Is the number system for $x$ assumed beforehand when proving the quadratic formula?

Question

When proving the quadratic formula (or any other mathematical equation, definition, formula, etc., from like all the way from basic math to advanced calculus), do we have to assume/declare the number system of $x$ beforehand, or do we determine that afterwards? Like, is #1 or #2 correct below?

1. We already have to assume/declare that $x$ is a real number or a complex number before we solve. This ensures that we know what number system it belongs to and what operations are valid for it. Also, after we solve for $x$, we can determine the solutions for $x$ in that number system (i.e., we find the quadratic formula and it gives the solutions for $x$ in the number system that $x$ already exists in).

2. We determine that $x$ must be a real or complex number after proving and using the quadratic formula (i.e., if the formula evaluated gives a real or complex number, or if the discriminant is positive or negative). So basically, we start by not assuming anything about $x$ (so it can be ANY type of number). And then after we solve for $x$ and evaluate the formula (this would require choosing the number system we are working in for at least the operations. For example, we must choose our operations to take place in $\Bbb{R}$ or $\Bbb{C}$, so then we can apply basic arithmetic operations, and we must also choose either $\Bbb{R}$ or $\Bbb{C}$ so we know if square roots will exist or not for negative numbers), we can determine the number system for $x$ based on what answer we get from the formula (i.e., whether or not the value is real or complex).

I feel like #1 is correct, but I'm not fully sure. Because we at least need to know what something represents, so like we need to know what number $x$ is even supposed to be. So then we would also need to know what number system we are even trying to solve it for (like if we want a solution that is real, complex, integer, positive, etc.). And also, if we have a function $f(x)$ (like a quadratic), then we also need to define its domain and codomain, which includes determining the number systems for $x$ and $f(x)$ beforehand. And also, we need to know what number system $x$ is part of so that we know what operations are valid on it.

Also, I have added links to similar questions (related to whether or not we need to assume that x exists in a specific number system when solving algebraic equations) that I have asked before, in case they may help anyone answer my question and understand it better. Links: Q1, Q2, Q3, Q4, Q5

Any help regarding these assumptions about variables in proofs would be greatly appreciated! Thank you!

Answer 1

This is maybe a worthwhile reminder of seeing how the quadratic formula is derived. Here goes. Assume $a \neq 0$.

$$ax^2+bx+c =0 \\ \Leftrightarrow x^2+\frac{b}{a}x +\frac{c}{a} = 0 \\ \Leftrightarrow (x^2+\frac{b}{a}x + (\frac{b}{2a})^2)-(\frac{b}{2a})^2 +\frac{c}{a}=0 \\ \Leftrightarrow (x+\frac{b}{2a})^2 = (\frac{b}{2a})^2-\frac{c}{a} \\ \Leftrightarrow x+ \frac{b}{2a} = \pm \sqrt{\dfrac{b^2-4ac}{4a^2}} =\pm\dfrac{\sqrt{b^2-4ac}}{2a}$$

And now let's check what we need for all these things to truly be equivalences:

• First equivalence: We need to divide by $a$. Possible in fields. If we are just in a commutative domain, we can extend to its field of fractions and get this equivalence.
• Second equivalence: For the extra term to make sense, we need to divide by $2$. Possible (in a possibly enlarged ring) as long as we are in a domain where $2 \neq 0$.
• Third equivalence: True in any commutative ring (binomial formula).
• Fourth equivalence: True if square roots of the term $b^2-4ac$ exist at all and come in a $\pm$ pair, possibly in a larger ring than ours.

So that's all you need.

In a nutshell, the quadratic formula will be true as soon as it makes sense at all -- if you can divide the terms you have by $2a$, and if there are two distinct square roots of $b^2-4ac$ which are each other's negatives. Notice that in commutative domains of characteristic $\neq 2$, this is always doable in some extension field of the field of fraction of our domain.

---

For a more in-depth answer to some of your good follow-up questions, note the following: what the quadratic formula a.k.a. completing the square achieves in spirit is that it reduces the solution of an arbitrary quadratic equation to (up to basic algebra) finding a certain (pair of) square root(s).

So once you say, you accept the division by $2a$ thing, all insight into your questions must come from (and already apply to) the special case

$$x^2 = d \qquad (*)$$

where $d$ is an element of your "number system", let's call that "number system" $D$. What are solutions to that? If you say, well, the square roots of $d$, then it is hard to argue, but that is just a name for the thing we are looking for. It does not tell us what that thing is, or how many there could be, or what "number system" (possibly bigger than the $D$ we started with) the answer(s) are in. I think these are your questions, and once they are solved for this special case, it's straightforward to go to the general case as per above.

