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NotesMath AA HLTopic 2.7Quadratic inequalities
Back to Math AA HL Topics
2.7.32 min read

Quadratic inequalities

IB Mathematics: Analysis and Approaches • Unit 2

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Contents

  • Roots first, then regions
  • Between or outside? Use the shape
  • IB-style question — solve it fully
  • Writing the answer
Find where it equals zero first: To solve a quadratic inequality, first find the roots (solve = 0). The roots split the number line into regions; the inequality is true in some of them.

IB-style question — get the roots

For the inequality x² − x − 6 > 0, find the roots that mark the regions.

Step by step

  1. First swap the > for an = and solve the matching equation. Factorise the left side.
  2. Each bracket = 0 gives a root. These two x-values are the ONLY places the curve crosses the x-axis — so the only places it can switch between positive and negative.
  3. Mark both roots on the number line. Two marks chop the line into THREE stretches (just like two snips in a ribbon give three pieces): left of −2, between −2 and 3, and right of 3.

Final answer

The roots are x = −2 and x = 3 — the two points where the curve meets the x-axis. They split the number line into THREE regions; the inequality is true in some of them (we test which next).

Rearrange to one side first: Always get the inequality into the form (quadratic) > 0 (or < 0, ≥, ≤) with zero on the right before finding roots.
The shape tells you the sign — no testing needed: Once you have the roots, just picture the parabola.

An upward parabola (a > 0, a U-shape) dips below the x-axis between the roots and sits above it outside them. So:

• below zero (< 0) → BETWEEN the roots

• above zero (> 0) → OUTSIDE the roots

A downward parabola (a < 0, a ∩-shape) is the mirror image: above between, below outside.

Upward parabola (a > 0)

  • f(x) < 0 → between the roots.
  • f(x) > 0 → outside (x < p or x > q).
  • It dips below the axis in the middle.

Downward parabola (a < 0)

  • f(x) > 0 → between the roots.
  • f(x) < 0 → outside the roots.
  • It rises above the axis in the middle.
Sketch a quick parabola: Draw the U (or ∩), mark the roots, and shade where it's above/below the x-axis — the inequality reads straight off.

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IB-style question — an upward parabola, > 0

Solve x² − x − 6 > 0.

Step by step

  1. Roots: (x − 3)(x + 2) = 0 ⇒ x = −2, 3.
  2. a = 1 > 0 (upward); > 0 means OUTSIDE the roots.

Final answer

x < −2 or x > 3.

IB-style question — the ≤ version

Solve x² − x − 6 ≤ 0.

Step by step

  1. Same roots −2 and 3; ≤ 0 means BETWEEN (and including) the roots.

Final answer

−2 ≤ x ≤ 3.

≤ includes the roots; < does not: Use closed ends (≤, ≥) when the inequality includes equality, and open ends (<, >) when it doesn't.
Two pieces or one? Match the picture: Look at which part of the axis your sketch shades:

• Shaded the two outside arms → write two inequalities joined by or: x < p or x > q.

• Shaded the single middle stretch → write one chain: p ≤ x ≤ q.

That's the whole choice — 'outside' uses or, 'between' is one chain.

IB-style question — a downward parabola

Solve 8 − 2x − x² ≥ 0.

Step by step

  1. Step 1 — find the roots. To find them, solve the matching equation 8 − 2x − x² = 0. The leading −x² is awkward to factorise, so multiply every term by −1. The right side is 0 and 0 × −1 = 0, so this is totally safe — there's no inequality to flip yet, we're just solving = 0.
  2. Now it factorises easily — two numbers that multiply to −8 and add to +2 are +4 and −2.
  3. Step 2 — which way does it open? Read the ORIGINAL 8 − 2x − x²: the x² term is −x², so a = −1 < 0. The parabola opens DOWN — a ∩ arch.
  4. Step 3 — apply the shape. We want ≥ 0, i.e. ON or ABOVE the axis. A ∩ arch is above the axis BETWEEN its roots, so take that middle stretch with the ends included.

Final answer

−4 ≤ x ≤ 2 — the arch sits on or above the x-axis between its two roots.

Flip carefully if you multiply by −1: Multiplying an equation (= 0) by −1 is always safe. But if you ever multiply or divide an inequality by a negative, you must reverse the inequality sign. Keeping a > 0 by rearranging avoids the trap.

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(x − 1)(x − 5) > 0. [2 marks]

Related Math AA HL Topics

Continue learning with these related topics from the same unit:

2.1.1Equations of lines
2.1.2Parallel lines
2.1.3Perpendicular lines
2.1.4Perpendicular bisector
View all Math AA HL topics

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