Processing Math: Done

From Förberedande kurs i matematik 1

Revision as of 09:38, 9 September 2008 (edit) Tekbot (Talk | contribs) m (Lösning 4.1:7d moved to Solution 4.1:7d: Robot: moved page) ← Previous diff		Current revision (11:42, 8 October 2008) (edit) (undo) Tek (Talk | contribs) m
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-	{{NAVCONTENT_START}}	+	We rewrite the equation in standard form by completing the square for the ''x''- and ''y''-terms,
-	<center> [[Image:4_1_7_d.gif]] </center>	+
-	<center> [[Image:4_1_7d.gif]] </center>	+	{{Displayed math||<math>\begin{align}
-	{{NAVCONTENT_STOP}}	+	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
		+
		+	{{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>}}
		+
		+	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.
		+
		+
		+	<center> [[Image:4_1_7_d.gif]] </center>

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Current revision

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

x2−2xy2+2y=(x−1)2−12=(y+1)2−12.

Now, the equation is

(x−1)2−1+(y+1)2−1(x−1)2+(y+1)2=−2=0.

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