Lösung 4.1:4c - Versionsgeschichte

Tek: Replaced figure with metapost figure

Tek — Thu, 20 Aug 2009 07:42:06 GMT

Replaced figure with metapost figure

← Nächstältere Version		Version vom 07:42, 20. Aug. 2009
Zeile 21:		Zeile 21:


-	~~[[Image~~:~~4_1_4_c-~~1(2)~~.gif\|~~center]]	+	<center>{{:4.1.4c - Solution - Two line segments from (2,0) to (3,3) and from (2,0) to (5,1)}}</center>

	Zum Schluss kontrollieren wir, ob die beiden Abstände auch wirklich gleich sind. Der Abstand zwischen <math>(2,0)</math> und <math>(3,3)</math>		Zum Schluss kontrollieren wir, ob die beiden Abstände auch wirklich gleich sind. Der Abstand zwischen <math>(2,0)</math> und <math>(3,3)</math>

Bayerer um 15:37, 23. Jul. 2009

Bayerer — Thu, 23 Jul 2009 15:37:09 GMT

← Nächstältere Version		Version vom 15:37, 23. Jul. 2009
Zeile 32:		Zeile 32:
	{{Abgesetzte Formel\|\|<math>\sqrt{(5-2)^2 + (1-0)^2} = \sqrt{3^2 + 1^2} = \sqrt{9+1} = \sqrt{10}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\sqrt{(5-2)^2 + (1-0)^2} = \sqrt{3^2 + 1^2} = \sqrt{9+1} = \sqrt{10}\,\textrm{.}</math>}}

-	Hinweis: Obwohl wir unsere Gleichung quadriert haben, besteht kein Risiko ~~Scheingleichungen~~ zu erhalten, ~~nachdem~~ die Quadrate in der Wurzel immer positiv sind.	+	Hinweis: Obwohl wir unsere Gleichung quadriert haben, besteht kein Risiko Scheinlösungen zu erhalten, da die Quadrate in der Wurzel immer positiv sind.

Markus.bez: Sprache und Formulierung

Markus.bez — Tue, 16 Jun 2009 13:45:00 GMT

Sprache und Formulierung

Martin um 17:17, 2. Apr. 2009

Martin — Thu, 02 Apr 2009 17:17:19 GMT

← Nächstältere Version		Version vom 17:17, 2. Apr. 2009
Zeile 1:		Zeile 1:
-	~~Let the point on the~~ ''x''-~~axis have coordinates~~ <math>(x,0)</math>, ~~where~~ ''x'' ~~is an unknown number~~. ~~Then~~, ~~using the distance formula, the distance from the point~~	+	Die ''x''-Achse besteht aus allen Punkten <math>(x,0)</math>, wo ''x'' unbekannt ist. Mit der Formel für den Abstand zwischen zwei Punkten, erhalten wir den Abstand zwischen <math>(x,0)</math> und <math>(3,3)</math> und zwischen <math>(5,1)</math> und <math>(x,0)</math>,
-	<math>(x,0)</math> to <math>(3,3)</math> ~~and~~ <math>(5,1)</math> ~~is given by~~	+

-	{{Abgesetzte Formel\|\|<math>\sqrt{(x-3)^2+(0-3)^2}\qquad\text{~~and~~}\qquad \sqrt{(x-5)^2+(0-1)^2}\,\textrm{,}</math>}}	+	{{Abgesetzte Formel\|\|<math>\sqrt{(x-3)^2+(0-3)^2}\qquad\text{und}\qquad \sqrt{(x-5)^2+(0-1)^2}\,\textrm{,}</math>}}

-	~~respectively. Because these two distances should be the same~~, ~~we get the equation~~	+	Nachdem die beiden Abstände gleich sein sollen, erhalten wir

	{{Abgesetzte Formel\|\|<math>\sqrt{(x-3)^2+9} = \sqrt{(x-5)^2+1}</math>}}		{{Abgesetzte Formel\|\|<math>\sqrt{(x-3)^2+9} = \sqrt{(x-5)^2+1}</math>}}

