Solution 2.3:5a - Revision history

Tek at 10:57, 29 September 2008

Tek — Mon, 29 Sep 2008 10:57:26 GMT

←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

Ian — Sun, 21 Sep 2008 09:44:04 GMT

←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

Tekbot — Tue, 09 Sep 2008 09:01:03 GMT

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:)

Tekbot — Thu, 21 Aug 2008 06:47:39 GMT

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

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Oskar: Ny sida: {{NAVCONTENT_START}} Bild:2_3_5a.gif {{NAVCONTENT_STOP}}

Oskar — Tue, 01 Apr 2008 10:10:10 GMT

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