The Square Root Symbol Picks Only the Non-Negative Root — The ± Comes from the Equation

A lot of students carry around a quiet contradiction. In one breath they say "\sqrt{9} = \pm 3," and in the next they say "\sqrt{9} = 3." Both cannot be right. And when they solve x^2 = 9, the answer is x = \pm 3 — and it becomes unclear where that \pm actually came from.

The resolution is short, sharp, and saves a surprising amount of grief: the square root symbol \sqrt{\,\,} has exactly one output, and it is never negative. The \pm enters the scene only when you are solving an equation of the form x^2 = k — and the "plus-or-minus" comes from the step of applying the square root, not from the symbol itself.

The rule, stated crisply

For any non-negative real number a:

\sqrt{a} \;\text{is defined to be the non-negative number whose square is } a.

So \sqrt{9} = 3 and nothing else. Not \pm 3. Not -3. Just 3.

This is a definition, not a theorem — the mathematics community agreed, a long time ago, that the symbol would be single-valued. The alternative would be to let \sqrt{9} be a "multi-valued" expression that returns the set \{3, -3\}, but that breaks lots of algebra. A symbol that produces a unique number is easier to compute with than one that produces a set.

Why single-valued: a function is a rule with one input and one output. \sqrt{\,\,} is treated as a function (it has a graph, a domain, a range). If it returned two values, it would not be a function at all — and statements like \sqrt{a} \cdot \sqrt{b} = \sqrt{ab} would be ambiguous. The convention keeps the algebra clean.

The formal identity that cements it

The identity that enforces the single-value rule is

\sqrt{x^2} = |x|.

For any real x, the left side is the non-negative number whose square is x^2. That number is always |x| — positive if x > 0, zero if x = 0, the positive version of x if x < 0. It is never x itself for negative x, because that would give a negative output from the \sqrt{\,} symbol, violating its non-negativity.

So: the symbol outputs a non-negative value, always, regardless of the sign of whatever went into the square.

Where does the \pm come from when solving x^2 = k?

Here is the step that matters. You have the equation

x^2 = 9

and you want to solve for x. What you actually do, carefully, is take the square root of both sides:

\sqrt{x^2} = \sqrt{9}.

The right side is 3 (by the symbol's definition). The left side, by the identity above, is |x|. So

|x| = 3.

Now, |x| = 3 means x is a number whose distance from zero is 3 — so x is 3 or x is -3:

x = 3 \quad\text{or}\quad x = -3, \quad\text{written}\quad x = \pm 3.

The \pm came from solving the absolute-value equation |x| = 3, not from the \sqrt{\,\,} symbol. Every time you see \pm appearing in "x^2 = k \implies x = \pm\sqrt{k}," what has really happened under the hood is \sqrt{x^2} = |x|, then |x| = \sqrt{k}, then x = \pm \sqrt{k}.

A worked example where the distinction matters

Consider: find all x satisfying \sqrt{x^2} = 5.

Wrong reasoning. Treat \sqrt{x^2} as x. Then x = 5 is the only answer. Miss x = -5.

Right reasoning. Use \sqrt{x^2} = |x|. So |x| = 5, giving x = 5 or x = -5.

The wrong reasoning "loses" the negative root because it pretended \sqrt{x^2} was just x, which is only true when x \geq 0. When x could be negative, you have to keep the |x| until you know the sign.

Solve $(x - 4)^2 = 25$ step by step.

Apply square root to both sides, carefully:

\sqrt{(x - 4)^2} = \sqrt{25}.

Left side: |x - 4|. Right side: 5.

|x - 4| = 5.

Absolute-value equation. Two cases:

x - 4 = 5 \Rightarrow x = 9.
x - 4 = -5 \Rightarrow x = -1.

So x \in \{9, -1\}, often written x - 4 = \pm 5, i.e. x = 4 \pm 5.

Why write |x-4| and not just x-4: because you do not know yet whether x-4 is positive or negative. Keeping the absolute value bars is what lets both solutions come out naturally. The \pm sign in the compact notation is a shorthand for "take both cases."

A visual that makes the rule stick

Picture the graph of y = x^2. Draw a horizontal line at y = 9. It crosses the parabola at two points: (3, 9) and (-3, 9). These are the two values of x whose square is 9.

The \sqrt{\,} symbol, applied to 9, returns only the right-hand crossing, namely 3. The left-hand crossing, -3, is not returned by the symbol; it is recovered separately by noticing that the equation x^2 = 9 is symmetric about the y-axis.

Where this intuition pays off

Carrying this distinction sharpens four common tasks:

Quadratic formula. The \pm in x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} is written in because it came from the step "take the square root of both sides of (2ax + b)^2 = b^2 - 4ac." The \sqrt{\,\,} inside still gives a single non-negative value; the \pm in front is what makes it cover both roots.
Radical equations. When you square both sides of something like \sqrt{2x + 1} = x - 2, you can introduce extraneous solutions — values of x that satisfy the squared equation but not the original, because the squaring step does not care about signs. Always substitute back to check.
Simplifying \sqrt{x^2} under an assumption. If the problem says x \geq 0, you can write \sqrt{x^2} = x directly. If the problem does not say so, write \sqrt{x^2} = |x| and carry the bars until you have more information.
JEE trap questions. Problems are often designed around the difference between "\sqrt{\text{something}}" (one value, non-negative) and "a number whose square is X" (two values, \pm\sqrt{X}). Reading the question carefully and spotting which one is being asked is worth marks.

The one-line rule

The square root symbol always returns a non-negative number. The \pm you sometimes see around a square root is a consequence of taking the root of both sides of an equation — specifically, from the step \sqrt{x^2} = |x| and the absolute-value split that follows. Never from the symbol itself.