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für 3.2 Wurzelgleichungen

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{| border="0" cellspacing="0" cellpadding="0" height="30" width="100%"
| style="border-bottom:1px solid #797979" width="5px" | &nbsp;
{{Selected tab|[[3.2 Equations with roots|Theory]]}}
{{Not selected tab|[[3.2 Exercises|Exercises]]}}
| style="border-bottom:1px solid #797979"  width="100%"| &nbsp;
|}

{{Info|
'''Contents: '''
* Equations of the type  <math>\sqrt{ax+b}= cx +d </math>  
* Spurious roots
}}

{{Info|
'''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

{{Displayed math||<math>\sqrt{x} + 3x = 2\,,</math>}}
{{Displayed math||<math>\sqrt{x - 1} - 2x = x^2\,,</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. 

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

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

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

If we square both sides of this equation, we get 

{{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. 
</div>

<div class="exempel">
''' Example 2'''

Solve the equation  <math>\ 2\sqrt{x - 1} = 1 - x</math>.


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 
{{Displayed math||<math>4(x - 1) = (1 - x)^2</math>}}
and we expand the square on the right-hand side giving 
{{Displayed math||<math>4(x - 1)= 1 - 2x + x^2\,\mbox{.}</math>}}
This is a quadratic equation, which can be written as
{{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>. 


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: 

* <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 = 5</math>  gives that  <math>\mbox{LHS} = 2\sqrt{5 - 1} = 2\cdot2 = 4</math> and <math>\mbox{RHS} = 1 - 5 = -4</math>. So  <math>\mbox{LHS} \ne \mbox{RHS}</math> and the equation is not satisfied! 

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

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

</div>


[[3.2 Exercises|Exercises]]

<div class="inforuta" style="width:580px;">
'''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  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. 


'''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 reference

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


'''Useful web sites'''

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


</div>

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