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Solving Radical Equations: Introduction (page 1 of 6)

A "radical" equation is an equation in which at least one variable expression is stuck inside a radical, usually a square root.

For instance, this is a radical equation:


sqrt(x) + 2 = 6


...but this is not:


x + sqrt(2) = 6

The "radical" in "radical equations" can be any root, whether a square root, a cube root, or some other root. Most of the examples in what follows use square roots as the radical, but (warning!) you should not be surprised to see an occasional cube root or fourth root in your homework or on a test.

In general, you "solve" equations by "isolating" the variable; you isolate the variable by "undoing" whatever had been done to it. Copyright © Elizabeth Stapel 2002-2011 All Rights Reserved

For instance, given x + 2 = 5, you would solve by undoing the addition of the 2. That is, the addition undone by applying the opposite: subtraction:


x = 3


In the same manner, given something like –3x = 12, you would solve by undoing the multiplication by applying the opposite operation; namely, division:


 x = –4


When you have a variable inside a square root, you undo the root by doing the opposite: squaring. For instance, given sqrt(x) = 4, you would square both sides:


 x = 16

Issue 1:




There are a couple of issues that frequently arise when solving radical equations. The first is that you must square sides, not terms. Here is a classic example of why this is so:

    I start with a true equation and then square both sides:

      3 + 4 = 7  
      (3 + 4)2 = 72
      49 = 49

    ...but if I square the terms on the left-hand side:

      3 + 4 = 7  
      32 + 42 "=" 72
      9 + 16 "=" 49
      25 "=" 49

In each case, I started with a true statement; namely, 3 + 4 = 7. When I squared both sides, I also ended with a true statement: 49 = 49. But when I squared the terms, 32 + 42, I ended up with something that was not true: 25 "=" 49. The most common mistake that students make when solving radical equations is squaring terms instead of sides. Don't make this mistake! You should always remember to:


Issue 2:

The other issue is that you will need to check your answers. You can always check your answers in a solved equation by plugging your answer back into the original equation and making sure that it fits.

For instance, in my first example above, you can check that I got the correct answer by plugging 3 in for x and verifying that the equation is still true:


   x + 2 = 5
(3) + 2 = 5
         5 = 5

You probably did some of this type of checking when you first starting solving linear equations. But eventually you developed your skills, and you quit checking. The difficulty with radical equations is that you may have done every step correctly, but your answer may still be wrong. This is because the very act of squaring the sides can create solutions that never existed before.

For instance, I could say "–2 = 2", and you would know that this is false. But look what happens when I square both sides:


(–2)2 = 22
4 = 4

I started with something that was not true, squared both sides of it, and ended with something that was true. This is not good!

A more pertinent example would be this:


sqrt(x) = –3

This "equation" is no more true than the "–2 = 2" "equation" above, because no positive square root can ever equal a negative number.

But suppose I hadn't noticed that this equation has no solution, and had proceeded to square both sides:


x = 9


By squaring, I created a solution ("x = 9") that had not existed before and is in fact not valid. But I won't discover this error unless I remembered to check my solution:



sqrt(x) = ?3...., but 3 is not equal to –3

So the actual answer for the equation sqrt(x) = –3 is "no solution".

There is another way to look at this "no solution" difficulty: When you are solving an equation, you can view the process as trying to find where two lines intersect on a graph.

For instance, when I was solving "x + 2 = 5" above,
you could also say that I was trying to find the intersection of
y1 = x + 2 (from the left-hand side of
x + 2 = 5) and y2 = 5 (from the right-hand side):



As you can see in the graph above, the two lines intersect at x = 3, which was the solution we had already found. Similarly, when I was solving the equation sqrt(x) = 4, I was also trying to find the intersection of y1 = sqrt(x) and y2 = 4:


As the above graph displays, the solution is at x = 16


But when I was trying to solve the equation:

    sqrt(x) = -3

...I was trying to find the intersection of y1 = sqrt(x)
and y2 = –3, which do not intersect.


(Note: If you don't know how I got the curvy blue line in the graph at right, then review how to graphradical equations.)





So what happened when I squared both sides? I also "squared" both line equations, and got the two new lines y1 = x and y2 = 9. And, as the graph shows, these two lines actually do intersect!



This is how squaring created a solution where there hadn't been a solution before. But the after-squaring solution did not work in the before-squaring equation, because the original lines had not intersected. This illustrates why checking the solution showed that the real answer was "no solution".

Warning: Many instructors do not to show many examples (in class or in the homework) of radical equations for which the solutions don't actually work. But then they'll put one of these on the test. You should expect a "no solution" radical equation on the test, so you do not want to forget to check your solutions!


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Cite this article as:

Stapel, Elizabeth. "Solving Radical Equations: Introduction." Purplemath. Available from Accessed



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