Solution 2.2:3d - Revision history

Tek at 10:18, 24 September 2008

Tek — Wed, 24 Sep 2008 10:18:38 GMT

←Older revision		Revision as of 10:18, 24 September 2008
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-	~~(The exercise is taken from an actual exam in Spring Term 1945!)~~	+	There are no common factors on the left-hand side which we can take out, so we choose to expand the three terms on the left-hand side
-		+
-	There are no common factors on the left-hand side which we can take out, so we choose to expand the three terms on the left-hand side:	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& \left( \frac{2}{x}-3 \right)\left( \frac{1}{4x}+\frac{1}{2} \right)=\frac{2}{x}\centerdot \frac{1}{4x}-\frac{2}{x}\centerdot \frac{1}{2}-3\centerdot \frac{1}{4x}-3\centerdot \frac{1}{2} \\~~	+
-	~~& =\frac{1}{2x^{2}}+\frac{1}{x}-\frac{3}{4x}-\frac{3}{2}=\frac{1}{2x^{2}}+\frac{1}{4x}-\frac{3}{2}, \\~~	+
-	~~& \\~~	+
-	~~& \left( \frac{1}{2x}-\frac{2}{3} \right)^{2}=\frac{1}{\left( 2x \right)^{2}}-2\centerdot \frac{1}{2x}\centerdot \frac{2}{3}+\left( \frac{2}{3} \right)^{2} \\~~	+
-	~~& =\frac{1}{4x^{2}}-\frac{2}{3x}+\frac{4}{9}, \\~~	+
-	~~& \\~~	+
-	~~& \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& \left( \frac{1}{2x}+\frac{1}{3} \right)\left( \frac{1}{2x}-\frac{1}{3} \right)=\left\{ \text{conjugate rule} \right\}=\frac{1}{\left( 2x \right)^{2}}-\frac{1}{3^{2}} \\~~	+
-	~~& =\frac{1}{4x^{2}}-\frac{1}{9} \\~~	+
-	~~\end{align}</math>~~	+
-		+

		+	{{Displayed math\|\|<math>\begin{align}
		+	\biggl(\frac{2}{x}-3\biggr)\biggl(\frac{1}{4x}+\frac{1}{2}\biggr)
		+	&= \frac{2}{x}\cdot\frac{1}{4x} - \frac{2}{x}\cdot\frac{1}{2} - 3\cdot\frac{1}{4x} - 3\cdot\frac{1}{2}\\[5pt]
		+	&= \frac{1}{2x^{2}} + \frac{1}{x} - \frac{3}{4x} - \frac{3}{2}\\[5pt]
		+	&= \frac{1}{2x^{2}} + \frac{1}{4x} - \frac{3}{2}\,,\\[15pt]
		+	\biggl(\frac{1}{2x}-\frac{2}{3}\biggr)^{2}
		+	&= \frac{1}{(2x)^{2}} - 2\cdot\frac{1}{2x}\cdot\frac{2}{3} + \biggl(\frac{2}{3}\biggr)^{2}\\[5pt]
		+	&= \frac{1}{4x^{2}}-\frac{2}{3x}+\frac{4}{9}\,,\\[15pt]
		+	\biggl(\frac{1}{2x}+\frac{1}{3}\biggr)\biggl(\frac{1}{2x}-\frac{1}{3}\biggr)
		+	&= \bigl\{\,\text{difference of two squares}\,\}\\[5pt]
		+	&= \frac{1}{(2x)^{2}}-\frac{1}{3^{2}}\\[5pt]
		+	&= \frac{1}{4x^{2}}-\frac{1}{9}\,\textrm{.}
		+	\end{align}</math>}}

	Collecting up terms, the left-hand side becomes		Collecting up terms, the left-hand side becomes

