Solution 2.3:2e - Revision history

Tek at 08:13, 29 September 2008

2008-09-29T08:13:15Z

←Older revision		Revision as of 08:13, 29 September 2008
Line 1:		Line 1:
-	Write the equation in normalized form by dividing both sides by	+	Write the equation in normalized form by dividing both sides by 5,
-	~~<math>~~5~~</math>~~,	+

		+	{{Displayed math\|\|<math>x^{2}+\frac{2}{5}x-\frac{3}{5}=0\,\textrm{.}</math>}}

-	~~<math>x^{2}+\frac{2}{5}x-\frac{3}{5}=0</math>~~	+	Complete the square on the left-hand side,
-		+
-		+
-	~~Completing~~ the square on the left-hand side,	+
-	~~<math></math>~~	+
-		+
-	~~<math></math>~~	+
-		+
-	~~<math></math>~~	+
-		+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& x^{2}+\frac{2}{5}x-\frac{3}{5}=\left( x+\frac{\frac{2}{5}}{2} \right)^{2}-\left( \frac{\frac{2}{5}}{2} \right)^{2}-\frac{3}{5} \\~~	+
-	~~& =\left( x+\frac{1}{5} \right)^{2}-\left( \frac{1}{5} \right)^{2}-\frac{3}{5}=\left( x+\frac{1}{5} \right)^{2}-\frac{1}{25}-\frac{3\centerdot 5}{25}=\left( x+\frac{1}{5} \right)^{2}-\frac{16}{25} \\~~	+
-	~~\end{align}</math>~~	+

		+	{{Displayed math\|\|<math>\begin{align}
		+	x^{2}+\frac{2}{5}x-\frac{3}{5}
		+	&= \Bigl(x+\frac{2/5}{2}\Bigr)^{2} - \Bigl(\frac{2/5}{2}\Bigr)^{2} - \frac{3}{5}\\[5pt]
		+	&= \Bigl(x+\frac{1}{5}\Bigr)^{2} - \Bigl(\frac{1}{5}\Bigr)^{2} - \frac{3}{5}\\[5pt]
		+	&= \Bigl(x+\frac{1}{5}\Bigr)^{2} - \frac{1}{25} - \frac{3\cdot 5}{25}\\[5pt]
		+	&= \Bigl(x+\frac{1}{5}\Bigr)^{2} - \frac{16}{25}\,\textrm{.}
		+	\end{align}</math>}}

	The equation is now rewritten as		The equation is now rewritten as

-		+	{{Displayed math\|\|<math>\left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}\,\textrm{,}</math>}}
-	<math>\left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}</math>	+
-		+

	and taking the root gives the solutions		and taking the root gives the solutions

		+	:*<math>x+\tfrac{1}{5} = \sqrt{\tfrac{16}{25}} = \tfrac{4}{5}</math> because <math>\bigl(\tfrac{4}{5}\bigr)^{2} = \tfrac{16}{25}\,,</math> which gives <math>x=-\tfrac{1}{5}+\tfrac{4}{5}=\tfrac{3}{5},</math>

-	<math>x+\~~frac~~{1}{5}=\sqrt{\~~frac~~{16}{25}}~~=\frac{4}{5}</math>~~	+	:*<math>x+\tfrac{1}{5} = -\sqrt{\tfrac{16}{25}} = -\tfrac{4}{5}\,,</math> which gives <math>x = -\tfrac{1}{5}-\tfrac{4}{5}=-1\,\textrm{.}</math>
-	~~because~~	+
-	~~<math>\left( \frac{4}{5}^{2} \right)=\frac{16}{25}</math>~~	+
-	~~which gives~~	+
-	~~<math>x~~=-\~~frac{1}{5}+\frac~~{4}{5}=\~~frac{3}{5}~~,</math>	+
-		+
-		+
-	~~<math>x+\frac{1}{5}=-\sqrt{\frac{16}{25}}=-\frac{4}{5}</math>~~	+
-	which gives	+
-	<math>x=-\~~frac~~{1}{5}-\~~frac~~{4}{5}=-1,~~</math>~~	+
-		+
-		+
-	~~Finally, we check the answer by substituting~~	+
-	~~<math>x=-~~\~~text~~{1 }</math>	+
-	~~and~~	+
-	~~<math>x=\text{3}/\text{5 }</math>~~	+
-	~~into the equation:~~	+

		+	Finally, we check the answer by substituting <math>x=-1</math> and <math>x=3/5</math> into the equation:

