From Förberedande kurs i matematik 1

Revision as of 12:06, 27 September 2008 (edit) Ian (Talk | contribs) ← Previous diff		Current revision (11:31, 8 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	By completing the square, we can rewrite the	+	By completing the square, we can rewrite the ''x''- and ''y''-terms as quadratic expressions,
-	<math>x</math>	+
-	- and	+
-	<math>y</math>	+
-	-terms as quadratic expressions,	+
-		+	{{Displayed math||<math>\begin{align}
-	<math>x^{2}-2x=\left( x-1 \right)^{2}-1^{2}</math>	+	x^2 - 2x &= (x-1)^2 - 1^2\,,\\[5pt]
-		+	y^2 + 6y &= (y+3)^2 - 3^2\,,
-		+	\end{align}</math>}}
-	<math>y^{2}+6y=\left( y+3 \right)^{2}-3^{2}</math>	+
	and the whole equation then has standard form,		and the whole equation then has standard form,
		+	{{Displayed math||<math>\begin{align}
		+	(x-1)^2 - 1 + (y+3)^2 - 9 &= -3\,,\\[5pt]
		+	(x-1)^2 + (y+3)^2 &= 7\,\textrm{.}
		+	\end{align}</math>}}
-	<math>\begin{align}	+	From this, we see that the circle has its centre at (1,-3) and radius <math>\sqrt{7}\,</math>.
-	& \left( x-1 \right)^{2}-1+\left( y+3 \right)^{2}-9=-3 \\	+
-	& \Leftrightarrow \quad \left( x-1 \right)^{2}+\left( y+3 \right)^{2}=7 \\	+
-	\end{align}</math>	+
-
-	From this, we see that the circle has its centre at
-	<math>\left( 1 \right.,\left. -3 \right)</math>
-	and radius
-	<math>\sqrt{7}</math>.
-
-
-	{{NAVCONTENT_START}}
	<center> [[Image:4_1_7_c.gif]] </center>		<center> [[Image:4_1_7_c.gif]] </center>
-
-	{{NAVCONTENT_STOP}}

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Current revision

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}\,.