Lösung 3.1:3a

Aus Online Mathematik Brückenkurs 1

(Unterschied zwischen Versionen)
Wechseln zu: Navigation, Suche
K
Zeile 1: Zeile 1:
First expand the expression
First expand the expression
 +
{{Displayed math||<math>\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}</math>}}
-
<math>\begin{align}
+
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
-
& \left( \sqrt{5}-\sqrt{2} \right)\left( \sqrt{5}3\sqrt{2} \right)=\sqrt{5}\centerdot \sqrt{5}+\sqrt{5}\centerdot \sqrt{2}-\sqrt{2}\centerdot \sqrt{5}-\sqrt{2}\centerdot \sqrt{2} \\
+
-
& =\sqrt{5}\centerdot \sqrt{5}-\sqrt{2}\centerdot \sqrt{2} \\
+
-
\end{align}</math>
+
 +
{{Displayed math||<math>\sqrt{5}\cdot\sqrt{5} - \sqrt{2}\cdot\sqrt{2} = 5-2 = 3\,\textrm{.}</math>}}
-
Because
 
-
<math>\sqrt{5}</math>
 
-
and
 
-
<math>\sqrt{2}</math>
 
-
are defined as those numbers which, when multiplied with themselves give
 
-
<math>\text{5}</math>
 
-
and
 
-
<math>2</math> respectively,
 
-
 
+
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>.
-
<math>\sqrt{5}\centerdot \sqrt{5}-\sqrt{2}\centerdot \sqrt{2}=5-2=3</math>
+
-
 
+
-
 
+
-
NOTE: The expansion of
+
-
<math>\left( \sqrt{5}-\sqrt{2} \right)\left( \sqrt{5}3\sqrt{2} \right)</math>
+
-
can also be done directly with the conjugate rule
+
-
<math>\left( a-b \right)(a+b)=a^{\text{2}}-b^{\text{2}}</math>
+
-
using
+
-
<math>a=\sqrt{5}</math>
+
-
and
+
-
<math>b=\sqrt{2}</math>.
+

Version vom 08:43, 30. Sep. 2008

First expand the expression

Vorlage:Displayed math

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

Vorlage:Displayed math


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“