In short

|x| = 0 has one solution: x = 0. Only zero itself sits at distance zero from zero. |x| = 5 has two solutions: x = 5 and x = -5, because two different points (one on each side of the origin) are exactly five units away. Zero is the degenerate case — the two solutions you would normally get from \pm k collapse onto the same point because +0 and -0 are the same number.

The first time you solved |x| = 5 in class, the rule felt almost mechanical: "drop the bars, write x = 5 or x = -5, done." Then someone wrote |x| = 0 on the board, you applied the same rule, got x = 0 or x = -0 — and realised both answers are the same number. So is it one solution or two?

It is genuinely one. And understanding why it is one — not as a memorised exception, but as a natural consequence of what absolute value means — is the kind of small clarity that the CBSE Class 11 syllabus quietly assumes you have. Without it, absolute value inequalities, piecewise functions, and even later questions about the roots of |f(x)| = 0 feel arbitrary.

The trichotomy: three cases, never more

Every equation of the form |x| = k falls into exactly one of three buckets, depending on the sign of k. There are no other possibilities, because k is a real number and a real number is either positive, zero, or negative.

The trichotomy of $|x| = k$

For the equation |x| = k where k is a real constant:

  • If k > 0: two solutions, x = k and x = -k.
  • If k = 0: one solution, x = 0.
  • If k < 0: no solution, because |x| is never negative.

The first case is the one you meet most often. If k = 5, you are asking "which numbers are 5 units from zero?" — and there are two such numbers, sitting symmetrically on either side of the origin.

The third case (k < 0) is the one most students remember as a trick. |x| = -3 has no solution because the bars guarantee a non-negative output, and a non-negative number can never equal -3. You do not even need algebra to see it.

The middle case is the subtle one. Why: when k = 0 the equation does not stop being a "two-case" equation — the two cases just happen to give the same answer. The structure \pm k still applies; it is only that +0 and -0 are not different numbers.

Why k = 0 is special: the two roads meet

The standard solving move for |x| = k is to unfold the bars into two equations:

x = k \quad \text{or} \quad x = -k

Try k = 5:

x = 5 \quad \text{or} \quad x = -5

Two different numbers. Two solutions. Now try k = 0:

x = 0 \quad \text{or} \quad x = -0

But -0 is just 0. So both branches give the same equation, and the same single answer. Why: the additive identity has no opposite distinct from itself. Negation reflects every other real number across the origin to a different point, but it sends zero to zero. Geometrically, zero is the only fixed point of the map x \mapsto -x.

This is what a mathematician means by degeneracy: a structure that normally produces two distinct things produces only one because the two things have collided. It is the same flavour of degeneracy that makes a quadratic have a "repeated root" when its discriminant is zero, or a circle shrink to a single point when its radius hits zero.

So the equation |x| = 0 does not have a different rule from |x| = 5. It has the same rule, applied to a value of k where the rule's two outputs happen to coincide.

The geometric view: a V meeting a horizontal line

The cleanest way to see all three cases at once is to draw the graph y = |x| — a perfect V with its vertex at the origin — and ask: where does the horizontal line y = k cut it?

The V graph y equals absolute value of x meeting horizontal linesA coordinate plane with the V-shaped graph of y equals absolute value of x. A horizontal line at y equals zero touches the V at exactly one point, the vertex at the origin. A second horizontal line at y equals five crosses the V at two points, x equals minus five and x equals five. x y 0 −5 5 5 y = 5 y = 0 one touch two crossings
The line $y = 0$ kisses the V at exactly one point — its vertex. The line $y = 5$ slices clean through both arms, giving two crossings. Move the line up and you always get two; bring it down to the vertex and the two collapse to one; push it below zero and you get none.

Three pictures, three cases, no rule to memorise. The V's vertex sits on the x-axis, so the horizontal line y = 0 touches it at a single point. Any positive line y = k slides up and intersects both arms of the V — one on the left, one on the right. Any negative y = k misses the V entirely, because the V never dips below the axis.

The vertex is precisely the place where the V's two arms meet. That is the geometric picture of degeneracy: Why: the two solutions of |x| = k live one on each arm of the V. As k \to 0, those two intersection points slide toward each other along the arms and fuse at the vertex. Zero is the unique value of k at which the fusion happens.

Three worked examples

$|x| = 0$ — exactly one solution

Solve |x| = 0.

Unfold into the two cases:

x = 0 \quad \text{or} \quad x = -0

But -0 = 0, so both branches give the same equation. The unique solution is \boxed{x = 0}.

Check. |0| = 0. ✓

$|x| = 5$ — two solutions

Solve |x| = 5.

Unfold:

x = 5 \quad \text{or} \quad x = -5

Both are valid because 5 \ne -5. The solution set is \boxed{\{-5, \, 5\}}.

Check. |5| = 5 ✓ and |-5| = 5 ✓.

Generalising to a shifted V: $|x - 3| = 0$ vs $|x - 3| = 5$

The same logic works when the expression inside the bars is more interesting than a bare x. The graph of y = |x - 3| is the same V, just shifted three units to the right so that its vertex sits at x = 3.

First, |x - 3| = 0. The expression inside the bars must equal zero:

x - 3 = 0 \implies x = 3

The horizontal line y = 0 kisses the shifted V at its vertex. One solution: x = 3.

Now, |x - 3| = 5. Unfold:

x - 3 = 5 \quad \text{or} \quad x - 3 = -5
x = 8 \quad \text{or} \quad x = -2

The line y = 5 slices both arms of the shifted V. Two solutions: x = -2 and x = 8.

Notice the symmetry: the two answers -2 and 8 are equidistant from the vertex x = 3, sitting five units to either side. That is the geometry of absolute value, untouched by the shift.

Why this matters past Class 11

Recognising the degenerate case looks like a small thing — one solution instead of two, big deal. But the same pattern reappears constantly in Class 11 and beyond:

  • A quadratic equation ax^2 + bx + c = 0 has two roots when its discriminant is positive, one repeated root when the discriminant is zero, no real roots when it is negative. Same trichotomy.
  • A circle of radius r has infinitely many points; a "circle" of radius zero is a single point. Same collapse.
  • In calculus, f(x) = 0 usually has isolated solutions, but at a double root two solutions have fused into one — exactly why the derivative also vanishes there.

Each of these is the same idea wearing a different costume: a parameter (the right side of an equation, the discriminant, the radius) crosses a critical value, and a structure that produced two outputs starts producing one. Once you see it in |x| = 0, you will see it everywhere.

So the next time someone asks you why |x| = 0 has one solution, do not say "because the rule says so." Say: because zero is its own opposite, and the V touches the axis at a single point.

References

  1. NCERT Class 11 Mathematics — Linear Inequalities — official CBSE textbook chapter that assumes the absolute-value-as-distance idea.
  2. Wolfram MathWorld — Absolute Value — formal definition and basic properties.
  3. Khan Academy — Solving absolute value equations — worked examples at the introductory level.
  4. Paul's Online Math Notes — Absolute Value Equations — clear breakdown of the three cases.