# Why are equations written by equating something to zero?

A linear equation is $$ax + b = 0 ; \,\, \,\, a\neq 0$$

A quadratic equation is $$ax^2 + bx + c = 0 ; \,\, a\neq 0$$

And so on...

Why are all these equations written as $\dots = 0$? Why do mathematicians do it this way?

• Not sure why the downvotes. It seems like a reasonable question?
– John
Mar 6, 2014 at 1:48
• In fact, before negative numbers were invented/accepted, there were a handful of different forms of the quadratic formula, since without negative numbers one cannot generally normalize every equation into such a standard form. Mar 6, 2014 at 1:48
• possible duplicate of Why a quadratic equations always equals zero? Mar 6, 2014 at 1:50
• ok nvm then. Just saying so you can look at the other question for possible answers. Mar 6, 2014 at 1:53
• And it's not that they always must be written that way, it's that they always can be written that way. It characterizes quadratic equations.
– MPW
Mar 7, 2014 at 1:13

At least for quadratics, if you want to solve (for example) $x^2 + 5x +8 = 2$, it is much easier to subtract 2 from each side, and factor:

$x^2 +5x +8-2 =0$

$x^2 + 5x +6 =0$

$(x+2)(x+3)=0$

Here is the key: the only way for a product of numbers ($x+2$) and ($x+3$) to be equal to zero is for one to be zero. This is a property unique to zero, and explains (at least in part) why we often set equations equal to zero.

• +1 Aaaahhhhhhhh the zero divisor property of the reals and the complex plane. Mar 6, 2014 at 4:34
• That's true, if we start introducing more complicated rings, then we need to start worrying about zero divisors. (Read: if we start introducing more interesting rings, we get to start thinking about zero divisors.) Mar 6, 2014 at 6:39
• I would have thought $x^2=100$ might be easier to solve than $x^2-100=0$, and similarly for problems of the completing the square type so $x^2+6x -91=0$ gets easier if you write it as $x^2+6x +9=100$ Mar 6, 2014 at 8:23
• Before formulas were introduced, and before negative numbers were invented (outside India), there were many cases of (what is today called) the quadratic equation. See this important book from whose name the word "algebra" comes. (The word "algorithm" comes from its author's name, Algorismus (al-Khwārizmī).) Mar 6, 2014 at 12:53

Once upon a time, mathematicians studied three different kinds of quadratic equations:

• $ax^2 + bx = c$
• $ax^2 + c = bx$
• $ax^2 = bx + c$

(I'm not sure if they studied the fourth case, since the solutions would be negative numbers)

Correspondingly, you had to learn three different methods for solving a quadratic equation! Quite annoying! By normalizing the equation to just a single form,

$$ax^2 + bx + c = 0$$

you only have to learn one one method to solve all quadratic equations! I think this choice of the four possibilities is the least ad-hoc choice: many different sorts equations forms share an "something equals zero" version, when they might not have anything else in common.

It's because of the constant term. Look at it this way:

If you have $ax+b=c$, then $ax+(b-c)=0$ and there's some $d=b-c$ so that $ax+d=0$. Likewise, when $ax^2+bx+c=d$, $ax^2+cx+e=0$, where $e=c-d$. It's just a standard way of writing equations so that they are easier to deal with, categorize, and solve.

It's simply a way of putting an equation into a standard form. You can always add and subtract the same quantities from both sides so that one of the sides becomes zero without changing the solution(s) of the equation.

By equating the polynomial equation to zero and factoring the polynomial, we can find its roots. (For a product to be equal to zero, at least one of its factors must be equal to zero.) This procedure was first done by Thomas Harriot (1560-1621). According to this website, "Harriot was the first mathematician to set an equation equal to zero and then factor it."

