Rings and Polynomials

This is a stub article; the stuff on transformations and completing the square should be expanded.

The goal of this post is to show (algebraically) that all quadratics have the "same basic form", or equivalently (geometrically), that all parabolas have the "same basic shape".

Specifically, we'll show that any quadratic $aX^2+bX+c$ can be easily transformed to a "simpler" quadratic $X^2 - d$ (which has roots at $\pm\sqrt{d}$, (almost) by definition of the square root of $d$, at least if $d\ge0$). So in order to understand the "entire family" of quadratics, it's "enough" to understand the "simpler family" of quadratics of the form $X^2 - d$. As long as we keep track of how the more "complicated" quadratics are related to their simpler basic forms, we'll usually be good to go.

For example, the quadratic formula is a special case of this idea: we can relate the roots of a "complicated" quadratic to the roots of a "simpler" quadratic $X^2 - d$. In some sense, that's all the quadratic formula really says; we're just "solving for the roots of a general quadratic in terms of the roots of simpler quadratics $X^2 - d$ (which are $\pm\sqrt{d}$, essentially by definition)".

It is often thought that the imaginary unit $i$, and complex numbers by extension, were created by "dreaming up" a solution to the equation $x^2 = -1$, for the purposes of solving quadratics with no real solutions.

Actually, this is not the case, as Noah Snyder comments on Math StackExchange:

I believe the historical context was in solving cubic equations. If you try to solve a cubic equation that has a real solution, sometimes you have to use complex numbers in intermediate steps in the calculation. But then when you get to the end you find a real number that really is a solution to a real cubic equation!

Indeed, people were motivated by Cardano's method for solving cubics. People knew (using tools like the intermediate value theorem or "graphing" without graphing calculators (e.g. Newton's method); nowadays you can just use WolframAlpha) that the real polynomial $P(x) = x^3 - 6x + 2$ (example courtesy of MIT) had all real roots, but Cardano's formula gave gave the roots in terms of the complex cube roots $A = (-1 + \sqrt{-7})^{1/3}$ and $B = (-1 - \sqrt{-7})^{1/3}$ (whatever these mean). For example, using the "nonsensical rule" $\sqrt{-7}^2 = -7$ (from the definition of square root), people checked that $$(A+B)^3 - 6(A+B) + 2 = 0.$$ So somehow, these "nonsensical" expressions $A = (-1 + \sqrt{-7})^{1/3}$ and $B = (-1 - \sqrt{-7})^{1/3}$ (admittedly ambiguous—more on that later) arose as intermediate steps in solving a totally real problem (finding the roots of the real polynomial $P(x) = x^3 - 6x + 2$), and $(-1 + \sqrt{-7})^{1/3} + (-1 - \sqrt{-7})^{1/3}$ is the simplest exact algebraic expression (try putting it in WolframAlpha!) for the irrational (but still "algebraic") real root $A+B = 2.26180224525997\ldots$ (which is just a numerical approximation from algorithms like Newton's method or the bisection method). By the way, as WolframAlpha tells you, we've glossed over the fact that the cube roots $(-1 \pm \sqrt{-7})^{1/3}$ are ambiguous (see here and Euler's formula on Expii), because the fundamental theorem of algebra tells you there are three possible interpretations of a cube root (you can get even more precise information from complex analysis). And the mere existence of a root with an "algebraic" expression (in terms of radicals) like $(-1 + \sqrt{-7})^{1/3} + (-1 - \sqrt{-7})^{1/3}$ has connections with Galois theory (based on the symmetry of roots from Vieta's formulas, and algebraic number theory).

Stub explanation

The quadratic discriminant $b^2-4ac$ of a real quadratic $P(x) = ax^2+bx+c$ has a nice interpretation in terms of the roots $r_1,r_2$ (possibly equal) of $P$. The story starts with factored form of $P$: $$P(x) = a(x-r_1)(x-r_2) = ax^2 - a(r_1+r_2)x + a(r_1r_2).$$ By the method of undetermined coefficients (or more precisely, Vieta's formulas for quadratics), we have $b = -a(r_1+r_2)$ and $c = a(r_1r_2)$. Substituting, we find the discriminant of $P$ is $$b^2 - 4ac = [-a(r_1+r_2)]^2 - 4a[a(r_1r_2)] = a^2[(r_1+r_2)^2 - 4r_1r_2],$$ where we have factored out $a^2$ in the second step. But we may expand to get $(r_1+r_2)^2 = r_1^2+2r_1r_2+r_2^2$ (the $n=2$ case of the binomial theorem), so our discriminant expression simplifies to $$a^2[r_1^2-2r_1r_2+r_2^2] = a^2(r_1 - r_2)^2.$$ So $b^2 - 4ac = a^2(r_1-r_2)^2$ (which you can also check using the quadratic formula for $r_1,r_2$).

This "squares of differences of roots" type expression can also be thought of as an alternative definition of the quadratic discriminant. In fact, it generalizes to discriminants of higher degree polynomials as well, like the cubic discriminant.