Solution 4.1:4c - Revision history

Tek at 10:43, 7 October 2008

2008-10-07T10:43:54Z

←Older revision		Revision as of 10:43, 7 October 2008
Line 1:		Line 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 at 10:15, 27 September 2008

2008-09-27T10:15:57Z

←Older revision		Revision as of 10:15, 27 September 2008
Line 1:		Line 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

2008-09-09T09:35:43Z

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

←Older revision

Revision as of 09:35, 9 September 2008

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

2008-08-21T07:04:59Z

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

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Martin at 13:26, 15 August 2008

2008-08-15T13:26:10Z

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Oskar: Ny sida: {{NAVCONTENT_START}} Bild:4_1_4c-1(2).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:4_1_4c-2(2).gif {{NAVCONTENT_STOP}}

2008-04-02T13:00:30Z

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