Solution 3.1:5d - Revision history

Tek at 11:36, 30 September 2008

2008-09-30T11:36:45Z

←Older revision		Revision as of 11:36, 30 September 2008
Line 1:		Line 1:
-	We can get rid of both square roots in the denominator if we multiply the top and bottom of the fraction by the conjugate expression	+	We can get rid of both square roots in the denominator if we multiply the top and bottom of the fraction by the conjugate expression <math>\sqrt{17}+\sqrt{13}</math>, and use the difference of two squares
-	<math>\~~left( a-b \right)\left( a~~+b \~~right)=a^{2}-b^~~{2}</math>, and use the ~~conjugate rule~~	+

		+	{{Displayed math\|\|<math>(a-b)(a+b) = a^2-b^2</math>}}

-	with	+	with <math>a=\sqrt{17}</math> and <math>b=\sqrt{13}</math>. Both roots are squared away and we get
-	<math>a=\sqrt{17}</math>	+
-	and	+
-	<math>b=\sqrt{13}</math>. Both roots are squared away and we get	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\frac{1}{\sqrt{17}-\sqrt{13}}
		+	&= \frac{1}{\sqrt{17}-\sqrt{13}}\cdot\frac{\sqrt{17}+\sqrt{13}}{\sqrt{17}+\sqrt{13}}\\[5pt]
		+	&= \frac{\sqrt{17}+\sqrt{13}}{(\sqrt{17})^{2}-(\sqrt{13})^{2}}\\[5pt]
		+	&= \frac{\sqrt{17}+\sqrt{13}}{17-13}\\[5pt]
		+	&= \frac{\sqrt{17}+\sqrt{13}}{4}\,\textrm{.}
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	This expression cannot be simplified any further because neither 17 nor 13 contain any squares as factors.
-	~~& \frac{1}{\sqrt{17}-\sqrt{13}}=\frac{1}{\sqrt{17}-\sqrt{13}}\centerdot \frac{\sqrt{17}+\sqrt{13}}{\sqrt{17}+\sqrt{13}} \\~~	+
-	~~& =\frac{\sqrt{17}+\sqrt{13}}{\left( \sqrt{17} \right)^{2}-\left( \sqrt{13} \right)^{2}}=\frac{\sqrt{17}+\sqrt{13}}{17-13}=\frac{\sqrt{17}+\sqrt{13}}{4}. \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	This expression cannot be simplified any further because neither	+
-	~~<math>\text{~~17~~}</math>~~	+
-	nor	+
-	~~<math>\text{~~13~~}</math>~~	+
-	contain any squares as factors.	+

Ian at 14:51, 22 September 2008

2008-09-22T14:51:45Z

←Older revision		Revision as of 14:51, 22 September 2008
Line 1:		Line 1:
-	{{~~NAVCONTENT_START~~}}	+	We can get rid of both square roots in the denominator if we multiply the top and bottom of the fraction by the conjugate expression
-	<~~center~~> ~~[[Image:3_1_5d.gif]]~~ </~~center~~>	+	<math>\left( a-b \right)\left( a+b \right)=a^{2}-b^{2}</math>, and use the conjugate rule
-	{{~~NAVCONTENT_STOP~~}}	+
		+
		+	with
		+	<math>a=\sqrt{17}</math>
		+	and
		+	<math>b=\sqrt{13}</math>. Both roots are squared away and we get
		+
		+
		+	<math>\begin{align}
		+	& \frac{1}{\sqrt{17}-\sqrt{13}}=\frac{1}{\sqrt{17}-\sqrt{13}}\centerdot \frac{\sqrt{17}+\sqrt{13}}{\sqrt{17}+\sqrt{13}} \\
		+	& =\frac{\sqrt{17}+\sqrt{13}}{\left( \sqrt{17} \right)^{2}-\left( \sqrt{13} \right)^{2}}=\frac{\sqrt{17}+\sqrt{13}}{17-13}=\frac{\sqrt{17}+\sqrt{13}}{4}. \\
		+	\end{align}</math>
		+
		+
		+	This expression cannot be simplified any further because neither
		+	<math>\text{17}</math>
		+	nor
		+	<math>\text{13}</math>
		+	contain any squares as factors.

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

2008-09-09T09:13:23Z

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

←Older revision

Revision as of 09:13, 9 September 2008

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

2008-08-21T06:53:49Z

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

←Older revision		Revision as of 06:53, 21 August 2008
Line 1:		Line 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_1_5d.gif]] </center>	+	<center> [[Image:3_1_5d.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:3_1_5d.gif {{NAVCONTENT_STOP}}

2008-04-01T14:07:42Z

Ny sida: {{NAVCONTENT_START}} <center> Bild:3_1_5d.gif </center> {{NAVCONTENT_STOP}}

New page

{{NAVCONTENT_START}}
<center> [[Bild:3_1_5d.gif]] </center>
{{NAVCONTENT_STOP}}