		+	{{Displayed math\|\|<math>\begin{align}
		+	&\biggl(\frac{1}{2x^{2}}+\frac{1}{4x}-\frac{3}{2}\biggr)-\biggl(\frac{1}{4x^{2}}-\frac{2}{3x}+\frac{4}{9}\biggr)-\biggl(\frac{1}{4x^{2}}-\frac{1}{9}\biggr)\\[7pt]
		+	&\qquad\quad{}= \biggl(\frac{1}{2}-\frac{1}{4}-\frac{1}{4}\biggr)\frac{1}{x^{2}} + \biggl(\frac{1}{4}+\frac{2}{3}\biggr)\frac{1}{x} + \biggl(-\frac{3}{2}-\frac{4}{9}+\frac{1}{9}\biggr)\\[7pt]
		+	&\qquad\quad{}= \frac{2-1-1}{4}\,\frac{1}{x^{2}} + \frac{3+2\cdot 4}{3\cdot 4}\,\frac{1}{x} + \frac{-3\cdot 9 - 4\cdot 2 + 1\cdot 2}{2\cdot 9}\\[7pt]
		+	&\qquad\quad{}=\frac{11}{3\cdot 4}\cdot \frac{1}{x}-\frac{33}{2\cdot 9}
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	and because <math>33=3\cdot 11</math>, <math>9=3\cdot 3</math> and <math>4=2\cdot 2</math>, the whole equation can rewritten as
-	~~& \left( \frac{1}{2x^{2}}+\frac{1}{4x}-\frac{3}{2} \right)-\left( \frac{1}{4x^{2}}-\frac{2}{3x}+\frac{4}{9} \right)-\left( \frac{1}{4x^{2}}-\frac{1}{9} \right) \\~~	+
-	~~& =\left( \frac{1}{2}-\frac{1}{4}-\frac{1}{4} \right)\frac{1}{x^{2}}+\left( \frac{1}{4}+\frac{2}{3} \right)\frac{1}{x}+\left( -\frac{3}{2}-\frac{4}{9}+\frac{1}{9} \right) \\~~	+
-	~~& =\frac{2-1-1}{4}\frac{1}{x^{2}}+\frac{3+2\centerdot 4}{3\centerdot 4}\frac{1}{x}+\frac{-3\centerdot 9-4\centerdot 2+1\centerdot 2}{2\centerdot 9} \\~~	+
-	~~& =\frac{11}{3\centerdot 4}\centerdot \frac{1}{x}-\frac{33}{2\centerdot 9} \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-		+
-	and because	+
-	<math>33=3\~~centerdot~~ 11</math>,	+
-	<math>9=3\~~centerdot~~ 3</math>	+
-	and	+
-	<math>4=2\~~centerdot~~ 2</math>, the whole equation can rewritten as	+
-		+
-		+
-	~~<math>\frac{11}{3\centerdot 2\centerdot 2}\centerdot \frac{1}{x}-\frac{3\centerdot 11}{2\centerdot 3\centerdot 3}=0</math>~~	+

		+	{{Displayed math\|\|<math>\frac{11}{3\cdot 2\cdot 2}\cdot \frac{1}{x}-\frac{3\cdot 11}{2\cdot 3\cdot 3}=0\,\textrm{.}</math>}}

	Taking out common factors, we get		Taking out common factors, we get

		+	{{Displayed math\|\|<math>\frac{11}{3\cdot 2}\biggl(\frac{1}{2x}-1\biggr)=0</math>}}

-	~~<math>\frac{11}{3\centerdot 2}\left( \frac{1}{2x}-1 \right)=0</math>~~	+	and then we see that the equation has the solution <math>x=1/2</math>.
-		+
-		+
-	and then we see that the equation has the solution	+
-	<math>x={1}/{2~~}\;~~</math>.	+
-		+
-	~~Finally, we substitute~~	+
-	~~<math>x={1}/{2}\;</math>~~	+
-	~~into the original equation to check that we have calculated correctly~~.	+

		+	Finally, we substitute <math>x=1/2</math> into the original equation to check that we have calculated correctly.

