Most inequalities you meet in a Class 11 textbook have a nice, tidy answer — one interval, one ray, done. Solve 2x - 5 < 7 and you get (-\infty, 6). Solve |x - 3| \le 4 and you get [-1, 7]. One shaded stretch of the number line is enough.

Then you hit a factored polynomial or an absolute-value inequality with cases, and suddenly the solution set is not one stretch — it is two, or three, separated by gaps. The final answer is a union of disjoint intervals, written with the \cup symbol: something like (-\infty, -2] \cup [3, 5).

This builder lets you assemble such a set piece by piece. Drag the endpoints, toggle each bracket between open and closed, watch the union readout rebuild itself in interval notation. The goal is to make the notation feel as concrete as the pictures you are shading.

The builder

A
B
C
Union:
Each row controls one piece of the solution set. Set the left endpoint (or type -Infinity) and the right endpoint (or Infinity), choose whether each bracket is open or closed, and toggle the piece on or off. The canvas shades every active piece in its own colour, and the readout combines them into the union with $\cup$.

Try it. The defaults shade (-\infty, -2] in red and [3, 5) in blue, and the readout shows (-\infty, -2] \cup [3, 5). Now turn on the green row C and set its left endpoint to 7 (open) and its right to \infty. The union becomes three disjoint pieces. Now pull B's right endpoint out to 8 and watch the notation collapse — pieces B and C touch and fuse into one.

Where disjoint solution sets come from

Two common situations force the answer to break into pieces separated by gaps.

Factored polynomial inequalities

A polynomial like (x + 2)(x - 3)(x - 5) changes sign every time it crosses a root. Between the roots, the sign is constant. If the inequality asks for the places where the polynomial is positive, you pick some of those intervals and reject others — and the picked ones are usually not adjacent. The answer is their union.

Absolute-value inequalities of the "greater than" flavour

|x - c| > r says the distance from x to c exceeds r. Geometrically, x is outside the interval of radius r around c — which means x < c - r or x > c + r. Two rays with a gap in the middle. The solution is (-\infty, c - r) \cup (c + r, \infty), a union you cannot compress into a single interval.

Walkthrough: solve (x + 2)(x - 3)(x - 5) > 0

This is the classic example the builder is designed for. You want every x for which the product of three factors is strictly positive.

Step 1. Find the roots. The product is zero when any factor is zero, so the roots are x = -2, x = 3, and x = 5. These three points cut the number line into four open intervals: (-\infty, -2), (-2, 3), (3, 5), (5, \infty).

Step 2. Build a sign chart. In each of the four intervals the product keeps a constant sign, because to change sign the product would have to pass through zero, which only happens at the roots. Pick a test point in each interval and record the sign of each factor.

Interval x + 2 x - 3 x - 5 Product
(-\infty, -2), test x = -3 - - - -
(-2, 3), test x = 0 + - - +
(3, 5), test x = 4 + + - -
(5, \infty), test x = 6 + + + +

The signs alternate as you step across each root — exactly what you expect, because crossing a simple (odd-multiplicity) root flips one factor's sign and therefore flips the product's sign.

Step 3. Select the intervals where the product is positive. From the table, the product is + on (-2, 3) and on (5, \infty). The inequality is strict (> 0), so the roots themselves are excluded — the brackets stay open at every endpoint.

Step 4. Write the union.

\{x : (x+2)(x-3)(x-5) > 0\} \; = \; (-2, 3) \; \cup \; (5, \infty)

This answer cannot be collapsed. The two pieces are separated by a genuine gap — the interval [3, 5] is not in the solution set. If you try typing these two rows into the builder (A = (-2, 3), B = (5, \infty)), you will see exactly this shape: two shaded pieces with a visible hole between them.

If the inequality had been (x+2)(x-3)(x-5) \ge 0, the roots would be included, and the answer would be [-2, 3] \cup [5, \infty) — brackets close, but the structure (two disjoint pieces) stays the same.

Reading the union notation aloud

A compact way to hear what (-\infty, -2] \cup [3, 5) is saying: "every real number that is at most -2, together with every real number from 3 up to but not including 5." The word "together with" is the union. The reader can mentally walk the number line and mark each included stretch without ever trying to fit the two pieces into a single bracket.

The opposite operation, intersection \cap, is usually pronounced "and also" — "every number that is both in piece A and in piece B." The builder above does not compute intersections; it is a union machine by design, because disjoint-piece answers are the ones students most often struggle to write down correctly.

A small caution: union versus simplification

Watch what happens in the builder if you set A to [1, 4] and B to [3, 6]. The readout does not show [1, 4] \cup [3, 6] — it shows [1, 6], because the two pieces overlap and merge. Union is an operation on sets, not on symbols, and once two intervals share any point their union is a single interval.

Disjoint solution sets — the reason you need the \cup symbol at all — only appear when the pieces are genuinely separated. One way to check: if you can list the endpoints in order and the right end of one piece is strictly less than the left end of the next (with at least one of the two brackets open at the meeting point), the pieces are disjoint. Otherwise they collapse.

Where to go next

For the mechanical technique of using sign charts to attack polynomial inequalities — including repeated roots, strict versus non-strict, and the subtle rules for even-multiplicity factors — see Solving Inequalities Using the Location of Roots. That article is the systematic companion to this visual one: it gives you the algorithm, and this builder gives you the mental picture of what the algorithm is producing.

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