-	~~or, if we take the square, so as to get rid of the square root sign,~~	+	Wir quadrieren diese Gleichung und erhalten

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

-	~~Now~~, ~~expand the squares and collect together all the terms onto one side~~,	+	Wir erweitern alle Terme, und sammeln danach alle Terme auf einer Seite,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 19:		Zeile 18:
	\end{align}</math>}}		\end{align}</math>}}

-	~~This gives that~~ <math>x=2</math>, ~~i.e. the point on the~~ ''x''-~~axis is~~ <math>(2,0)\,</math>.	+	Dies ergibt <math>x=2</math>, und also ist der Punkt auf der ''x''-Achse <math>(2,0)\,</math>.


	[[Image:4_1_4_c-1(2).gif\|center]]		[[Image:4_1_4_c-1(2).gif\|center]]

-		+	Zum Schluss kontrollieren wir ob die beiden Abstände auch wirklich gleich sind. Der Abstand zwischen <math>(2,0)</math> und <math>(3,3)</math>
-	~~As a final step, we check that we have calculated correctly and that the distances really are the same~~. ~~The distance between~~ <math>(2,0)</math> ~~and~~ <math>(3,3)</math>	+	ist
-	is	+

	{{Abgesetzte Formel\|\|<math>\sqrt{(3-2)^2 + (3-0)^2} = \sqrt{1^2+3^2} = \sqrt{1+9} = \sqrt{10}</math>}}		{{Abgesetzte Formel\|\|<math>\sqrt{(3-2)^2 + (3-0)^2} = \sqrt{1^2+3^2} = \sqrt{1+9} = \sqrt{10}</math>}}

-	~~and the distance between~~ <math>(2,0)</math> ~~and~~ <math>(5,1)</math> is	+	und der Abstand zwischen <math>(2,0)</math> und <math>(5,1)</math> ist

	{{Abgesetzte Formel\|\|<math>\sqrt{(5-2)^2 + (1-0)^2} = \sqrt{3^2 + 1^2} = \sqrt{9+1} = \sqrt{10}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\sqrt{(5-2)^2 + (1-0)^2} = \sqrt{3^2 + 1^2} = \sqrt{9+1} = \sqrt{10}\,\textrm{.}</math>}}

-		+	Hinweis: Obwohl wir unsere Gleichung quadriert haben, besteht kein Risiko Scheingleichungen zu erhalten, nachdem die Quadraten in der Wurzel immer positiv sind.
-	~~Note~~: ~~Although we squared our root equation~~, ~~it is not in fact necessary to test the solution for that reason~~, ~~because the expressions under the root signs are sums of squares which are never negative and therefore cannot give rise to so-called spurious roots~~.	+

Tekbot: hat „Solution 4.1:4c“ nach „Lösung 4.1:4c“ verschoben: Robot: moved page

Tekbot — Wed, 22 Oct 2008 14:46:04 GMT

hat „Solution 4.1:4c“ nach „Lösung 4.1:4c“ verschoben: Robot: moved page

← Nächstältere Version

Version vom 14:46, 22. Okt. 2008

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

Tekbot — Wed, 22 Oct 2008 08:48:18 GMT

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

Tek um 10:43, 7. Okt. 2008

Tek — Tue, 07 Oct 2008 10:43:54 GMT

← Nächstältere Version		Version vom 10:43, 7. Okt. 2008
Zeile 1:		Zeile 1:
-	Let the point on the x-axis have coordinates	+	Let the point on the ''x''-axis have coordinates <math>(x,0)</math>, where ''x'' is an unknown number. Then, using the distance formula, the distance from the point
-	<math>~~\left~~( x ~~\right.~~,~~\left.~~ 0 ~~\right~~)</math>, where x is an unknown number. Then, using the distance formula, the distance from the point	+	<math>(x,0)</math> to <math>(3,3)</math> and <math>(5,1)</math> is given by
-	<math>~~\left~~( x ~~\right.~~,~~\left.~~ 0 ~~\right~~)</math>	+
-	to	+
-	<math>~~\left~~( 3 ~~\right.~~,~~\left.~~ 3 ~~\right~~)</math>	+
-	and	+
-	<math>~~\left~~( 5 ~~\right.~~,~~\left.~~ 1 ~~\right~~)</math>	+
-	is given by	+