Now I want to convince you that in most cases you are looking at, there can either only be one or two solutions to that. How so? Well, assume I have two solutions $\alpha$ and $\beta$ to (*). Behold:

$$\alpha^2=d=\beta^2 \Rightarrow \alpha^2-\beta^2=0 \stackrel{A}\Leftrightarrow (\alpha-\beta)(\alpha+\beta) = 0 \stackrel{B}\Leftrightarrow (\alpha=\beta \vee\alpha=-\beta)$$

Equivalence A holds iff $\alpha \beta =\beta \alpha$, in particular if you are dealing with commutative rings.

Equivalence B holds if $x\cdot y=0 \implies x=0 \vee y=0$, in particular if you are dealing with rings without zero divisors, a.k.a. "domains".

This shows that in any commutative domain containing the element $d$, equation $(*)$ can have at most two solutions. (In fact, it has exactly two, which are each others negatives -- unless $d=0$, or the characteristic of the domain is $2$: in both these cases, any seemingly two solutions are actually identical to each other)

These solutions, if they exist, are denoted $\pm \sqrt{d}$, but indeed that is a slight abuse of notation. Because for different extensions of $D$, these can be different things. (That is one of the main points of my answer https://math.stackexchange.com/a/4007515/96384: E.g. the two things denoted $\pm \sqrt 2$ in the field $\mathbb Q_7$ are not the same as the two things denoted $\pm \sqrt 2$ in the field $\mathbb R$. The fields $\mathbb Q_p$ and $\mathbb R$ are two different extensions of the "original" number system (here, $\mathbb Q$ or arguably $\mathbb Z$), and in that answer, I warn against this symbol which, if we wanted to be really precise, should always include a notation about what field we are in now. (Like $\pm \sqrt 2 _{\mathbb R}$ or so.)

A good reason why we do not always carry around such symbolism "square root in the domain so-and-so" is that for many practical purposes, it is redundant. In undergraduate math, mostly everything is tacitly understood to be in $\mathbb C$, where anything has square roots, so it's those.

Also, if you find square roots in some domain $E \supset D$, and then you look at an even bigger domain $F\supset E$, the two ones you already had in $E$ will still do the job, and according to what was proven above there cannot be any new ones "on top of that".

Also also, here's something to learn in higher algebra: In our situation ($D$ commutative domain) there is a "universal" solution to $(*)$ in the sense that I can always construct an extension $D'$ of $D$ which does contain solutions to $(*)$, and so that if $E$ is another extension of $D$ containing such solutions, there will be a pair of homomorphisms from $D'$ to $E$ which map the solutions in $D'$ to the ones in $E$. (In a certain sense, this provides us with a "minimal" extension in which we can solve the equation; and any other domain which contains solutions can be seen as a "further" extension of that "minimal" one, where, see last paragraph, nothing new of interest for our equation happens.)

Summing up: If you have never dealt with non-commutative rings, or rings with zero-divisors, it is great that you are a bit careful about what the symbol $\pm\sqrt{d}$ is supposed to mean (because it indeed can mean different things in different rings), but in the sense of the previous three paragraphs, the people who are a bit less precise than you have some good points.

---

If, on the other hand, you are allowing "number systems" which are not commutative, and/or have zero-divisors, then all hell can break loose, and the notation $\pm\sqrt{}$ should just not be used at all.

Example 1: In the ring $\mathbb Z/12 \mathbb Z$ ("clock numbers", modular arithmetic) the equation $x^2=\bar 1$ has four solutions: $\bar 1, \bar 5, \bar 7, \bar 11$). No one should use the notation $\sqrt{\bar 1}$ for any, or all, of them.

Example 2: In the quaternions, the equation $x^2=-1$ has infinitely many solutions, cf. Solving an equation in quaternions. No one should use the notation $\sqrt{-1}$ for any, or all, of them. (It is still true though that for each individual solution, I can find a commutative subfield of the quaternions which contains that solution and its negative, and none of the others. That is the universality alluded to above. Once one chooses such a subfield, one could call that pair $\pm\sqrt{-1}$ or with a related abuse of notation $\pm i$. But it's not the $i$ in the standard definition of the quaternions: it could be $j$, or $k$, or $\frac{5}{13}i - \frac{12}{13}k$, or...)

Answer 2

As a general rule, one should always define precisely the variables they are using. In the context of quadratic equations, this means saying explicitly if one is interested in $x$ being a real number, a complex number, or something else (an integer, a matrix, etc); and similarly, $a$, $b$ and $c$ in $ax^2+bx+c=0$ have to be clearly defined.

