
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.
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 20022011 All Rights Reserved
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 ...but if I square the terms on the lefthand side: 3 + 4 = 7 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, 3^{2} + 4^{2}, 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: ** SQUARE SIDES, NOT TERMS ** 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.
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.
I started with something that was not true, squared both sides of it, and ended with something that was true. This is not good!
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.
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_{1} = sqrt(x) and y_{2} = 4: As the above graph displays, the solution is at x = 16.
This is how squaring created a solution where there hadn't been a solution before. But the aftersquaring solution did not work in the beforesquaring 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! ** CHECK ALL SOLUTIONS ** Top  1  2  3  4  5  6  Return to Index Next >>


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