Lösung 3.3:4c - Versionsgeschichte

Sekretariat1 um 13:07, 3. Sep. 2009

2009-09-03T13:07:16Z

← Nächstältere Version		Version vom 13:07, 3. Sep. 2009
Zeile 1:		Zeile 1:
-	Durch ~~quadratischer~~ Ergänzung ~~von~~ der linken Seite erhalten wir	+	Durch quadratische Ergänzung der linken Seite erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 8:		Zeile 8:
	Und die Wurzeln sind <math>z+1=\pm i\sqrt{2}</math>, also <math>z=-1+i\sqrt{2}</math> und <math>z=-1-i\sqrt{2}</math>.		Und die Wurzeln sind <math>z+1=\pm i\sqrt{2}</math>, also <math>z=-1+i\sqrt{2}</math> und <math>z=-1-i\sqrt{2}</math>.

-	Wir substituieren die Wurzeln in der ursprünglichen Gleichung und erhalten,	+	Wir substituieren die Wurzeln in der ursprünglichen Gleichung und erhalten

	<math>\begin{align}		<math>\begin{align}

Martin um 16:04, 18. Mai 2009

2009-05-18T16:04:52Z

← Nächstältere Version		Version vom 16:04, 18. Mai 2009
Zeile 1:		Zeile 1:
-	~~We complete the square on the left-hand side,~~	+	Durch quadratischer Ergänzung von der linken Seite erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 6:		Zeile 6:
	\end{align}</math>}}		\end{align}</math>}}

-	~~Taking the root now gives~~ <math>z+1=\pm i\sqrt{2}</math>, ~~i.e.~~ <math>z=-1+i\sqrt{2}</math> ~~and~~ <math>z=-1-i\sqrt{2}</math>.	+	Und die Wurzeln sind <math>z+1=\pm i\sqrt{2}</math>, also <math>z=-1+i\sqrt{2}</math> und <math>z=-1-i\sqrt{2}</math>.

-	~~We test the solutions~~ in ~~the equation to ascertain that we have calculated correctly.~~	+	Wir substituieren die Wurzeln in der ursprünglichen Gleichung und erhalten,

	<math>\begin{align}		<math>\begin{align}

Tekbot: Solution 3.3:4c moved to Lösung 3.3:4c: Robot: moved page

2009-03-11T10:47:55Z

Solution 3.3:4c moved to Lösung 3.3:4c: Robot: moved page

← Nächstältere Version

Version vom 10:47, 11. Mär. 2009

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

2009-03-10T13:13:24Z

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

← Nächstältere Version		Version vom 13:13, 10. Mär. 2009
Zeile 1:		Zeile 1:
	We complete the square on the left-hand side,		We complete the square on the left-hand side,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	(z+1)^2-1^2+3 &= 0\,,\\[5pt]		(z+1)^2-1^2+3 &= 0\,,\\[5pt]
	(z+1)^2+2 &= 0\,\textrm{.}		(z+1)^2+2 &= 0\,\textrm{.}

Tek um 14:44, 30. Okt. 2008

2008-10-30T14:44:34Z

← Nächstältere Version		Version vom 14:44, 30. Okt. 2008
Zeile 20:		Zeile 20:
	&= (-1)^2 + 2\cdot i\sqrt{2} + i^2\bigl(\sqrt{2}\,\bigr)^2 - 2 - 2i\sqrt{2} + 3\\[5pt]		&= (-1)^2 + 2\cdot i\sqrt{2} + i^2\bigl(\sqrt{2}\,\bigr)^2 - 2 - 2i\sqrt{2} + 3\\[5pt]
	&= 1+2\cdot i\sqrt{2} - 2 - 2 - 2\sqrt{2}i + 3\\[5pt]		&= 1+2\cdot i\sqrt{2} - 2 - 2 - 2\sqrt{2}i + 3\\[5pt]
-	&= 0.	+	&= 0\,\textrm{.}
	\end{align}</math>		\end{align}</math>

Tek um 14:44, 30. Okt. 2008

2008-10-30T14:44:08Z

← Nächstältere Version		Version vom 14:44, 30. Okt. 2008
Zeile 20:		Zeile 20:
	&= (-1)^2 + 2\cdot i\sqrt{2} + i^2\bigl(\sqrt{2}\,\bigr)^2 - 2 - 2i\sqrt{2} + 3\\[5pt]		&= (-1)^2 + 2\cdot i\sqrt{2} + i^2\bigl(\sqrt{2}\,\bigr)^2 - 2 - 2i\sqrt{2} + 3\\[5pt]
	&= 1+2\cdot i\sqrt{2} - 2 - 2 - 2\sqrt{2}i + 3\\[5pt]		&= 1+2\cdot i\sqrt{2} - 2 - 2 - 2\sqrt{2}i + 3\\[5pt]
-	&= 0,	+	&= 0.
	\end{align}</math>		\end{align}</math>

Tek um 14:43, 30. Okt. 2008

2008-10-30T14:43:47Z

← Nächstältere Version		Version vom 14:43, 30. Okt. 2008
Zeile 1:		Zeile 1:
-	We complete the square on the left-hand side:	+	We complete the square on the left-hand side,

		+	{{Displayed math\|\|<math>\begin{align}
		+	(z+1)^2-1^2+3 &= 0\,,\\[5pt]
		+	(z+1)^2+2 &= 0\,\textrm{.}
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	Taking the root now gives <math>z+1=\pm i\sqrt{2}</math>, i.e. <math>z=-1+i\sqrt{2}</math> and <math>z=-1-i\sqrt{2}</math>.
-	~~& \left( z+1 \right)^{\text{2}}-1^{2}+3=0 \\~~	+
-	~~& \left( z+1 \right)^{\text{2}}+2=0 \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	Taking the root now gives	+
-	<math>z+1=\pm i\sqrt{2}</math>	+
-	i.e.	+
-	<math>z=-1+i\sqrt{2}</math>	+
-	and	+
-	<math>z=-1-i\sqrt{2}</math>.	+

