√(x²) Is Always |x| — Drop the Bars Only Once You Know the Sign

There is a two-symbol identity that quietly catches out students every year on the JEE. It is

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

Not x. Never just x, unless you already know x is non-negative. The vertical bars are there on purpose, and dropping them prematurely is one of the most common ways to lose a mark or miss a solution.

This article argues for a habit: write |x| first, then drop the bars only when the sign is forced. The habit costs you one symbol per line. It saves you a whole class of mistakes.

The identity, quickly justified

The symbol \sqrt{\,\,} returns the non-negative number whose square is what is under the radical. For any real x:

If x \geq 0, then x^2 \geq 0, and the non-negative number whose square is x^2 is x itself. So \sqrt{x^2} = x.
If x < 0, then x^2 > 0, and the non-negative number whose square is x^2 is -x (which is positive, since x is negative). So \sqrt{x^2} = -x.

Combining: \sqrt{x^2} is always the non-negative version of x, which is exactly |x|:

|x| = \begin{cases} x & \text{if } x \geq 0 \\ -x & \text{if } x < 0 \end{cases}.

So \sqrt{x^2} = |x| for every real x, without exception.

Why not just "equals x": the symbol \sqrt{\,\,} outputs a non-negative value. If x is negative, then x itself is negative — but the square root symbol cannot output a negative number. The output has to be -x (the positive version). The one notation that covers both cases at once is |x|.

A quick numeric check

Plug in x = -5:

Left side: \sqrt{(-5)^2} = \sqrt{25} = 5.
Right side: |-5| = 5. Match.

Plug in x = 7:

Left side: \sqrt{(7)^2} = \sqrt{49} = 7.
Right side: |7| = 7. Match.

If you (wrongly) wrote \sqrt{x^2} = x, the x = -5 case would give \sqrt{25} = -5, which is false. That is the mistake.

The habit: carry the bars until the sign is forced

There are two kinds of situations where you meet \sqrt{x^2}:

Situation A: the problem tells you x \geq 0, or the context forces it.

Example: "For x \geq 0, simplify \sqrt{x^2}." You can drop the bars immediately: \sqrt{x^2} = x. The problem has given you permission.

Example: x represents a length (in geometry), a mass, a probability, or a number of seconds. These are always non-negative by context, so \sqrt{x^2} = x directly.

Situation B: x could be negative, or you just do not know yet.

Example: "Simplify \sqrt{(2a - 5)^2}" with no restriction on a. Here 2a - 5 could be negative (if a < 5/2) or positive (if a > 5/2). Write \sqrt{(2a - 5)^2} = |2a - 5| and leave it at that. Dropping the bars here would be wrong whenever a < 5/2.

Example: solving \sqrt{x^2} = 7. Do not write x = 7. Write |x| = 7, then x = 7 or x = -7.

A worked example where dropping bars costs you

Problem: how many real solutions does \sqrt{(x - 3)^2} + \sqrt{(x + 1)^2} = 4 have?

Sloppy, bar-dropping reasoning. Write (x - 3) + (x + 1) = 4, so 2x - 2 = 4, x = 3. One solution.

Careful reasoning. Write |x - 3| + |x + 1| = 4. This is an absolute-value equation. Case-split by the sign of each piece:

If x \geq 3: |x - 3| = x - 3 and |x + 1| = x + 1, summing to 2x - 2 = 4, so x = 3. Check 3 \geq 3: yes. Valid.
If -1 \leq x < 3: |x - 3| = 3 - x and |x + 1| = x + 1, summing to 4. Always 4. Every x in [-1, 3] works.
If x < -1: |x - 3| = 3 - x and |x + 1| = -x - 1, summing to -2x + 2 = 4, so x = -1. Not in the range x < -1, so no new solution.

The equation has infinitely many solutions: the entire interval [-1, 3]. The sloppy method found only one of them.

The absolute-value bars were not decoration — they were carrying information about the sign of each piece, and splitting the problem into cases is what found the full solution set.

A graph that makes the identity unforgettable

Plot y = \sqrt{x^2}. It is a V-shape — identical to the graph of y = |x|.

The left branch comes from negative x values, where \sqrt{x^2} spits out the positive version -x. The right branch comes from non-negative x, where \sqrt{x^2} = x. At x = 0, both branches meet at the origin.

If the identity were \sqrt{x^2} = x (without the absolute value), the graph would be the line y = x — going negative on the left. But the square root symbol's output cannot go below zero. The V-shape is what reconciles "output is non-negative" with "input can be anything."

Simplify $\sqrt{4x^2 - 12x + 9}$ (with no restriction on $x$).

First notice that 4x^2 - 12x + 9 = (2x - 3)^2, a perfect square. So

\sqrt{4x^2 - 12x + 9} = \sqrt{(2x - 3)^2} = |2x - 3|.

Do not write 2x - 3. The bars are needed because 2x - 3 could be negative (whenever x < 3/2).

If later the problem tells you x \geq 2, then 2x - 3 \geq 1 > 0, and you can drop the bars: |2x - 3| = 2x - 3. But do not drop them before.

Why spot the perfect square first: once you recognise that the radicand is (2x-3)^2, the simplification is forced — \sqrt{(\text{anything})^2} = |\text{anything}|. The harder half of these problems is always the algebra step, not the absolute-value step.

Where this habit pays the most

Four settings where bar-carrying directly saves marks:

Solving equations with \sqrt{f(x)^2} on one side. Always rewrite as |f(x)| = \ldots first. Split into cases afterwards.
Definite integrals over symmetric intervals. When integrating \sqrt{f(x)^2} over [-a, a], the integrand is |f(x)|, which may behave very differently from f(x) on the left half of the interval. Pretending it is just f(x) can produce wrong answers.
Domain of a function like g(x) = \sqrt{x^2 - 4}. The domain is \{x : x^2 \geq 4\} = (-\infty, -2] \cup [2, \infty), which comes from solving |x| \geq 2. If you carelessly write \sqrt{x^2} = x, you miss the left branch.
Simplifying \sqrt{a^2 b} with a unknown. Write |a|\sqrt{b}, not a\sqrt{b}. JEE answer keys often accept either only when the problem explicitly says a > 0.

The one-line rule

\sqrt{x^2} = |x|, always. Write the bars. Drop them only when the problem or context guarantees the sign is non-negative. This tiny habit catches a category of lost solutions that would otherwise quietly eat your marks.