Solution 3.1:6d - Revision history

Tek at 13:57, 30 September 2008

Tek — Tue, 30 Sep 2008 13:57:13 GMT

←Older revision		Revision as of 13:57, 30 September 2008
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-	The problem with this expression is that the denominator contains three roots and so there is no simple way to get rid of all root signs at once; rather, we need to work step by step. In the first step, we view the numerator as	+	The problem with this expression is that the denominator contains three roots and so there is no simple way to get rid of all root signs at once; rather, we need to work step by step. In the first step, we view the numerator as <math>(\sqrt{2}+\sqrt{3})+\sqrt{6}</math> and multiply the top and bottom of the fraction by the conjugate-like expression <math>(\sqrt{2}+\sqrt{3})-\sqrt{6}\,</math>. Then, at least <math>\sqrt{6}</math> will be squared away using the formula for the difference of two squares
-	<math>~~\left~~( \sqrt{2}+\sqrt{3} ~~\right~~)+\sqrt{6}</math>	+
-	and multiply the top and bottom of the fraction by the conjugate-like expression	+
-	<math>~~\left~~( \sqrt{2}+\sqrt{3} ~~\right~~)-\sqrt{6}</math>	+
-	Then, at least	+
-	<math>\sqrt{6}</math>	+
-	will be squared away using the ~~conjugate rule~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\frac{1}{(\sqrt{2}+\sqrt{3})+\sqrt{6}}\cdot \frac{(\sqrt{2}+\sqrt{3})-\sqrt{6}}{(\sqrt{2}+\sqrt{3})-\sqrt{6}}
		+	&= \frac{(\sqrt{2}+\sqrt{3})-\sqrt{6}}{(\sqrt{2}+\sqrt{3})^{2}-(\sqrt{6})^{2}}\\[10pt]
		+	&= \frac{(\sqrt{2}+\sqrt{3})-\sqrt{6}}{(\sqrt{2}+\sqrt{3})^{2}-6}\,\textrm{.}
		+	\end{align}</math>}}

-	<math>~~\begin{align}~~	+	We expand the remaining quadratic, <math>(\sqrt{2}+\sqrt{3})^{2}</math>, using the formula for the difference of two squares,
-	~~& \frac{1}{\left~~( \sqrt{2}+\sqrt{3} ~~\right)+\sqrt{6}}\centerdot \frac{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}=\frac{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}{\left( \sqrt{2}+\sqrt{3} \right~~)^{2~~}-\left( \sqrt{6} \right)^{2}} \\~~	+
-	~~& =\frac{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}{\left( \sqrt{2}+\sqrt{3} \right)^{2}-6} \\~~	+
-	~~\end{align~~}</math>	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\frac{(\sqrt{2}+\sqrt{3})-\sqrt{6}}{(\sqrt{2}+\sqrt{3})^{2}-6}
		+	&= \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{(\sqrt{2})^{2}+2\sqrt{2}\sqrt{3}+(\sqrt{3})^{2}-6}\\[10pt]
		+	&= \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2+2\sqrt{2\cdot 3}+3-6}\\[10pt]
		+	&= \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1}\,\textrm{.}
		+	\end{align}</math>}}

-	~~We expand~~ the ~~remaining quadratic,~~	+	This expression has only a root sign in the denominator and we can then complete the calculation by multiplying top and bottom by the conjugate <math>2\sqrt{6}+1</math>,
-	<math>~~\left(~~ \sqrt{2}+~~\sqrt{3} \right)^{2}~~</math>, ~~using the squaring rule~~	+

