Where does the plus-minus come from in the quadratic formula?

Question

In the formula

$$x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

for solving quadratic equations, where does the $\pm$ come from?

The square root already results in both a positive and negative term, is the $\pm$ not therefore extraneous?

"The square root already results in both a positive and negative term" -- Usually, $\sqrt{z}$ implicitly means "the positive square root." Otherwise, $\sqrt{z}$ would not even represent a function in the usual sense (because we don't want multiple outputs for a single input). — PrincessEev, Mar 28 '21 at 00:33
See "Square roots — positive and negative", "Would the answer of the square root of a square root be positive or negative?", "Is $\sqrt{64}$ considered $8$? or is it $8,-8$?", "Square root positive/negative", and various other similar questions. — David K, Mar 28 '21 at 00:38
Thanks for the clarification, the implicit meaning was not clear to me. Should the question be removed, as it is inherently a misunderstanding of notation, rather than an issue specific to the quadratic formula? — Minos, Mar 28 '21 at 00:39
@Minos A questions which is based on misunderstanding the notation is a valid question, in my opinion. So I wouldn´t delete the question. — callculus42, Mar 28 '21 at 00:42
Actually one of my arguments to explain how we know that people define $\sqrt{x}$ as a function of $x$ (rather than something with multiple values) is that we always see $\pm$ in the solution of the quadratic equation, indicating that $\sqrt{x}$ is only one of the square roots and we need $-\sqrt{x}$ to get the other one. So I think the quadratic equation is very relevant, if only for the reason it is where we often see $\pm$ in front of a square root. — David K, Mar 28 '21 at 00:43
Does this answer your question? Square roots -- positive and negative — Siong Thye Goh, Mar 28 '21 at 01:16

score 1 · Answer 1 · answered Mar 28 '21 at 00:35

What about completing squares? \begin{align*} ax^{2} + bx + c & = a\left(x^{2} + \frac{bx}{a}\right) + c\\\\ & = a\left(x^{2} + \frac{bx}{a} + \frac{b^{2}}{4a^{2}}\right) + c - \frac{b^{2}}{4a}\\\\ & = a\left(x + \frac{b}{2a}\right)^{2} + \frac{4ac - b^{2}}{4a} \end{align*} Hence we deduce that \begin{align*} ax^{2} + bx + c = 0 & \Longleftrightarrow a\left(x + \frac{b}{2a}\right)^{2} + \frac{4ac - b^{2}}{4a} = 0\\\\ & \Longleftrightarrow \left(x + \frac{b}{2a}\right)^{2} = \frac{b^{2} - 4ac}{4a^{2}}\\\\\ & \Longleftrightarrow x = \frac{-b \pm \sqrt{b^{2}-4ac}}{2a} \end{align*}

and we are done.

Hopefully this helps!

score 1 · Answer 2 · answered Mar 28 '21 at 01:10

Given a general quadratic \begin{eqnarray*} ax^2+bx+c=0. \end{eqnarray*} Multiply through by $4a$, add $b^2$ & this will allow us to "complete the square" \begin{eqnarray*} 4a^2x^2+4abx+b^2 &=& b^2-4ac\\ (2ax+b)^2 &=& b^2-4ac.\\ \end{eqnarray*} Now take the square root, this is where the plus-minus comes from ...

score 1 · Accepted Answer · answered Mar 28 '21 at 01:20

Per the comments (and making this CW to avoid reputation gain for others' work), the issue is how we interpret the symbol "$\sqrt\cdot$" - is it the function outputting the unique nonnegative square root of a nonnegative input, or is it describing the set of all square roots of the input?

The standard meaning is the former, so e.g. $\sqrt{4}=2$. Regardless, the notation "$\pm$" only adds clarity to the situation.

Incidentally, thinking ahead to the complex numbers, note that while $\{x: x^2=a\}$ always makes sense there are serious problems with trying to produce a "square root function" in the context of $\mathbb{C}$, and more generally with exponentiation in the complex numbers in general. There are lots of posts relating to this on the site - see e.g. here or here.

Where does the plus-minus come from in the quadratic formula?

3 Answers3