Solving a quadratic inequality means finding the set of x-values where the quadratic is positive (or negative). The trick is to find the roots first, then read off the right region from a sketch. Get the regions right and these become almost automatic.
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What you need to know
Multiplying or dividing by a negative number flips the inequality sign
Always rearrange so the x2 coefficient is positive and one side is zero
The 4-step method: rearrange, find roots, sketch, identify region
For > 0 (above x-axis): the solution is the two intervals OUTSIDE the roots
For < 0 (below x-axis): the solution is the interval BETWEEN the roots
For (x − h)2 < n: never write ±√n — use absolute value instead
Quick Recap: When Does the Sign Flip?
Inequalities behave a lot like equations — but with one critical difference:
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Sign stays the same
Add or subtract any term: x > 9 → x + 3 > 12
Multiply or divide by a positive: x > 9 → 3x > 27
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Sign flips!
Multiply or divide by a negative:
x > 9 → −3x < −27
The > becomes < (and vice versa)
The 4-Step Method
For any quadratic inequality:
Rearrange into the form ax2 + bx + c > 0 (or <, ≥, ≤) with a positivex2 term and zero on one side.
Find the roots of the corresponding equation ax2 + bx + c = 0. Call them x1 < x2.
Sketch the parabola — since a > 0 it’ll be ∪-shaped — and label the roots.
Identify the region on the graph that satisfies the inequality (above or below the x-axis).
Reading the Regions
Here’s the key visual — a ∪-shaped parabola with two roots, and the regions where it sits above or below the x-axis:
Above or Below the x-axis
For > 0 (or ≥ 0)
Take the regions OUTSIDE the roots: x < x1 or x > x2
For < 0 (or ≤ 0)
Take the region BETWEEN the roots: x1 < x < x2
Memory hook: A ∪-shaped parabola is positive on the outside, negative in the middle. So if you want positive (> 0), pick the outside; if you want negative (< 0), pick the middle.
Strict vs Inclusive Inequalities
The signs < and > are strict — the boundary points (the roots themselves) are not included. The signs ≤ and ≥ are inclusive — the roots are included:
ax2 + bx + c> 0: x < x1 or x > x2
ax2 + bx + c≥ 0: x ≤ x1 or x ≥ x2
ax2 + bx + c< 0: x1 < x < x2
ax2 + bx + c≤ 0: x1 ≤ x ≤ x2
Special Case: (x − h)² inequalities
If you see (x − h)2 < n or (x − h)2 > n, the safest method is to expand and follow the 4-step method. But there’s also a quick way using absolute values:
Less than
(x − h)2 < n
means | x − h | < √n
h − √n < x < h + √n
Between bounds
Greater than
(x − h)2 > n
means | x − h | > √n
x < h − √n or x > h + √n
Outside bounds
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DON’T write x − h < ±√n
This is a very common mistake. ±√n on its own doesn’t mean anything as a single inequality. Always use the absolute-value approach (or expand and use the 4-step method).
Worked Examples
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Example 1 — Below x-axis (between roots)
Solve x2 − x − 6 < 0.
Answer:
Step 1: already in form ax² + bx + c < 0 with a > 0.Step 2: find the roots.x² − x − 6 = (x − 3)(x + 2) = 0x = −2 or x = 3Step 3: sketch a ∪ shape with roots at −2 and 3.Step 4: we want < 0 (below x-axis), so pick the BETWEEN region.−2 < x < 3
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Example 2 — Above x-axis with ≥ (inclusive)
Solve x2 − 5x + 4 ≥ 0.
Answer:
Step 1: already in form ax² + bx + c ≥ 0 with a > 0.Step 2: find the roots.x² − 5x + 4 = (x − 1)(x − 4) = 0x = 1 or x = 4Step 3: sketch a ∪ shape with roots at 1 and 4.Step 4: we want ≥ 0 (on or above x-axis), so pick OUTSIDE — and INCLUDE the roots.x ≤ 1 or x ≥ 4Use ≤ and ≥ (not strict) because the inequality includes equality.
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Example 3 — Need to rearrange first
Find the set of values which satisfy 3x2 + 2x − 6 > x2 + 4x − 2.
Answer:
Step 1: rearrange so one side is 0 (and a > 0).(3x² + 2x − 6) − (x² + 4x − 2) > 02x² − 2x − 4 > 0x² − x − 2 > 0 (divided by 2)Step 2: find the roots.(x − 2)(x + 1) = 0x = −1 or x = 2Step 3: sketch a ∪ shape with roots at −1 and 2.Step 4: we want > 0 (above x-axis), so pick OUTSIDE.x < −1 or x > 2
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Example 4 — Negative leading coefficient (sign flip)
Solve −x2 + 4x − 3 > 0.
Answer:
Step 1: multiply both sides by −1 to make a > 0 — sign FLIPS!x² − 4x + 3 < 0Step 2: find the roots.(x − 1)(x − 3) = 0x = 1 or x = 3Step 3: sketch a ∪ shape with roots at 1 and 3.Step 4: we want < 0 (below x-axis), so pick BETWEEN.1 < x < 3Forgetting to flip the sign is the #1 mistake here. Always make a positive first.
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Example 5 — Special case (x − h)²
Solve (x − 3)2 > 16.
Answer:
Method 1: use absolute value.|x − 3| > 4x − 3 < −4 or x − 3 > 4x < −1 or x > 7Method 2: expand and use the 4-step method.x² − 6x + 9 > 16x² − 6x − 7 > 0(x − 7)(x + 1) = 0 → roots −1, 7Want > 0, pick OUTSIDE.x < −1 or x > 7Both methods give the same answer ✓
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Tips
Always sketch the parabola — even a tiny scribble. The picture makes the regions obvious.
Use your GDC on Paper 2 — graph the quadratic and read off the roots, then pick the region. Some GDCs even solve inequalities directly.
Memory hook for ∪-shape: “positive outside, negative in the middle.” That’s all you need to pick the right region.
Make a positive first. Always multiply through by −1 (and flip the sign!) if needed. It’s much harder to read regions off a ∩-shape.
Watch the boundaries. Strict (<, >) excludes the roots; inclusive (≤, ≥) includes them.
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Common mistakes
Forgetting to flip the sign when multiplying or dividing by a negative. This is by far the biggest source of wrong answers.
Writing x − h < ±√n for the special case. Wrong! Use absolute value or expand and rearrange.
Mixing up the regions. If you want > 0, pick OUTSIDE the roots. If you want < 0, pick BETWEEN. Sketch the parabola to be sure.
Writing < instead of ≤ (or vice versa). Pay attention to whether the inequality is strict or inclusive.
Reporting the answer as x = … instead of an inequality. Quadratic inequalities have a range of solutions, not specific values.
Stopping at the roots without going on to identify the region. The roots aren’t the answer — they’re the boundaries of the answer.
Final word: Get it to ax² + bx + c, find the roots, sketch, pick the region. Four steps — and the sketch carries you through every time.
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