Lösung 2.3:2f - Versionsgeschichte

Ombtut3 um 12:14, 9. Sep. 2009

Ombtut3 — Wed, 09 Sep 2009 12:14:17 GMT

← Nächstältere Version		Version vom 12:14, 9. Sep. 2009
Zeile 17:		Zeile 17:

	:*<math>x-\tfrac{5}{3} = -\sqrt{\tfrac{1}{9}} = -\tfrac{1}{3}\,,\quad</math> also <math>x = \tfrac{5}{3} - \tfrac{1}{3} = \tfrac{4}{3}\,\textrm{.}</math>		:*<math>x-\tfrac{5}{3} = -\sqrt{\tfrac{1}{9}} = -\tfrac{1}{3}\,,\quad</math> also <math>x = \tfrac{5}{3} - \tfrac{1}{3} = \tfrac{4}{3}\,\textrm{.}</math>
		+

	Kontrolle:		Kontrolle:
Zeile 23:		Zeile 24:

	:*''x'' = 2: <math>\ \text{Linke Seite} = 3\cdot 2^{2} - 10\cdot 2 + 8 = 12 - 20 + 8 = 0 = \text{Rechte Seite.}</math>		:*''x'' = 2: <math>\ \text{Linke Seite} = 3\cdot 2^{2} - 10\cdot 2 + 8 = 12 - 20 + 8 = 0 = \text{Rechte Seite.}</math>
		+
		+	Alternativer Lösungsweg: [[2.3:2f_alt6\|p-q_Formel]]

Markus.bez um 12:31, 9. Jun. 2009

Markus.bez — Tue, 09 Jun 2009 12:31:53 GMT

← Nächstältere Version		Version vom 12:31, 9. Jun. 2009
Zeile 10:		Zeile 10:
	Wir bekommen so die Gleichung		Wir bekommen so die Gleichung

-	{{Abgesetzte Formel\|\|<math>\left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}\,\textrm{,}</math>}}	+	{{Abgesetzte Formel\|\|<math>\left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}\,\textrm{.}</math>}}

-	~~Wir bekommen die~~ Wurzeln	+	Als Wurzeln erhalten wir

	:*<math>x-\tfrac{5}{3} = \sqrt{\tfrac{1}{9}} = \tfrac{1}{3}\,,\quad</math> also <math>x = \tfrac{5}{3} + \tfrac{1}{3} = \tfrac{6}{3} = 2\,\textrm{,}</math>		:*<math>x-\tfrac{5}{3} = \sqrt{\tfrac{1}{9}} = \tfrac{1}{3}\,,\quad</math> also <math>x = \tfrac{5}{3} + \tfrac{1}{3} = \tfrac{6}{3} = 2\,\textrm{,}</math>

Markus.bez um 12:31, 9. Jun. 2009

Markus.bez — Tue, 09 Jun 2009 12:31:30 GMT

← Nächstältere Version		Version vom 12:31, 9. Jun. 2009
Zeile 8:		Zeile 8:
	\end{align}</math>}}		\end{align}</math>}}

-	~~und~~ bekommen so die Gleichung	+	Wir bekommen so die Gleichung

	{{Abgesetzte Formel\|\|<math>\left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}\,\textrm{,}</math>}}		{{Abgesetzte Formel\|\|<math>\left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}\,\textrm{,}</math>}}

Martin um 15:04, 16. Mär. 2009

Martin — Mon, 16 Mar 2009 15:04:25 GMT

← Nächstältere Version		Version vom 15:04, 16. Mär. 2009
Zeile 1:		Zeile 1:
-	~~We divide both sides by~~ 3 ~~and complete the square on the left-hand side~~,	+	Wir dividieren zuerst alle Terme mit 3, und führen die quadratische Ergänzung auf der linken Seite aus

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 8:		Zeile 8:
	\end{align}</math>}}		\end{align}</math>}}

-	~~The equation then becomes~~	+	und bekommen so die Gleichung

	{{Abgesetzte Formel\|\|<math>\left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}\,\textrm{,}</math>}}		{{Abgesetzte Formel\|\|<math>\left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}\,\textrm{,}</math>}}

-	~~and taking the square root gives the solutions as~~	+	Wir bekommen die Wurzeln

-	:*<math>x-\tfrac{5}{3} = \sqrt{\tfrac{1}{9}} = \tfrac{1}{3}\,,\quad</math> ~~i.e.~~ <math>x = \tfrac{5}{3} + \tfrac{1}{3} = \tfrac{6}{3} = 2\,\textrm{,}</math>	+	:*<math>x-\tfrac{5}{3} = \sqrt{\tfrac{1}{9}} = \tfrac{1}{3}\,,\quad</math> also <math>x = \tfrac{5}{3} + \tfrac{1}{3} = \tfrac{6}{3} = 2\,\textrm{,}</math>

