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

##### 2 Answers

#### Answer:

If you mean is there a *closed formula* for solutions of polynomial equations of degree

#### Explanation:

Basically there is a formula for roots of

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

For example, the roots of

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 (

As regards degree

**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

Then our equation becomes:

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

Let

Then

Now write

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

Then:

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

Multiplying through by

#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

#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

#### Answer:

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^4+pt^2+qt+r = 0#

This must factorise in the form

#B+C-A^2 = p#

#A(C-B) = q#

#BC = r#

Now

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

We can solve the cubic to derive

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.