3.2 Equations with roots

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After this section, you will have learned to:
After this section, you will have learned to:
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*Solve by squaring, simple equations containing roots.
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*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 ==
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There are many different types of equations containing roots, some such examples are
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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>}}
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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 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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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'''
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The minus disappears when squaring. Consider a simple (trivial) equation
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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>}}
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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.
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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> 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>
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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
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If we square both sides of the equation we get
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{{Displayed math||<math>4(x - 1) = (1 - x)^2</math>}}
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{{Displayed math||<math>4(x - 1) = (1 - x)^2.</math>}}
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and we expand the square on the right-hand side giving
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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: '''
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'''Keep in mind that...'''
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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.
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'''You should always test the solution in the original equation containing roots.'''
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'''You should always test the solution in the original equation.'''
'''Reviews'''
'''Reviews'''
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For those of you who want to deepen your studies or need more detailed explanations consider the following reference
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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'''
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[http://www.webmath.com/simpsqrt.html What is the root of -? Webmath.com helps you to simplify root expressions.]
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[http://www.webmath.com/simpsqrt.html Webmath.com can help you to simplify root expressions.]
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</div>

Revision as of 14:20, 26 October 2008

       Theory          Exercises      

Contents:

  • Equations of the type \displaystyle \sqrt{ax+b}= cx +d
  • Spurious roots

Learning outcomes

After this section, you will have learned to:

  • Solve 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 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 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.

Example 1

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 = -2 only satisfies the original equation. So we see that the squared equation has more solutions than the original equation. In this case \displaystyle x = -2 is a spurious root.

Example 2

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


If we square both sides of the equation we get

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

Expanding the square on the right-hand side we see that

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

This is a quadratic equation, 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 formula. 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 check whether \displaystyle x=1 and \displaystyle x=5 are also solutions to the original equation:

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

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

[Image]


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 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.


Reviews

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

Understanding Algebra - English online book for pre-university studies


Useful web sites

Webmath.com can help you to simplify root expressions.


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