3.2 Equations with roots - Revision history

Phil: Minor alterations to language.

2009-01-02T16:14:40Z

Minor alterations to language.

←Older revision		Revision as of 16:14, 2 January 2009
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	{{Displayed math\|\|<math>\sqrt[\scriptstyle3]{x + 2} = x\,\mbox{.}</math>}}		{{Displayed math\|\|<math>\sqrt[\scriptstyle3]{x + 2} = x\,\mbox{.}</math>}}

-	To solve equations with roots we need to get rid of the root sign. The strategy is to reformulate the equation so that the root sign only appears on one side of the equals sign. Then we can square both sides of the equation (in the case of quadratic roots), so that the root sign disappears, and solve the resulting (squared) equation. We have to be careful though, since a solution to the resulting equation might not be a solution to the original equation. We lose information when squaring, since both positive and negative quantities become positive ~~after squaring~~. Therefore~~, upon squaring the equation, we~~ may ~~generate extra roots which solve~~ the squared equation but not the original equation. These extra roots are called spurious roots. ~~Thus we~~ must ~~examine the solutions that appear, and~~ identify ~~the~~ spurious roots by checking whether or not ~~they solve~~ the original equation.	+	To solve equations with roots we need to get rid of the root sign. The strategy is to reformulate the equation so that the root sign only appears on one side of the equals sign. Then we can square both sides of the equation (in the case of quadratic roots), so that the root sign disappears, and solve the resulting (squared) equation. We have to be careful though, since a solution of the resulting equation might not be a solution of the original equation. We lose information when squaring, since both positive and negative quantities become positive. Therefore there may be values that satisfy the squared equation but not the original equation. These extra roots are called spurious roots. We must always identify any spurious roots by checking whether or not each of the solutions we have obtained satisfies the original equation.

	<div class="exempel">		<div class="exempel">
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-	Since we squared the original equation, there is a risk that spurious roots have been introduced, and therefore we need to check whether <math>x=1</math> and <math>x=5</math> are also solutions to the original equation:	+	Since we squared the original equation, there is a risk that spurious roots have been introduced, and therefore we need to check whether <math>x=1</math> and <math>x=5</math> are also solutions of the original equation:

	* <math>x = 1</math> gives that <math>\mbox{LHS} = 2\sqrt{1 - 1} = 0</math> and <math>\mbox{RHS} = 1 - 1 = 0</math>. So <math>\mbox{LHS} = \mbox{RHS}</math> and the equation is satisfied!		* <math>x = 1</math> gives that <math>\mbox{LHS} = 2\sqrt{1 - 1} = 0</math> and <math>\mbox{RHS} = 1 - 1 = 0</math>. So <math>\mbox{LHS} = \mbox{RHS}</math> and the equation is satisfied!
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	'''Keep in mind that...'''		'''Keep in mind that...'''

-	When squaring an equation remember that the solutions obtained might not be ~~the~~ solutions to the original equation ~~i.e.~~ we might generate spurious roots. This is because we lose information when squaring, since minus signs disappear. Therefore, one must verify that the solutions obtained are indeed solutions to the original equation.	+	When squaring an equation remember that the solutions obtained might not all be solutions of the original equation; that is, we might generate spurious roots. This is because we lose information when squaring, since minus signs disappear. Therefore, one must verify that the solutions obtained are indeed solutions of the original equation.

Jdinglis at 14:21, 26 October 2008

2008-10-26T14:21:52Z

←Older revision		Revision as of 14:21, 26 October 2008
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	{{Displayed math\|\|<math>x^2 = 4\mbox{.}</math>}}		{{Displayed math\|\|<math>x^2 = 4\mbox{.}</math>}}

-	This new equation has two solutions <math>x = 2</math> or <math>x = -2</math>. The solution <math>x = 2</math> satisfies the original equation, while <math> x = -2</math> only satisfies the ~~original~~ equation. So we see that the squared equation has more solutions than the original equation. In this case <math> x = -2</math> is a spurious root.	+	This new equation has two solutions <math>x = 2</math> or <math>x = -2</math>. The solution <math>x = 2</math> satisfies the original equation, while <math> x = -2</math> only satisfies the squared equation. So we see that the squared equation has more solutions than the original equation. In this case <math> x = -2</math> is a spurious root.
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Jdinglis at 14:20, 26 October 2008

2008-10-26T14:20:10Z

←Older revision		Revision as of 14:20, 26 October 2008
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	After this section, you will have learned to:		After this section, you will have learned to:
-	*Solve ~~by squaring,~~ simple equations containing roots.	+	*Solve simple equations containing roots.
	*Manage spurious roots, and know when they might appear.		*Manage spurious roots, and know when they might appear.
	}}		}}
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	== Equations with roots ==		== Equations with roots ==

-	There are many different types of equations containing roots~~, some such~~ examples are	+	There are many different types of equations containing roots. Some examples are

	{{Displayed math\|\|<math>\sqrt{x} + 3x = 2\,,</math>}}		{{Displayed math\|\|<math>\sqrt{x} + 3x = 2\,,</math>}}
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	{{Displayed math\|\|<math>\sqrt[\scriptstyle3]{x + 2} = x\,\mbox{.}</math>}}		{{Displayed math\|\|<math>\sqrt[\scriptstyle3]{x + 2} = x\,\mbox{.}</math>}}

