Solution 2.3:2c - Revision history

Tek at 07:38, 29 September 2008

Tek — Mon, 29 Sep 2008 07:38:28 GMT

←Older revision		Revision as of 07:38, 29 September 2008
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-	We start by completing the square of the left-hand side:	+	We start by completing the square of the left-hand side,
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& y^{2}+3y+4=\left( y+\frac{3}{2} \right)^{2}-\left( \frac{3}{2} \right)^{2}+4 \\~~	+
-	~~& =\left( y+\frac{3}{2} \right)^{2}-\frac{9}{4}+\frac{16}{4}=\left( y+\frac{3}{2} \right)^{2}+\frac{7}{4} \\~~	+
-	~~\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	y^{2}+3y+4 &= \Bigl(y+\frac{3}{2}\Bigr)^{2} - \Bigl(\frac{3}{2}\Bigr)^{2}+4\\[5pt]
		+	&= \Bigl(y+\frac{3}{2}\Bigr)^{2} - \frac{9}{4} + \frac{16}{4}\\[5pt]
		+	&= \Bigl(y+\frac{3}{2}\Bigr)^{2} + \frac{7}{4}\,\textrm{.}
		+	\end{align}</math>}}

	The equation is then		The equation is then

		+	{{Displayed math\|\|<math>\left( y+\frac{3}{2} \right)^{2}+\frac{7}{4}=0\,\textrm{.}</math>}}

-	~~<math>\left( y+\frac{3}{2} \right)^{2}+\frac{7}{4}=0</math>~~	+	The first term <math>\bigl(y+\tfrac{3}{2}\bigr)^{2}</math> is always greater than or equal to zero because it is a square and <math>\tfrac{7}{4}</math> is a positive number. This means that the left hand side cannot be zero, regardless of how ''y'' is chosen. The equation has no solution.
-		+
-	The first term	+
-	<math>\~~left~~( y+\~~frac~~{3}{2} \~~right~~)^{2}</math>	+
-	is always greater than or equal to zero because it is a square and	+
-	<math>\~~frac~~{7}{4}</math>	+
-	is a positive number. This means that the left hand side cannot be zero, regardless of how	+
-	~~<math>~~y~~</math>~~	+
-	is chosen. The equation has no solution.	+

Ian at 13:37, 20 September 2008

Ian — Sat, 20 Sep 2008 13:37:43 GMT

←Older revision		Revision as of 13:37, 20 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	We start by completing the square of the left-hand side:
-	<~~center~~> ~~[[Image:2_3_2c.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	<math>\begin{align}
		+	& y^{2}+3y+4=\left( y+\frac{3}{2} \right)^{2}-\left( \frac{3}{2} \right)^{2}+4 \\
		+	& =\left( y+\frac{3}{2} \right)^{2}-\frac{9}{4}+\frac{16}{4}=\left( y+\frac{3}{2} \right)^{2}+\frac{7}{4} \\
		+	\end{align}</math>
		+
		+
		+	The equation is then
		+
		+
		+	<math>\left( y+\frac{3}{2} \right)^{2}+\frac{7}{4}=0</math>
		+
		+	The first term
		+	<math>\left( y+\frac{3}{2} \right)^{2}</math>
		+	is always greater than or equal to zero because it is a square and
		+	<math>\frac{7}{4}</math>
		+	is a positive number. This means that the left hand side cannot be zero, regardless of how
		+	<math>y</math>
		+	is chosen. The equation has no solution.

Tekbot: Lösning 2.3:2c moved to Solution 2.3:2c: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:56:43 GMT

Lösning 2.3:2c moved to Solution 2.3:2c: Robot: moved page

←Older revision

Revision as of 08:56, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 06:45:29 GMT

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

←Older revision		Revision as of 06:45, 21 August 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_3_2c.gif]] </center>	+	<center> [[Image:2_3_2c.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:2_3_2c.gif {{NAVCONTENT_STOP}}

Oskar — Tue, 01 Apr 2008 09:57:52 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:2_3_2c.gif </center> {{NAVCONTENT_STOP}}

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{{NAVCONTENT_START}}
<center> [[Bild:2_3_2c.gif]] </center>
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