Lösung 3.2:5 - Versionsgeschichte

Bayerer um 12:33, 23. Jul. 2009

Bayerer — Thu, 23 Jul 2009 12:33:28 GMT

← Nächstältere Version		Version vom 12:33, 23. Jul. 2009
Zeile 34:		Zeile 34:
	Schließlich kontrollieren wir die Wurzeln in der ursprünglichen Gleichung, um Scheinlösungen zu entdecken.		Schließlich kontrollieren wir die Wurzeln in der ursprünglichen Gleichung, um Scheinlösungen zu entdecken.

-	:*''x'' = 1: <math>\ \text{Linke Seite} = \sqrt{3\cdot 1-2} = 1\ </math> ~~and~~ <math>\ \text{Rechte Seite} = 2-1 = 1</math>	+	:*''x'' = 1: <math>\ \text{Linke Seite} = \sqrt{3\cdot 1-2} = 1\ </math> und <math>\ \text{Rechte Seite} = 2-1 = 1</math>

-	:*''x'' = 6: <math>\ \text{Linke Seite} = \sqrt{3\cdot 6-2} = 4\ </math> ~~and~~ <math>\ \text{Rechte Seite} = 2-6 = -4</math>	+	:*''x'' = 6: <math>\ \text{Linke Seite} = \sqrt{3\cdot 6-2} = 4\ </math> und <math>\ \text{Rechte Seite} = 2-6 = -4</math>

	Also hat die Gleichung die Lösung <math>x=1\,</math>.		Also hat die Gleichung die Lösung <math>x=1\,</math>.

Markus.bez um 11:47, 10. Jun. 2009

Markus.bez — Wed, 10 Jun 2009 11:47:54 GMT

← Nächstältere Version		Version vom 11:47, 10. Jun. 2009
Zeile 1:		Zeile 1:
	Nachdem wir beide Seiten quadrieren, erhalten wir die Gleichung		Nachdem wir beide Seiten quadrieren, erhalten wir die Gleichung

-	{{Abgesetzte Formel\|\|<math>3x-2 = (2-x)^2</math>\|(*)}}	+	{{Abgesetzte Formel\|\|<math>3x-2 = (2-x)^2\,.</math>\|(*)}}

-	Wir erweitern die rechte Seite, und erhalten	+	Wir erweitern die rechte Seite und erhalten

	{{Abgesetzte Formel\|\|<math>x^{2}-7x+6=0\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>x^{2}-7x+6=0\,\textrm{.}</math>}}
Zeile 16:		Zeile 16:
	\end{align}</math>}}		\end{align}</math>}}

-	~~Und also~~ kann die Gleichung ~~wir~~	+	Also kann die Gleichung wie

	{{Abgesetzte Formel\|\|<math>\Bigl(x-\frac{7}{2}\Bigr)^2 = \frac{25}{4}</math>}}		{{Abgesetzte Formel\|\|<math>\Bigl(x-\frac{7}{2}\Bigr)^2 = \frac{25}{4}</math>}}

-	geschrieben werden, und hat ~~also~~ die Wurzeln	+	geschrieben werden und hat deshalb die Wurzeln

	:*<math>x = \frac{7}{2} + \sqrt{\frac{25}{4}} = \frac{7}{2} + \frac{5}{2} = \frac{12}{2} = 6\,,</math>		:*<math>x = \frac{7}{2} + \sqrt{\frac{25}{4}} = \frac{7}{2} + \frac{5}{2} = \frac{12}{2} = 6\,,</math>
Zeile 26:		Zeile 26:
	:*<math>x = \frac{7}{2} - \sqrt{\frac{25}{4}} = \frac{7}{2} - \frac{5}{2} = \frac{2}{2} = 1\,\textrm{.}</math>		:*<math>x = \frac{7}{2} - \sqrt{\frac{25}{4}} = \frac{7}{2} - \frac{5}{2} = \frac{2}{2} = 1\,\textrm{.}</math>

-	~~Ersetzten~~ wir <math>x=1</math> und <math>x=6</math> in der quadratischen Gleichung (*), sehen wir dass die Wurzeln richtig sind.	+	Ersetzen wir <math>x=1</math> und <math>x=6</math> in der quadratischen Gleichung (*), sehen wir dass die Wurzeln richtig sind.

