As every student has experienced some time in their school and university mathematics courses, the “general” quadratic equation is

$ax^2+bx+c=0$

and it can be solved by using the “quadratic formula”:

$\displaystyle{x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}}.$

For the struggling student, this formula can be difficult to remember – all those coefficients! But it can easily be simplified. First, note that even though

$f(x)=ax^2+bx+c$

is a general quadratic function, the general quadratic equation can in fact be simplified by dividing through by $a$ (which is assumed to be non-zero) to obtain

$\displaystyle{x^2+\frac{b}{a}x+\frac{c}{a}=0}.$

And in fact this step is the beginning of most derivations of the quadratic formula. Writing the $x$ coefficient as $2a$ and the constant term as $b$ produces an equivalent general quadratic equation

$x^2+2ax+b=0$

for which the solution is

$x=-a\pm\sqrt{a^2-b}$.

Isn’t that simpler?

A similar approach can be taken to the general “reduced cubic equation”

$x^3+3ax+2b=0.$

This is in fact completely general as any cubic equation can be put into this form by a linear transformation.

Writing

$x=p^{1/3}+q^{1/3}$

and cubing both sides produces:

$x^3=p+q+3p^{1/3}q^{1/3}x.$

Comparing coefficients with the cubic equation:

$a = -p^{1/3}q^{1/3}$

or

$pq=-a^3$

and

$p+q=-2b$

These can be easily solved to produce

$\displaystyle{p,q=-b\pm\sqrt{b^2+a^3}}.$

This gives a solution to the cubic:

$x=\displaystyle{\left(-b+\sqrt{b^2+a^3}\right)^{\!1/3}+\left(-b-\sqrt{b^2+a^3}\right)^{\!1/3}}$

which is nearly simple enough to memorize. The other solutions are obtained by multiplying each term by various powers of $\omega$, where $\omega^3=1$ and $\omega^2+\omega+1=0$:

$x=\displaystyle{\omega\left(-b+\sqrt{b^2+a^3}\right)^{\!1/3}+\omega^2\left(-b-\sqrt{b^2+a^3}\right)^{\!1/3}}$

and

$x=\displaystyle{\omega^2\left(-b+\sqrt{b^2+a^3}\right)^{\!1/3}+\omega\left(-b-\sqrt{b^2+a^3}\right)^{\!1/3}}$.