Solution 2.3:1c - Revision history

Tek at 14:08, 26 September 2008

2008-09-26T14:08:32Z

←Older revision		Revision as of 14:08, 26 September 2008
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	<math>2x-x^{2}</math>, which we also can write as <math>-(x^{2}-2x)</math>. If we neglect the minus sign, we can complete square of the expression <math>2x-x^{2}</math> by using the formula		<math>2x-x^{2}</math>, which we also can write as <math>-(x^{2}-2x)</math>. If we neglect the minus sign, we can complete square of the expression <math>2x-x^{2}</math> by using the formula

-	{{Displayed math\|\|<math>x^{2}-ax = \~~biggl~~(x-\frac{a}{2}\~~biggr~~)^{2} - \~~biggl~~(\frac{a}{2}\~~biggr~~)^{2}</math>}}	+	{{Displayed math\|\|<math>x^{2}-ax = \Bigl(x-\frac{a}{2}\Bigr)^{2} - \Bigl(\frac{a}{2}\Bigr)^{2}</math>}}

	and we obtain		and we obtain

-	{{Displayed math\|\|<math>x^{2}-2x = \~~biggl~~(x-\frac{2}{2}\~~biggr~~)^{2} - \~~biggl~~(\frac{2}{2}\~~biggr~~)^{2} = (x-1)^{2}-1\,\textrm{.}</math>}}	+	{{Displayed math\|\|<math>x^{2}-2x = \Bigl(x-\frac{2}{2}\Bigr)^{2} - \Bigl(\frac{2}{2}\Bigr)^{2} = (x-1)^{2}-1\,\textrm{.}</math>}}

	This means that		This means that

Tek at 13:53, 26 September 2008

2008-09-26T13:53:06Z

←Older revision		Revision as of 13:53, 26 September 2008
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	As always when completing the square, we focus on the quadratic and linear terms		As always when completing the square, we focus on the quadratic and linear terms
-	<math>2x-x^{2}</math>, which we also can write as	+	<math>2x-x^{2}</math>, which we also can write as <math>-(x^{2}-2x)</math>. If we neglect the minus sign, we can complete square of the expression <math>2x-x^{2}</math> by using the formula
-	<math>-~~\left~~( x^{2}-2x ~~\right~~)</math>	+
-	. If we neglect the minus sign, we can complete square of the expression	+
-	<math>2x-x^{2}</math>	+
-	by using the formula	+
-		+
-		+
-	~~<math>x^{2}-ax=\left( x-\frac{a}{2} \right)^{2}-\left( \frac{a}{2} \right)^{2}</math>~~	+

		+	{{Displayed math\|\|<math>x^{2}-ax = \biggl(x-\frac{a}{2}\biggr)^{2} - \biggl(\frac{a}{2}\biggr)^{2}</math>}}

	and we obtain		and we obtain

-		+	{{Displayed math\|\|<math>x^{2}-2x = \biggl(x-\frac{2}{2}\biggr)^{2} - \biggl(\frac{2}{2}\biggr)^{2} = (x-1)^{2}-1\,\textrm{.}</math>}}
-	<math>x^{2}-2x=\~~left~~( x-\frac{2}{2} \~~right~~)^{2}-\~~left~~( \frac{2}{2} \~~right~~)^{2}=~~\left~~( x-1 ~~\right~~)^{2}-1</math>	+
-		+

	This means that		This means that

		+	{{Displayed math\|\|<math>\begin{align}
		+	5+2x-x^{2} &= 5-(x^{2}-2x) = 5-\bigl((x-1)^{2}-1\bigr)\\[5pt]
		+	&= 5-(x-1)^{2}+1 = 6-(x-1)^{2}\textrm{.}
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	A quick check shows that we have completed the square correctly
-	~~& 5+2x-x^{2}=5-\left( x^{2}-2x \right)=5-\left( \left( x-1 \right)^{2}-1 \right) \\~~	+
-	~~& \\~~	+
-	~~& =5-\left( x-1 \right)^{2}+1=6-\left( x-1 \right)^{2} \\~~	+
-	~~& \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	A quick check shows that we have completed the square correctly.:	+
-		+

-	<math>\begin{align}	+	{{Displayed math\|\|<math>\begin{align}
-	& 6-~~\left~~( x-1 ~~\right~~)^{2}=6-~~\left~~( x^{2}-2x+1 \~~right)~~=6-x^{2}+2x-1 \\	+	6-(x-1)^{2}
-	~~& \\~~	+	&= 6-(x^{2}-2x+1)\\[5pt]
-	& =5+2x-x^{2} \\	+	&= 6-x^{2}+2x-1\\[5pt]
-	~~& \\~~	+	& =5+2x-x^{2}\textrm{.}
-	\end{align}</math>	+	\end{align}</math>}}

Ian at 11:20, 12 September 2008

2008-09-12T11:20:54Z

←Older revision		Revision as of 11:20, 12 September 2008
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	As always when completing the square, we focus on the quadratic and linear terms		As always when completing the square, we focus on the quadratic and linear terms
-	<math>2x-x^{2}</math>	+	<math>2x-x^{2}</math>, which we also can write as
-	, which we also can write as	+
	<math>-\left( x^{2}-2x \right)</math>		<math>-\left( x^{2}-2x \right)</math>
	. If we neglect the minus sign, we can complete square of the expression		. If we neglect the minus sign, we can complete square of the expression

Ian at 11:19, 12 September 2008

2008-09-12T11:19:48Z

←Older revision		Revision as of 11:19, 12 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	As always when completing the square, we focus on the quadratic and linear terms
-	<~~center~~> ~~[[Image:2_3_1c.gif]]~~ </~~center~~>	+	<math>2x-x^{2}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	, which we also can write as
		+	<math>-\left( x^{2}-2x \right)</math>
		+	. If we neglect the minus sign, we can complete square of the expression
		+	<math>2x-x^{2}</math>
		+	by using the formula
		+
		+
		+	<math>x^{2}-ax=\left( x-\frac{a}{2} \right)^{2}-\left( \frac{a}{2} \right)^{2}</math>
		+
		+
		+	and we obtain
		+
		+
		+	<math>x^{2}-2x=\left( x-\frac{2}{2} \right)^{2}-\left( \frac{2}{2} \right)^{2}=\left( x-1 \right)^{2}-1</math>
		+
		+
		+	This means that
		+
		+
		+	<math>\begin{align}
		+	& 5+2x-x^{2}=5-\left( x^{2}-2x \right)=5-\left( \left( x-1 \right)^{2}-1 \right) \\
		+	& \\
		+	& =5-\left( x-1 \right)^{2}+1=6-\left( x-1 \right)^{2} \\
		+	& \\
		+	\end{align}</math>
		+
		+
		+	A quick check shows that we have completed the square correctly.:
		+
		+
		+	<math>\begin{align}
		+	& 6-\left( x-1 \right)^{2}=6-\left( x^{2}-2x+1 \right)=6-x^{2}+2x-1 \\
		+	& \\
		+	& =5+2x-x^{2} \\
		+	& \\
		+	\end{align}</math>

Tekbot: Lösning 2.3:1c moved to Solution 2.3:1c: Robot: moved page

2008-09-09T08:55:23Z

Lösning 2.3:1c moved to Solution 2.3:1c: Robot: moved page

←Older revision

Revision as of 08:55, 9 September 2008

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

2008-08-21T06:44:49Z

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

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2008-04-01T09:54:49Z

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