath}
\usepackage{amsthm}
\newtheorem{thm}{Theorem}
\usepackage[english]{babel}
\usepackage[utf8]{inputenc}
\usepackage{graphicx}
\usepackage[colorinlistoftodos]{todonotes}
\title{The Quadratic Formula}
\author{Amanda King}
\date{\today}
\begin{document}
\maketitle
\begin{abstract}
This paper will state and prove the quadratic formula.
\end{abstract}
\section{Introduction}
We have all, at some point, learned the quadratic formula. Usually, we learn how to use it to find solutions of a quadratic equation. We might have seen the proof at this point, but most of us did not commit it to memory since the proof does not help us to use the formula.
\section{The Quadratic Formula}
\label{sec:examples}
\begin{thm} Let a, b, and c be real number. The solutions of $ax^2+bx+c=0$ are
$$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}.$$
%\begin{array}${*{20}c}{x = \frac{{ - b \pm \sqrt {b^2 - 4ac} }}{{2a}}} & {{\rm{when}}} & {ax^2 + bx + c = 0}$ \\ \end{array}
\end{thm}
\begin{proof} In order to prove the quadratic formula, we use the process of completing the square. Starting with $ax^2+bx+c=c$, arithmetic gives
\begin{align*}
\\ax^2+bx&=-c,
\\x^2+\frac{b}{a}x&=-\frac{c}{a},
\\x^2+\frac{b}{a}x+\frac{b^2}{4a^2}&=-\frac{c}{a}+\frac{b^2}{4a^2},
\\\Bigg(x+\frac{b}{2a}\Bigg)^2&=-\frac{4ac}{4a^2}+\frac{b^2}{4a^2},
\\x+\frac{b}{2a}&=\pm\sqrt[]{\frac{b-4ac}{4a^2}}
\\x&=\frac{-b\pm\sqrt[]{b^2-4ac}}{2a}
\end{align*}
\end{proof}
\end{document}
Theorem 1
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\centering
\includegraphics[width=0.3\textwidth]{frog.jpg}
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\end{tabular}
\caption{\label{tab:widgets}An example table.}
\end{table}
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$$S_n = \frac{X_1 + X_2 + \cdots + X_n}{n}
= \frac{1}{n}\sum_{i}^{n} X_i$$
denote their mean. Then as $n$ approaches infinity, the random variables $\sqrt{n}(S_n - \mu)$ converge in distribution to a normal $\mathcal{N}(0, \sigma^2)$.
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\end{document}