Solution 3.2:3 - Revision history

Tek at 12:06, 1 October 2008

Tek — Wed, 01 Oct 2008 12:06:04 GMT

←Older revision		Revision as of 12:06, 1 October 2008
Line 1:		Line 1:
-	First, we move the	+	First, we move the 2 to the right-hand side to get <math>\sqrt{3x-8}=x-2</math> and then square away the root sign,
-	~~<math>\text{~~2~~}</math>~~	+
-	to the right-hand side to get	+
-	<math>\sqrt{3x-8}=x-2</math>	+
-	and then square away the root sign,	+

-	~~(*)~~	+	{{Displayed math\|\|<math>3x-8 = (x-2)^{2}</math>\|(*)}}
-	<math>3x-8=~~\left~~( x-2 ~~\right~~)^{2}</math>	+

-
	or, with the right-hand side expanded		or, with the right-hand side expanded

-		+	{{Displayed math\|\|<math>3x-8=x^{2}-4x+4\,\textrm{.}</math>}}
-	<math>3x-8=x^{2}-4x+4</math>	+
-		+

	If we move over all the terms to the left-hand side, we get		If we move over all the terms to the left-hand side, we get

-		+	{{Displayed math\|\|<math>x^{2}-7x+12=0\,\textrm{.}</math>}}
-	<math>x^{2}-7x+12=0</math>	+
-		+

	If we complete the square of the left-hand side,		If we complete the square of the left-hand side,

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	x^2-7x+12 &= \Bigl(x-\frac{7}{2}\Bigr)^2 - \Bigl(\frac{7}{2}\Bigr)^2 + 12\\[5pt]
-	& x^{2}-7x+12=\~~left~~( x-\frac{7}{2} \~~right~~)^{2}-\~~left~~( \frac{7}{2} \~~right~~)^{2}+12 \\	+	&= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{49}{4} + \frac{48}{4}\\[5pt]
-	& =\~~left~~( x-\frac{7}{2} \~~right~~)^{2}-\frac{49}{4}+\frac{48}{4} \\	+	&= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{1}{4}
-	& =\~~left~~( x-\frac{7}{2} \~~right~~)^{2}-\frac{1}{4} \\	+	\end{align}</math>}}
-	\end{align}</math>	+
-		+

	the equation can be written as		the equation can be written as

-		+	{{Displayed math\|\|<math>\Bigl(x-\frac{7}{2}\Bigr)^{2} = \frac{1}{4}</math>}}
-	<math>\~~left~~( x-\frac{7}{2} \~~right~~)^{2}=\frac{1}{4}</math>	+
-		+

	and the solutions are		and the solutions are

		+	:*<math>x = \frac{7}{2} + \sqrt{\frac{1}{4}} = \frac{7}{2} + \frac{1}{2} = \frac{8}{2} = 4\,,</math>

-	~~<math>x=\frac{7}{2}+\sqrt{\frac{1}{4}}=\frac{7}{2}+\frac{1}{2}=\frac{8}{2}=4</math>~~	+	:*<math>x = \frac{7}{2} - \sqrt{\frac{1}{4}} = \frac{7}{2} - \frac{1}{2} = \frac{6}{2} = 3\,\textrm{.}</math>
-		+
-		+
-	<math>x=\frac{7}{2}-\sqrt{\frac{1}{4}}=\frac{7}{2}-\frac{1}{2}=\frac{6}{2}=3~~</math>~~	+
-		+
-	~~To be on the safe side~~, ~~we verify that~~	+
-	~~<math>x=~~\~~text~~{3 }</math>	+
-	~~and~~	+
-	~~<math>x=\text{4}</math>~~	+
-	~~satisfy the squared equation (*)~~	+
-		+
-		+
-	~~<math>x=\text{3 }</math>: LHS~~	+
-	~~<math>=3\centerdot 3-8=9-8=1</math>~~	+
-	~~and RHS~~	+
-	~~<math>=\left( 3-2 \right)^{2}=1</math>~~	+
-		+
-		+
-	~~<math>x=\text{4}</math>: LHS~~	+
-	~~<math>=3\centerdot 4-8=12-8=4</math>~~	+
-	~~and RHS~~	+
-	~~<math>=\left( 4\centerdot 2 \right)^{2}=4</math>~~	+
-		+
-		+
-	~~Because we squared the root equation, possible false roots turn up and we therefore have to verify the solutions when we go back to the original root equation:~~	+
-		+

-	<math>x=~~\text{~~3 }</math>~~: LHS~~	+	To be on the safe side, we verify that <math>x=3</math> and <math>x=4</math> satisfy the squared equation (*)
-	<math>=~~\sqrt{3\centerdot 3-8}+2=\sqrt{9-8}+2=1+2=3</math>~~	+
-		+
-	~~RHS~~	+
-	~~<math>=3~~</math>	+

		+	{\|
		+	\|\|<ul><li>''x'' = 3:</li></ul>
		+	\|\|<math>\ \text{LHS} = 3\cdot 3-8 = 9-8 = 1</math> and
		+	\|-
		+	\|\|
		+	\|\|<math>\ \text{RHS} = (3-2)^2 = 1</math>
		+	\|-
		+	\|\|<ul><li>''x'' = 4:</li></ul>
		+	\|\|<math>\ \text{LHS} = 3\cdot 4-8 = 12-8 = 4</math> and
		+	\|-
		+	\|\|
		+	\|\|<math>\ \text{RHS} = (4-2)^2 = 4</math>
		+	\|}

