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### Solving Polynomials

```
Date: 6/16/96 at 17:50:5
From: Anonymous
Subject: Solving polynomials

x^2 + 2x  +     3          26
--------     --------  =  ----
3       x^2 + 2x        5

I've tried doing this by finding a common denominator, but after that
everything gets confusing and starts looking like this:

5x^4 + 20x^3 - 58x^2 - 156x + 45
--------------------------------  = 0
15x^2 + 30

Lisa
```

```
Date: 6/16/96 at 19:2:25
From: Doctor Anthony
Subject: Re: Solving polynomials

You could continue and simply put the numerator of above expression
equal to zero and solve the quartic, but that would be extremely
tedious unless you have Maple or a TI-92.  I expect you went through
unnecessary steps in deriving the above equation, but I would not do
it the way you started out.

Look at the expression again, and you will see that there is a
similarity of the terms on the left hand side.  The second term is
just the reciprocal of the first term.  I am going to let y = x^2+2x,
and so we get:

y/3 + 3/y = 26/5   or  (y^2+9)/3y = 26/5

Multiply out  5y^2 + 45 = 78y

5y^2 - 78y + 45 = 0

Factorize   (5y-3)(y-15) = 0  and so

y = 3/5   or y = 15

Now go back to the x variable and we have:

x^2 + 2x = 3/5      and   x^2 + 2x = 15

Solve these two quadratics to get the four x values that you need.

The second quadratic is x^2 + 2x - 15 = 0

(x+5)(x-3) = 0
so -5 and 3 will be two of the solutions.  The other quadratic doesn't
factorize, so use the quadratic formula to solve  5x^2 + 10x - 3 = 0

I get x = -2.264911 and 0.264911

-Doctor Anthony,  The Math Forum
Check out our web site!  http://mathforum.org/dr.math/
```
Associated Topics:
High School Polynomials

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