-	<math>\begin{align}	+	{{Displayed math\|\|<math>\begin{align}
-	& \~~left~~( \frac{2}{\frac{1}{2}}-3 \~~right~~)\~~left~~( \frac{1}{4\~~centerdot~~ \frac{1}{2}}+\frac{1}{2} \~~right~~)-\~~left~~( \frac{1}{2\~~centerdot~~ \frac{1}{2}}-\frac{2}{3} \~~right~~)^{2}-\~~left~~( \frac{1}{2\~~centerdot~~ \frac{1}{2}}+\frac{1}{3} \~~right~~)\~~left~~( \frac{1}{2\~~centerdot~~ \frac{1}{2}}-\frac{1}{3} \~~right~~) \\	+	& \biggl(\frac{2}{\frac{1}{2}}-3\biggr)\biggl(\frac{1}{4\cdot\frac{1}{2}} + \frac{1}{2}\biggr) - \biggl(\frac{1}{2\cdot\frac{1}{2}}-\frac{2}{3}\biggr)^{2} - \biggl(\frac{1}{2\cdot\frac{1}{2}}+\frac{1}{3}\biggr)\biggl(\frac{1}{2\cdot\frac{1}{2}}-\frac{1}{3}\biggr)\\[5pt]
-	& \\	+	&\qquad\quad{}= (4-3)\biggl(\frac{1}{2}+\frac{1}{2}\biggr) - \biggl(1-\frac{2}{3}\biggr)^{2} - \biggl(1+\frac{1}{3}\biggr)\biggl(1-\frac{1}{3}\biggr)\\[5pt]
-	& =~~\left~~( 4-3 ~~\right~~)\~~left~~( \frac{1}{2}+\frac{1}{2} \~~right~~)-\~~left~~( 1-\frac{2}{3} \~~right~~)^{2}-\~~left~~( 1+\frac{1}{3} \~~right~~)\~~left~~( 1-\frac{1}{3} \~~right~~) \\	+	&\qquad\quad{}= 1 - \biggl(\frac{1}{3}\biggr)^{2} - \frac{4}{3}\cdot\frac{2}{3}\\[5pt]
-	& \\	+	&\qquad\quad{}= 1 - \frac{1}{9} - \frac{8}{9}\\[5pt]
-	& =1-\~~left~~( \frac{1}{3} \~~right~~)^{2}-\frac{4}{3}\~~centerdot~~ \frac{2}{3}=1-\frac{1}{9}-\frac{8}{9}=0 \\	+	&\qquad\quad{}= 0\,\textrm{.}
-	\end{align}</math>	+	\end{align}</math>}}

