quadratic formula

Consider the equation

(1)$a{x}^2 + b{x} + c = 0 ,$

which we wish to solve for $x$. In certain contexts, the solutions are given by one or more versions of the quadratic formula.

The coefficients $a, b, c$ are commonly taken from an algebraically closed field $K$ of characteristic $0$, such as the field $\mathbb{C}$ of complex numbers, although any quadratically closed field whose characteristic is not $2$ would work just as well. Alternatively, the coefficients can be taken from a real closed field $K$, such as the field $\mathbb{R}$ of real numbers; then the solutions belong to $K[\mathrm{i}]$. (Of course, $\mathbb{R}[\mathrm{i}]$ is simply $\mathbb{C}$ again.) More generally, starting from any integral domain $K$ whose characteristic is not $2$, the solutions belong to some splitting field of $K$. (Of course, there are solutions in *some* splitting field, regardless of the characteristic, but they are not given by the quadratic formula if the characteristic is $2$.)

Explicitly, the solutions of (1) may be given by the **usual quadratic formula**

(2)$x_\pm = \frac{-b \pm \sqrt{b^2 - 4a{c}}}{2a} ,$

which works as long as $a \ne 0$. There is also an **alternate quadratic formula**

(3)$x_\pm = \frac{2c}{-b \mp \sqrt{b^2 - 4a{c}}} ,$

which may be obtained from (2) by rationalizing the numerator; this works as long as $c \ne 0$. (Note that $\pm$ and $\mp$ appear here simply to indicate the two square roots of the determinant $b^2 - 4a{c}$ and how they correspond to the two solutions $x_\pm$; we do not need to have a function $\sqrt{}$ which always chooses a ‘principal’ square root.)

These two formulas are reconciled in the projective line of $K$. As long as $(a, b, c) \ne (0, 0, 0)$, there are two solutions (which might happen to be equal) in the projective line. If $a = 0$, then one of these solutions is $\infty$, and (2) correctly gives us that solution (as long as $b \ne 0$) for one choice of square root, although it gives $0/0$ for the other choice. Similarly, (3) correctly gives us $x = 0$ when $c = 0$ and $b \ne 0$, but it does not give us the other root when $c = 0$. Note that if $a, c = 0$ but $b \ne 0$, then (2) gives us one root ($\infty$) while (3) gives us the other ($0$).

So in general, we should be given $a \ne 0$, $b \ne 0$, or $c \ne 0$ for a nondegenerate equation (1). If $a \ne 0$, then we use (2); if $c \ne 0$, then we use (3). Finally, if $b \ne 0$, then we use both; each root will be successfully given by at least one formula for some choice of square root of $b^2 - 4a{c}$.

When the coefficients come from an ordered field $K$ (which we assume real closed), then we can write down a formula specially for the case when $b \ne 0$. This is the **numerical analysts' quadratic formula**

(4)$\begin {gathered}
\displaystyle x_{\hat{b}} = \frac{2c}{-b - \hat{b}\sqrt{b^2 - 4a{c}}} ;\\
\displaystyle x_{-\hat{b}} = \frac{-b - \hat{b}\sqrt{b^2 - 4a{c}}}{2a} .\\
\end {gathered}$

In this formula, $\hat{b}$ is the sign of $b$, that is ${|b|}/b$; also, we must choose a nonnegative principal square root, so that $\sqrt{b^2 - 4a{c}} \lt 0$ in $K$ is avoided (and thus the common denominator of $x_{\hat{b}}$ and numerator of $x_{-\hat{b}}$ is nonzero even if not imaginary). Despite the name, this formula is not sufficient for all purposes in numerical analysis?; one still needs all three formulas and chooses between them based on whether $a \ne 0$, $b \ne 0$, or $c \ne 0$ is best established.

There is also an interesting issue about whether $b^2 - 4a{c} \ne 0$. Everything above is valid in weak forms of constructive mathematics, except for the statement that $\mathbb{C}$ is algebraically closed. That claim follows from weak countable choice ($WCC$), which in turn will follow from either excluded middle or countable choice, which is accepted by most constructive mathematicians. Nevertheless, the statement

$\forall\, a, b, c\colon \mathbb{C},\; \exists\, r\colon \mathbb{C},\; r^2 = b^2 - 4a{c}$

is false in (for example) the internal language of the sheaf topos over the real line. (Essentially, this is because there is no continuous map $\sqrt{}$ on any neighbourhood of $0$ in $\mathbb{C}$.) If we are given that $a, b, c$ are real, or if we are given that $b^2 \ne 4a{c}$, then there is no problem. But in general, we cannot define this square root, which appears in every version of the quadratic formula.

However, there is a more subtle sense in which $\mathbb{C}$ is algebraically closed even without $WCC$; essentially, this allows us to approximate the subset of $\mathbb{C}$ whose elements are the two solutions of (1) (using two-element subsets of the field of, say, Gaussian numbers) even if we can't approximate any one solution (using individual, say, Gaussian numbers); see Richman (1998) for details. The quadratic formula can then be interpreted as indicating this approximated subset.

Sometimes one considers the equation

$a{x}^2 + 2p{x} + c = 0$

instead of (1); then (2) simplifies to

$x_\pm = \frac{-p \pm \sqrt{p^2 - a{c}}}a .$

This is valid even in characteristic $2$, but unfortunately then it is fairly useless, since $2p = 0$. In fact, (1) is solvable in characteristic $2$ only if $b = 0$.

- Fred Richman; 1998;
*The fundamental theorem of algebra: a constructive development without choice*; Fred Richman’s Documents

Revised on August 6, 2015 23:43:06
by Toby Bartels
(98.16.172.201)