Solution 3.1:5c - Revision history

Tek at 11:29, 30 September 2008

2008-09-30T11:29:48Z

←Older revision		Revision as of 11:29, 30 September 2008
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-	The trick is to use the ~~conjugate rule~~	+	The trick is to use the formula for the difference of two squares
-	<math>~~\left~~( a-b ~~\right~~)(a+b)=a^~~{\text~~{2}}-b^~~{\text~~{2}}</math>	+	<math>(a-b)(a+b) = a^{2}-b^{2}</math> and multiply the top and bottom of the fraction by <math>3-\sqrt{7}</math> (note the minus sign), since then the new denominator will be <math>(3+\sqrt{7})(3-\sqrt{7}) = 3^{2} - (\sqrt{7})^{2} = 9-7 = 2</math> (the formula with <math>a=3</math> and <math>b=\sqrt{7}\,</math>), i.e. the root sign is squared away.
-	and multiply the top and bottom of the fraction by	+
-	<math>3-\sqrt{7}</math>	+
-	(note the minus sign), since then the new denominator will be	+
-	<math>~~\left~~( 3+\sqrt{7} ~~\right~~)~~\left~~( 3-\sqrt{7} ~~\right~~)=3^{2}-~~\left~~( \sqrt{7} ~~\right~~)^{2}=9-7=2</math>	+
-	(~~conjugate rule~~ with	+
-	<math>a=~~\text{~~3 }</math>	+
-	and	+
-	<math>b=\sqrt~~{\text~~{7}}</math>	+
-	), i.e. the root sign is squared away.	+

	The whole calculation is		The whole calculation is

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

Ian at 14:43, 22 September 2008

2008-09-22T14:43:41Z

←Older revision		Revision as of 14:43, 22 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	The trick is to use the conjugate rule
-	<~~center~~> ~~[[Image:3_1_5c.gif]]~~ </~~center~~>	+	<math>\left( a-b \right)(a+b)=a^{\text{2}}-b^{\text{2}}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and multiply the top and bottom of the fraction by
		+	<math>3-\sqrt{7}</math>
		+	(note the minus sign), since then the new denominator will be
		+	<math>\left( 3+\sqrt{7} \right)\left( 3-\sqrt{7} \right)=3^{2}-\left( \sqrt{7} \right)^{2}=9-7=2</math>
		+	(conjugate rule with
		+	<math>a=\text{3 }</math>
		+	and
		+	<math>b=\sqrt{\text{7}}</math>
		+	), i.e. the root sign is squared away.
		+
		+	The whole calculation is
		+
		+
		+	<math>\begin{align}
		+	& \frac{2}{3+\sqrt{7}}=\frac{2}{3+\sqrt{7}}\centerdot \frac{3-\sqrt{7}}{3-\sqrt{7}}=\frac{2\left( 3-\sqrt{7} \right)}{3^{2}-\left( \sqrt{7} \right)^{2}} \\
		+	& =\frac{2\centerdot 3-2\sqrt{7}}{2}=3-\sqrt{7} \\
		+	\end{align}</math>

Tekbot: Lösning 3.1:5c moved to Solution 3.1:5c: Robot: moved page

2008-09-09T09:13:03Z

Lösning 3.1:5c moved to Solution 3.1:5c: Robot: moved page

←Older revision

Revision as of 09:13, 9 September 2008

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

2008-08-21T06:53:39Z

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

←Older revision		Revision as of 06:53, 21 August 2008
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Oskar: Ny sida: {{NAVCONTENT_START}} Bild:3_1_5c.gif {{NAVCONTENT_STOP}}

2008-04-01T14:07:08Z

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