Why Does √9 Equal Just 3, Not ±3?

You write \sqrt{9} = \pm 3 on a test. The teacher puts a red cross through the \pm and writes "just 3." You push back: "But (-3)^2 = 9 too! Why can't \sqrt{9} be -3?" The teacher shrugs and says, "That is just how the symbol is defined." Unsatisfying. You know the negative root exists — you can square -3 and land on 9 — so why does the radical sign pretend it does not?

The answer is that the symbol \sqrt{\phantom{x}} is a very specific thing with a very specific job. It is not shorthand for "all numbers that square to what is inside." It is a function, and a function is allowed only one output. By universal mathematical convention, \sqrt{\phantom{x}} returns the principal (non-negative) square root — the positive one — and only that one. The negative root is real, and it matters, but it lives somewhere else: in the step where you solve an equation, not in the evaluation of the symbol.

This page is about separating those two jobs cleanly so you stop getting marked wrong for something that is really a notation issue.

What the \sqrt{\phantom{x}} symbol means

Here is the definition, spelled out:

Read that again. "The non-negative real number whose square is x." There is exactly one such number for every x \geq 0, and the radical sign names it. For x = 9, that number is 3, not -3. So \sqrt{9} = 3. Full stop. There is nothing to break, no rule to bend — the statement \sqrt{9} = -3 is not wrong in some interpretive sense; it is wrong in the same way 2 + 2 = 5 is wrong. The symbol on the left does not name the number on the right.

The same goes for \sqrt{16} = 4 (not -4), \sqrt{25} = 5 (not -5), \sqrt{100} = 10 (not -10). The radical is a one-output machine, and the output is always the positive root.

Why would we define it this way?

You might wonder whose idea this was and whether it could have gone differently. The answer is that it really could not — not if you want mathematics to work.

The radical sign is meant to be a function. A function is a rule that takes an input and produces exactly one output. "Take any number and give me the set of things that square to it" is a rule too, but it is a multi-valued rule, giving \{+3, -3\} for input 9. Multi-valued rules are allowed in mathematics, but they break almost all the machinery you rely on. You cannot differentiate a multi-valued rule cleanly. You cannot compose multi-valued rules without getting combinatorial explosions. Graphing becomes awkward — the graph of a multi-valued "square root" would fail the vertical line test, because each positive input would have two outputs.

So mathematicians made a choice: the radical sign returns one root, the positive one. The other root is still there, still real, still has its uses — but it is accessed explicitly, by writing -\sqrt{x} when you need it. The convention is universal. Every textbook, every calculator, every programming language agrees: \sqrt{\phantom{x}} means the principal root.

Where \pm 3 comes from — solving x^2 = 9

Now the reason students keep getting confused. The equation

genuinely has two solutions. That is not an opinion or a convention — it is a fact about the equation. Substitute x = 3: 3^2 = 9. True. Substitute x = -3: (-3)^2 = 9. Also true. Both values satisfy the equation. An equation is a question, and this question has two answers.

To solve it, you take the square root of both sides. Here is the step that causes all the trouble, written carefully:

Notice where the \pm shows up. It is not inside the radical. It is not attached to the 3 on the right-hand side as a property of the symbol \sqrt{9}. The \pm appears because |x| = 3 means "x is a number whose distance from zero is 3" — and that allows two possibilities, x = 3 and x = -3.

The \sqrt{\phantom{x}} itself gave a single number, 3. The \pm is a separate piece of information added to acknowledge that the original squaring could have gone either way.

The distinction in one line

Burn this into your head:

• \sqrt{9} is an evaluation of a symbol. It equals 3. One value.
• x^2 = 9 is an equation. Its solutions are 3 and -3. Two values.

These are two different mathematical objects. The first is a number. The second is a question about numbers. The \pm belongs to the second, never to the first.

Worked example — solving x^2 = 16

Why \sqrt{x^2} = |x| and not x

This is the technical heart of the matter. Suppose x = -5. Then x^2 = 25. Now compute \sqrt{x^2} = \sqrt{25} = 5. Is that equal to x? No — x was -5, and you got 5. They differ in sign.

