In short

Squaring both sides of |f(x)| = c to get f(x)^2 = c^2 is a perfectly valid algebraic move — but it is one-way. Squaring kills the sign of both sides, so any equation where the original right-hand side is negative (think |f(x)| = -5, which has no solution) will still produce candidate roots after squaring. Those candidates are extraneous — they satisfy the squared equation but not the original. The same trap appears whenever the right-hand side is a variable expression that could turn out negative. Rule: square if you must, but always plug every candidate back into the original absolute-value equation. Why: a^2 = b^2 is true whenever a = \pm b, so squaring throws away the sign distinction that made the original equation specific.

You have just learned the case-split method for absolute-value equations, and somewhere a friend or YouTube tutorial whispers a faster trick: "just square both sides — the bars vanish." It feels efficient. |x - 3| = 5 becomes (x - 3)^2 = 25, no cases, no fuss, expand and solve.

The trick really does work — most of the time. But it has a hidden cost. Squaring is a non-invertible operation: 5^2 and (-5)^2 both equal 25, so once you have squared, you cannot tell which sign you started with. Any equation that depended on the sign — and absolute-value equations are exactly those that depend on signs — can sneak in fake answers that the squared form happily endorses.

This article is a quick tour through three scenarios: when squaring is harmless, when it spawns ghosts, and how to spot the difference before you write your final answer in a CBSE Class 11 paper.

The technique — when it works cleanly

Take the textbook equation |x - 3| = 5.

Square both sides:

(x - 3)^2 = 25

Why this is legal: if two non-negative numbers are equal, their squares are equal. Both sides of |x - 3| = 5 are non-negative (|x-3| \ge 0 and 5 > 0), so the implication runs in both directions here.

Expand and solve:

x^2 - 6x + 9 = 25 \;\;\Longrightarrow\;\; x^2 - 6x - 16 = 0

Or, equivalently, (x - 3)^2 = 25 gives x - 3 = \pm 5, so x = 8 or x = -2.

Verify in the original:

  • x = 8: |8 - 3| = |5| = 5. Matches.
  • x = -2: |-2 - 3| = |-5| = 5. Matches.

Both candidates survive. No extraneous roots. Same answer as the case-split method gives. So far, so calm.

When extraneous roots appear

Now change one symbol. Look at |x - 3| = -5.

You already know from the no-solution rule that this equation has no solution — the left side is non-negative, the right side is negative, end of story.

But watch what happens if you blindly square:

(x - 3)^2 = (-5)^2 = 25

The square wiped out the minus sign on the right. The equation has become identical to the one in the previous section, with the same candidate roots x = 8 and x = -2.

Verify in the original |x - 3| = -5:

  • x = 8: |8 - 3| = 5 \ne -5. Fails.
  • x = -2: |-2 - 3| = 5 \ne -5. Fails.

Both candidates are extraneous. The algebra says "here are two roots," the original equation says "neither of them works." If you skipped the verification step you would have confidently written down \{8, -2\} on your answer sheet — and lost the marks completely. Why squaring lied: the squared equation (x-3)^2 = 25 is satisfied by any x for which |x-3| = 5 or |x-3| = -5. The first equation has solutions; the second does not. Squaring merged both equations into one, and the solutions of the first slipped into the answer set for the second.

Squaring trap flowchart — original equation, square, candidate roots, verify, final solution setA flowchart with five boxes connected by arrows. Box 1 on the left says "Original |f(x)| = c". An arrow labelled "square" points to Box 2 "f(x) squared = c squared". An arrow labelled "solve" points to Box 3 "Candidate roots". An arrow labelled "VERIFY in original" points down to Box 4 which is split into two halves: a green half labelled "passes — keep" and a red half labelled "fails — discard". A final arrow points to Box 5 "Solution set = verified candidates only". A red warning box at the bottom reads "Skipping verify is how extraneous roots get into your answer". Original |f(x)| = c square f(x)² = c² sign info lost solve Candidate roots x₁, x₂, … VERIFY in original passes keep ✓ fails discard ✗ Solution set verified candidates only Skipping the verify step is exactly how extraneous roots get into your final answer — and how marks vanish in CBSE Class 11 papers
The squaring trap, drawn as a one-way street. Squaring is fine on the way down, but signs are erased — so every candidate must walk back up through the verification gate before being admitted to the final solution set.

The procedure — non-negotiable

Once you accept that squaring can lie, the fix is mechanical:

  1. Square both sides to remove the bars.
  2. Solve the resulting polynomial equation for all candidate roots.
  3. Verify every candidate in the original absolute-value equation.
  4. The solution set is exactly the verified candidates. Discard the rest as extraneous.

That third step is not a "good practice" or a "double-check." It is the price of admission. Skip it once and you will lose marks — or worse, build a wrong answer into a longer derivation and not notice until the final answer is nonsense.

