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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

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.

For quadratic, you can use the much loved quadratic formula.

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
quadratic equation that we could solve by the quadratic formula.

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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