Solution 2.3:1d - Revision history

Tek at 14:06, 26 September 2008

Tek — Fri, 26 Sep 2008 14:06:28 GMT

←Older revision		Revision as of 14:06, 26 September 2008
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	We apply the standard formula for completing the square,		We apply the standard formula for completing the square,

-		+	{{Displayed math\|\|<math>x^{2}+ax = \Bigl(x+\frac{a}{2}\Bigr)^{2} - \Bigl(\frac{a}{2}\Bigr)^{2}\,\textrm{,}</math>}}
-	<math>x^{2}+ax=\~~left~~( x+\frac{a}{2} \~~right~~)^{2}-\~~left~~( \frac{a}{2} \~~right~~)^{2}</math>	+

	on our expression and this gives		on our expression and this gives

-		+	{{Displayed math\|\|<math>x^{2}+5x = \Bigl(x+\frac{5}{2}\Bigr)^{2} - \Bigl(\frac{5}{2}\Bigr)^{2} = \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4}\,\textrm{.}</math>}}
-	<math>x^{2}+5x=\~~left~~( x+\frac{5}{2} \~~right~~)^{2}-\~~left~~( \frac{5}{2} \~~right~~)^{2}=\~~left~~( x+\frac{5}{2} \~~right~~)^{2}-\frac{25}{4}</math>	+

	The whole expression becomes		The whole expression becomes

		+	{{Displayed math\|\|<math>\begin{align}
		+	x^{2}+5x+3
		+	&= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4}+3\\[5pt]
		+	&= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{25}{4} + \frac{12}{4}\\[5pt]
		+	&= \Bigl(x+\frac{5}{2}\Bigr)^{2} + \frac{12-25}{4}\\[5pt]
		+	&= \Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{13}{4}\,\textrm{.}
		+	\end{align}</math>}}

-	~~<math>\begin{align}~~	+	A quick check shows that we have calculated correctly
-	~~& x^{2}+5x+3=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}+3=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{12}{4} \\~~	+
-	~~& =\left( x+\frac{5}{2} \right)^{2}+\frac{12-25}{4}=\left( x+\frac{5}{2} \right)^{2}-\frac{13}{4} \\~~	+
-	~~\end{align}</math>~~	+
-		+
-		+
-	A quick check shows that we have calculated correctly.	+
-		+

-	<math>\begin{align}	+	{{Displayed math\|\|<math>\begin{align}
-	& \~~left~~( x+\frac{5}{2} \~~right~~)^{2}-\frac{13}{4}=x^{2}+2\~~centerdot~~ \frac{5}{2}\~~centerdot~~ x+\~~left~~( \frac{5}{2} \~~right~~)^{2}-\frac{13}{4}=x^{2}+5x+\frac{25}{4}-\frac{13}{4} \\	+	\Bigl(x+\frac{5}{2}\Bigr)^{2} - \frac{13}{4}
-	& =x^{2}+5x+\frac{12}{4}=x^{2}+5x+3 \\	+	&= x^{2} + 2\cdot\frac{5}{2}\cdot x + \Bigl(\frac{5}{2}\Bigr)^{2} - \frac{13}{4}\\[5pt]
-	\end{align}</math>	+	&= x^{2} + 5x + \frac{25}{4} - \frac{13}{4}\\[5pt]
		+	&= x^{2} + 5x + \frac{12}{4}\\[5pt]
		+	&= x^{2}+5x+3\,\textrm{.}
		+	\end{align}</math>}}

Ian at 13:02, 20 September 2008

Ian — Sat, 20 Sep 2008 13:02:18 GMT

←Older revision		Revision as of 13:02, 20 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	We apply the standard formula for completing the square,
-	<~~center~~> ~~[[Image:2_3_1d.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>x^{2}+ax=\left( x+\frac{a}{2} \right)^{2}-\left( \frac{a}{2} \right)^{2}</math>
		+
		+	on our expression and this gives
		+
		+
		+	<math>x^{2}+5x=\left( x+\frac{5}{2} \right)^{2}-\left( \frac{5}{2} \right)^{2}=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}</math>
		+
		+	The whole expression becomes
		+
		+
		+	<math>\begin{align}
		+	& x^{2}+5x+3=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}+3=\left( x+\frac{5}{2} \right)^{2}-\frac{25}{4}+\frac{12}{4} \\
		+	& =\left( x+\frac{5}{2} \right)^{2}+\frac{12-25}{4}=\left( x+\frac{5}{2} \right)^{2}-\frac{13}{4} \\
		+	\end{align}</math>
		+
		+
		+	A quick check shows that we have calculated correctly.
		+
		+
		+	<math>\begin{align}
		+	& \left( x+\frac{5}{2} \right)^{2}-\frac{13}{4}=x^{2}+2\centerdot \frac{5}{2}\centerdot x+\left( \frac{5}{2} \right)^{2}-\frac{13}{4}=x^{2}+5x+\frac{25}{4}-\frac{13}{4} \\
		+	& =x^{2}+5x+\frac{12}{4}=x^{2}+5x+3 \\
		+	\end{align}</math>

Tekbot: Lösning 2.3:1d moved to Solution 2.3:1d: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:55:43 GMT

Lösning 2.3:1d moved to Solution 2.3:1d: Robot: moved page

←Older revision

Revision as of 08:55, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 06:44:59 GMT

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

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Oskar — Tue, 01 Apr 2008 09:55:31 GMT

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