		+	{{Displayed math\|\|<math>\sqrt{(x-3)^2+(0-3)^2}\qquad\text{and}\qquad \sqrt{(x-5)^2+(0-1)^2}\,\textrm{,}</math>}}

-	~~<math>\sqrt{\left( x-3 \right)^{2}+\left( 0-3 \right)^{2}}</math>~~	+	respectively. Because these two distances should be the same, we get the equation
-	~~and~~	+
-	~~<math>\sqrt{\left( x-5 \right)^{2}+\left( 0-1 \right)^{2}}</math>~~	+
-		+
-		+
-	Because these two distances should be the same, we get the equation	+
-		+
-		+
-	~~<math>\sqrt{\left( x-3 \right)^{2}+9}=\sqrt{\left( x-5 \right)^{2}+1}</math>~~	+

		+	{{Displayed math\|\|<math>\sqrt{(x-3)^2+9} = \sqrt{(x-5)^2+1}</math>}}

	or, if we take the square, so as to get rid of the square root sign,		or, if we take the square, so as to get rid of the square root sign,

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

-	~~<math>\left( x-3 \right)^{2}+9=\left( x-5 \right)^{2}+1</math>~~	+	Now, expand the squares and collect together all the terms onto one side,

		+	{{Displayed math\|\|<math>\begin{align}
		+	& x^2-6x+9+9 = x^2-10x+25+1\\[5pt]
		+	&\quad\Leftrightarrow\quad 4x-8=0\,\textrm{.}
		+	\end{align}</math>}}

-	~~Now~~, ~~expand~~ the ~~squares and collect together all~~ the ~~terms onto one side:~~	+	This gives that <math>x=2</math>, i.e. the point on the ''x''-axis is <math>(2,0)\,</math>.


-	<math>\begin{align}
-	& x^{2}-6x+9+9=x^{2}-10x+25+1 \\
-	& \Leftrightarrow \quad 4x-8=0 \\
-	\end{align}</math>
-
-
-	This gives that
-	<math>x=\text{2}</math>, i.e. the point on the x-axis is
-	<math>\left( 2 \right.,\left. 0 \right)</math>.
-
-
-	{{NAVCONTENT_START}}
	[[Image:4_1_4_c-1(2).gif\|center]]		[[Image:4_1_4_c-1(2).gif\|center]]


-	~~{{NAVCONTENT_STOP}}~~	+	As a final step, we check that we have calculated correctly and that the distances really are the same. The distance between <math>(2,0)</math> and <math>(3,3)</math>
-		+
-	As a final step, we check that we have calculated correctly and that the distances really are the same. The distance between	+
-	<math>~~\left~~( 2 ~~\right.~~,~~\left.~~ 0 ~~\right~~)</math>	+
-	and	+
-	<math>~~\left~~( 3 ~~\right.~~,~~\left.~~ 3 ~~\right~~)</math>	+
	is		is

		+	{{Displayed math\|\|<math>\sqrt{(3-2)^2 + (3-0)^2} = \sqrt{1^2+3^2} = \sqrt{1+9} = \sqrt{10}</math>}}

-	~~<math>\sqrt{\left( 3-2 \right)^{2}+\left( 3-0 \right)^{2}}=\sqrt{1^{2}+3^{2}}=\sqrt{1+9}=\sqrt{10}</math>~~	+	and the distance between <math>(2,0)</math> and <math>(5,1)</math> is
-		+
-		+
-	and the distance between	+
-	<math>~~\left~~( 2 ~~\right.~~,~~\left.~~ 0 ~~\right~~)</math>	+
-	and	+
-	<math>~~\left~~( 5 ~~\right.~~,~~\left.~~ 1 ~~\right~~)</math>	+
-	is	+
-		+

