Solution 3.1:6b - Revision history

Tek at 11:53, 30 September 2008

2008-09-30T11:53:16Z

←Older revision		Revision as of 11:53, 30 September 2008
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-	The root sign in the denominator lies in a quadratic term and we therefore expand first the quadratic	+	The root sign in the denominator lies in a quadratic term and we therefore expand first the quadratic
-	~~<math>\left( \sqrt{3}-2 \right)^{2}</math>~~	+
-	~~using the square rule~~	+

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	(\sqrt{3}-2)^2
-	~~& \left~~( \sqrt{3}-2 ~~\right~~)^{2}=~~\left~~( \sqrt{3} \~~right~~)^{2}-2\~~centerdot~~ \sqrt{3}\~~centerdot~~ 2+2^{2} \\	+	&= (\sqrt{3}\,)^{2} - 2\cdot\sqrt{3}\cdot 2 + 2^{2}\\[5pt]
-	& =3-4\sqrt{3}+4=7-4\sqrt{3} \\	+	&= 3-4\sqrt{3}+4\\[5pt]
-	\end{align}</math>	+	&= 7-4\sqrt{3}\,\textrm{.}
		+	\end{align}</math>}}

	Thus,		Thus,

		+	{{Displayed math\|\|<math>\frac{1}{(\sqrt{3}-2)^{2}-2} = \frac{1}{7-4\sqrt{3}-2} = \frac{1}{5-4\sqrt{3}}</math>}}

-	~~<math>\begin{align}~~	+	and in the expression we can get rid of the root sign from the denominator by multiplying the top and bottom of the equation by the conjugate <math>5+4\sqrt{3}</math>,
-	~~& \left( \sqrt{3}-2 \right)^{2}=\left( \sqrt{3} \right)^{2}-2\centerdot \sqrt{3}\centerdot 2+2^{2} \\~~	+
-	~~& =3-4\sqrt{3}+4=7-4\sqrt{3} \\~~	+
-	~~& \frac{1}{\left( \sqrt{3}-2 \right)^{2}-2}=\frac{1}{7-4\sqrt{3}-2}=\frac{1}{5-4\sqrt{3}} \\~~	+
-	~~\end{align}</math>~~	+
-		+
-	and in the expression we can get rid of the root sign from the denominator by multiplying the top and bottom of the equation by the conjugate	+
-	<math>5+4\sqrt{3}</math>,	+
-		+

-	<math>\begin{align}	+	{{Displayed math\|\|<math>\begin{align}
-	& \frac{1}{5-4\sqrt{3}}=\frac{1}{5-4\sqrt{3}}\~~centerdot~~ \frac{5+4\sqrt{3}}{5+4\sqrt{3}}=\frac{5+4\sqrt{3}}{5^{2}-~~\left~~( 4\sqrt{3} ~~\right~~)^{2}} \\	+	\frac{1}{5-4\sqrt{3}}
-	& =\frac{5+4\sqrt{3}}{5^{2}-4^{2}~~\left~~( \sqrt{3} ~~\right~~)^{2}}=\frac{5+4\sqrt{3}}{25-16\~~centerdot~~ 3} \\	+	&= \frac{1}{5-4\sqrt{3}}\cdot \frac{5+4\sqrt{3}}{5+4\sqrt{3}}\\[5pt]
-	& =\frac{5+4\sqrt{3}}{-23}=-\frac{5+4\sqrt{3}}{23} \\	+	&= \frac{5+4\sqrt{3}}{5^{2}-(4\sqrt{3})^{2}}\\[5pt]
-	\end{align}</math>	+	&= \frac{5+4\sqrt{3}}{5^{2}-4^{2}(\sqrt{3})^{2}}\\[5pt]
		+	&= \frac{5+4\sqrt{3}}{25-16\cdot 3}\\[5pt]
		+	&= \frac{5+4\sqrt{3}}{-23}\\[5pt]
		+	&= -\frac{5+4\sqrt{3}}{23}\,\textrm{.}
		+	\end{align}</math>}}

Ian at 09:00, 23 September 2008

2008-09-23T09:00:46Z

←Older revision		Revision as of 09:00, 23 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	The root sign in the denominator lies in a quadratic term and we therefore expand first the quadratic
-	<~~center~~> ~~[[Image:3_1_6b.gif]]~~ </~~center~~>	+	<math>\left( \sqrt{3}-2 \right)^{2}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	using the square rule
		+
		+
		+	<math>\begin{align}
		+	& \left( \sqrt{3}-2 \right)^{2}=\left( \sqrt{3} \right)^{2}-2\centerdot \sqrt{3}\centerdot 2+2^{2} \\
		+	& =3-4\sqrt{3}+4=7-4\sqrt{3} \\
		+	\end{align}</math>
		+
		+	Thus,
		+
		+
		+	<math>\begin{align}
		+	& \left( \sqrt{3}-2 \right)^{2}=\left( \sqrt{3} \right)^{2}-2\centerdot \sqrt{3}\centerdot 2+2^{2} \\
		+	& =3-4\sqrt{3}+4=7-4\sqrt{3} \\
		+	& \frac{1}{\left( \sqrt{3}-2 \right)^{2}-2}=\frac{1}{7-4\sqrt{3}-2}=\frac{1}{5-4\sqrt{3}} \\
		+	\end{align}</math>
		+
		+	and in the expression we can get rid of the root sign from the denominator by multiplying the top and bottom of the equation by the conjugate
		+	<math>5+4\sqrt{3}</math>,
		+
		+
		+	<math>\begin{align}
		+	& \frac{1}{5-4\sqrt{3}}=\frac{1}{5-4\sqrt{3}}\centerdot \frac{5+4\sqrt{3}}{5+4\sqrt{3}}=\frac{5+4\sqrt{3}}{5^{2}-\left( 4\sqrt{3} \right)^{2}} \\
		+	& =\frac{5+4\sqrt{3}}{5^{2}-4^{2}\left( \sqrt{3} \right)^{2}}=\frac{5+4\sqrt{3}}{25-16\centerdot 3} \\
		+	& =\frac{5+4\sqrt{3}}{-23}=-\frac{5+4\sqrt{3}}{23} \\
		+	\end{align}</math>

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

2008-09-09T09:14:03Z

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

←Older revision

Revision as of 09:14, 9 September 2008

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

2008-08-21T06:54:09Z

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

←Older revision		Revision as of 06:54, 21 August 2008
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2008-04-01T14:09:27Z

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