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	\frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1}
-	& \frac{~~\left(~~ \sqrt{2}+\sqrt{3} ~~\right)~~-\sqrt{6}}~~{\left( \sqrt~~{2}+\sqrt{~~3} \right)^{2}-~~6~~}=\frac{\sqrt{2}+\sqrt{3~~}-~~\sqrt{6~~}~~}{\left( \sqrt{2} \right)^{2}+2\sqrt{2}\sqrt{3}+\left( \sqrt{3} \right)^{2}-6} \\~~	+	&= \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1}\cdot\frac{2\sqrt{6}+1}{2\sqrt{6}+1}\\[10pt]
-	& ~~=\frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2+2\sqrt{2\centerdot 3}+3-6}~~=\frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1} \\	+	&= \frac{(\sqrt{2}+\sqrt{3}-\sqrt{6})(2\sqrt{6}+1)}{(2\sqrt{6})^{2}-1^{2}}\\[10pt]
-	~~\end~~{~~align}</math>~~	+	&= \frac{\sqrt{2}\cdot 2\sqrt{6}+\sqrt{2}\cdot 1+\sqrt{3}\cdot 2\sqrt{6}+\sqrt{3}\cdot 1-\sqrt{6}\cdot 2\sqrt{6}-\sqrt{6}\cdot 1}{2^{2}(\sqrt{6})^{2}-1^{2}}\\[10pt]
-		+	&= \frac{\sqrt{2}\cdot 2\sqrt{2\cdot 3}+\sqrt{2}+\sqrt{3}\cdot 2\sqrt{2\cdot 3}+\sqrt{3}-2(\sqrt{6})^{2}-\sqrt{6}}{4\cdot 6-1^{2}}\\[10pt]
-		+	&= \frac{2(\sqrt{2})^{2}\sqrt{3}+\sqrt{2}+2(\sqrt{3})^{2}\sqrt{2}+\sqrt{3}-2\cdot 6-\sqrt{6}}{24-1}\\[10pt]
-	~~This expression has only a root sign in the denominator and we can then complete the calculation by multiplying top and bottom by the conjugate~~	+	&= \frac{2\cdot 2\cdot \sqrt{3}+\sqrt{2}+2\cdot 3\cdot \sqrt{2}+\sqrt{3}-12-\sqrt{6}}{23}\\[10pt]
-	~~<math>~~2\sqrt{6}+1~~</math>,~~	+	&= \frac{(1+2\cdot 3)\sqrt{2}+(2\cdot 2+1)\sqrt{3}-12-\sqrt{6}}{23}\\[10pt]
-		+	&= \frac{7\sqrt{2}+5\sqrt{3}-\sqrt{6}-12}{23}\,\textrm{.}
-	~~<math>\begin{align~~}	+	\end{align}</math>}}
-	~~& \frac{\sqrt~~{2~~}+\sqrt{3}-~~\sqrt{6}~~}{2\sqrt{6}-~~1}\~~centerdot~~ \~~frac{2\sqrt{6}+1}{2\sqrt{6}+1}~~=\frac{~~\left~~( \sqrt{2}+\sqrt{3}-\sqrt{6} ~~\right~~)~~\left~~( 2\sqrt{6}+1 ~~\right~~)}{~~\left~~( 2\sqrt{6} ~~\right~~)^{2}-1^{2}} \\	+
-	& =\frac{\sqrt{2}\~~centerdot~~ 2\sqrt{6}+\sqrt{2}\~~centerdot~~ 1+\sqrt{3}\~~centerdot~~ 2\sqrt{6}+\sqrt{3}\~~centerdot~~ 1-\sqrt{6}\~~centerdot~~ 2\sqrt{6}-\sqrt{6}\~~centerdot~~ 1}{2^{2}~~\left~~( \sqrt{6} ~~\right~~)^{2}-1^{2}} \\	+
-	& =\frac{\sqrt{2}\~~centerdot~~ 2\sqrt{2\~~centerdot~~ 3}+\sqrt{2}+\sqrt{3}\~~centerdot~~ 2\sqrt{2\~~centerdot~~ 3}+\sqrt{3}-2~~\left~~( \sqrt{6} ~~\right~~)^{2}-\sqrt{6}}{4\~~centerdot~~ 6-1^{2}} \\	+
-	& =\frac{2~~\left~~( \sqrt{2} ~~\right~~)^{2}\sqrt{3}+\sqrt{2}+2~~\left~~( \sqrt{3} ~~\right~~)^{2}\sqrt{2}+\sqrt{3}-2\~~centerdot~~ 6-\sqrt{6}}{24-1} \\	+
-	& =\frac{2\~~centerdot~~ 2\~~centerdot~~ \sqrt{3}+\sqrt{2}+2\~~centerdot~~ 3\~~centerdot~~ \sqrt{2}+\sqrt{3}-12-\sqrt{6}}{23} \\	+
-	& =\frac{~~\left~~( 1+2\~~centerdot~~ 3 ~~\right~~)\sqrt{2}+~~\left~~( 2\~~centerdot~~ 2+1 ~~\right~~)\sqrt{3}-12-\sqrt{6}}{23} \\	+
-	& =\frac{7\sqrt{2}+5\sqrt{3}-\sqrt{6}-12}{23} \\	+
-	\end{align}</math>	+

