Lösung 3.1:3a

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Zeile 1: Zeile 1:
First expand the expression
First expand the expression
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{{Displayed math||<math>\begin{align}
+
{{Abgesetzte Formel||<math>\begin{align}
\bigl(\sqrt{5}-\sqrt{2}\bigr)\bigl(\sqrt{5}+\sqrt{2}\bigr)
\bigl(\sqrt{5}-\sqrt{2}\bigr)\bigl(\sqrt{5}+\sqrt{2}\bigr)
&= \sqrt{5}\cdot\sqrt{5} + \sqrt{5}\cdot\sqrt{2} - \sqrt{2}\cdot\sqrt{5} - \sqrt{2}\cdot\sqrt{2}\\[5pt]
&= \sqrt{5}\cdot\sqrt{5} + \sqrt{5}\cdot\sqrt{2} - \sqrt{2}\cdot\sqrt{5} - \sqrt{2}\cdot\sqrt{2}\\[5pt]
Zeile 9: Zeile 9:
Because <math>\sqrt{5}</math> and <math>\sqrt{2}</math> are defined as those numbers which, when multiplied with themselves give 5 and 2 respectively, we have that
Because <math>\sqrt{5}</math> and <math>\sqrt{2}</math> are defined as those numbers which, when multiplied with themselves give 5 and 2 respectively, we have that
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{{Displayed math||<math>\sqrt{5}\cdot\sqrt{5} - \sqrt{2}\cdot\sqrt{2} = 5-2 = 3\,\textrm{.}</math>}}
+
{{Abgesetzte Formel||<math>\sqrt{5}\cdot\sqrt{5} - \sqrt{2}\cdot\sqrt{2} = 5-2 = 3\,\textrm{.}</math>}}
Note: The expansion of <math>\bigl(\sqrt{5}-\sqrt{2}\bigr)\bigl(\sqrt{5}+\sqrt{2}\bigr)</math> can also be done directly with the formula for difference of two squares <math>(a-b)(a+b) = a^{2} - b^{2}</math> using <math>a=\sqrt{5}</math> and <math>b=\sqrt{2}</math>.
Note: The expansion of <math>\bigl(\sqrt{5}-\sqrt{2}\bigr)\bigl(\sqrt{5}+\sqrt{2}\bigr)</math> can also be done directly with the formula for difference of two squares <math>(a-b)(a+b) = a^{2} - b^{2}</math> using <math>a=\sqrt{5}</math> and <math>b=\sqrt{2}</math>.

Version vom 08:36, 22. Okt. 2008

First expand the expression

\displaystyle \begin{align}

\bigl(\sqrt{5}-\sqrt{2}\bigr)\bigl(\sqrt{5}+\sqrt{2}\bigr) &= \sqrt{5}\cdot\sqrt{5} + \sqrt{5}\cdot\sqrt{2} - \sqrt{2}\cdot\sqrt{5} - \sqrt{2}\cdot\sqrt{2}\\[5pt] &= \sqrt{5}\cdot\sqrt{5} - \sqrt{2}\cdot\sqrt{2}\,\textrm{.} \end{align}

Because \displaystyle \sqrt{5} and \displaystyle \sqrt{2} are defined as those numbers which, when multiplied with themselves give 5 and 2 respectively, we have that

\displaystyle \sqrt{5}\cdot\sqrt{5} - \sqrt{2}\cdot\sqrt{2} = 5-2 = 3\,\textrm{.}


Note: The expansion of \displaystyle \bigl(\sqrt{5}-\sqrt{2}\bigr)\bigl(\sqrt{5}+\sqrt{2}\bigr) can also be done directly with the formula for difference of two squares \displaystyle (a-b)(a+b) = a^{2} - b^{2} using \displaystyle a=\sqrt{5} and \displaystyle b=\sqrt{2}.

Von „http://wiki.sommarmatte.se/wikis/2009/bridgecourse1-TU-Berlin/index.php/L%C3%B6sung_3.1:3a“