Solution 2.3:4b - Revision history

Tek at 10:33, 29 September 2008

2008-09-29T10:33:06Z

←Older revision		Revision as of 10:33, 29 September 2008
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-	A first-degree equation which	+	A first-degree equation which has <math>x=1+\sqrt{3}</math> as a root is <math>x-(1+\sqrt{3}\,)=0</math>, which we can also write as <math>x-1-\sqrt{3} = 0</math>. In the same way, we have that <math>x-(1-\sqrt{3}\,)=0</math>, i.e., <math>x-1+\sqrt{3}=0</math> is a first-degree equation that has <math>x=1-\sqrt{3}</math> as a root. If we multiply these two first-degree equations together, we get a second-degree equation with <math>x=1+\sqrt{3}</math> and <math>x=1-\sqrt{3}</math> as roots,
-	<math>x=~~\text{~~ 1}+\sqrt~~{\text~~{3}}</math>	+
-	as a root is	+
-	<math>x-~~\left~~( ~~\text{~~1}+\sqrt~~{\text~~{3}} \~~right~~)=0</math>, which we can also write as	+

-	~~<math>x-\text~~{~~1-}\sqrt~~{~~\text{3}}=0</~~math~~>. In the same way, we have that~~	+	{{Displayed math\|\|<math>(x-1-\sqrt{3}\,)(x-1+\sqrt{3}\,) = 0\,\textrm{.}</math>}}
-	<math>x-~~\left( \text{~~1-}\sqrt~~{\text~~{3}} \~~right~~)~~=0</math>, i.e.,~~	+
-	~~<math>~~x-~~\text{~~1+}\sqrt~~{\text~~{3}}=0~~</math>~~	+
-	~~is a first-degree equation that has~~	+
-	~~<math>x=~~\~~text{ 1-}~~\~~sqrt~~{~~\text{3}}</math>~~	+
-	~~as a root~~. ~~If we multiply these two first-degree equations together, we get a second-degree equation with~~	+
-	~~<math>x=\text{ 1}+\sqrt{\text{3}~~}</math>	+
-	~~and~~	+
-	~~<math>x=\text{ 1-~~}~~\sqrt{\text{3~~}~~}</math>. as roots:~~	+

		+	The first factor become zero when <math>x=1+\sqrt{3}</math> and the second factor becomes zero when <math>x=1-\sqrt{3}\,</math>.

-	<math>~~\left~~( x-~~\text{~~1-}\sqrt~~{\text~~{3}} \~~right~~)~~\left~~( x-~~\text{~~1+}\sqrt~~{\text~~{3}} \~~right~~)=0</math>	+	Nothing really prevents us from answering with <math>(x-1-\sqrt{3}\,)(x-1+\sqrt{3}\,) = 0</math>, but if we want to give the equation in standard form, we need to expand the left-hand side,

		+	{{Displayed math\|\|<math>\begin{align}
		+	(x-1-\sqrt{3}\,)(x-1+\sqrt{3}\,)
		+	&= x^{2} - x + \sqrt{3}x - x + 1 - \sqrt{3} - \sqrt{3}x + \sqrt{3} - (\sqrt{3}\,)^{2}\\[5pt]
		+	&= x^{2} + (-x+\sqrt{3}x-x-\sqrt{3}x) + (1-\sqrt{3}+\sqrt{3}-3)\\[5pt]
		+	&= x^{2}-2x-2
		+	\end{align}</math>}}

-	~~The first factor become zero when~~	+	to get the equation <math>x^{2}-2x-2=0\,</math>.
-	<math>x~~=\text~~{ 1}~~+\sqrt{\text{3}}</math>~~	+
-	~~and the second factor becomes zero when~~	+
-	~~<math>x~~=\~~text{ 1-}\sqrt{\text{3}}~~</math>.	+

-	Nothing really prevents us from answering with
-	<math>\left( x-\text{1-}\sqrt{\text{3}} \right)\left( x-\text{1+}\sqrt{\text{3}} \right)=0</math>, but if we want to give the equation in standard form, we need to expand the left-hand side,

		+	Note: Exactly as in exercise a, we can multiply the equation by a non-zero constant ''a''

