IB Maths AA HLTopic 2 — FunctionsPaper 1 & 2~7 min read
Quadratic Inequalities
Same quadratic, different question. Instead of “where is it equal to zero?”, you’re now asked “where is it positive” or “negative?”. The whole technique is one picture: find the roots, sketch the parabola, pick the correct region. That’s it. The rest is just being careful with signs and inequality directions.
📘 What you need to know
4-step process: rearrange to get zero on one side → find the roots → sketch the parabola → read off the region.
For ∪-shape (positive a): the curve is above the x-axis outside the roots, and below the x-axis between the roots.
“> 0” ⟹ outside the roots: x < x1 or x > x2.
“< 0” ⟹ between the roots: x1 < x < x2.
If the leading coefficient is negative, multiply through by −1 and flip the inequality sign first. Then proceed as ∪.
Use ≤ and ≥ if the original inequality includes “or equals” — the roots themselves are then included.
(x − h)2 < n ⟹ h − √n < x < h + √n. The “shortcut” x − h < ±√n is wrong — always expand or use the modulus version.
The 4-step recipe
🧭 Recipe — solving any quadratic inequality
Rearrange so the inequality is in the form (positive coefficient quadratic) > 0, < 0, ≥ 0, or ≤ 0. If a < 0, multiply by −1 and flip the sign.
Find the roots by solving the matching equation = 0 (factorise, formula, or GDC).
Sketch the parabola as a ∪ on the x-axis with the two roots marked.
Pick the region: above the x-axis (> 0) means outside the roots; below (< 0) means between.
Above vs below — pick the region
∪-parabola — outside vs between the roots
If the inequality has ≤ or ≥, the roots themselves are part of the solution — switch to closed brackets and use ≤/≥ in your final answer.
If a is negative — flip first
The whole “outside vs between” rule assumes a ∪-shaped parabola (positive a). When a < 0, the curve is ∩-shaped — above the axis between the roots, below outside. Easiest fix: multiply the inequality by −1 and flip the sign. Then it’s a standard ∪ problem.
Sign flip rule
multiply by negative number ⟹ flip the inequality
e.g. −x2 + 4x + 5 > 0 ⟺ x2 − 4x − 5 < 0
(x − h)2 < n form
Tempting to write x − h < ±√n. Don’t. That’s wrong. The correct interpretations:
(x − h)2 < n
h − √n < x < h + √n
“close to h” — between the two ±√n shifts
(x − h)2 > n
x < h − √n or x > h + √n
“far from h” — outside both shifts
Safest method: expand back to ax2 + bx + c form and use the standard 4-step recipe.
Worked examples
WE 1
Solve a basic “> 0” inequality
Solve x2 − 5x + 6 > 0.
Step 1: Already in standard form, a > 0 ✓Step 2: Find rootsx² − 5x + 6 = 0 → (x − 2)(x − 3) = 0 → x = 2, 3Step 3: Sketch ∪ with roots at 2, 3Step 4: Want > 0 → outside the rootsx < 2 or x > 3strict inequality > → strict signs <, > in answer (no equals)
WE 2
Non-monic with “≤ 0”
Solve 2x2 + 5x − 3 ≤ 0.
Step 1: Standard form, a = 2 > 0 ✓Step 2: Roots — split middle, m + n = 5, mn = −66 + (−1) = 5 ✓ → 2x² + 6x − x − 32x(x + 3) − 1(x + 3) = (2x − 1)(x + 3)roots: x = 1/2 and x = −3Step 3: Sketch ∪ with roots at −3, 1/2Step 4: Want ≤ 0 → between the roots, including endpoints−3 ≤ x ≤ 1/2≤ keeps the roots in the answer; if it had been < 0 the answer would be −3 < x < 1/2
WE 3
Negative leading coefficient — flip first
Solve −x2 + 4x + 5 > 0.
Step 1: Multiply by −1 and FLIP inequality signx² − 4x − 5 < 0Step 2: Find roots(x − 5)(x + 1) = 0 → x = −1, 5Step 3: Sketch ∪ with roots at −1, 5Step 4: Want < 0 → between the roots−1 < x < 5flipping the sign on multiplying by −1 is the most-missed step in this whole topic
WE 4
Rearrange before solving
Find the values of x for which x2 + 8 < 6x.
Step 1: Move everything to the leftx² − 6x + 8 < 0Step 2: Find roots(x − 2)(x − 4) = 0 → x = 2, 4Step 3: Sketch ∪ with roots at 2, 4Step 4: Want < 0 → between the roots2 < x < 4always rearrange to “(quadratic) compared to 0” first — never tackle inequalities with stuff on both sides
WE 5
(x − h)2 ≤ n form
Solve (x − 4)2 ≤ 9.
Method: expand and use the standard recipe(x − 4)² ≤ 9x² − 8x + 16 ≤ 9x² − 8x + 7 ≤ 0Find roots(x − 1)(x − 7) = 0 → x = 1, 7Want ≤ 0 → between the roots, inclusive1 ≤ x ≤ 7shortcut check: h = 4, √n = 3 → 4 − 3 ≤ x ≤ 4 + 3 → 1 ≤ x ≤ 7 ✓
WE 6
Inequality requiring rearrangement and “≥”
Find all values of x satisfying 3x2 − x ≥ 14.
Step 1: Rearrange3x² − x − 14 ≥ 0Step 2: Roots — split middle, m + n = −1, mn = −42−7 + 6 = −1 ✓ → 3x² − 7x + 6x − 14x(3x − 7) + 2(3x − 7) = (x + 2)(3x − 7)roots: x = −2 and x = 7/3Step 3: Sketch ∪ with roots at −2, 7/3Step 4: Want ≥ 0 → outside the roots, inclusivex ≤ −2 or x ≥ 7/3≥ keeps the roots in; the connecting word is “or” since the regions don’t overlap
💡 Top tips
Always sketch. A quick ∪ on the x-axis with the roots marked makes the region obvious — no guessing.
Get a positive leading coefficient first. If a < 0, multiply by −1 AND flip the sign. Then proceed as ∪.
Keep things on one side. Always rearrange to “(quadratic) compared to 0” before doing anything else.
“> 0 outside, < 0 between” is the rhyme that saves you. Memorise it.
Strict vs non-strict. < or > means the roots are excluded; ≤ or ≥ includes them.
Connecting word matters. “Outside” gives an “or” answer; “between” gives a single chained inequality.
Use your GDC on Paper 2. Graph the quadratic and read off where it’s above or below the x-axis directly.
⚠ Common mistakes
Forgetting to flip the inequality when multiplying or dividing by a negative. Single biggest error in this topic.
Writing x − h < ±√n for (x − h)2 < n. This isn’t a valid statement — expand or use the modulus version instead.
Connecting the two regions with “and” instead of “or” when the answer is outside the roots.
Mixing up < and ≤ in the final answer. If the question used ≤, your answer must include the roots.
Reporting a single number as the answer. The answer to a quadratic inequality is always an interval (or a union of two), never a single value.
Trying to factorise without rearranging. 2x2 + 7 < 9x needs the −9x moved over before you can do anything useful.
Writing the smaller root second in a “between” answer. x1 < x < x2 requires x1 to actually be smaller.
Quadratic inequalities show up everywhere later — domain restrictions, range of functions, calculus optimisation problems with constraints. Get the 4-step recipe under your belt and they’re free marks. The next note is the final one for Section 2.2: discriminants, where we use b2 − 4ac to figure out how many roots a quadratic has — often without solving the equation at all.
Need help with Quadratic Inequalities?
Get 1-on-1 help from an IB examiner who knows exactly what Paper 1 & 2 are looking for.