-	~~<math>x=\text{4}</math>~~: ~~LHS~~	+	Because we squared the root equation, possible spurious roots turn up and we therefore have to verify the solutions when we go back to the original root equation:
-	~~<math>=\sqrt{3\centerdot 4-8}-2=\sqrt{12-8}-2=2+2=4</math>~~	+
-		+
-	~~RHS~~	+
-	~~<math>=4</math>~~	+

		+	{\|
		+	\|\|<ul><li>''x'' = 3:</li></ul>
		+	\|\|<math>\ \text{LHS} = \sqrt{3\cdot 3-8} + 2 = \sqrt{9-8} + 2 = 1+2 = 3</math> and
		+	\|-
		+	\|\|
		+	\|\|<math>\ \text{RHS} = 3</math>
		+	\|-
		+	\|\|<ul><li>''x'' = 4:</li></ul>
		+	\|\|<math>\ \text{LHS} = \sqrt{3\cdot 4-8}-2 = \sqrt{12-8}+2 = 2+2 = 4</math> and
		+	\|-
		+	\|\|
		+	\|\|<math>\ \text{RHS} = 4</math>
		+	\|}

-	The solutions to the root equation are	+	The solutions to the root equation are <math>x=3</math> and <math>x=4</math>.
-	<math>x=~~\text{~~3 }</math>	+
-	and	+
-	<math>x=~~\text{~~4}</math>.	+

Ian at 13:15, 23 September 2008

Ian — Tue, 23 Sep 2008 13:15:03 GMT

←Older revision		Revision as of 13:15, 23 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	First, we move the
-	<~~center~~> ~~[[Image:3_2_3~~-1(2)~~.gif]]~~ </~~center~~>	+	<math>\text{2}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	to the right-hand side to get
-	{{~~NAVCONTENT_START~~}}	+	<math>\sqrt{3x-8}=x-2</math>
-	<~~center~~> ~~[[Image~~:~~3_2_3~~-2(2)~~.gif]]~~ </~~center~~>	+	and then square away the root sign,
-	{{~~NAVCONTENT_STOP~~}}	+
		+	(*)
		+	<math>3x-8=\left( x-2 \right)^{2}</math>
		+
		+
		+	or, with the right-hand side expanded
		+
		+
		+	<math>3x-8=x^{2}-4x+4</math>
		+
		+
		+	If we move over all the terms to the left-hand side, we get
		+
		+
		+	<math>x^{2}-7x+12=0</math>
		+
		+
		+	If we complete the square of the left-hand side,
		+
		+
		+	<math>\begin{align}
		+	& x^{2}-7x+12=\left( x-\frac{7}{2} \right)^{2}-\left( \frac{7}{2} \right)^{2}+12 \\
		+	& =\left( x-\frac{7}{2} \right)^{2}-\frac{49}{4}+\frac{48}{4} \\
		+	& =\left( x-\frac{7}{2} \right)^{2}-\frac{1}{4} \\
		+	\end{align}</math>
		+
		+
		+	the equation can be written as
		+
		+
		+	<math>\left( x-\frac{7}{2} \right)^{2}=\frac{1}{4}</math>
		+
		+
		+	and the solutions are
		+
		+
		+	<math>x=\frac{7}{2}+\sqrt{\frac{1}{4}}=\frac{7}{2}+\frac{1}{2}=\frac{8}{2}=4</math>
		+
		+
		+	<math>x=\frac{7}{2}-\sqrt{\frac{1}{4}}=\frac{7}{2}-\frac{1}{2}=\frac{6}{2}=3</math>
		+
		+	To be on the safe side, we verify that
		+	<math>x=\text{3 }</math>
		+	and
		+	<math>x=\text{4}</math>
		+	satisfy the squared equation (*)
		+
		+
		+	<math>x=\text{3 }</math>: LHS
		+	<math>=3\centerdot 3-8=9-8=1</math>
		+	and RHS
		+	<math>=\left( 3-2 \right)^{2}=1</math>
		+
		+
		+	<math>x=\text{4}</math>: LHS
		+	<math>=3\centerdot 4-8=12-8=4</math>
		+	and RHS
		+	<math>=\left( 4\centerdot 2 \right)^{2}=4</math>
		+
		+
		+	Because we squared the root equation, possible false roots turn up and we therefore have to verify the solutions when we go back to the original root equation:
		+
		+
		+	<math>x=\text{3 }</math>: LHS
		+	<math>=\sqrt{3\centerdot 3-8}+2=\sqrt{9-8}+2=1+2=3</math>
		+
		+	RHS
		+	<math>=3</math>
		+
		+
		+	<math>x=\text{4}</math>: LHS
		+	<math>=\sqrt{3\centerdot 4-8}-2=\sqrt{12-8}-2=2+2=4</math>
		+
		+	RHS
		+	<math>=4</math>
		+
		+
		+	The solutions to the root equation are
		+	<math>x=\text{3 }</math>
		+	and
		+	<math>x=\text{4}</math>.

Tekbot: Lösning 3.2:3 moved to Solution 3.2:3: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:18:03 GMT

Lösning 3.2:3 moved to Solution 3.2:3: Robot: moved page

←Older revision

Revision as of 09:18, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 06:56:09 GMT

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

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Oskar — Tue, 01 Apr 2008 15:05:04 GMT

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