Lösung 3.3:5b - Versionsgeschichte

Sekretariat1 um 13:20, 3. Sep. 2009

2009-09-03T13:20:14Z

← Nächstältere Version		Version vom 13:20, 3. Sep. 2009
Zeile 1:		Zeile 1:
-	Durch ~~quadratischer~~ Ergänzung der linken Seite erhalten wir	+	Durch quadratische Ergänzung der linken Seite erhalten wir

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
-	\Bigl(z-\frac{2-i}{2}\Bigr)^2-\Bigl(\frac{2-i}{2}\Bigr)^2+3-i &= 0~~\,,~~\\[5pt]	+	\Bigl(z-\frac{2-i}{2}\Bigr)^2-\Bigl(\frac{2-i}{2}\Bigr)^2+3-i &= 0\\[5pt]
-	\Bigl(z-\frac{2-i}{2}\Bigr)^2-\Bigl(1-i+\frac{1}{4}i^2\Bigr)+3-i&=0~~\,,~~\\[5pt]	+	\Bigl(z-\frac{2-i}{2}\Bigr)^2-\Bigl(1-i+\frac{1}{4}i^2\Bigr)+3-i&=0\\[5pt]
-	\Bigl(z-\frac{2-i}{2}\Bigr)^2-1+i+\frac{1}{4}+3-i&=0~~\,,~~\\[5pt]	+	\Bigl(z-\frac{2-i}{2}\Bigr)^2-1+i+\frac{1}{4}+3-i&=0\\[5pt]
	\Bigl(z-\frac{2-i}{2}\Bigr)^2+\frac{9}{4}&=0\,\textrm{.}		\Bigl(z-\frac{2-i}{2}\Bigr)^2+\frac{9}{4}&=0\,\textrm{.}
	\end{align}</math>}}		\end{align}</math>}}
Zeile 15:		Zeile 15:
	\end{align}\right.</math>}}		\end{align}\right.</math>}}

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

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

Martin um 16:51, 18. Mai 2009

2009-05-18T16:51:12Z

← Nächstältere Version		Version vom 16:51, 18. Mai 2009
Zeile 1:		Zeile 1:
-	~~Complete the square of the left-hand side,~~	+	Durch quadratischer Ergänzung der linken Seite erhalten wir

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

-	~~Taking the square root then gives that the solutions are~~	+	Die Wurzeln sind

	{{Abgesetzte Formel\|\|<math>z-\frac{2-i}{2} = \pm\frac{3}{2}\,i\quad \Leftrightarrow \quad z=\left\{ \begin{align}		{{Abgesetzte Formel\|\|<math>z-\frac{2-i}{2} = \pm\frac{3}{2}\,i\quad \Leftrightarrow \quad z=\left\{ \begin{align}
Zeile 15:		Zeile 15:
	\end{align}\right.</math>}}		\end{align}\right.</math>}}

-	~~Finally~~, ~~we substitute the solutions into the equation and check that it is satisfied~~	+	Wir substituieren die Wurzeln in der ursprünglichen Gleichung,

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

Tekbot: Solution 3.3:5b moved to Lösung 3.3:5b: Robot: moved page

2009-03-11T10:48:55Z

Solution 3.3:5b moved to Lösung 3.3:5b: Robot: moved page

← Nächstältere Version

Version vom 10:48, 11. Mär. 2009

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

2009-03-10T13:13:54Z

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

← Nächstältere Version		Version vom 13:13, 10. Mär. 2009
Zeile 1:		Zeile 1:
	Complete the square of the left-hand side,		Complete the square of the left-hand side,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	\Bigl(z-\frac{2-i}{2}\Bigr)^2-\Bigl(\frac{2-i}{2}\Bigr)^2+3-i &= 0\,,\\[5pt]		\Bigl(z-\frac{2-i}{2}\Bigr)^2-\Bigl(\frac{2-i}{2}\Bigr)^2+3-i &= 0\,,\\[5pt]
	\Bigl(z-\frac{2-i}{2}\Bigr)^2-\Bigl(1-i+\frac{1}{4}i^2\Bigr)+3-i&=0\,,\\[5pt]		\Bigl(z-\frac{2-i}{2}\Bigr)^2-\Bigl(1-i+\frac{1}{4}i^2\Bigr)+3-i&=0\,,\\[5pt]
Zeile 10:		Zeile 10:
	Taking the square root then gives that the solutions are		Taking the square root then gives that the solutions are

