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