Solution 2.2:2d - Revision history

Tek at 14:39, 23 September 2008

2008-09-23T14:39:13Z

←Older revision		Revision as of 14:39, 23 September 2008
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	First, we move all the terms over to the left-hand side		First, we move all the terms over to the left-hand side

		+	{{Displayed math\|\|<math>(x^{2}+4x+1)^{2}+3x^{4}-2x^{2}-(2x^{2}+2x+3)^{2}=0\,\textrm{.}</math>}}

-	~~<math>\left( x^{2}+4x+1 \right)^{2}+3x^{4}-2x^{2}-\left( 2x^{2}+2x+3 \right)^{2}=0</math>~~	+	As the equation stands now, it seems that the best approach for solving the equation is to expand the squares, simplify and see what it leads to.

		+	When the squares are expanded, each term inside a square is multiplied by itself and all other terms

-	~~As the equation stands now, it seems that the best approach for solving the equation is to expand the~~	+	{{Displayed math\|\|<math>\begin{align}
-	~~squares, simplify and see what it leads to.~~	+	(x^{2}+4x+1)^{2} &= (x^{2}+4x+1)(x^{2}+4x+1)\\[5pt]
-		+	&= x^{2}\cdot x^{2}+x^{2}\cdot 4x+x^{2}\cdot 1+4x\cdot x^{2}+4x\cdot 4x+4x\cdot 1\\[5pt]
-	~~When the squares are expanded, each term inside a square is multiplied by itself and all other terms:~~	+	&\qquad\quad{}+1\cdot x^{2}+1\cdot 4x+1\cdot 1\\[5pt]
-		+	&= x^{4}+4x^{3}+x^{2}+4x^{3}+16x^{2}+4x+x^{2}+4x+1\\[5pt]
-		+	&= x^{4}+8x^{3}+18x^{2}+8x+1\,,\\[10pt]
-	<math>\begin{align}	+	(2x^{2}+2x+3)^{2} &= (2x^{2}+2x+3)(2x^{2}+2x+3)\\[5pt]
-	~~& \left~~( x^{2}+4x+1 ~~\right~~)^{2}=~~\left~~( x^{2}+4x+1 ~~\right~~)~~\left~~( x^{2}+4x+1 ~~\right~~) \\	+	&= 2x^{2}\cdot 2x^{2}+2x^{2}\cdot 2x+2x^{2}\cdot 3+2x\cdot 2x^{2}+2x\cdot 2x\\[5pt]
-	& =x^{2}\~~centerdot~~ x^{2}+x^{2}\~~centerdot~~ 4x+x^{2}\~~centerdot~~ 1+4x\~~centerdot~~ x^{2}+4x\~~centerdot~~ 4x+4x\~~centerdot~~ 1+1\~~centerdot~~ x^{2}+1\~~centerdot~~ 4x+1\~~centerdot~~ 1 \\	+	&\qquad\quad{}+2x\cdot 3+3\cdot 2x^{2}+3\cdot 2x+3\cdot 3\\[5pt]
-	& =x^{4}+4x^{3}+x^{2}+4x^{3}+16x^{2}+4x+x^{2}+4x+1 \\	+	&= 4x^{4}+4x^{3}+6x^{2}+4x^{3}+4x^{2}+6x+6x^{2}+6x+9 \\[5pt]
-	& =x^{4}+8x^{3}+18x^{2}+8x+1 \\	+	&= 4x^{4}+8x^{3}+16x^{2}+12x+9\,\textrm{.}
-	\~~end{align}</math>~~	+	\end{align}</math>}}
-		+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& \left~~( 2x^{2}+2x+3 ~~\right~~)^{2}=~~\left~~( 2x^{2}+2x+3 ~~\right~~)~~\left~~( 2x^{2}+2x+3 ~~\right~~) \\	+
-	& =2x^{2}\~~centerdot~~ 2x^{2}+2x^{2}\~~centerdot~~ 2x+2x^{2}\~~centerdot~~ 3+2x\~~centerdot~~ 2x^{2}+2x\~~centerdot~~ 2x+2x\~~centerdot~~ 3+3\~~centerdot~~ 2x^{2}+3\~~centerdot~~ 2x+3\~~centerdot~~ 3 \\	+
-	& =4x^{4}+4x^{3}+6x^{2}+4x^{3}+4x^{2}+6x+6x^{2}+6x+9 \\	+
-	& =4x^{4}+8x^{3}+16x^{2}+12x+9 \\	+
-	\end{align}</math>	+
-		+