Ian at 09:50, 23 September 2008

Ian — Tue, 23 Sep 2008 09:50:09 GMT

←Older revision		Revision as of 09:50, 23 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	The problem with this expression is that the denominator contains three roots and so there is no simple way to get rid of all root signs at once; rather, we need to work step by step. In the first step, we view the numerator as
-	<~~center~~> ~~[[Image:3_1_6d~~-1(2)~~.gif]]~~ </~~center~~>	+	<math>\left( \sqrt{2}+\sqrt{3} \right)+\sqrt{6}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and multiply the top and bottom of the fraction by the conjugate-like expression
-	{{~~NAVCONTENT_START~~}}	+	<math>\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}</math>
-	<~~center~~> ~~[[Image:3_1_6d~~-2(2)~~.gif]]~~ </~~center~~>	+	Then, at least
-	{{~~NAVCONTENT_STOP~~}}	+	<math>\sqrt{6}</math>
		+	will be squared away using the conjugate rule
		+
		+
		+	<math>\begin{align}
		+	& \frac{1}{\left( \sqrt{2}+\sqrt{3} \right)+\sqrt{6}}\centerdot \frac{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}=\frac{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}{\left( \sqrt{2}+\sqrt{3} \right)^{2}-\left( \sqrt{6} \right)^{2}} \\
		+	& =\frac{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}{\left( \sqrt{2}+\sqrt{3} \right)^{2}-6} \\
		+	\end{align}</math>
		+
		+
		+	We expand the remaining quadratic,
		+	<math>\left( \sqrt{2}+\sqrt{3} \right)^{2}</math>, using the squaring rule
		+
		+
		+	<math>\begin{align}
		+	& \frac{\left( \sqrt{2}+\sqrt{3} \right)-\sqrt{6}}{\left( \sqrt{2}+\sqrt{3} \right)^{2}-6}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{\left( \sqrt{2} \right)^{2}+2\sqrt{2}\sqrt{3}+\left( \sqrt{3} \right)^{2}-6} \\
		+	& =\frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2+2\sqrt{2\centerdot 3}+3-6}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1} \\
		+	\end{align}</math>
		+
		+
		+	This expression has only a root sign in the denominator and we can then complete the calculation by multiplying top and bottom by the conjugate
		+	<math>2\sqrt{6}+1</math>,
		+
		+	<math>\begin{align}
		+	& \frac{\sqrt{2}+\sqrt{3}-\sqrt{6}}{2\sqrt{6}-1}\centerdot \frac{2\sqrt{6}+1}{2\sqrt{6}+1}=\frac{\left( \sqrt{2}+\sqrt{3}-\sqrt{6} \right)\left( 2\sqrt{6}+1 \right)}{\left( 2\sqrt{6} \right)^{2}-1^{2}} \\
		+	& =\frac{\sqrt{2}\centerdot 2\sqrt{6}+\sqrt{2}\centerdot 1+\sqrt{3}\centerdot 2\sqrt{6}+\sqrt{3}\centerdot 1-\sqrt{6}\centerdot 2\sqrt{6}-\sqrt{6}\centerdot 1}{2^{2}\left( \sqrt{6} \right)^{2}-1^{2}} \\
		+	& =\frac{\sqrt{2}\centerdot 2\sqrt{2\centerdot 3}+\sqrt{2}+\sqrt{3}\centerdot 2\sqrt{2\centerdot 3}+\sqrt{3}-2\left( \sqrt{6} \right)^{2}-\sqrt{6}}{4\centerdot 6-1^{2}} \\
		+	& =\frac{2\left( \sqrt{2} \right)^{2}\sqrt{3}+\sqrt{2}+2\left( \sqrt{3} \right)^{2}\sqrt{2}+\sqrt{3}-2\centerdot 6-\sqrt{6}}{24-1} \\
		+	& =\frac{2\centerdot 2\centerdot \sqrt{3}+\sqrt{2}+2\centerdot 3\centerdot \sqrt{2}+\sqrt{3}-12-\sqrt{6}}{23} \\
		+	& =\frac{\left( 1+2\centerdot 3 \right)\sqrt{2}+\left( 2\centerdot 2+1 \right)\sqrt{3}-12-\sqrt{6}}{23} \\
		+	& =\frac{7\sqrt{2}+5\sqrt{3}-\sqrt{6}-12}{23} \\
		+	\end{align}</math>

Tekbot: Lösning 3.1:6d moved to Solution 3.1:6d: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:14:43 GMT

Lösning 3.1:6d moved to Solution 3.1:6d: Robot: moved page

←Older revision

Revision as of 09:14, 9 September 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

Tekbot — Thu, 21 Aug 2008 06:54:29 GMT

Robot: Automated text replacement (-[[Bild: +[[Image:)

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Oskar — Tue, 01 Apr 2008 14:11:25 GMT

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