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### Rational Root Rigging

```Date: 12/17/2013 at 22:39:08
From: Roger
Subject: submitted earlier about rational equations

Consider this equation:

x/(x + 4) = -3/(x - 2) + 18/((x + 4)(x - 2).

There are 2 solutions:

x = -3
x = 2

But x = 2 is extraneous because it makes a denominator equal 0.

With rational equations, sometimes you can get 2 solutions, both of which
work; and sometimes neither works.

Is there a way to knowingly generate rational equations so as to result in
a pre-determined set of outcomes? I cannot find a reliable algorithm.

Thanks.

```

```
Date: 12/17/2013 at 23:50:07
From: Doctor Peterson
Subject: Re: submitted earlier about rational equations

Hi, Roger.

It's technically wrong to say that there are two solutions; it is the
quadratic equation you obtained in the process of solving it that has two
solutions. Your equation has ONE solution.

Here's a problem I just made up:

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

The standard way to solve this is to multiply by the least common
denominator (LCD), x^2 - 1, yielding

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

The solution for this is

x^2 + 4x - 5 = 0
(x + 5)(x - 1) = 0
x = -5 or 1

Of these, 1 is extraneous because the factor (x - 1) is found in the
denominator of the original equation. So the only actual solution is -5.

Here's how I came up with that equation.

I wanted the equation to have the general form

A     C     E
--- + --- = ---
B     D     BD

Here, A, B, C, D, and E are either monomials or binomials. Multiplying by
the LCD, I get

I want one solution to be a zero of B or D, let's say B, so that it will
be extraneous. This means that B must be a factor of

Since B is already a factor of BC, it has to be a factor of the polynomial

Now I pick specific values for A, B and D, arbitrarily. I choose

A = x
B = x - 1
D = x + 1

(Interestingly, C doesn't come into this at all, so I will be free to
choose it at will later!)

Now I want to choose E to make B a factor of AD - E; that is, I can choose
E = AD - BP, where P can be whatever makes things look good:

E = AD - BP = x(x + 1) - P(x - 1)

I want E to have degree less than 2, so I'd like P to have the form
(x + p) in order to cancel out the x^2 term:

E = x(x + 1) - (x + p)(x - 1)
= x^2 + x - (x^2 + (p - 1)x - p)
= x^2 + x - x^2 - (p - 1)x + p
= -(p - 2)x + p

To make it simple, I'll take p = 2 so that E is just a constant:

E = -(2 - 2)x + 2 = 2

Now my equation (with C still unchosen) is

x       C        2
----- + ----- = -------
x - 1   x + 1   x^2 - 1

Now I can choose C in order to make the solution a nice number. I'll make
C a constant c rather than a binomial. Our equation after multiplying by
the LCD is

AD + BC - E = 0
(AD - E) + BC = 0
BP + BC = 0
B(P + C) = 0

So the non-extraneous solution will be the solution of

P + C = 0
x + 2 + C = 0
x = -(C + 2)

I can pick any integer value for c and I'll get a nice solution. I picked
3, so the other solution is -5. And there's my equation:

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

The whole process is similar in some ways to partial fractions. (I tried
doing that explicitly, but I found it gave me less control over the
results.) But there may well be a way to adapt and improve upon that
technique ...

I'll try to make another equation with the same general pattern, to see if
my easy success that time was a fluke. I'll pick

A = x + 1
B = x - 2
D = x + 3

Again, letting P = x + p to cancel x^2, we have

= (x + 1)(x + 3) - (x + p)(x - 2)
= x^2 + 4x + 3 - x^2 + 2x - px + 2p
= (6 - p)x + (2p + 3)

I'll take p = -1 and P = x - 1, so

E = 7x + 1

The other solution is the solution of

P + C = 0

I'll make C a binomial x + c this time, just to be different:

x - 1 + x + c = 0
2x = 1 - c

I'll take c = 2, so the solution to the equation will be -1/2 (which is
not extraneous as it would be for c = -3 or -5).

The equation is

x + 1   x + 2      7x + 1
----- + ----- = -----------
x - 2   x + 3   x^2 + x - 6

Let's solve it and see that it works:

(x + 1)(x + 3) + (x + 2)(x - 2) = 7x + 1
x^2 + 4x + 3 + x^2 - 4 = 7x + 1
2x^2 - 3x - 2 = 0
(2x + 1)(x - 2) = 0

As expected, the solutions of this are x = -1/2 and 2, but the latter is
extraneous.

Looks like the technique works.

I've had to do this before in making tests, but hadn't worked out a
routine method, so this has been an interesting and useful exercise.
Still, I'll leave this question open in case anyone else has a better
method.

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

