Read the answer straight off the parabola: where is the curve above the axis, and where below?
An equation freezes the unknown to a single point. A quadratic equation freezes it to one or two points — the places a parabola crosses the axis. But change that = to a > or a < and the question opens wide: now you want every x where the parabola climbs above the line, or every one where it dips below. The answer is a stretch of the number line again — but this time it can be the middle piece, or the two outside pieces, depending on which way you point the symbol.
Here is the good news, and it is the whole strategy: you already drew this picture in Stage 11. A quadratic inequality is solved by looking at the parabola you can already sketch. By the end of this lesson you will be able to find a quadratic's roots, split the number line at them, decide the sign of the quadratic in each piece, and read off the solution set. You will handle the three cases the discriminant gives you — two roots, one root, or none — and you will tame product and fraction inequalities by tracking signs. We keep our colors honest throughout: green marks the solution set (where your inequality is true), red marks what is not in the solution (and the no-solution case), amber marks the boundary (root) values, and blue is the number line and the structure that holds it all up.
Take the upward-opening parabola y = x² − x − 6. Factor the right side and the crossing points fall right out:
x² − x − 6 = (x − 3)(x + 2), so it crosses at x = −2 and x = 3.
(Check the middle term: −3x + 2x = −x ✓, and the constant −3 · 2 = −6 ✓.) Now picture the curve. It is a valley — it opens upward — with its low point dipping below the axis between those two crossings, and both arms reaching up high outside them. So the value of x² − x − 6 tells you exactly where the curve is:
Read it straight off the hero figure at the top. Outside the roots the curve is up high, so
x² − x − 6 > 0 ⇒ x < −2 or x > 3 (the two outside pieces).
Between the roots the curve dips under, so
x² − x − 6 < 0 ⇒ −2 < x < 3 (the one middle piece).
For an upward parabola: above the axis = outside the roots (the > 0 answer is two rays), and below the axis = between the roots (the < 0 answer is the middle interval). The roots are the boundary; the relation just tells you which stretches to keep.
The most common slip is getting inside and outside backward. Burn this in: < 0 is between (the dip), > 0 is outside (the arms). If you are ever unsure, sketch the valley for two seconds — the picture never lies.
You do not always have a graph in front of you, and you do not need one. The same idea becomes a clean four-step routine called the sign chart:
Run it once on x² − x − 6 < 0. The roots are −2 and 3, which cut the line into three pieces. Test one point in each:
| interval | test x | value of x²−x−6 | sign |
|---|---|---|---|
| x < −2 | x = −3 | 9 + 3 − 6 = 6 | + |
| −2 < x < 3 | x = 0 | 0 − 0 − 6 = −6 | − |
| x > 3 | x = 4 | 16 − 4 − 6 = 6 | + |
Now the watch-point in step 1. If the x² coefficient is negative, the parabola opens downward and "above/below" gets confusing. The fix is the flip rule you met in 12.2 and used in 12.3: multiply the whole inequality by −1 and reverse the symbol. For example,
−x² + x + 6 > 0 ×(−1), flip ⇒ x² − x − 6 < 0 ⇒ −2 < x < 3.
Same answer, but now you are reading a familiar upward valley instead of a confusing hill. Always orient upward first.
Solve x² − x − 6 ≥ 0 (note the ≥). Same roots, same sign chart; we keep the + pieces, the outside. Because the relation now includes equality, the roots themselves count: at x = −2 and x = 3 the value is exactly 0, which satisfies "≥ 0." So the answer is x ≤ −2 or x ≥ 3 — filled dots at both ends.
How many times the parabola meets the axis is decided by the discriminant Δ = b² − 4ac from Stage 11. For an upward parabola there are exactly three pictures, and each gives a clean rule. Once you know which picture you are in, the answer almost writes itself.
The familiar case. The curve crosses twice, so > 0 gives the outside x < x₁ or x > x₂ and < 0 gives the middle x₁ < x < x₂. Our running example x² − x − 6 is exactly this, with roots −2 and 3.
Take (x − 2)² = x² − 4x + 4. Here Δ = (−4)² − 4·1·4 = 16 − 16 = 0. A perfect square is never negative, so the curve rides along the axis, dipping down only to touch it at x = 2 and never going below. That single touch point is where it equals 0; everywhere else it is positive. So:
x² − 4x + 4 > 0 ⇒ all x ≠ 2 · x² − 4x + 4 < 0 ⇒ no solution.
