Question

Is there a quadratic equation for polynomials of degree 3 or higher?

Answer 1

If you mean is there a closed formula for solutions of polynomial equations of degree `3` and higher, the answer is yes for `3` and `4`, 'sort of' for degree `5` and probably no for `6` and higher.

Explanation:

Basically there is a formula for roots of `ax^3+bx^2+cx+d = 0` and a horribly complex one for `ax^4+bx^3+cx^2+dx+e = 0`.

When you get to quintic equations, in general the roots are not expressible as ordinary `n`th roots.

For example, the roots of `x^5+4x+2 = 0` are not expressible in terms of ordinary radicals.

You can use something called a Bring Radical to express solutions of the general quintic equation, but getting the quintic into Bring Jerrard normal form (`x^5+ax+b = 0`) is way too complex.

As regards degree `6` and above, there may be some generalisations of the advanced methods used for quintics, but I'm not sure.

Cubic

Note: I'm doing this from scratch and not checking my answer carefully, so there may be some errors in the following, but it should illustrate the principles:

`ax^3+bx^2+cx+d = 0`

First use a Tschirnhaus transformation, letting `t = x+b/(3a)` to eliminate the term of degree `2`.

Then our equation becomes:

`at^3+(c-b^2/(3a))t+(d+b^3/(9a^2)-(bc)/(3a)) = 0`

Let `p = c/a-b^2/(3a^2)` and `q = d/a+b^3/(9a^3)-(bc)/(3a^2)`

Then `t^3+pt+q = 0`

Now write `t = u+v`

Then:

`0 = t^3+pt+q = (u+v)^3+p(u+v) + q`

`=u^3+v^3+(3uv+p)(u+v) + q`

Next add the constraint that `v = -p/(3u)`, so that `3uv+p = 0`.

Then:

`u^3-p^3/(27u^3)+q = 0`

Multiplying through by `27u^3`, we get:

`27(u^3)^2+27q(u^3)-p^3 = 0`

Use the quadratic formula to deduce:

`u^3 = (-27q+-sqrt((27q)^2+4*27p^3))/(2*27)`

`=-q/2+-sqrt(81q^2+12p^3)/18`

Due to the symmetry in `u` and `v`, one of these roots will be `u^3` and the other `v^3`, so we can deduce:

`t = root(3)(-q/2+sqrt(81q^2+12p^3)/18) + root(3)(-q/2-sqrt(81q^2+12p^3)/18)`

Then:

`x = t-b/(3a)`

`= -b/(3a) + root(3)(-q/2+sqrt(81q^2+12p^3)/18) + root(3)(-q/2-sqrt(81q^2+12p^3)/18)`

If the cubic has one Real root or a repeated root then `81q^2+12p^3 >= 0` resulting in a Real square root and this formula works well. Otherwise, you have to deal with cube roots of Complex numbers, which is not so nice.

Answer 2

Quartic equations can be reduced to solving cubics and hence potentially to a closed formula solution that way...

Explanation:

General Quartic Equation

`ax^4+bx^3+cx^2+dx+e = 0`

First use a Tschirnhaus transformation `t = x+b/(4a)` to get a quartic with no cube term and divide that through by `a` to get a quartic equation in the form:

`t^4+pt^2+qt+r = 0`

This must factorise in the form `(t^2+At+B)(t^2-At+C)` since there is no cube term. Equating coefficients, we get a system of equations:

`B+C-A^2 = p`

`A(C-B) = q`

`BC = r`

Now `(B+C)^2 = (B-C)^2 + 4BC`

So:

`(p+A^2)^2 = q^2/A^2 + 4r`

Hence:

`(A^2)^3+2p(A^2)^2+(p^2-4r)(A^2)-q^2 = 0`

...which is a cubic in `A^2`

We can solve the cubic to derive `3` possible roots for `A^2`, at least one of which is Real (though it may be negative). Choose one of the roots of the cubic and take its square root to get a value for `A`, hence `B` and `C`. This leaves you with a couple of quadratic factors to solve.

Putting this all together can give you a formula for the roots of a general quartic in terms of square and cube roots, but it is horribly complex.