Solution 3.2:2 - Revision history

Tek at 11:39, 1 October 2008

Tek — Wed, 01 Oct 2008 11:39:27 GMT

←Older revision		Revision as of 11:39, 1 October 2008
Line 1:		Line 1:
	The first thing we do is to square both sides of the equation		The first thing we do is to square both sides of the equation

		+	{{Displayed math\|\|<math>2x+7 = (x+2)^2</math>}}

-	~~<math>2x+7=\left( x+2 \right)^{2}</math>~~	+	to obtain an equation without a root sign. It is possible that we thereby introduce so-called spurious roots (solutions to the new equation which are not solutions to the old equation), so we need to test the solutions in the original root equation before we answer.
-		+
-		+
-	to ~~get~~ an equation without a root sign. It is possible that we thereby introduce so-called ~~false~~ roots (solutions to the new equation which are not solutions to the old equation), so we need to test the solutions in the original root equation before we answer.	+

	If we expand the right-hand side in the squared equation, we get		If we expand the right-hand side in the squared equation, we get

-		+	{{Displayed math\|\|<math>2x+7=x^{2}+4x+4</math>\|(*)}}
-	~~(*)~~	+
-	<math>2x+7=x^{2}+4x+4</math>	+
-		+

	which we also can write as		which we also can write as

-		+	{{Displayed math\|\|<math>x^{2}+2x-3=0\,\textrm{.}</math>}}
-	<math>x^{2}+2x-3=0</math>	+
-		+

	Completing the square of the left-hand side gives		Completing the square of the left-hand side gives

		+	{{Displayed math\|\|<math>x^2+2x-3 = (x+1)^2-1^2-3 = (x+1)^2-4\,\textrm{.}</math>}}

-	<math>x^{2}+2x-3=\left( x+1 \right)^{2}-1^{2}-3=\left( x+1 \right)^{2}-4.</math>
-
-
	The equation then becomes		The equation then becomes

-		+	{{Displayed math\|\|<math>(x+1)^2 = 4</math>}}
-	<math>~~\left~~( x+1 ~~\right~~)^{2}=4</math>	+
-		+

	which has solutions		which has solutions

		+	:*<math>x=-1+\sqrt{4}=-1+2=1</math>

-	~~<math>x=-1+\sqrt{4}=-1+2=1</math>~~	+	:*<math>x=-1-\sqrt{4}=-1-2=-3</math>
-		+
-		+
-	<math>x=-1-\sqrt{4}=-1-2=-3~~</math>~~	+
-		+
-		+
-	~~A quick check shows also that~~	+
-	~~<math>x=-\text{3 }</math>~~	+
-	~~and~~	+
-	~~<math>x=\text{1 }</math>~~	+
-	~~are solutions to the squared equation (*)~~	+
-		+
-		+
-	~~<math>x=-\text{3}</math>: LHS~~	+
-	~~<math>=2\centerdot \left( -3 \right)+7=-6+7=1</math>~~	+
-	~~and~~	+
-	~~RHS~~	+
-	~~<math>=\left( -3+2 \right)^{2}=1</math>~~	+
-		+
-		+
-	~~<math>x=\text{1}</math>: LHS~~	+
-	~~<math>=2\centerdot 1+7=2+7=9</math>~~	+
-	~~and~~	+
-	~~RHS~~	+
-	~~<math>=\left( 1+2 \right)^{2}=9</math>~~	+
-		+
-		+
-	~~When we test solution in the root equation, we get that~~	+
-		+
-		+
-	~~<math>x=-\text{3}</math>: LHS~~	+
-	~~<math>=\sqrt{2\centerdot \left( -3 \right)+7}=\sqrt{-6+7}=1~~</math>	+

-	~~RHS~~	+	A quick check shows also that <math>x=-3</math> and <math>x=1</math> are solutions to the squared equation (*):
-	<math>=-3+2=-1</math>	+

		+	:{\|
		+	\|\|<ul><li>''x'' = -3:</li></ul>
		+	\|\|<math>\ \text{LHS} = 2\cdot (-3)+7 = -6+7 = 1</math> and
		+	\|-
		+	\|\|
		+	\|\|<math>\ \text{RHS} = (-3+2)^{2} = 1</math>
		+	\|-
		+	\|\|<ul><li>''x'' = 1:</li></ul>
		+	\|\|<math>\ \text{LHS} = 2\cdot 1+7 = 2+7 = 9</math> and
		+	\|-
		+	\|\|
		+	\|\|<math>\ \text{RHS} = (1+2)^2 = 9</math>
		+	\|}

