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Solving quartic equations
By The Editor:

Michael originally posted a new method of solving cubic equations, which turned out to contain an error. The post below, however, is about a method for turning a quartic equation into a cubic equation. This does seem to work, and you can find plenty about solving cubic equations here.

By Michael Doré (P904) on Tuesday, July 4, 2000 - 08:57 pm:

We take a equation:

x4 + ax3 + bx2 + cx + d = 0

We aim to reach an equation eventually where a = c and d = 1. But first we transform the above equation into the form:

x4 + ax3 + bx2 + cx + (c/a)2 = 0

Again by linear substitution. Let x = y + r.

(y+r)4 + a(y+r)3 + b(y+r)2 + c(y+r) + d = 0

If the constant term is going to be equal to (coefficient of y/coefficient of y3)2 then:

d = [(4r3 + 3ar2 + 2br + c)/(4r+a)]2

We then solve this cubic equation for r.

So without loss of generality we can transform

x4 + ax3 + bx2 + cx + d = 0

into:

x4 + ax3 + bx2 + cx + c2/a2 = 0

(where a,b,c might be different in the two equations)

Now we want to make coefficient of x4 = constant term, and coefficient of x3 = coefficient of x.

Let x = Au. If the two conditions are to be satisfied then:

A4 = c2/a2

and

aA3 = cA.

Again because of the special form of the equation, these two equations are consistent. Take A = sqrt(c/a). So dividing through by the leading coefficient we have got to:

[x4+1] + p[x3+x] + qx2 = 0

for some p,q (which can be determined).

Unfortunately we can't factorise this like before. But we can do the following:

Divide through by x2:

[x2 + x-2] + p[x + x-1] + q = 0

Now let v = x + x-1. x2 + x-2 = v2 - 2. So:

(v2-2) + pv + q = 0

This can now be solved quadratically for v. Then you can use this value to solve for x quadratically. So this looks as if it can be used to solve the quartic...

Next thing to try I suppose is to see why this fails for quintics etc.

Yours,

Michael