Iteration & Inequalities

Iteration & Inequalities

Inequalities

Inequalities use the symbols < (less than), > (greater than), ≤ (less than or equal to) and ≥ (greater than or equal to).

Solving Linear Inequalities

Solve like equations, but reverse the sign when multiplying or dividing by a negative number.

Worked Example: Solve 3x − 7 > 5.

• 3x > 12
• x > 4

Worked Example: Solve 4 − 2x ≤ 10.

• −2x ≤ 6
• x ≥ −3 (inequality reversed because we divided by −2)

Double Inequalities

Worked Example: Solve −3 < 2x + 1 ≤ 9.

• Subtract 1: −4 < 2x ≤ 8
• Divide by 2: −2 < x ≤ 4

Representing on a Number Line

• Open circle (○) for < or > (value not included)
• Closed circle (●) for ≤ or ≥ (value included)
• Draw a line between or extending from the circles

Quadratic Inequalities (Higher)

Worked Example: Solve x² − 5x + 6 < 0.

1. Solve x² − 5x + 6 = 0: (x − 2)(x − 3) = 0, so x = 2 and x = 3

2. Sketch the quadratic (U-shaped parabola crossing at x = 2 and x = 3)

3. The quadratic is below zero (< 0) between the roots

4. 2 < x < 3

Worked Example: Solve x² − 9 ≥ 0.

1. x² − 9 = 0: (x + 3)(x − 3) = 0, so x = −3 and x = 3

2. The parabola is above or on zero outside the roots

3. x ≤ −3 or x ≥ 3

Key rule: For x² > a², the solution is x < −√a or x > √a (two separate regions). For x² < a², the solution is −√a < x < √a (one region between the roots).

Inequalities on Graphs (Higher)

Shading regions defined by inequalities:

• Solid line for ≤ or ≥
• Dashed line for < or >
• Test a point (like the origin) to determine which side to shade

To find the region satisfying multiple inequalities, shade the unwanted regions — the feasible region is left unshaded.

Iteration (Higher)

Iteration is a method for finding approximate solutions to equations that cannot be solved exactly. You rearrange the equation into the form x = g(x) and repeatedly substitute.

Worked Example: Show that x³ + 2x − 7 = 0 has a solution between x = 1 and x = 2, then use the iteration formula xₙ₊₁ = ∛(7 − 2xₙ) with x₀ = 1.5 to find this solution to 2 d.p.

Step 1: Show a sign change (change of sign indicates a root).

• f(1) = 1 + 2 − 7 = −4 (negative)
• f(2) = 8 + 4 − 7 = 5 (positive)
• Sign change between f(1) and f(2), so there is a root between 1 and 2 ✓

Step 2: Iterate.

• x₁ = ∛(7 − 2(1.5)) = ∛4 = 1.5874...
• x₂ = ∛(7 − 2(1.5874)) = ∛3.8252 = 1.5651...
• x₃ = ∛(7 − 2(1.5651)) = ∛3.8698 = 1.5709...
• x₄ = ∛(7 − 2(1.5709)) = ∛3.8582 = 1.5694...
• x₅ = ∛(7 − 2(1.5694)) = ∛3.8612 = 1.5698...
• x₅ and x₄ both round to 1.57, so x ≈ 1.57 (to 2 d.p.)

Exam Tips

• The most common error is forgetting to flip the inequality when dividing by a negative — practise this
• For quadratic inequalities, always sketch the graph to identify the correct region
• In iteration, show enough steps that two consecutive values agree to the required accuracy
• For sign-change questions, state the values clearly AND conclude ("there is a sign change so a root exists between...")