Solution 2.1:1e - Revision history

Tek at 07:53, 23 September 2008

2008-09-23T07:53:07Z

←Older revision		Revision as of 07:53, 23 September 2008
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-	~~{{NAVCONTENT_START}}~~	+	If we use the rule for squaring <math>(a-b)^2 = a^2-2ab+b^2 </math> with <math> a=x </math> and <math> b=7</math>, we obtain directly that
-	~~<!--center> [[Image:2_1_1e.gif]] </center-->~~	+
-	If we use the rule for squaring <math>(a-b)^2 = a^2-2ab+b^2 </math> with <math> a=x </math> and <math> b=7, </math> we obtain directly that	+

-	:<math> (x-7)^2=x^2-2 \cdot x \cdot 7 + 7^2 = x^2-14x+49.</math>	+	{{Displayed math\|\|<math> (x-7)^2=x^2-2 \cdot x \cdot 7 + 7^2 = x^2-14x+49\,\textrm{.}</math>}}

	An alternative is to write the square as <math> (x-7)\cdot (x-7)</math> and then multiply the brackets in two steps		An alternative is to write the square as <math> (x-7)\cdot (x-7)</math> and then multiply the brackets in two steps

-	<math>	+	{{Displayed math\|\|<math>\begin{align}
-	~~\qquad~~	+	(x-7)\cdot (x-7) &= (x-7)\cdot x - (x-7)\cdot 7 \\[3pt]
-	\begin{align}	+	&= x\cdot x-7 \cdot x -(x\cdot 7 - 7\cdot 7) \\[3pt]
-	(x-7)\cdot (x-7) &= (x-7)\cdot x - (x-7)\cdot 7 \\	+	&= x^2 -7x-(7x-49)\\[3pt]
-	&= x\cdot x-7 \cdot x -(x\cdot 7 - 7\cdot 7) \\	+	& \stackrel{*}= x^2-7x-7x+49 \\[3pt]
-	&= x^2 -7x-(7x-49)\\	+	&= x^2-(7+7)x+49\\[3pt]
-	& \stackrel{*}= x^2-7x-7x+49 \\	+	&= x^2-14x+49\,\textrm{.}
-	&= x^2-(7+7)x+49\\	+	\end{align}</math>}}
-	&= x^2-14x+49	+
-	\end{align}	+
-	</math>	+

	In the line that has been marked with an asterisk, we have removed the bracket and at the same time changed signs on all terms inside the bracket.		In the line that has been marked with an asterisk, we have removed the bracket and at the same time changed signs on all terms inside the bracket.
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-	{{NAVCONTENT_STOP}}

Tekbot: Lösning 2.1:1e moved to Solution 2.1:1e: Robot: moved page

2008-09-09T08:31:43Z

Lösning 2.1:1e moved to Solution 2.1:1e: Robot: moved page

←Older revision

Revision as of 08:31, 9 September 2008

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

2008-08-21T06:32:59Z

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

←Older revision		Revision as of 06:32, 21 August 2008
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-	<!--center> [[~~Bild~~:2_1_1e.gif]] </center-->	+	<!--center> [[Image:2_1_1e.gif]] </center-->
	If we use the rule for squaring <math>(a-b)^2 = a^2-2ab+b^2 </math> with <math> a=x </math> and <math> b=7, </math> we obtain directly that		If we use the rule for squaring <math>(a-b)^2 = a^2-2ab+b^2 </math> with <math> a=x </math> and <math> b=7, </math> we obtain directly that

Madde at 09:35, 13 August 2008

2008-08-13T09:35:24Z

←Older revision		Revision as of 09:35, 13 August 2008
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-	<center> [[Bild:2_1_1e.gif]] </center>	+	<!--center> [[Bild:2_1_1e.gif]] </center-->
		+	If we use the rule for squaring <math>(a-b)^2 = a^2-2ab+b^2 </math> with <math> a=x </math> and <math> b=7, </math> we obtain directly that
		+
		+	:<math> (x-7)^2=x^2-2 \cdot x \cdot 7 + 7^2 = x^2-14x+49.</math>
		+
		+	An alternative is to write the square as <math> (x-7)\cdot (x-7)</math> and then multiply the brackets in two steps
		+
		+	<math>
		+	\qquad
		+	\begin{align}
		+	(x-7)\cdot (x-7) &= (x-7)\cdot x - (x-7)\cdot 7 \\
		+	&= x\cdot x-7 \cdot x -(x\cdot 7 - 7\cdot 7) \\
		+	&= x^2 -7x-(7x-49)\\
		+	& \stackrel{*}= x^2-7x-7x+49 \\
		+	&= x^2-(7+7)x+49\\
		+	&= x^2-14x+49
		+	\end{align}
		+	</math>
		+
		+	In the line that has been marked with an asterisk, we have removed the bracket and at the same time changed signs on all terms inside the bracket.
		+
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2008-03-31T09:25:31Z

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