-	<math>~~x=-~~\text{1 }~~</math>: LHS~~	+	:*''x'' = 1: <math>\ \text{LHS} = 5\cdot (-1)^{2} + 2\cdot (-1) - 3 = 5 - 2 - 3 = 0 = \text{RHS,}</math>
-	~~<math>~~=5\~~centerdot \left~~( -1 ~~\right~~)^{2}+2\~~centerdot \left~~( -1 ~~\right~~)-3=5-2-3=0=</math>	+
-	~~RHS~~	+

-	~~<math>~~x=~~\text{~~3}/~~\text{~~5 }</math>: LHS	+	:*''x'' = 3/5: <math>\ \text{LHS} = 5\cdot\bigl(\tfrac{3}{5}\bigr)^{2} + 2\cdot\bigl(\tfrac{3}{5}\bigr) - 3 = 5\cdot\tfrac{9}{25} + \tfrac{6}{5} - \tfrac{3\cdot 5}{5} = 0 = \text{RHS.}</math>
-	~~<math>~~=5\~~centerdot~~ \~~left~~( \~~frac~~{3}{5} \~~right~~)^{2}+2\~~centerdot~~ \~~left~~( \~~frac~~{3}{5} \~~right~~)-3=5\~~centerdot~~ \~~frac~~{9}{25}+\~~frac~~{6}{5}-\~~frac~~{3\~~centerdot~~ 5}{5}=\~~frac~~{~~9+6-15~~}~~{5}=0=~~</math>	+
-	~~RHS~~	+

Ian at 14:17, 20 September 2008

2008-09-20T14:17:04Z

←Older revision		Revision as of 14:17, 20 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	Write the equation in normalized form by dividing both sides by
-	<~~center~~> ~~[[Image~~:~~2_3_2e.gif]]~~ </~~center~~>	+	<math>5</math>,
-	{{~~NAVCONTENT_STOP~~}}	+
		+
		+	<math>x^{2}+\frac{2}{5}x-\frac{3}{5}=0</math>
		+
		+
		+	Completing the square on the left-hand side,
		+	<math></math>
		+
		+	<math></math>
		+
		+	<math></math>
		+
		+
		+
		+	<math>\begin{align}
		+	& x^{2}+\frac{2}{5}x-\frac{3}{5}=\left( x+\frac{\frac{2}{5}}{2} \right)^{2}-\left( \frac{\frac{2}{5}}{2} \right)^{2}-\frac{3}{5} \\
		+	& =\left( x+\frac{1}{5} \right)^{2}-\left( \frac{1}{5} \right)^{2}-\frac{3}{5}=\left( x+\frac{1}{5} \right)^{2}-\frac{1}{25}-\frac{3\centerdot 5}{25}=\left( x+\frac{1}{5} \right)^{2}-\frac{16}{25} \\
		+	\end{align}</math>
		+
		+
		+	The equation is now rewritten as
		+
		+
		+	<math>\left( x+\frac{1}{5} \right)^{2}=\frac{16}{25}</math>
		+
		+
		+	and taking the root gives the solutions
		+
		+
		+	<math>x+\frac{1}{5}=\sqrt{\frac{16}{25}}=\frac{4}{5}</math>
		+	because
		+	<math>\left( \frac{4}{5}^{2} \right)=\frac{16}{25}</math>
		+	which gives
		+	<math>x=-\frac{1}{5}+\frac{4}{5}=\frac{3}{5},</math>
		+
		+
		+	<math>x+\frac{1}{5}=-\sqrt{\frac{16}{25}}=-\frac{4}{5}</math>
		+	which gives
		+	<math>x=-\frac{1}{5}-\frac{4}{5}=-1,</math>
		+
		+
		+	Finally, we check the answer by substituting
		+	<math>x=-\text{1 }</math>
		+	and
		+	<math>x=\text{3}/\text{5 }</math>
		+	into the equation:
		+
		+
		+	<math>x=-\text{1 }</math>: LHS
		+	<math>=5\centerdot \left( -1 \right)^{2}+2\centerdot \left( -1 \right)-3=5-2-3=0=</math>
		+	RHS
		+
		+	<math>x=\text{3}/\text{5 }</math>: LHS
		+	<math>=5\centerdot \left( \frac{3}{5} \right)^{2}+2\centerdot \left( \frac{3}{5} \right)-3=5\centerdot \frac{9}{25}+\frac{6}{5}-\frac{3\centerdot 5}{5}=\frac{9+6-15}{5}=0=</math>
		+	RHS

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

2008-09-09T08:57:23Z

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

←Older revision

Revision as of 08:57, 9 September 2008

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

2008-08-21T06:45:49Z

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

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2008-04-01T09:59:19Z

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