Aus Online Mathematik Brückenkurs 1

Version vom 09:38, 9. Sep. 2008 (bearbeiten) Tekbot (Diskussion | Beiträge) K (Lösning 4.1:7d moved to Solution 4.1:7d: Robot: moved page) ← Zum vorherigen Versionsunterschied		Version vom 12:12, 27. Sep. 2008 (bearbeiten) (rückgängig) Ian (Diskussion | Beiträge) Zum nächsten Versionsunterschied →
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		+	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.
		+
		+
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	<center> [[Image:4_1_7_d.gif]] </center>		<center> [[Image:4_1_7_d.gif]] </center>
-	<center> [[Image:4_1_7d.gif]] </center>	+
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Version vom 12:12, 27. Sep. 2008

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

\displaystyle x^{2}-2x=\left( x-1 \right)^{2}-1^{2}

\displaystyle y^{2}+2y=\left( y+1 \right)^{2}-1^{2}

Now, the equation is

\displaystyle \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}

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