-	<math>\sqrt{~~\left~~( 5-2 ~~\right~~)^{2}+~~\left~~( 1-0 ~~\right~~)^{2}}=\sqrt{3^{2}+1^{2}}=\sqrt{9+1}=\sqrt{10}</math>	+	{{Displayed math\|\|<math>\sqrt{(5-2)^2 + (1-0)^2} = \sqrt{3^2 + 1^2} = \sqrt{9+1} = \sqrt{10}\,\textrm{.}</math>}}


-	~~NOTE~~: Although we squared our root equation, it is not in fact necessary to test the solution for that reason, because the expressions under the root signs are sums of squares which are never negative and therefore cannot give rise to so-called ~~false~~ roots.	+	Note: Although we squared our root equation, it is not in fact necessary to test the solution for that reason, because the expressions under the root signs are sums of squares which are never negative and therefore cannot give rise to so-called spurious roots.

Ian um 10:15, 27. Sep. 2008

Ian — Sat, 27 Sep 2008 10:15:57 GMT

← Nächstältere Version		Version vom 10:15, 27. Sep. 2008
Zeile 1:		Zeile 1:
		+	Let the point on the x-axis have coordinates
		+	<math>\left( x \right.,\left. 0 \right)</math>, where x is an unknown number. Then, using the distance formula, the distance from the point
		+	<math>\left( x \right.,\left. 0 \right)</math>
		+	to
		+	<math>\left( 3 \right.,\left. 3 \right)</math>
		+	and
		+	<math>\left( 5 \right.,\left. 1 \right)</math>
		+	is given by
		+
		+
		+	<math>\sqrt{\left( x-3 \right)^{2}+\left( 0-3 \right)^{2}}</math>
		+	and
		+	<math>\sqrt{\left( x-5 \right)^{2}+\left( 0-1 \right)^{2}}</math>
		+
		+
		+	Because these two distances should be the same, we get the equation
		+
		+
		+	<math>\sqrt{\left( x-3 \right)^{2}+9}=\sqrt{\left( x-5 \right)^{2}+1}</math>
		+
		+
		+	or, if we take the square, so as to get rid of the square root sign,
		+
		+
		+	<math>\left( x-3 \right)^{2}+9=\left( x-5 \right)^{2}+1</math>
		+
		+
		+	Now, expand the squares and collect together all the terms onto one side:
		+
		+
		+	<math>\begin{align}
		+	& x^{2}-6x+9+9=x^{2}-10x+25+1 \\
		+	& \Leftrightarrow \quad 4x-8=0 \\
		+	\end{align}</math>
		+
		+
		+	This gives that
		+	<math>x=\text{2}</math>, i.e. the point on the x-axis is
		+	<math>\left( 2 \right.,\left. 0 \right)</math>.
		+
		+
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
	[[Image:4_1_4_c-1(2).gif\|center]]		[[Image:4_1_4_c-1(2).gif\|center]]

-	~~<center> [[Image:4_1_4c-1(2).gif]] </center>~~	+
-	~~{{NAVCONTENT_STOP}}~~	+
-	~~{{NAVCONTENT_START}}~~	+
-	~~<center> [[Image:4_1_4c-2(2).gif]] </center>~~	+
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
		+
		+	As a final step, we check that we have calculated correctly and that the distances really are the same. The distance between
		+	<math>\left( 2 \right.,\left. 0 \right)</math>
		+	and
		+	<math>\left( 3 \right.,\left. 3 \right)</math>
		+	is
		+
		+
		+	<math>\sqrt{\left( 3-2 \right)^{2}+\left( 3-0 \right)^{2}}=\sqrt{1^{2}+3^{2}}=\sqrt{1+9}=\sqrt{10}</math>
		+
		+
		+	and the distance between
		+	<math>\left( 2 \right.,\left. 0 \right)</math>
		+	and
		+	<math>\left( 5 \right.,\left. 1 \right)</math>
		+	is
		+
		+
		+	<math>\sqrt{\left( 5-2 \right)^{2}+\left( 1-0 \right)^{2}}=\sqrt{3^{2}+1^{2}}=\sqrt{9+1}=\sqrt{10}</math>
		+
		+
		+	NOTE: Although we squared our root equation, it is not in fact necessary to test the solution for that reason, because the expressions under the root signs are sums of squares which are never negative and therefore cannot give rise to so-called false roots.

