Solution 3.2:1 - Revision history

Tek at 11:17, 1 October 2008

2008-10-01T11:17:25Z

←Older revision		Revision as of 11:17, 1 October 2008
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-	In order to get rid of the root sign in the equation, we square both sides:	+	In order to get rid of the root sign in the equation, we square both sides
-		+
-		+
-	~~<math>x-4=\left( 6-x \right)^{2}</math>~~	+

		+	{{Displayed math\|\|<math>x-4 = (6-x)^2\,\textrm{.}</math>}}

	It is important to remember that this step means that we are now working with a new equation which may have solutions which the equation that we started with did not have. At the end, it will therefore be necessary for us to check that the solutions that we calculate satisfy the original equation also.		It is important to remember that this step means that we are now working with a new equation which may have solutions which the equation that we started with did not have. At the end, it will therefore be necessary for us to check that the solutions that we calculate satisfy the original equation also.
Line 9:		Line 7:
	If we continue with the squared equation and expand the right-hand side, we get a second order equation		If we continue with the squared equation and expand the right-hand side, we get a second order equation

-		+	{{Displayed math\|\|<math>x-4=36-12x+x^{2}\,\textrm{,}</math>\|(*)}}
-	~~(*)~~	+
-	<math>x-4=36-12x+x^{2}</math>	+
-		+

	i.e.		i.e.

-		+	{{Displayed math\|\|<math>x^{2}-13x+40=0\,\textrm{.}</math>}}
-	<math>x^{2}-~~12x~~+40=0</math>	+
-		+

	We complete the square of the left-hand side		We complete the square of the left-hand side

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	x^2-13x+40
-	& x^{2}-~~12x~~+40=\~~left~~( x-\frac{13}{2} \~~right~~)^{2}-\~~left~~( \frac{13}{2} \~~right~~)^{2}+40 \\	+	&= \Bigl(x-\frac{13}{2}\Bigr)^{2} - \Bigl(\frac{13}{2}\Bigr)^2 + 40\\[5pt]
-	& =\~~left~~( x-\frac{13}{2} \~~right~~)^{2}-\frac{169}{4}+\frac{160}{4}=\~~left~~( x-\frac{13}{2} \~~right~~)^{2}-\frac{9}{4} \\	+	&= \Bigl(x-\frac{13}{2}\Bigr)^{2} - \frac{169}{4} + \frac{160}{4}\\[5pt]
-	\end{align}</math>	+	&= \Bigl(x-\frac{13}{2}\Bigr)^{2} - \frac{9}{4}\,\textrm{,}
-		+	\end{align}</math>}}

	which gives the equation		which gives the equation

-		+	{{Displayed math\|\|<math>\Bigl(x-\frac{13}{2}\Bigr)^2 = \frac{9}{4}</math>}}
-	<math>\~~left~~( x-\frac{13}{2} \~~right~~)^{2}=\frac{9}{4}</math>	+
-		+

	which has the solutions		which has the solutions

		+	:*<math>x = \frac{13}{2} + \sqrt{\frac{9}{4}} = \frac{13}{2} + \frac{3}{2} = \frac{16}{2} = 8\,,</math>

-	<math>x=\frac{13}{2}+\sqrt{\frac{9}{4}}=\frac{13}{2}+\frac{3}{2}=\frac{16}{2}=~~8</math>~~	+	:*<math>x = \frac{13}{2} - \sqrt{\frac{9}{4}} = \frac{13}{2} - \frac{3}{2} = \frac{10}{2} = 5\,\textrm{.}</math>
-		+
-		+
-	~~<math>x=~~\~~frac{13}{2}-~~\~~sqrt~~{~~\frac{9~~}~~{4}}=\frac{13}{2}-\frac{3}{2}=\frac{10}{2}=5~~</math>	+
-		+
-		+
-	~~Just to be on the safe side, we check whether we have solved the second-order equation (*) correctly by substituting x=5 and x=8 into it.~~	+
-		+

-	<math>x=~~\text{~~5}</math>~~: LHS~~	+	Just to be on the safe side, we check whether we have solved the second-order equation (*) correctly by substituting <math>x=5</math> and <math>x=8</math> into it:
-	~~<math>=5-4=1</math>~~	+
-	and ~~RHS~~	+
-	<math>=~~\left( 6-5 \right)^{2}=1~~</math>	+

		+	:*''x'' = 5: <math>\ \text{LHS} = 5-4 = 1\ </math> and <math>\ \text{RHS} = (6-5)^2 = 1</math>

-	<math>x=\text{8}~~</math>: LHS~~	+	:*''x'' = 8: <math>\ \text{LHS} = 8-4 = 4\ </math> and <math>\ \text{RHS} = (6-8)^{2} = 4</math>
-	~~<math>~~=8-4=4</math>	+
-	and ~~RHS~~	+
-	<math>=~~\left~~( 6-8 ~~\right~~)^{2}=4</math>	+

		+	Then, we must test the solutions <math>x=5</math> and <math>x=8</math> in the original equation:

-	~~Then, we must test the solutions~~ x=5 and x=~~8 in the original equation:~~	+	:*''x'' = 5: <math>\ \text{LHS} = \sqrt{5-4} = 1\ </math> and <math>\ \text{RHS} = 6-5 = 1</math>

