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### Avoiding the Final Step of Checking for Extraneous Solutions

```Date: 01/10/2010 at 17:40:19
From: Ahmed
Subject: Avoiding extraneous solutions

I want to know is there a method, other than checking at the end,
that avoids extraneous solutions?  Is there an initial way to know?

For example, say I want to solve this equation for x:

sqrt(x + 1) + 3 = 2x

I can solve this in four steps.

1) Isolate the radical on the left side of the equation.
2) Square both sides, which results in:

x + 1 = (2x - 3)^2

Now, (a - b)^2 = a^2 - 2ab + b^2, so the right side expands to:

x + 1 = 4x^2 - 12x + 9

3) Changing that to standard form, we get

4x^2 - 13x + 8 = 0

x = 2.43 and x = 0.82 -- two different answers

But when we check, we find that only one solution is appropriate:

x = 2.43

So the algebraic steps are easy.  But during a quiz I might forget to
check my solution at the end.  I want a way to know the extraneous
solution BEFORE solving.

```

```
Date: 01/10/2010 at 23:13:35
From: Doctor Peterson
Subject: Re: Avoiding extraneous solutions

Hi, Ahmed.

I would normally do just what you did; but for some kinds of
problems, it is possible to circumvent a step that would potentially
introduce extraneous solutions.

sqrt(x + 1) + 3 = 2x

(Note that the parentheses replace the bar, called a vinculum, that
we usually put over the radicand.)

This equation has the possibility of an extraneous solution even
before you do any work, because of the extra meaning implied by the
radical.  The square root symbol represents only the principal root,
but there are really two roots; the other root would change the
equation to

-sqrt(x + 1) + 3 = 2x

So the radical in the equation means not only that you are getting a
number whose square is x + 1, but also that that number is positive.

In your work, you isolate the radical, then square both sides.  That
step introduces the possibility of an extraneous solution because
squaring eliminates the difference between these two equations; the
sign of the number being squared is ignored in the new equation.  In
essence, from here on you are solving both equations at once and then
sorting out which solution belongs to which equation!

I know of no way to eliminate the need for checking -- but there may
be a way to keep the check very simple.  One that would work here is
to solve not by squaring the equation as it stands, but by making a
substitution.  The dangerous part of the equation is sqrt(x + 1),
which has to be positive in order to obtain a valid solution.

Consider this substitution:

Let u = sqrt(x + 1), which means that u >= 0, and
u^2 = x + 1 (so that x = u^2 - 1).

Now the equation becomes

u + 3 = 2(u^2 - 1)     where u >= 0

(Note that this equation, including the restriction on u, IS fully
equivalent to the original.)

Solving,

u + 3 = 2u^2 - 2

2u^2 - u - 5 = 0

1 +/- sqrt(41)
u = --------------
4

Now we can eliminate the subtraction case, which would produce a
negative value for u; this is the equivalent of the check at the end
of your work!  (In other problems, we might find that both or neither
of the solutions are negative, and therefore invalid.  We can't know
before this point whether there are any extraneous solutions.)

So our only solution is u = (1 + sqrt(41))/4, from which we find that

x = [(1 + sqrt(41))/4]^2 - 1

At this point, you could either simplify this, or just calculate its
value.

In fact, this uglier result may explain why I haven't seen this
approach taught; but perhaps it will interest you.

I'll assume you will continue to use the usual method.  The key to
not forgetting to check your answer is to learn well that the check
is an essential part of the work of solving in a case like this; once
you have squared both sides, you should already have taken note that
the radical requires you to check that the positive root is the one
that works.  Alternately, if you prefer, keep in mind that squaring
both sides does not produce an equivalent equation; so the check is a
necessary part of the process.

Of course, you should make a habit of checking all solutions anyway,
because you could always make a mistake!

If you have any further questions, feel free to write back.

- Doctor Peterson, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 01/11/2010 at 11:31:07
From: Ahmed
Subject: Thank you (Avoiding extraneous solutions)

Thank you very much indeed for your fast response and detailed
explanation!  THANK YOU.
```
Associated Topics:
High School Polynomials

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