AlgebraLAB    
About Us Activities Career Profiles Connections Site Directions Glossary
Lessons Reading Comprehension Passages Practice Exercises Search AlgebraLAB StudyAids: Recipes Word Problems
   
   
Solving Radical Equations
We examine ways to solve equations such as . In solving any equation, we always “undo” the operation so that we can get x isolated. In this case we have to square both sides of the equation. The examples cited below describe some of the methods needed to solve equations which include radicals.

#1. To solve we first square both sides of the equation.
The result is x - 2 = 81. This equation is simple to solve. We have x = 83.
#2. A more complicated situation is .
In this case we still begin by squaring both sides of the equation. The result is .

To finish solving this requires us to set all terms equal to zero and either factor or use the quadratic formula. We get .

This factors: and the solutions are x = 2 or x = - 1.

We must check each of these solutions in the original equation to see if the value of x gives a solution.
x = 2 gives which is correct.

x = - 1 gives which is impossible since for n > 0 is always positive.
Therefore the only solution is x = 2.
Keeping in mind the basic idea of squaring both sides of the equation in order to solve for x when there is a radical in the equation, we will now show some additional examples.

#3. Solve for x if .
Squaring both sides of the equation gives us .

Setting terms equal to zero gives .

The expression factors: .

Setting each factor equal to 0 gives two possible answers: x = 3 or x = – 1.

We check each answer in the original equation:
If x = 3 we have which is correct.

If x = – 1 we have which is impossible since the square root cannot be negative.
Therefore the only answer is x = 3.
#4. Suppose .
We begin again by squaring both sides of the equation. The result is 3x – 2 = x + 7.

We solve for x getting 2x = 9 or .

We check this possible answer in the original equation.
If we have .

This is the same as which is correct.
Therefore the solution is .
#5. Solve for x if .
We square both sides. This gives x + 2 = 2 – x.

Solving for x gives 2x = 0. The only solution is x = 0.
Checking this in the original equation gives which is correct.
Therefore the solution is x = 0.
#6. Let .
Our first step is to separate the radicals. The result is .

Now we square both sides. The result is .

We next isolate the radical term on the right side. The result is .

We square both sides again. The result is .

We set the terms equal to zero. The result is .

We solve this by factoring or using the quadratic formula. This does factor: (x – 5)(x + 3) = 0. This gives two possible solutions: x = 5 or x = – 3.

We check these possible solutions in the original equation.
If x = 5 we have . This is the same as 4 – 3 = 1.

If x = – 3 we have . This is impossible since .
Therefore the only solution is x = 5.
Another way to obtain the same result in example #6 is by using the Intersect feature on a graphing calculator. The graphs of Y1= and Y2 = 1 are shown below intersecting where x = 5.

Examples
Example
What is your answer?
 
Example
What is your answer?
 
Example
What is your answer?
 
Example
What is your answer?
 



M Ransom

Show Related AlgebraLab Documents



AlgebraLAB
Project Manager
Copyright © 2003-2014
Mainland High School
All rights reserved.