Solution 4.1:7d - Revision history

Tek at 11:42, 8 October 2008

Tek — Wed, 08 Oct 2008 11:42:50 GMT

←Older revision		Revision as of 11:42, 8 October 2008
Line 1:		Line 1:
-	We rewrite the equation in standard by completing the square for the x- and y-terms:	+	We rewrite the equation in standard form by completing the square for the ''x''- and ''y''-terms,
-		+
-		+
-	~~<math>x^{2}-2x=\left( x-1 \right)^{2}-1^{2}</math>~~	+
-		+
-		+
-	~~<math>y^{2}+2y=\left( y+1 \right)^{2}-1^{2}</math>~~	+

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

	Now, the equation is		Now, the equation is

		+	{{Displayed math\|\|<math>\begin{align}
		+	(x-1)^2 - 1 + (y+1)^2 - 1 &= -2\\
		+	\Leftrightarrow\quad (x-1)^2 + (y+1)^2 &= 0\,\textrm{.}
		+	\end{align}</math>}}

-	<math>~~\begin{align}~~	+	The only point which satisfies this equation is <math>(x,y) = (1,-1)</math> because, for all other values of ''x'' and ''y'', the left-hand side is strictly positive and therefore not zero.
-	~~& \left~~( x~~-1 \right)^{2}-1+\left(~~ y~~+1 \right~~)~~^{2}-1~~=~~-2 \\~~	+
-	~~& \Leftrightarrow \quad \left~~( x-1 ~~\right~~)~~^{2}+\left( y+1 \right)^{2}=0 \\~~	+
-	~~\end{align}~~</math>	+


-	The only point which satisfies this equation is
-	<math>\left( x \right.,\left. y \right)=\left( 1 \right.,\left. -1 \right)</math>
-	because, for all other values of
-	<math>x</math>
-	and
-	<math>y</math> , the left-hand side is strictly positive and therefore not zero.
-
-
-	{{NAVCONTENT_START}}
	<center> [[Image:4_1_7_d.gif]] </center>		<center> [[Image:4_1_7_d.gif]] </center>
-
-
-	{{NAVCONTENT_STOP}}

Ian — Sat, 27 Sep 2008 12:12:49 GMT

←Older revision		Revision as of 12:12, 27 September 2008
Line 1:		Line 1:
		+	We rewrite the equation in standard by completing the square for the x- and y-terms:
		+
		+
		+	<math>x^{2}-2x=\left( x-1 \right)^{2}-1^{2}</math>
		+
		+
		+	<math>y^{2}+2y=\left( y+1 \right)^{2}-1^{2}</math>
		+
		+
		+	Now, the equation is
		+
		+
		+	<math>\begin{align}
		+	& \left( x-1 \right)^{2}-1+\left( y+1 \right)^{2}-1=-2 \\
		+	& \Leftrightarrow \quad \left( x-1 \right)^{2}+\left( y+1 \right)^{2}=0 \\
		+	\end{align}</math>
		+
		+
		+	The only point which satisfies this equation is
		+	<math>\left( x \right.,\left. y \right)=\left( 1 \right.,\left. -1 \right)</math>
		+	because, for all other values of
		+	<math>x</math>
		+	and
		+	<math>y</math> , the left-hand side is strictly positive and therefore not zero.
		+
		+
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
	<center> [[Image:4_1_7_d.gif]] </center>		<center> [[Image:4_1_7_d.gif]] </center>

-	~~<center> [[Image:4_1_7d.gif]] </center>~~	+
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Tekbot — Tue, 09 Sep 2008 09:38:43 GMT

←Older revision

Revision as of 09:38, 9 September 2008

Tekbot — Thu, 21 Aug 2008 07:06:29 GMT

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

Martin — Fri, 15 Aug 2008 14:29:19 GMT

Oskar — Wed, 02 Apr 2008 13:32:16 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:4_1_7d.gif </center> {{NAVCONTENT_STOP}}

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