Solution 2.3:4a - Revision history

Tek at 10:18, 29 September 2008

Tek — Mon, 29 Sep 2008 10:18:18 GMT

←Older revision		Revision as of 10:18, 29 September 2008
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	A first thought is perhaps to write the equation as		A first thought is perhaps to write the equation as

		+	{{Displayed math\|\|<math>x^{2}+ax+b=0</math>}}

-	<math>x~~^{2}+ax+b~~=0</math>	+	and then try to choose the constants ''a'' and ''b'' in some way so that
		+	<math>x=-1</math> and <math>x=2</math> are solutions. But a better way is to start with a factorized form of a second-order equation,

		+	{{Displayed math\|\|<math>(x+1)(x-2)=0\,\textrm{.}</math>}}

-	~~and then try to choose the constants~~	+	If we consider this equation, we see that both <math>x=-1</math> and <math>x=2</math> are solutions to the equation, since <math>x=-1</math> makes the first factor on the left-hand side zero, whilst <math>x=2</math> makes the second factor zero. Also, it really is a second order equation, because if we multiply out the left-hand side, we get
-	<math>a</math>	+
-	and	+
-	<math>b</math>	+
-	~~in some way so that~~	+
-	<math>x=-~~\text{~~1 }</math>	+
-	~~and~~	+
-	<math>x=~~\text{~~2 }</math>	+
-	~~are solutions~~. ~~But a better way~~ is ~~to start with a factorized form of~~ a second-order equation,	+

		+	{{Displayed math\|\|<math>x^{2}-x-2=0\,\textrm{.}</math>}}

-	<math>~~\left~~( x+1 ~~\right~~)~~\left~~( x-2 ~~\right~~)=0</math>	+	One answer is thus the equation <math>(x+1)(x-2)=0</math>, or <math>x^{2}-x-2=0\,</math>.


-	~~If we consider~~ this ~~equation~~, ~~we see~~ that ~~both~~	+	Note: There are actually many answers to this exercise, but what all second-degree equations that have <math>x=-1</math> and <math>x=2</math> as roots have in common is that they can be written in the form
-	<math>x=-~~\text{~~1 }</math>	+
-	and	+
-	<math>x=~~\text{~~2 }</math>	+
-	~~are solutions to the equation, since~~	+
-	~~<math>x=-\text{1 }</math>~~	+
-	~~makes the first factor on the left-hand side zero, whilst~~	+
-	~~<math>x=\text{2 }</math>~~	+
-	~~makes the second factor zero. Also, it really~~ is ~~a second order equation, because if we multiply out~~ the ~~left-hand side, we get~~	+

		+	{{Displayed math\|\|<math>ax^{2}-ax-2a=0\,,</math>}}

-	~~<math>x^{2}-x-2=0</math>~~	+	where ''a'' is a non-zero constant.
-		+
-		+
-	~~One answer is thus the equation~~	+
-	~~<math>\left( x+1 \right)\left( x-2 \right)=0</math>, or~~	+
-	~~<math>x^{2}-x-2=0</math>.~~	+
-		+
-	~~NOTE: There are actually many answers to this exercise, but what all second-degree equations that have~~	+
-	~~<math>x=-\text{1 }</math>~~	+
-	~~and~~	+
-	~~<math>x=\text{2 }</math>~~	+
-	~~as roots have in common is that they can be written in the form~~	+
-		+
-		+
-	~~<math>ax^{2}-ax-2a=0</math>~~	+
-		+
-		+
-	where	+
-	~~<math>~~a~~</math>~~	+
-	is a non-zero constant.	+

Ian at 09:14, 21 September 2008

Ian — Sun, 21 Sep 2008 09:14:48 GMT

←Older revision		Revision as of 09:14, 21 September 2008
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-	{{~~NAVCONTENT_START}~~}	+	A first thought is perhaps to write the equation as
-	<~~center~~> ~~[[Image:2_3_4a~~-1(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
-	{~~{NAVCONTENT_START}~~}	+	<math>x^{2}+ax+b=0</math>
-	<~~center~~> ~~[[Image:2_3_4a~~-2(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	and then try to choose the constants
		+	<math>a</math>
		+	and
		+	<math>b</math>
		+	in some way so that
		+	<math>x=-\text{1 }</math>
		+	and
		+	<math>x=\text{2 }</math>
		+	are solutions. But a better way is to start with a factorized form of a second-order equation,
		+
		+
		+	<math>\left( x+1 \right)\left( x-2 \right)=0</math>
		+
		+
		+	If we consider this equation, we see that both
		+	<math>x=-\text{1 }</math>
		+	and
		+	<math>x=\text{2 }</math>
		+	are solutions to the equation, since
		+	<math>x=-\text{1 }</math>
		+	makes the first factor on the left-hand side zero, whilst
		+	<math>x=\text{2 }</math>
		+	makes the second factor zero. Also, it really is a second order equation, because if we multiply out the left-hand side, we get
		+
		+
		+	<math>x^{2}-x-2=0</math>
		+
		+
		+	One answer is thus the equation
		+	<math>\left( x+1 \right)\left( x-2 \right)=0</math>, or
		+	<math>x^{2}-x-2=0</math>.
		+
		+	NOTE: There are actually many answers to this exercise, but what all second-degree equations that have
		+	<math>x=-\text{1 }</math>
		+	and
		+	<math>x=\text{2 }</math>
		+	as roots have in common is that they can be written in the form
		+
		+
		+	<math>ax^{2}-ax-2a=0</math>
		+
		+
		+	where
		+	<math>a</math>
		+	is a non-zero constant.

Tekbot: Lösning 2.3:4a moved to Solution 2.3:4a: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:00:03 GMT

Lösning 2.3:4a moved to Solution 2.3:4a: Robot: moved page

←Older revision

Revision as of 09:00, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 06:47:09 GMT

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

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Oskar — Tue, 01 Apr 2008 10:07:14 GMT

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