Solution 3.1:3a

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
m (Lösning 3.1:3a moved to Solution 3.1:3a: Robot: moved page)
Line 1: Line 1:
-
{{NAVCONTENT_START}}
+
First expand the expression
-
<center> [[Image:3_1_3a.gif]] </center>
+
 
-
{{NAVCONTENT_STOP}}
+
 
 +
<math>\begin{align}
 +
& \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>
 +
 
 +
 
 +
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,
 +
 
 +
 
 +
<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>.

Revision as of 12:33, 22 September 2008

First expand the expression


\displaystyle \begin{align} & \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}


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


\displaystyle \sqrt{5}\centerdot \sqrt{5}-\sqrt{2}\centerdot \sqrt{2}=5-2=3


NOTE: The expansion of \displaystyle \left( \sqrt{5}-\sqrt{2} \right)\left( \sqrt{5}3\sqrt{2} \right) can also be done directly with the conjugate rule \displaystyle \left( a-b \right)(a+b)=a^{\text{2}}-b^{\text{2}} using \displaystyle a=\sqrt{5} and \displaystyle b=\sqrt{2}.

Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_3.1:3a"