Begin with a general quartic equation

depress it using the substitution

creates a new quartic equation (without the term) in

where

When i.e., the quartic equation in becomes a quadratic equation in

It follows that

For example, to solve quartic equation , depress it using the substitution we obtain

This is a quadratic equation in

Therefore,

Consequently,

In general, we solve the depressed quartic equation as follows:

Substituting for in the left side of (**) yields

By

If

Since

If

It means that is a solution of if

Moreover, squaring (4-3) gives

By Vieta’s theorem, satisfying (5-1, 5-2, 5-3) are the three solutions of cubic equation

(See “How to solve a cubic equation“)

Suppose the three solutions are we have

Clearly, there are eight combinations of

Among them, only four are valid due to constraint (4-3) placed on the product

From (5-3), we see

If the valid ones are:

since

since

since

since

Otherwise , we have

Consequently, a solution to the general quartic equation (*) is

There are four such solutions.

*Exercise-1 *Show that (5-3)