	:*''x'' = 1: <math>\ \text{Linke Seite} = 3\cdot 1-2 = 1\ </math> und <math>\ \text{Rechte Seite} = (2-1)^2 = 1</math>		:*''x'' = 1: <math>\ \text{Linke Seite} = 3\cdot 1-2 = 1\ </math> und <math>\ \text{Rechte Seite} = (2-1)^2 = 1</math>

Martin um 13:27, 26. Mär. 2009

Martin — Thu, 26 Mar 2009 13:27:35 GMT

← Nächstältere Version		Version vom 13:27, 26. Mär. 2009
Zeile 1:		Zeile 1:
-	~~After squaring both sides~~, ~~we obtain the equation~~	+	Nachdem wir beide Seiten quadrieren, erhalten wir die Gleichung

	{{Abgesetzte Formel\|\|<math>3x-2 = (2-x)^2</math>\|(*)}}		{{Abgesetzte Formel\|\|<math>3x-2 = (2-x)^2</math>\|(*)}}

-	~~and if we expand the right-hand side and then collect the terms~~, ~~we get~~	+	Wir erweitern die rechte Seite, und erhalten

	{{Abgesetzte Formel\|\|<math>x^{2}-7x+6=0\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>x^{2}-7x+6=0\,\textrm{.}</math>}}

-	~~Completing the square of the left-hand side, we obtain~~	+	Durch quadratische Ergänzung erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 16:		Zeile 16:
	\end{align}</math>}}		\end{align}</math>}}

-	~~which means that the equation can be written as~~	+	Und also kann die Gleichung wir

	{{Abgesetzte Formel\|\|<math>\Bigl(x-\frac{7}{2}\Bigr)^2 = \frac{25}{4}</math>}}		{{Abgesetzte Formel\|\|<math>\Bigl(x-\frac{7}{2}\Bigr)^2 = \frac{25}{4}</math>}}

-	~~and the solutions are therefore~~	+	geschrieben werden, und hat also die Wurzeln

	:*<math>x = \frac{7}{2} + \sqrt{\frac{25}{4}} = \frac{7}{2} + \frac{5}{2} = \frac{12}{2} = 6\,,</math>		:*<math>x = \frac{7}{2} + \sqrt{\frac{25}{4}} = \frac{7}{2} + \frac{5}{2} = \frac{12}{2} = 6\,,</math>
Zeile 26:		Zeile 26:
	:*<math>x = \frac{7}{2} - \sqrt{\frac{25}{4}} = \frac{7}{2} - \frac{5}{2} = \frac{2}{2} = 1\,\textrm{.}</math>		:*<math>x = \frac{7}{2} - \sqrt{\frac{25}{4}} = \frac{7}{2} - \frac{5}{2} = \frac{2}{2} = 1\,\textrm{.}</math>

-	~~Substituting~~ <math>x=1</math> ~~and~~ <math>x=6</math> ~~into the quadratic equation~~ (*) ~~shows that we have solved the equation correctly~~.	+	Ersetzten wir <math>x=1</math> und <math>x=6</math> in der quadratischen Gleichung (*), sehen wir dass die Wurzeln richtig sind.

-	:*''x'' = 1: <math>\ \text{~~LHS~~} = 3\cdot 1-2 = 1\ </math> ~~and~~ <math>\ \text{~~RHS~~} = (2-1)^2 = 1</math>	+	:*''x'' = 1: <math>\ \text{Linke Seite} = 3\cdot 1-2 = 1\ </math> und <math>\ \text{Rechte Seite} = (2-1)^2 = 1</math>

-	:*''x'' = 6: <math>\ \text{~~LHS~~} = 3\cdot 6-2 = 16\ </math> ~~and~~ <math>\ \text{~~RHS~~} = (2-6)^2 = 16</math>	+	:*''x'' = 6: <math>\ \text{Linke Seite} = 3\cdot 6-2 = 16\ </math> und <math>\ \text{Rechte Seite} = (2-6)^2 = 16</math>

-	~~Finally~~, ~~we need to sort away possible spurious roots to the root equation by verifying the solutions~~.	+	Schließlich kontrollieren wir die Wurzeln in der ursprünglichen Gleichung, um Scheinlösungen zu entdecken.