-	~~<math>x=\text{1}</math>: LHS~~	+	When we test the solutions in the root equation, we get that
-	~~<math>=\sqrt{2\centerdot 1+7}=\sqrt{2+7}=3</math>~~	+

-	RHS	+	:{\|
-	<math>=1+2=3</math>	+	\|\|<ul><li>''x'' = -3:</li></ul>
		+	\|\|<math>\ \text{LHS} = \sqrt{2\cdot (-3)+7} = \sqrt{-6+7} = \sqrt{1} = 1</math> and
		+	\|-
		+	\|\|
		+	\|\|<math>\ \text{RHS} = -3+2 = -1</math>
		+	\|-
		+	\|\|<ul><li>''x'' = 1:</li></ul>
		+	\|\|<math>\ \text{LHS} = \sqrt{2\cdot 1+7} = \sqrt{2+7} = \sqrt{9} = 3</math>
		+	\|-
		+	\|\|
		+	\|\|<math>\ \text{RHS} = 1+2 = 3</math>
		+	\|}

		+	and therefore <math>x=1</math> is the only solution to the root equation (<math>x=-3</math> is a spurious root).

-	and therefore
-	<math>x=\text{1}</math>
-	is the only solution to the root equation (
-	<math>x=-\text{3}</math>
-	is a false root).

-	~~NOTE~~: The check we carry out when substituting the solutions into equation (*) is not strictly speaking necessary, but more for seeing that we haven't calculated incorrectly. On the other hand, testing in the root equation is necessary.	+	Note: The check we carry out when substituting the solutions into equation (*) is not strictly speaking necessary, but more for seeing that we haven't calculated incorrectly. On the other hand, testing in the root equation is necessary.

Ian at 10:30, 25 September 2008

Ian — Thu, 25 Sep 2008 10:30:24 GMT

←Older revision		Revision as of 10:30, 25 September 2008
Line 10:		Line 10:


		+	(*)
	<math>2x+7=x^{2}+4x+4</math>		<math>2x+7=x^{2}+4x+4</math>

Ian at 12:34, 23 September 2008

Ian — Tue, 23 Sep 2008 12:34:19 GMT

←Older revision		Revision as of 12:34, 23 September 2008
Line 1:		Line 1:
-	{{~~NAVCONTENT_START~~}}	+	The first thing we do is to square both sides of the equation
-	<~~center~~> ~~[[Image:3_2_2_1~~(3).~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
-	{~~{NAVCONTENT_START~~}}	+	<math>2x+7=\left( x+2 \right)^{2}</math>
-	<~~center~~> ~~[[Image:3_2_2~~-2(3)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP}~~}	+
-	{~~{NAVCONTENT_START}~~}	+	to get an equation without a root sign. It is possible that we thereby introduce so-called false roots (solutions to the new equation which are not solutions to the old equation), so we need to test the solutions in the original root equation before we answer.
-	<~~center~~> ~~[[Image:3_2_2~~-3(3)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	If we expand the right-hand side in the squared equation, we get
		+
		+
		+	<math>2x+7=x^{2}+4x+4</math>
		+
		+
		+	which we also can write as
		+
		+
		+	<math>x^{2}+2x-3=0</math>
		+
		+
		+	Completing the square of the left-hand side gives
		+
		+
		+	<math>x^{2}+2x-3=\left( x+1 \right)^{2}-1^{2}-3=\left( x+1 \right)^{2}-4.</math>
		+
		+
		+	The equation then becomes
		+
		+
		+	<math>\left( x+1 \right)^{2}=4</math>
		+
		+
		+	which has solutions
		+
		+
		+	<math>x=-1+\sqrt{4}=-1+2=1</math>
		+
		+
		+	<math>x=-1-\sqrt{4}=-1-2=-3</math>
		+
		+
		+	A quick check shows also that
		+	<math>x=-\text{3 }</math>
		+	and
		+	<math>x=\text{1 }</math>
		+	are solutions to the squared equation (*)
		+
		+
		+	<math>x=-\text{3}</math>: LHS
		+	<math>=2\centerdot \left( -3 \right)+7=-6+7=1</math>
		+	and
		+	RHS
		+	<math>=\left( -3+2 \right)^{2}=1</math>
		+
		+
		+	<math>x=\text{1}</math>: LHS
		+	<math>=2\centerdot 1+7=2+7=9</math>
		+	and
		+	RHS
		+	<math>=\left( 1+2 \right)^{2}=9</math>
		+
		+
		+	When we test solution in the root equation, we get that
		+
		+
		+	<math>x=-\text{3}</math>: LHS
		+	<math>=\sqrt{2\centerdot \left( -3 \right)+7}=\sqrt{-6+7}=1</math>
		+
		+	RHS
		+	<math>=-3+2=-1</math>
		+
		+
		+	<math>x=\text{1}</math>: LHS
		+	<math>=\sqrt{2\centerdot 1+7}=\sqrt{2+7}=3</math>
		+
		+	RHS
		+	<math>=1+2=3</math>
		+
		+
		+	and therefore
		+	<math>x=\text{1}</math>
		+	is the only solution to the root equation (
		+	<math>x=-\text{3}</math>
		+	is a false root).
		+
		+	NOTE: The check we carry out when substituting the solutions into equation (*) is not strictly speaking necessary, but more for seeing that we haven't calculated incorrectly. On the other hand, testing in the root equation is necessary.

