From Förberedande kurs i matematik 1

Revision as of 12:12, 27 September 2008 (edit) Ian (Talk | contribs) ← Previous diff		Current revision (11:42, 8 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	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.
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-
-	{{NAVCONTENT_START}}
	<center> [[Image:4_1_7_d.gif]] </center>		<center> [[Image:4_1_7_d.gif]] </center>
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-	{{NAVCONTENT_STOP}}

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Current revision

We rewrite the equation in standard form by completing the square for the x- and y-terms,

\displaystyle \begin{align} x^{2} - 2x &= (x-1)^2 - 1^2\,,\\[5pt] y^{2} + 2y &= (y+1)^2 - 1^2\,\textrm{.} \end{align}

Now, the equation is

\displaystyle \begin{align} (x-1)^2 - 1 + (y+1)^2 - 1 &= -2\\ \Leftrightarrow\quad (x-1)^2 + (y+1)^2 &= 0\,\textrm{.} \end{align}

The only point which satisfies this equation is \displaystyle (x,y) = (1,-1) because, for all other values of x and y, the left-hand side is strictly positive and therefore not zero.