Solution 2.3:1b - Revision history

Tek at 14:07, 26 September 2008

Tek — Fri, 26 Sep 2008 14:07:35 GMT

←Older revision		Revision as of 14:07, 26 September 2008
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	When we complete the square, it is only the first two terms, <math>x^{2}+2x</math>, that are involved. The general formula for completing the square states that <math>x^{2}+ax</math> equals		When we complete the square, it is only the first two terms, <math>x^{2}+2x</math>, that are involved. The general formula for completing the square states that <math>x^{2}+ax</math> equals

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

	Note how the coefficient ''a'' in front of the ''x'' turns up halved in two places.		Note how the coefficient ''a'' in front of the ''x'' turns up halved in two places.
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	If we use this formula, we obtain		If we use this formula, 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</math>}}	+	{{Displayed math\|\|<math>x^{2}+2x = \Bigl(x+\frac{2}{2}\Bigr)^{2} - \Bigl(\frac{2}{2}\Bigr)^{2} = (x+1)^{2}-1</math>}}

	and if we subtract the last "1", we obtain		and if we subtract the last "1", we obtain

Tek at 13:39, 26 September 2008

Tek — Fri, 26 Sep 2008 13:39:30 GMT

←Older revision		Revision as of 13:39, 26 September 2008
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-	When we complete the square, it is only the first two terms,	+	When we complete the square, it is only the first two terms, <math>x^{2}+2x</math>, that are involved. The general formula for completing the square states that <math>x^{2}+ax</math> equals
-	<math>x^{2}+2x</math>	+
-	, that are involved. The general	+
-	formula for completing the square states that	+
-	<math>x^{2}+ax</math>	+
-	equals	+

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

-	~~<math>\left( x+\frac{a}{2} \right)^{2}-\left( \frac{a}{2} \right)^{2}</math>~~	+	Note how the coefficient ''a'' in front of the ''x'' turns up halved in two places.
-		+
-		+
-	Note how the coefficient	+
-	~~<math>~~a~~</math>~~	+
-	in front of the	+
-	~~<math>~~x~~</math>~~	+
-	turns up halved in two places.	+

	If we use this formula, we obtain		If we use this formula, 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</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>~~	+	and if we subtract the last "1", we obtain
-		+
-		+
-	and if we subtract the last "	+
-	~~<math>~~1~~</math>~~	+
-	" , we obtain	+
-		+
-		+
-	~~<math>x^{2}+2x-1=\left( x+1 \right)^{2}-1-1=\left( x+1 \right)^{2}-2</math>~~	+

		+	{{Displayed math\|\|<math>x^{2}+2x-1 = (x+1)^{2}-1-1 = (x+1)^{2}-2\,\textrm{.}</math>}}

	To be completely certain that we have used the correct formula, we can expand the quadratic on the right-hand side,		To be completely certain that we have used the correct formula, we can expand the quadratic on the right-hand side,

-		+	{{Displayed math\|\|<math>(x+1)^{2}-2 = x^{2}+2x+1-2 = x^{2}+2x-1</math>}}
-	<math>~~\left~~( x+1 ~~\right~~)^{2}-2=x^{2}+2x+1-2=x^{2}+2x-1</math>	+
-		+

	and see that the relation really holds.		and see that the relation really holds.

Ian at 10:11, 12 September 2008

Ian — Fri, 12 Sep 2008 10:11:04 GMT

←Older revision		Revision as of 10:11, 12 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	When we complete the square, it is only the first two terms,
-	<~~center~~> ~~[[Image:2_3_1b~~.~~gif]]~~ </~~center~~>	+	<math>x^{2}+2x</math>
-	{{~~NAVCONTENT_STOP~~}}	+	, that are involved. The general
		+	formula for completing the square states that
		+	<math>x^{2}+ax</math>
		+	equals
		+
		+
		+	<math>\left( x+\frac{a}{2} \right)^{2}-\left( \frac{a}{2} \right)^{2}</math>
		+
		+
		+	Note how the coefficient
		+	<math>a</math>
		+	in front of the
		+	<math>x</math>
		+	turns up halved in two places.
		+
		+	If we use this formula, 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>
		+
		+
		+	and if we subtract the last "
		+	<math>1</math>
		+	" , we obtain
		+
		+
		+	<math>x^{2}+2x-1=\left( x+1 \right)^{2}-1-1=\left( x+1 \right)^{2}-2</math>
		+
		+
		+	To be completely certain that we have used the correct formula, we can expand the quadratic on the right-hand side,
		+
		+
		+	<math>\left( x+1 \right)^{2}-2=x^{2}+2x+1-2=x^{2}+2x-1</math>
		+
		+
		+	and see that the relation really holds.

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

Tekbot — Tue, 09 Sep 2008 08:55:03 GMT

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

←Older revision

Revision as of 08:55, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 06:44:39 GMT

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

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Oskar — Tue, 01 Apr 2008 09:54:11 GMT

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