-	:*<math>x-\tfrac{5}{3} = -\sqrt{\tfrac{1}{9}} = -\tfrac{1}{3}\,,\quad</math> ~~i.e.~~ <math>x = \tfrac{5}{3} - \tfrac{1}{3} = \tfrac{4}{3}\,\textrm{.}</math>	+	:*<math>x-\tfrac{5}{3} = -\sqrt{\tfrac{1}{9}} = -\tfrac{1}{3}\,,\quad</math> also <math>x = \tfrac{5}{3} - \tfrac{1}{3} = \tfrac{4}{3}\,\textrm{.}</math>

-	~~Check~~:	+	Kontrolle:

-	:*''x'' = 4/3: <math>\ \text{~~LHS~~} = 3\cdot\bigl(\tfrac{4}{3}\bigr)^{2} - 10\cdot\tfrac{4}{3} + 8 = 3\cdot\tfrac{16}{9} - \tfrac{40}{3} + \tfrac{8\cdot 3}{3} = 0 = \text{~~RHS~~,}</math>	+	:*''x'' = 4/3: <math>\ \text{Linke Seite} = 3\cdot\bigl(\tfrac{4}{3}\bigr)^{2} - 10\cdot\tfrac{4}{3} + 8 = 3\cdot\tfrac{16}{9} - \tfrac{40}{3} + \tfrac{8\cdot 3}{3} = 0 = \text{Rechte Seite,}</math>

-	:*''x'' = 2: <math>\ \text{~~LHS~~} = 3\cdot 2^{2} - 10\cdot 2 + 8 = 12 - 20 + 8 = 0 = \text{~~RHS~~.}</math>	+	:*''x'' = 2: <math>\ \text{Linke Seite} = 3\cdot 2^{2} - 10\cdot 2 + 8 = 12 - 20 + 8 = 0 = \text{Rechte Seite.}</math>

Tekbot: hat „Solution 2.3:2f“ nach „Lösung 2.3:2f“ verschoben: Robot: moved page

Tekbot — Wed, 22 Oct 2008 14:08:05 GMT

hat „Solution 2.3:2f“ nach „Lösung 2.3:2f“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 14:08, 22. Okt. 2008

Tekbot: Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

Tekbot — Wed, 22 Oct 2008 08:32:28 GMT

Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)

← Nächstältere Version		Version vom 08:32, 22. Okt. 2008
Zeile 1:		Zeile 1:
	We divide both sides by 3 and complete the square on the left-hand side,		We divide both sides by 3 and complete the square on the left-hand side,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	x^{2}-\frac{10}{3}x+\frac{8}{3}		x^{2}-\frac{10}{3}x+\frac{8}{3}
	&= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \Bigl(\frac{5}{3}\Bigr)^{2} + \frac{8}{3}\\[5pt]		&= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \Bigl(\frac{5}{3}\Bigr)^{2} + \frac{8}{3}\\[5pt]
Zeile 10:		Zeile 10:
	The equation then becomes		The equation then becomes

-	{{~~Displayed math~~\|\|<math>\left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}\,\textrm{,}</math>}}	+	{{Abgesetzte Formel\|\|<math>\left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}\,\textrm{,}</math>}}

	and taking the square root gives the solutions as		and taking the square root gives the solutions as

Tek um 08:24, 29. Sep. 2008

Tek — Mon, 29 Sep 2008 08:24:28 GMT

← Nächstältere Version		Version vom 08:24, 29. Sep. 2008
Zeile 1:		Zeile 1:
-	We divide both sides by	+	We divide both sides by 3 and complete the square on the left-hand side,
-	~~<math>~~3~~</math>~~	+
-	and complete the square on the left-hand side:	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& x^{2}-\frac{10}{3}x+\frac{8}{3}=\left( x-\frac{5}{3} \right)^{2}-\left( -\frac{5}{3} \right)^{2}+\frac{8}{3}=\left( x-\frac{5}{3} \right)^{2}-\frac{25}{9}+\frac{24}{9} \\~~	+
-	~~& =\left( x-\frac{5}{3} \right)^{2}-\frac{1}{9} \\~~	+
-	~~\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	x^{2}-\frac{10}{3}x+\frac{8}{3}
		+	&= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \Bigl(\frac{5}{3}\Bigr)^{2} + \frac{8}{3}\\[5pt]
		+	&= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \frac{25}{9} + \frac{24}{9}\\[5pt]
		+	&= \Bigl(x-\frac{5}{3}\Bigr)^{2} - \frac{1}{9}\,\textrm{.}
		+	\end{align}</math>}}

	The equation then becomes		The equation then becomes

		+	{{Displayed math\|\|<math>\left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}\,\textrm{,}</math>}}

-	~~<math>\left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}</math>~~	+	and taking the square root gives the solutions as

