3.2 Equations with roots

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{{Vald flik|[[3.2 Rotekvationer|Teori]]}}
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{{Vald flik|[[3.2 Rotekvationer|Theory]]}}
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{{Ej vald flik|[[3.2 Övningar|Övningar]]}}
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{{Ej vald flik|[[3.2 Övningar|Exercise]]}}
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{{Info|
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'''Innehåll:'''
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'''Contents: '''
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*Rotekvationer av typen <math>\sqrt{ax+b}= cx +d </math>
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*equations of the type <math>\sqrt{ax+b}= cx +d </math> containing roots
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*Falska rötter
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* spurious roots
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{{Info|
{{Info|
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'''Lärandemål:'''
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'''Learning outcomes'''
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Efter detta avsnitt ska du ha lärt dig att:
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After this section, you will have learned to:
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*Lösa enkla rotekvationer med kvadrering.
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*Solve, by squaring, simple equations containing roots.
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*Hantera falska rötter och veta när de uppstår.
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*Manage spurious roots, and know when they might appear.
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== Rotekvationer ==
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== Equations with roots ==
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Det finns många olika varianter av rotekvationer, t.ex.
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There are many different types of equations containing roots, some such examples are
{{Fristående formel||<math>\sqrt{x} + 3x = 2\,,</math>}}
{{Fristående formel||<math>\sqrt{x} + 3x = 2\,,</math>}}
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{{Fristående formel||<math>\sqrt[\scriptstyle3]{x + 2} = x\,\mbox{.}</math>}}
{{Fristående formel||<math>\sqrt[\scriptstyle3]{x + 2} = x\,\mbox{.}</math>}}
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För att lösa rotekvationer vill man bli av med rottecknet. Strategin för att uppnå detta är att skriva ekvationen så att rottecknet blir ensamt kvar på ena sidan av likhetstecknet. Sedan kvadrerar man båda led i ekvationen (om det handlar om en kvadratrot), så att rottecknet försvinner och löser sedan den nya, kvadrerade, ekvationen. När man kvadrerar en ekvation måste man tänka på att de lösningar som man får fram kanske inte är lösningar till den ursprungliga ekvationen. Detta beror på att eventuella minustecken försvinner. Man tappar information när man kvadrerar. Oavsett om man hade något positivt eller negativt så har man alltid något positivt efter en kvadrering. Därför måste man pröva de lösningar som man får fram. Man behöver verifiera att de inte bara är lösningar till den kvadrerade ekvationen, utan också till den ursprungliga ekvationen.
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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 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.
<div class="exempel">
<div class="exempel">
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'''Exempel 1'''
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''' Example 1'''
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Minustecken försvinner vid kvadrering. Betrakta en enkel (trivial) ekvation
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The minus disappears when squaring. Consider a simple (trivial) equation
{{Fristående formel||<math>x = 2\mbox{.}</math>}}
{{Fristående formel||<math>x = 2\mbox{.}</math>}}
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Om vi kvadrerar båda led i denna ekvation får vi
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If we square both sides of this equation, we get
{{Fristående formel||<math>x^2 = 4\mbox{.}</math>}}
{{Fristående formel||<math>x^2 = 4\mbox{.}</math>}}
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Denna nya ekvation har två lösningar <math>x = 2</math> eller <math>x = -2</math>. Lösningen <math>x = 2</math> uppfyller den ursprungliga ekvationen medan <math>x = -2</math> är en lösning som uppstod i den kvadrerade ekvationen.
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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.
</div>
</div>
<div class="exempel">
<div class="exempel">
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'''Exempel 2'''
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''' Example 2'''
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Lös ekvationen <math>\ 2\sqrt{x - 1} = 1 - x</math>.
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Solve the equation <math>\ 2\sqrt{x - 1} = 1 - x</math>.
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Tvåan framför rottecknet är en faktor. Vi kan dividera vänster- och högerled med 2, men vi kan också låta tvåan stå kvar. Om vi kvadrerar ekvationen som den är får vi
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The two in 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>}}
{{Fristående formel||<math>4(x - 1) = (1 - x)^2</math>}}
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och utvecklar vi kvadraten fås
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and we expand the square on the right-hand side giving
{{Fristående formel||<math>4(x - 1)= 1 - 2x + x^2\,\mbox{.}</math>}}
{{Fristående formel||<math>4(x - 1)= 1 - 2x + x^2\,\mbox{.}</math>}}
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Detta är en andragradsekvation, som kan skrivas
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This is a quadratic equations, which can be written as
{{Fristående formel||<math>x^2 - 6x + 5 = 0\,\mbox{.}</math>}}
{{Fristående formel||<math>x^2 - 6x + 5 = 0\,\mbox{.}</math>}}
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Denna kan lösas med kvadratkomplettering eller med den allmänna lösningsformeln. Lösningarna blir <math>x = 3 \pm 2</math>, dvs. <math>x = 1</math> eller <math>x = 5</math>.
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This can be solved by completing the square or by using the general solution. Either way the solutions are <math>x = 3 \pm 2</math>, i.e. <math>x = 1</math> or <math>x = 5</math>.
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Eftersom vi kvadrerar ekvationen finns risken att detta introducerar falska rötter och därför behöver vi pröva om <math>x=1</math> och <math>x=5</math> också är lösningarna till den ursprungliga rotekvationen:
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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 consider whether <math>x=1</math> and <math>x=5</math> are also solutions to the original equation:
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* <math>x = 1</math> medför att <math>\mbox{VL} = 2\sqrt{1 - 1} = 0</math> och <math>\mbox{HL} = 1 - 1 = 0</math>. Alltså är <math>\mbox{VL} = \mbox{HL}</math> och ekvationen är uppfylld!
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* <math>x = 1</math> gives that <math>\mbox{LS} = 2\sqrt{1 - 1} = 0</math> and <math>\mbox{RS} = 1 - 1 = 0</math>. So <math>\mbox{LS} = \mbox{RS}</math> and the equation is satisfied!
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* <math>x = 5</math> medför att <math>\mbox{VL} = 2\sqrt{5 - 1} = 2\cdot2 = 4</math> och <math>\mbox{HL} = 1 - 5 = -4</math>. Alltså är <math>\mbox{VL} \ne \mbox{HL}</math> och ekvationen är ''inte'' uppfylld!
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* <math>x = 5</math> gives that <math>\mbox{LS} = 2\sqrt{5 - 1} = 2\cdot2 = 4</math> and <math>\mbox{RS} = 1 - 5 = -4</math>. So <math>\mbox{LS} \ne \mbox{RS}</math> and the equation is not satisfied!
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Ekvationen har därmed bara en lösning <math>x = 1</math>.
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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 - Figur - Kurvorna y = 2√(x - 1) och y = 1 - x}}</center>
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[[3.2 Övningar|Övningar]]
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[[3.2 Övningar|Exercises]]
<div class="inforuta" style="width:580px;">
<div class="inforuta" style="width:580px;">
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'''Råd för inläsning'''
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'''Study advice'''
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'''Grund- och slutprov'''
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'''The basic and final tests'''
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Efter att du har läst texten och arbetat med övningarna ska du göra grund- och slutprovet för att bli godkänd på detta avsnitt. Du hittar länken till proven i din student lounge.
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After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.
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'''Tänk på att:'''
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'''Keep in mind that: '''
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När man kvadrerar en ekvation måste man tänka på att de lösningar som man får fram kanske inte är lösningar till den ursprungliga ekvationen, s. k. '''falska rötter'''. Detta beror på att eventuella minustecken försvinner. Man tappar information när man kvadrerar. Därför måste man verifiera att de lösningar man får fram, inte bara är lösningar till den kvadrerade ekvationen, utan också är lösningar till den ursprungliga ekvationen.
+
When squaring an equation bear in mind that the solutions obtain edmight 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.
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'''Du ska alltid pröva lösningen i rotekvationer.'''
 