-	~~<math>\begin~~{~~align}~~	+	{{Displayed math\|\|<math>ax^{2}-2ax-2a=0</math>}}
-	~~& \left( x-\text~~{~~1-}\sqrt{\text{3}} \right)\left( x-\text{1+}\sqrt{\text{3}} \right)=x^{2}-x+\sqrt{3}-x+1-\sqrt{3}-\sqrt{3}x+\sqrt{3}-\left( \sqrt{3} \right)^{2} \\~~	+
-	~~& =x^{2}+\left( -x+\sqrt{3}x-x-\sqrt{3}x \right)+\left( 1-\sqrt{3}+\sqrt{3}-3 \right) \\~~	+
-	~~& =x^{2}-2x-2 \\~~	+
-	~~\end{align}</~~math>	+
-		+
-		+
-	~~to get the equation~~	+
-	~~<math>x^{2}-2x-2=0</math>~~	+
-		+
-		+
-	~~NOTE: Exactly as in exercise~~	+
-	~~<math>\text{a}</math>, we can multiply the equation by a non-zero constant~~	+
-	~~<math>a</math>,~~	+
-		+
-		+
-	<math>ax^{2}-2ax-2a=0</math>	+
-		+

	and still have a second-degree equation with the same roots.		and still have a second-degree equation with the same roots.

Ian at 09:29, 21 September 2008

2008-09-21T09:29:54Z

←Older revision		Revision as of 09:29, 21 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	A first-degree equation which
-	<~~center~~> ~~[[Image:2_3_4b~~-1(2).~~gif]]~~ </~~center~~>	+	<math>x=\text{ 1}+\sqrt{\text{3}}</math>
-	{{~~NAVCONTENT_STOP~~}}	+	as a root is
-	{{~~NAVCONTENT_START~~}}	+	<math>x-\left( \text{1}+\sqrt{\text{3}} \right)=0</math>, which we can also write as
-	<~~center~~> ~~[[Image~~:~~2_3_4b~~-2(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+	<math>x-\text{1-}\sqrt{\text{3}}=0</math>. In the same way, we have that
		+	<math>x-\left( \text{1-}\sqrt{\text{3}} \right)=0</math>, i.e.,
		+	<math>x-\text{1+}\sqrt{\text{3}}=0</math>
		+	is a first-degree equation that has
		+	<math>x=\text{ 1-}\sqrt{\text{3}}</math>
		+	as a root. If we multiply these two first-degree equations together, we get a second-degree equation with
		+	<math>x=\text{ 1}+\sqrt{\text{3}}</math>
		+	and
		+	<math>x=\text{ 1-}\sqrt{\text{3}}</math>. as roots:
		+
		+
		+	<math>\left( x-\text{1-}\sqrt{\text{3}} \right)\left( x-\text{1+}\sqrt{\text{3}} \right)=0</math>
		+
		+
		+	The first factor become zero when
		+	<math>x=\text{ 1}+\sqrt{\text{3}}</math>
		+	and the second factor becomes zero when
		+	<math>x=\text{ 1-}\sqrt{\text{3}}</math>.
		+
		+	Nothing really prevents us from answering with
		+	<math>\left( x-\text{1-}\sqrt{\text{3}} \right)\left( x-\text{1+}\sqrt{\text{3}} \right)=0</math>, but if we want to give the equation in standard form, we need to expand the left-hand side,
		+
		+
		+	<math>\begin{align}
		+	& \left( x-\text{1-}\sqrt{\text{3}} \right)\left( x-\text{1+}\sqrt{\text{3}} \right)=x^{2}-x+\sqrt{3}-x+1-\sqrt{3}-\sqrt{3}x+\sqrt{3}-\left( \sqrt{3} \right)^{2} \\
		+	& =x^{2}+\left( -x+\sqrt{3}x-x-\sqrt{3}x \right)+\left( 1-\sqrt{3}+\sqrt{3}-3 \right) \\
		+	& =x^{2}-2x-2 \\
		+	\end{align}</math>
		+
		+
		+	to get the equation
		+	<math>x^{2}-2x-2=0</math>
		+
		+
		+	NOTE: Exactly as in exercise
		+	<math>\text{a}</math>, we can multiply the equation by a non-zero constant
		+	<math>a</math>,
		+
		+
		+	<math>ax^{2}-2ax-2a=0</math>
		+
		+
		+	and still have a second-degree equation with the same roots.

Tekbot: Lösning 2.3:4b moved to Solution 2.3:4b: Robot: moved page

2008-09-09T09:00:23Z

Lösning 2.3:4b moved to Solution 2.3:4b: Robot: moved page

←Older revision

Revision as of 09:00, 9 September 2008

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

2008-08-21T06:47:19Z

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

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2008-04-01T10:08:16Z

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