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### Formula for Finding Roots

```
Date: Fri, 20 Jan 1995 16:48:45 -0600
From: Betty Thompson
Subject: formulas for roots

I need to know the formula for finding the roots of any quadratic,
cubic or quartic polynomial. Also who is credited with the
discovery of the last two formulas.

Betty Thompson
thompson_be@mailserver.capnet.state.tx.us
```

```
Date: 31 Jan 1995 01:48:35 GMT
From: Dr. Ethan
Subject: Re: formulas for roots

Here are a few answers; but for the quartic, I think I'm just going
to send you to the library (I expect that you have a card).  This isn't
just to give you researching practice, it is also because the method for
solving the quartic is extremely long and involved and to type it in
would be a nightmare.

Given ax^2 + bx +c = 0 the roots are

x = (-b +- (b^2 - 4ac)^.5)/2a

For cubics, things get a little nastier.  Here is an explanation given by
one of my buddies, Ken, Dr. Stud:

"Okay, here's how you do it.  Let's say you have the equation
ax^3 + bx^2 + cx + d = 0.  The first thing you do is to get rid of the a
out in front by dividing the whole equation by it.  Then we get
something in the form of x^3 + ex^2 + fx + g = 0.  The next thing we do
is to get rid of the x^2 term by replacing x with (x - e/3).  That will
give us something of the form of x^3 + px + q = 0.  This is much easier
to solve, although it's still hard.

Now introduce two new variables, t and u, defined by
u - t = q   and   tu = (p/3)^3.

Then x = CubeRoot{t} - CubeRoot{u} will be a solution of x^3 + px + q = 0.
Verify this result now, and make sure you see why it works.
To find the other two solutions (if there are any) we could divide
x^3 + px + q by its known factor (x - CubeRoot{t} + CubeRoot{u}), getting a

So that's the basic idea behind the cubic.  If you wanted to find the actual
expression for t and u in terms of p and q, you could solve those two
equations defining p and q (substitution would probably be easiest).  Then
you could obtain a real formula for x in terms of p and q."

As you can see the complexity has greatly increased.  To try to go
backwards and come up with a closed form in the original a, b, c, d would
be a real pain.

Well the quartics are even worse.  But you should be able to find them in
most modern algebra texts.(And if you happen to find one that doesn't give
them, then they certainly will cite a place where they can be found.)

Hope this helps,

Ethan Doctor On Call
```
Associated Topics:
High School Polynomials

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