Lösung 2.3:4a - Versionsgeschichte

Silke um 16:25, 23. Aug. 2009

Silke — Sun, 23 Aug 2009 16:25:38 GMT

← Nächstältere Version		Version vom 16:25, 23. Aug. 2009
Zeile 10:		Zeile 10:


-	Hinweis: Eigentlich gibt es unendlich viele Funktionen mit den ~~Wurzeln~~ -1 und 2, nämlich alle Funktionen	+	Hinweis: Eigentlich gibt es unendlich viele Funktionen mit den Nullstellen -1 und 2, nämlich alle Funktionen

	{{Abgesetzte Formel\|\|<math>ax^{2}-ax-2a=0\,,</math>}}		{{Abgesetzte Formel\|\|<math>ax^{2}-ax-2a=0\,,</math>}}

	wobei <math>a\ne 0</math> eine beliebige Konstante ist.		wobei <math>a\ne 0</math> eine beliebige Konstante ist.

Markus.bez um 12:53, 9. Jun. 2009

Markus.bez — Tue, 09 Jun 2009 12:53:49 GMT

← Nächstältere Version		Version vom 12:53, 9. Jun. 2009
Zeile 3:		Zeile 3:
	{{Abgesetzte Formel\|\|<math>(x+1)(x-2)=0\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>(x+1)(x-2)=0\,\textrm{.}</math>}}

-	~~und~~ sehen dass die Gleichung erfüllt ist wenn <math>x=-1</math> oder <math>x=2</math>. Erweitern wir die linke Seite erhalten wir	+	Wir sehen, dass die Gleichung erfüllt ist, wenn <math>x=-1</math> oder <math>x=2</math>. Erweitern wir die linke Seite, erhalten wir

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

-	Eine Funktion ist also <math>(x+1)(x-2)=0</math>, oder <math>x^{2}-x-2=0\,</math>.	+	Eine Funktion ist also <math>(x+1)(x-2)=0</math> oder <math>x^{2}-x-2=0\,</math>.


-	~~Hinweit~~: Eigentlich gibt es unendlich viele Funktionen mit den Wurzeln -1 und 2, nämlich alle Funktionen	+	Hinweis: Eigentlich gibt es unendlich viele Funktionen mit den Wurzeln -1 und 2, nämlich alle Funktionen

	{{Abgesetzte Formel\|\|<math>ax^{2}-ax-2a=0\,,</math>}}		{{Abgesetzte Formel\|\|<math>ax^{2}-ax-2a=0\,,</math>}}

-	Wo <math>a\ne 0</math> eine beliebige Konstante ist.	+	wobei <math>a\ne 0</math> eine beliebige Konstante ist.

Martin um 16:41, 16. Mär. 2009

Martin — Mon, 16 Mar 2009 16:41:53 GMT

← Nächstältere Version		Version vom 16:41, 16. Mär. 2009
Zeile 14:		Zeile 14:
	{{Abgesetzte Formel\|\|<math>ax^{2}-ax-2a=0\,,</math>}}		{{Abgesetzte Formel\|\|<math>ax^{2}-ax-2a=0\,,</math>}}

-	Wo <math>a\ne 0</math> ~~ein beliebiger Konstant~~ ist.	+	Wo <math>a\ne 0</math> eine beliebige Konstante ist.

Martin um 15:47, 16. Mär. 2009

Martin — Mon, 16 Mar 2009 15:47:47 GMT

← Nächstältere Version		Version vom 15:47, 16. Mär. 2009
Zeile 1:		Zeile 1:
-	~~A first thought is perhaps to write the equation as~~	+	Wir schreiben die quadratische Gleichung faktorisiert
-		+
-	~~{{Abgesetzte Formel\|\|<math>x^{2}+ax+b=0</math>}}~~	+
-		+
-	~~and then try to choose the constants ''a'' and ''b'' in some way so that~~	+
-	~~<math>x=-1</math> and <math>x=2</math> are solutions. But a better way is to start with a factorized form of a second-order equation,~~	+

	{{Abgesetzte Formel\|\|<math>(x+1)(x-2)=0\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>(x+1)(x-2)=0\,\textrm{.}</math>}}

-	~~If we consider this equation, we see that both~~ <math>x=-1</math> ~~and~~ <math>x=2</math> ~~are solutions to the equation, since <math>x=-1</math> makes the first factor on the left-hand side zero, whilst <math>x=2</math> makes the second factor zero~~. ~~Also, it really is a second order equation, because if we multiply out the left-hand side, we get~~	+	und sehen dass die Gleichung erfüllt ist wenn <math>x=-1</math> oder <math>x=2</math>. Erweitern wir die linke Seite erhalten wir

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

-	~~One answer is thus the equation~~ <math>(x+1)(x-2)=0</math>, or <math>x^{2}-x-2=0\,</math>.	+	Eine Funktion ist also <math>(x+1)(x-2)=0</math>, oder <math>x^{2}-x-2=0\,</math>.