-	{{~~Displayed math~~\|\|<math>z-\frac{2-i}{2} = \pm\frac{3}{2}\,i\quad \Leftrightarrow \quad z=\left\{ \begin{align}	+	{{Abgesetzte Formel\|\|<math>z-\frac{2-i}{2} = \pm\frac{3}{2}\,i\quad \Leftrightarrow \quad z=\left\{ \begin{align}
	&1+i\,,\\		&1+i\,,\\
	&1-2i\,\textrm{.}		&1-2i\,\textrm{.}

Tek um 15:32, 30. Okt. 2008

2008-10-30T15:32:17Z

← Nächstältere Version		Version vom 15:32, 30. Okt. 2008
Zeile 1:		Zeile 1:
-	Complete the square of the left-hand side:	+	Complete the square of the left-hand side,

		+	{{Displayed math\|\|<math>\begin{align}
		+	\Bigl(z-\frac{2-i}{2}\Bigr)^2-\Bigl(\frac{2-i}{2}\Bigr)^2+3-i &= 0\,,\\[5pt]
		+	\Bigl(z-\frac{2-i}{2}\Bigr)^2-\Bigl(1-i+\frac{1}{4}i^2\Bigr)+3-i&=0\,,\\[5pt]
		+	\Bigl(z-\frac{2-i}{2}\Bigr)^2-1+i+\frac{1}{4}+3-i&=0\,,\\[5pt]
		+	\Bigl(z-\frac{2-i}{2}\Bigr)^2+\frac{9}{4}&=0\,\textrm{.}
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	Taking the square root then gives that the solutions are
-	~~& \left( z-\frac{2-i}{2} \right)^{2}-\left( \frac{2-i}{2} \right)^{2}+3-i=0 \\~~	+
-	~~& \left( z-\frac{2-i}{2} \right)^{2}-\left( 1-i+\frac{1}{4}i^{2} \right)+3-i=0 \\~~	+
-	~~& \left( z-\frac{2-i}{2} \right)^{2}-1+i+\frac{1}{4}+3-i=0 \\~~	+
-	~~& \left( z-\frac{2-i}{2} \right)^{2}+\frac{9}{4}=0 \\~~	+
-	~~& \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	Taking the root then gives that the solutions are	+
-		+
-		+
-	~~<math>z-\frac{2-i}{2}=\pm \frac{3}{2}i\quad \Leftrightarrow \quad z=\left\{ \begin{array}{*{35}l}~~	+
-	~~1+i \\~~	+
-	~~1-2i\text{ } \\~~	+
-	~~\end{array} \right.</math>~~	+
-		+
-		+
-	~~Finally, we substitute the solutions into the equation and check that it is satisfied:~~	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& z=1+i:\quad z^{2}-\left( 2-i \right)z+\left( 3-i \right) \\~~	+
-	~~& =\left( 1+i \right)^{2}-\left( 2-i \right)\left( 1+i \right)+3-i \\~~	+
-	~~& =1+2i+i^{2}-\left( 2+2i-i-i^{2} \right)+3-i \\~~	+
-	~~& =1+2i-1-2-i-1+3-i=0, \\~~	+
-	~~\end{align}</math>~~	+
-		+

		+	{{Displayed math\|\|<math>z-\frac{2-i}{2} = \pm\frac{3}{2}\,i\quad \Leftrightarrow \quad z=\left\{ \begin{align}
		+	&1+i\,,\\
		+	&1-2i\,\textrm{.}
		+	\end{align}\right.</math>}}

		+	Finally, we substitute the solutions into the equation and check that it is satisfied

