Solution 3.2:5 - Revision history

Tek at 12:58, 1 October 2008

Tek — Wed, 01 Oct 2008 12:58:37 GMT

←Older revision		Revision as of 12:58, 1 October 2008
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	After squaring both sides, we obtain the equation		After squaring both sides, we obtain the equation

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

-	~~<math>3x-2=\left( 2-x \right)^{2}\quad \quad (*)</math>~~	+	and if we expand the right-hand side and then collect the terms, we get
-		+
-		+
-	and if we expand the right-hand side and then collect ~~gather~~ the terms, we get	+
-		+
-		+
-	~~<math>x^{2}-7x+6=0</math>~~	+

		+	{{Displayed math\|\|<math>x^{2}-7x+6=0\,\textrm{.}</math>}}

	Completing the square of the left-hand side, we obtain		Completing the square of the left-hand side, we obtain

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	x^{2}-7x+6
-	& x^{2}-7x+6=\~~left~~( x-\frac{7}{2} \~~right~~)^{2}-\~~left~~( \frac{7}{2} \~~right~~)^{2}+6 \\	+	&= \Bigl(x-\frac{7}{2}\Bigr)^2 - \Bigl(\frac{7}{2}\Bigr)^2+6\\[5pt]
-	& =\~~left~~( x-\frac{7}{2} \~~right~~)^{2}-\frac{49}{4}+\frac{24}{4} \\	+	&= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{49}{4} + \frac{24}{4}\\[5pt]
-	& =\~~left~~( x-\frac{7}{2} \~~right~~)^{2}-\frac{25}{4} \\	+	&= \Bigl(x-\frac{7}{2}\Bigr)^2 - \frac{25}{4}
-	\end{align}</math>	+	\end{align}</math>}}

	which means that the equation can be written as		which means that the equation can be written as

-		+	{{Displayed math\|\|<math>\Bigl(x-\frac{7}{2}\Bigr)^2 = \frac{25}{4}</math>}}
-	<math>\~~left~~( x-\frac{7}{2} \~~right~~)^{2}=\frac{25}{4}</math>	+
-		+

	and the solutions are therefore		and the solutions are therefore

		+	:*<math>x = \frac{7}{2} + \sqrt{\frac{25}{4}} = \frac{7}{2} + \frac{5}{2} = \frac{12}{2} = 6\,,</math>

-	<math>~~\begin{align}~~	+	:*<math>x = \frac{7}{2} - \sqrt{\frac{25}{4}} = \frac{7}{2} - \frac{5}{2} = \frac{2}{2} = 1\,\textrm{.}</math>
-	& x=\frac{7}{2}+\sqrt{\frac{25}{4}~~=}\frac{7}{2}+\frac{5}{2~~}~~=\frac{12}{2}=6 \\~~	+
-	~~& x~~=~~\frac{7}{2}-\sqrt{\frac{25}{4}=}~~\frac{7}{2}-\frac{5}{2}=\frac{2}{2}=1 \\	+
-	~~\end~~{~~align~~}</math>	+
-	~~EQ6~~	+

-	Substituting	+	Substituting <math>x=1</math> and <math>x=6</math> into the quadratic equation (*) shows that we have solved the equation correctly.
-	<math>x=~~\text{~~1 }</math>	+
-	and	+
-	<math>x=~~\text{~~6 }</math>	+
-	into the quadratic equation (*) shows that we have solved the equation correctly.	+

-		+	:*''x'' = 1: <math>\ \text{LHS} = 3\cdot 1-2 = 1\ </math> and <math>\ \text{RHS} = (2-1)^2 = 1</math>
-	<math>x=\text{1 }~~</math>: LHS~~	+
-	~~<math>~~=3\~~centerdot~~ 1-2=1</math>	+
-	and ~~RHS~~	+
-	<math>=~~\left~~( 2-1 ~~\right~~)^{2}=1</math>	+

		+	:*''x'' = 6: <math>\ \text{LHS} = 3\cdot 6-2 = 16\ </math> and <math>\ \text{RHS} = (2-6)^2 = 16</math>

-	~~<math>x=\text{6 }</math>: LHS~~	+	Finally, we need to sort away possible spurious roots to the root equation by verifying the solutions.
-	~~<math>=3\centerdot 6-2=16</math>~~	+
-	~~and RHS~~	+
-	~~<math>=\left( 2-6 \right)^{2}=16</math>~~	+