-	~~Note~~: ~~There are actually many answers to this exercise, but what all second-degree equations that have <math>x=~~-1~~</math> and <math>x=~~2~~</math> as roots have in common is that they can be written in the form~~	+	Hinweit: Eigentlich gibt es unendlich viele Funktionen mit den Wurzeln -1 und 2, nämlich alle Funktionen

	{{Abgesetzte Formel\|\|<math>ax^{2}-ax-2a=0\,,</math>}}		{{Abgesetzte Formel\|\|<math>ax^{2}-ax-2a=0\,,</math>}}

-	~~where ''~~a~~'' is a non-zero constant~~.	+	Wo <math>a\ne 0</math> ein beliebiger Konstant ist.

Tekbot: hat „Solution 2.3:4a“ nach „Lösung 2.3:4a“ verschoben: Robot: moved page

Tekbot — Wed, 22 Oct 2008 14:10:25 GMT

hat „Solution 2.3:4a“ nach „Lösung 2.3:4a“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 14:10, 22. Okt. 2008

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

Tekbot — Wed, 22 Oct 2008 08:33:08 GMT

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

← Nächstältere Version		Version vom 08:33, 22. Okt. 2008
Zeile 1:		Zeile 1:
	A first thought is perhaps to write the equation as		A first thought is perhaps to write the equation as

-	{{~~Displayed math~~\|\|<math>x^{2}+ax+b=0</math>}}	+	{{Abgesetzte Formel\|\|<math>x^{2}+ax+b=0</math>}}

	and then try to choose the constants ''a'' and ''b'' in some way so that		and then try to choose the constants ''a'' and ''b'' in some way so that
	<math>x=-1</math> and <math>x=2</math> are solutions. But a better way is to start with a factorized form of a second-order equation,		<math>x=-1</math> and <math>x=2</math> are solutions. But a better way is to start with a factorized form of a second-order equation,

-	{{~~Displayed math~~\|\|<math>(x+1)(x-2)=0\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>(x+1)(x-2)=0\,\textrm{.}</math>}}

	If we consider this equation, we see that both <math>x=-1</math> and <math>x=2</math> are solutions to the equation, since <math>x=-1</math> makes the first factor on the left-hand side zero, whilst <math>x=2</math> makes the second factor zero. Also, it really is a second order equation, because if we multiply out the left-hand side, we get		If we consider this equation, we see that both <math>x=-1</math> and <math>x=2</math> are solutions to the equation, since <math>x=-1</math> makes the first factor on the left-hand side zero, whilst <math>x=2</math> makes the second factor zero. Also, it really is a second order equation, because if we multiply out the left-hand side, we get

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

	One answer is thus the equation <math>(x+1)(x-2)=0</math>, or <math>x^{2}-x-2=0\,</math>.		One answer is thus the equation <math>(x+1)(x-2)=0</math>, or <math>x^{2}-x-2=0\,</math>.
Zeile 17:		Zeile 17:
	Note: There are actually many answers to this exercise, but what all second-degree equations that have <math>x=-1</math> and <math>x=2</math> as roots have in common is that they can be written in the form		Note: There are actually many answers to this exercise, but what all second-degree equations that have <math>x=-1</math> and <math>x=2</math> as roots have in common is that they can be written in the form

-	{{~~Displayed math~~\|\|<math>ax^{2}-ax-2a=0\,,</math>}}	+	{{Abgesetzte Formel\|\|<math>ax^{2}-ax-2a=0\,,</math>}}

	where ''a'' is a non-zero constant.		where ''a'' is a non-zero constant.

Tek um 10:18, 29. Sep. 2008

Tek — Mon, 29 Sep 2008 10:18:18 GMT

← Nächstältere Version		Version vom 10:18, 29. Sep. 2008
Zeile 1:		Zeile 1:
	A first thought is perhaps to write the equation as		A first thought is perhaps to write the equation as

		+	{{Displayed math\|\|<math>x^{2}+ax+b=0</math>}}

-	<math>x~~^{2}+ax+b~~=0</math>	+	and then try to choose the constants ''a'' and ''b'' in some way so that
		+	<math>x=-1</math> and <math>x=2</math> are solutions. But a better way is to start with a factorized form of a second-order equation,

		+	{{Displayed math\|\|<math>(x+1)(x-2)=0\,\textrm{.}</math>}}

-	~~and then try to choose the constants~~	+	If we consider this equation, we see that both <math>x=-1</math> and <math>x=2</math> are solutions to the equation, since <math>x=-1</math> makes the first factor on the left-hand side zero, whilst <math>x=2</math> makes the second factor zero. Also, it really is a second order equation, because if we multiply out the left-hand side, we get
-	<math>a</math>	+
-	and	+
-	<math>b</math>	+
-	~~in some way so that~~	+
-	<math>x=-~~\text{~~1 }</math>	+
-	~~and~~	+
-	<math>x=~~\text{~~2 }</math>	+
-	~~are solutions~~. ~~But a better way~~ is ~~to start with a factorized form of~~ a second-order equation,	+

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

-	<math>~~\left~~( x+1 ~~\right~~)~~\left~~( x-2 ~~\right~~)=0</math>	+	One answer is thus the equation <math>(x+1)(x-2)=0</math>, or <math>x^{2}-x-2=0\,</math>.


