Can √16 Be Both 4 and −4, or Just 4?

The student's question is classic, and it arrives every year from perhaps every class. "Sir, if I square 4, I get 16. But if I square -4, I also get 16. So shouldn't \sqrt{16} be both 4 and -4? Why do you insist on just 4?"

The answer has two parts, and both are important. Part one: 16 really does have two square roots, +4 and -4, and the student is absolutely right to notice that. Part two: the symbol \sqrt{16} refers specifically to the non-negative one, by convention. It is not that -4 is wrong — it is that \sqrt{\,\,} is defined to pick the positive one.

Untangling those two ideas resolves the question completely.

Two square roots exist; the symbol picks one

A square root of 16 is a number whose square is 16. By plain arithmetic, there are two such numbers:

4, because 4^2 = 16.
-4, because (-4)^2 = 16.

Both are equally legitimate as "square roots of 16" in the common-language sense. If you are solving the equation x^2 = 16, both x = 4 and x = -4 are correct solutions, and you write them together as x = \pm 4.

But now look at the symbol \sqrt{16}. This is not the same as the English phrase "a square root of 16." It is a specific mathematical symbol with a specific definition:

\sqrt{a} (for a \geq 0) denotes the principal square root of a — the unique non-negative number whose square is a.

With this definition, \sqrt{16} is the unique non-negative number whose square is 16. That number is +4. Not -4. The symbol has picked one of the two square roots, and by convention it always picks the non-negative one.

So:

"The square roots of 16 are \pm 4." — True.
"\sqrt{16} = \pm 4." — False.
"\sqrt{16} = 4." — True.
"-\sqrt{16} = -4." — True.

The word "root" and the symbol \sqrt{\,\,} are not synonyms, even though they feel like they should be. The word is about a property ("a number with this property"); the symbol is about a function ("the non-negative number with this property"). The two meanings diverge for every positive number.

Why the convention matters

You might ask: why not just let \sqrt{16} mean "both +4 and -4"? Then the student's objection would evaporate.

Because then \sqrt{\,\,} would not be a function, and that would break almost every formula that uses it.

A function, by definition, assigns exactly one output to each input. f(x) = x + 1 gives one output, not two. f(x) = 3x^2 - 7 gives one output, not two. If \sqrt{\,\,} returned two values, then every expression containing \sqrt{\,\,} would branch into two expressions, every equation containing \sqrt{\,\,} would split into two equations, and the whole algebraic machinery would become tangled.

Consider: the quadratic formula is

x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}

The \pm is explicit, and it is outside the \sqrt{\,\,}. If the \sqrt{\,\,} itself were multi-valued, you would have to write \pm\sqrt{\,\,} inside the formula as well — \pm \pm — which would give four things instead of two, most of which are redundant. The quadratic formula works because \sqrt{\,\,} is single-valued and the \pm is written explicitly. Convention makes the formula clean.

Why one output, not two: mathematical notation is built on functions, and functions demand a unique output per input. The choice to make \sqrt{\,\,} the non-negative root is a convention, but once chosen, every identity involving the symbol respects it: \sqrt{a}\cdot\sqrt{b} = \sqrt{ab}, \sqrt{a^2} = |a|, (\sqrt{a})^2 = a for a \geq 0, and so on. Change the convention, and every one of those identities would need rewriting.

The correct way to get the negative one

You do not get -4 by saying "\sqrt{16} is -4." You get it by writing -\sqrt{16} — the negative of the principal root.

-\sqrt{16} = -(4) = -4

So when a problem genuinely asks for "both square roots," you write \pm\sqrt{16}, which is the pair (+\sqrt{16}, -\sqrt{16}) = (4, -4). The \pm is external to the radical; the radical itself is always the non-negative one.

This is why the solutions of x^2 = 16 are written x = \pm 4, not x = \sqrt{16}. The \pm makes explicit that you want both square roots; the single \sqrt{16} would only give you one.

Why this convention is the one we chose

The choice of "non-negative" for the principal root is not arbitrary. It has two major advantages:

Compatibility with the natural ordering. For positive a, \sqrt{a} is the positive side of the square-root relationship. Since positive numbers come up more often in real applications — lengths, areas, times, probabilities — it is convenient for \sqrt{\,\,} to default to the positive answer. A physicist writing "length = \sqrt{A}" for the side of a square with area A wants a positive length, not a negative one.
Compatibility with exponent notation. The identity \sqrt{a} = a^{1/2} extends the laws of exponents to fractional powers. Under the convention a^{1/2} \geq 0 (for a \geq 0), the exponent laws work out cleanly — a^{1/2} \cdot a^{1/2} = a^{1/2 + 1/2} = a^1 = a, as you would expect. If a^{1/2} could be negative, the exponent laws would behave inconsistently.

Both arguments point the same way: picking the non-negative root makes the rest of algebra work smoothly.

The convention for other roots

The principal-root convention for square roots generalises:

For even roots (\sqrt[4]{\,\,}, \sqrt[6]{\,\,}, etc.), the same convention applies: \sqrt[2k]{a} is the unique non-negative 2k-th root of a \geq 0. So \sqrt[4]{16} = 2, not -2, even though (-2)^4 = 16.
For odd roots (\sqrt[3]{\,\,}, \sqrt[5]{\,\,}, etc.), there is no sign ambiguity — every real number has exactly one real odd root. \sqrt[3]{-8} = -2, no "positive version" convention needed.

Odd roots preserve sign; even roots destroy it, so the convention only matters for even roots. See Roots and Radicals for the general rule.

The pitfall: losing solutions

Here is where the confusion costs marks. When you take the square root of both sides of an equation, you must insert the \pm explicitly — otherwise you silently lose a solution.

Wrong: x^2 = 16 \implies x = \sqrt{16} = 4. (Loses the solution x = -4.)

Right: x^2 = 16 \implies |x| = \sqrt{16} = 4 \implies x = \pm 4. (Both solutions present.)

Or equivalently: x^2 = 16 \implies x = \pm\sqrt{16} = \pm 4.

The absolute-value version is the most honest: \sqrt{x^2} = |x| by the identity in Why √(x²) Draws the Same V as |x|, and |x| = 4 gives two cases, x = 4 or x = -4.

The \pm version is the slick version used in the quadratic formula and most derivations. Whichever you prefer, the key is the \pm (or the absolute value) — without it, the single-valued \sqrt{\,\,} silently discards the negative root.

A quick sanity check: the calculator

Punch \sqrt{16} into any calculator. It returns 4. Not \pm 4, just 4. Every calculator in the world obeys the principal-root convention, because if they returned both roots they would have to return two numbers per calculation, and that is not how display screens work. So the physical calculator confirms the convention: \sqrt{16} is 4.

The takeaway

The number 16 has two square roots, +4 and -4. The symbol \sqrt{16} refers to the principal square root, which is the non-negative one — so \sqrt{16} = 4. To refer to the negative root, write -\sqrt{16}. To refer to both, write \pm\sqrt{16}. The convention exists to make \sqrt{\,\,} a function, which is what makes algebra clean, and dropping the convention (or forgetting the \pm when solving x^2 = 16) is the quickest way to lose a solution and some marks.