Hi everyone,

This post completes the solution to all differential equations idea I had.

We use the quadratic formula iteratively. From my notes:

Set:

A = 1
B = ((highest degree term coefficient) times x^((highest degree term power)-1))
C

On Sunday, July 10, 2022 at 11:51:32 AM UTC-4, B.H. wrote:
> Hi everyone,
>
> This post completes the solution to all differential equations idea I had.
>
> We use the quadratic formula iteratively. From my notes:
>
> Set:
>
> A = 1
> B = ((highest degree term coefficient) times x^((highest degree term power)-1))
> C

Oops, that got posted too early. Let me finish the post.... -Philip White

Complete idea:

We are given a polynomial. (This is the correct version, from my notes.)

Then, construct a new polynomial with these values:

(Let k be the degree of the polynomial.)

A' = the coefficient of x^k in the previous polynomial times x^(k-2)
B' = 0
C' = everything else summed together

Now, use the quadratic formula on that. Then apply it again, on the resulting solutions taken as polynomials, until you have your full set of answers, with cardinality governed by the fundamental theorem of algebra.

The goal is to use the quadratic formula and solve equations and inequalities so that, e.g., b^2 - 4ac >= 0, and so that the degree of the key polynomial decreases by at least 1 each time we apply this process. We'll get more and more values to check, and eventually, the degree of the polynomial will only be 2 (or less), and then we can use the quadratic formula one last time to find the full set of solutions. Obviously, many of the solutions will be irrational.

Note, if you have something of the form sqrt(f(x)) = 0, you can square both sides to check for solutions.

-Philip White (philipjwhite@yahoo.com)