-	x=5: ~~LHS~~	+	:*''x'' = 8: <math>\ \text{LHS} = \sqrt{8-4} = 2\ </math> and <math>\ \text{RHS} = 6-8 = -2</math>
-	<math>=\sqrt{5-4}=1</math>	+
-	and ~~RHS~~	+
-	<math>=6-5=1</math>	+

-	~~x=8: LHS~~	+	This shows that <math>x=5</math> is a solution to the original equation, whilst
-	<math>=~~\sqrt{8-4}=2~~</math>	+	<math>x=8</math> is a spurious root.
-	~~and RHS~~	+
-	<math>=6-8~~=-2~~</math>	+
-		+

-	This shows that
-	<math>x=\text{5 }</math>
-	is a solution to the original equation, whilst
-	<math>x=\text{8 }</math>
-	is a false root.

-	~~NOTE~~: That we get a ~~false~~ root does not mean that we calculated incorrectly, but is totally the result of squaring the equation.	+	Note: That we get a spurious root does not mean that we calculated incorrectly, but is totally the result of squaring the equation.

Ian at 10:29, 25 September 2008

2008-09-25T10:29:10Z

←Older revision		Revision as of 10:29, 25 September 2008
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		+	(*)
	<math>x-4=36-12x+x^{2}</math>		<math>x-4=36-12x+x^{2}</math>

Ian at 12:16, 23 September 2008

2008-09-23T12:16:07Z

←Older revision		Revision as of 12:16, 23 September 2008
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-	~~{{NAVCONTENT_START}}~~	+	In order to get rid of the root sign in the equation, we square both sides:
-	<~~center~~> ~~[[Image:3_2_1~~-1(3).~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+	<math>x-4=\left( 6-x \right)^{2}</math>
-	<~~center~~> ~~[[Image:3_2_1~~-2(3)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+	It is important to remember that this step means that we are now working with a new equation which may have solutions which the equation that we started with did not have. At the end, it will therefore be necessary for us to check that the solutions that we calculate satisfy the original equation also.
-	<~~center~~> ~~[[Image:3_2_1~~-3(3).~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	If we continue with the squared equation and expand the right-hand side, we get a second order equation
		+
		+
		+	<math>x-4=36-12x+x^{2}</math>
		+
		+
		+	i.e.
		+
		+
		+	<math>x^{2}-12x+40=0</math>
		+
		+
		+	We complete the square of the left-hand side
		+
		+
		+	<math>\begin{align}
		+	& x^{2}-12x+40=\left( x-\frac{13}{2} \right)^{2}-\left( \frac{13}{2} \right)^{2}+40 \\
		+	& =\left( x-\frac{13}{2} \right)^{2}-\frac{169}{4}+\frac{160}{4}=\left( x-\frac{13}{2} \right)^{2}-\frac{9}{4} \\
		+	\end{align}</math>
		+
		+
		+	which gives the equation
		+
		+
		+	<math>\left( x-\frac{13}{2} \right)^{2}=\frac{9}{4}</math>
		+
		+
		+	which has the solutions
		+
		+
		+	<math>x=\frac{13}{2}+\sqrt{\frac{9}{4}}=\frac{13}{2}+\frac{3}{2}=\frac{16}{2}=8</math>
		+
		+
		+	<math>x=\frac{13}{2}-\sqrt{\frac{9}{4}}=\frac{13}{2}-\frac{3}{2}=\frac{10}{2}=5</math>
		+
		+
		+	Just to be on the safe side, we check whether we have solved the second-order equation (*) correctly by substituting x=5 and x=8 into it.
		+
		+
		+	<math>x=\text{5}</math>: LHS
		+	<math>=5-4=1</math>
		+	and RHS
		+	<math>=\left( 6-5 \right)^{2}=1</math>
		+
		+
		+	<math>x=\text{8}</math>: LHS
		+	<math>=8-4=4</math>
		+	and RHS
		+	<math>=\left( 6-8 \right)^{2}=4</math>
		+
		+
		+	Then, we must test the solutions x=5 and x=8 in the original equation:
		+
		+	x=5: LHS
		+	<math>=\sqrt{5-4}=1</math>
		+	and RHS
		+	<math>=6-5=1</math>
		+
		+	x=8: LHS
		+	<math>=\sqrt{8-4}=2</math>
		+	and RHS
		+	<math>=6-8=-2</math>
		+
		+
		+	This shows that
		+	<math>x=\text{5 }</math>
		+	is a solution to the original equation, whilst
		+	<math>x=\text{8 }</math>
		+	is a false root.
		+
		+	NOTE: That we get a false root does not mean that we calculated incorrectly, but is totally the result of squaring the equation.

Tekbot: Lösning 3.2:1 moved to Solution 3.2:1: Robot: moved page

2008-09-09T09:17:23Z

Lösning 3.2:1 moved to Solution 3.2:1: Robot: moved page

←Older revision

Revision as of 09:17, 9 September 2008

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

2008-08-21T06:55:49Z

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

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2008-04-01T14:55:57Z

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