Extraneous Roots Checked Less Tediously

Date: 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.