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		<title>1.3 Lösning 4b - Versionshistorik</title>
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		<updated>2026-11-05T14:14:27Z</updated>
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		<id>https://wiki.mattekollen.se/index.php?title=1.3_L%C3%B6sning_4b&amp;diff=13135&amp;oldid=prev</id>
		<title>Taifun den 15 augusti 2014 kl. 11.33</title>
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				<updated>2014-08-15T11:33:18Z</updated>
		
		<summary type="html">&lt;p&gt;&lt;/p&gt;
&lt;p&gt;&lt;b&gt;Ny sida&lt;/b&gt;&lt;/p&gt;&lt;div&gt;Nollproduktmetoden ger:&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; \begin{align} (3\,x-1)\;\cdot\; &amp;amp; (2\,x+1) &amp;amp;      = \ \ \,\, 0            \\&lt;br /&gt;
                     (3\,x-1)\;\;\,\,  &amp;amp;          &amp;amp;      = \ \ \,\, 0            \\&lt;br /&gt;
                                       &amp;amp;          &amp;amp; 3\,x = \ \ \,\, 1            \\&lt;br /&gt;
                                       &amp;amp;          &amp;amp; x_1  = \ \ \, {1 \over 3}    \\&lt;br /&gt;
                                       &amp;amp; (2\,x+1) &amp;amp;      = \ \ \,\, 0            \\&lt;br /&gt;
                                       &amp;amp;          &amp;amp; 2\,x = \, -1                 \\&lt;br /&gt;
                                       &amp;amp;          &amp;amp; x_2  = -{1 \over 2}          \\&lt;br /&gt;
       \end{align}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&amp;lt;math&amp;gt; (3\,x-1) \cdot (2\,x+1) &amp;lt;/math&amp;gt; har två nollställen:&lt;br /&gt;
&lt;br /&gt;
:::&amp;lt;math&amp;gt;x_1 = {1 \over 3} \; {\rm och} \; x_2 = -{1 \over 2} &amp;lt;/math&amp;gt;.&lt;/div&gt;</summary>
		<author><name>Taifun</name></author>	</entry>

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