Both Sides Could Be Negative? Squaring Is Illegal — and Square-Rooting Needs an Absolute Value

Students square inequalities the way they square equations — as a reflex. For equations the reflex is (mostly) safe; for inequalities, it is safe only when both sides are non-negative. Miss that check, and the inequality you "simplified" now says something different from the one you started with. The recognition cue, the one you should drill until it fires before you pick up the pen, is simple: look at the sign of each side. If either side could be negative, put the squaring hammer down.

This satellite is about that cue — and its partner, the fact that square-rooting an inequality always pulls an absolute value out of the x^2. Squaring and square-rooting are inverses only on the non-negative reals. The rest of algebra pays the price of forgetting that.

The cue, stated cleanly

Before you square both sides of an inequality, run this mental check:

"Can the left side be negative? Can the right side be negative? If I cannot answer no to both, squaring is illegal."

That is it. One sentence, two sign checks. If one of the two questions is "maybe" or "yes," you do not square. You either split into cases based on the sign, or you rewrite the inequality using an absolute value and solve from there.

The concrete example that burns it in

Start with an inequality that is obviously true:

-3 < 2.

Now square both sides without thinking:

(-3)^2 < 2^2 \quad\Longrightarrow\quad 9 < 4.

That is false. The inequality flipped. You did not make an arithmetic slip — you did exactly what you would do to an equation. One side (-3) is negative, and squaring a negative number throws away the minus sign while keeping the magnitude. The arrow had pointed from -3 to 2 (smaller to larger); after squaring, it points the wrong way because |-3| = 3 > |2| = 2.

The rule the example exposes:

$a < b$ implies $a^2 < b^2$ **only when $0 \le a \le b$**. If either side is negative, squaring can preserve the arrow, flip it, or destroy it — you do not know which without more work.

Why the cue fires — the three possibilities

When you square a < b, three things can happen depending on the signs:

Both non-negative (0 \le a < b): squaring preserves the arrow. 2 < 5 gives 4 < 25. Safe.
Both non-positive (a < b \le 0): squaring flips the arrow. -5 < -2 gives 25 > 4. Arrow reversed — because the number further left on the line has larger magnitude, which survives squaring while the sign does not.
Mixed signs (a < 0 \le b): the result depends on magnitudes, not position on the line. -3 < 2 gives 9 > 4 (flipped), but -1 < 2 gives 1 < 4 (preserved). Squaring is neither safe nor reliably unsafe — it is unpredictable.

Equations do not have this problem because = is symmetric — no "arrow" to preserve or flip. Squaring an equation can invent new solutions (extraneous roots you check for), but it cannot silently swap the meaning. Inequalities are more fragile.

The companion fact: \sqrt{x^2} = |x|, not x

The other half of the recognition reflex is what happens when you undo a squaring. If you have x^2 < 9 and want to take square roots, the correct move is not "x < 3." The correct move is:

x^2 < 9 \iff \sqrt{x^2} < \sqrt{9} \iff |x| < 3.

And |x| < 3 is the interval (-3, 3), not the ray x < 3. The absolute value is forced by the identity

\sqrt{x^2} = |x|

which holds for every real x. The principal square-root symbol \sqrt{\cdot} always returns a non-negative result, so \sqrt{x^2} cannot equal x when x is negative — it has to equal |x| instead.

Once the absolute value is there, the standard rewrite from Intervals and Inequalities Preview finishes the problem: |x| < 3 \iff -3 < x < 3, so the solution set is (-3, 3). If you had written "x < 3" instead, you would have claimed that x = -100 satisfies x^2 < 9. It does not — (-100)^2 = 10000, enormously greater than 9.

A worked problem from end to end

Problem. Solve (x - 2)^2 < 16.

The trap answer is to "undo the square": x - 2 < 4, giving x < 6. That answer claims x = -100 works. Check: (-100 - 2)^2 = 10404, far greater than 16. The trap answer is wrong.

Correct solution. Take the square root, keeping the absolute value:

\sqrt{(x - 2)^2} < \sqrt{16} \iff |x - 2| < 4.

Now apply the absolute-value rewrite:

|x - 2| < 4 \iff -4 < x - 2 < 4 \iff -2 < x < 6.

Solution set: (-2, 6). Spot-check at the new left endpoint: x = -2 gives (-4)^2 = 16, which is not strictly less than 16 — correctly excluded by the open interval. x = 0 gives (-2)^2 = 4 < 16. Good. x = -100 gives 10404 \not< 16. Correctly excluded by the interval.

The trap answer — "x < 6" — is what you get when you skip the absolute value and pretend \sqrt{x^2} = x. The absolute value is not a pedantic flourish; it is the only thing that saves you from losing the entire left boundary.

How the case-split alternative works

Sometimes an absolute-value rewrite is awkward, and a case-split on the sign is cleaner. The cue is identical — "can this side be negative?" — but the resolution is different.

Example. Solve x - 1 < \sqrt{x + 5}.

You may be tempted to square both sides. Ask the cue: is \sqrt{x + 5} ever negative? No — principal square roots are non-negative. Is x - 1 ever negative? Yes, whenever x < 1. So mixed signs are possible. You cannot square straight away.

Case 1: x - 1 < 0 (i.e., x < 1). The left side is negative, the right side is non-negative (and defined only where x + 5 \ge 0, i.e., x \ge -5). Any negative is less than any non-negative, so the inequality holds automatically. Intersect with -5 \le x < 1: the case contributes [-5, 1).

Case 2: x - 1 \ge 0 (i.e., x \ge 1). Now both sides are non-negative, so squaring is legal and preserves the arrow:

(x - 1)^2 < x + 5 \iff x^2 - 3x - 4 < 0 \iff (x - 4)(x + 1) < 0,

which holds for -1 < x < 4. Intersect with x \ge 1: the case contributes [1, 4).

Combine. [-5, 1) \cup [1, 4) = [-5, 4). The boundary x = -5 is included because the domain allows it and the original inequality -6 < 0 holds. The squaring was only legal in Case 2, and only because the case assumption pinned down the signs.

When the cue fires

Any of these patterns should trigger the recognition:

Squaring appears in your first instinct, and at least one side involves a subtraction (e.g., x - 5) that might be negative.
A square root is embedded in one side and a polynomial on the other.
You see x^2 or (\text{expression})^2 and reach for \sqrt{\cdot}.
You are comparing two quantities without knowing their relative magnitudes or signs.

In every one of these, stop, run the two-question check, and pick: squaring (only if both sides are provably non-negative), case-splitting (if a side's sign varies), or rewriting with an absolute value (if you are undoing a square).

The compressed version

Before squaring, check both signs. If either side can be negative, do not square — split into cases or rewrite with an absolute value. When square-rooting, \sqrt{x^2} is |x|, never x, so an inequality like x^2 < 9 becomes |x| < 3 and then (-3, 3) — not x < 3. The squaring hammer is not banned; it is conditional, and the condition is "no negatives in sight."