Solution 2.3:3f - Revision history

Tek at 08:57, 29 September 2008

Tek — Mon, 29 Sep 2008 08:57:38 GMT

←Older revision		Revision as of 08:57, 29 September 2008
Line 1:		Line 1:
-	We can split up the first term on the left-hand side,	+	We can split up the first term on the left-hand side, <math>x(x^{2}-2x)</math>, into factors by taking <math>x</math> outside the bracket, <math>x(x^{2}-2x) = x\cdot x\cdot (x-2)</math> and writing the other term as <math>x\cdot (2-x) = -x(x-2)</math>. From this we see that both terms contain <math>x(x-2)</math> as common factors and, if we take out those, the left-hand side becomes
-	<math>x~~\left~~( x^{2}-2x ~~\right~~)</math>	+
-	, into factors by taking	+
-	<math>x</math>	+
-	outside the bracket,	+
-	<math>x~~\left~~( x^{2}-2x ~~\right~~)=x\~~centerdot~~ x\~~centerdot \left~~( x-2 ~~\right~~)</math>	+
-	and writing the other term as	+
-	<math>x\~~centerdot \left~~( 2-x ~~\right~~)=-x~~\left~~( x-2 ~~\right~~)</math>. From this we see that both terms contain	+
-	<math>x~~\left~~( x-2 ~~\right~~)</math>	+
-	as common factors and, if we take out those, the left-hand side becomes	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& x\left( x^{2}-2x \right)+x\left( 2-x \right)=x^{2}\left( x-2 \right)-x\left( x-2 \right) \\~~	+
-	~~& =x\left( x\left( x-2 \right)-\left( x-2 \right) \right)=x\left( x-2 \right)\left( x-1 \right). \\~~	+
-	~~\end{align}</math>~~	+

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

	The whole equation can be written as		The whole equation can be written as

		+	{{Displayed math\|\|<math>x(x-2)(x-1) = 0</math>}}

-	~~<math>x\left( x-2 \right)\left( x-1 \right)=0</math>~~	+	and this equation is satisfied only when one of the three factors <math>x</math>, <math>x-2</math> or <math>x-1</math> is zero, i.e. the solutions are <math>x=0</math>, <math>x=2</math> and <math>x=1</math>.
-		+
-		+
-	and this equation is satisfied only when one of the three factors	+
-	<math>x</math>,	+
-	<math>x-~~\text{~~2}</math>	+
-	or	+
-	<math>x-~~\text{~~1}</math>	+
-	is zero, i.e. the solutions are	+
-	<math>x=0</math>,	+
-	<math>x=~~\text{~~2 }</math>	+
-	and	+
-	<math>x=~~\text{~~1}</math>.	+

-	Because it is not completely obvious that x x=1 is a solution of the equation, we check that x=1 satisfies the equation, i.e. that we haven't calculated incorrectly:	+	Because it is not completely obvious that <math>x=1</math> is a solution of the equation, we check that <math>x=1</math> satisfies the equation, i.e. that we haven't calculated incorrectly:

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

Ian at 15:28, 20 September 2008

Ian — Sat, 20 Sep 2008 15:28:58 GMT

←Older revision		Revision as of 15:28, 20 September 2008
Line 1:		Line 1:
-	{{~~NAVCONTENT_START~~}}	+	We can split up the first term on the left-hand side,
-	<~~center~~> ~~[[Image:2_3_3f~~.~~gif]]~~ </~~center~~>	+	<math>x\left( x^{2}-2x \right)</math>
-	{{~~NAVCONTENT_STOP~~}}	+	, into factors by taking
		+	<math>x</math>
		+	outside the bracket,
		+	<math>x\left( x^{2}-2x \right)=x\centerdot x\centerdot \left( x-2 \right)</math>
		+	and writing the other term as
		+	<math>x\centerdot \left( 2-x \right)=-x\left( x-2 \right)</math>. From this we see that both terms contain
		+	<math>x\left( x-2 \right)</math>
		+	as common factors and, if we take out those, the left-hand side becomes
		+
		+
		+	<math>\begin{align}
		+	& x\left( x^{2}-2x \right)+x\left( 2-x \right)=x^{2}\left( x-2 \right)-x\left( x-2 \right) \\
		+	& =x\left( x\left( x-2 \right)-\left( x-2 \right) \right)=x\left( x-2 \right)\left( x-1 \right). \\
		+	\end{align}</math>
		+
		+
		+	The whole equation can be written as
		+
		+
		+	<math>x\left( x-2 \right)\left( x-1 \right)=0</math>
		+
		+
		+	and this equation is satisfied only when one of the three factors
		+	<math>x</math>,
		+	<math>x-\text{2}</math>
		+	or
		+	<math>x-\text{1}</math>
		+	is zero, i.e. the solutions are
		+	<math>x=0</math>,
		+	<math>x=\text{2 }</math>
		+	and
		+	<math>x=\text{1}</math>.
		+
		+	Because it is not completely obvious that x x=1 is a solution of the equation, we check that x=1 satisfies the equation, i.e. that we haven't calculated incorrectly:
		+
		+	x=1: LHS
		+	<math>=1\centerdot \left( 1^{2}-2\centerdot 1 \right)+1\centerdot \left( 2-1 \right)=1\centerdot \left( -1 \right)+1\centerdot 1=0=</math>
		+	RHS

Tekbot: Lösning 2.3:3f moved to Solution 2.3:3f: Robot: moved page

Tekbot — Tue, 09 Sep 2008 08:59:43 GMT

Lösning 2.3:3f moved to Solution 2.3:3f: Robot: moved page

←Older revision

Revision as of 08:59, 9 September 2008

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

Tekbot — Thu, 21 Aug 2008 06:46:59 GMT

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

←Older revision		Revision as of 06:46, 21 August 2008
Line 1:		Line 1:
	{{NAVCONTENT_START}}		{{NAVCONTENT_START}}
-	<center> [[~~Bild~~:2_3_3f.gif]] </center>	+	<center> [[Image:2_3_3f.gif]] </center>
	{{NAVCONTENT_STOP}}		{{NAVCONTENT_STOP}}

Oskar: Ny sida: {{NAVCONTENT_START}} Bild:2_3_3f.gif {{NAVCONTENT_STOP}}

Oskar — Tue, 01 Apr 2008 10:05:22 GMT

Ny sida: {{NAVCONTENT_START}} <center> Bild:2_3_3f.gif </center> {{NAVCONTENT_STOP}}

New page

{{NAVCONTENT_START}}
<center> [[Bild:2_3_3f.gif]] </center>
{{NAVCONTENT_STOP}}