<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom" xmlns:content="http://purl.org/rss/1.0/modules/content/"><channel><title>Ecuaciones De Segundo Grado on Karpoke - Just Another Blog</title><link>http://karpoke.ignaciocano.com/tags/ecuaciones-de-segundo-grado/</link><description>Recent content in Ecuaciones De Segundo Grado on Karpoke - Just Another Blog</description><generator>Hugo -- 0.159.0</generator><language>es</language><lastBuildDate>Tue, 29 Mar 2011 18:54:00 +0100</lastBuildDate><atom:link href="http://karpoke.ignaciocano.com/tags/ecuaciones-de-segundo-grado/index.xml" rel="self" type="application/rss+xml"/><item><title>LaTeX en Wordpress</title><link>http://karpoke.ignaciocano.com/2011/03/29/latex-en-wordpress/</link><pubDate>Tue, 29 Mar 2011 18:54:00 +0100</pubDate><guid>http://karpoke.ignaciocano.com/2011/03/29/latex-en-wordpress/</guid><description>&lt;p&gt;Descarga el &lt;a href="http://wordpress.org/extend/plugins/latex/"&gt;plugin de LaTeX para Wordpress&lt;/a&gt;. Luego, escribe:&lt;/p&gt;
&lt;div class="highlight"&gt;&lt;pre tabindex="0" style="color:#f8f8f2;background-color:#272822;-moz-tab-size:4;-o-tab-size:4;tab-size:4;-webkit-text-size-adjust:none;"&gt;&lt;code class="language-latex" data-lang="latex"&gt;&lt;span style="display:flex;"&gt;&lt;span&gt;&lt;span style="color:#66d9ef"&gt;\begin&lt;/span&gt;{align*}
&lt;/span&gt;&lt;/span&gt;&lt;span style="display:flex;"&gt;&lt;span&gt;ax^2+bx+c &amp;amp;= 0 &lt;span style="color:#66d9ef"&gt;\\&lt;/span&gt;
&lt;/span&gt;&lt;/span&gt;&lt;span style="display:flex;"&gt;&lt;span&gt;x^2+&lt;span style="color:#66d9ef"&gt;\frac&lt;/span&gt;{b}{a}x+&lt;span style="color:#66d9ef"&gt;\frac&lt;/span&gt;{c}{a} &amp;amp;= 0 &lt;span style="color:#66d9ef"&gt;\\&lt;/span&gt;
&lt;/span&gt;&lt;/span&gt;&lt;span style="display:flex;"&gt;&lt;span&gt;x^2+&lt;span style="color:#66d9ef"&gt;\frac&lt;/span&gt;{b}{a}x &amp;amp;= -&lt;span style="color:#66d9ef"&gt;\frac&lt;/span&gt;{c}{a} &lt;span style="color:#66d9ef"&gt;\\&lt;/span&gt;
&lt;/span&gt;&lt;/span&gt;&lt;span style="display:flex;"&gt;&lt;span&gt;x^2+&lt;span style="color:#66d9ef"&gt;\frac&lt;/span&gt;{b}{a}x+&lt;span style="color:#66d9ef"&gt;\frac&lt;/span&gt;{b^2}{4a^2} &amp;amp;= &lt;span style="color:#66d9ef"&gt;\frac&lt;/span&gt;{b^2}{4a^2} - &lt;span style="color:#66d9ef"&gt;\frac&lt;/span&gt;{c}{a} &lt;span style="color:#66d9ef"&gt;\\&lt;/span&gt;
&lt;/span&gt;&lt;/span&gt;&lt;span style="display:flex;"&gt;&lt;span&gt;(x+&lt;span style="color:#66d9ef"&gt;\frac&lt;/span&gt;{b}{2a})^2 &amp;amp;= &lt;span style="color:#66d9ef"&gt;\frac&lt;/span&gt;{b^2}{4a^2} - &lt;span style="color:#66d9ef"&gt;\frac&lt;/span&gt;{4ac}{4a^2} &lt;span style="color:#66d9ef"&gt;\\&lt;/span&gt;
&lt;/span&gt;&lt;/span&gt;&lt;span style="display:flex;"&gt;&lt;span&gt;x+&lt;span style="color:#66d9ef"&gt;\frac&lt;/span&gt;{b}{2a} &amp;amp;= &lt;span style="color:#66d9ef"&gt;\pm\sqrt&lt;/span&gt;{&lt;span style="color:#66d9ef"&gt;\frac&lt;/span&gt;{b^2-4ac}{4a^2}} &lt;span style="color:#66d9ef"&gt;\\&lt;/span&gt;
&lt;/span&gt;&lt;/span&gt;&lt;span style="display:flex;"&gt;&lt;span&gt;x+&lt;span style="color:#66d9ef"&gt;\frac&lt;/span&gt;{b}{2a} &amp;amp;= &lt;span style="color:#66d9ef"&gt;\frac&lt;/span&gt;{&lt;span style="color:#66d9ef"&gt;\pm\sqrt&lt;/span&gt;{b^2-4ac}}{2a} &lt;span style="color:#66d9ef"&gt;\\&lt;/span&gt;
&lt;/span&gt;&lt;/span&gt;&lt;span style="display:flex;"&gt;&lt;span&gt;x &amp;amp;= &lt;span style="color:#66d9ef"&gt;\frac&lt;/span&gt;{-b&lt;span style="color:#66d9ef"&gt;\pm\sqrt&lt;/span&gt;{b^2-4ac}}{2a}
&lt;/span&gt;&lt;/span&gt;&lt;span style="display:flex;"&gt;&lt;span&gt;&lt;span style="color:#66d9ef"&gt;\end&lt;/span&gt;{align*}
&lt;/span&gt;&lt;/span&gt;&lt;/code&gt;&lt;/pre&gt;&lt;/div&gt;&lt;p&gt;El resultado será parecido a éste:&lt;/p&gt;
&lt;div&gt;
$$
\begin{align*}
ax^2+bx+c &amp;= 0 \\
x^2+\frac{b}{a}x+\frac{c}{a} &amp;= 0 \\
x^2+\frac{b}{a}x &amp;= -\frac{c}{a} \\
x^2+\frac{b}{a}x+\frac{b^2}{4a^2} &amp;= \frac{b^2}{4a^2} - \frac{c}{a} \\
(x+\frac{b}{2a})^2 &amp;= \frac{b^2}{4a^2} - \frac{4ac}{4a^2} \\
x+\frac{b}{2a} &amp;= \pm\sqrt{\frac{b^2-4ac}{4a^2}} \\
x+\frac{b}{2a} &amp;= \frac{\pm\sqrt{b^2-4ac}}{2a} \\
x &amp;= \frac{-b\pm\sqrt{b^2-4ac}}{2a}
\end{align*}
$$
&lt;/div&gt;</description></item></channel></rss>