	We test the solutions in the equation to ascertain that we have calculated correctly.		We test the solutions in the equation to ascertain that we have calculated correctly.
-
-
-	<math>\begin{align}
-	& z=-1+i\sqrt{2}:\quad z^{2}+2z+3=\left( -1+i\sqrt{2} \right)^{2}+2\left( -1+i\sqrt{2} \right)+3 \\
-	& =\left( -1 \right)^{2}-2\centerdot i\sqrt{2}+i^{2}\left( \sqrt{2} \right)^{2}-2+2i\sqrt{2}+3 \\
-	& =1-2\centerdot i\sqrt{2}-2-2+2i\sqrt{2}+3=0, \\
-	\end{align}</math>
-
-

	<math>\begin{align}		<math>\begin{align}
-	& z=-1-i\sqrt{2}:\quad z^{2}+2z+3=\~~left~~( -1-i\sqrt{2} \~~right~~)^{2}+2\~~left~~( -1-i\sqrt{2} \~~right~~)+3 \\	+	z=-1+i\sqrt{2}:\quad z^2+2z+3
-	& =~~\left~~( -1 \~~right~~)^{2}+2\~~centerdot~~ i\sqrt{2}+i^{2}\~~left~~( \sqrt{2} \~~right~~)^{2}-2-2i\sqrt{2}+3 \\	+	&= \bigl(-1+i\sqrt{2}\,\bigr)^2 + 2\bigl(-1+i\sqrt{2}\bigr) + 3\\[5pt]
-	& =1+2\~~centerdot~~ i\sqrt{2}-2-2-2\sqrt{2}i+3=0, \\	+	&= (-1)^2 - 2\cdot i\sqrt{2} + i^2\bigl(\sqrt{2}\,\bigr)^2 - 2 + 2i\sqrt{2} + 3\\[5pt]
		+	&= 1-2\cdot i\sqrt{2}-2-2+2i\sqrt{2}+3\\[5pt]
		+	&= 0,\\[10pt]
		+	z={}\rlap{-1-i\sqrt{2}:}\phantom{-1+i\sqrt{2}:}{}\quad z^2+2z+3
		+	&= \bigl(-1-i\sqrt{2}\,\bigr)^2 + 2\bigl(-1-i\sqrt{2}\,\bigr) + 3\\[5pt]
		+	&= (-1)^2 + 2\cdot i\sqrt{2} + i^2\bigl(\sqrt{2}\,\bigr)^2 - 2 - 2i\sqrt{2} + 3\\[5pt]
		+	&= 1+2\cdot i\sqrt{2} - 2 - 2 - 2\sqrt{2}i + 3\\[5pt]
		+	&= 0,
	\end{align}</math>		\end{align}</math>

Ian um 10:32, 25. Okt. 2008

2008-10-25T10:32:16Z

← Nächstältere Version		Version vom 10:32, 25. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	We complete the square on the left-hand side:
-	<~~center~~> ~~[[Image:3_3_4c~~.~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\begin{align}
		+	& \left( z+1 \right)^{\text{2}}-1^{2}+3=0 \\
		+	& \left( z+1 \right)^{\text{2}}+2=0 \\
		+	\end{align}</math>
		+
		+
		+	Taking the root now gives
		+	<math>z+1=\pm i\sqrt{2}</math>
		+	i.e.
		+	<math>z=-1+i\sqrt{2}</math>
		+	and
		+	<math>z=-1-i\sqrt{2}</math>.
		+
		+	We test the solutions in the equation to ascertain that we have calculated correctly.
		+
		+
		+	<math>\begin{align}
		+	& z=-1+i\sqrt{2}:\quad z^{2}+2z+3=\left( -1+i\sqrt{2} \right)^{2}+2\left( -1+i\sqrt{2} \right)+3 \\
		+	& =\left( -1 \right)^{2}-2\centerdot i\sqrt{2}+i^{2}\left( \sqrt{2} \right)^{2}-2+2i\sqrt{2}+3 \\
		+	& =1-2\centerdot i\sqrt{2}-2-2+2i\sqrt{2}+3=0, \\
		+	\end{align}</math>
		+
		+
		+
		+	<math>\begin{align}
		+	& z=-1-i\sqrt{2}:\quad z^{2}+2z+3=\left( -1-i\sqrt{2} \right)^{2}+2\left( -1-i\sqrt{2} \right)+3 \\
		+	& =\left( -1 \right)^{2}+2\centerdot i\sqrt{2}+i^{2}\left( \sqrt{2} \right)^{2}-2-2i\sqrt{2}+3 \\
		+	& =1+2\centerdot i\sqrt{2}-2-2-2\sqrt{2}i+3=0, \\
		+	\end{align}</math>

Tekbot: Lösning 3.3:4c moved to Solution 3.3:4c: Robot: moved page

2008-09-16T15:08:21Z

Lösning 3.3:4c moved to Solution 3.3:4c: Robot: moved page

← Nächstältere Version

Version vom 15:08, 16. Sep. 2008

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

2008-08-21T08:13:54Z

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

← Nächstältere Version		Version vom 08:13, 21. Aug. 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_3_4c.gif]] </center>	+	<center> [[Image:3_3_4c.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}