-	:*''x'' = 1: <math>\ \text{~~LHS~~} = \sqrt{3\cdot 1-2} = 1\ </math> and <math>\ \text{~~RHS~~} = 2-1 = 1</math>	+	:*''x'' = 1: <math>\ \text{Linke Seite} = \sqrt{3\cdot 1-2} = 1\ </math> and <math>\ \text{Rechte Seite} = 2-1 = 1</math>

-	:*''x'' = 6: <math>\ \text{~~LHS~~} = \sqrt{3\cdot 6-2} = 4\ </math> and <math>\ \text{~~RHS~~} = 2-6 = -4</math>	+	:*''x'' = 6: <math>\ \text{Linke Seite} = \sqrt{3\cdot 6-2} = 4\ </math> and <math>\ \text{Rechte Seite} = 2-6 = -4</math>

-	~~This shows that the root equation has the solution~~ <math>x=1\,</math>.	+	Also hat die Gleichung die Lösung <math>x=1\,</math>.

Tekbot: hat „Solution 3.2:5“ nach „Lösung 3.2:5“ verschoben: Robot: moved page

Tekbot — Wed, 22 Oct 2008 14:29:05 GMT

hat „Solution 3.2:5“ nach „Lösung 3.2:5“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 14:29, 22. Okt. 2008

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

Tekbot — Wed, 22 Oct 2008 08:41:08 GMT

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

← Nächstältere Version		Version vom 08:41, 22. Okt. 2008
Zeile 1:		Zeile 1:
	After squaring both sides, we obtain the equation		After squaring both sides, we obtain the equation

-	{{~~Displayed math~~\|\|<math>3x-2 = (2-x)^2</math>\|(*)}}	+	{{Abgesetzte Formel\|\|<math>3x-2 = (2-x)^2</math>\|(*)}}

	and if we expand the right-hand side and then collect the terms, we get		and if we expand the right-hand side and then collect the terms, we get

-	{{~~Displayed math~~\|\|<math>x^{2}-7x+6=0\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>x^{2}-7x+6=0\,\textrm{.}</math>}}

	Completing the square of the left-hand side, we obtain		Completing the square of the left-hand side, we obtain

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	x^{2}-7x+6		x^{2}-7x+6
	&= \Bigl(x-\frac{7}{2}\Bigr)^2 - \Bigl(\frac{7}{2}\Bigr)^2+6\\[5pt]		&= \Bigl(x-\frac{7}{2}\Bigr)^2 - \Bigl(\frac{7}{2}\Bigr)^2+6\\[5pt]
Zeile 18:		Zeile 18:
	which means that the equation can be written as		which means that the equation can be written as

-	{{~~Displayed math~~\|\|<math>\Bigl(x-\frac{7}{2}\Bigr)^2 = \frac{25}{4}</math>}}	+	{{Abgesetzte Formel\|\|<math>\Bigl(x-\frac{7}{2}\Bigr)^2 = \frac{25}{4}</math>}}

	and the solutions are therefore		and the solutions are therefore

Tek um 12:58, 1. Okt. 2008

Tek — Wed, 01 Oct 2008 12:58:37 GMT

← Nächstältere Version		Version vom 12:58, 1. Okt. 2008
Zeile 1:		Zeile 1:
	After squaring both sides, we obtain the equation		After squaring both sides, we obtain the equation

		+	{{Displayed math\|\|<math>3x-2 = (2-x)^2</math>\|(*)}}

-	~~<math>3x-2=\left( 2-x \right)^{2}\quad \quad (*)</math>~~	+	and if we expand the right-hand side and then collect the terms, we get
-		+
-		+
-	and if we expand the right-hand side and then collect ~~gather~~ the terms, we get	+
-		+
-		+
-	~~<math>x^{2}-7x+6=0</math>~~	+

		+	{{Displayed math\|\|<math>x^{2}-7x+6=0\,\textrm{.}</math>}}

	Completing the square of the left-hand side, we obtain		Completing the square of the left-hand side, we obtain