That said,

1. Human beings like to save their energy and the rule is not often followed to the letter. The "nature" of a variable (real number, etc) is often omitted if it can be easily inferred by context. For example, it is common to write things like "we prove $P(n)$ by induction on $n\ge 2$" since everybody who understands induction will understand that $n$ is assumed to be an integer.
2. How to solve quadratic equations is usually told to students who have never heard about complex numbers. Complex numbers would appear a little bit further down the line in their study of precalculus. The only numbers they know are real numbers, so at that point, every number is probably implicitly assumed to be real, except otherwise stated (rational, integer, etc)

Answer 3

The quadratic formula applies to a wide range of number systems, even to ones you have not seen on a algebra-precalculus level. (For those who know what this means: it applies at the minimum to fields which are not of characteristic $2$.)

Thus as a general theorem the quadratic formula is valid for different number systems and we don't need to specify exactly the number system to state it.

In any case where it is applied, the quadratic formula tells us not only that a certain expression is a solution, but also that IF there are any solutions they to have a certain form, including a certain square root. So if this square root doesn't exist in your number system, there are no solutions (but there will always exist solutions in a larger system).

Your question is like asking: When we divide by two, do we have to declare the number system beforehand? Because in some number systems like $\Bbb Z$ you can't always divide by two, you need a larger number system like $\Bbb Q$. But this is not an issue: if we divide an integer by two, the result may not be an integer, but it makes sense unambigously as a rational number. The situation with the quadratic formula is similar: we can apply it and even if we started with real coefficients and get a complex solution, the set of solutions is uniquely determined and makes sense. So we don't need to declare beforehand that $x$ is allowed to be complex.

Answer 4

The answer depends on the formula in question.

Any general formula should be derived in a way that makes it hold in any number system that has the right properties to derive the formula. Most number systems in which we might want to use it, won't have those properties. But that's OK as long as some extension of the number system does. Then the answer will work as long as the output of the formula is within the original number system.

So, for instance, the quadratic formula requires a general square root. This may in general force us to operate in the complex numbers. If the final answer involves a square root of a negative number, we can then take this formula for the complex numbers, and then use it to show that there is no real answer. If it involves the square root of a rational that has no rational square root, then there is no rational answer.

In this case, when the formula can be evaluated within a number system, the answer is correct for that number system.

This gets more complicated with Cardano's famous solution of the cubic equation. We start with a cubic equation like $ax^3 + bx^2 + cx + d = 0$. Divide by $a$ and substitute $x = y - \frac{b}{3a}$ to put it in the form $y^3 + py + q = 0$. And now we solve for $y$ with:

$$ y = \sqrt[3]{-\tfrac{q}{2} + \sqrt{(\tfrac{q}{2})^2 + (\tfrac{p}{3})^3}} + \sqrt[3]{-\tfrac{q}{2} - \sqrt{(\tfrac{q}{2})^2 + (\tfrac{p}{3})^3}} $$

Even when you start with real coefficients and all roots are real, the intermediate terms in this calculation are often complex. (This is one of the things that drove mathematicians to start working with the complex numbers.) This is a good example of a useful formula for finding answers in a number system, where the calculations can't even be done in that number system!

If we are being formal, we need to care about such details. But informally we just assume that we're in "a strong enough" system. And that system is usually either the reals or the complex numbers.

Answer 5

For the quadratic formula to apply, you need the expressions $ax^2 + bx + c$ and $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ to be well-defined. That is, you need a type of “number” where the following operations obey reasonable expectations:

• Addition (in order to add the terms of the polynomial, and to make the plus branch of the $\pm$).
• Additive inverse (to make the expressions $-b$ and $-4ac$)
• Multiplication (to make the expressions $ax^2$, $bx$, $b^2$, $4ac$, and $2a$).
• Multiplicative inverse (to make $\frac{1}{2a}$)
• Square root (to make $\sqrt{b^2 - 4ac}$).

The first four operations are satisfied if $a$, $b$, $c$, and $x$ belong to any field, including $\mathbb{Q}$, $\mathbb{R}$, and $\mathbb{C}$.

The complication is the square root. Rationals and reals are not closed under this operation, but complex numbers are.

So, to make the formula as general as possible, we can say that $x \in \mathbb{C}$. It's possible that for a particular quadratic, $x$ happens to be a real number, rational number, or even an integer, but since all of these are subsets of complex numbers, $\boxed{x \in \mathbb{C}}$ covers everything.

It's possible to define a field with a square root that isn't $\mathbb{C}$, but $\mathbb{C}$ is the only one you're likely to encounter in high school.