-	To solve equations with roots we need to get rid of the root sign. The strategy ~~to achieve this~~ is to ~~rewrite~~ the equation so that the root sign only appears on one side of the equals sign. Then ~~one squares~~ both sides of the equation (in the case of quadratic roots), so that the root sign disappears and ~~solves~~ the resulting (squared) equation. ~~When one squares an equation~~, ~~one must bear in mind that~~ a solution to the resulting equation might not be a solution to the original equation. ~~This is because some minus signs might disappear. One loses~~ information when squaring~~. Both~~ positive and negative quantities become positive after squaring. Therefore, we must examine the solutions that appear~~. We need to verify that~~ they ~~are not only solutions to the squared equation, but even to~~ the original equation.	+	To solve equations with roots we need to get rid of the root sign. The strategy is to reformulate the equation so that the root sign only appears on one side of the equals sign. Then we can square both sides of the equation (in the case of quadratic roots), so that the root sign disappears, and solve the resulting (squared) equation. We have to be careful though, since a solution to the resulting equation might not be a solution to the original equation. We lose information when squaring, since both positive and negative quantities become positive after squaring. Therefore, upon squaring the equation, we may generate extra roots which solve the squared equation but not the original equation. These extra roots are called spurious roots. Thus we must examine the solutions that appear, and identify the spurious roots by checking whether or not they solve the original equation.

	<div class="exempel">		<div class="exempel">
	''' Example 1'''		''' Example 1'''

-	~~The minus disappears when squaring.~~ Consider a simple (trivial) equation	+	Consider a simple (trivial) equation

	{{Displayed math\|\|<math>x = 2\mbox{.}</math>}}		{{Displayed math\|\|<math>x = 2\mbox{.}</math>}}
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	{{Displayed math\|\|<math>x^2 = 4\mbox{.}</math>}}		{{Displayed math\|\|<math>x^2 = 4\mbox{.}</math>}}

-	This new equation has two solutions <math>x = 2</math> or <math>x = -2</math>. The solution <math>x = 2</math> satisfies the original equation, while <math> x = -2</math> ~~is a solution that arose because~~ we squared the original equation.	+	This new equation has two solutions <math>x = 2</math> or <math>x = -2</math>. The solution <math>x = 2</math> satisfies the original equation, while <math> x = -2</math> only satisfies the original equation. So we see that the squared equation has more solutions than the original equation. In this case <math> x = -2</math> is a spurious root.
	</div>		</div>

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-	~~The two in front of the root is a factor. We can divide~~ both sides of the equation ~~by 2, but we can also let the two remain where it is. If we square the equation as it is,~~ we get	+	If we square both sides of the equation we get
-	{{Displayed math\|\|<math>4(x - 1) = (1 - x)^2</math>}}	+	{{Displayed math\|\|<math>4(x - 1) = (1 - x)^2.</math>}}
-	~~and we expand~~ the square on the right-hand side ~~giving~~	+	Expanding the square on the right-hand side we see that
	{{Displayed math\|\|<math>4(x - 1)= 1 - 2x + x^2\,\mbox{.}</math>}}		{{Displayed math\|\|<math>4(x - 1)= 1 - 2x + x^2\,\mbox{.}</math>}}
	This is a quadratic equation, which can be written as		This is a quadratic equation, which can be written as
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-	'''Keep in mind that: '''	+	'''Keep in mind that...'''

-	When squaring an equation ~~bear in mind~~ that the solutions obtained might not be the solutions to the original equation~~, so called~~ spurious roots. This is because ~~potential~~ minus signs disappear~~. One loses information when squaring~~. Therefore, one must verify that the solutions obtained~~, not only are solutions to the squared equation, but also~~ are solutions to the original equation.	+	When squaring an equation remember that the solutions obtained might not be the solutions to the original equation i.e. we might generate spurious roots. This is because we lose information when squaring, since minus signs disappear. Therefore, one must verify that the solutions obtained are indeed solutions to the original equation.


-	'''You should always test the solution in the original equation ~~containing roots~~.'''	+	'''You should always test the solution in the original equation.'''


	'''Reviews'''		'''Reviews'''

-	For those of you who want to deepen your ~~studies~~ or need more detailed explanations consider the following reference	+	For those of you who want to deepen your understanding or need more detailed explanations consider the following reference:

	[http://www.jamesbrennan.org/algebra/ Understanding Algebra - English online book for pre-university studies ]		[http://www.jamesbrennan.org/algebra/ Understanding Algebra - English online book for pre-university studies ]
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	'''Useful web sites'''		'''Useful web sites'''

-	[http://www.webmath.com/simpsqrt.html ~~What is the~~ ~~root of -?~~ Webmath.com ~~helps~~ you to simplify root expressions.]	+	[http://www.webmath.com/simpsqrt.html Webmath.com can help you to simplify root expressions.]


