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3.2 Equations with roots

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       Theory          Exercises      

Contents:

  • Equations of the type 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

x+3x=2 
x−1−2x=x2 
3x+2=x. 

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.

Example 1

Consider a simple (trivial) equation

x=2.

If we square both sides of this equation, we get

x2=4.

This new equation has two solutions x=2 or x=−2. The solution x=2 satisfies the original equation, while x=−2 only satisfies the squared equation. So we see that the squared equation has more solutions than the original equation. In this case x=−2 is a spurious root.

Example 2

Solve the equation  2x−1=1−x .


If we square both sides of the equation we get

4(x−1)=(1−x)2

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

4(x−1)=1−2x+x2.

This is a quadratic equation, which can be written as

x2−6x+5=0.

This can be solved by completing the square or by using the general solution formula. Either way the solutions are x=32, i.e. x=1 or 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 x=1 and x=5 are also solutions of the original equation:

  • x=1 gives that LHS=21−1=0  and RHS=1−1=0. So LHS=RHS and the equation is satisfied!
  • x=5 gives that LHS=25−1=22=4  and RHS=1−5=−4. So LHS=RHS and the equation is not satisfied!

Thus the equation has only one solution x=1.

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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 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.


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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