Tekbot: Lösning 4.1:4c moved to Solution 4.1:4c: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:35:43 GMT

Lösning 4.1:4c moved to Solution 4.1:4c: Robot: moved page

← Nächstältere Version

Version vom 09:35, 9. Sep. 2008

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

Tekbot — Thu, 21 Aug 2008 07:04:59 GMT

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

← Nächstältere Version		Version vom 07:04, 21. Aug. 2008
Zeile 1:		Zeile 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	[[~~Bild~~:4_1_4_c-1(2).gif\|center]]	+	[[Image:4_1_4_c-1(2).gif\|center]]

-	<center> [[~~Bild~~:4_1_4c-1(2).gif]] </center>	+	<center> [[Image:4_1_4c-1(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:4_1_4c-2(2).gif]] </center>	+	<center> [[Image:4_1_4c-2(2).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

← Nächstältere Version		Version vom 13:45, 16. Jun. 2009
Zeile 1:		Zeile 1:
-	Die ''x''-Achse besteht aus allen Punkten <math>(x,0)</math>, wo ''x'' ~~unbekannt ist~~. Mit der Formel für den Abstand zwischen zwei Punkten, erhalten wir den Abstand zwischen <math>(x,0)</math> und <math>(3,3)</math> ~~und~~ zwischen <math>(5,1)</math> und <math>(x,0)</math>,	+	Die ''x''-Achse besteht aus allen Punkten <math>(x,0)</math>, wobei ''x'' die Unbekannte sei. Mit der Formel für den Abstand zwischen zwei Punkten erhalten wir den Abstand zwischen <math>(x,0)</math> und <math>(3,3)</math> sowie zwischen <math>(5,1)</math> und <math>(x,0)</math>:

-	{{Abgesetzte Formel\|\|<math>\sqrt{(x-3)^2+(0-3)^2}\qquad\text{und}\qquad \sqrt{(x-5)^2+(0-1)^2}\,\textrm{,}</math>}}	+	{{Abgesetzte Formel\|\|<math>\sqrt{(x-3)^2+(0-3)^2}\qquad\text{und}\qquad \sqrt{(x-5)^2+(0-1)^2}\,\textrm{.}</math>}}

-	Nachdem die beiden Abstände gleich sein sollen, erhalten wir	+	Nachdem die beiden Abstände gleich sein sollen, erhalten wir:

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

	Wir quadrieren diese Gleichung und erhalten		Wir quadrieren diese Gleichung und erhalten
Zeile 11:		Zeile 11:
	{{Abgesetzte Formel\|\|<math>(x-3)^2 + 9 = (x-5)^2+1\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>(x-3)^2 + 9 = (x-5)^2+1\,\textrm{.}</math>}}

-	Wir erweitern alle Terme, und ~~sammeln~~ danach alle ~~Terme~~ auf ~~einer~~ Seite,	+	Wir erweitern alle Terme und schreiben danach alle auf eine Seite,

	{{Abgesetzte Formel\|\|<math>\begin{align}		{{Abgesetzte Formel\|\|<math>\begin{align}
Zeile 18:		Zeile 18:
	\end{align}</math>}}		\end{align}</math>}}

-	Dies ergibt <math>x=2</math>, ~~und~~ also ist der Punkt auf der ''x''-Achse <math>(2,0)\,</math>.	+	Dies ergibt <math>x=2</math>, also ist der Punkt auf der ''x''-Achse <math>(2,0)\,</math>.