	After we collect together all terms of the same order, the left hand side becomes		After we collect together all terms of the same order, the left hand side becomes

-		+	{{Displayed math\|\|<math>\begin{align}
-	<math>\begin{align}	+	&(x^{2}+4x+1)^{2}+3x^{4}-2x^{2}-(2x^{2}+2x+3)^{2}\\[5pt]
-	& ~~\left~~( x^{2}+4x+1 ~~\right~~)^{2}+3x^{4}-2x^{2}-~~\left~~( 2x^{2}+2x+3 ~~\right~~)^{2} \\	+	&\qquad{}= (x^{4}+8x^{3}+18x^{2}+8x+1)+3x^{4}-2x^{2}\\[5pt]
-	& =\~~left~~( x^{4}+8x^{3}+18x^{2}+8x+1 ~~\right~~)+3x^{4}-2x^{2}-\~~left~~( 4x^{4}+8x^{3}+16x^{2}+12x+9 ~~\right~~) \\	+	&\qquad\qquad{}-(4x^{4}+8x^{3}+16x^{2}+12x+9)\\[5pt]
-	& =\~~left~~( x^{4}+3x^{4}-4x^{4} ~~\right~~)+~~\left~~( 8x^{3}-8x^{3} ~~\right~~)+~~\left~~( 18x^{2}-2x^{2}-16x^{2} ~~\right~~)+\~~left~~( 8x-12x ~~\right~~)+~~\left~~( 1-9 ~~\right~~) \\	+	&\qquad{}= (x^{4}+3x^{4}-4x^{4})+(8x^{3}-8x^{3})+(18x^{2}-2x^{2}-16x^{2})\\[5pt]
-	& =-4x-8 \\	+	&\qquad\qquad{}+(8x-12x)+(1-9)\\[5pt]
-	\end{align}</math>	+	&\qquad{}= -4x-8\,\textrm{.}
-		+	\end{align}</math>}}
-	~~<math></math>~~	+
-		+

	After all simplifications, the equation becomes		After all simplifications, the equation becomes

		+	{{Displayed math\|\|<math>-4x-8=0\quad \Leftrightarrow \quad x=-2\,\textrm{.}</math>}}

-	~~<math>-4x-8=0\quad \Leftrightarrow \quad x=-2</math>~~	+	Finally, we check that <math>x=-2</math> is the correct answer by substituting
-		+	<math>x=-2</math> into the equation
-	Finally, we check that	+
-	<math>x=-2</math>	+
-	is the correct answer by substituting	+
-	<math>x=-2</math>	+
-	into the equation	+
-		+
-		+
-	~~<math>\begin{align}~~	+
-	~~& \text{LHS}\quad =\quad \left( \left( -2 \right)^{2}+4\left( -2 \right)+1 \right)^{2}+3\centerdot \left( -2 \right)^{4}-2\centerdot \left( -2 \right)^{2} \\~~	+
-	~~& =\left( 4-8+1 \right)^{2}+3\centerdot 16-2\centerdot 4=\left( -3 \right)^{2}+48-8=9+48-8=49 \\~~	+
-	~~\end{align}</math>~~	+
-		+

-	<math>\~~text~~{~~RHS~~}\~~quad~~ =\~~quad~~ \~~left~~( 2\~~left~~( -2 \~~right~~)^{2}+2\~~centerdot~~ \~~left~~( -2 \~~right~~)+3 \~~right~~)^{2}=~~\left~~( 2\~~centerdot~~ 4-4+3 ~~\right~~)^{2}=7^{2}=49</math>	+	{{Displayed math\|\|<math>\begin{align}
		+	\text{LHS} &= \bigl((-2)^{2}+4\cdot(-2)+1\bigr)^{2}+3\cdot (-2)^{4}-2\cdot (-2)^{2}\\[5pt]
		+	&= (4-8+1)^{2} + 3\cdot 16 - 2\cdot 4 = (-3)^{2} + 48 - 8 = 9 + 48 - 8 = 49\,,\\[10pt]
		+	\text{RHS} &= \bigl(2\cdot(-2)^{2}+2\cdot (-2)+3\bigr)^{2} = (2\cdot 4-4+3)^{2} = 7^{2} = 49\,\textrm{.}
		+	\end{align}</math>}}