-	~~and taking the root gives the solutions as~~	+	:*<math>x-\tfrac{5}{3} = \sqrt{\tfrac{1}{9}} = \tfrac{1}{3}\,,\quad</math> i.e. <math>x = \tfrac{5}{3} + \tfrac{1}{3} = \tfrac{6}{3} = 2\,\textrm{,}</math>
-		+
-		+
-	<math>x-\~~frac~~{5}{3}=\sqrt{\~~frac~~{1}{9}}=\~~frac~~{1}{3}</math>	+
-	i.e.	+
-	<math>x=\~~frac~~{5}{3}+\~~frac~~{1}{3}=\~~frac~~{6}{3}=2~~.</math>~~	+
-		+
-		+
-	~~<math>x-~~\~~frac{5}{3}=-~~\~~sqrt~~{~~\frac{1~~}~~{9}}=-\frac{1}{3}</math>~~	+
-	~~i.e.~~	+
-	~~<math>x=\frac{5}{3}-\frac{1}{3}=\frac{4}{3}.~~</math>	+

		+	:*<math>x-\tfrac{5}{3} = -\sqrt{\tfrac{1}{9}} = -\tfrac{1}{3}\,,\quad</math> i.e. <math>x = \tfrac{5}{3} - \tfrac{1}{3} = \tfrac{4}{3}\,\textrm{.}</math>

	Check:		Check:

		+	:*''x'' = 4/3: <math>\ \text{LHS} = 3\cdot\bigl(\tfrac{4}{3}\bigr)^{2} - 10\cdot\tfrac{4}{3} + 8 = 3\cdot\tfrac{16}{9} - \tfrac{40}{3} + \tfrac{8\cdot 3}{3} = 0 = \text{RHS,}</math>

-	~~<math>~~x=~~\text{4}/\text{3}</math>~~: ~~LHS~~	+	:*''x'' = 2: <math>\ \text{LHS} = 3\cdot 2^{2} - 10\cdot 2 + 8 = 12 - 20 + 8 = 0 = \text{RHS.}</math>
-	<math>=3\~~centerdot \left( \frac{4}{3} \right)^{2}-10\centerdot \frac{4}{3}+8=3\centerdot \frac{16}{9}-\frac{40}{3}+\frac{8\centerdot 3}{3}=\frac{16-40+24}{3}=0=</math>~~	+
-	~~RHS~~	+
-		+
-	~~<math>x=~~\text{2}~~</math>: LHS~~	+
-	~~<math>~~=3\~~centerdot~~ 2^{2}-10\~~centerdot~~ 2+8=12-20+8=0=</math>	+
-	~~RHS~~	+

Ian um 14:31, 20. Sep. 2008

Ian — Sat, 20 Sep 2008 14:31:03 GMT

← Nächstältere Version		Version vom 14:31, 20. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	We divide both sides by
-	<~~center~~> ~~[[Image:2_3_2f~~.~~gif]]~~ </~~center~~>	+	<math>3</math>
-	{{~~NAVCONTENT_STOP~~}}	+	and complete the square on the left-hand side:
		+
		+
		+	<math>\begin{align}
		+	& x^{2}-\frac{10}{3}x+\frac{8}{3}=\left( x-\frac{5}{3} \right)^{2}-\left( -\frac{5}{3} \right)^{2}+\frac{8}{3}=\left( x-\frac{5}{3} \right)^{2}-\frac{25}{9}+\frac{24}{9} \\
		+	& =\left( x-\frac{5}{3} \right)^{2}-\frac{1}{9} \\
		+	\end{align}</math>
		+
		+
		+	The equation then becomes
		+
		+
		+	<math>\left( x-\frac{5}{3} \right)^{2}=\frac{1}{9}</math>
		+
		+	and taking the root gives the solutions as
		+
		+
		+	<math>x-\frac{5}{3}=\sqrt{\frac{1}{9}}=\frac{1}{3}</math>
		+	i.e.
		+	<math>x=\frac{5}{3}+\frac{1}{3}=\frac{6}{3}=2.</math>
		+
		+
		+	<math>x-\frac{5}{3}=-\sqrt{\frac{1}{9}}=-\frac{1}{3}</math>
		+	i.e.
		+	<math>x=\frac{5}{3}-\frac{1}{3}=\frac{4}{3}.</math>
		+
		+
		+	Check:
		+
		+
		+	<math>x=\text{4}/\text{3}</math>: LHS
		+	<math>=3\centerdot \left( \frac{4}{3} \right)^{2}-10\centerdot \frac{4}{3}+8=3\centerdot \frac{16}{9}-\frac{40}{3}+\frac{8\centerdot 3}{3}=\frac{16-40+24}{3}=0=</math>
		+	RHS
		+
		+	<math>x=\text{2}</math>: LHS
		+	<math>=3\centerdot 2^{2}-10\centerdot 2+8=12-20+8=0=</math>
		+	RHS

Tekbot: Lösning 2.3:2f moved to Solution 2.3:2f: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:57:43 GMT

Lösning 2.3:2f moved to Solution 2.3:2f: Robot: moved page

← Nächstältere Version

Version vom 08:57, 9. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 06:45:59 GMT

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

← Nächstältere Version		Version vom 06:45, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_3_2f.gif]] </center>	+	<center> [[Image:2_3_2f.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}