Aus Online Mathematik Brückenkurs 1

Version vom 11:42, 8. Okt. 2008 (bearbeiten) Tek (Diskussion | Beiträge) K ← Zum vorherigen Versionsunterschied		Version vom 08:49, 22. Okt. 2008 (bearbeiten) (rückgängig) Tekbot (Diskussion | Beiträge) K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) Zum nächsten Versionsunterschied →
Zeile 1:		Zeile 1:
	We rewrite the equation in standard form 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,
-	{{Displayed math||<math>\begin{align}	+	{{Abgesetzte Formel||<math>\begin{align}
	x^{2} - 2x &= (x-1)^2 - 1^2\,,\\[5pt]		x^{2} - 2x &= (x-1)^2 - 1^2\,,\\[5pt]
	y^{2} + 2y &= (y+1)^2 - 1^2\,\textrm{.}		y^{2} + 2y &= (y+1)^2 - 1^2\,\textrm{.}
Zeile 8:		Zeile 8:
	Now, the equation is		Now, the equation is
-	{{Displayed math||<math>\begin{align}	+	{{Abgesetzte Formel||<math>\begin{align}
	(x-1)^2 - 1 + (y+1)^2 - 1 &= -2\\		(x-1)^2 - 1 + (y+1)^2 - 1 &= -2\\
	\Leftrightarrow\quad (x-1)^2 + (y+1)^2 &= 0\,\textrm{.}		\Leftrightarrow\quad (x-1)^2 + (y+1)^2 &= 0\,\textrm{.}

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Version vom 08:49, 22. Okt. 2008

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.