Ian at 13:48, 17 September 2008

2008-09-17T13:48:53Z

←Older revision		Revision as of 13:48, 17 September 2008
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-	{{~~NAVCONTENT_START~~}}	+	First, we move all the terms over to the left-hand side
-	<~~center~~> ~~[[Image~~:~~2_2_2d-~~1(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
-	{{~~NAVCONTENT_START~~}}	+	<math>\left( x^{2}+4x+1 \right)^{2}+3x^{4}-2x^{2}-\left( 2x^{2}+2x+3 \right)^{2}=0</math>
-	<~~center~~> ~~[[Image:2_2_2d~~-2(2)~~.gif]]~~ </~~center~~>	+
-	{{~~NAVCONTENT_STOP~~}}	+
		+	As the equation stands now, it seems that the best approach for solving the equation is to expand the
		+	squares, simplify and see what it leads to.
		+
		+	When the squares are expanded, each term inside a square is multiplied by itself and all other terms:
		+
		+
		+	<math>\begin{align}
		+	& \left( x^{2}+4x+1 \right)^{2}=\left( x^{2}+4x+1 \right)\left( x^{2}+4x+1 \right) \\
		+	& =x^{2}\centerdot x^{2}+x^{2}\centerdot 4x+x^{2}\centerdot 1+4x\centerdot x^{2}+4x\centerdot 4x+4x\centerdot 1+1\centerdot x^{2}+1\centerdot 4x+1\centerdot 1 \\
		+	& =x^{4}+4x^{3}+x^{2}+4x^{3}+16x^{2}+4x+x^{2}+4x+1 \\
		+	& =x^{4}+8x^{3}+18x^{2}+8x+1 \\
		+	\end{align}</math>
		+
		+
		+
		+	<math>\begin{align}
		+	& \left( 2x^{2}+2x+3 \right)^{2}=\left( 2x^{2}+2x+3 \right)\left( 2x^{2}+2x+3 \right) \\
		+	& =2x^{2}\centerdot 2x^{2}+2x^{2}\centerdot 2x+2x^{2}\centerdot 3+2x\centerdot 2x^{2}+2x\centerdot 2x+2x\centerdot 3+3\centerdot 2x^{2}+3\centerdot 2x+3\centerdot 3 \\
		+	& =4x^{4}+4x^{3}+6x^{2}+4x^{3}+4x^{2}+6x+6x^{2}+6x+9 \\
		+	& =4x^{4}+8x^{3}+16x^{2}+12x+9 \\
		+	\end{align}</math>
		+
		+
		+	After we collect together all terms of the same order, the left hand side becomes
		+
		+
		+	<math>\begin{align}
		+	& \left( x^{2}+4x+1 \right)^{2}+3x^{4}-2x^{2}-\left( 2x^{2}+2x+3 \right)^{2} \\
		+	& =\left( x^{4}+8x^{3}+18x^{2}+8x+1 \right)+3x^{4}-2x^{2}-\left( 4x^{4}+8x^{3}+16x^{2}+12x+9 \right) \\
		+	& =\left( x^{4}+3x^{4}-4x^{4} \right)+\left( 8x^{3}-8x^{3} \right)+\left( 18x^{2}-2x^{2}-16x^{2} \right)+\left( 8x-12x \right)+\left( 1-9 \right) \\
		+	& =-4x-8 \\
		+	\end{align}</math>
		+
		+	<math></math>
		+
		+
		+	After all simplifications, the equation becomes
		+
		+
		+	<math>-4x-8=0\quad \Leftrightarrow \quad x=-2</math>
		+
		+	Finally, we check that
		+	<math>x=-2</math>
		+	is the correct answer by substituting
		+	<math>x=-2</math>
		+	into the equation
		+
		+
		+	<math>\begin{align}
		+	& \text{LHS}\quad =\quad \left( \left( -2 \right)^{2}+4\left( -2 \right)+1 \right)^{2}+3\centerdot \left( -2 \right)^{4}-2\centerdot \left( -2 \right)^{2} \\
		+	& =\left( 4-8+1 \right)^{2}+3\centerdot 16-2\centerdot 4=\left( -3 \right)^{2}+48-8=9+48-8=49 \\
		+	\end{align}</math>
		+
		+
		+	<math>\text{RHS}\quad =\quad \left( 2\left( -2 \right)^{2}+2\centerdot \left( -2 \right)+3 \right)^{2}=\left( 2\centerdot 4-4+3 \right)^{2}=7^{2}=49</math>

Tekbot: Lösning 2.2:2d moved to Solution 2.2:2d: Robot: moved page

2008-09-09T08:44:43Z

Lösning 2.2:2d moved to Solution 2.2:2d: Robot: moved page

←Older revision

Revision as of 08:44, 9 September 2008

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

2008-08-21T06:39:29Z

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

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