Three worked examples

Example 1: $|x - 3| = 5$ — both candidates survive

Step 1. Square both sides.

(x - 3)^2 = 25

Why allowed: both sides of the original were non-negative, so squaring is a true biconditional here.

Step 2. Solve.

x - 3 = \pm 5 \;\;\Longrightarrow\;\; x = 8 \text{ or } x = -2

Step 3. Verify in the original |x - 3| = 5.

  • x = 8: |8 - 3| = 5. ✓
  • x = -2: |-2 - 3| = |-5| = 5. ✓

Result. x = 8 or x = -2. No extraneous roots. Squaring was safe because the right-hand side was a positive constant.

Example 2: $|x - 3| = x - 5$ — the squaring trap in action

The right-hand side is now a variable expression, x - 5, which can be positive or negative depending on x. This is exactly where squaring most often lies.

Step 1. Square both sides.

(x - 3)^2 = (x - 5)^2

Step 2. Expand and solve.

x^2 - 6x + 9 = x^2 - 10x + 25

The x^2 terms cancel, leaving -6x + 9 = -10x + 25, so 4x = 16 and x = 4.

Step 3. Verify in the original |x - 3| = x - 5.

  • LHS at x = 4: |4 - 3| = 1.
  • RHS at x = 4: 4 - 5 = -1.
  • 1 \ne -1. ✗

Result. No solution. The single candidate x = 4 is extraneous. Why it failed: squaring matched (x-3)^2 = (x-5)^2, which holds whenever x - 3 = x - 5 (impossible) or x - 3 = -(x - 5) (which gave x = 4). The second case corresponds to the original equation being |x - 3| = -(x - 5), not |x - 3| = x - 5. The squaring secretly solved the wrong equation.

You can also see this directly: |x - 3| is non-negative, but x - 5 is negative for x < 5. So at x = 4 the right-hand side is negative — the equation never had a chance there to begin with.

Example 3: $|x| = -3$ — every candidate is a ghost

A short one. The original equation is |x| = -3, which has no solution by inspection (left side \ge 0, right side < 0). Watch squaring pretend otherwise.

Step 1. Square: x^2 = 9.

Step 2. Solve: x = \pm 3.

Step 3. Verify in the original |x| = -3.

  • x = 3: |3| = 3 \ne -3. ✗
  • x = -3: |-3| = 3 \ne -3. ✗

Result. No solution. Both candidates are extraneous. The squaring step turned the impossible equation |x| = -3 into the perfectly solvable x^2 = 9 — and the verification step caught the lie.

When NOT to square

Squaring is a sledgehammer. Sometimes you do not need one.

  • If the right-hand side is a non-negative constant, just unfold into two cases the way you learned in the parent article on absolute-value equations. Writing |2x + 3| = 7 as "2x + 3 = 7 or 2x + 3 = -7" is faster than squaring, expanding, and re-factoring — and it cannot create extraneous roots.

  • If the right-hand side is a negative constant, do not even unfold. Write no solution immediately. (See the misconception killer for that case.)

  • If absolute values appear on both sides — like |3x - 1| = |x + 5| — squaring is genuinely cleaner than case-splitting, because |a|^2 = a^2 for all real a. The squared equation (3x - 1)^2 = (x + 5)^2 is equivalent to the original; no extraneous roots can appear. Why safe here: both sides of the original equation are guaranteed non-negative, so the squaring biconditional holds. Sign information is lost on each side individually, but no false signs sneak in.

  • If the right-hand side is a variable expression — like |x - 3| = x - 5 or |2x + 1| = 3x — squaring is allowed but dangerous. Either restrict the domain in advance (require the RHS \ge 0 before squaring) or accept that you must verify every candidate ruthlessly. The case-split method is often safer here.

A short heuristic: square only when both sides are obviously non-negative. Otherwise, split into cases — or at least pre-screen the right-hand side.

Why this is a CBSE Class 11 trap

In Class 11, absolute-value equations show up as a warm-up before quadratic equations and inequalities, and examiners love a question whose squared form has clean roots that don't satisfy the original. Two of the most common patterns:

  1. |x - a| = bx + c with a variable RHS — designed so that one of the two squared roots makes the RHS negative.
  2. \sqrt{(x - a)^2} = c disguised as a square-root identity — really an absolute-value equation in costume, with the same extraneous-root risk.

If you see either pattern in a board paper or a JEE Main question, the verification step is where the marks live. Skipping it is like driving without a seatbelt: usually fine, occasionally catastrophic.

In one line

Squaring is a fast way to remove absolute-value bars. It is also a fast way to invent answers that aren't there. Always finish with a plug-back into the original — that is the only step that distinguishes a true root from an extraneous one.

References