-		+	:*''x'' = 1: <math>\ \text{LHS} = \sqrt{3\cdot 1-2} = 1\ </math> and <math>\ \text{RHS} = 2-1 = 1</math>
-	~~Finally, we need to sort away possible false roots to the root equation by verifying the solutions.~~	+
-		+
-		+
-	~~<math>~~x=~~\text{~~1 ~~}</math>~~: ~~LHS~~	+
-	<math>=\sqrt{3\~~centerdot~~ 1-2}=1</math>	+
-	and ~~RHS~~	+
-	<math>=2-1=1</math>	+

		+	:*''x'' = 6: <math>\ \text{LHS} = \sqrt{3\cdot 6-2} = 4\ </math> and <math>\ \text{RHS} = 2-6 = -4</math>

-	~~<math>x=\text{6 }</math>: LHS~~	+	This shows that the root equation has the solution <math>x=1\,</math>.
-	~~<math>=\sqrt{3\centerdot 6-2}=4</math>~~	+
-	~~and RHS =~~	+
-	~~<math>=2-6=-4</math>~~	+
-		+
-	This shows that the root equation has the solution	+
-	<math>x=~~\text{~~1}</math>.	+

Ian at 10:54, 25 September 2008

Ian — Thu, 25 Sep 2008 10:54:45 GMT

←Older revision		Revision as of 10:54, 25 September 2008
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-	{{~~NAVCONTENT_START}~~}	+	After squaring both sides, we obtain the equation
-	<~~center~~> ~~[[Image:3_2_5~~-1(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+	<math>3x-2=\left( 2-x \right)^{2}\quad \quad (*)</math>
-	<~~center~~> ~~[[Image~~:~~3_2_5~~-2(2).~~gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	and if we expand the right-hand side and then collect gather the terms, we get
		+
		+
		+	<math>x^{2}-7x+6=0</math>
		+
		+
		+	Completing the square of the left-hand side, we obtain
		+
		+
		+	<math>\begin{align}
		+	& x^{2}-7x+6=\left( x-\frac{7}{2} \right)^{2}-\left( \frac{7}{2} \right)^{2}+6 \\
		+	& =\left( x-\frac{7}{2} \right)^{2}-\frac{49}{4}+\frac{24}{4} \\
		+	& =\left( x-\frac{7}{2} \right)^{2}-\frac{25}{4} \\
		+	\end{align}</math>
		+
		+	which means that the equation can be written as
		+
		+
		+	<math>\left( x-\frac{7}{2} \right)^{2}=\frac{25}{4}</math>
		+
		+
		+	and the solutions are therefore
		+
		+
		+	<math>\begin{align}
		+	& x=\frac{7}{2}+\sqrt{\frac{25}{4}=}\frac{7}{2}+\frac{5}{2}=\frac{12}{2}=6 \\
		+	& x=\frac{7}{2}-\sqrt{\frac{25}{4}=}\frac{7}{2}-\frac{5}{2}=\frac{2}{2}=1 \\
		+	\end{align}</math>
		+	EQ6
		+
		+	Substituting
		+	<math>x=\text{1 }</math>
		+	and
		+	<math>x=\text{6 }</math>
		+	into the quadratic equation (*) shows that we have solved the equation correctly.
		+
		+
		+	<math>x=\text{1 }</math>: LHS
		+	<math>=3\centerdot 1-2=1</math>
		+	and RHS
		+	<math>=\left( 2-1 \right)^{2}=1</math>
		+
		+
		+	<math>x=\text{6 }</math>: LHS
		+	<math>=3\centerdot 6-2=16</math>
		+	and RHS
		+	<math>=\left( 2-6 \right)^{2}=16</math>
		+
		+
		+	Finally, we need to sort away possible false roots to the root equation by verifying the solutions.
		+
		+
		+	<math>x=\text{1 }</math>: LHS
		+	<math>=\sqrt{3\centerdot 1-2}=1</math>
		+	and RHS
		+	<math>=2-1=1</math>
		+
		+
		+	<math>x=\text{6 }</math>: LHS
		+	<math>=\sqrt{3\centerdot 6-2}=4</math>
		+	and RHS =
		+	<math>=2-6=-4</math>
		+
		+	This shows that the root equation has the solution
		+	<math>x=\text{1}</math>.

Tekbot: Lösning 3.2:5 moved to Solution 3.2:5: Robot: moved page

Tekbot — Tue, 09 Sep 2008 09:18:43 GMT

Lösning 3.2:5 moved to Solution 3.2:5: 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 06:56:29 GMT

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

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Oskar at 15:07, 1 April 2008

Oskar — Tue, 01 Apr 2008 15:07:47 GMT

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Oskar — Tue, 01 Apr 2008 15:07:25 GMT

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