The quadratic formula

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{-bpmsqrt{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=-apmsqrt{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=-bpmsqrt{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{omegaleft(-b+sqrt{b^2+a^3}right)^{!1/3}+omega^2left(-b-sqrt{b^2+a^3}right)^{!1/3}}

and

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

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