• For example, say we have the equation $x^2-5x=-6$. How do we solve it? If we factor the left side we get $x(x-5)=-6$. The number $-6$ has many factors. Is $x=-2$ a solution? But if we represent it as $x^2-5x+6=0$, we can rewrite the left side as a product and get $(x-2)(x-3)=0$ and it becomes very easy to see that this will only be true if either $(x-2)=0$ or $(x-3)=0$, that is, if $x=2$ or $x=3$. Mar 6, 2014 at 2:02
• This is the only answer of any use... the answer is the no zero divisors theorem. Mar 31, 2014 at 19:04

You might be more accustomed to the notation $y = ax^2 + bx + c$. Essentially, the zero is stating where the equation intersects with the x axis, because when y = 0, the the equation is on the x axis. Also, it makes it really convenient for equations like $y = 8x^2 - 16x - 8$ because when finding the root (or solution) (or value of x when = 0), we can divide out the 8.

You can consider functions with multiple independent functions instead of a single independent variable. When doing so a function is the set of solution points (in multivariable space) that satisfies the equation or a system of equations.

Consider

$$y_1(x_1,x_2) = x_2^3 - x_1x_2 + x_1 + 2 = 0$$

and

$$y_2(x_1,x_2) = x_1^3 - x_1x_2 - 2x_1 + x_2 - 5 = 0$$

By satisfying the equation, I mean that $(x_{1_0}, x_{2_0})$ satisfies $(y_1, y_2)$ if and only if $y_1(x_{1_0},x_{2_0}) = y_2(x_{1_0},x_{2_0}) = 0$.

This works for $1$, $2$, ... , $n$ variables. In the univariate case, points on the graph visually represent your solution set that satisfies the equation.

So, for a quadratic, we write it as $y_1 - (ax_1^2 + bx_1 + c) = 0$ and so the points on the graph of a quadratic satisfy the most previous equation. Note that the $y_i$ are the dependent variables and they are functions that depend on the independent variables $x_i$.

The short answer is that it's just a standard form that can be nicely extended to any degree (as G Watt notices). We could have used another one. For example, as Hurkyl already mentioned, Babylonians used up to three different normal forms. This, together with the fact that they didn't use negative numbers, probably made the subject a lot harder to learn than it is today.

This standard form makes it easier to factorize a quadratic polynomials when the factors are of the form $x + u$ with $u$ a whole number or fraction. If you see $x^2 + ax + b$, you can ask yourself "What are numbers $u$ and $v$ such that $u + v = a$ and $uv = b$?". If you found numbers $u$ and $v$ that satifsy, $(x+u)(x+v)$ is a solution.

When you try to find irrational roots, I personally like the form $$x^2 = ax + b$$

Because this has a nice geometric interpretation (you find the intersections of a parabola and a line), and makes for a very simple variant of the abc-formula:

$$x = \frac{a}{2} \pm \sqrt{(\frac{a}{2})^2+b}$$

which is much easier to remember and calculate than the one I learned in high school (honestly, I feel like they have been making it harder than necessary). Additionally, the geometric interpretation can help tracking mistakes (both numerical or conceptual),

Please also note that it's not true that "all are equal to zero". While it is common to use factoring to solve equations in reals or complex numbers (as explained in other answers), in case of e.g. integral equations (ones in which the unknown is not a number – real or otherwise – but a function, and which involve some kind of integral of this function – see e.g. http://en.wikipedia.org/wiki/Integral_equation), a common way to solve or analyse is to write them down in the form $F(x)=x$ (where $x$ is the unknown function and $F$ is some operator, i.e. a mapping transforming functions into other functions) and apply one of many fixed point theorems.

In principle, while there is nothing to stop one from applying fixed point theory to equations in $\mathbb{R}$ or $\mathbb{C}$ (see the quoted Wikipedia article for an example), it is not a usual way – the usual ones being either using algebraic transformations, or some approximate methods, like the Newton–Raphson method (which, by the way, is in fact closely related to fixed-point methods, but this fact is not often highlighted in elementary courses).

Mathematicians love to generalize and any polynomial can be written in the form:

$ax^n + bx^{n-1} + ....... + cx^2 + dx^1 +cx^0 = 0$

(remember $x^1 = x$ and $x^0 = 1$) This is the prettiest way of writing it

• Any polynomial can be written as $a_nx^n + a_{n-1}x^{n-1} + \ldots + a_0x^0$. Just that. No $=0$. Mar 6, 2014 at 6:48