Tekbot: Lösning 3.2:2 moved to Solution 3.2:2: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:17:43 GMT

Lösning 3.2:2 moved to Solution 3.2:2: Robot: moved page

←Older revision

Revision as of 09:17, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 06:55:59 GMT

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

←Older revision		Revision as of 06:55, 21 August 2008
Line 1:		Line 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_2_2_1(3).gif]] </center>	+	<center> [[Image:3_2_2_1(3).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_2_2-2(3).gif]] </center>	+	<center> [[Image:3_2_2-2(3).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:3_2_2-3(3).gif]] </center>	+	<center> [[Image:3_2_2-3(3).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar at 15:03, 1 April 2008

Oskar — Tue, 01 Apr 2008 15:03:12 GMT

←Older revision		Revision as of 15:03, 1 April 2008
Line 1:		Line 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[Bild:~~3_2_2-1~~(3).gif]] </center>	+	<center> [[Bild:3_2_2_1(3).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}

Oskar at 15:00, 1 April 2008

Oskar — Tue, 01 Apr 2008 15:00:44 GMT

←Older revision	Revision as of 15:00, 1 April 2008
Line 1:	Line 1:
-	+	{{NAVCONTENT_START}}
	+	<center> [[Bild:3_2_2-1(3).gif]] </center>
	+	{{NAVCONTENT_STOP}}
	+	{{NAVCONTENT_START}}
	+	<center> [[Bild:3_2_2-2(3).gif]] </center>
	+	{{NAVCONTENT_STOP}}
	+	{{NAVCONTENT_START}}
	+	<center> [[Bild:3_2_2-3(3).gif]] </center>
	+	{{NAVCONTENT_STOP}}

Oskar: Tar bort sidans innehåll

Oskar — Tue, 01 Apr 2008 15:00:13 GMT

Tar bort sidans innehåll

←Older revision		Revision as of 15:00, 1 April 2008
Line 1:		Line 1:
-	~~{{NAVCONTENT_START}}~~	+
-	~~<center> [[Bild:3_2_2-1(3).gif]] </center>~~	+
-	~~{{NAVCONTENT_STOP}}~~	+
-	~~{{NAVCONTENT_START}}~~	+
-	~~<center> [[Bild:3_2_2-2(3).gif]] </center>~~	+
-	~~{{NAVCONTENT_STOP}}~~	+
-	~~{{NAVCONTENT_START}}~~	+
-	~~<center> [[Bild:3_2_2-3(3).gif]] </center>~~	+
-	~~{{NAVCONTENT_STOP}}~~	+

Oskar at 14:59, 1 April 2008

Oskar — Tue, 01 Apr 2008 14:59:11 GMT

←Older revision		Revision as of 14:59, 1 April 2008
Line 3:		Line 3:
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[Bild:3_2_2-1(3).gif]] </center>	+	<center> [[Bild:3_2_2-2(3).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[Bild:3_2_2-1(3).gif]] </center>	+	<center> [[Bild:3_2_2-3(3).gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:3_2_2-1(3).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} Bild:3_2_2-1(3).gif {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} ...

Oskar — Tue, 01 Apr 2008 14:57:30 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:3_2_2-1(3).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} <center> Bild:3_2_2-1(3).gif </center> {{NAVCONTENT_STOP}} {{NAVCONTENT_START}} ...

New page

{{NAVCONTENT_START}}
<center> [[Bild:3_2_2-1(3).gif]] </center>
{{NAVCONTENT_STOP}}
{{NAVCONTENT_START}}
<center> [[Bild:3_2_2-1(3).gif]] </center>
{{NAVCONTENT_STOP}}
{{NAVCONTENT_START}}
<center> [[Bild:3_2_2-1(3).gif]] </center>
{{NAVCONTENT_STOP}}