Solution 2.3:2b - Revision history

Tek at 14:30, 26 September 2008

2008-09-26T14:30:35Z

←Older revision		Revision as of 14:30, 26 September 2008
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-	The first step when we solve the second-degree equation is to complete the square on the left-hand side:	+	The first step when we solve the second-degree equation is to complete the square on the left-hand side

		+	{{Displayed math\|\|<math>y^{2}+2y-15 = (y+1)^{2}-1^{2}-15 = (y+1)^{2}-16\,\textrm{.}</math>}}

-	~~<math>y^{2}+2y-15=\left( y+1 \right)^{2}-1^{2}-15=\left( y+1 \right)^{2}-16.</math>~~	+	The equation can now be written as

-	~~The equation can now be written as~~	+	{{Displayed math\|\|<math>(y+1)^{2} = 16</math>}}

		+	and has, after taking the square root, the solutions:

-	<math>~~\left(~~ y+1 \~~right)^~~{2}=16</math>	+	:*<math>y+1 = \sqrt{16} = 4\,\textrm{,}\ </math> which gives <math>y=-1+4=3\,\textrm{,}</math>

		+	:*<math>y+1 = -\sqrt{16} = -4\,\textrm{,}\ </math> which gives <math>y=-1-4=-5\,\textrm{.}</math>

-	and has, after taking the square root, the solutions

		+	A quick check shows that <math>y=-5</math> and <math>y=3</math> satisfy the equation:

-	<math>~~y+1=~~\~~sqrt~~{16}=~~4</math>~~	+	:*''y'' = -5: <math>\ \text{LHS} = (-5)^{2} + 2\cdot (-5)-15 = 25-10-15 = 0 = \text{RHS,}</math>
-	~~which gives~~	+
-	~~<math>y=~~-1+4=3</math>	+

-		+	:*''y'' = 3: <math>\ \text{LHS} = 3^{2} + 2\cdot 3 - 15 = 9+6-15 = 0 = \text{RHS.}</math>
-	~~<math>~~y+1=~~-\sqrt{16}=-4</math>~~	+
-	~~which gives~~	+
-	~~<math>y=-1-4=-5</math>~~	+
-		+
-		+
-	~~A quick check shows that~~	+
-	~~<math>y=-\text{5 }</math>~~	+
-	~~and~~	+
-	~~<math>y=\text{~~3 ~~}</math>~~	+
-	~~satisfy the equation~~:	+
-		+
-		+
-	<math>~~y=-~~\text{~~5 }</math>:~~ LHS=	+
-	~~<math>\left( -5 \right)^{2~~}~~+2\centerdot \left( -5 \right)-15~~=~~25-10-15=0</math>~~	+
-	~~= RHS~~	+
-		+
-	~~<math>y=\text{3 }</math>: LHS=~~	+
-	~~<math>~~3^{2}+2\~~centerdot~~ 3-15=9+6-15=0</math>	+
-	~~= RHS~~	+

Ian at 13:31, 20 September 2008

2008-09-20T13:31:18Z

←Older revision		Revision as of 13:31, 20 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	The first step when we solve the second-degree equation is to complete the square on the left-hand side:
-	<~~center~~> ~~[[Image~~:~~2_3_2b.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>y^{2}+2y-15=\left( y+1 \right)^{2}-1^{2}-15=\left( y+1 \right)^{2}-16.</math>
		+
		+	The equation can now be written as
		+
		+
		+	<math>\left( y+1 \right)^{2}=16</math>
		+
		+
		+	and has, after taking the square root, the solutions
		+
		+
		+	<math>y+1=\sqrt{16}=4</math>
		+	which gives
		+	<math>y=-1+4=3</math>
		+
		+
		+	<math>y+1=-\sqrt{16}=-4</math>
		+	which gives
		+	<math>y=-1-4=-5</math>
		+
		+
		+	A quick check shows that
		+	<math>y=-\text{5 }</math>
		+	and
		+	<math>y=\text{3 }</math>
		+	satisfy the equation:
		+
		+
		+	<math>y=-\text{5 }</math>: LHS=
		+	<math>\left( -5 \right)^{2}+2\centerdot \left( -5 \right)-15=25-10-15=0</math>
		+	= RHS
		+
		+	<math>y=\text{3 }</math>: LHS=
		+	<math>3^{2}+2\centerdot 3-15=9+6-15=0</math>
		+	= RHS

Tekbot: Lösning 2.3:2b moved to Solution 2.3:2b: Robot: moved page

2008-09-09T08:56:23Z

Lösning 2.3:2b moved to Solution 2.3:2b: Robot: moved page

←Older revision

Revision as of 08:56, 9 September 2008

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

2008-08-21T06:45:19Z

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

←Older revision		Revision as of 06:45, 21 August 2008
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-	<center> [[~~Bild~~:2_3_2b.gif]] </center>	+	<center> [[Image:2_3_2b.gif]] </center>
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Oskar: Ny sida: {{NAVCONTENT_START}} Bild:2_3_2b.gif {{NAVCONTENT_STOP}}

2008-04-01T09:56:51Z

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