Solution 3.1:6a - Revision history

Tek at 11:41, 30 September 2008

Tek — Tue, 30 Sep 2008 11:41:44 GMT

←Older revision		Revision as of 11:41, 30 September 2008
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-	We use the standard method and augment the fraction with the conjugate of the denominator	+	We use the standard method and augment the fraction with the conjugate of the denominator <math>\sqrt{5}+2</math>. Then the formula for the difference of two squares gives
-	<math>\sqrt{5}+2</math>. Then the ~~conjugate rule~~ gives	+

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\frac{\sqrt{2}+3}{\sqrt{5}-2}
-	& \frac{\sqrt{2}+3}{\sqrt{5}-2}=\frac{\sqrt{2}+3}{\sqrt{5}-2}\~~centerdot~~ \frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{~~\left~~( \sqrt{2}+3 ~~\right~~)~~\left~~( \sqrt{5}+2 ~~\right~~)}{~~\left~~( \sqrt{5} ~~\right~~)^{2}-2^{2}} \\	+	&= \frac{\sqrt{2}+3}{\sqrt{5}-2}\cdot\frac{\sqrt{5}+2}{\sqrt{5}+2}\\[5pt]
-	& =\frac{\sqrt{2}\~~centerdot~~ \sqrt{5}+\sqrt{2}\~~centerdot~~ 2+3\~~centerdot~~ \sqrt{5}+3\~~centerdot~~ 2}{5-4}=\sqrt{2\~~centerdot~~ 5}+2\sqrt{2}+3\sqrt{5}+6 \\	+	&= \frac{(\sqrt{2}+3)(\sqrt{5}+2)}{(\sqrt{5})^{2}-2^{2}}\\[5pt]
-	& =6+2\sqrt{2}+3\sqrt{5}+10 \\	+	&= \frac{\sqrt{2}\cdot\sqrt{5}+\sqrt{2}\cdot 2+3\cdot \sqrt{5}+3\cdot 2}{5-4}\\[5pt]
-	\end{align}</math>	+	&= \sqrt{2\cdot 5} + 2\sqrt{2} + 3\sqrt{5} + 6\\[5pt]
		+	&= 6+2\sqrt{2}+3\sqrt{5}+\sqrt{10}\,\textrm{.}
		+	\end{align}</math>}}

Ian — Tue, 23 Sep 2008 08:50:01 GMT

←Older revision		Revision as of 08:50, 23 September 2008
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-	{~~{NAVCONTENT_START~~}}	+	We use the standard method and augment the fraction with the conjugate of the denominator
-	<~~center~~> ~~[[Image:3_1_6a~~.~~gif]]~~ <~~/center~~>	+	<math>\sqrt{5}+2</math>. Then the conjugate rule gives
-	{{~~NAVCONTENT_STOP~~}}	+
		+
		+	<math>\begin{align}
		+	& \frac{\sqrt{2}+3}{\sqrt{5}-2}=\frac{\sqrt{2}+3}{\sqrt{5}-2}\centerdot \frac{\sqrt{5}+2}{\sqrt{5}+2}=\frac{\left( \sqrt{2}+3 \right)\left( \sqrt{5}+2 \right)}{\left( \sqrt{5} \right)^{2}-2^{2}} \\
		+	& =\frac{\sqrt{2}\centerdot \sqrt{5}+\sqrt{2}\centerdot 2+3\centerdot \sqrt{5}+3\centerdot 2}{5-4}=\sqrt{2\centerdot 5}+2\sqrt{2}+3\sqrt{5}+6 \\
		+	& =6+2\sqrt{2}+3\sqrt{5}+10 \\
		+	\end{align}</math>

Tekbot — Tue, 09 Sep 2008 09:13:43 GMT

←Older revision

Revision as of 09:13, 9 September 2008

Tekbot — Thu, 21 Aug 2008 06:53:59 GMT

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Oskar — Tue, 01 Apr 2008 14:08:54 GMT

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