Solution 2.1:2d - Revision history

Tek at 08:22, 23 September 2008

Tek — Tue, 23 Sep 2008 08:22:25 GMT

←Older revision		Revision as of 08:22, 23 September 2008
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-	~~{{NAVCONTENT_START}}~~	+	The three brackets can be multiplied together in any order, but just in this case it seems appropriate to start by multiplying the first two brackets because we can carry out the multiplication directly by using the conjugate rule,
-	~~<!--center> [[Image:2_1_2d.gif]] </center-->~~	+
-	The three brackets can be multiplied together in any order, but just in this case it seems appropriate to start by multiplying the first two brackets because we can carry out the multiplication directly by using the conjugate rule.	+

-	<math>	+	{{Displayed math\|\|<math>\begin{align}
-	~~\qquad~~	+
-	\begin{align}	+
	(3x^2+2)(3x^2-2)(9x^4+4) &= \big( (3x^2)^2-2^2 \big) (9x^4+4)\\		(3x^2+2)(3x^2-2)(9x^4+4) &= \big( (3x^2)^2-2^2 \big) (9x^4+4)\\
-	&= (9x^4-4)(9x^4+4)	+	&= (9x^4-4)(9x^4+4)\,\textrm{.}
-	\end{align}	+	\end{align}</math>}}
-	</math>	+

-	We happend to obtain an expression which can also be multiplied using the conjugate rule.	+	We happend to obtain an expression which can also be multiplied using the conjugate rule,

-	<math>	+	{{Displayed math\|\|<math>(9x^4-4)(9x^4+4)=(9x^4)^2-4^2=81x^8-16\,\textrm{.}</math>}}
-	~~\qquad~~	+
-	~~\begin{align}~~	+
-	(9x^4-4)(9x^4+4)=(9x^4)^2-4^2=81x^8-16	+
-	\~~end~~{~~align~~}	+
-	</math>	+

-	~~NOTE~~: If we had instead multiplied together the terms in the first and third brackets first, we would also have got the right answer, but the calculations would have been lengthier.	+	Note: If we had instead multiplied together the terms in the first and third brackets first, we would also have got the right answer, but the calculations would have been lengthier.
-		+
-	~~{{NAVCONTENT_STOP}}~~	+

Tekbot: Lösning 2.1:2d moved to Solution 2.1:2d: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:34:03 GMT

Lösning 2.1:2d moved to Solution 2.1:2d: Robot: moved page

←Older revision

Revision as of 08:34, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 06:34:09 GMT

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

←Older revision		Revision as of 06:34, 21 August 2008
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-	<!--center> [[~~Bild~~:2_1_2d.gif]] </center-->	+	<!--center> [[Image:2_1_2d.gif]] </center-->
	The three brackets can be multiplied together in any order, but just in this case it seems appropriate to start by multiplying the first two brackets because we can carry out the multiplication directly by using the conjugate rule.		The three brackets can be multiplied together in any order, but just in this case it seems appropriate to start by multiplying the first two brackets because we can carry out the multiplication directly by using the conjugate rule.

Madde at 12:53, 13 August 2008

Madde — Wed, 13 Aug 2008 12:53:29 GMT

←Older revision		Revision as of 12:53, 13 August 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[Bild:2_1_2d.gif]] </center>	+	<!--center> [[Bild:2_1_2d.gif]] </center-->
	The three brackets can be multiplied together in any order, but just in this case it seems appropriate to start by multiplying the first two brackets because we can carry out the multiplication directly by using the conjugate rule.		The three brackets can be multiplied together in any order, but just in this case it seems appropriate to start by multiplying the first two brackets because we can carry out the multiplication directly by using the conjugate rule.

Madde at 12:53, 13 August 2008

Madde — Wed, 13 Aug 2008 12:53:14 GMT

←Older revision		Revision as of 12:53, 13 August 2008
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	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
	<center> [[Bild:2_1_2d.gif]] </center>		<center> [[Bild:2_1_2d.gif]] </center>
		+	The three brackets can be multiplied together in any order, but just in this case it seems appropriate to start by multiplying the first two brackets because we can carry out the multiplication directly by using the conjugate rule.
		+
		+	<math>
		+	\qquad
		+	\begin{align}
		+	(3x^2+2)(3x^2-2)(9x^4+4) &= \big( (3x^2)^2-2^2 \big) (9x^4+4)\\
		+	&= (9x^4-4)(9x^4+4)
		+	\end{align}
		+	</math>
		+
		+	We happend to obtain an expression which can also be multiplied using the conjugate rule.
		+
		+	<math>
		+	\qquad
		+	\begin{align}
		+	(9x^4-4)(9x^4+4)=(9x^4)^2-4^2=81x^8-16
		+	\end{align}
		+	</math>
		+
		+	NOTE: If we had instead multiplied together the terms in the first and third brackets first, we would also have got the right answer, but the calculations would have been lengthier.
		+
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Lina: Ny sida: {{NAVCONTENT_START}} Bild:2_1_2d.gif {{NAVCONTENT_STOP}}

Lina — Mon, 31 Mar 2008 09:49:29 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:2_1_2d.gif </center> {{NAVCONTENT_STOP}}

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