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

2 Answers
Oct 22, 2015

Answer:

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.

Oct 22, 2015

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 = 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.