Solution 3.1:7b

From Förberedande kurs i matematik 1

(Difference between revisions)
Jump to: navigation, search
Current revision (14:15, 30 September 2008) (edit) (undo)
m
 
Line 1: Line 1:
We multiply the top and bottom of the fraction by the conjugate of the denominator,
We multiply the top and bottom of the fraction by the conjugate of the denominator,
-
<math>\sqrt{7}+\sqrt{5}</math>
+
<math>\sqrt{7}+\sqrt{5}</math>, and see what it leads to,
-
, and see what it leads to:
+
-
 
+
{{Displayed math||<math>\begin{align}
-
<math>\begin{align}
+
\frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}}
-
& \frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}}\centerdot \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}=\frac{\left( 5\sqrt{7}-7\sqrt{5} \right)\centerdot \left( \sqrt{7}+\sqrt{5} \right)}{\left( \sqrt{7} \right)^{2}-\left( \sqrt{5} \right)^{2}} \\
+
&= \frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}}\cdot \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}\\[10pt]
-
& =\frac{5\sqrt{7}\centerdot \sqrt{7}+5\sqrt{5}\centerdot \sqrt{7}-7\sqrt{5}\centerdot \sqrt{7}-7\sqrt{5}\centerdot \sqrt{5}}{7-5} \\
+
&= \frac{(5\sqrt{7}-7\sqrt{5})(\sqrt{7}+\sqrt{5})}{(\sqrt{7})^{2}-(\sqrt{5})^{2}}\\[10pt]
-
& =\frac{5\left( \sqrt{7} \right)^{2}+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7\left( \sqrt{5} \right)^{2}}{2} \\
+
&= \frac{5\sqrt{7}\cdot\sqrt{7}+5\sqrt{5}\cdot\sqrt{7}-7\sqrt{5}\cdot\sqrt{7}-7\sqrt{5}\cdot\sqrt{5}}{7-5}\\[10pt]
-
& =\frac{5\centerdot 7+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7\centerdot 5}{2} \\
+
&= \frac{5(\sqrt{7})^{2}+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7(\sqrt{5})^{2}}{2}\\[10pt]
-
& =\frac{5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}}{2}=\frac{\left( 5-7 \right)\sqrt{5}\sqrt{7}}{2}=\frac{-2\sqrt{5}\sqrt{7}}{2} \\
+
&= \frac{5\cdot 7+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7\cdot 5}{2}\\[10pt]
-
& =-\sqrt{35} \\
+
&= \frac{5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}}{2}\\[10pt]
-
\end{align}</math>
+
&= \frac{(5-7)\sqrt{5}\sqrt{7}}{2}\\[10pt]
 +
&= \frac{-2\sqrt{5\cdot 7}}{2}\\[10pt]
 +
&= -\sqrt{35}\,\textrm{.}
 +
\end{align}</math>}}

Current revision

We multiply the top and bottom of the fraction by the conjugate of the denominator, \displaystyle \sqrt{7}+\sqrt{5}, and see what it leads to,

\displaystyle \begin{align}

\frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}} &= \frac{5\sqrt{7}-7\sqrt{5}}{\sqrt{7}-\sqrt{5}}\cdot \frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}}\\[10pt] &= \frac{(5\sqrt{7}-7\sqrt{5})(\sqrt{7}+\sqrt{5})}{(\sqrt{7})^{2}-(\sqrt{5})^{2}}\\[10pt] &= \frac{5\sqrt{7}\cdot\sqrt{7}+5\sqrt{5}\cdot\sqrt{7}-7\sqrt{5}\cdot\sqrt{7}-7\sqrt{5}\cdot\sqrt{5}}{7-5}\\[10pt] &= \frac{5(\sqrt{7})^{2}+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7(\sqrt{5})^{2}}{2}\\[10pt] &= \frac{5\cdot 7+5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}-7\cdot 5}{2}\\[10pt] &= \frac{5\sqrt{5}\sqrt{7}-7\sqrt{5}\sqrt{7}}{2}\\[10pt] &= \frac{(5-7)\sqrt{5}\sqrt{7}}{2}\\[10pt] &= \frac{-2\sqrt{5\cdot 7}}{2}\\[10pt] &= -\sqrt{35}\,\textrm{.} \end{align}

Retrieved from "http://wiki.sommarmatte.se/wikis/2008/bridgecourse1-ImperialCollege/index.php/Solution_3.1:7b"