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'''You should always test the solution in the original equation containing roots..'''
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'''Lästips'''
 
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för dig som vill fördjupa dig ytterligare eller skulle vilja ha en längre förklaring
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'''Reviews'''
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[http://www.jamesbrennan.org/algebra/ Understanding Algebra - engelsk bok på nätet för högskoleförberedande studier]
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For those of you who want to deepen your studies or need more detailed explanations consider the following references
 +
 
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[http://www.jamesbrennan.org/algebra/ Understanding Algebra - English online book for pre-university studies ]
'''Länktips'''
'''Länktips'''
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[http://www.webmath.com/simpsqrt.html Vad är roten ur -- ? Webmath.com hjälper dig att förenkla rotuttryck]
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[http://www.webmath.com/simpsqrt.html What is the root of -? Webmath.com helps you to simplify root expressions.
 +
Exercises]
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Revision as of 13:21, 13 July 2008

       Theory          Exercise      

Contents:

  • equations of the type \displaystyle \sqrt{ax+b}= cx +d containing roots
  • spurious roots

Learning outcomes

After this section, you will have learned to:

  • Solve, by squaring, simple equations containing roots.
  • Manage spurious roots, and know when they might appear.

Equations with roots

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

\displaystyle \sqrt{x} + 3x = 2\,,
\displaystyle \sqrt{x - 1} - 2x = x^2\,,
\displaystyle \sqrt[\scriptstyle3]{x + 2} = x\,\mbox{.}

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 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.

Example 1

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

\displaystyle x = 2\mbox{.}

If we square both sides of this equation, we get

\displaystyle x^2 = 4\mbox{.}

This new equation has two solutions \displaystyle x = 2 or \displaystyle x = -2. The solution \displaystyle x = 2 satisfies the original equation, while \displaystyle x = -2is a solution that arose because we squared the original equation.

Example 2

Solve the equation \displaystyle \ 2\sqrt{x - 1} = 1 - x.


The two in 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

\displaystyle 4(x - 1) = (1 - x)^2

and we expand the square on the right-hand side giving

\displaystyle 4(x - 1)= 1 - 2x + x^2\,\mbox{.}

This is a quadratic equations, which can be written as

\displaystyle x^2 - 6x + 5 = 0\,\mbox{.}

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


Since we squared the original equation, there is a risk that spurious roots have been introduced, and therefore we need to consider whether \displaystyle x=1 and \displaystyle x=5 are also solutions to the original equation:

  • \displaystyle x = 1 gives that \displaystyle \mbox{LS} = 2\sqrt{1 - 1} = 0 and \displaystyle \mbox{RS} = 1 - 1 = 0. So \displaystyle \mbox{LS} = \mbox{RS} and the equation is satisfied!
  • \displaystyle x = 5 gives that \displaystyle \mbox{LS} = 2\sqrt{5 - 1} = 2\cdot2 = 4 and \displaystyle \mbox{RS} = 1 - 5 = -4. So \displaystyle \mbox{LS} \ne \mbox{RS} and the equation is not satisfied!

Thus the equation has only one solution \displaystyle x = 1.

3.2 - Figur - Kurvorna y = 2√(x - 1) och y = 1 - x


Exercises

Study advice

The basic and final tests

After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.


Keep in mind that:

When squaring an equation bear in mind that the solutions obtain edmight 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.


You should always test the solution in the original equation containing roots..


Reviews

For those of you who want to deepen your studies or need more detailed explanations consider the following references

Understanding Algebra - English online book for pre-university studies


Länktips

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

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