```

```
Date: 12/20/2013 at 22:11:59
From: Doctor Peterson
Subject: Re: submitted earlier about rational equations

Hi, Roger.

No one else has added any better ideas, but I've thought a little more
about how to make this easier to carry out in practice.

Here is another example, worked out in a way that is easier to follow,
without needing to refer to A, B, ... and magical formulas.

I start by choosing my denominators; they must have a common factor, so
I'll choose

2x - 1, x + 2, and (2x - 1)(x + 2)

I'll put all terms on the left for now.

I have to choose the extraneous solution of the equation I'm going to
write (which will be the zero of one of my factors). I'll pick 1/2. I'll
hold off on deciding what the actual solution will be until I have more to
go on.

I can pick anything for the first numerator; I'll take x:

x       A            B
------ + ----- + --------------- = 0
2x - 1   x + 2   (2x - 1)(x + 2)

Now in order for the factor 2x - 1 to yield an extraneous solution, it
must be a factor of the new numerator,

x(x + 2) + A(2x - 1) + B

This implies that x(x + 2) + B must be a multiple of (2x - 1); let's do
that by choosing B in this way:

B = (2x - 1)(...) - x(x + 2)

I can pick anything I like for that missing factor; if I wanted B to be
linear, I'd want it to start with x/2 so the x^2 terms would cancel. But
let's go ahead and let B be quadratic, since I didn't do one like that
before. I'll take this:

B = (2x - 1)(x - 1) - x(x + 2)
= (2x^2 - 3x + 1) - (x^2 + 2x)
= x^2 - 5x + 1

So,

x(x + 2) + B = (2x - 1)(x - 1)

Now the left side above will factor as

x(x + 2) + A(2x - 1) + B = [x(x + 2) + B]  + A(2x - 1)
= (2x - 1)(x - 1) + A(2x - 1)
= (2x - 1)(x - 1 + A)

Again, I can pick anything I want for A (not necessarily a constant), with
the goal of making the root of this a nice number (since that will be the
actual solution of my equation):

x - 1 + A = 0

Let's arbitrarily take A = 2x + a, so the equation becomes

3x - 1 + a = 0

This must be satisfied by the actual solution x; given that x, I'll take

a = 1 - 3x

I'll pick x = 2, so a = -5, and A = 2x - 5.

Here's my equation:

x^2 - 5x + 1       x     2x - 5
--------------- + ------ + ------ = 0
(2x - 1)(x + 2)   2x - 1    x + 2

For presentation to students, I might rearrange it this way:

x     2x - 5   -x^2 + 5x - 1
------ + ------ = -------------
2x - 1    x + 2   2x^2 + 3x - 2

I'll "leave it as an exercise for the reader" to solve this and confirm
that it does what I wanted, with the only solution being x = 2.

Note that in this process, we arbitrarily specified the factors of the
denominators, one numerator, the extraneous root, and the genuine root,
and also made a partially arbitrary choice of the complexity of the other
two numerators, though the specifics were then forced on us.

I think this is a reasonable procedure for writing this kind of equation.

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

```

```
Date: 12/20/2013 at 22:25:21
From: Roger
Subject: Thank you (submitted earlier about rational equations)

THANK YOU for a great solution.

I hope you will submit this to NCTM's "Mathematics Teacher" journal,
because a lot of people could use it.

Your explanations are great and easy to follow.

Have a Merry Christmas and happy holiday season.

- rg
```
Associated Topics:
High School Linear Equations

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