Ian at 15:48, 17 September 2008

Ian — Wed, 17 Sep 2008 15:48:05 GMT

←Older revision		Revision as of 15:48, 17 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	(The exercise is taken from an actual exam in Spring Term 1945!)
-	~~<center> [[Image:2_2_3d~~-1(3)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	There are no common factors on the left-hand side which we can take out, so we choose to expand the three terms on the left-hand side:
-	{{~~NAVCONTENT_START~~}}	+
-	<~~center~~> ~~[[Image:2_2_3d~~-2(3)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	<math>\begin{align}
-	{{~~NAVCONTENT_START~~}}	+	& \left( \frac{2}{x}-3 \right)\left( \frac{1}{4x}+\frac{1}{2} \right)=\frac{2}{x}\centerdot \frac{1}{4x}-\frac{2}{x}\centerdot \frac{1}{2}-3\centerdot \frac{1}{4x}-3\centerdot \frac{1}{2} \\
-	<~~center~~> ~~[[Image:2_2_3d-~~3(3).~~gif]]~~ </~~center~~>	+	& =\frac{1}{2x^{2}}+\frac{1}{x}-\frac{3}{4x}-\frac{3}{2}=\frac{1}{2x^{2}}+\frac{1}{4x}-\frac{3}{2}, \\
-	{{~~NAVCONTENT_STOP~~}}	+	& \\
		+	& \left( \frac{1}{2x}-\frac{2}{3} \right)^{2}=\frac{1}{\left( 2x \right)^{2}}-2\centerdot \frac{1}{2x}\centerdot \frac{2}{3}+\left( \frac{2}{3} \right)^{2} \\
		+	& =\frac{1}{4x^{2}}-\frac{2}{3x}+\frac{4}{9}, \\
		+	& \\
		+	& \\
		+	\end{align}</math>
		+
		+
		+	<math>\begin{align}
		+	& \left( \frac{1}{2x}+\frac{1}{3} \right)\left( \frac{1}{2x}-\frac{1}{3} \right)=\left\{ \text{conjugate rule} \right\}=\frac{1}{\left( 2x \right)^{2}}-\frac{1}{3^{2}} \\
		+	& =\frac{1}{4x^{2}}-\frac{1}{9} \\
		+	\end{align}</math>
		+
		+
		+
		+	Collecting up terms, the left-hand side becomes
		+
		+
		+	<math>\begin{align}
		+	& \left( \frac{1}{2x^{2}}+\frac{1}{4x}-\frac{3}{2} \right)-\left( \frac{1}{4x^{2}}-\frac{2}{3x}+\frac{4}{9} \right)-\left( \frac{1}{4x^{2}}-\frac{1}{9} \right) \\
		+	& =\left( \frac{1}{2}-\frac{1}{4}-\frac{1}{4} \right)\frac{1}{x^{2}}+\left( \frac{1}{4}+\frac{2}{3} \right)\frac{1}{x}+\left( -\frac{3}{2}-\frac{4}{9}+\frac{1}{9} \right) \\
		+	& =\frac{2-1-1}{4}\frac{1}{x^{2}}+\frac{3+2\centerdot 4}{3\centerdot 4}\frac{1}{x}+\frac{-3\centerdot 9-4\centerdot 2+1\centerdot 2}{2\centerdot 9} \\
		+	& =\frac{11}{3\centerdot 4}\centerdot \frac{1}{x}-\frac{33}{2\centerdot 9} \\
		+	\end{align}</math>
		+
		+
		+
		+	and because
		+	<math>33=3\centerdot 11</math>,
		+	<math>9=3\centerdot 3</math>
		+	and
		+	<math>4=2\centerdot 2</math>, the whole equation can rewritten as
		+
		+
		+	<math>\frac{11}{3\centerdot 2\centerdot 2}\centerdot \frac{1}{x}-\frac{3\centerdot 11}{2\centerdot 3\centerdot 3}=0</math>
		+
		+
		+	Taking out common factors, we get
		+
		+
		+	<math>\frac{11}{3\centerdot 2}\left( \frac{1}{2x}-1 \right)=0</math>
		+
		+
		+	and then we see that the equation has the solution
		+	<math>x={1}/{2}\;</math>.
		+
		+	Finally, we substitute
		+	<math>x={1}/{2}\;</math>
		+	into the original equation to check that we have calculated correctly.
		+
		+
		+	<math>\begin{align}
		+	& \left( \frac{2}{\frac{1}{2}}-3 \right)\left( \frac{1}{4\centerdot \frac{1}{2}}+\frac{1}{2} \right)-\left( \frac{1}{2\centerdot \frac{1}{2}}-\frac{2}{3} \right)^{2}-\left( \frac{1}{2\centerdot \frac{1}{2}}+\frac{1}{3} \right)\left( \frac{1}{2\centerdot \frac{1}{2}}-\frac{1}{3} \right) \\
		+	& \\
		+	& =\left( 4-3 \right)\left( \frac{1}{2}+\frac{1}{2} \right)-\left( 1-\frac{2}{3} \right)^{2}-\left( 1+\frac{1}{3} \right)\left( 1-\frac{1}{3} \right) \\
		+	& \\
		+	& =1-\left( \frac{1}{3} \right)^{2}-\frac{4}{3}\centerdot \frac{2}{3}=1-\frac{1}{9}-\frac{8}{9}=0 \\
		+	\end{align}</math>

Tekbot: Lösning 2.2:3d moved to Solution 2.2:3d: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:46:03 GMT

Lösning 2.2:3d moved to Solution 2.2:3d: Robot: moved page

←Older revision

Revision as of 08:46, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 06:40:09 GMT

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

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Lina — Mon, 31 Mar 2008 14:41:40 GMT

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