The identity that works for every real x, positive or negative, is

because |x| = 5 when x = -5, which is exactly what the radical returned. The absolute value absorbs the sign difference — whatever sign x had, |x| strips it away, matching the non-negativity that the radical sign insists on.

This is why, whenever you "take the square root of both sides" of an equation, you must write the absolute value if the variable could be negative. Skipping that step is how the \pm disappears by accident.

The quadratic formula — where the \pm lives

The quadratic formula is the cleanest example of the radical's discipline in action:

Look at what is inside the radical: b^2 - 4ac. That expression, when positive, has a single non-negative square root, and that is what the \sqrt{\phantom{x}} computes. The symbol is single-valued, as always.

The \pm sits outside the radical, in front of it. It is the formula's way of saying "and there are two solutions, one for each sign choice." It is not part of the square root. It is an explicit acknowledgement that solving a quadratic equation — an equation of degree two — gives up to two solutions.

Any time you see \pm next to a radical, the pattern is the same: the radical itself gives one value; the \pm is bookkeeping from the equation-solving step.

The cube root is different

Something worth noticing: the whole \pm confusion is specific to even roots. For cube roots (and fifth roots, seventh roots — any odd root), every real number has exactly one real cube root, no sign ambiguity.

You can check: 2^3 = 8, and that is the only real number cubed that gives 8. Similarly (-2)^3 = -8, and that is the only real number cubed that gives -8. There is no "other" cube root hiding with an opposite sign, because (+2)^3 and (-2)^3 already disagree — one is +8, the other is -8. Cubing preserves sign, so cube-rooting does too.

The \pm nonsense is purely an even-root phenomenon, arising because squaring destroys sign information (both +3 and -3 square to the same 9).

Common confusions

• "Writing \sqrt{9} = 3 loses the negative solution." No, it does not. The "negative solution" is not part of the symbol \sqrt{9} — it is a solution to the equation x^2 = 9, which is a different object entirely. When you solve that equation properly, both solutions appear via the \pm. Nothing is lost.

• "The \pm goes inside the radical." No. You never write \sqrt{\pm 9} to mean "both roots." The \pm always sits in front: \pm \sqrt{9}, meaning "either +3 or -3." The radical itself is single-valued; the \pm is a separate modifier.

• "The \pm in the quadratic formula is baked into the radical." No. Look at the formula: \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a}. The \pm is a stand-alone sign; the \sqrt{\phantom{x}} computes one non-negative value. The \pm arises from solving the equation, not from the symbol.

• "\sqrt{x} is undefined for negative x, so the negative root doesn't exist at all." In real numbers, \sqrt{x} is indeed undefined for x < 0. But this is about the radicand being negative, not about the output being negative. \sqrt{9} is perfectly defined — it equals 3 — and the existence of a negative number that also squares to 9 is a fact about the equation x^2 = 9, not the symbol. (In complex numbers, \sqrt{-1} is defined as the principal root i, extending the same principle: one output, the "principal" one, chosen by convention.)

Why this convention is worth accepting

Two reasons this choice of definition is not just a teacher being picky:

1. Functions must be single-valued. All of calculus — derivatives, limits, integrals — is built on functions that return exactly one output per input. A multi-valued \sqrt{\phantom{x}} would not fit into that machinery. The derivative of a multi-valued function is not a well-defined object.

2. The \pm belongs in the open, not hidden in a symbol. If \sqrt{9} secretly meant "either 3 or -3," you would have to carry around both possibilities through every subsequent step, branching your calculations. By making the symbol single-valued and writing \pm explicitly where it is actually needed — in the equation-solving step — the sign ambiguity is contained, visible, and under your control.

Closing

The radical sign \sqrt{\phantom{x}} is a symbol with one output. When you see it, it gives you the principal (non-negative) square root. That is what the symbol means, not a rule you can break. The \pm is a different creature — it lives in the step where you solve an equation like x^2 = 9, acknowledging that two values could have been squared to produce the 9. Separate the two jobs — evaluating the symbol versus solving the equation — and the confusion clears up permanently. \sqrt{9} = 3. The -3 is waiting for you in the equation, not in the symbol.