	<math>\begin{align}		<math>\begin{align}
-	& z=1-2i:\quad z^{2}-~~\left~~( 2-i ~~\right~~)z+~~\left~~( 3-i \~~right~~) \\	+	z={}\rlap{1+i:}\phantom{1-2i:}{}\quad z^2-(2-i)z+(3-i)
-	& =\~~left(~~ 1-2i \~~right~~)^{2}-~~\left~~( 2-i ~~\right~~)~~\left~~( 1-2i ~~\right~~)+3-i \\	+	&= (1+i)^2-(2-i)(1+i)+3-i\\[5pt]
-	& =1-4i+4i^{2}-~~\left~~( 2-4i-i+2i^{2~~} \right~~)+3-i \\	+	&= 1+2i+i^2-(2+2i-i-i^2)+3-i\\[5pt]
-	& =1-4i-4-2+5i+2+3-i=0. \\	+	&= 1+2i-1-2-i-1+3-i\\[5pt]
		+	&=0\,,\\[10pt]
		+	z=1-2i:\quad z^2-(2-i)z+(3-i)
		+	&= (1-2i)^2-(2-i)(1-2i)+3-i\\[5pt]
		+	&= 1-4i+4i^2-(2-4i-i+2i^2)+3-i\\[5pt]
		+	&= 1-4i-4-2+5i+2+3-i\\[5pt]
		+	&= 0\,\textrm{.}
	\end{align}</math>		\end{align}</math>

Ian um 11:30, 25. Okt. 2008

2008-10-25T11:30:23Z

← Nächstältere Version		Version vom 11:30, 25. Okt. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START~~}}	+	Complete the square of the left-hand side:
-	<~~center~~> ~~[[Image:3_3_5b~~.~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\begin{align}
		+	& \left( z-\frac{2-i}{2} \right)^{2}-\left( \frac{2-i}{2} \right)^{2}+3-i=0 \\
		+	& \left( z-\frac{2-i}{2} \right)^{2}-\left( 1-i+\frac{1}{4}i^{2} \right)+3-i=0 \\
		+	& \left( z-\frac{2-i}{2} \right)^{2}-1+i+\frac{1}{4}+3-i=0 \\
		+	& \left( z-\frac{2-i}{2} \right)^{2}+\frac{9}{4}=0 \\
		+	& \\
		+	\end{align}</math>
		+
		+
		+	Taking the root then gives that the solutions are
		+
		+
		+	<math>z-\frac{2-i}{2}=\pm \frac{3}{2}i\quad \Leftrightarrow \quad z=\left\{ \begin{array}{*{35}l}
		+	1+i \\
		+	1-2i\text{ } \\
		+	\end{array} \right.</math>
		+
		+
		+	Finally, we substitute the solutions into the equation and check that it is satisfied:
		+
		+
		+	<math>\begin{align}
		+	& z=1+i:\quad z^{2}-\left( 2-i \right)z+\left( 3-i \right) \\
		+	& =\left( 1+i \right)^{2}-\left( 2-i \right)\left( 1+i \right)+3-i \\
		+	& =1+2i+i^{2}-\left( 2+2i-i-i^{2} \right)+3-i \\
		+	& =1+2i-1-2-i-1+3-i=0, \\
		+	\end{align}</math>
		+
		+
		+
		+
		+	<math>\begin{align}
		+	& z=1-2i:\quad z^{2}-\left( 2-i \right)z+\left( 3-i \right) \\
		+	& =\left( 1-2i \right)^{2}-\left( 2-i \right)\left( 1-2i \right)+3-i \\
		+	& =1-4i+4i^{2}-\left( 2-4i-i+2i^{2} \right)+3-i \\
		+	& =1-4i-4-2+5i+2+3-i=0. \\
		+	\end{align}</math>

Tekbot: Lösning 3.3:5b moved to Solution 3.3:5b: Robot: moved page

2008-09-16T15:09:21Z

Lösning 3.3:5b moved to Solution 3.3:5b: Robot: moved page

← Nächstältere Version

Version vom 15:09, 16. Sep. 2008

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

2008-08-21T08:14:24Z

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

← Nächstältere Version		Version vom 08:14, 21. Aug. 2008
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Oskar: Ny sida: {{NAVCONTENT_START}} Bild:3_3_5b.gif {{NAVCONTENT_STOP}}

2008-04-07T13:44:54Z

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