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	x^{2}-7x+6
-	& x^{2}-7x+6=\~~left~~( x-\frac{7}{2} \~~right~~)^{2}-\~~left~~( \frac{7}{2} \~~right~~)^{2}+6 \\	+	&= \Bigl(x-\frac{7}{2}\Bigr)^2 - \Bigl(\frac{7}{2}\Bigr)^2+6\\[5pt]
-	& =\~~left~~( x-\frac{7}{2} \~~right~~)^{2}-\frac{49}{4}+\frac{24}{4} \\	+	&= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{49}{4} + \frac{24}{4}\\[5pt]
-	& =\~~left~~( x-\frac{7}{2} \~~right~~)^{2}-\frac{25}{4} \\	+	&= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{25}{4}
-	\end{align}</math>	+	\end{align}</math>}}

	which means that the equation can be written as		which means that the equation can be written as

-		+	{{Displayed math\|\|<math>\Bigl(x-\frac{7}{2}\Bigr)^2 = \frac{25}{4}</math>}}
-	<math>\~~left~~( x-\frac{7}{2} \~~right~~)^{2}=\frac{25}{4}</math>	+
-		+

	and the solutions are therefore		and the solutions are therefore

		+	:*<math>x = \frac{7}{2} + \sqrt{\frac{25}{4}} = \frac{7}{2} + \frac{5}{2} = \frac{12}{2} = 6\,,</math>

-	<math>~~\begin{align}~~	+	:*<math>x = \frac{7}{2} - \sqrt{\frac{25}{4}} = \frac{7}{2} - \frac{5}{2} = \frac{2}{2} = 1\,\textrm{.}</math>
-	& x=\frac{7}{2}+\sqrt{\frac{25}{4}~~=}\frac{7}{2}+\frac{5}{2~~}~~=\frac{12}{2}=6 \\~~	+
-	~~& x~~=~~\frac{7}{2}-\sqrt{\frac{25}{4}=}~~\frac{7}{2}-\frac{5}{2}=\frac{2}{2}=1 \\	+
-	~~\end~~{~~align~~}</math>	+
-	~~EQ6~~	+

-	Substituting	+	Substituting <math>x=1</math> and <math>x=6</math> into the quadratic equation (*) shows that we have solved the equation correctly.
-	<math>x=~~\text{~~1 }</math>	+
-	and	+
-	<math>x=~~\text{~~6 }</math>	+
-	into the quadratic equation (*) shows that we have solved the equation correctly.	+

-		+	:*''x'' = 1: <math>\ \text{LHS} = 3\cdot 1-2 = 1\ </math> and <math>\ \text{RHS} = (2-1)^2 = 1</math>
-	<math>x=\text{1 }~~</math>: LHS~~	+
-	~~<math>~~=3\~~centerdot~~ 1-2=1</math>	+
-	and ~~RHS~~	+
-	<math>=~~\left~~( 2-1 ~~\right~~)^{2}=1</math>	+

		+	:*''x'' = 6: <math>\ \text{LHS} = 3\cdot 6-2 = 16\ </math> and <math>\ \text{RHS} = (2-6)^2 = 16</math>

-	~~<math>x=\text{6 }</math>: LHS~~	+	Finally, we need to sort away possible spurious roots to the root equation by verifying the solutions.
-	~~<math>=3\centerdot 6-2=16</math>~~	+
-	~~and RHS~~	+
-	~~<math>=\left( 2-6 \right)^{2}=16</math>~~	+

-		+	:*''x'' = 1: <math>\ \text{LHS} = \sqrt{3\cdot 1-2} = 1\ </math> and <math>\ \text{RHS} = 2-1 = 1</math>
-	~~Finally, we need to sort away possible false roots to the root equation by verifying the solutions.~~	+
-		+
-		+
-	~~<math>~~x=~~\text{~~1 ~~}</math>~~: ~~LHS~~	+
-	<math>=\sqrt{3\~~centerdot~~ 1-2}=1</math>	+
-	and ~~RHS~~	+
-	<math>=2-1=1</math>	+

		+	:*''x'' = 6: <math>\ \text{LHS} = \sqrt{3\cdot 6-2} = 4\ </math> and <math>\ \text{RHS} = 2-6 = -4</math>

