In short

|x|^2 = x^2 because squaring already destroys the sign all by itself. The expression |x|^2 reads as "(absolute value of x) squared" — a non-negative number, then squared. The expression x^2 is just "x squared" — also non-negative no matter what x was. Both routes end at the same place: a non-negative number equal to the magnitude of x multiplied by itself. No information is lost — squaring and absolute-value are both sign-killing operations, so doing one after the other is no different from doing one alone. This little identity is what lets you square both sides of an absolute-value equation and turn it into a polynomial — a standard JEE Main / Advanced trick.

When you first see the equation |x|^2 = x^2, it can feel like a magic trick — or worse, a cheat. The whole point of |x| is that it discards the sign of x. So how can it just disappear when you square it? Did the absolute value matter at all?

The answer is: yes, it mattered, but only until the moment you squared. Once you square, the sign was going to vanish anyway. The absolute value bars are doing the same job that squaring does, except they leave the magnitude alone instead of multiplying it by itself.

1. The algebraic proof

The cleanest way to see this is to use the case definition of absolute value from the parent article:

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

Now compute |x|^2 in each case.

Case 1: x \ge 0. Then |x| = x, so

|x|^2 = x \cdot x = x^2

Why: |x| is literally x here. Squaring x gives x^2.

Case 2: x < 0. Then |x| = -x, so

|x|^2 = (-x)(-x) = x^2

Why: (-x)(-x) = (-1)(-1) \cdot x \cdot x = 1 \cdot x^2 = x^2. Two negatives make a positive.

Either way — positive x, negative x, or x = 0 — the result is x^2. So the identity

\boxed{|x|^2 = x^2}

holds for every real number x, with no exceptions.

There is also a one-line proof using the fact (from the parent article) that |x| = \sqrt{x^2}. Squaring both sides:

|x|^2 = (\sqrt{x^2})^2 = x^2

Why: squaring undoes the square root, since x^2 \ge 0 already.

2. The intuition — squaring already kills the sign

Forget the algebra for a moment and think about what these two operations do.

  • Absolute value takes a number, throws away its sign, and keeps the magnitude. So |-5| becomes 5, and |5| stays 5.
  • Squaring takes a number and multiplies it by itself. Two negatives make a positive, so the sign is destroyed automatically. (-5)^2 = 25, and 5^2 = 25.

Both operations are sign-killers. They both promise that the output is non-negative regardless of what the input was. So if you apply absolute value first and squaring second, the absolute value is doing work that squaring would have done anyway. It is like washing your hands twice — perfectly fine, but the second wash adds nothing the first did not.

Squaring kills the sign — both negative five and positive five map to twenty-fiveA horizontal number line at the top shows the points negative five, zero, and positive five. Two arrows curve down from negative five and from positive five, both pointing to a single point labelled twenty-five on a second horizontal line below. Each arrow is labelled "square it". The diagram makes the point that squaring sends both signs to the same non-negative output. −5 0 5 25 square it square it
Both $-5$ and $+5$ get squared to the same number, $25$. Squaring already collapses the two signs into a single non-negative output. Wrapping $|·|$ around $x$ before squaring just collapses the sign one step earlier — the destination is identical.

This is why |x|^2 and x^2 are the same function. They both take a real number and return its squared magnitude. The absolute value bars in |x|^2 are not wrong — they are just redundant. You can erase them without changing anything.

3. Why this is useful — the squaring trick

Here is where the identity earns its keep. Suppose you are solving an equation like

|f(x)| = |g(x)|

The case-split method (from the parent article) tells you to write two cases: f(x) = g(x) and f(x) = -g(x). That works, but it is bookkeeping-heavy.

The squaring trick is faster. Square both sides:

|f(x)|^2 = |g(x)|^2 \quad \implies \quad [f(x)]^2 = [g(x)]^2

Why: |f|^2 = f^2 and |g|^2 = g^2 by the identity we just proved. The bars vanish.

