Solving Radical Equations: Introduction


Print-friendly page

Solving Radical Equations: Introduction (page 1 of 6)

A "radical" equation is an equation in which the variable is stuck inside a radical.

For instance, this is a radical equation:

...but this is not:

The "radical" in "radical equations" can be any root, whether a square root, a cube root, or some other root. The topic of "solving radical equations" usually involves mostly or only square roots, so most of the examples in what follows use square roots as the radical, but you should not be surprised to see a cube root or fourth root in your homework or on a test.

For instance, given x + 2 = 5, you solved this by undoing the addition of the 2. You undid this addition by doing the opposite: subtraction:

x = 3

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

x = –4

When you have a variable inside a square root, you undo the root by doing the opposite: squaring. For instance, given , you would square both sides:

There are a couple of sticking points with 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 = 7²49 = 49

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

3² + 4² = 9 + 16 = 25 = 49 ...............Oops!

As you can see, I started with a true statement ("3 + 4 = 7"), and 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"). This is the most common mistake that students make: squaring terms instead of sides. Don't make this mistake! You should always remember to:

SQUARE SIDES, NOT TERMS

The other sticking point 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 problem with radical equations, however, 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 = 2²4 = 4

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

A more pertinent example would be this:
This "equation" is no more true than the "–2 = 2" "equation" above, because no positive square root can ever possibly equal a negative number.
But suppose I hadn't noticed that this equation has no solution, and proceeded to square both sides:
By squaring, I created a solution ("x = 9") that has not existed before and is not valid. But I won't discover this error unless I remembered to check my solution:

So the actual answer 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 y = x + 2 (from the left-hand side of x + 2 = 5)and y = 5 (from the right-hand side):

graph

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 , I was also trying to find the intersection of y = sqrt(x) and y = 4:

graph

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

But when I was trying to solve the equation:

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

(Note: If you do not understand why the graphs of radical equations tend to tail off at one end, as shown at right, then review graphing radical equations.)

graph

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

graph

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 is why checking the solution showed that the real answer was "no solution".

By the way, it is very common for instructors not to give many examples (in class or in the homework) of these equations for which the solutions don't actually work, but then to 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!

Top | 1 | 2 | 3 | 4 | 5 | 6 | Return to Index Next >>

Cite this article as:

Stapel, Elizabeth. "Solving Radical Equations: Introduction." Purplemath. Available from
http://www.purplemath.com/modules/solverad.htm. Accessed

Lessons index

Lessons CD

Purplemath:
  Linking to this site
  Printing pages
  Donating
  School licensing

Reviews of
Internet Sites:
   Free Help
   Practice
   Et Cetera

The "Homework
Guidelines"

Study Skills Survey

Tutoring ($$)

This lesson may be printed out for your personal use.

Copyright © 2006-2008 Elizabeth Stapel \| About \| Terms of Use		Feedback \| Error?