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1Removing the Radical

2Finding all Possible Solutions

3Discarding the Extraneous Solutions

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**Co-authored by**JohnK Wright V

Last Updated: March 29, 2019

A radical equation is an equation that contains a square root, cube root, or other higher root of the variable in the original problem. “Radical” is the term used for the symbol, so the problem is called a “radical equation.”^{[1]} To solve a radical equation, you have to eliminate the root by isolating it, squaring or cubing the equation, and then simplifying to find your answer. However, this procedure can create answers that appear to be correct, but are not, because of the squaring process. These are called extraneous solutions. You must learn to identify and discard the extraneous solutions.

Part 1

Part 1 of 3:

### Removing the Radical

1

**Isolate the radical term.**The first step to solving a radical equation is to move the radical term to stand alone on one side of the equation. Move all other terms to the opposite side. In this step, if possible, combine any other like terms that may exist.^{[2]}- Consider the sample problem . Your first step is to isolate the radical on the left side of the equation, as follows:
- ………. (subtract 4 from both sides)
- ………. (combine like terms)

- Consider the sample problem . Your first step is to isolate the radical on the left side of the equation, as follows:
2

**Square both sides of the equation.**To remove the radical sign from the problem, you need to perform its opposite function. The opposite of the square root function is to square both sides of the equation. Be careful, when squaring both sides of the equation, to do so correctly. Recall, for example, that is NOT . You need to treat the term as a binomial and square it accordingly.^{[3]}- Continue working with the sample problem and square both sides of it as follows:
- If you need help with this step, you may want to review Multiply Binomials.

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- Continue working with the sample problem and square both sides of it as follows:
3

**Repeat the previous steps if necessary.**If your original problem contained two or more radical terms, then the first round of isolating and squaring may not have removed all the radicals. If that is the case, then you should, once again, manipulate your equation to isolate the radical that remains and square each side again.^{[4]}- An example of such a problem would be something like . Because of the two radicals, you will need to do this procedure twice.

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Part 2

Part 2 of 3:

### Finding all Possible Solutions

1

**Consolidate and combine like terms.**After you have eliminated all the radicals from the problem, move all the terms to one side of the equation and combine like terms.^{[5]}- Returning to the working sample problem, this looks as follows:

- Returning to the working sample problem, this looks as follows:
2

**Solve the equation.**In most cases, this step will create a quadratic polynomial. This is an equation that contains an term as its highest variable. If the original radical was something other than a square root (such as a cube root or fourth root, for example), then you may have a more difficult problem. We will focus on the quadratic for this article. You may be able to solve the quadratic equation by factoring, or you can go directly to the quadratic formula.^{[6]}- In this case, the sample problem, , can be factored into the two binomial factors of and .

3

**Determine your solutions.**Factoring the quadratic equation in this case suggests two possible solutions. Because the quadratic equation is equal to 0, you find the solutions by setting each factor equal to 0 and then solve.^{[7]}- In the working problem, the two factors are and .
- Set each of these equal to 0 to get the solutions and .
- With another problem, you may not be able to factor and would then have to use the quadratic formula to find the solution.

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Part 3

Part 3 of 3:

### Discarding the Extraneous Solutions

1

**Recognize the potential for an extraneous solution.**Recall that after isolating the radical on one side of the equation, you then squared both sides to remove the radical sign. This is a necessary step to solving the problem. However, the squaring operation is what creates the extraneous solutions.^{[8]}- Remember some basic mathematics, that both a negative and a positive number, when squared, will give the same result. For example, and both give the answer of . However, both the negative and positive numbers might not be solutions to whatever problem you are solving. The one that does not work is called the extraneous solution.

2

**Test each of your solutions in the original problem.**After you have found the solutions to your problem, you may have found one, two or more different possible values for the variable. You need to check each of these in the original problem to see which work. Remember that the original problem here was .^{[9]}- First check the solution :
- ………. (substitute 5 for x)
- .
- Because your result is an incorrect statement, the original solution of must be an extraneous solution that was caused by the squaring process.

- Check the second solution :
- In this case, you get a true statement. This shows that the solution is a true solution to the original problem.

- First check the solution :
3

**Discard the extraneous solution and report your result.**The extraneous solution is incorrect and can be discarded. Whatever remains is the answer to your problem. In this case, you would report that .^{[10]}Advertisement

## Community Q&A

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What is an example of an extraneous solution?

Sergeantpro

Community Answer

For example, say that you want to find the length of a square with the given area of 25 cm squared. You do that by taking the square root of 25. If you take the square root of 25, you will have two answers, which are 5 and -5. However, -5 cm would not work since negative lengths do not really make much sense in geometry. -5 cm would be, in this case, the extraneous solution, since it is a valid answer but doesn't solve the problem.

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## References

- ↑ http://www.mathsisfun.com/definitions/radical.html
- ↑ http://www.purplemath.com/modules/solverad2.htm
- ↑ http://www.purplemath.com/modules/solverad2.htm
- ↑ http://www.purplemath.com/modules/solverad2.htm
- ↑ http://www.purplemath.com/modules/solverad2.htm
- ↑ http://www.purplemath.com/modules/solverad2.htm
- ↑ http://www.purplemath.com/modules/solverad2.htm
- ↑ http://www.purplemath.com/modules/solverad2.htm
- ↑ http://www.purplemath.com/modules/solverad2.htm

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## About this article

Co-authored by:

JohnK Wright V

Texas Certified Math Teacher

This article was co-authored by JohnK Wright V. JohnK Wright V is a Certified Math Teacher at Bridge Builder Academy in Plano, Texas. With over 20 years of teaching experience, he is a Texas SBEC Certified 8-12 Mathematics Teacher. He has taught in six different schools and has taught pre-algebra, algebra 1, geometry, algebra 2, pre-calculus, statistics, math reasoning, and math models with applications. He was a Mathematics Major at Southeastern Louisiana and he has a Bachelor of Science from The University of the State of New York (now Excelsior University) and a Master of Science in Computer Information Systems from Boston University. This article has been viewed 19,389 times.

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Co-authors: 2

Updated: March 29, 2019

Views:19,389

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