	[[Image:4_1_4_c-1(2).gif\|center]]		[[Image:4_1_4_c-1(2).gif\|center]]

-	Zum Schluss kontrollieren wir ob die beiden Abstände auch wirklich gleich sind. Der Abstand zwischen <math>(2,0)</math> und <math>(3,3)</math>	+	Zum Schluss kontrollieren wir, ob die beiden Abstände auch wirklich gleich sind. Der Abstand zwischen <math>(2,0)</math> und <math>(3,3)</math>
	ist		ist

Zeile 32:		Zeile 32:
	{{Abgesetzte Formel\|\|<math>\sqrt{(5-2)^2 + (1-0)^2} = \sqrt{3^2 + 1^2} = \sqrt{9+1} = \sqrt{10}\,\textrm{.}</math>}}		{{Abgesetzte Formel\|\|<math>\sqrt{(5-2)^2 + (1-0)^2} = \sqrt{3^2 + 1^2} = \sqrt{9+1} = \sqrt{10}\,\textrm{.}</math>}}

-	Hinweis: Obwohl wir unsere Gleichung quadriert haben, besteht kein Risiko Scheingleichungen zu erhalten, nachdem die ~~Quadraten~~ in der Wurzel immer positiv sind.	+	Hinweis: Obwohl wir unsere Gleichung quadriert haben, besteht kein Risiko Scheingleichungen zu erhalten, nachdem die Quadrate in der Wurzel immer positiv sind.

← Nächstältere Version		Version vom 08:48, 22. Okt. 2008
Zeile 2:		Zeile 2:
	<math>(x,0)</math> to <math>(3,3)</math> and <math>(5,1)</math> is given by		<math>(x,0)</math> to <math>(3,3)</math> and <math>(5,1)</math> is given by

-	{{~~Displayed math~~\|\|<math>\sqrt{(x-3)^2+(0-3)^2}\qquad\text{and}\qquad \sqrt{(x-5)^2+(0-1)^2}\,\textrm{,}</math>}}	+	{{Abgesetzte Formel\|\|<math>\sqrt{(x-3)^2+(0-3)^2}\qquad\text{and}\qquad \sqrt{(x-5)^2+(0-1)^2}\,\textrm{,}</math>}}

	respectively. Because these two distances should be the same, we get the equation		respectively. Because these two distances should be the same, we get the equation

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

	or, if we take the square, so as to get rid of the square root sign,		or, if we take the square, so as to get rid of the square root sign,

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

	Now, expand the squares and collect together all the terms onto one side,		Now, expand the squares and collect together all the terms onto one side,

-	{{~~Displayed math~~\|\|<math>\begin{align}	+	{{Abgesetzte Formel\|\|<math>\begin{align}
	& x^2-6x+9+9 = x^2-10x+25+1\\[5pt]		& x^2-6x+9+9 = x^2-10x+25+1\\[5pt]
	&\quad\Leftrightarrow\quad 4x-8=0\,\textrm{.}		&\quad\Leftrightarrow\quad 4x-8=0\,\textrm{.}
Zeile 28:		Zeile 28:
	is		is

-	{{~~Displayed math~~\|\|<math>\sqrt{(3-2)^2 + (3-0)^2} = \sqrt{1^2+3^2} = \sqrt{1+9} = \sqrt{10}</math>}}	+	{{Abgesetzte Formel\|\|<math>\sqrt{(3-2)^2 + (3-0)^2} = \sqrt{1^2+3^2} = \sqrt{1+9} = \sqrt{10}</math>}}

	and the distance between <math>(2,0)</math> and <math>(5,1)</math> is		and the distance between <math>(2,0)</math> and <math>(5,1)</math> is

-	{{~~Displayed math~~\|\|<math>\sqrt{(5-2)^2 + (1-0)^2} = \sqrt{3^2 + 1^2} = \sqrt{9+1} = \sqrt{10}\,\textrm{.}</math>}}	+	{{Abgesetzte Formel\|\|<math>\sqrt{(5-2)^2 + (1-0)^2} = \sqrt{3^2 + 1^2} = \sqrt{9+1} = \sqrt{10}\,\textrm{.}</math>}}


	Note: Although we squared our root equation, it is not in fact necessary to test the solution for that reason, because the expressions under the root signs are sums of squares which are never negative and therefore cannot give rise to so-called spurious roots.		Note: Although we squared our root equation, it is not in fact necessary to test the solution for that reason, because the expressions under the root signs are sums of squares which are never negative and therefore cannot give rise to so-called spurious roots.