-	~~<math>x=\text{6 }</math>: LHS~~	+	This shows that the root equation has the solution <math>x=1\,</math>.
-	~~<math>=\sqrt{3\centerdot 6-2}=4</math>~~	+
-	~~and RHS =~~	+
-	~~<math>=2-6=-4</math>~~	+
-		+
-	This shows that the root equation has the solution	+
-	<math>x=~~\text{~~1}</math>.	+

Ian um 10:54, 25. Sep. 2008

Ian — Thu, 25 Sep 2008 10:54:45 GMT

← Nächstältere Version		Version vom 10:54, 25. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START}~~}	+	After squaring both sides, we obtain the equation
-	<~~center~~> ~~[[Image:3_2_5~~-1(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+	<math>3x-2=\left( 2-x \right)^{2}\quad \quad (*)</math>
-	<~~center~~> ~~[[Image~~:~~3_2_5~~-2(2).~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	and if we expand the right-hand side and then collect gather the terms, we get
		+
		+
		+	<math>x^{2}-7x+6=0</math>
		+
		+
		+	Completing the square of the left-hand side, we obtain
		+
		+
		+	<math>\begin{align}
		+	& x^{2}-7x+6=\left( x-\frac{7}{2} \right)^{2}-\left( \frac{7}{2} \right)^{2}+6 \\
		+	& =\left( x-\frac{7}{2} \right)^{2}-\frac{49}{4}+\frac{24}{4} \\
		+	& =\left( x-\frac{7}{2} \right)^{2}-\frac{25}{4} \\
		+	\end{align}</math>
		+
		+	which means that the equation can be written as
		+
		+
		+	<math>\left( x-\frac{7}{2} \right)^{2}=\frac{25}{4}</math>
		+
		+
		+	and the solutions are therefore
		+
		+
		+	<math>\begin{align}
		+	& x=\frac{7}{2}+\sqrt{\frac{25}{4}=}\frac{7}{2}+\frac{5}{2}=\frac{12}{2}=6 \\
		+	& x=\frac{7}{2}-\sqrt{\frac{25}{4}=}\frac{7}{2}-\frac{5}{2}=\frac{2}{2}=1 \\
		+	\end{align}</math>
		+	EQ6
		+
		+	Substituting
		+	<math>x=\text{1 }</math>
		+	and
		+	<math>x=\text{6 }</math>
		+	into the quadratic equation (*) shows that we have solved the equation correctly.
		+
		+
		+	<math>x=\text{1 }</math>: LHS
		+	<math>=3\centerdot 1-2=1</math>
		+	and RHS
		+	<math>=\left( 2-1 \right)^{2}=1</math>
		+
		+
		+	<math>x=\text{6 }</math>: LHS
		+	<math>=3\centerdot 6-2=16</math>
		+	and RHS
		+	<math>=\left( 2-6 \right)^{2}=16</math>
		+
		+
		+	Finally, we need to sort away possible false roots to the root equation by verifying the solutions.
		+
		+
		+	<math>x=\text{1 }</math>: LHS
		+	<math>=\sqrt{3\centerdot 1-2}=1</math>
		+	and RHS
		+	<math>=2-1=1</math>
		+
		+
		+	<math>x=\text{6 }</math>: LHS
		+	<math>=\sqrt{3\centerdot 6-2}=4</math>
		+	and RHS =
		+	<math>=2-6=-4</math>
		+
		+	This shows that the root equation has the solution
		+	<math>x=\text{1}</math>.

Tekbot: Lösning 3.2:5 moved to Solution 3.2:5: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:18:43 GMT

Lösning 3.2:5 moved to Solution 3.2:5: Robot: moved page

← Nächstältere Version

Version vom 09:18, 9. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 06:56:29 GMT

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

← Nächstältere Version		Version vom 06:56, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_2_5-1(2).gif]] </center>	+	<center> [[Image:3_2_5-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_2_5-2(2).gif]] </center>	+	<center> [[Image:3_2_5-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar um 15:07, 1. Apr. 2008

Oskar — Tue, 01 Apr 2008 15:07:47 GMT

← Nächstältere Version		Version vom 15:07, 1. Apr. 2008
Zeile 3:		Zeile 3:
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[Bild:3_2_5-1(2).gif]] </center>	+	<center> [[Bild:3_2_5-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}