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2008-09-12T12:53:44Z

Robot: Automated text replacement (-{{:3.2 - Figur - Kurvorna y = 2√(x - 1) och y = 1 - x}} +{{:3.2 - Figure - The curves y = 2√(x - 1) and y = 1 - x}})

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	Thus the equation has only one solution <math>x = 1</math>.		Thus the equation has only one solution <math>x = 1</math>.

-	<center>{{:3.2 - ~~Figur~~ - ~~Kurvorna~~ y = 2√(x - 1) ~~och~~ y = 1 - x}}</center>	+	<center>{{:3.2 - Figure - The curves y = 2√(x - 1) and y = 1 - x}}</center>

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-	[[3.2 ~~Övningar~~\|Exercises]]	+	[[3.2 Exercises\|Exercises]]

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	There are many different types of equations containing roots, some such examples are		There are many different types of equations containing roots, some such examples are

-	{{~~Fristående formel~~\|\|<math>\sqrt{x} + 3x = 2\,,</math>}}	+	{{Displayed math\|\|<math>\sqrt{x} + 3x = 2\,,</math>}}
-	{{~~Fristående formel~~\|\|<math>\sqrt{x - 1} - 2x = x^2\,,</math>}}	+	{{Displayed math\|\|<math>\sqrt{x - 1} - 2x = x^2\,,</math>}}
-	{{~~Fristående formel~~\|\|<math>\sqrt[\scriptstyle3]{x + 2} = x\,\mbox{.}</math>}}	+	{{Displayed math\|\|<math>\sqrt[\scriptstyle3]{x + 2} = x\,\mbox{.}</math>}}

	To solve equations with roots we need to get rid of the root sign. The strategy to achieve this is to rewrite the equation so that the root sign only appears on one side of the equals sign. Then one squares both sides of the equation (in the case of quadratic roots), so that the root sign disappears and solves the resulting (squared) equation. When one squares an equation, one must bear in mind that a solution to the resulting equation might not be a solution to the original equation. This is because some minus signs might disappear. One loses information when squaring. Both positive and negative quantities become positive after squaring. Therefore, we must examine the solutions that appear. We need to verify that they are not only solutions to the squared equation, but even to the original equation.		To solve equations with roots we need to get rid of the root sign. The strategy to achieve this is to rewrite the equation so that the root sign only appears on one side of the equals sign. Then one squares both sides of the equation (in the case of quadratic roots), so that the root sign disappears and solves the resulting (squared) equation. When one squares an equation, one must bear in mind that a solution to the resulting equation might not be a solution to the original equation. This is because some minus signs might disappear. One loses information when squaring. Both positive and negative quantities become positive after squaring. Therefore, we must examine the solutions that appear. We need to verify that they are not only solutions to the squared equation, but even to the original equation.
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	The minus disappears when squaring. Consider a simple (trivial) equation		The minus disappears when squaring. Consider a simple (trivial) equation

-	{{~~Fristående formel~~\|\|<math>x = 2\mbox{.}</math>}}	+	{{Displayed math\|\|<math>x = 2\mbox{.}</math>}}

	If we square both sides of this equation, we get		If we square both sides of this equation, we get

-	{{~~Fristående formel~~\|\|<math>x^2 = 4\mbox{.}</math>}}	+	{{Displayed math\|\|<math>x^2 = 4\mbox{.}</math>}}

	This new equation has two solutions <math>x = 2</math> or <math>x = -2</math>. The solution <math>x = 2</math> satisfies the original equation, while <math> x = -2</math> is a solution that arose because we squared the original equation.		This new equation has two solutions <math>x = 2</math> or <math>x = -2</math>. The solution <math>x = 2</math> satisfies the original equation, while <math> x = -2</math> is a solution that arose because we squared the original equation.
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	The two in front of the root is a factor. We can divide both sides of the equation by 2, but we can also let the two remain where it is. If we square the equation as it is, we get		The two in front of the root is a factor. We can divide both sides of the equation by 2, but we can also let the two remain where it is. If we square the equation as it is, we get
-	{{~~Fristående formel~~\|\|<math>4(x - 1) = (1 - x)^2</math>}}	+	{{Displayed math\|\|<math>4(x - 1) = (1 - x)^2</math>}}
	and we expand the square on the right-hand side giving		and we expand the square on the right-hand side giving
-	{{~~Fristående formel~~\|\|<math>4(x - 1)= 1 - 2x + x^2\,\mbox{.}</math>}}	+	{{Displayed math\|\|<math>4(x - 1)= 1 - 2x + x^2\,\mbox{.}</math>}}
	This is a quadratic equation, which can be written as		This is a quadratic equation, which can be written as
-	{{~~Fristående formel~~\|\|<math>x^2 - 6x + 5 = 0\,\mbox{.}</math>}}	+	{{Displayed math\|\|<math>x^2 - 6x + 5 = 0\,\mbox{.}</math>}}

	This can be solved by completing the square or by using the general solution formula. Either way the solutions are <math>x = 3 \pm 2</math>, i.e. <math>x = 1</math> or <math>x = 5</math>.		This can be solved by completing the square or by using the general solution formula. Either way the solutions are <math>x = 3 \pm 2</math>, i.e. <math>x = 1</math> or <math>x = 5</math>.