Aus Online Mathematik Brückenkurs 1

Version vom 11:31, 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:
	By completing the square, we can rewrite the ''x''- and ''y''-terms as quadratic expressions,		By completing the square, we can rewrite the ''x''- and ''y''-terms as quadratic expressions,
-	{{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 + 6y &= (y+3)^2 - 3^2\,,		y^2 + 6y &= (y+3)^2 - 3^2\,,
Zeile 8:		Zeile 8:
	and the whole equation then has standard form,		and the whole equation then has standard form,
-	{{Displayed math||<math>\begin{align}	+	{{Abgesetzte Formel||<math>\begin{align}
	(x-1)^2 - 1 + (y+3)^2 - 9 &= -3\,,\\[5pt]		(x-1)^2 - 1 + (y+3)^2 - 9 &= -3\,,\\[5pt]
	(x-1)^2 + (y+3)^2 &= 7\,\textrm{.}		(x-1)^2 + (y+3)^2 &= 7\,\textrm{.}

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Version vom 08:49, 22. Okt. 2008

By completing the square, we can rewrite the x- and y-terms as quadratic expressions,

\displaystyle \begin{align} x^2 - 2x &= (x-1)^2 - 1^2\,,\\[5pt] y^2 + 6y &= (y+3)^2 - 3^2\,, \end{align}

and the whole equation then has standard form,

\displaystyle \begin{align} (x-1)^2 - 1 + (y+3)^2 - 9 &= -3\,,\\[5pt] (x-1)^2 + (y+3)^2 &= 7\,\textrm{.} \end{align}

From this, we see that the circle has its centre at (1,-3) and radius \displaystyle \sqrt{7}\,.