-	~~If we consider~~ this ~~equation~~, ~~we see~~ that ~~both~~	+	Note: There are actually many answers to this exercise, but what all second-degree equations that have <math>x=-1</math> and <math>x=2</math> as roots have in common is that they can be written in the form
-	<math>x=-~~\text{~~1 }</math>	+
-	and	+
-	<math>x=~~\text{~~2 }</math>	+
-	~~are solutions to the equation, since~~	+
-	~~<math>x=-\text{1 }</math>~~	+
-	~~makes the first factor on the left-hand side zero, whilst~~	+
-	~~<math>x=\text{2 }</math>~~	+
-	~~makes the second factor zero. Also, it really~~ is ~~a second order equation, because if we multiply out~~ the ~~left-hand side, we get~~	+

		+	{{Displayed math\|\|<math>ax^{2}-ax-2a=0\,,</math>}}

-	~~<math>x^{2}-x-2=0</math>~~	+	where ''a'' is a non-zero constant.
-		+
-		+
-	~~One answer is thus the equation~~	+
-	~~<math>\left( x+1 \right)\left( x-2 \right)=0</math>, or~~	+
-	~~<math>x^{2}-x-2=0</math>.~~	+
-		+
-	~~NOTE: There are actually many answers to this exercise, but what all second-degree equations that have~~	+
-	~~<math>x=-\text{1 }</math>~~	+
-	~~and~~	+
-	~~<math>x=\text{2 }</math>~~	+
-	~~as roots have in common is that they can be written in the form~~	+
-		+
-		+
-	~~<math>ax^{2}-ax-2a=0</math>~~	+
-		+
-		+
-	where	+
-	~~<math>~~a~~</math>~~	+
-	is a non-zero constant.	+

Ian um 09:14, 21. Sep. 2008

Ian — Sun, 21 Sep 2008 09:14:48 GMT

← Nächstältere Version		Version vom 09:14, 21. Sep. 2008
Zeile 1:		Zeile 1:
-	{{~~NAVCONTENT_START}~~}	+	A first thought is perhaps to write the equation as
-	<~~center~~> ~~[[Image:2_3_4a~~-1(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
-	{~~{NAVCONTENT_START}~~}	+	<math>x^{2}+ax+b=0</math>
-	<~~center~~> ~~[[Image:2_3_4a~~-2(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	and then try to choose the constants
		+	<math>a</math>
		+	and
		+	<math>b</math>
		+	in some way so that
		+	<math>x=-\text{1 }</math>
		+	and
		+	<math>x=\text{2 }</math>
		+	are solutions. But a better way is to start with a factorized form of a second-order equation,
		+
		+
		+	<math>\left( x+1 \right)\left( x-2 \right)=0</math>
		+
		+
		+	If we consider this equation, we see that both
		+	<math>x=-\text{1 }</math>
		+	and
		+	<math>x=\text{2 }</math>
		+	are solutions to the equation, since
		+	<math>x=-\text{1 }</math>
		+	makes the first factor on the left-hand side zero, whilst
		+	<math>x=\text{2 }</math>
		+	makes the second factor zero. Also, it really is a second order equation, because if we multiply out the left-hand side, we get
		+
		+
		+	<math>x^{2}-x-2=0</math>
		+
		+
		+	One answer is thus the equation
		+	<math>\left( x+1 \right)\left( x-2 \right)=0</math>, or
		+	<math>x^{2}-x-2=0</math>.
		+
		+	NOTE: There are actually many answers to this exercise, but what all second-degree equations that have
		+	<math>x=-\text{1 }</math>
		+	and
		+	<math>x=\text{2 }</math>
		+	as roots have in common is that they can be written in the form
		+
		+
		+	<math>ax^{2}-ax-2a=0</math>
		+
		+
		+	where
		+	<math>a</math>
		+	is a non-zero constant.

Tekbot: Lösning 2.3:4a moved to Solution 2.3:4a: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:00:03 GMT

Lösning 2.3:4a moved to Solution 2.3:4a: Robot: moved page

← Nächstältere Version

Version vom 09:00, 9. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 06:47:09 GMT

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

← Nächstältere Version		Version vom 06:47, 21. Aug. 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_3_4a-1(2).gif]] </center>	+	<center> [[Image:2_3_4a-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_3_4a-2(2).gif]] </center>	+	<center> [[Image:2_3_4a-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}