Solution 3.2:6 - Revision history

Tek at 13:17, 1 October 2008

2008-10-01T13:17:58Z

←Older revision		Revision as of 13:17, 1 October 2008
Line 3:		Line 3:
	Squaring once gives		Squaring once gives

		+	{{Displayed math\|\|<math>\bigl(\sqrt{x+1}+\sqrt{x+5}\,\bigr)^2 = 4^2</math>}}

-	~~<math>\~~left~~( \sqrt{x+1}+\sqrt{x+5} \right)^{2}=4^{2}</math>~~	+	and expanding the left-hand side gives

-	~~and expanding the left-hand side with the squaring rule gives~~	+	{{Displayed math\|\|<math>\bigl(\sqrt{x+1}\bigr)^2 + 2\sqrt{x+1}\sqrt{x+5} + \bigl(\sqrt{x+5}\bigr)^2 = 16</math>}}

		+	which, after simplification, results in the equation

-	<math>~~\left( \sqrt{~~x+1~~} \right)^{2}~~+2\sqrt{x+1}\sqrt{x+5}+~~\left( \sqrt{~~x+5~~} \right)^{2}~~=16~~</math>~~	+	{{Displayed math\|\|<math>x+1+2\sqrt{x+1}\sqrt{x+5}+x+5=16\,\textrm{.}</math>}}
-		+
-		+
-	~~which~~, ~~after simplification, results in the equation:~~	+
-		+
-		+
-	~~<math>x+1+2~~\~~sqrt~~{~~x+1~~}~~\sqrt{x+5}+x+5=16~~</math>	+
-		+

	Moving all the terms, other than the root term, over to the right-hand side,		Moving all the terms, other than the root term, over to the right-hand side,

-		+	{{Displayed math\|\|<math>2\sqrt{x+1}\sqrt{x+5}=-2x+10</math>}}
-	<math>2\sqrt{x+1}\sqrt{x+5}=-2x+10</math>	+
-		+

	and squaring once again,		and squaring once again,

		+	{{Displayed math\|\|<math>\bigl(2\sqrt{x+1}\sqrt{x+5}\bigr)^2 = (-2x+10)^2</math>}}

-	~~<math>\left( 2\sqrt{x+1}\sqrt{x+5} \right)^{2}=\left( -2x+10 \right)^{2}</math>~~	+	at last gives an equation that is completely free of root signs,
-		+
-		+
-	at last gives an equation that is completely free of root signs:	+
-		+
-		+
-	~~<math>4\left( x+1 \right)\left( x+5 \right)=\left( -2x+10 \right)^{2}</math>~~	+

		+	{{Displayed math\|\|<math>4(x+1)(x+5) = (-2x+10)^{2}\,\textrm{.}</math>}}

	Expand both sides		Expand both sides

		+	{{Displayed math\|\|<math>4(x^{2}+6x+5) = 4x^2-40x+100</math>}}

-	~~<math>4\left(~~ x~~^{2}+6x+5 \right)=4x^{2}~~-~~40x+100</math>~~	+	and then cancel the common ''x''²-term,

		+	{{Displayed math\|\|<math>24x+20=-40x+100\,\textrm{.}</math>}}

-	~~and then cancel the common x2-term,~~	+	We can write this equation as <math>64x = 80</math>, which gives
-		+
-		+
-	~~<math>24x+20=-40x+100</math>~~	+
-		+
-		+
-	We can write this equation as	+
-	<math>~~\text{64}x~~=~~\text{8}0~~</math>, which gives	+
-		+
-		+
-	~~<math>x=\frac{80}{64}=\frac{2^{4}\centerdot 5}{2^{6}}=\frac{5}{2^{2}}=\frac{5}{4}</math>~~	+

		+	{{Displayed math\|\|<math>x = \frac{80}{64} = \frac{2^{4}\cdot 5}{2^{6}} = \frac{5}{2^{2}} = \frac{5}{4}\,\textrm{.}</math>}}

	Because we squared our original equation (twice), we have to verify the solution		Because we squared our original equation (twice), we have to verify the solution
-	<math>x={5}/{4~~}\;~~</math>	+	<math>x=5/4</math> in order to be able to rule out that we have a spurious root:
-	in order to be able to rule out that we have a ~~false~~ root:	+
-		+
-	~~LHS=~~	+
-		+
-	~~<math>\begin{align}~~	+
-	~~& =\sqrt{\frac{5}{4}+1}+\sqrt{\frac{5}{4}+5}=\sqrt{\frac{9}{4}}+\sqrt{\frac{25}{4}} \\~~	+
-	~~& =\frac{3}{2}+\frac{5}{2}=\frac{8}{2}=4= \\~~	+
-	~~\end{align}</math>~~	+

