Solution 2.3:5a - Revision history

Tek at 10:57, 29 September 2008

2008-09-29T10:57:26Z

←Older revision		Revision as of 10:57, 29 September 2008
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	In this exercise we can use the technique for writing equations in factorized form. Consider in our case the equation		In this exercise we can use the technique for writing equations in factorized form. Consider in our case the equation

		+	{{Displayed math\|\|<math>(x+7)(x+7)=0\,\textrm{.}</math>}}

-	<math>~~\left(~~ x+7 ~~\right)\left(~~ x~~+7 \right)~~=0</math>	+	This equation has only <math>x=-7</math> as a root because both factors become zero only when <math>x=-7</math>. In addition, it is an second-degree equation, which we can clearly see if the left-hand side is expanded,

-		+	{{Displayed math\|\|<math>(x+7)(x+7) = x^{2}+14x+49\,\textrm{.}</math>}}
-	~~This equation has only~~	+
-	<math>x=~~-\text~~{7}~~</math>~~	+
-	~~as a root because both factors become zero only when~~	+
-	~~<math>x=-~~\~~text~~{7}</math>~~. In addition, it is an second-degree equation, which we can clearly see if the left-hand side is expanded:~~	+

		+	Thus, one answer is the equation <math>x^{2}+14x+49=0\,</math>.

-	<math>\left( x+7 \right)\left( x+7 \right)=x^{2}+14x+49</math>

		+	Note: All second-degree equations which have <math>x=-7</math> as its sole root can be written as

-	~~Thus, one answer is the equation~~	+	{{Displayed math\|\|<math>ax^{2}+14ax+49a=0\,,</math>}}
-	<math>x^{2}+~~14x~~+49=0</math>.	+

-	~~NOTE: All second-degree equations which have~~	+	where ''a'' is a non-zero constant.
-	~~<math>x=-\text{7}</math>~~	+
-	~~as a root can be written as~~	+
-		+
-		+
-	~~<math>ax^{2}+14ax+49a=0</math>~~	+
-		+
-		+
-	where	+
-	~~<math>~~a~~</math>~~	+
-	is a non-zero constant.	+

Ian at 09:44, 21 September 2008

2008-09-21T09:44:04Z

←Older revision		Revision as of 09:44, 21 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	In this exercise we can use the technique for writing equations in factorized form. Consider in our case the equation
-	<~~center~~> ~~[[Image:2_3_5a~~.~~gif]]~~ </~~center~~>	+
-	{~~{NAVCONTENT_STOP}~~}	+
		+	<math>\left( x+7 \right)\left( x+7 \right)=0</math>
		+
		+
		+	This equation has only
		+	<math>x=-\text{7}</math>
		+	as a root because both factors become zero only when
		+	<math>x=-\text{7}</math>. In addition, it is an second-degree equation, which we can clearly see if the left-hand side is expanded:
		+
		+
		+	<math>\left( x+7 \right)\left( x+7 \right)=x^{2}+14x+49</math>
		+
		+
		+	Thus, one answer is the equation
		+	<math>x^{2}+14x+49=0</math>.
		+
		+	NOTE: All second-degree equations which have
		+	<math>x=-\text{7}</math>
		+	as a root can be written as
		+
		+
		+	<math>ax^{2}+14ax+49a=0</math>
		+
		+
		+	where
		+	<math>a</math>
		+	is a non-zero constant.

Tekbot: Lösning 2.3:5a moved to Solution 2.3:5a: Robot: moved page

2008-09-09T09:01:03Z

Lösning 2.3:5a moved to Solution 2.3:5a: Robot: moved page

←Older revision

Revision as of 09:01, 9 September 2008

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

2008-08-21T06:47:39Z

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

←Older revision		Revision as of 06:47, 21 August 2008
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Oskar: Ny sida: {{NAVCONTENT_START}} Bild:2_3_5a.gif {{NAVCONTENT_STOP}}

2008-04-01T10:10:10Z

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