The whole lesson in one picture. The parabola is above the axis outside the roots and below it between them. A quadratic inequality just asks which of those stretches you want.
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.
12.5.1 Read the sign from the parabola
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:
Where the curve sits above the axis, the value is positive — that is the > 0 region.
Where the curve dips below the axis, the value is negative — that is the < 0 region.
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).
Key idea
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.
Watch out
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.
🎮 Try it PARABOLA SIGN READER
Slide the point along y = x² − x − 6. Watch whether it is above or below the axis — that is the sign. Then choose a relation and the matching stretch lights up green on the line below.
Point at x =1.0
Shade where
12.5.2 Roots first, then the intervals
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:
Orient upward. Make sure the x² coefficient is positive. (More on that in a moment.)
Find the roots. Solve the matching equation — by factoring, or with the quadratic formula from Stage 11. These are the boundary values.
Split and test. The roots slice the number line into pieces. Pick one easy test number in each piece and check the sign of the quadratic there.
Keep the matching pieces. Choose the pieces whose sign matches your relation, and read off the interval.
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
+
The signs go +, −, + across the three pieces — exactly the shape of an upward valley. For < 0 we keep the middle piece, giving −2 < x < 3.
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.
Worked example
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.
🎮 Try it SIGN-CHART BUILDER
Pick a quadratic and a relation. The widget finds the roots, splits the line, prints the +/− sign in each piece, and shades the solution green.
Quadratic
Relation
12.5.3 The three discriminant cases
How many times the parabola meets the axis is decided by the discriminantΔ = 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 three upward pictures: cross twice, just kiss the axis once, or float clear above it.
Δ > 0 — two roots x₁ < x₂
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.
Δ = 0 — one double root, the curve just touches
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.
Δ < 0 — the curve floats above, no roots at all
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.
A quick map (upward parabola)
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
Watch out
"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.
🎮 Try it DISCRIMINANT EXPLORER
Tap a case to load one of the three pinned quadratics. See where the parabola sits relative to the axis, and the solution sets for both > 0 and < 0.
Case
12.5.4 Products and fractions by sign-tracking
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
+
−
+
Two negatives multiply to a positive; a positive and a negative give a negative. For > 0 we keep the + columns: x < −2 or x > 1.
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.
Watch out
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.
🎮 Try it PRODUCT / FRACTION SIGN-TRACKER
Toggle between the product and the fraction. The factor signs appear in each interval, then the combined sign, then the green solution. In the fraction case, x = −2 is punched out as a red open hole.
Form
★ The big ideas, in one breath
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.
What's next
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.
✎ Exercises 12.5
Factor and solve x² − x − 6 > 0.
Show answer
x² − x − 6 = (x − 3)(x + 2), roots −2 and 3. Since > 0 wants the outside of an upward parabola: x < −2 or x > 3.
Solve x² − x − 6 < 0.
Show answer
Same roots −2 and 3. Now < 0 wants the middle (the dip below the axis): −2 < x < 3.
Solve x² − x − 6 ≥ 0. How does the ≥ change the endpoints?
Show answer
Outside the roots again, but because ≥ includes equality, the roots themselves count (at −2 and 3 the value is exactly 0). Answer: x ≤ −2 or x ≥ 3 — filled dots at both ends.
Solve (x − 2)² > 0.
Show answer
This is x² − 4x + 4 with Δ = 16 − 16 = 0 — a perfect square, so it is ≥ 0 always and equals 0 only at x = 2. Strictly greater than 0 everywhere except that touch point: all x ≠ 2.
Solve x² − 4x + 4 < 0.
Show answer
A perfect square is never negative, so the curve never dips below the axis. No solution.
Solve x² + x + 1 > 0 and x² + x + 1 < 0.
Show answer
Δ = 1 − 4 = −3 < 0, so the upward parabola floats entirely above the axis (always positive). Therefore > 0 gives all real numbers, and < 0 gives no solution.
Multiply by −1 and flip: x² − x − 6 < 0. Roots −2 and 3, and < 0 is the middle. Answer: −2 < x < 3.
Solve (x − 1)(x + 2) > 0 by tracking signs.
Show answer
Sign changes at x = 1 and x = −2. Signs of the product go +, −, + across the pieces. Keep the + pieces: x < −2 or x > 1.
Solve (x − 1)/(x + 2) > 0. What is different from Exercise 8?
Show answer
Same sign pattern, so the shape matches: x < −2 or x > 1. The difference: the denominator forbids x = −2, so we add the exclusion x ≠ −2 (an open hole). Here it lies outside the solution anyway, but it must still be flagged as never allowed.
A ball's height (in metres) after t seconds is h = −5t² + 10t. For which times is the ball above the ground, h > 0? (Use only the physically meaningful times t ≥ 0.)
Show answer
h = −5t² + 10t = −5t(t − 2), roots t = 0 and t = 2. Orient upward: divide by −5 and flip to get t(t − 2) < 0, the middle interval 0 < t < 2. So the ball is above the ground from launch until it lands at t = 2 seconds.
🎯 Quick check
Six questions to lock it in. Tap the answer you think is right.
§ For teachers and parents
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.