-	=RHS	+	{{Displayed math\|\|<math>\begin{align}
		+	\text{LHS} &= \sqrt{\frac{5}{4}+1} + \sqrt{\frac{5}{4}+5} = \sqrt{\frac{9}{4}} + \sqrt{\frac{25}{4}}\\[10pt]
		+	&= \frac{3}{2} + \frac{5}{2} = \frac{8}{2} = 4 = \text{RHS}\,\textrm{.}
		+	\end{align}</math>}}

-	Thus, the equation has the solution	+	Thus, the equation has the solution <math>x=5/4\,</math>.
-	<math>x={5}/{4}\;</math>.	+

Ian at 11:12, 25 September 2008

2008-09-25T11:12:16Z

←Older revision		Revision as of 11:12, 25 September 2008
Line 1:		Line 1:
-	{{~~NAVCONTENT_START~~}}	+	This equation differs from earlier examples in that it contains two root terms, in which case it is not possible to get rid of all square roots at once with one squaring, but rather we need to work in two steps.
-	<~~center~~> ~~[[Image:3_2_6~~-1(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	Squaring once gives
-	{{~~NAVCONTENT_START~~}}	+
-	<~~center~~> ~~[[Image~~:~~3_2_6~~-2(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	<math>\left( \sqrt{x+1}+\sqrt{x+5} \right)^{2}=4^{2}</math>
		+
		+	and expanding the left-hand side with the squaring rule gives
		+
		+
		+	<math>\left( \sqrt{x+1} \right)^{2}+2\sqrt{x+1}\sqrt{x+5}+\left( \sqrt{x+5} \right)^{2}=16</math>
		+
		+
		+	which, after simplification, results in the equation:
		+
		+
		+	<math>x+1+2\sqrt{x+1}\sqrt{x+5}+x+5=16</math>
		+
		+
		+	Moving all the terms, other than the root term, over to the right-hand side,
		+
		+
		+	<math>2\sqrt{x+1}\sqrt{x+5}=-2x+10</math>
		+
		+
		+	and squaring once again,
		+
		+
		+	<math>\left( 2\sqrt{x+1}\sqrt{x+5} \right)^{2}=\left( -2x+10 \right)^{2}</math>
		+
		+
		+	at last gives an equation that is completely free of root signs:
		+
		+
		+	<math>4\left( x+1 \right)\left( x+5 \right)=\left( -2x+10 \right)^{2}</math>
		+
		+
		+	Expand both sides
		+
		+
		+	<math>4\left( x^{2}+6x+5 \right)=4x^{2}-40x+100</math>
		+
		+
		+	and then cancel the common x2-term,
		+
		+
		+	<math>24x+20=-40x+100</math>
		+
		+
		+	We can write this equation as
		+	<math>\text{64}x=\text{8}0</math>, which gives
		+
		+
		+	<math>x=\frac{80}{64}=\frac{2^{4}\centerdot 5}{2^{6}}=\frac{5}{2^{2}}=\frac{5}{4}</math>
		+
		+
		+	Because we squared our original equation (twice), we have to verify the solution
		+	<math>x={5}/{4}\;</math>
		+	in order to be able to rule out that we have a false root:
		+
		+	LHS=
		+
		+	<math>\begin{align}
		+	& =\sqrt{\frac{5}{4}+1}+\sqrt{\frac{5}{4}+5}=\sqrt{\frac{9}{4}}+\sqrt{\frac{25}{4}} \\
		+	& =\frac{3}{2}+\frac{5}{2}=\frac{8}{2}=4= \\
		+	\end{align}</math>
		+
		+	=RHS
		+
		+	Thus, the equation has the solution
		+	<math>x={5}/{4}\;</math>.

Tekbot: Lösning 3.2:6 moved to Solution 3.2:6: Robot: moved page

2008-09-09T09:19:03Z

Lösning 3.2:6 moved to Solution 3.2:6: Robot: moved page

←Older revision

Revision as of 09:19, 9 September 2008

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

2008-08-21T06:56:39Z

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

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2008-04-01T15:09:04Z

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