Quadratic Inequalities

When you solve a quadratic equation like x^2 - x - 6 = 0, you get two numbers: x = -2 and x = 3. Now change that = into a >. Ask instead

x^2 - x - 6 > 0

and the question quietly transforms. We're no longer asking "which two values hit zero?" — we're asking "for which whole range of x is this expression above zero?" Just like a linear inequality, the answer is not a point but a set of intervals on the number line.

This matters everywhere something can be positive or negative. A ball thrown upward has height h(t) = -5t^2 + 20t; asking "when is it in the air?" means solving h(t) > 0. A shop's profit is revenue minus cost — often a quadratic in the number of items — and "when do we make money?" is P(x) > 0. In every case the trick is the same: find where the quadratic crosses zero, then read off which side (or middle) you want.

The method: roots first, then pick the intervals

A quadratic inequality is any of ax^2 + bx + c > 0 (or <, \le, \ge 0). Solve it in three moves:

  1. Get zero on one side — rearrange so everything is compared to 0 (exactly as you'd tidy a quadratic before factorising).
  2. Find the roots — factorise (or use the formula) to find the critical values: the x where the parabola crosses the axis. These are the break-points that chop the number line into pieces.
  3. Pick the intervals by the parabola's shape. A parabola with a > 0 opens up — a smile — so it is positive OUTSIDE the roots and negative BETWEEN them. (If a < 0 it opens down and those flip.) Match that to the sign you want, > or <.

The whole thing leans on being able to sketch the parabola — even a rough smile with two crossing points tells you the answer at a glance.

Worked example 1: x^2 - x - 6 > 0

Zero is already on the right. Factorise the left side:

x^2 - x - 6 = (x - 3)(x + 2)

so the roots (critical values) are x = 3 and x = -2. The leading coefficient is a = 1 > 0, so the parabola opens up. We want where it is above zero — and an up-parabola is positive outside its roots. Reveal the sketch step by step:

So the solution is the two outside pieces:

x < -2 \quad\text{or}\quad x > 3

In interval notation that's (-\infty,\,-2)\,\cup\,(3,\,\infty) — two open intervals joined by a union. The circles at -2 and 3 are open (hollow) because the inequality is strict (>, not \ge), so the roots themselves — where the value is exactly 0 — are not included.

The same answer from a sign table

Don't like sketching? A sign table gives the identical answer. The roots -2 and 3 split the line into three zones. In each zone, check the sign of each factor of (x+2)(x-3), then multiply the signs:

Zone x < -2 -2 < x < 3 x > 3
(x+2) − + +
(x-3) − − +
product + − +

We wanted the product > 0 (positive), so we take the + zones: x < -2 or x > 3. Exactly what the parabola told us — a −, +, −, + pattern is the signature of an up-opening parabola: positive on the outside, negative in the middle.

Worked example 2: x^2 + 2x - 3 \le 0

Zero is on the right already. Factorise:

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

Roots are x = -3 and x = 1, and again a = 1 > 0 (opens up). This time we want where the curve is at or below zero — \le 0. An up-parabola is negative between its roots, so:

-3 \le x \le 1

In interval notation, [-3,\,1] — a single closed interval. The brackets are square (and the endpoints included) because the inequality is \le: at x = -3 and x = 1 the value is exactly 0, and 0 \le 0 is true, so the roots do belong to the solution. Notice the shape of the answer flipped completely from Example 1: a strict-greater gave two outside rays; this less-than-or-equal gives one inside interval.

Right here. It's easy to carry a habit from equations — "solve" means "find the number(s)" — but an inequality asks a different question. x^2 - x - 6 = 0 has the two-number answer \{-2, 3\}; x^2 - x - 6 > 0 has the range answer "everything to the left of -2 and everything to the right of 3" — infinitely many values.

The two numbers don't vanish, though — they become the fence-posts. Every quadratic inequality is really the equation-answer (the roots) plus a decision about which side of the fence you stand on. Find the posts by solving = 0; choose the pastures by looking at the parabola. That two-step rhythm — solve the equation, then read the sign — is the heart of every inequality you'll ever meet, right up to polynomial and rational inequalities.

A quadratic has two break-points, so the answer is almost always two intervals (or the one interval between them) — never just a lone "x > \text{root}". Writing x > 3 alone for x^2 - x - 6 > 0 throws away the entire left-hand ray x < -2. Two posts, two decisions.

And the killer: if the parabola opens DOWN (a < 0), the regions flip. For a down-parabola, "greater than zero" is the piece between the roots and "less than zero" is the two outside rays — the exact opposite of the up-case. For example, -x^2 + x + 6 > 0 factorises to -(x-3)(x+2) > 0, and its solution is -2 < x < 3 (between!), not the outside.

So never guess from the sign alone — sketch the parabola (or run a sign table). A ten-second smile-or-frown with two crossing points removes all doubt about which intervals to keep.

To solve ax^2 + bx + c \gtrless 0: