Question

What are the roots of `x^4+2x^2+x-25 = 0` ?

Answer 1

I'll show you how to reduce the problem from a quartic to a cubic, but basically this has no nice easy algebraic solution.

Explanation:

Let `f(x) = x^4+2x^2+x-25`

By the rational root theorem, the only possible rational roots of `f(x) = 0` are `+-1`, `+-5`, `+-25`.

None of these work, so `f(x) = 0` has no rational roots.

How about quadratic factors?

Since there's no term in `x^3`, we can write:

`x^4+2x^2+x-25 = (x^2+ax+b)(x^2-ax+c)`

Equating coefficients we get the following set of equations:

`b+c-a^2 = 2` hence: `b+c = a^2 + 2`

`a(c-b) = 1` hence: `c - b = 1/a`

`bc = -25`

Now `(b+c)^2 = (c-b)^2+4bc`

So: `(a^2+2)^2 = 1/a^2-100`

That is `a^4+4a^2+104-1/a^2 = 0`

Multiplying through by `a^2` we get:

`(a^2)^3 + 4(a^2)^2 + 104(a^2) - 1 = 0`

So we have managed to reduce the quartic equation we started with to a cubic in `a^2`.

It gets messier from here.

Let `t = a^2 + 4/3`

Then `t^3 + 296/3t - 3643/27 = 0`

Multiply through by `27` to get:

`27t^3 + 2664t - 3643 = 0`

Solve this cubic using Cardano's method or similar.

Hence find `a`, `b` and `c`.

Answer 2

To calculate approximate solutions you can use Newton's method to find Real roots:

`x_1 = 1.97522342578671`
`x_2 = -2.07326332952723`

Explanation:

Let `f(x) = x^4+2x^2+x-25`

Then `f'(x) = 4x^3+4x+1`

Given a first approximation `a_0`, you can derive better approximations by iterating using the formula:

`a_(i+1) = a_i - f(x)/(f'(x)) = a_i - (x^4+2x^2+x-25)/(4x^3+4x+1)`

A couple of good first approximations to try are `a_0 = 2` and `a_0 = -2`, since `f(2) = 16+8+2-25 = 1` and `f(-2) = 16+8-2-25 = -3` are both close to `0`.

Putting this formula into a spreadsheet, I get the following approximations:

`a_0 = 2`
`a_1 = 1.dot(9)756dot(0)`
`a_2 = 1.9752235212239`
`a_3 = 1.97522342578671`

`a_0 = -2`
`a_1 = -2.dot(0)7692dot(3)`
`a_2 = -2.07327197579025`
`a_3 = -2.07326332957561`
`a_4 = -2.07326332952723`