Solution 2.3:9b - Revision history

Tek at 14:11, 29 September 2008

2008-09-29T14:11:17Z

←Older revision		Revision as of 14:11, 29 September 2008
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-	The points of intersection are those points on the curve which also lie on the	+	The points of intersection are those points on the curve which also lie on the ''x''-axis, i.e. they are those points which satisfy both the equation of the curve
-	~~<math>~~~x~~</math>~~	+	<math>y=x^{2}-5x+6</math> and the equation of the ''x''-axis <math>y=0</math>,
-	-axis, i.e. they are those points which satisfy both the equation of the curve	+
-	<math>y=x^~~{\text~~{2}}-~~\text{5}x~~+~~\text{~~6}</math>	+
-	and the equation of the	+
-	~~<math>~~~x~~</math>~~	+
-	-axis	+
-	<math>y=0</math>,	+

		+	{{Displayed math\|\|<math>\left\{\begin{align}
		+	y&=x^{2}-5x+6\,,\\
		+	y&=0\,\textrm{.}
		+	\end{align}\right.</math>}}

-	<math>~~\left\{ \begin{matrix}~~	+	This system of equations gives directly that <math>y=0</math> and that <math>x</math> must satisfy the quadratic equation <math>x^{2}-5x+6=0\,</math>. By completing the square, we obtain that the left-hand side is
-	y=x^~~{\text~~{2}}-~~\text{5}x~~+~~\text{~~6~~} \\~~	+
-	y=0\~~quad \quad \quad \quad \\~~	+
-	~~\end{matrix} \right.~~</math>	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	x^{2} - 5x + 6 &= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} + 6\\[5pt]
		+	&= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{24}{4}\\[5pt]
		+	&= \Bigl(x-\frac{5}{2}\Bigr)^{2} - \frac{1}{4}
		+	\end{align}</math>}}

-	~~This system of equations~~ gives ~~directly~~ that	+	and this gives that the equation has solutions <math>x=\tfrac{5}{2}\pm\tfrac{1}{2}</math>, i.e. <math>x=\tfrac{5}{2}-\tfrac{1}{2}=\tfrac{4}{2}=2</math> and <math>x=\tfrac{5}{2}+\tfrac{1}{2}=\tfrac{6}{2}=3</math>.
-	<math>y=0</math>	+
-	~~and that~~	+
-	<math>~x</math>	+
-	~~must satisfy the second-order equation~~	+
-	<math>x^{\~~text~~{2}}-\~~text~~{5}x+\~~text~~{6}=0</math>	+
-	. ~~By completing the square, we obtain that the left-hand side is~~	+

		+	The intersection points are therefore (2,0) and (3,0).

-	<math>\begin{align}
-	& x^{\text{2}}-\text{5}x+\text{6}=\left( x-\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}+6 \\
-	& =\left( x-\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{24}{4}=\left( x-\frac{5}{2} \right)^{2}-\frac{1}{4} \\
-	\end{align}</math>

-		+	<center>[[Image:2_3_9b-2(2).gif]]</center>
-	~~and this gives that the equation has solutions~~	+
-		+
-	~~<math>x=\frac{5}{2}\pm \frac{1}{2}</math>, i.e.~~	+
-	~~<math>x=\frac{5}{2}-\frac{1}{2}=\frac{4}{2}=2</math>~~	+
-	~~and~~	+
-	~~<math>x=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3</math>.~~	+
-		+
-	~~The intersection points are therefore~~	+
-	~~<math>\left( 2 \right.,\left. 0 \right)</math>~~	+
-	~~and~~	+
-	~~<math>\left( 3 \right.,\left. 0 \right)</math>.~~	+
-		+
-		+
-		+
-	~~{{NAVCONTENT_START}}~~	+
-	<center> [[Image:2_3_9b-2(2).gif]] </center>	+
-	~~{{NAVCONTENT_STOP}}~~	+

Ian at 11:57, 21 September 2008

2008-09-21T11:57:08Z

←Older revision		Revision as of 11:57, 21 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	The points of intersection are those points on the curve which also lie on the
-	<~~center~~> ~~[[Image:2_3_9b~~-1(2).~~gif]]~~ </~~center~~>	+	<math>~x</math>
-	{{~~NAVCONTENT_STOP~~}}	+	-axis, i.e. they are those points which satisfy both the equation of the curve
		+	<math>y=x^{\text{2}}-\text{5}x+\text{6}</math>
		+	and the equation of the
		+	<math>~x</math>
		+	-axis
		+	<math>y=0</math>,
		+
		+
		+	<math>\left\{ \begin{matrix}
		+	y=x^{\text{2}}-\text{5}x+\text{6} \\
		+	y=0\quad \quad \quad \quad \\
		+	\end{matrix} \right.</math>
		+
		+
		+	This system of equations gives directly that
		+	<math>y=0</math>
		+	and that
		+	<math>~x</math>
		+	must satisfy the second-order equation
		+	<math>x^{\text{2}}-\text{5}x+\text{6}=0</math>
		+	. By completing the square, we obtain that the left-hand side is
		+
		+
		+	<math>\begin{align}
		+	& x^{\text{2}}-\text{5}x+\text{6}=\left( x-\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}+6 \\
		+	& =\left( x-\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{24}{4}=\left( x-\frac{5}{2} \right)^{2}-\frac{1}{4} \\
		+	\end{align}</math>
		+
		+
		+	and this gives that the equation has solutions
		+
		+	<math>x=\frac{5}{2}\pm \frac{1}{2}</math>, i.e.
		+	<math>x=\frac{5}{2}-\frac{1}{2}=\frac{4}{2}=2</math>
		+	and
		+	<math>x=\frac{5}{2}+\frac{1}{2}=\frac{6}{2}=3</math>.
		+
		+	The intersection points are therefore
		+	<math>\left( 2 \right.,\left. 0 \right)</math>
		+	and
		+	<math>\left( 3 \right.,\left. 0 \right)</math>.
		+
		+
		+
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Tekbot: Lösning 2.3:9b moved to Solution 2.3:9b: Robot: moved page

2008-09-09T09:05:23Z

Lösning 2.3:9b moved to Solution 2.3:9b: Robot: moved page

←Older revision

Revision as of 09:05, 9 September 2008

Tekbot: Robot: Automated text replacement (-[[Bild: +[[Image:)

2008-08-21T06:49:49Z

Robot: Automated text replacement (-[[Bild: +[[Image:)

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2008-04-01T10:28:51Z

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