Extraneous Roots Checked Less TediouslyDate: 06/06/2012 at 07:33:23 From: Muhammad Asif Subject: Checking Extraneous roots Given (x - 4)/x + x/3 = 6 I solved this for x and found two roots: x = [15 - SQRT(273)]/2 x = [15 + SQRT(273)]/2 Can anybody help me check whether one of these roots is extraneous or not? Plugging in these values of x and satisfying the left hand side and right hand side would take too long. For simple values of x, like 2 and 3, it's easier to find the extraneous roots. But values of x that have two parts and contain square roots, as in my case, cause difficulty in finding the extraneous roots. Other than plugging in values, is there a simple way of checking for extraneous roots? Thanks in advance. Date: 06/06/2012 at 09:42:11 From: Doctor Peterson Subject: Re: Checking Extraneous roots Hi, Muhammad. In this kind of equation, the source of extraneous roots is multiplication by x, which can introduce an extraneous root if x = 0 turns out to be a root of the new equation, because multiplication by 0 does not produce an equivalent equation. To put it another way, an extraneous root will prove to be extraneous only by making a denominator of the original equation zero. So all you need to do to test for an extraneous root is to make sure each solution is in the domain of the equation -- that is, none makes the denominator zero. In this case, it is clear that they don't. Of course, it's also wise to check your answer to make sure you didn't just make a mistake in your work. For this purpose, I would recommend using a calculator to make this less time-consuming, so that you will actually do it when you need to. The first root, for example, is approximately -0.76135582. I would store that value in my calculator to avoid having to retype it three times, and use it in the equation: (x - 4)/x + x/3 = (-0.76135582-4)/-0.76135582 + -0.76135582/3 = 6.25378527 + -0.25378527 = 6 If this had come out to 5.99999999, I would consider it verified! By the way, I have had students who took the idea of extraneous roots too far, so that whenever a check failed, they would say it was an extraneous root. That's another reason it's important to know HOW a root can be extraneous in each kind of equation -- so you can tell whether the problem lies in your work or in the root itself. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 06/06/2012 at 13:22:30 From: Muhammad Asif Subject: Thank you (Checking Extraneous roots) Thanks a lot, Dr. Math. Your answer helped me solve my problem in just two to three steps. Once again thanks for your answer. |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/