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

Oct 22, 2015

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 $a {x}^{3} + b {x}^{2} + c x + d = 0$ and a horribly complex one for $a {x}^{4} + b {x}^{3} + c {x}^{2} + \mathrm{dx} + e = 0$.

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

For example, the roots of ${x}^{5} + 4 x + 2 = 0$ 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 (${x}^{5} + a x + b = 0$) 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:

$a {x}^{3} + b {x}^{2} + c x + d = 0$

First use a Tschirnhaus transformation, letting $t = x + \frac{b}{3 a}$ to eliminate the term of degree $2$.

Then our equation becomes:

$a {t}^{3} + \left(c - {b}^{2} / \left(3 a\right)\right) t + \left(d + {b}^{3} / \left(9 {a}^{2}\right) - \frac{b c}{3 a}\right) = 0$

Let $p = \frac{c}{a} - {b}^{2} / \left(3 {a}^{2}\right)$ and $q = \frac{d}{a} + {b}^{3} / \left(9 {a}^{3}\right) - \frac{b c}{3 {a}^{2}}$

Then ${t}^{3} + p t + q = 0$

Now write $t = u + v$

Then:

$0 = {t}^{3} + p t + q = {\left(u + v\right)}^{3} + p \left(u + v\right) + q$

$= {u}^{3} + {v}^{3} + \left(3 u v + p\right) \left(u + v\right) + q$

Next add the constraint that $v = - \frac{p}{3 u}$, so that $3 u v + p = 0$.

Then:

${u}^{3} - {p}^{3} / \left(27 {u}^{3}\right) + q = 0$

Multiplying through by $27 {u}^{3}$, we get:

$27 {\left({u}^{3}\right)}^{2} + 27 q \left({u}^{3}\right) - {p}^{3} = 0$

Use the quadratic formula to deduce:

${u}^{3} = \frac{- 27 q \pm \sqrt{{\left(27 q\right)}^{2} + 4 \cdot 27 {p}^{3}}}{2 \cdot 27}$

$= - \frac{q}{2} \pm \frac{\sqrt{81 {q}^{2} + 12 {p}^{3}}}{18}$

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:

$t = \sqrt{- \frac{q}{2} + \frac{\sqrt{81 {q}^{2} + 12 {p}^{3}}}{18}} + \sqrt{- \frac{q}{2} - \frac{\sqrt{81 {q}^{2} + 12 {p}^{3}}}{18}}$

Then:

$x = t - \frac{b}{3 a}$

$= - \frac{b}{3 a} + \sqrt{- \frac{q}{2} + \frac{\sqrt{81 {q}^{2} + 12 {p}^{3}}}{18}} + \sqrt{- \frac{q}{2} - \frac{\sqrt{81 {q}^{2} + 12 {p}^{3}}}{18}}$

If the cubic has one Real root or a repeated root then $81 {q}^{2} + 12 {p}^{3} \ge 0$ 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

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

#### Explanation:

General Quartic Equation

$a {x}^{4} + b {x}^{3} + c {x}^{2} + \mathrm{dx} + e = 0$

First use a Tschirnhaus transformation $t = x + \frac{b}{4 a}$ to get a quartic with no cube term and divide that through by $a$ to get a quartic equation in the form:

${t}^{4} + p {t}^{2} + q t + r = 0$

This must factorise in the form $\left({t}^{2} + A t + B\right) \left({t}^{2} - A t + C\right)$ since there is no cube term. Equating coefficients, we get a system of equations:

$B + C - {A}^{2} = p$

$A \left(C - B\right) = q$

$B C = r$

Now ${\left(B + C\right)}^{2} = {\left(B - C\right)}^{2} + 4 B C$

So:

${\left(p + {A}^{2}\right)}^{2} = {q}^{2} / {A}^{2} + 4 r$

Hence:

${\left({A}^{2}\right)}^{3} + 2 p {\left({A}^{2}\right)}^{2} + \left({p}^{2} - 4 r\right) \left({A}^{2}\right) - {q}^{2} = 0$

...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.