Take x² + x + 1, where Δ = 1² − 4·1·1 = 1 − 4 = −3 < 0. The parabola never reaches the axis, so for an upward curve it stays entirely positive, every single x. So:
x² + x + 1 > 0 ⇒ all real numbers · x² + x + 1 < 0 ⇒ no solution.
| case | picture | > 0 solves to | < 0 solves to |
|---|---|---|---|
| Δ > 0 | crosses twice | outside roots | between roots |
| Δ = 0 | touches once at r | all x ≠ r | no solution |
| Δ < 0 | floats above | all reals | no solution |
"No solution" and "all reals" are real, correct answers — do not panic when the roots vanish. If Δ < 0 there is simply nothing between non-existent roots (so < 0 is empty) and the whole line is above the axis (so > 0 is everything). For the touching case Δ = 0, the only subtlety is whether the lone point counts: it is excluded for strict > 0 but included for ≥ 0.
Once a quadratic is factored, you can skip the graph entirely and just track signs. A product is positive when its factors agree in sign (both + or both −) and negative when they disagree. Take
(x − 1)(x + 2) > 0.
The factors change sign at x = 1 and x = −2. Make a tiny table of each factor's sign in the three pieces, then multiply down each column:
| x < −2 | −2 < x < 1 | x > 1 | |
|---|---|---|---|
| x − 1 | − | − | + |
| x + 2 | − | + | + |
| product | + | − | + |
So (x − 1)(x + 2) > 0 ⇒ x < −2 or x > 1 — same outside-the-roots pattern as before, reached by pure sign-counting.
A fraction inequality works the very same way, because a quotient follows the exact same sign rule as a product: same signs give positive, opposite signs give negative. There is just one extra thing to remember from the rational-expression work of Stage 9 — the denominator can never be zero. Take
x − 1x + 2 > 0.
The sign table is identical to the product above (a quotient flips sign at the same two places). So the answer is the same shape: x < −2 or x > 1. But at x = −2 the denominator is 0, so the fraction is undefined there — that value must be thrown out. We mark it with a red open hole:
x − 1x + 2 > 0 ⇒ x < −2 or x > 1, with x ≠ −2.
Never clear a fraction by multiplying both sides by x + 2 the way you would in an equation — its sign is unknown, so you would not know whether to flip the inequality (callback to 12.2). Track signs instead, and always exclude any value that makes a denominator 0. Here that excluded value, −2, happens to already be outside the solution shape, but you must still flag it: it can never be part of the answer.
A quadratic inequality asks which stretch of the line makes the parabola the sign you want. Orient the parabola upward first (multiply by −1 and flip if the x² coefficient is negative). Find the roots; they slice the line. Then for an upward curve, > 0 lives outside the roots and < 0 lives between them — or, when there are no two roots, the discriminant hands you "all reals," "all x ≠ r," or "no solution." Factored products and fractions need no graph at all: just multiply the column of signs, and in a fraction punch out any value that makes the denominator 0.
You have now solved every standard one-variable inequality. Next, 12.6 turns inequalities into a tool for finding the largest and smallest values — the basic (AM–GM) inequality — and 12.7 sends the whole stage out into real-world problems. You can also revisit where parabolas began in 12.1–12.4 if any earlier step feels shaky.
Six questions to lock it in. Tap the answer you think is right.
This lesson develops A-REI.B.4 (solving quadratic equations to obtain the boundary values), A-APR.B.3 and F-IF.C.7a (using zeros to sketch and read a quadratic graph), and especially A-REI.D.11 (relating an inequality's solution to where a graph lies above or below the x-axis). It builds directly on the parabola work of Stage 11 and reuses the sign-flip rule from 12.2. The #1 misconception is getting inside and outside backward — students often pair "< 0" with the outside and "> 0" with the middle — together with forgetting to orient the parabola upward (and flip) when the x² coefficient is negative, and wrongly including the excluded value in a fraction inequality. The antidote is mechanical and reliable: always sketch the upward valley and read it directly — below the axis is between the roots, above is outside — and in any fraction, punch out every value that zeroes a denominator before stating the answer.