Now you have a polynomial equation — no absolute values, no case-splitting. Move everything to one side, factor or use the quadratic formula, done. This trick shows up constantly in CBSE Class 11 algebra and is a standard JEE Main / Advanced shortcut for absolute-value problems.

The same trick works for |f(x)| = c when c \ge 0:

|f(x)|^2 = c^2 \quad \implies \quad [f(x)]^2 = c^2

A polynomial equation again.

The catch. If c could be negative, squaring will silently invent solutions that the original equation does not have. Squaring is a non-invertible operation — both 5 and -5 square to 25, so the squared equation cannot tell them apart. These phantoms are called extraneous solutions, and the only defence is to plug every answer back into the original equation and verify.

Example 1: $|x| = 5$ via squaring

Square both sides:

|x|^2 = 5^2 \quad \implies \quad x^2 = 25

Why: |x|^2 = x^2 by the identity, and 5^2 = 25.

Take the square root of both sides, remembering both signs:

x = \pm 5

Check. |5| = 5. Correct. |-5| = 5. Correct.

Result. x = 5 or x = -5 — exactly the same answer the case-split method gives, in fewer steps.

Example 2: $|x - 1| = |2x + 3|$ via squaring

Square both sides:

(x - 1)^2 = (2x + 3)^2

Why: |x - 1|^2 = (x - 1)^2 and |2x + 3|^2 = (2x + 3)^2. Bars gone.

Expand each side:

x^2 - 2x + 1 = 4x^2 + 12x + 9

Move everything to the right:

0 = 3x^2 + 14x + 8

Factor. You want two numbers that multiply to 3 \cdot 8 = 24 and add to 14. Those are 12 and 2:

3x^2 + 12x + 2x + 8 = 3x(x + 4) + 2(x + 4) = (3x + 2)(x + 4) = 0

So x = -\tfrac{2}{3} or x = -4.

Check x = -4. |-4 - 1| = 5 and |2(-4) + 3| = |-5| = 5. Equal. Good. Check x = -\tfrac{2}{3}. |-\tfrac{2}{3} - 1| = |-\tfrac{5}{3}| = \tfrac{5}{3} and |2(-\tfrac{2}{3}) + 3| = |\tfrac{5}{3}| = \tfrac{5}{3}. Equal. Good.

Result. x = -4 or x = -\tfrac{2}{3} — the same answers a careful case-split would give, found through one squaring move and a quadratic.

Example 3: The catch — $|x| = -3$

Square both sides:

|x|^2 = (-3)^2 \quad \implies \quad x^2 = 9 \quad \implies \quad x = \pm 3

So the squared equation says the answers are 3 and -3. But look at the original.

Check x = 3. |3| = 3. Is this equal to -3? No. Check x = -3. |-3| = 3. Is this equal to -3? No.

Neither candidate works. The original equation |x| = -3 has no solution at all — absolute values are never negative, so this equation was doomed from the start. The squaring step quietly turned -3 into 9 and let two phantom solutions sneak in.

Result. No solution. Both \pm 3 are extraneous.

This is why you must check every answer when you square. Squaring is a one-way street — it forgets sign, just like absolute value, and that forgetfulness can manufacture solutions that were never really there.

The squaring trick is one of the most-used identities in JEE algebra: it shows up in absolute-value equations, in solving radical equations like \sqrt{f(x)} = g(x), and in proving inequalities like the triangle inequality. It is short, elegant, and always-correct — provided you remember to verify your answers at the end.

References

  1. NCERT, Mathematics — Class 11, Chapter 1: Sets and Functions and Chapter 2 on relations introduce absolute value formally.
  2. R. D. Sharma, Mathematics for Class 11 — chapter on real numbers and the modulus function.
  3. Wikipedia, Absolute value — formal definition and properties.
  4. M. Spivak, Calculus, 4th ed., Chapter 1 — properties of absolute value derived from order axioms.
  5. Wikipedia, Extraneous and missing solutions — why squaring can introduce phantom solutions.